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326 files changed, 61573 insertions, 3294 deletions
diff --git a/buch/chapters/10-vektorenmatrizen/linear.tex b/buch/chapters/10-vektorenmatrizen/linear.tex index ac2b85d..3ad51f1 100644..100755 --- a/buch/chapters/10-vektorenmatrizen/linear.tex +++ b/buch/chapters/10-vektorenmatrizen/linear.tex @@ -33,7 +33,7 @@ aber mit Punkten kann man trotzdem noch nicht rechnen. Ein Vektor fasst die Koordinaten eines Punktes in einem Objekt zusammen, mit dem man auch rechnen und zum Beispiel Parallelverschiebungen algebraisieren kann. -Um auch Streckungen ausdrücken zu können, wird auch eine Menge von +Um auch Streckungen ausdrücken zu können, wird auch eine Menge von Streckungsfaktoren benötigt, mit denen alle Komponenten eines Vektors multipliziert werden können. Sie heissen auch {\em Skalare} und liegen in $\Bbbk$. @@ -73,7 +73,7 @@ a+b = \begin{pmatrix}\lambda a_1\\\vdots\\\lambda a_n\end{pmatrix}. \] -Die üblichen Rechenregeln sind erfüllt, nämlich +Die üblichen Rechenregeln sind erfüllt, nämlich \begin{equation} \begin{aligned} &\text{Kommutativität:} @@ -149,7 +149,7 @@ kann als (abstrakter) Vektor betrachtet werden. \begin{definition} Eine Menge $V$ von Objekten, auf der zwei Operationen definiert, nämlich die Addition, geschrieben $a+b$ für $a,b\in V$ und die -Multiplikation mit Skalaren, geschrieben $\lambda a$ für $a\in V$ und +Multiplikation mit Skalaren, geschrieben $\lambda a$ für $a\in V$ und $\lambda\in \Bbbk$, heisst ein {\em $\Bbbk$-Vektorraum} oder {\em Vektorraum über $\Bbbk$} (oder einfach nur {\em Vektorraum}, wenn $\Bbbk$ aus dem Kontext klar sind), @@ -172,7 +172,7 @@ $\mathbb{C}$ ein Vektorraum über $\mathbb{R}$. \end{beispiel} \begin{beispiel} -Die Menge $C([a,b])$ der stetigen Funktionen $[a,b]\to\mathbb{Re}$ +Die Menge $C([a,b])$ der stetigen Funktionen $[a,b]\to\mathbb{Re}$ bildet ein Vektorraum. Funktionen können addiert und mit reellen Zahlen multipliziert werden: \[ @@ -188,7 +188,7 @@ Die Vektorraum-Rechenregeln \end{beispiel} Die Beispiele zeigen, dass der Begriff des Vektorraums die algebraischen -Eigenschaften eine grosse Zahl sehr verschiedenartiger mathematischer +Eigenschaften eine grosse Zahl sehr verschiedenartiger mathematischer Objekte beschreiben kann. Alle Erkenntnisse, die man ausschliesslich aus Vekotorraumeigenschaften gewonnen hat, sind auf alle diese Objekte übertragbar. @@ -300,7 +300,7 @@ folgt, dass alle $\lambda_1,\dots,\lambda_n=0$ sind. Lineare Abhängigkeit der Vektoren $a_1,\dots,a_n$ bedeutet auch, dass man einzelne der Vektoren durch andere ausdrücken kann. Hat man nämlich eine -Linearkombination~\eqref{buch:vektoren-und-matrizen:eqn:linabhdef} und +Linearkombination~\eqref{buch:vektoren-und-matrizen:eqn:linabhdef} und ist der Koeffizient $\lambda_k\ne 0$, dann kann man nach $a_k$ auflösen: \[ a_k = -\frac{1}{\lambda_k}(\lambda_1a_1+\dots+\widehat{\lambda_ka_k}+\dots+\lambda_na_n). @@ -323,7 +323,7 @@ offenbar eine besondere Bedeutung. Eine linear unabhängig Menge von Vektoren $\mathcal{B}=\{a_1,\dots,a_n\}\subset V$ heisst {\em Basis} von $V$. -Die maximale Anzahl linear unabhängiger Vektoren in $V$ heisst +Die maximale Anzahl linear unabhängiger Vektoren in $V$ heisst {\em Dimension} von $V$. \end{definition} @@ -331,7 +331,7 @@ Die Standardbasisvektoren bilden eine Basis von $V=\Bbbk^n$. \subsubsection{Unterräume} Die Mengen $\langle a_1,\dots,a_n\rangle$ sind Teilmengen -von $V$, in denen die Addition von Vektoren und die Multiplikation mit +von $V$, in denen die Addition von Vektoren und die Multiplikation mit Skalaren immer noch möglich ist. \begin{definition} @@ -352,7 +352,7 @@ gilt. % \subsection{Matrizen \label{buch:grundlagen:subsection:matrizen}} -Die Koeffizienten eines linearen Gleichungssystems finden in einem +Die Koeffizienten eines linearen Gleichungssystems finden in einem Zeilen- oder Spaltenvektor nicht Platz. Wir erweitern das Konzept daher in einer Art, dass Zeilen- und Spaltenvektoren Spezialfälle sind. @@ -378,14 +378,14 @@ M_{m\times n}(\Bbbk) = \{ A\;|\; \text{$A$ ist eine $m\times n$-Matrix}\}. \] Falls $m=n$ gilt, heisst die Matrix $A$ auch {\em quadratisch} \index{quadratische Matrix}% -Man kürzt die Menge der quadratischen Matrizen als +Man kürzt die Menge der quadratischen Matrizen als $M_n(\Bbbk) = M_{n\times n}(\Bbbk)$ ab. \end{definition} -Die $m$-dimensionalen Spaltenvektoren $v\in \Bbbk^m$ sind $m\times 1$-Matrizen +Die $m$-dimensionalen Spaltenvektoren $v\in \Bbbk^m$ sind $m\times 1$-Matrizen $v\in M_{n\times 1}(\Bbbk)$, die $n$-dimensionalen Zeilenvetoren $u\in\Bbbk^n$ sind $1\times n$-Matrizen $v\in M_{1\times n}(\Bbbk)$. -Eine $m\times n$-Matrix $A$ mit den Koeffizienten $a_{ij}$ besteht aus +Eine $m\times n$-Matrix $A$ mit den Koeffizienten $a_{ij}$ besteht aus den $n$ Spaltenvektoren \[ a_1 = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix},\quad @@ -435,7 +435,7 @@ werden kann. \begin{definition} Eine $m\times n$-Matrix $A\in M_{m\times n}(\Bbbk)$ und eine $n\times l$-Matrix $B\in M_{n\times l}(\Bbbk)$ haben als Produkt -eine $n\times l$-Matrix $C=AB\in M_{n\times l}(\Bbbk)$ mit den +eine $m\times l$-Matrix $C=AB\in M_{m\times l}(\Bbbk)$ mit den Koeffizienten \begin{equation} c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}. @@ -483,7 +483,7 @@ I 1 &0 &\dots &0 \\ 0 &1 &\dots &0 \\[-2pt] \vdots&\vdots&\ddots&\vdots\\ -0 &0 &\dots &1 +0 &0 &\dots &1 \end{pmatrix}. \] @@ -521,10 +521,10 @@ Ein Gleichungssystem mit $0$ auf der rechten Seite ist also bereits ausreichend um zu entscheiden, ob die Lösung eindeutig ist. Ein Gleichungssystem mit rechter Seite $0$ heisst {\em homogen}. \index{homogenes Gleichungssystem}% -Zu jedem {\em inhomogenen} Gleichungssystem $Ax=b$ mit $b\ne 0$ +Zu jedem {\em inhomogenen} Gleichungssystem $Ax=b$ mit $b\ne 0$ ist $Ax=0$ das zugehörige homogene Gleichungssystem. -Ein homogenes Gleichungssytem $Ax=0$ hat immer mindestens die +Ein homogenes Gleichungssytem $Ax=0$ hat immer mindestens die Lösung $x=0$, man nennt sie auch die {\em triviale} Lösung. Eine Lösung $x\ne 0$ heisst auch eine nichttriviale Lösung. Die Lösungen eines inhomgenen Gleichungssystem $Ax=b$ ist also nur dann @@ -535,7 +535,7 @@ Lösung hat. Der Gauss-Algorithmus oder genauer Gausssche Eliminations-Algorithmus löst ein lineare Gleichungssystem der Form~\eqref{buch:vektoren-und-matrizen:eqn:vektorform}. -Die Koeffizienten werden dazu in das Tableau +Die Koeffizienten werden dazu in das Tableau \[ \begin{tabular}{|>{$}c<{$}>{$}c<{$}>{$}c<{$}|>{$}c<{$}|} \hline @@ -552,7 +552,7 @@ Der Algorithmus is so gestaltet, dass er nicht mehr Speicher als das Tableau benötigt, alle Schritte operieren direkt auf den Daten des Tableaus. -In jedem Schritt des Algorithmus wird zunächst eine Zeile $i$ und +In jedem Schritt des Algorithmus wird zunächst eine Zeile $i$ und Spalte $j$ ausgewählt, das Elemente $a_{ij}$ heisst das Pivotelement. \index{Pivotelement}% Die {\em Pivotdivision} @@ -646,7 +646,7 @@ In der Phase der {\em Vorwärtsreduktion} werden Pivotelemente von links nach rechts möglichst auf der Diagonale gewählt und mit Zeilensubtraktionen die darunterliegenden Spalten freigeräumt. \index{Vorwärtsreduktion}% -Während des Rückwärtseinsetzens werden die gleichen Pivotelemente von +Während des Rückwärtseinsetzens werden die gleichen Pivotelemente von rechts nach links genutzt, um mit Zeilensubtraktionen auch die Spalten über den Pivotelemnten frei zu räumen. \index{Rückwärtseinsetzen}% @@ -800,7 +800,7 @@ $x = b_1c_1+b_2c_2+\dots+b_nc_n$ konstruieren. Tatsächlich gilt \begin{align*} Ax -&= +&= A( b_1c_1+b_2c_2+\dots+b_nc_n) \\ &= @@ -837,7 +837,178 @@ Seite~\pageref{buch:vektorenmatrizen:satz:gruppenregeln} die Eigenschaft $A^{-1}A=I$ ganz allgemein gezeigt. \subsubsection{Determinante} -XXX TODO +Ein Gleichungssystem mit $n$ Gleichungen und $n$ Unbekannten ist genau +dann lösbar, wenn sich der Gauss-Algorithmus bis zum Ende durchführen lässt. +Das ist gleichbedeutend damit, dass keines der Pivot-Elemente verschwindet. +Das Produkt der Pivot-Elemente ist also eine aus der Koeffizientenmatrix +$A$ berechnete Kennzahl, die zu entscheiden erlaubt, ob ein Gleichungssystem +lösbar ist. + +\begin{definition} +\label{buch:linear:determinate:def} +Das Produkt der Pivot-Elemente bei der Durchführung des Gauss-Algorithmus +für eine Gleichungssystem mit quadratischer Koeffizientenmatrix $A$ +heisst die Determinante $\det(A)$ der Matrix $A$. +\end{definition} + +Aus den Regeln für die Durchführung des Gauss-Algorithmus kann man die +folgenden Regeln für die Determinante ableiten. +Wir stellen die Eigenschaften hier nur zusammen, detaillierte Herleitungen +kann man in jedem Kurs zur linearen Algebra finden, zum Beispiel im +Kapitel~2 des Skripts \cite{buch:linalg}. +\begin{enumerate} +\item +\label{buch:linear:determinante:einheitsmatrix} +Die Determinante der Einheitsmatrix ist $\det(I)=1$. +\item +Sind zwei Zeilen einer Matrix gleich, dann tritt beim Gauss-Algorithmus +eine Nullzweile auf, die Matrix kann also nicht regulär sein und die +Determinante ist $0$. +\item +\label{buch:linear:determinante:vorzeichen} +Vertauscht man zwei Zeilen einer Matrix, dann kehrt das Vorzeichen der +Determinante. +\item +Addiert man ein Vielfaches einer Zeile der Matrix zu einer anderen Zeile, +dann ändert der Wert der Determinante nicht. +\item +Wird eine Zeile der Matrix mit einer Zahl $\lambda$ multipliziert, dann +wird auch der Wert der Determinanten mit $\lambda$ multipliziert. +\item +\label{buch:linear:determinante:asymetrisch} +Die Determinante ist eine lineare Funktion der Zeilen von $A$. +Zusammen mit der Eigeschaft~\ref{buch:linear:determinante:vorzeichen} +folgt, dass die Determinante eine antisymmetrische lineare Funktion +der Zeilen ist. +\item +Die Determinante ist durch die Eigenschaften +\ref{buch:linear:determinante:einheitsmatrix} +und +\ref{buch:linear:determinante:asymetrisch} +eindeutig bestimmt. +\item +Der Entwicklungssatz von Laplace. +\index{Entwicklungssatz Laplace}% +Die Determinante der $n\times n$-Matrix $A$ kann mit der Formel +\begin{equation} +\det(A) += +\sum_{i=1}^n (-1)^{i+j} a_{ij} \cdot \det(A_{ij}) +\end{equation} +wobei die $(n-1)\times(n-1)$-Matrix $A_{ij}$ die Matrix $A$ ist, aus der +man Zeile $i$ und Spalte $j$ entfernt hat. +$A_{ij}$ heisst ein {\em Minor} der Matrix $A$. +\index{Minor einer Matrix}% +\end{enumerate} + +Die bekannte Formel $\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$ +ist ein Spezialfall des Entwicklungssatzes von Laplace. +Auch für $3\times 3$-Matrizen ist eine übersichtliche Form möglich, +die als die Sarrus-Formel bekannt ist. +\index{Sarrus-Formel}% + +\begin{satz}[Sarrus] +\label{buch:linear:determinate:sarrus} +Die Determinante einer $3\times 3$-Matrix ist +\[ +\left|\begin{matrix} +a&b&c\\ +d&e&f\\ +g&h&i +\end{matrix}\right| += +aei + bfg + cdh - ceg - bdi - afh. +\] +\end{satz} + +\subsubsection{Die Regel von Cramer} +Die Determinanten ermöglicht auch, eine Formel für die Lösung eines +Gleichungssystems zu geben. +Dies ist bekannt als die {\em Regel von Cramer}. + +\begin{satz} +\label{buch:linear:determinante:cramer} +Die Lösung $x_k$ eines $n\times n$-Gleichungssystem $Ax=b$ mit +Koeffizientenmatrix $A$ und rechter Seite $b$ hat die Lösungen +\begin{equation} +x_k += +\frac{ +\left|\begin{matrix} +a_{11}&a_{12}&\dots &b_1 &\dots &a_{1n}\\ +a_{21}&a_{22}&\dots &b_2 &\dots &a_{2n}\\ +\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ +a_{n1}&a_{n2}&\dots &b_n &\dots &a_{nn} +\end{matrix}\right| +}{ +\det(A), +} +\end{equation} +wobei im Zähler die Spalte $k$ der Matrix $A$ durch den Vektor $b$ +der rechten Seiten ersetzt worden ist. +\end{satz} + +Die Cramersche Formel ist besonders nützlich, wenn die Abhängigkeit +einer Lösungsvariablen von den Einträgen der Koeffizientenmatrix +untersucht werden soll. +Für die Details der Herleitung sei wieder auf \cite{buch:linalg} +verwiesen. + +\subsubsection{Die inverse Matrix mit Hilfe der Determinanten} +Die inverse Matrix löst ein quadratisches Gleichungssystem $Ax=b$ mit +Hilfe der Formel $x=A^{-1}b$. +Man kann daher auch erwarten, dass sich die inverse Matrix dank +der Cramerschen Regel mit Hilfe von Determinanten ausdrücken lässt. +Tatsächlich gilt der folgende Satz. + +\begin{satz} +\label{buch:linalg:inverse:adjunkte} +Die Inverse der $n\times n$-Matrix $A$ ist gegeben durch +\index{Formel für die inverse Matrix}% +\index{inverse Matrix, Formel für}% +\begin{equation} +(A^{-1})_{ij} += +\frac{1}{\det(A)} +\begin{pmatrix} +\det(A_{11}) & -\det(A_{21}) & \dots & (-1)^{i+1}\det(A_{i1}) & \dots + & (-1)^{1+n} \det(A_{n1}) \\ +-\det(A_{12}) & \det(A_{22}) & \dots & (-1)^{i+2}\det(A_{i2}) & \dots + & (-1)^{2+n} \det(A_{n2}) \\ +\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ +(-1)^{1+j}\det(A_{1j}) & (-1)^{2+j}\det(A_{2j}) & \dots + & (-1)^{i+j} \det(A_{ji}) + & \dots & (-1)^{j+n} \det(A_{nj}) \\ +\vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ +(-1)^{1+n}\det(A_{1n}) & (-1)^{2+n}\det(A_{2n}) & \dots + & (-1)^{i+n}\det(A_{in}) + & \dots & \det(A_{nn}) +\end{pmatrix} +\label{buch:linalg:inverse:formel} +\end{equation} +Die Transponierte der Matrix auf der rechten Seite (ohne den Vorfaktor +$1/\det(A)$ +heisst die {\em Adjunkte} $\operatorname{adj}A$ von $A$. +\index{Adjunkte}% +\end{satz} + +Der Satz~\ref{buch:linalg:inverse:adjoint} liefert eine algebraische +Formel für die Elemente der inversen Matrix. +Für kleine Matrizen wie im nachfolgenden Beispiel ist die +Formel~\eqref{buch:linalg:inverse:formel} oft einfachter anzuwenden. +Besonders einfach wird die Formel für eine $2\times 2$-Matrix, +wo man +\[ +\begin{pmatrix} +a&b\\c&d +\end{pmatrix}^{-1} += +\frac{1}{ad-bc}\begin{pmatrix} +d&-b\\ +-c&a +\end{pmatrix} +\] +erhält. \begin{beispiel} Die Inverse der Matrix @@ -852,21 +1023,22 @@ a&a&1 ist mit Hilfe von Determinanten besonders einfach zu invertieren. Die Determinante von $A$ ist nach der Sarrus-Formel \[ -\det A +\operatorname{adj}A = 1 + 2a^3 - 3a^2. \] -Die adjungiert Matrix ist +Die Adjunkte ist \begin{align*} -A^{-1} +(\operatorname{adj}A)^t &= -\frac{1}{\det{A}} -\begin{pmatrix} -\det A_{11} & \det A_{21} & \det A_{31} \\ -\det A_{12} & \det A_{22} & \det A_{32} \\ -\det A_{13} & \det A_{23} & \det A_{33} -\end{pmatrix} -\\ +%\frac{1}{\det{A}} +\begin{pmatrix*}[r] + \det A_{11} & -\det A_{21} & \det A_{31} \\ +-\det A_{12} & \det A_{22} & -\det A_{32} \\ + \det A_{13} & -\det A_{23} & \det A_{33} +\end{pmatrix*} +\intertext{und damit ist die inverse Matrix} +A^{-1} &= \frac{1}{2a^3-3a^2+1} \renewcommand\arraystretch{1.1} @@ -896,7 +1068,7 @@ A^{-1} 1-a^2 & a^2-a & a^2-a\\ a^2-a & 1-a^2 & a^2-a\\ a^2-a & a^2-a & 1-a^2 -\end{pmatrix} +\end{pmatrix}. \end{align*} Mit $1-a^2=(1+a)(1-a)$ und $a^2-a=a(a-1)$ kann man dies noch etwas vereinfachen, indem man den gemeinsamen Faktor $1-a$ ausklammern. @@ -912,10 +1084,19 @@ A^{-1} \end{pmatrix}. \label{buch:vektoren-und-matrizen:abeispiel:eqn2} \end{equation} -für die Inverse einer Matrix der Form +für die Inverse einer Matrix der Form \eqref{buch:vektoren-und-matrizen:abeispiel:eqn1}. \end{beispiel} +\subsubsection{Produktregel für die Determinante} +Aus der Charakterisierung der Determinanten kann man auch ableiten, +dass die Produktregel +\[ +\det (AB) = \det(A) \cdot \det(B) +\] +gilt. +Daraus folgt auch, dass $\det(A^{-1})=\det(A)^{-1}$. + % % Lineare Abbildungen % @@ -937,7 +1118,7 @@ Eine Abbildung $f\colon V\to U$ zwischen Vektorräumen $V$ und $U$ heisst linear, wenn \[ \begin{aligned} -f(v+w) &= f(v) + f(w)&&\forall v,w\in V +f(v+w) &= f(v) + f(w)&&\forall v,w\in V \\ f(\lambda v) &= \lambda f(v) &&\forall v\in V,\lambda \in \Bbbk \end{aligned} @@ -948,16 +1129,16 @@ gilt. Lineare Abbildungen sind in der Mathematik sehr verbreitet. \begin{beispiel} -Sie $V=C^1([a,b])$ die Menge der stetig differenzierbaren Funktionen +Sie $V=C^1([a,b])$ die Menge der stetig differenzierbaren Funktionen auf dem Intervall $[a,b]$ und $U=C([a,b])$ die Menge der -stetigen Funktion aif $[a,b]$. +stetigen Funktion aif $[a,b]$. Die Ableitung $\frac{d}{dx}$ macht aus einer Funktion $f(x)$ die Ableitung $f'(x)$. -Die Rechenregeln für die Ableitung stellen sicher, dass +Die Rechenregeln für die Ableitung stellen sicher, dass \[ \frac{d}{dx} \colon -C^1([a,b]) \to C([a,b]) +C^1([a,b]) \to C([a,b]) : f \mapsto f' \] @@ -976,7 +1157,7 @@ eine lineare Abbildung. \end{beispiel} \subsubsection{Matrix} -Um mit linearen Abbildungen rechnen zu können, ist eine Darstellung +Um mit linearen Abbildungen rechnen zu können, ist eine Darstellung mit Hilfe von Matrizen nötig. Sei also $\mathcal{B}=\{b_1,\dots,b_n\}$ eine Basis von $V$ und $\mathcal{C} = \{ c_1,\dots,c_m\}$ eine Basis von $U$. @@ -984,12 +1165,12 @@ Das Bild des Basisvektors $b_i$ kann als Linearkombination der Vektoren $c_1,\dots,c_m$ dargestellt werden. Wir verwenden die Bezeichnung \[ -f(b_i) +f(b_i) = a_{1i} c_1 + \dots + a_{mi} c_m. \] Die lineare Abbildung $f$ bildet den Vektor $x$ mit Koordinaten -$x_1,\dots,x_n$ ab auf +$x_1,\dots,x_n$ ab auf \begin{align*} f(x) &= @@ -1012,7 +1193,7 @@ x_n(a_{1n} c_1 + \dots + a_{mn} c_m) + ( a_{m1} x_1 + \dots + a_{mn} x_n ) c_m \end{align*} -Die Koordinaten von $f(x)$ in der Basis $\mathcal{C}$ in $U$ sind +Die Koordinaten von $f(x)$ in der Basis $\mathcal{C}$ in $U$ sind also gegeben durch das Matrizenprodukt $Ax$, wenn $x$ der Spaltenvektor aus den Koordinaten in der Basis $\mathcal{B}$ in $V$ ist. @@ -1050,7 +1231,7 @@ b_{m1}x_1&+& \dots &+&b_{mn}x_n&=&b_{m1}'x_1'&+& \dots &+&b_{mn}'x_n' \end{linsys} \] Dieses Gleichungssystem kann man mit Hilfe eines Gauss-Tableaus lösen. -Wir schreiben die zugehörigen Variablen +Wir schreiben die zugehörigen Variablen \[ \renewcommand{\arraystretch}{1.1} \begin{tabular}{|>{$}c<{$} >{$}c<{$} >{$}c<{$}|>{$}c<{$}>{$}c<{$}>{$}c<{$}|} @@ -1096,7 +1277,7 @@ Für zwei Vektoren $u$ und $w$ in $U$ gibt es daher Vektoren $a=g(u)$ und $b=g(w)$ in $V$ derart, dass $f(a)=u$ und $f(b)=w$. Weil $f$ linear ist, folgt daraus $f(a+b)=u+w$ und $f(\lambda a)=\lambda a$ für jedes $\lambda\in\Bbbk$. -Damit kann man jetzt +Damit kann man jetzt \begin{align*} g(u+w)&=g(f(a)+f(b)) = g(f(a+b)) = a+b = g(u)+g(w) \\ @@ -1134,7 +1315,7 @@ Der Kern oder Nullraum der Matrix $A$ ist die Menge \] \end{definition} -Der Kern ist ein Unterraum, denn für zwei Vektoren $u,w\in \ker f$ +Der Kern ist ein Unterraum, denn für zwei Vektoren $u,w\in \ker f$ \[ \begin{aligned} f(u+v)&=f(u) + f(v) = 0+0 = 0 &&\Rightarrow& u+v&\in\ker f\\ @@ -1150,7 +1331,7 @@ Wir definieren daher das Bild einer linearen Abbildung oder Matrix. \begin{definition} Ist $f\colon V\to U$ eine lineare Abbildung dann ist das Bild von $f$ -der Unterraum +der Unterraum \[ \operatorname{im}f = \{ f(v)\;|\;v\in V\} \subset U \] @@ -1194,7 +1375,7 @@ $\operatorname{def}A=\dim\ker A$. \end{definition} Da der Kern mit Hilfe des Gauss-Algorithmus bestimmt werden kann, -können Rang und Defekt aus dem Schlusstableau +können Rang und Defekt aus dem Schlusstableau eines homogenen Gleichungssystems mit $A$ als Koeffizientenmatrix abgelesen werden. @@ -1210,8 +1391,3 @@ n-\operatorname{def}A. \subsubsection{Quotient} TODO: $\operatorname{im} A \simeq \Bbbk^m/\ker A$ - - - - - diff --git a/buch/chapters/10-vektorenmatrizen/skalarprodukt.tex b/buch/chapters/10-vektorenmatrizen/skalarprodukt.tex index d951221..408bfeb 100644 --- a/buch/chapters/10-vektorenmatrizen/skalarprodukt.tex +++ b/buch/chapters/10-vektorenmatrizen/skalarprodukt.tex @@ -197,7 +197,7 @@ mit Gleichheit genau dann, wenn $x=ty$ ist für ein $t\ge 0$. &= (\|x\|_2 + \|y\|_2)^2 \\ -\|x\|_2 + \|y\|_2 +\|x + y\|_2 &\le \|x\|_2 + \|y\|_2, \end{align*} Gleichheit tritt genau dann ein, wenn diff --git a/buch/chapters/50-permutationen/determinante.tex b/buch/chapters/50-permutationen/determinante.tex index c440caf..805235d 100644 --- a/buch/chapters/50-permutationen/determinante.tex +++ b/buch/chapters/50-permutationen/determinante.tex @@ -7,3 +7,105 @@ \section{Determinante \label{buch:section:determinante}} \rhead{Determinante} +Das Signum einer Permutationsmatrizen lässt sich +gemäss~\eqref{buch:permutationen:determinante} +mit der Determinanten berechnen. +Umgekehrt sollte es auch möglich sein, eine Formel +für die Determinante zu finden. +Die Basis dafür ist der +Entwicklungssatz +\begin{equation} +\det(A) += +\sum_{i=1}^n (-1)^{i+j} a_{ij} \cdot \det(A_{ij}) +\label{buch:permutationen:entwicklungssatz} +\end{equation} +von Laplace für die Determinante. +In den Produkten $a_{ij}\cdot\det(A_{ij})$ enthält +die Untermatrix $A_{ij}$ weder Elemente der Zeile $i$ noch der +Zeile $j$. +Die Summanden auf der rechten Seite von +\eqref{buch:permutationen:entwicklungssatz} +sind daher Produkte der Form +\[ +a_{1i_1} +a_{2i_2} +a_{3i_3} +\dots +a_{ni_n}, +\] +in denen nur Faktoren aus verschiedenen Spalten der Matrix $A$ +vorkommen. +Das ist gleichbedeutend damit, dass unter den Spaltenindizes +$i_1,i_2,i_3,\dots,i_n$ keine zwei gleich sind, dass also +\[ +\sigma += +\begin{pmatrix} +1&2&3&\dots&n\\ +i_1&i_2&i_3&\dots&i_n +\end{pmatrix} +\] +eine Permutation ist. + +Die Determinante muss sich daher als Summe über alle Permutationen +in der Form +\begin{equation} +\det(A) += +\sum_{\sigma\in S_n} +c(\sigma) +a_{1\sigma(1)} +a_{2\sigma(2)} +\dots +a_{n\sigma(n)} +\label{buch:permutationen:cformel} +\end{equation} +schreiben lassen, wobei die Koeffizienten $c(\sigma)$ noch zu bestimmen +sind. +Setzt man in +\eqref{buch:permutationen:cformel} +eine Permutationsmatrix $P_\tau$ ein, dann verschwinden alle +Terme auf der rechten Seite ausser dem zur Permutation $\tau$, +also +\[ +\det(P_\tau) += +\sum_{\sigma \in S_n} +c(\sigma) +(P_\tau)_{1\sigma(1)} +(P_\tau)_{2\sigma(2)} +\dots +(P_\tau)_{n\sigma(n)} += +c(\tau) +1\cdot 1\cdot\dots\cdot 1 += +c(\tau). +\] +Der Koeffizientn $c(\tau)$ ist also genau das Vorzeichen +der Permutation $\tau$. +Damit erhalten wir den folgenden Satz: + +\begin{satz} +Die Determinante einer $n\times n$-Matrix $A$ kann berechnet werden als +\[ +\det(A) += +\sum_{\sigma\in S_n} +\operatorname{sgn}(\sigma) +a_{1\sigma(1)} +a_{2\sigma(2)} +\dots +a_{n\sigma(n)} += +\sum_{\tau\in S_n} +\operatorname{sgn}(\tau) +a_{\tau(1)1} +a_{\tau(2)2} +\dots +a_{\tau(n)n}. +\] +Insbesondere folgt auch $\det(A)=\det(A^t)$. +\end{satz} + diff --git a/buch/chapters/50-permutationen/matrizen.tex b/buch/chapters/50-permutationen/matrizen.tex index 7e55364..f7e9e31 100644 --- a/buch/chapters/50-permutationen/matrizen.tex +++ b/buch/chapters/50-permutationen/matrizen.tex @@ -181,7 +181,7 @@ Die Determinante einer solchen Permutationsmatrix ist Nach der Produktregel für die Determinante folgt für eine Darstellung der Permutation $\sigma=\tau_1\dots\tau_l$ als Produkt von Transpositionen, dass -\[ +\begin{equation} \det P_{\sigma} = \det P_{\tau_1} \dots \det P_{\tau_l} @@ -189,7 +189,8 @@ dass (-1)^l = \operatorname{sgn}(\sigma). -\] +\label{buch:permutationen:determinante} +\end{equation} Das Vorzeichen einer Permutation ist also identisch mit der Determinante der zugehörigen Permutationsmatrix. diff --git a/buch/chapters/60-gruppen/symmetrien.tex b/buch/chapters/60-gruppen/symmetrien.tex index 7364c85..aee3b41 100644 --- a/buch/chapters/60-gruppen/symmetrien.tex +++ b/buch/chapters/60-gruppen/symmetrien.tex @@ -714,8 +714,8 @@ Kurve so zu definieren, dass dabei Längen und Winkel erhalten bleiben. Dieser Ansatz ist die Basis der Theorie der Krümmung sogenannter Riemannscher Mannigfaltigkeiten. -\subsection{Der Satz von Noether -\label{buch:subsection:noether}} +%\subsection{Der Satz von Noether +%\label{buch:subsection:noether}} diff --git a/buch/chapters/70-graphen/chapter.tex b/buch/chapters/70-graphen/chapter.tex index 6def393..530d96c 100644 --- a/buch/chapters/70-graphen/chapter.tex +++ b/buch/chapters/70-graphen/chapter.tex @@ -19,7 +19,7 @@ erste Approximation dreidimensionaler Objekte dienen. Die Bedeutung des Graphenkozeptes wird unterstrichen von der Vielzahl von Fragestellungen, die über Graphen gestellt, und der -zugehöriten Lösungsalgorithmen, die zu ihrer Beantwortung gefunden +zugehörigen Lösungsalgorithmen, die zu ihrer Beantwortung gefunden worden sind. Die Komplexitätstheorie hat sogar gezeigt, dass sich jedes diskrete Problem in ein Graphenproblem umformulieren lässt. diff --git a/buch/chapters/70-graphen/wavelets.tex b/buch/chapters/70-graphen/wavelets.tex index ef1520e..8baa88c 100644 --- a/buch/chapters/70-graphen/wavelets.tex +++ b/buch/chapters/70-graphen/wavelets.tex @@ -10,7 +10,7 @@ In Abschnitt~\ref{buch:subsection:standardbasis-und-eigenbasis} wurde gezeigt dass die Standardbasis den Zusammenhang zwischen den einzelnen Teilen des Graphen völlig ignoriert, während die Eigenbasis Wellen beschreibt, die mit vergleichbarer Amplitude sich über den ganzen -Graphen entsprechen. +Graphen erstrecken. Die Eigenbasis unterdrückt also die ``Individualität'' der einzelnen Knoten fast vollständig. diff --git a/buch/chapters/90-crypto/Makefile.inc b/buch/chapters/90-crypto/Makefile.inc index 9543ce1..508add5 100644 --- a/buch/chapters/90-crypto/Makefile.inc +++ b/buch/chapters/90-crypto/Makefile.inc @@ -8,5 +8,4 @@ CHAPTERFILES = $(CHAPTERFILES) \ chapters/90-crypto/arith.tex \ chapters/90-crypto/ff.tex \ chapters/90-crypto/aes.tex \ - chapters/90-crypto/rs.tex \ chapters/90-crypto/chapter.tex diff --git a/buch/chapters/90-crypto/arith.tex b/buch/chapters/90-crypto/arith.tex index dcc31b8..b05110f 100644 --- a/buch/chapters/90-crypto/arith.tex +++ b/buch/chapters/90-crypto/arith.tex @@ -91,6 +91,7 @@ Die Berechnung der Quadratwurzel lässt sich in Hardware effizient implementieren. \begin{algorithmus} +\label{buch:crypto:teile-und-hersche} Der folgende Algorithmus berechnet $a^k$ in $O(\log_2(k))$ Multiplikationen \begin{enumerate} diff --git a/buch/chapters/90-crypto/ff.tex b/buch/chapters/90-crypto/ff.tex index 535b359..a1cb747 100644 --- a/buch/chapters/90-crypto/ff.tex +++ b/buch/chapters/90-crypto/ff.tex @@ -7,6 +7,15 @@ \section{Kryptographie und endliche Körper \label{buch:section:kryptographie-und-endliche-koerper}} \rhead{Kryptographie und endliche Körper} +In diesem Abschnitt soll illustriert werden, wie die Arithmetik in +endlichen Körpern Algorithmen zu konstruieren erlaubt, mit denen sich +zum Beispiel sehr effizient kryptographische Schlüssel aushandeln +lassen. +Der klassische Diffie-Hellmann-Algorithmus in einem Galois-Körper +$\mathbb{F}_p$ wird in Abschnitt~\ref{buch:subsection:elliptische-kurven} +verallgemeinert auf eine sogenannte elliptische Kurve. +Diese Version des Algorithmus ist sehr effizient was die Bitlänge der +Schlüssel betrifft. \subsection{Potenzen in $\mathbb{F}_p$ und diskreter Logarithmus \label{buch:subsection:potenzen-diskreter-logarithmus}} @@ -439,6 +448,7 @@ Das Polynom ist \[ p(t) = +XXX \] Nach Division durch $t(t-1)$ erhält man als den Quotienten \begin{align*} @@ -652,13 +662,44 @@ Diese Operationen machen $E_{a,b}(\mathbb{F}_{p^l})$ zu einer endlichen abelschen Gruppe. \end{satz} -\subsubsection{Beispiele} -% XXX -TODO: elliptische Kurven in IPsec: Oakley Gruppen - \subsubsection{Diffie-Hellman in einer elliptischen Kurve} -% XXX -TODO: $g^x$ in einer elliptischen Kurve +Der klassische Diffie-Hellmann-Schlüsselalgorithmus in einem Körper +$\mathbb{F}_p$ basiert darauf, dass man beliebige Potenzen eines +Elementes berechnen kann, und dass es schwierig ist, diese Operation +umzukehren. +Die Addition in $\mathbb{F}_p$ wird für diesen Algorithmus überhaupt +nicht benötigt. + +In einer elliptischen Kurve gibt es ebenfalls eine Multiplikation, +aus der sich mit dem +Algorithmus~\ref{buch:crypto:teile-und-hersche} eine effizienter +Potenzieralgorithmus konstruieren lässt. + +Die im Internet Key Exchange Protokol +in RFC 2409 +\cite{buch:rfc2409} +definierte Oakley-Gruppe 4 +zum Beispiel verwendet einen Galois-Körper $\mathbb{F}_{2^{185}}$ +mit dem Minimalpolynom $m(x)=x^{185}+x^{69}+1\in \mathbb{F}_2[x]$ +und den Koeffizienten +\begin{align*} +a&=0\\ +b&=x^{12}+x^{11} + x^{10} + x^9 + x^7 + x^6 + x^5 + x^3 +1, +\end{align*} +die die elliptische Kurve definieren. + +Als Elemente $g$ für den Diffie-Hellmann-Algorithmus wird ein Punkt +der elliptischen Kurve verwendet, dessen $X$-Koordinaten durch das +Polynom $g_x = x^4+x^3$ gegeben ist. +Der Standard spezifiziert die $Y$-Koordinate nicht, diese kann aus +den gegebenen Daten abgeleitet werden. +Die entstehende Gruppe hat etwa $4.9040\cdot10^{55}$ Elemente, die +für einen brute-force-Angriff durchprobiert werden müssten. + + + + + diff --git a/buch/chapters/90-crypto/rs.tex b/buch/chapters/90-crypto/rs.tex deleted file mode 100644 index ec8ec8c..0000000 --- a/buch/chapters/90-crypto/rs.tex +++ /dev/null @@ -1,41 +0,0 @@ -% -% rs.tex -- Reed-Solomon-Code -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Fehlerkorrigierende Codes nach Reed-Solomon -\label{buch:section:reed-solomon}} -\rhead{Fehlerkorrigierende Codes} -Jede Art von Datenübertragung muss sich mit dem Problem der Fehler befassen, -die auf dem Übertragungskanal entstehen können. -Die einfachste Lösung dieses Problem versucht, Fehler zu erkennen und -dann eine erneute Übermittelung zu veranlassen. -Dies ist zum Beispiel bei der Datenübertragung von einer Raumsonde -wie Voyager~1 nicht möglich, die Signallaufzeit von der Sonde und wieder -zurück ist über 40 Stunden. -Es ist auch nicht sinnvoll beim Lesen eines optischen Mediums wie einer -CD oder DVD, wenn ein Fehler durch eine Beschädigung der Oberfläche -des Mediums verursacht wird. -Erneutes Lesen würde das Resultat auch nicht ändern. -Es wird also eine Möglichkeit gesucht, die Daten so zu codieren, dass -ein Fehler nicht nur erkannt sondern auch korrigiert werden kann. - -In diesem Abschnitt werden die algebraisch besonders interessanten -Reed-Solmon-Codes beschrieben. -Ihren ersten Einsatz hatten Sie bei den Voyager-Raumsonden, die 1977 -gestartet wurden. -Sie befinden sich im Moment in einer Entfernung von -Zum ersten mal kommerziell verwendet wurden sie für die optischen -Medien CD und DVD. - -% https://www.youtube.com/watch?v=uOLW43OIZJ0 -% https://www.youtube.com/watch?v=4BfCmZgOKP8 - -\subsection{Was ist ein Code? -\label{buch:subsection:was-ist-ein-code}} - -\subsection{Reed-Solomon-Code -\label{buch:subsection:reed-solomon-code}} - -\subsection{Decodierung -\label{buch:subsection:decodierung}} diff --git a/buch/chapters/95-homologie/Makefile.inc b/buch/chapters/95-homologie/Makefile.inc index 7e6f1e7..3b2b50c 100644 --- a/buch/chapters/95-homologie/Makefile.inc +++ b/buch/chapters/95-homologie/Makefile.inc @@ -8,8 +8,11 @@ CHAPTERFILES = $(CHAPTERFILES) \ chapters/95-homologie/simplex.tex \ chapters/95-homologie/komplex.tex \ chapters/95-homologie/homologie.tex \ - chapters/95-homologie/mayervietoris.tex \ + chapters/95-homologie/homologieketten.tex \ + chapters/95-homologie/basiswahl.tex \ chapters/95-homologie/fixpunkte.tex \ + chapters/95-homologie/eulerchar.tex \ + chapters/95-homologie/induzierteabb.tex \ chapters/95-homologie/chapter.tex diff --git a/buch/chapters/95-homologie/basiswahl.tex b/buch/chapters/95-homologie/basiswahl.tex new file mode 100644 index 0000000..aacfa9f --- /dev/null +++ b/buch/chapters/95-homologie/basiswahl.tex @@ -0,0 +1,817 @@ +\subsection{Basiswahl +\label{buch:subsection:basiswahl}} +Die Definition der Homologiegruppen $H_k(C)$ als Quotient von +Vektorräumen ist ziemlich abstrakt. +Sie besteht aus Klassen von Zyklen, die sich höchstens um einen +Rand unterscheiden. +Indem wir eine geeignete Basis wählen, können wir konkrete Zyklen +identifizieren, die eine Basis für den Vektorraum $H_k(C)$ bilden. +Dies soll im Folgenden schrittweise durchgeführt werden. + +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/gausshomoex.pdf} +\caption{Beispiel für die Berechnung von Basisvektoren und Homologieklassen +mit Hilfe des Gauss-Algorithmus +\label{buch:homologie:fig:gausshomoex}} +\end{figure} + +\subsubsection{Basis von $Z_k(C)$} +Um eine Basis für $H_k(C)$ zu konstruieren, ist es zunächst nötig, +eine Basis der Zyklen $Z_k(C)$ zu bestimmen. +Ausgehend von einer beliebigen Basis der $C_k$ und einer +zugehörigen Darstellung des Randoperators $\partial_k$ als +Matrix, kann eine Basis von Zyklen mit Hilfe des Gauss-Algorithmus +gefunden werden. +Wir bezeichnen die Menge dieser Zyklen mit +\[ +\mathcal{Z}_k += +\{ +z_1^{(k)}, +z_2^{(k)}, +\dots, +z_l^{(k)} +\}. +\] +$\mathcal{Z}_k$ erzeugt den $l$-dimensionalen Vektorraum $Z_k(C)$. + +\begin{beispiel} +\label{buch:homologie:beispiel:gausshomo} +In Abbildung~\ref{buch:homologie:fig:gausshomoex} ist ein Polyeder +dargestellt, dessen Homologiegruppe $H_1$ berechnet werden soll. +Um eine Basis für die Zyklen zu berechnen, wird zunächst die Matrix +des Randoperators $\partial_1$ aufgestellt. +Sie ist +\[ +\setcounter{MaxMatrixCols}{27} +\partial_1 += +\footnotesize +\setlength\arraycolsep{2pt} +\begin{pmatrix*}[r] +%1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 +-1& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 1 + 1&-1& 0& 0& 0&-1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 2 + 0& 1&-1& 0& 0& 0& 0&-1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 3 + 0& 0& 1&-1& 0& 0& 0& 0& 0&-1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 4 + 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 5 + 0& 0& 0& 0& 1& 1& 0& 0& 0& 0& 0& 0&-1& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 6 + 0& 0& 0& 0& 0& 0&-1& 1& 0& 0& 0& 0& 1& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 7 + 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 0& 0& 0&-1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 8 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 0& 1& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ % 9 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 0& 0&-1& 0& 1& 0& 0& 0& 0& 0& 0\\ %10 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1&-1& 0&-1& 1& 0& 0& 0& 0\\ %11 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 0& 1& 0& 0& 0&-1& 0& 0& 0\\ %12 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 0& 0&-1& 1& 0\\ %13 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1& 1& 0&-1\\ %14 + 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&-1& 1\\ %15 +\end{pmatrix*} +\] +Die reduzierte Zeilenstufenform von $\partial_1$ ist +(Pivotpositionen in {\color{red}rot}, frei wählbare Variablen +in {\color{darkgreen}grün}) +\begin{center} +%\tiny +\scriptsize +%\footnotesize +\setlength\tabcolsep{3pt} +\begin{tabular}{|>{$}r<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|} +\hline + & 1& 2& 3& 4& 5&{\color{darkgreen}6}& 7&{\color{darkgreen}8}& 9&{\color{darkgreen}10}&11&{\color{darkgreen}12}&{\color{darkgreen}13}&{\color{darkgreen}14}&15&{\color{darkgreen}16}&17&{\color{darkgreen}18}&19&{\color{darkgreen}20}&21&{\color{darkgreen}22}&23&{\color{darkgreen}24}&{\color{darkgreen}25}&26&{\color{darkgreen}27}\\ +\hline + 1&\phantom{-}{\color{red}1}& 0& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 1& 0& 0&-1& 0& 0& 0& 1& 0& 0& 0&-1& 0& 0& 0\\ + 2& 0&\phantom{-}{\color{red}1}& 0& 0& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0&-1& 0& 0& 0\\ + 3& 0& 0&\phantom{-}{\color{red}1}& 0& 0& 0& 0& 0& 0&-1& 0& 0& 0& 1& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ + 4& 0& 0& 0&\phantom{-}{\color{red}1}& 0& 0& 0& 0& 0& 0& 0&-1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ + 5& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0& 0& 0& 0& 0& 0& 1& 0& 0&-1& 0& 0& 0& 1& 0& 0& 0&-1& 0& 0& 0\\ + 6& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0& 0& 0& 0&-1& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ + 7& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0& 0& 0& 1& 0& 0& 0&-1& 0&-1& 0& 0& 0& 1& 0& 0& 0\\ + 8& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0&-1& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ + 9& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0& 0& 0& 1& 0& 0& 0&-1& 0& 0& 0\\ +10& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0&-1& 0& 0& 0& 1& 0& 0& 0\\ +11& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0&-1& 0& 1& 1& 0&-1\\ +12& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1& 0& 0& 1& 0&-1\\ +13& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1&-1& 0& 1\\ +14& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0&\phantom{-}{\color{red}1}&-1\\ +15& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ +\hline +\end{tabular}. +\end{center} +Daraus kann man die Zyklen wie folgt ablesen, indem man jeweils +genau eine frei wählbare Variable auf $1$ setzt: +\begin{align*} +z_1 +&= +\tiny +\begin{pmatrix*}[r] +\phantom{-} + 1\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_2 +&= +\tiny +\begin{pmatrix*}[r] +\phantom{-} + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_3 +&= +\tiny +\begin{pmatrix*}[r] +\phantom{-} + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_4 % variable 12 = 1 +&= +\tiny +\begin{pmatrix*}[r] +\phantom{-} + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_5 % variable 13 = 1 +&= +\tiny +\begin{pmatrix*}[r] +-1\\ + 0\\ + 0\\ + 0\\ +-1\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_6 % variable 14 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 0\\ + 0\\ +-1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ +-1\\ + 0\\ + 1\\ + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_7 % variable 16 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 1\\ + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ +-1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*},\\ +z_8 % variable 18 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ +-1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_9 % variable 20 = 1 +&= +\tiny +\begin{pmatrix*}[r] +-1\\ +-1\\ + 0\\ + 0\\ +-1\\ + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ +-1\\ + 0\\ + 1\\ + 0\\ + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0 +\end{pmatrix*}, +&z_{10} % variable 22 = 1 +&= +\tiny +\begin{pmatrix*}[r] +\phantom{-} + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ %5 + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ %10 + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ %15 + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ %20 + 1\\ + 1\\ + 0\\ + 0\\ + 0\\ %25 + 0\\ + 0 +\end{pmatrix*}, +&z_{11} % variable 24 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 1\\ + 1\\ + 0\\ + 0\\ + 1\\ %5 + 0\\ + 0\\ + 0\\ +-1\\ + 0\\ %10 + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ %15 + 0\\ +-1\\ + 0\\ +-1\\ + 0\\ %20 + 0\\ + 0\\ + 1\\ + 1\\ + 0\\ %25 + 0\\ + 0 +\end{pmatrix*}, +&z_{12} % variable 25 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ %10 + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ %15 + 0\\ + 0\\ + 0\\ +-1\\ + 0\\ %20 +-1\\ + 0\\ + 1\\ + 0\\ + 1\\ %25 + 0\\ + 0 +\end{pmatrix*}, +&z_{13} % variable 27 = 1 +&= +\tiny +\begin{pmatrix*}[r] + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 0\\ + 1\\ + 0\\ %20 + 1\\ + 0\\ +-1\\ + 0\\ + 0\\ %25 + 1\\ + 1 +\end{pmatrix*} +\end{align*} +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/homocycles.pdf} +\caption{Zyklen des Randoperators $\partial_1$ im Beispiel von +Seite~\pageref{buch:homologie:beispiel:gausshomo}. +\label{buch:homologie:fig:homocycles}} +\end{figure}% +Die Zyklen sind in Abbildung~\ref{buch:homologie:fig:homocycles} {\color{red}rot} dargestellt. +\end{beispiel} + +\subsubsection{Basis für $B_k(C)$} +Da $B_k(C)\subset Z_k(C)$ gilt, lässt sich für jedes $c_{k+1}\in C_{k+1}$ +der Rand $\partial_{k+1}c_{k+1}$ als Linearkombination der im +vorangegangenen Schritt gefundenen Basiszyklen finden. +Wir können also aus der Standardbasis $e^{(k+1)}_i\in C_{k+1}$ eine Menge +von Vektoren $\partial_{k+1}e^{(k+1)}_i$ gewinnen, die mit Sicherheit +ganz $B_k(C)$ aufspannen. +Es ist aber davon auszugehen, dass diese Vektoren nicht linear unabhängig +sind. +Es ist also nötig, eine Teilmenge +\[ +\mathcal{B}_k += +\{ +\partial_{k+1}e^{(k+1)}_{i_1}, +\partial_{k+1}e^{(k+1)}_{i_2}, +\dots, +\partial_{k+1}e^{(k+1)}_{i_m} +\} +\] +von Vektoren auszuwählen, die linear +unabhängig sind. +Diese bilden eine Basis von $B_k(C)$. + +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/homoboundaries.pdf} +\caption{Die Ränder $\partial_2e_i^{(2)}$ für das Beispiel von +Seite~\pageref{buch:homologie:beispiel:gausshomo}. +Die grauen Dreiecke bilden die Standardbasis $e_i^{(2)}$ von $C_2$, +die blauen Dreiecke sind die Ränder $\partial_2e_i^{(2)}$ dieser +Dreiecke. +\label{buch:homologie:fig:homoboundaries}} +\end{figure} + +Aus den Abbildungen~\ref{buch:homologie:fig:homocycles} und +\ref{buch:homologie:fig:homoboundaries} kann man auch ablesen, +wie die Ränder $\partial_2e_i^{(2)}$ aus den Zyklen von $\mathcal{Z}_1$ +linear kombiniert werden können. +Man erhält so die Beziehungen +\begin{equation} +\setcounter{MaxMatrixCols}{29} +\setlength\arraycolsep{1pt} +\begin{array}{lcrcrcrcrcrcrcrcrcrcrcrcrcr} +\partial_2e_1^{(2)} &=&z_1& & & & & & & & & & & & & & & & & & & & & & & & \\ +\partial_2e_2^{(2)} &=& & &z_2& & & & & & & & & & & & & & & & & & & & & & \\ +\partial_2e_3^{(2)} &=& & & & &z_3& & & & & & & & & & & & & & & & & & & & \\ +\partial_2e_4^{(2)} &=& & & & & & &z_4& & & & & & & & & & & & & & & & & & \\ +\partial_2e_5^{(2)} &=& & & & & & & & &z_5& & &+&z_7& & & & & & & & & & & & \\ +\partial_2e_6^{(2)} &=& & & & & & & & & & &z_6& & &+&z_8& & & & & & & & & & \\ +\partial_2e_7^{(2)} &=& & & & & & & & & & & & & & & & & & &z_{10}& & & & & & \\ +\partial_2e_8^{(2)} &=& & & & & & & & & & & & & & & & & & & & &z_{11}& & & & \\ +\partial_2e_9^{(2)} &=& &\phantom{+}& &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & &\phantom{+} & & &z_{12}&+&z_{13} +\end{array} +\end{equation} +Dies reicht jedoch nicht, um herauszufinden, welche der blauen Dreiecke +linear unabhängig sind. +Im vorliegenden Fall ist dies einfach: jedes blaue Dreieck besteht aus +Kanten, die in keinem anderen blauen Dreieck vorkommen, daher müssen +sie alle linear unabhängig sein. + +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/gausshomobasis.pdf} +\caption{Bestimmung einer Basis für die Homologiegruppe $H_k(C)$ mit +Hilfe der Vorwärtsreduktion des Gaussalgorithmus. +Die schwarzen Nullzeilen zeigen an, welche Zeilenvektoren zusammen mit +den darüberliegenden Vektoren nicht linear unabhängig sind und damit nicht +in Frage kommen für die besuchte Basis. +Übrig bleiben die {\color{red}rot} und {\color{darkgreen}grün} hervorgehobenen +Vektoren. +\label{buch:homologie:fig:gausshomobasis}} +\end{figure} + +Diese Auswahl lässt sich sehr leicht mit Hilfe der folgenden +Variante des Gauss-Algorithmus realisieren. +Dazu werden die $n_{k+1}$ Zeilen Gauss-Tableau zunächst mit den Vektoren +$\partial_{k+1}{e_i^{(k+1)}}^t$ gefüllt. +Führt man in diesem Tableau die Vorwärtsreduktion durch, wobei man +entstehende Nullzeilen einfach überspringt, bleiben nur noch Zeilen +übrig, die linear unabhängig sind. +Diese Zeilen entsprechen den linear unabhängigen Vektoren von $\mathcal{B}_k$, +die Zeilennummern sind $i_1,i_2,\dots,i_m$. +Dieses Vorgehen ist schematisch im oberen Teil der +Abbildung~\ref{buch:homologie:fig:gausshomobasis} dargestellt. + +\subsubsection{Basis für die Homologiegruppe $H_k(C)$} +Um eine Basis von $H_k(C)$ zu konstruieren, müssen wir jetzt eine +Basis von Zyklen finden, die sich nicht nur um einen Rand unterscheiden, +die also zu verschiedenen Homologie-Klassen in $H_k(C)$ gehören. +Gesucht sind jetzt also Vektoren $\mathcal{Z}'_k$ derart, dass +die Vektoren von $\mathcal{Z}'_k\cup\mathcal{B}_k$ immer noch $Z_k(C)$ +aufspannen, aber zusätzlich linear unabhängig sind. + +Dazu kann man wie folgt vorgehen. +\begin{enumerate} +\item +Man beginnt mit $\mathcal{D}_0=\emptyset$ und setzt $j=0$. +\item +Dann testet man der Reihe nach alle noch nicht getesteten Vektoren +von $z_i^{(k)}\in\mathcal{Z}_k$ daraufhin, ob sie von den Vektoren +$\mathcal{B}_k\cup \mathcal{D}_j$ linear unabhängig sind. +Wenn ja, bildet man $\mathcal{D}_{j+1} = \mathcal{D}\cup\{z^{(k)}_i\}$ und +setzt $j=1$. +Andernfalls ignoriert man $z^{(k)}_i$. +\item +Schritt 2 wird wiederholt, bis man alle Vektoren von $\mathcal{Z}_k$ +getestet hat. +Die gesuchte Basis setzt sich zusammen aus $\mathcal{B}_k$ und +$\mathcal{D}_l$, +also +$ +\mathcal{Z}_k' += +\mathcal{B}_k +\cup +\mathcal{D}_l. +$ +\end{enumerate} + +Dieser Algorithmus kann ebenfalls mit der oben angesprochenen Variante +des Gauss-Algorithmus durchgeführt werden. +Dazu werden die Zeilen $n_k+1$ bis $n_k+1+|\mathcal{Z}_k|$ mit den +Vektoren $z_i^t$. +Dann führt man die Vorwärtsreduktion im ganzen Tableau durch, wobei +man wieder die Nullzeilen stehen lässt. +Nullzeilen zeigen wieder Vektoren an, die sich linear durch die darüber +liegenden Vektoren ausdrücken lassen. +Die auszuwählenden Vektoren sind daher genau diejenigen, die für +$\mathcal{Z}_k'$ ausgewählt werden müssen. + +Um den Algorithmus durchzuführen, bilden wir daher das Gauss-Tableau +in Abbildung~\ref{buch:homologie:beispiel:gausstableau}, +bestehend aus den Vektoren $\partial_2e_i^{(2)}$ in den ersten 9 +Zeilen und den Zyklen $z_1,\dots,z_{13}$ in den folgenden 13 Zeilen. +Das reduzierte Tableau nach der Vorwärtsreduktion ist in +Abbildung~\ref{buch:homologie:beispiel:gausstableaureduziert} +dargestellt, amn erkennt, dass die Zyklen $z_1$ bis $z_4$, $z_7$ und $z_8$, +$z_9$ und $z_{10}$ sowie $z_{13}$ weggelassen werden müssen. +Es bleiben die folgenden Zyklen: +\begin{center} +\begin{tabular}{>{$}l<{$}l} +\text{Zyklus}&Eigenschaft\\ +\hline +z_5 &Zyklus umschliesst das kleine weisse Dreieck links unten\\ +z_6 &Zyklus umschliesst das kleine weisse Dreieck rechts unten\\ +z_9 &Zyklus umschliesst das grosse weisse Dreieck\\ +z_{12}&Zyklus umschliesst das kleine weisse Dreicke oben\\ +\hline +\end{tabular} +\end{center} +Die Zyklen, die nach der Reduktion übrig bleiben, sind in +Abbildung~\ref{buch:homologie:beispiel:homoclasses} zusammengestellt. +Jede solche Klasse entspricht genau einem der ``Löcher'', der weissen +Dreiecke. +Die Homologie kann man also als eine exakte Version der Idee eines +Vektorraums erzeugt von den ``Löchern'' eines Polygons verstehen. + +\begin{figure} +\centering +\setlength\tabcolsep{1pt} +\begin{tabular}{|>{$}c<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|} +\hline +&\scriptstyle 1&\scriptstyle 2&\scriptstyle 3&\scriptstyle 4 &\scriptstyle 5 +&\scriptstyle 6 &\scriptstyle 7 &\scriptstyle 8 &\scriptstyle 9 &\scriptstyle 10 +&\scriptstyle 11 &\scriptstyle 12 &\scriptstyle 13 &\scriptstyle 14 &\scriptstyle 15 +&\scriptstyle 16 &\scriptstyle 17 &\scriptstyle 18 &\scriptstyle 19 &\scriptstyle 20 +&\scriptstyle 21 &\scriptstyle 22 &\scriptstyle 23 &\scriptstyle 24 &\scriptstyle 25 +&\scriptstyle 26 &\scriptstyle 27 +\\ +% 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 +\hline +\scriptstyle\partial_2e_1^{(2)}& 1& & & & 1&\phantom{-}1& & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_2^{(2)}& & 1& & & & & 1&\phantom{-}1& & & & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_3^{(2)}& & & 1& & & & & &\phantom{-}1&\phantom{-}1& & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_4^{(2)}& & & &\phantom{-}1& & & & & & & 1&\phantom{-}1& & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_5^{(2)}& & & & & & & & & & & & & 1& & 1&\phantom{-}1& & & & & & & & & & & \\ +\scriptstyle\partial_2e_6^{(2)}& & & & & & & & & & & & & &\phantom{-}1& & & 1&\phantom{-}1& & & & & & & & & \\ +\scriptstyle\partial_2e_7^{(2)}& & & & & & & & & & & & & & & & & & & 1& &\phantom{-}1& 1& & & & & \\ +\scriptstyle\partial_2e_8^{(2)}& & & & & & & & & & & & & & & & & & & &\phantom{-}1& & & 1&\phantom{-}1& & & \\ +\scriptstyle\partial_2e_9^{(2)}& & & & & & & & & & & & & & & & & & & & & & & & &\phantom{-}1&\phantom{-}1&\phantom{-}1\\ +\hline +% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 +\scriptstyle z_{ 1}& 1& & & & 1& 1& & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 2}& & 1& & & & & 1& 1& & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 3}& & & 1& & & & & & 1& 1& & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 4}& & & & 1& & & & & & & 1& 1& & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 5}&-1& & & &-1& & 1& & & & & & 1& & & & & & & & & & & & & & \\ +\scriptstyle z_{ 6}& & &-1& & & & & &-1& & 1& & & 1& & & & & & & & & & & & & \\ +\scriptstyle z_{ 7}& 1& & & & 1& &-1& & & & & & & & 1& 1& & & & & & & & & & & \\ +\scriptstyle z_{ 8}& & & 1& & & & & & 1& &-1& & & & & & 1& 1& & & & & & & & & \\ +\scriptstyle z_{ 9}&-1&-1& & & 1& & & & 1& & & & & &-1& & 1& 1& 1& & & & & & & & \\ +\scriptstyle z_{10}& & & & & & & & & & & & & & & & & & 1& & 1& 1& & & & & & \\ +\scriptstyle z_{11}& 1& 1& & & 1& & & &-1& & & & & & 1& &-1& &-1& & & & 1& 1& & & \\ +\scriptstyle z_{12}& & & & & & & & & & & & & & & & & & &-1& &-1& & 1& & 1& & \\ +\scriptstyle z_{13}& & & & & & & & & & & & & & & & & & & 1& & 1& &-1& & & 1& 1\\ +\hline +\end{tabular} +\caption{Gauss-Tableau für die Bestimmung einer Basis von +$H_1$ für das Beispiel. +Die ersten neuen Zeilen bestehen aus den Bildern der +Basisvektoren von $C_2$. +Im vorliegenden Fall kann man sofort sehen, dass alle diese +Zeilen linear unabhängig sind. +Die folgenden Zeilen sind die Zyklen in $\mathbb{Z}_2$, sie +sind ebenfalls linear unabhängig. +Mit Hilfe der Vorwärtsreduktion müssen jetzt diejenigen +Zeilen elminiert werden, die bereits aus anderen Zyklen +mit Hilfe von Rändern der Zeilen 1--9 kombiniert werden können. +\label{buch:homologie:beispiel:gausstableau}} +\end{figure} + +\begin{figure} +\centering +\setlength\tabcolsep{1pt} +\begin{tabular}{|>{$}c<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}>{$}r<{$}|} +\hline +&\scriptstyle 1&\scriptstyle 2&\scriptstyle 3&\scriptstyle 4 &\scriptstyle 5 +&\scriptstyle 6 &\scriptstyle 7 &\scriptstyle 8 &\scriptstyle 9 &\scriptstyle 10 +&\scriptstyle 11 &\scriptstyle 12 &\scriptstyle 13 &\scriptstyle 14 &\scriptstyle 15 +&\scriptstyle 16 &\scriptstyle 17 &\scriptstyle 18 &\scriptstyle 19 &\scriptstyle 20 +&\scriptstyle 21 &\scriptstyle 22 &\scriptstyle 23 &\scriptstyle 24 &\scriptstyle 25 +&\scriptstyle 26 &\scriptstyle 27 +\\ +% 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 +\hline +\scriptstyle\partial_2e_1^{(2)}&\phantom{-}1& & & &\phantom{-}1&\phantom{-}1& & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_2^{(2)}& &\phantom{-}1& & & & &\phantom{-}1&\phantom{-}1& & & & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_3^{(2)}& & &\phantom{-}1& & & & & &\phantom{-}1&\phantom{-}1& & & & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_4^{(2)}& & & &\phantom{-}1& & & & & & &\phantom{-}1&\phantom{-}1& & & & & & & & & & & & & & & \\ +\scriptstyle\partial_2e_5^{(2)}& & & & & & & & & & & & & 1& & 1&\phantom{-}1& & & & & & & & & & & \\ +\scriptstyle\partial_2e_6^{(2)}& & & & & & & & & & & & & &\phantom{-}1& & & 1&\phantom{-}1& & & & & & & & & \\ +\scriptstyle\partial_2e_7^{(2)}& & & & & & & & & & & & & & & & & & & 1& &\phantom{-}1& 1& & & & & \\ +\scriptstyle\partial_2e_8^{(2)}& & & & & & & & & & & & & & & & & & & &\phantom{-}1& & & 1&\phantom{-}1& & & \\ +\scriptstyle\partial_2e_9^{(2)}& & & & & & & & & & & & & & & & & & & & & & & & &\phantom{-}1&\phantom{-}1&\phantom{-}1\\ +\hline +% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 +\scriptstyle z_{ 1}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 2}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 3}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 4}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 5}'& & & & & & 1& 1& & & & & & & &-1&-1& & & & & & & & & & & \\ +\scriptstyle z_{ 6}'& & & & & & & & & & 1& 1& & & & & &-1&-1& & & & & & & & & \\ +\scriptstyle z_{ 7}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 8}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{ 9}'& & & & & & & & 1& 1& & & & & & & 1& 1& & & &-1&-1&-1&-1& & & \\ +\scriptstyle z_{10}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{11}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\scriptstyle z_{12}'& & & & & & & & & & & & & & & & & & & & & & 1& 1& & &-1&-1\\ +\scriptstyle z_{13}'& & & & & & & & & & & & & & & & & & & & & & & & & & & \\ +\hline +\end{tabular} +\caption{Nach Durchführung der Vorwärtsreduktion kann man die Zyklen +ablesen, die nicht für eine Basis von $H_1$ gebraucht werden. +Die resultierenden Zyklen sind in Abbildung~\ref{buch:homologie:beispiel:homoclasses} +dargestellt. +\label{buch:homologie:beispiel:gausstableaureduziert}} +\end{figure} + +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/homoclasses.pdf} +\caption{Repräsentanten für die reduzierten Klassen aus dem +Tableau von +Abbildung~\ref{buch:homologie:beispiel:gausstableaureduziert}, +sie bilden eine Basis der Homologie-Gruppe $H_1$. +Jeder dieser Repräsentanten umschliesst genau ein ``Loch'', +also genau ein weisses Dreieck. +\label{buch:homologie:beispiel:homoclasses}} +\end{figure} + +\subsubsection{Basis von $H_k(C)$} +Die im vorangegangenen Abschnitt konstruierte Basis kann jetzt auch +dazu verwendet werden, eine Basis von $H_k(C)$ zu finden. +Die Vektoren in $\mathcal{B}_k$ bilden eine Basis von $B_k(C)$ +und die Vektoren in $\mathcal{Z}_k'$ sind davon unabhängig. +Die Klassen der Vektoren von $\mathcal{Z}_k'$ in $H_k(C)$ sind +daher ebenfalls linear unabhängig und bilden damit eine Basis +von $H_k(C)$. +Die von obigem Algorithmus ausgewählten Zyklen bilden also automatisch +eine Basis von Zyklen, die nicht Rand irgend einer Kette in $C_{k+1}$ +sein können. diff --git a/buch/chapters/95-homologie/chapter.tex b/buch/chapters/95-homologie/chapter.tex index eaa56c4..e25188c 100644 --- a/buch/chapters/95-homologie/chapter.tex +++ b/buch/chapters/95-homologie/chapter.tex @@ -3,9 +3,9 @@ % % (c) 2021 Prof Dr Andreas Müller, Hochschule Rapperswil % -\chapter{Homologie +\chapter{Kettenkomplexe und Homologie \label{buch:chapter:homologie}} -\lhead{Homologie} +\lhead{Kettenkomplexe und Homologie} \rhead{} Mit der Inzidenzmatrix war es möglich, einen Graphen zu beschreiben und verschiedene interessante Eigenschaften desselben zu berechnen. @@ -35,10 +35,20 @@ Der sogenannte Randoperator ordnet jedem Dreieck, Tetraeder oder allgemein jedem Simplex seinen Rand zu. Damit wird es möglich, das Dreieck vom Rand des Dreiecks zu unterschieden. +Die Verallgemeinerung dieser Idee liefert eine algebraische Konstruktion +zu jedem topologischen Raum, die sogenannten Homologie-Gruppen. +Sie formalisieren ein mögliches Konzept der Dimension und der +Idee von ``Löchern'' in einem topologischen Raum. +Sie können dabei helfen, die topologische Struktur verschiedener +Räume zu unterscheiden. +Das Ziel dieses Kapitels ist nicht, die Homologietheorie +vollständig zu entwickeln, sondern zu zeigen, wie man Matrizen +verwenden kann, um konkrete Rechnungen durchzuführen. + \input{chapters/95-homologie/simplex.tex} \input{chapters/95-homologie/komplex.tex} \input{chapters/95-homologie/homologie.tex} -\input{chapters/95-homologie/mayervietoris.tex} +%\input{chapters/95-homologie/mayervietoris.tex} \input{chapters/95-homologie/fixpunkte.tex} diff --git a/buch/chapters/95-homologie/eulerchar.tex b/buch/chapters/95-homologie/eulerchar.tex new file mode 100644 index 0000000..03e389b --- /dev/null +++ b/buch/chapters/95-homologie/eulerchar.tex @@ -0,0 +1,139 @@ +\subsection{Euler-Charakteristik} +Die Homologiegruppen fassen die Idee, die ``Löcher'' in +Dimension $k$ eines Polyeders zu zählen, algebraisch exakt. +Dazu ist aber die algebraische Struktur von $H_k(C)$ gar +nicht nötig, nur schon die Dimension des Vektorraumes $H_k(C)$ +liefert bereits die verlange Information. + +Dies ist auch der Ansatz, den der eulersche Polyedersatz verfolgt. +Euler hat für dreidimensionale Polyeder eine Invariante gefunden, +die unabhängig ist von der Triangulation. + +\begin{definition} +\label{buch:homologie:def:eulerchar0} +Ist $E$ die Anzahl der Ecken, $K$ die Anzahl der Kanten und $F$ +die Anzahl der Flächen eines dreidimensionalen Polyeders $P$, dann +heisst +\[ +\chi(P) = E-K+F +\] +die {\em Euler-Charakteristik} des Polyeders $P$. +\end{definition} + +Der Eulersche Polyedersatz, den wir nicht gesondert beweisen +wollen, besagt, dass $\chi(P)$ unabhängig ist von der +Triangulation. +Alle regelmässigen Polyeder sind verschiedene Triangulationen +einer Kugel, sie haben alle den gleichen Wert $2$ +der Euler-Charakteristik. + +Ändert man die Triangulation, dann wird die Dimension der +Vektorräume $B_k(C)$ und $Z_k(C)$ grösser werden. +Kann man eine Grösse analog zu $\chi(P)$ finden, die sich nicht ändert? + +\begin{definition} +\label{buch:homologie:def:eulerchar} +Sei $C$ ein Kettenkomplex, dann heisst +\[ +\chi(C) = \sum_{k=0}^n (-1)^k\dim H_k(C) +\] +die Euler-Charakteristik von $C$. +\end{definition} + +Die Summe in Definition~\ref{buch:homologie:def:eulerchar} erstreckt +sich bis zum Index $n$, der Dimension des Simplexes höchster Dimension +in einem Polyeder. +Für $k>n$ ist $H_k(C)=0$, es ändert sich also nichts, wenn wir +die Summe bis $\infty$ erstrecken, da die zusätzlichen Terme alle +$0$ sind. +Wir werden dies im folgenden zur Vereinfachung der Notation tun. + +Die Definition verlangt, dass man erst die Homologiegruppen +berechnen muss, bevor man die Euler-Charakteristik bestimmen +kann. +Dies ist aber in vielen Fällen gar nicht nötig, da dies nur +eine Frage der Dimensionen ist, die man direkt aus den +$C_k$ ablesen kann, wie wir nun zeigen wollen. + +Die Dimension der Homologiegruppen ist +\begin{equation} +\dim H_k(C) += +\dim \bigl(Z_k(C) / B_k(C)\bigr) += +\dim Z_k(C) - \dim B_k(C). +\label{buch:homologie:eqn:dimHk} +\end{equation} +Die Bestimmung der Dimensionen der Zyklen und Ränder erfordert +aber immer noch, dass wir dafür Basen bestimmen müssen, es ist +also noch nichts eingespart. +Die Zyklen bilden den Kern von $\partial$, also +\[ +\dim Z_k(C) = \dim\ker \partial_k. +\] +Die Ränder $B_k(C)$ sind die Bilder von $\partial_{k+1}$, also +\[ +\dim B_k(C) += +\dim C_{k+1} - \ker\partial_{k+1} += +\dim C_{k+1} - \dim Z_{k+1}(C). +\] +Daraus kann man jetzt eine Formel für die Euler-Charakteristik +gewinnen. +Sie ist +\begin{align*} +\chi(C) +&= +\sum_{k=0}^\infty (-1)^k \dim H_k(C) +\\ +&= +\sum_{k=0}^\infty (-1)^k \bigl(\dim Z_k(C) - \dim B_k(C)\bigr) +\\ +&= +\sum_{k=0}^\infty (-1)^k \dim Z_k(C) +- +\sum_{k=0}^\infty (-1)^k \bigl(\dim C_{k+1} - \dim_{k+1}(C)\bigr) +\\ +&= +-\sum_{k=0}^\infty (-1)^k \dim C_{k+1} ++ +\sum_{k=0}^\infty (-1)^k \dim Z_k(C) ++ +\sum_{k=0}^\infty (-1)^k \dim Z_{k+1}(C). +\intertext{Indem wir in der letzten Summe den Summationsindex $k$ durch +$k-1$ ersetzen, können wir bis auf den ersten Term die Summen +der $\dim Z_k(C)$ zum Verschwinden bringen:} +&= +-\sum_{k=0}^\infty (-1)^k \dim C_{k+1} ++ +\sum_{k=0}^\infty (-1)^k \dim Z_k(C) +- +\sum_{k=1}^\infty (-1)^k \dim Z_k(C) +\\ +&= +\sum_{k=1}^\infty (-1)^k \dim C_{k} ++ +\dim \underbrace{Z_0(C)}_{\displaystyle =C_0}. +\intertext{In der letzten Umformung haben wir auch in der ersten +Summe den Summationsindex $k$ durch $k-1$ ersetzt. +Damit beginnt die Summation bei $k=1$. +Der fehlende Term ist genau der Term, der von den Summen der +$\dim Z_k(C)$ übrig bleibt. +Damit erhalten wir} +&= +\sum_{k=0}^\infty (-1)^k \dim C_{k}. +\end{align*} + +\begin{satz} +Für die Euler-Charakteristik eines endlichdimensionalen Kettenkomplexes $C$ gilt +\[ +\chi(C) += +\sum_{k=0}^\infty (-1)^k \dim H_k(C) += +\sum_{k=0}^\infty (-1)^k \dim C_k. +\] +\end{satz} +Im nächsten Abschnitt wird gezeigt, dass die Euler-Charakteristik +als Spezialfall der Lefshetz-Zahl verstanden werden kann. diff --git a/buch/chapters/95-homologie/fixpunkte.tex b/buch/chapters/95-homologie/fixpunkte.tex index 1ed51ef..b3b184e 100644 --- a/buch/chapters/95-homologie/fixpunkte.tex +++ b/buch/chapters/95-homologie/fixpunkte.tex @@ -11,15 +11,127 @@ selbst gehört die zugehörige lineare Abbildung $f_*\colon H_*(X)\to H_*(X)$ der Homologiegruppen. Diese linearen Abbildungen sind im Allgemeinen viel einfacher zu analysieren. -Zum Beispiel soll in Abschnitt~\ref{buch:subsection:lefshetz} -die Lefshetz-Spurformel abgeleitet werden, die eine Aussagen darüber -ermöglicht, ob eine Abbildung einen Fixpunkt haben kann. -In Abschnitt~\ref{buch:subsection:brower} wird gezeigt wie man damit -den Browerschen Fixpunktsatz beweisen kann, der besagt, dass jede -Abbildung eines Einheitsballs in sich selbst immer einen Fixpunkt hat. - -\subsection{Lefshetz-Spurformel -\label{buch:subsection:lefshetz}} - -\subsection{Brower-Fixpunktsatz -\label{buch:subsection:brower}} +%Zum Beispiel soll in Abschnitt~\ref{buch:subsection:lefshetz} +%die Lefshetz-Spurformel abgeleitet werden, die eine Aussagen darüber +%ermöglicht, ob eine Abbildung einen Fixpunkt haben kann. +%In Abschnitt~\ref{buch:subsection:brower} wird gezeigt wie man damit +%den Browerschen Fixpunktsatz beweisen kann, der besagt, dass jede +%Abbildung eines Einheitsballs in sich selbst immer einen Fixpunkt hat. + +%\subsection{Brower-Fixpunktsatz +%\label{buch:subsection:brower}} +% +%\begin{satz}[Brower] +%\end{satz} + +%\subsection{Lefshetz-Fixpunktsatz +%\label{buch:subsection:lefshetz}} +Eine Selbstabbildung $f_*\colon C_*\to C_*$ von Kettenkomplexen führt auf +eine Selbstabbiludng der Homologiegruppen $H(f)\colon H(C)\to H(C)$. +Da sowohl $H_k$ wie auch $C_k$ endlichdimensionale Vektorräume sind, +ist die Spur von $H_k(f)$ wohldefiniert. + +\begin{definition} +Die {\em Lefshetz-Zahl} einer Abbildung $f$ von Kettenkomplexen ist +\begin{equation} +\lambda(f) += +\sum_{k=0}^\infty +(-1)^k \operatorname{Spur}f_k += +\sum_{k=0}^\infty +(-1)^k \operatorname{Spur}(H_k(f)). +\label{buch:homologie:lefschetz-zahl} +\end{equation} +\end{definition} + +Die zweite Darstellung der Lefshetz-Zahl auf der rechten Seite ist +meistens viel leichter zu berechnen als die erste. +Die einzelnen Vektorräume eines Kettenkomplexes können haben typischerweise +eine hohe Dimension, so hoch wie die Anzahl der Simplizes der Triangulation. +Die Homologiegruppen dagegen haben typischerweise sehr viel kleinere +Dimension, die Matrizen $H_k(f)$ sind also relativ klein. +Es ist aber nicht klar, dass beide Berechnungsmethoden für die +Lefshetz-Zahl auf das gleiche Resultat führen müssen. + +\begin{figure} +\centering +\includegraphics[width=\textwidth]{chapters/95-homologie/images/approximation.pdf} +\caption{Stückweise lineare Approximation einer Abbildung derart, +dass die Bildpunkt von Knoten auf Gitterpunkte fallen. +Die Abbildung wird damit zu einer Abbildung von Polyedern und +die induzierte Abbildung der Kettenkomplexe lässt sich direkt berechnen. +Wenn die Auflösung des Gitters klein genug ist, hat die Approximation +einer Abbildung ohne Fixpunkte immer noch keine Fixpunkte. +\label{buch:homologie:fig:simplapprox}} +\end{figure}% + +\begin{proof}[Beweis] +Im Abschnitt~\ref{buch:subsection:induzierte-abbildung} wurde gezeigt, +dass die Basis des Komplexes immer so gewählt werden kann, dass für +die Spuren der Teilmatrizen von $f_k$ die +Formel~\eqref{buch:homologie:eqn:spur} gilt. +Damit kann jetzt die alternierenierden Summe der Spuren von $f_k$ ermittelt +werden: +\begin{align*} +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_k) +&= +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,B}) ++ +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,Z}) ++ +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k-1,B}) +\\ +&= +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,B}) ++ +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,Z}) +- +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,B}) +\\ +&= +\sum_{k=0}^\infty (-1)^k\operatorname{Spur}(f_{k,Z}). +\intertext{Die Abbildung $H_k(f)$ hat $f_{k,Z}$ als Matrix, also ist +die letzte Form gleichbedeutend mit} +&= +\sum_{k=0} (-1)^k\operatorname{Spur} H_k(f). +\end{align*} +Damit ist die Formel +\eqref{buch:homologie:lefschetz-zahl} +bewiesen. +\end{proof} + +Die Lefshetz-Zahl ist eine Invariante einer topologischen Abbildung, +die Aussagen über Fixpunkte zu machen erlaubt. + +\begin{satz} +Ist $f\colon X\to X$ eine Selbstabbildung eines kompakten Polyeders und +ist $\lambda(f) \ne 0$, dann hat $f$ einen Fixpunkt. +\end{satz} + +Im Folgenden soll nur ein heuristisches Argument gegeben werden, warum +ein solcher Satz wahr sein könnte. + + +Wenn eine Abbildung keinen Fixpunkt hat, dann ist $f(x) \ne x$ für alle +Punkte von $X$. +Da $X$ kompakt ist, gibt es einen minimalen Abstand $d$ zwischen $f(x)$ und $x$. +Wenn man also für $X$ eine Triangulation wählt, die wesentlich feiner ist +als dieser minimale Abstand, dann wird kein Simplex der Triangulation auf +Punkte im selben Simplex oder in einem Nachbarsimplex abgebildet wird. +Indem man nötigenfalls die Triangulation nochmals verfeinert, kann man auch +genügend Platz schaffen, dass man die Abbildung $f$ etwas modifizieren kann, +so dass auch die deformierte Abbildung immer noch diese Eigenschaft hat. +Die Abbildung~\ref{buch:homologie:fig:simplapprox} illustriert, wie eine +Abbildung durch eine andere approximiert werden kann, die die Triangulation +im Bildraum respektiert. + +Die zugehörige Abbildung des Kettenkomplexes der Triangulation hat damit +die Eigenschaft, dass kein Basisvektor auf sich selbst abgebildet wird. +Die Matrix der Abbildung hat daher keine Nullen auf der Diagonalen, und +damit ist auch die Spur dieser Abbildung Null: $\operatorname{Spur}(H_k(f))=0$ +für alle $k$. +Erst recht ist die Lefshetz-Zahl $\lambda(f)=0$. +Wenn also die Lefshetz-Zahl verschieden ist von Null, dann muss $f$ +notwendigerweise einen Fixpunkt haben. + diff --git a/buch/chapters/95-homologie/homologie.tex b/buch/chapters/95-homologie/homologie.tex index 2b80a17..747c00f 100644 --- a/buch/chapters/95-homologie/homologie.tex +++ b/buch/chapters/95-homologie/homologie.tex @@ -6,13 +6,36 @@ \section{Homologie \label{buch:section:homologie}} \rhead{Homologie} +Die Idee der Trangulation ermöglicht, komplizierte geometrische +Objekte mit einem einfachen ``Gerüst'' auszustatten und so zu +analysieren. +Projiziert man ein mit einer Kugel konzentrisches Tetraeder auf die +Kugel, entsteht eine Triangulation der Kugeloberfläche. +Statt eine Kugel zu studieren, kann man also auch ein Tetraeder untersuchen. -\subsection{Homologie eines Kettenkomplexes -\label{buch:subsection:homologie-eines-kettenkomplexes}} +Das Gerüst kann natürlich nicht mehr alle Eigenschaften des ursprünglichen +Objektes wiedergeben. +Im Beispiel der Kugel geht die Information darüber, dass es sich um eine +glatte Mannigfaltigkeit handelt, verloren. +Was aber bleibt, sind Eigenschaften des Zusammenhangs. +Wenn sich zwei Punkte mit Wegen verbinden lassen, dann gibt es auch eine +Triangulation mit eindimensionalen Simplices, die diese Punkte als Ecken +enthalten, die sich in der Triangulation mit einer Folge von Kanten +verbinden lassen. +Algebraisch bedeutet dies, dass die beiden Punkte der Rand eines +Weges sind. +Fragen der Verbindbarkeit von Punkten mit Wegen lassen sich also +dadurch studieren, dass man das geometrische Objekt auf einen Graphen +reduziert. -\subsection{Induzierte Abbildung -\label{buch:subsection:induzierte-abbildung}} +In diesem Abschnitt soll gezeigt werden, wie diese Idee auf höhere +Dimensionen ausgedehnt werden. +Es soll möglich werden, kompliziertere Fragen des Zusammenhangs, zum +Beispiel das Vorhandensein von Löchern mit algebraischen Mitteln +zu analysieren. -\subsection{Homologie eines simplizialen Komplexes -\label{buch:subsection:simplizialekomplexe}} +\input{chapters/95-homologie/homologieketten.tex} +\input{chapters/95-homologie/basiswahl.tex} +\input{chapters/95-homologie/eulerchar.tex} +\input{chapters/95-homologie/induzierteabb.tex} diff --git a/buch/chapters/95-homologie/homologieketten.tex b/buch/chapters/95-homologie/homologieketten.tex new file mode 100644 index 0000000..1b40147 --- /dev/null +++ b/buch/chapters/95-homologie/homologieketten.tex @@ -0,0 +1,286 @@ +% +% homologieketten.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\subsection{Homologie eines Kettenkomplexes +\label{buch:subsection:homologie-eines-kettenkomplexes}} +Wegzusammenhang lässt sich untersuchen, indem man in der Triangulation +nach Linearkombinationen von Kanten sucht, die als Rand die beiden Punkte +haben. +Zwei Punkte sind also nicht verbindbar und liegen damit in verschiedenen +Komponenten, wenn die beiden Punkte nicht Rand irgend einer +Linearkombination von Kanten sind. +Komponenten können also identifiziert werden, indem man unter allen +Linearkombinationen von Punkten, also $C_0$ all diejenigen ignoriert, +die Rand einer Linearkombinationv on Kanten sind, also $\partial_1C_1$. +Der Quotientenraum $H_0=C_0/\partial_1C_1$ enthält also für jede Komponente +eine Dimension. + +Eine Dimension höher könnten wir danach fragen, ob sich ein geschlossener +Weg zusammenziehen lässt. +In der Triangulation zeichnet sich ein geschlossener Weg dadurch aus, +dass jedes Ende einer Kante auch Anfang einer Folgekante ist, dass also +der Rand der Linearkombination von Kanten 0 ist. +Algebraisch bedeutet dies, dass wir uns für diejenigen Linearkombinationen +$z\in C_1$ interessieren, die keinen Rand haben, für die also $\partial_1z=0$ +gilt. + +\begin{definition} +Die Elemente von +\[ +Z_k += +Z_k^C += +\{z\in C_k\;|\; \partial_k z = 0\} += +\ker \partial_k +\] +heissen die {\em ($k$-dimensionalen) Zyklen} von $C_*$. +\end{definition} + +In einem Dreieck ist der Rand ein geschlossener Weg, der sich zusammenziehen +lässt, indem man ihn durch die Dreiecksfläche deformiert. +Entfernt man aber die Dreiecksfläche, ist diese Deformation nicht mehr +möglich. +Einen zusammenziehbaren Weg kann man sich also als den Rand eines Dreiecks +einer vorstellen. +``Löcher'' sind durch geschlossene Wege erkennbar, die nicht Rand eines +Dreiecks sein können. +Wir müssen also ``Ränder'' ignorieren. + +\begin{definition} +Die Elemente von +\[ +B_k += +B_k^C += +\{\partial_{k+1}z\;|\; C_{k+1}\} += +\operatorname{im} \partial_{k+1} +\] +heissen die {\em ($k$-dimensionalen) Ränder} von $C_*$. +\end{definition} + +Algebraisch ausgedrückt interessieren uns also nur Zyklen, die selbst +keine Ränder sind. +Der Quotientenraum $Z_1/B_1$ ignoriert unter den Zyklen diejenigen, die +Ränder sind, drückt also algebraisch die Idee des eindimensionalen +Zusammenhangs aus. +Wir definieren daher + +\begin{definition} +Die $k$-dimensionale Homologiegruppe des Kettenkomplexes $C_*$ ist +\[ +H_k(C) = Z_k/B_k = \ker \partial_k / \operatorname{im} \partial_{k+1}. +\] +Wenn nur von einem Kettenkomplex die Rede ist, kann auch $H_k(C)=H_k$ +abgekürzt werden. +\end{definition} + +% XXX Visualisierung Zyklen/Ränder, Klassen von Zyklen, die sich um einen +% XXX Rand unterscheiden + +Die folgenden zwei ausführlichen Beispiele sollen zeigen, wie die +Homologiegruppe $H_2$ die Anwesenheit eines Hohlraumes detektieren kann, +der entsteht, wenn man aus einem Tetraeder das innere entfernt. + +\begin{beispiel} +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/tetraeder.pdf} +\caption{Triangulation eines Tetraeders, die Orientierung von Kanten +und Seitenflächen ist immer so gewählt, dass die Nummern der Ecken +aufsteigend sind. +\label{buch:homologie:tetraeder:fig}} +\end{figure} +Ein Tetraeder ist ein zweidmensionales Simplex, wir untersuchen seinen +Kettenkomplex und bestimmen die zugehörigen Homologiegruppen. +Zunächst müssen wir die einzelnen Mengen $C_k$ beschreiben und verwenden +dazu die Bezeichnungen gemäss Abbildung~\ref{buch:homologie:tetraeder:fig}. +$C_0$ ist der vierdimensionale Raum aufgespannt von den vier Ecken +$0$, $1$, $2$ und $3$ des Tetraeders. +$C_1$ ist der sechsdimensionale Vektorraum der Kanten +\[ +k_0 = [0,1],\quad +k_1 = [0,2],\quad +k_2 = [0,3],\quad +k_3 = [1,2],\quad +k_4 = [1,3],\quad +k_5 = [2,3] +\] +Der Randoperator $\partial_1$ hat die Matrix +\[ +\partial_1 += +\begin{pmatrix*}[r] +-1&-1&-1& 0& 0& 0\\ + 1& 0& 0&-1&-1& 0\\ + 0& 1& 0& 1& 0&-1\\ + 0& 0& 1& 0& 1& 1 +\end{pmatrix*}. +\] + +Wir erwarten natürlich, dass sich zwei beliebige Ecken verbinden lassen, +dass es also nur eine Komponente gibt und dass damit $H_1=\Bbbk$ ist. +Dazu beachten wir, dass das Bild von $\partial_1$ genau aus den Vektoren +besteht, deren Komponentensumme $0$ ist. +Das Bild $B_0$ von $\partial_1$ ist daher die Lösungsmenge der einen +Gleichung +\( +x_0+x_1+x_2+x_3=0. +\) +Der Quotientenraum $H_0=Z_0/B_0 = C_0/\operatorname{im}\partial_1$ +ist daher wie erwartet eindimensional. + +Wir bestimmen jetzt die Homologiegruppe $H_1$. +Da sich im Tetraeder jeder geschlossene Weg zusammenziehen lässt, +erwarten wir $H_1=0$. + +Die Menge der Zyklen $Z_1$ wird bestimmt, indem man die Lösungsmenge +des Gleichungssystems $\partial_1z=0$ bestimmt. +Der Gauss-Algorithmus für die Matrix $\partial_1$ liefert das +Schlusstableau +\[ +\begin{tabular}{|>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}|} +\hline +k_0&k_1&k_2&k_3&k_4&k_5\\ +\hline + 1& 0& 0& -1& -1& 0\\ + 0& 1& 0& 1& 0& -1\\ + 0& 0& 1& 0& 1& 1\\ + 0& 0& 0& 0& 0& 0\\ +\hline +\end{tabular} +\] +Daraus lassen sich drei linear unabhängig eindimensionale Zyklen ablesen, +die zu den Lösungsvektoren +\[ +z_1 += +\begin{pmatrix*}[r] +1\\ +-1\\ +0\\ +1\\ +0\\ +0 +\end{pmatrix*}, +\qquad +z_2 += +\begin{pmatrix*}[r] +1\\ +0\\ +-1\\ +0\\ +1\\ +0 +\end{pmatrix*}, +\qquad +z_3 += +\begin{pmatrix*}[r] +0\\ +1\\ +-1\\ +0\\ +0\\ +1 +\end{pmatrix*} +\] +gehören. + +$C_2$ hat die vier Seitenflächen +\[ +f_0=[0,1,2],\quad +f_1=[0,1,3],\quad +f_2=[0,2,3],\quad +f_3=[1,2,3] +\] +als Basis. +Der zweidimensionale Randoperator ist die $6\times 4$-Matrix +\[ +\partial_2 += +\begin{pmatrix*}[r] + 1& 1& 0& 0\\ +-1& 0& 1& 0\\ + 0&-1&-1& 0\\ + 1& 0& 0& 1\\ + 0& 1& 0&-1\\ + 0& 0& 1& 1 +\end{pmatrix*}. +\] +Man kann leicht nachrechnen, dass $\partial_1\partial_2=0$ ist, wie es +für einen Kettenkomplex sein muss. + +Um nachzurechnen, dass die Homologiegruppe $H_1=0$ ist, müssen wir jetzt +nachprüfen, ob jeder Zyklus in $Z_1$ auch Bild der Randabbildung $\partial_2$ +ist. +Die ersten drei Spalten von $\partial_2$ sind genau die drei Zyklen +$z_1$, $z_2$ und $z_3$. +Insbesondere lassen sich alle Zyklen als Ränder darstellen, die +Homologiegruppe $H_1=0$ verschwindet. + +Die Zyklen in $C_2$ sind die Lösungen von $\partial_2z=0$. +Der Gauss-Algorithmus für $\partial_2$ liefert das -Tableau +\[ +\begin{tabular}{|>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}|} +\hline +f_0&f_1&f_2&f_3\\ +\hline +1&0&0& 1\\ +0&1&0&-1\\ +0&0&1& 1\\ +0&0&0& 0\\ +0&0&0& 0\\ +0&0&0& 0\\ +\hline +\end{tabular} +\] +Daraus liest man ab, dass es genau einen Zyklus nämlich +\[ +z += +\begin{pmatrix} +-1\\1\\-1\\1 +\end{pmatrix} +\] +$Z_2$ besteht also aus Vielfachen des Vektors $z$. + +Da es nur ein zweidimensionales Simplex gibt, ist $C_3$ eindimensional. +Die Randabbildung $\partial_3$ hat die Matrix +\[ +\partial_3 += +\begin{pmatrix} +1\\ +-1\\ +1\\ +-1 +\end{pmatrix}. +\] +Die Zyklen $Z_2$ und die Ränder $B_2$ bilden also dieselbe Menge, auch +die Homologie-Gruppe $H_2$ ist $0$. + +Da es keine vierdimensionalen Simplizes gibt, ist $B_3=0$. +Die Zyklen $Z_3$ bestehen aus den Lösungen von $\partial_3w=0$, da +aber $\partial_3$ injektiv ist, ist $Z_3=0$. +Daher ist auch $H_3=0$. +\end{beispiel} + +\begin{beispiel} +Für dieses Beispiel entfernen wir das Innere des Tetraeders, es entsteht +ein Hohlraum. +Am Kettenkomplex der Triangulation ändert sich nur, dass $C_3$ jetzt +nur noch den $0$-Vektor enthält. +Das Bild $B_2=\operatorname{im}\partial_3$ wird damit auch $0$-dimensional, +während es im vorigen Beispiel eindimensional war. +Die einzige Änderung ist also in der Homologiegruppe +$H_2 = Z_2/B_2 = Z_2 / \{0\} \simeq \Bbbk$. +Die Homologiegruppe $H_2$ hat jetzt Dimension $1$ und zeigt damit den +Hohlraum an. +\end{beispiel} diff --git a/buch/chapters/95-homologie/hx.m b/buch/chapters/95-homologie/hx.m new file mode 100644 index 0000000..0003e76 --- /dev/null +++ b/buch/chapters/95-homologie/hx.m @@ -0,0 +1,129 @@ +split_long_rows(0) + +d = [ +#1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 +-1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 1 + 1,-1, 0, 0, 0,-1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 2 + 0, 1,-1, 0, 0, 0, 0,-1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 3 + 0, 0, 1,-1, 0, 0, 0, 0, 0,-1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 4 + 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 5 + 0, 0, 0, 0,-1, 1, 0, 0, 0, 0, 0, 0,-1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 6 + 0, 0, 0, 0, 0, 0,-1, 1, 0, 0, 0, 0, 1, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 7 + 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 0, 0,-1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 8 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 1, 0, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 9 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 0,-1, 0, 1, 0, 0, 0, 0, 0, 0; # 10 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,-1, 0,-1, 1, 0, 0, 0, 0; # 11 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 1, 0, 0, 0,-1, 0, 0, 0; # 12 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 0, 0,-1, 1, 0; # 13 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1, 1, 0,-1; # 14 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 1 # 15 +] + +rref(d) + +B = [ +#1 2 3 4 5 6 7 8 9101112131415161718192021222324252627 + 1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; + 0,1,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; + 0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; + 0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0; + 0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,0,0,0,0,0,0,0,0,0,0,0; + 0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0; + 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,0,0,0,0,0; + 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0; + 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1 +]'; + +d*B + +Z = [ +#1 2 3 4 5 6 7 8 9 10 11 12 13 + 1, 0, 0, 0,-1, 0, 1, 0,-1, 0, 1, 0, 0; # 1 + 0, 1, 0, 0, 0, 0, 0, 0,-1, 0, 1, 0, 0; # 2 + 0, 0, 1, 0, 0,-1, 0, 1, 0, 0, 0, 0, 0; # 3 + 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 4 + 1, 0, 0, 0,-1, 0, 1, 0,-1, 0, 1, 0, 0; # 5 + 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 6 + 0, 1, 0, 0, 1, 0,-1, 0, 0, 0, 0, 0, 0; # 7 + 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 8 + 0, 0, 1, 0, 0,-1, 0, 1, 1, 0,-1, 0, 0; # 9 + 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 10 + 0, 0, 0, 1, 0, 1, 0,-1, 0, 0, 0, 0, 0; # 11 + 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0; # 12 + 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0; # 13 + 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0; # 14 + 0, 0, 0, 0, 0, 0, 1, 0,-1, 0, 1, 0, 0; # 15 + 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0; # 16 + 0, 0, 0, 0, 0, 0, 0, 1, 1, 0,-1, 0, 0; # 17 + 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0; # 18 + 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,-1,-1, 1; # 19 + 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0; # 20 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0,-1, 1; # 21 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0; # 22 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,-1; # 23 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0; # 24 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0; # 25 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1; # 26 + 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 # 27 +] + +d * Z + +T = zeros(22, 9); +T(1:9,1:27) = B'; +T(10:22,1:27) = Z'; + +T + +for i = (2:22) + T(i,:) = T(i,:) - T(i,1) * T(1,:); +end +for i = (3:22) + T(i,:) = T(i,:) - T(i,2) * T(2,:); +end +for i = (4:22) + T(i,:) = T(i,:) - T(i,3) * T(3,:); +end +for i = (5:22) + T(i,:) = T(i,:) - T(i,4) * T(4,:); +end + +T + +for i = (15:22) + T(i,:) = T(i,:) - T(i,6) * T(14,:); +end +T +for i = (19:22) + T(i,:) = T(i,:) - T(i,8) * T(18,:); +end +T +for i = (16:22) + T(i,:) = T(i,:) - T(i,10) * T(15,:); +end +T +for i = (6:22) + T(i,:) = T(i,:) - T(i,13) * T(5,:); +end +T +for i = (7:22) + T(i,:) = T(i,:) - T(i,14) * T(6,:); +end +T +for i = (8:22) + T(i,:) = T(i,:) - T(i,19) * T(7,:); +end +T +for i = (9:22) + T(i,:) = T(i,:) - T(i,20) * T(8,:); +end +T +for i = (22:22) + T(i,:) = T(i,:) - T(i,22) * T(21,:); +end +T +for i = (10:22) + T(i,:) = T(i,:) - T(i,25) * T(9,:); +end +# +T diff --git a/buch/chapters/95-homologie/images/Makefile b/buch/chapters/95-homologie/images/Makefile index 82f1285..0a3979e 100644 --- a/buch/chapters/95-homologie/images/Makefile +++ b/buch/chapters/95-homologie/images/Makefile @@ -3,8 +3,44 @@ # # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule # -all: dreieck.pdf +all: complexbasis.pdf homocycles.pdf homoboundaries.pdf homoclasses.pdf \ + gausshomoex.pdf gausshomobasis.pdf dreieck.pdf polyeder.pdf \ + approximation.pdf tetraeder.pdf dreieck.pdf: dreieck.tex pdflatex dreieck.tex +polyeder.pdf: polyeder.tex + pdflatex polyeder.tex + +gausshomobasis.pdf: gausshomobasis.tex + pdflatex gausshomobasis.tex + +gausshomoex.pdf: gausshomoex.tex + pdflatex gausshomoex.tex + +homocycles.pdf: homocycles.tex + pdflatex homocycles.tex + +homoboundaries.pdf: homoboundaries.tex + pdflatex homoboundaries.tex + +homoclasses.pdf: homoclasses.tex + pdflatex homoclasses.tex + +complexbasis.pdf: complexbasis.tex + pdflatex complexbasis.tex + +approximation.pdf: approximation.tex approx.tex + pdflatex approximation.tex + +approx.tex: approx.m + octave approx.m + +tetraeder.png: tetraeder.pov + povray +A0.1 -W1920 -H1080 -O$@ $< +tetraeder.jpg: tetraeder.png Makefile + convert -extract 1080x1080+520 tetraeder.png tetraeder.jpg +tetraeder.pdf: tetraeder.tex tetraeder.jpg + pdflatex tetraeder.tex + diff --git a/buch/chapters/95-homologie/images/approx.m b/buch/chapters/95-homologie/images/approx.m new file mode 100644 index 0000000..0db41c2 --- /dev/null +++ b/buch/chapters/95-homologie/images/approx.m @@ -0,0 +1,77 @@ +# +# approx.m +# +# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +x = zeros(7,7); +y = zeros(7,7); + +s = 1.05; + +for i = (1:7) + winkel = (i-1) * 8.333333 + 20; + for j = (1:7) + radius = (j-1) * 0.5 + 3; + x(i,j) = 1.05 * radius * cosd(winkel); + y(i,j) = 1.05 * radius * sind(winkel); + endfor +endfor + +X = x; +Y = y; +for i = (1:7) + for j = (1:7) + X(i,j) = round(2 * x(i,j)) / 2; + Y(i,j) = round(2 * y(i,j)) / 2; + endfor +endfor + +fn = fopen("approx.tex", "w"); + + +for i = (1:6) + for j = (1:6) + winkel = (i-1+0.6666) * 8.33333 + 20; + radius = (j-1+0.3333) * 0.5 + 3; + fprintf(fn, "\\definecolor{mycolor}{rgb}{%.2f,%.2f,%.2f};\n", + (winkel - 20) / 50, 0.8, (radius-3)/3); + fprintf(fn, "\\fill[color=mycolor] (%.3f,%.3f) -- (%.3f,%.3f) -- (%.3f,%.3f) -- cycle;\n", + X(i,j), Y(i,j), + X(i+1,j+1), Y(i+1,j+1), + X(i+1,j), Y(i+1,j)); + winkel = (i-1+0.3333) * 8.33333 + 20; + radius = (j-1+0.6666) * 0.5 + 3; + fprintf(fn, "\\definecolor{mycolor}{rgb}{%.2f,%.2f,%.2f};\n", + (winkel - 20) / 50, 0.8, (radius-3)/3); + fprintf(fn, "\\fill[color=mycolor] (%.3f,%.3f) -- (%.3f,%.3f) -- (%.3f,%.3f) -- cycle;\n", + X(i,j), Y(i,j), + X(i,j+1), Y(i,j+1), + X(i+1,j+1), Y(i+1,j+1)); + endfor +endfor + +linewidth = 0.4; + +fprintf(fn, "\\gitter\n"); + +for i = (1:7) + for j = (1:6) + fprintf(fn, "\\draw[color=darkred,line width=%.1fpt] (%.3f,%.3f) -- (%.3f,%.3f);\n", linewidth, + X(i,j), Y(i,j), X(i,j+1), Y(i,j+1)); + endfor +endfor +for i = (1:6) + for j = (1:7) + fprintf(fn, "\\draw[color=darkred,line width=%.1fpt] (%.3f,%.3f) -- (%.3f,%.3f);\n", linewidth, + X(i,j), Y(i,j), X(i+1,j), Y(i+1,j)); + endfor +endfor +for i = (1:6) + for j = (1:6) + fprintf(fn, "\\draw[color=darkred,line width=%.1fpt] (%.3f,%.3f) -- (%.3f,%.3f);\n", linewidth, + X(i,j), Y(i,j), X(i+1,j+1), Y(i+1,j+1)); + endfor +endfor + +fclose(fn) + diff --git a/buch/chapters/95-homologie/images/approximation.pdf b/buch/chapters/95-homologie/images/approximation.pdf Binary files differnew file mode 100644 index 0000000..8bdd2e7 --- /dev/null +++ b/buch/chapters/95-homologie/images/approximation.pdf diff --git a/buch/chapters/95-homologie/images/approximation.tex b/buch/chapters/95-homologie/images/approximation.tex new file mode 100644 index 0000000..042f0e2 --- /dev/null +++ b/buch/chapters/95-homologie/images/approximation.tex @@ -0,0 +1,69 @@ +% +% approximation.tex -- Approximation einer Abbildung durch eine simpliziale +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1.3} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\definecolor{darkred}{rgb}{0.8,0,0} +\definecolor{darkred}{rgb}{0,0,0} + +\def\gitter{ + \foreach \x in {1,1.5,...,6}{ + \draw[color=gray] (\x,1) -- (\x,6); + \draw[color=gray] (1,\x) -- (6,\x); + } +} + +\def\s{1.05} + +\def\colorsector{ + \foreach \r in {3,3.2,...,5.8}{ + \foreach \a in {20,...,69}{ + \pgfmathparse{(\a-20)/50} + \xdef\rot{\pgfmathresult} + \pgfmathparse{(\r-3)/3} + \xdef\blau{\pgfmathresult} + \definecolor{mycolor}{rgb}{\rot,0.8,\blau} + \fill[color=mycolor] + (\a:{\s*\r}) -- (\a:{\s*(\r+0.2)}) -- ({\a+1}:{\s*(\r+0.2)}) -- ({\a+1}:{\s*\r}) -- cycle; + } + } +} + +\begin{scope}[xshift=0cm] +\colorsector +\gitter +\foreach \r in {3,3.5,...,6.0}{ + \draw[color=black,line width=0.4pt] (20:{\s*\r}) arc (20:70:{\s*\r}); +} +\foreach \a in {20,28.3333,...,70}{ + \draw[color=black,line width=0.4pt] (\a:{\s*3}) -- (\a:{\s*6}); +} +\begin{scope} +\clip (20:{\s*3}) -- (20:{\s*6}) arc (20:70:{\s*6}) -- (70:{\s*3}); +\foreach \a in {-5,...,5}{ + \draw[color=black,line width=0.4pt] + plot[domain={20+8.33333*\a}:{70+8.3333*\a},samples=100] + (\x:{\s*(3+3*(\x-(20+8.3333*\a))/50)}); +} +\end{scope} + +\end{scope} + +\begin{scope}[xshift=5.5cm] +\input{approx.tex} +\end{scope} + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/complexbasis.pdf b/buch/chapters/95-homologie/images/complexbasis.pdf Binary files differnew file mode 100644 index 0000000..9ff6709 --- /dev/null +++ b/buch/chapters/95-homologie/images/complexbasis.pdf diff --git a/buch/chapters/95-homologie/images/complexbasis.tex b/buch/chapters/95-homologie/images/complexbasis.tex new file mode 100644 index 0000000..bab89d2 --- /dev/null +++ b/buch/chapters/95-homologie/images/complexbasis.tex @@ -0,0 +1,158 @@ +% +% complexbasis.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\usetikzlibrary{decorations.pathreplacing} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\definecolor{darkgreen}{rgb}{0,0.6,0} + +\def\s{0.5} +\def\h{0.02} + +\def\rechteck#1#2#3{ + \fill[color=#3!20,rounded corners=2pt] + ({-\s+\h},{(2*(#1)-1)*\s+\h}) + rectangle + ({\s-\h},{(2*(#2)+1)*\s-\h}); + \draw[color=#3,rounded corners=2pt] + ({-\s+\h},{(2*(#1)-1)*\s+\h}) + rectangle + ({\s-\h},{(2*(#2)+1)*\s-\h}); + \foreach \y in {{#1},...,{#2}}{ + \fill[color=#3] (0,{2*\y*\s}) circle[radius=0.05]; + } +} +\def\Rechteck#1#2{ + \draw[rounded corners=3pt] + ({-\s-\h},{(2*(#1)-1)*\s-\h}) + rectangle + ({\s+\h},{(2*(#2)+1)*\s+\h}); +} + +\def\abbildung#1#2#3#4{ + \fill[color=gray!20] + ({\s+\h},{(2*(#1)+1)*\s}) + -- + ({3.5-\s-\h},{-\s}) + -- + ({3.5-\s-\h},{(2*(#3)+1)*\s}) + -- + ({\s+\h},{(2*(#1)+1)*\s}) + -- + cycle; + \fill[color=gray!40] + ({\s+\h},{(2*(#1+1)-1)*\s}) + -- + ({3.5-\s-\h},{(2*(#3+1)-1)*\s}) + -- + ({3.5-\s-\h},{(2*(#4)+1)*\s}) + -- + ({\s+\h},{(2*(#2)+1)*\s}) + -- + cycle; + \draw[<-,color=gray] + ({\s+\h},{(2*(#1+1)-1)*\s}) + -- + ({3.5-\s-\h},{(2*(#3+1)-1)*\s}); + \draw[->,color=gray] + ({3.5-\s-\h},{(2*(#4)+1)*\s}) + -- + ({\s+\h},{(2*(#2)+1)*\s}); + \draw[<-,color=gray!40] + ({\s+\h},{(2*(#1)+1)*\s}) + -- + ({3.5-\s-\h},{-\s}); +} + +\clip ({-3.5-1.7},-1.2) rectangle ({7+1.7},11.7); + +\begin{scope}[xshift=-7cm] + \abbildung{6}{7}{10}{11} +\end{scope} + +\begin{scope}[xshift=-3.5cm] + \abbildung{6}{10}{7}{11} + \rechteck{0}{6}{red} + \rechteck{7}{10}{darkgreen} + \rechteck{11}{11}{blue} + \Rechteck{0}{11} + \node[color=darkgreen] at ({0},{(9*2-1)*\s}) {$B_{k-2\mathstrut}$}; + \node at (1.75,{9*2*\s}) {$\Delta_{k-1}$}; + \node at (1.75,{-\s}) [above] {$\partial_{k-1\mathstrut}$}; + \draw[decorate,decoration={brace,amplitude=4pt}] + ({-\s-0.1},{-\s}) -- ({-\s-0.1},{(2*10+1)*\s}); + \node at ({-\s-0.17},{10*\s}) [left] {$Z_{k-2\mathstrut}$}; + \node at (0,{-\s}) [below] {$C_{k-2\mathstrut}$}; +\end{scope} + +\begin{scope} + \abbildung{2}{7}{5}{10} + \rechteck{8}{11}{blue} + \rechteck{3}{7}{darkgreen} + \rechteck{0}{2}{red} + \Rechteck{0}{11} + \node at (0,{-\s}) [below] {$C_{k-1\mathstrut}$}; + \node[color=darkgreen] at ({0},{(5*2-1)*\s}) {$B_{k-1\mathstrut}$}; + \node at (1.75,{6.5*2*\s}) {$\Delta_k$}; + \node at (1.75,{-\s}) [above] {$\partial_{k\mathstrut}$}; + \draw[decorate,decoration={brace,amplitude=4pt}] + ({-\s-0.1},{-\s}) -- ({-\s-0.1},{(2*7+1)*\s}); + \node at ({-\s-0.17},{7*\s}) [left] {$Z_{k-1\mathstrut}$}; +\end{scope} + +\begin{scope}[xshift=3.5cm] + \abbildung{3}{5}{5}{7} + \rechteck{6}{10}{blue} + \rechteck{4}{5}{darkgreen} + \rechteck{0}{3}{red} + \Rechteck{0}{10} + \node at (0,{-\s}) [below] {$C_{k\mathstrut}$}; + \node[color=darkgreen] at ({-0.25},{9*\s}) + {$B_{k\mathstrut}$}; + \node[color=darkgreen] at (0.24,{2*4*\s}) {$b_1$}; + \node[color=darkgreen] at (0.24,{2*4.5*\s+0.1}) {$\vdots$}; + \node[color=darkgreen] at (0.24,{2*5*\s}) {$b_r$}; + \node[color=red] at (0.24,{2*0*\s}) {$z_1$}; + \node[color=red] at (0.24,{2*1*\s}) {$z_2$}; + \node[color=red] at (0.24,{2*2*\s+0.1}) {$\vdots$}; + \node[color=red] at (0.24,{2*3*\s}) {$z_l$}; + \node[color=blue] at (0.24,{2*6*\s}) {$c_1$}; + \node[color=blue] at (0.24,{2*7*\s}) {$c_2$}; + \node[color=blue] at (0.24,{2*8*\s}) {$c_3$}; + \node[color=blue] at (0.24,{2*9*\s}) {$\vdots$}; + \node[color=blue] at (0.24,{2*10*\s}) {$c_s$}; + \node at (1.75,{5.5*2*\s}) {$\Delta_{k+1}$}; + \node at (1.75,{-\s}) [above] {$\partial_{k+1\mathstrut}$}; + \draw[decorate,decoration={brace,amplitude=4pt}] + ({-\s-0.1},{-\s}) -- ({-\s-0.1},{(2*5+1)*\s}); + \node at ({-\s-0.17},{5*\s}) [left] {$Z_{k\mathstrut}$}; +\end{scope} + +\begin{scope}[xshift=7cm] + \abbildung{0}{5}{4}{8} + \rechteck{5}{7}{blue} + \rechteck{1}{5}{darkgreen} + \rechteck{0}{0}{red} + \Rechteck{0}{7} + \node at (0,{-\s}) [below] {$C_{k+1\mathstrut}$}; + \node[color=darkgreen] at ({0},{(2.0*2+1)*\s}) + {$B_{k+1\mathstrut}$}; + \draw[decorate,decoration={brace,amplitude=4pt}] + ({-\s-0.1},{-\s}) -- ({-\s-0.1},{(2*5+1)*\s}); + \node at ({-\s-0.17},{5*\s}) [left] {$Z_{k+1\mathstrut}$}; +\end{scope} + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/gausshomobasis.pdf b/buch/chapters/95-homologie/images/gausshomobasis.pdf Binary files differnew file mode 100644 index 0000000..07414bb --- /dev/null +++ b/buch/chapters/95-homologie/images/gausshomobasis.pdf diff --git a/buch/chapters/95-homologie/images/gausshomobasis.tex b/buch/chapters/95-homologie/images/gausshomobasis.tex new file mode 100644 index 0000000..ba21f54 --- /dev/null +++ b/buch/chapters/95-homologie/images/gausshomobasis.tex @@ -0,0 +1,109 @@ +% +% gaushomobasis.tex -- Bestimmung einer Basis der Homologiegruppen +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\definecolor{darkgreen}{rgb}{0,0.6,0} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\s{0.5} +\def\inset{0.05} +\def\w{8} + +\def\zeile#1#2{ + \fill[color=#2] ({0+\inset},{(12-#1)*\s+\inset}) + rectangle ({\w*\s-\inset},{(13-#1)*\s-\inset}); +} +\def\marke#1#2{ +\node at ({0.5*\w*\s},{12.5-#1)*\s}) {$#2\mathstrut$}; +} + +\def\gauss{ +\draw (0,0) rectangle ({\w*\s},{12*\s}); +\draw (0,{7*\s}) -- ({\w*\s},{7*\s}); +} + +\draw[->,color=red,line width=1pt] ({0.1*\s},{(12.5-1)*\s}) + to[out=180,in=90] (-3.6,-2); +\draw[->,color=red,line width=1pt] ({0.1*\s},{(12.5-2)*\s}) + to[out=180,in=90] (-2.2,-2); +\draw[->,color=red,line width=1pt] ({0.1*\s},{(12.5-4)*\s}) + to[out=180,in=90] (-0.7,-2); + +\draw[->,color=darkgreen,line width=1pt] ({0.1*\s},{(12.5-7)*\s}) + to[out=180,in=90] (0.9,-2); +\draw[->,color=darkgreen,line width=1pt] ({0.1*\s},{(12.5-8)*\s}) + to[out=180,in=90] (1.6,-2); +\draw[->,color=darkgreen,line width=1pt] ({(\w-0.1)*\s},{(12.5-12)*\s}) + to[out=0,in=90] (2.6,-2); + +\draw[->,line width=2pt] ({\w*\s+0.1},{6*\s}) -- (5.4,{6*\s}); +\node at ({0.5*(\w*\s+5.5)},{6*\s}) [above] {Gauss}; + +\begin{scope} +\zeile{1}{red!30} +\zeile{2}{red!30} +\zeile{4}{red!30} +\zeile{7}{darkgreen!30} +\zeile{8}{darkgreen!30} +%\zeile{10}{darkgreen!30} +\zeile{12}{darkgreen!30} +\marke{1}{\scriptstyle\partial_{k+1}e_1^{(k+1)}} +\marke{2}{\scriptstyle\partial_{k+1}e_2^{(k+1)}} +\marke{3}{\scriptstyle\partial_{k+1}e_3^{(k+1)}} +\marke{4}{\vdots} +\marke{5}{\scriptstyle\partial_{k+1}e_{n_{k+1}}^{(k+1)}} +\marke{6}{\scriptstyle z_1^{(k)}} +\marke{7}{\scriptstyle z_2^{(k)}} +\marke{8}{\scriptstyle z_3^{(k)}} +\marke{9}{\scriptstyle z_4^{(k)}} +\marke{10}{\vdots} +\marke{11}{\scriptstyle z_{l-1}^{(k)}} +\marke{12}{\scriptstyle z_{l}^{(k)}} +\gauss +\end{scope} + +\begin{scope}[xshift=5.5cm] +\zeile{1}{black!20} +\zeile{2}{black!20} +\zeile{3}{black} +\marke{3}{\color{white}0} +\zeile{4}{black!20} +\zeile{5}{black} +\marke{5}{\color{white}0} +\zeile{6}{black} +\marke{6}{\color{white}0} +\zeile{7}{black!20} +\zeile{8}{black!20} +\zeile{9}{black} +\marke{9}{\color{white}0} +\zeile{10}{black} +\marke{10}{\color{white}0} +\zeile{11}{black} +\marke{11}{\color{white}0} +\zeile{12}{black!20} +\gauss +\end{scope} + +\node at (-4.4,-2) [below right] {$\{ +{\color{red}\partial_{k+1}e_1^{(k+1)}}, +{\color{red}\partial_{k+1}e_2^{(k+1)}}, +{\color{red}\partial_{k+1}e_{i_3}^{(k+1)}},\dots, +{\color{darkgreen}z_2^{(k)}}, +{\color{darkgreen}z_3^{(k)}}, +\dots +{\color{darkgreen}z_l^{(k)}} +\} = {\color{red}\mathcal{B}_k} \cup {\color{darkgreen}\mathcal{Z}_k'}$}; + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/gausshomoex.pdf b/buch/chapters/95-homologie/images/gausshomoex.pdf Binary files differnew file mode 100644 index 0000000..bc0b766 --- /dev/null +++ b/buch/chapters/95-homologie/images/gausshomoex.pdf diff --git a/buch/chapters/95-homologie/images/gausshomoex.tex b/buch/chapters/95-homologie/images/gausshomoex.tex new file mode 100644 index 0000000..df53f70 --- /dev/null +++ b/buch/chapters/95-homologie/images/gausshomoex.tex @@ -0,0 +1,120 @@ +% +% gausshomoex.tex -- Beispiel für die Bestimmung einer Basis von H_1 +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\s{2.0} + +\def\punkt#1#2{({((#1)+0.5*(#2))*\s},{(#2)*\s*sqrt(3)/2})} + +\def\knoten#1#2#3{ + \fill[color=white] \punkt{#1}{#2} circle[radius=0.3]; + \node at \punkt{#1}{#2} {$#3$\strut}; + \draw \punkt{#1}{#2} circle[radius=0.3]; +} +\def\dreieck#1#2#3{ + \fill[color=gray] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1}{(#2)+1} -- cycle; + \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; + \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] + \punkt{#1}{#2} -- \punkt{#1+1}{#2}; + \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] + \punkt{#1+1}{#2} -- \punkt{#1}{#2+1}; + \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] + \punkt{#1}{#2+1} -- \punkt{#1}{#2}; +} + +\def\Dreieck#1#2#3{ + \fill[color=gray!50] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1+1}{(#2)-1} -- cycle; + \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; +} + +\def\kante#1#2#3{ + \fill[color=white,opacity=0.8] \punkt{#1}{#2} circle[radius=0.15]; + \node at \punkt{#1}{#2} {$\scriptstyle #3$}; +} + +\dreieck{0}{0}{1} +\dreieck{1}{0}{2} +\dreieck{2}{0}{3} +\dreieck{3}{0}{4} + +\dreieck{0}{1}{5} +\dreieck{2}{1}{6} + +\dreieck{0}{2}{7} +\dreieck{1}{2}{8} + +\dreieck{0}{3}{9} + + +\knoten{0}{0}{1} +\knoten{1}{0}{2} +\knoten{2}{0}{3} +\knoten{3}{0}{4} +\knoten{4}{0}{5} + +\knoten{0}{1}{6} +\knoten{1}{1}{7} +\knoten{2}{1}{8} +\knoten{3}{1}{9} + +\knoten{0}{2}{10} +\knoten{1}{2}{11} +\knoten{2}{2}{12} + +\knoten{0}{3}{13} +\knoten{1}{3}{14} + +\knoten{0}{4}{15} + +\kante{0.5}{0}{1} +\kante{1.5}{0}{2} +\kante{2.5}{0}{3} +\kante{3.5}{0}{4} + +\kante{0}{0.5}{5} +\kante{0.5}{0.5}{6} +\kante{1}{0.5}{7} +\kante{1.5}{0.5}{8} +\kante{2}{0.5}{9} +\kante{2.5}{0.5}{10} +\kante{3}{0.5}{11} +\kante{3.5}{0.5}{12} + +\kante{0.5}{1}{13} +\kante{2.5}{1}{14} + +\kante{0}{1.5}{15} +\kante{0.5}{1.5}{16} +\kante{2}{1.5}{17} +\kante{2.5}{1.5}{18} + +\kante{0.5}{2}{19} +\kante{1.5}{2}{20} + +\kante{0}{2.5}{21} +\kante{0.5}{2.5}{22} +\kante{1}{2.5}{23} +\kante{1.5}{2.5}{24} + +\kante{0.5}{3}{25} + +\kante{0}{3.5}{26} +\kante{0.5}{3.5}{27} + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/homoboundaries.pdf b/buch/chapters/95-homologie/images/homoboundaries.pdf Binary files differnew file mode 100644 index 0000000..fb94ec8 --- /dev/null +++ b/buch/chapters/95-homologie/images/homoboundaries.pdf diff --git a/buch/chapters/95-homologie/images/homoboundaries.tex b/buch/chapters/95-homologie/images/homoboundaries.tex new file mode 100644 index 0000000..53087fa --- /dev/null +++ b/buch/chapters/95-homologie/images/homoboundaries.tex @@ -0,0 +1,115 @@ +% +% tikztemplate.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\s{0.55} + +\def\punkt#1#2{({((#1)+0.5*(#2))*\s},{(#2)*\s*sqrt(3)/2})} +\def\A{\punkt{0}{0}} +\def\B{\punkt{1}{0}} +\def\C{\punkt{2}{0}} +\def\D{\punkt{3}{0}} +\def\E{\punkt{4}{0}} +\def\F{\punkt{0}{1}} +\def\G{\punkt{1}{1}} +\def\H{\punkt{2}{1}} +\def\I{\punkt{3}{1}} +\def\J{\punkt{0}{2}} +\def\K{\punkt{1}{2}} +\def\L{\punkt{2}{2}} +\def\M{\punkt{0}{3}} +\def\N{\punkt{1}{3}} +\def\O{\punkt{0}{4}} + +\def\dreieck#1#2#3{ + \fill[color=gray] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1}{(#2)+1} -- cycle; +} + +\def\blau#1#2{ + \draw[color=blue] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1}{(#2)+1} -- cycle; + \draw[->,color=blue] \punkt{#1}{#2} -- \punkt{#1+1}{#2}; +} + +\def\gebiet{ + \dreieck{0}{0}{1} + \dreieck{1}{0}{2} + \dreieck{2}{0}{3} + \dreieck{3}{0}{4} + \dreieck{0}{1}{5} + \dreieck{2}{1}{6} + \dreieck{0}{2}{7} + \dreieck{1}{2}{8} + \dreieck{0}{3}{9} +} + +\begin{scope} +\gebiet +\blau{0}{0} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_1^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=3cm] +\gebiet +\blau{1}{0} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_2^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=6cm] +\gebiet +\blau{2}{0} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_3^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=9cm] +\gebiet +\blau{3}{0} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_4^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=1.5cm,yshift=2.59cm] +\gebiet +\blau{0}{1} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_5^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=7.5cm,yshift=2.59cm] +\gebiet +\blau{2}{1} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_6^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=3cm,yshift=5.19cm] +\gebiet +\blau{0}{2} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_7^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=6cm,yshift=5.19cm] +\gebiet +\blau{1}{2} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_8^{(2)}$}; +\end{scope} + +\begin{scope}[xshift=4.5cm,yshift=7.79cm] +\gebiet +\blau{0}{3} +\node[color=blue] at ({2*\s},-0.5) {$\partial_2e_9^{(2)}$}; +\end{scope} + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/homoclasses.pdf b/buch/chapters/95-homologie/images/homoclasses.pdf Binary files differnew file mode 100644 index 0000000..fbbaedd --- /dev/null +++ b/buch/chapters/95-homologie/images/homoclasses.pdf diff --git a/buch/chapters/95-homologie/images/homoclasses.tex b/buch/chapters/95-homologie/images/homoclasses.tex new file mode 100644 index 0000000..4467f08 --- /dev/null +++ b/buch/chapters/95-homologie/images/homoclasses.tex @@ -0,0 +1,109 @@ +% +% homoclasses.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1.4} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\definecolor{darkgreen}{rgb}{0,0.6,0} +\def\s{0.4} +\def\h{-0.3} + +\def\punkt#1#2{({((#1)+0.5*(#2))*\s},{(#2)*\s*sqrt(3)/2})} +\def\A{\punkt{0}{0}} +\def\B{\punkt{1}{0}} +\def\C{\punkt{2}{0}} +\def\D{\punkt{3}{0}} +\def\E{\punkt{4}{0}} +\def\F{\punkt{0}{1}} +\def\G{\punkt{1}{1}} +\def\H{\punkt{2}{1}} +\def\I{\punkt{3}{1}} +\def\J{\punkt{0}{2}} +\def\K{\punkt{1}{2}} +\def\L{\punkt{2}{2}} +\def\M{\punkt{0}{3}} +\def\N{\punkt{1}{3}} +\def\O{\punkt{0}{4}} + +%\def\knoten#1#2#3{ +% \fill[color=white] \punkt{#1}{#2} circle[radius=0.3]; +% \node at \punkt{#1}{#2} {$#3$\strut}; +% \draw \punkt{#1}{#2} circle[radius=0.3]; +%} +\def\dreieck#1#2#3{ + \fill[color=gray] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1}{(#2)+1} -- cycle; +% \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1}{#2} -- \punkt{#1+1}{#2}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1+1}{#2} -- \punkt{#1}{#2+1}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1}{#2+1} -- \punkt{#1}{#2}; +} + +%\def\Dreieck#1#2#3{ +% \fill[color=gray!50] \punkt{#1}{#2} -- \punkt{#1+1}{#2} +% -- \punkt{#1+1}{(#2)-1} -- cycle; +% \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; +%} + +%\def\kante#1#2#3{ +% \fill[color=white,opacity=0.8] \punkt{#1}{#2} circle[radius=0.15]; +% \node at \punkt{#1}{#2} {$\scriptstyle #3$}; +%} + +\def\gebiet{ + \dreieck{0}{0}{1} + \dreieck{1}{0}{2} + \dreieck{2}{0}{3} + \dreieck{3}{0}{4} + \dreieck{0}{1}{5} + \dreieck{2}{1}{6} + \dreieck{0}{2}{7} + \dreieck{1}{2}{8} + \dreieck{0}{3}{9} +} + +\begin{scope} +\gebiet +\draw[color=darkgreen] \B -- \G -- \J -- \F -- cycle; +\draw[->,color=darkgreen] \B -- \G; +\node[color=darkgreen] at ({2*\s},{\h}) {$z_5'$}; +\end{scope} + +\begin{scope}[xshift=2cm] +\gebiet +\draw[color=darkgreen] \D -- \I -- \L -- \H -- cycle; +\draw[->,color=darkgreen] \D -- \I; +\node[color=darkgreen] at ({2*\s},{\h}) {$z_6'$}; +\end{scope} + +\begin{scope}[xshift=4cm] +\gebiet +\draw[color=darkgreen] \C -- \L -- \N -- \K -- \M -- \J -- cycle; +\draw[->,color=darkgreen] \C -- \L; +\node[color=darkgreen] at ({2*\s},{\h}) {$z_9'$}; +\end{scope} + +\begin{scope}[xshift=6cm] +\gebiet +\draw[color=darkgreen] \K -- \N -- \O -- \M -- cycle; +\draw[->,color=darkgreen] \K -- \N; +\node[color=darkgreen] at ({2*\s},{\h}) {$z_{12}'$}; +\end{scope} + + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/homocycles.pdf b/buch/chapters/95-homologie/images/homocycles.pdf Binary files differnew file mode 100644 index 0000000..b68519e --- /dev/null +++ b/buch/chapters/95-homologie/images/homocycles.pdf diff --git a/buch/chapters/95-homologie/images/homocycles.tex b/buch/chapters/95-homologie/images/homocycles.tex new file mode 100644 index 0000000..8f20a0c --- /dev/null +++ b/buch/chapters/95-homologie/images/homocycles.tex @@ -0,0 +1,170 @@ +% +% tikztemplate.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\s{0.4} + +\def\punkt#1#2{({((#1)+0.5*(#2))*\s},{(#2)*\s*sqrt(3)/2})} +\def\A{\punkt{0}{0}} +\def\B{\punkt{1}{0}} +\def\C{\punkt{2}{0}} +\def\D{\punkt{3}{0}} +\def\E{\punkt{4}{0}} +\def\F{\punkt{0}{1}} +\def\G{\punkt{1}{1}} +\def\H{\punkt{2}{1}} +\def\I{\punkt{3}{1}} +\def\J{\punkt{0}{2}} +\def\K{\punkt{1}{2}} +\def\L{\punkt{2}{2}} +\def\M{\punkt{0}{3}} +\def\N{\punkt{1}{3}} +\def\O{\punkt{0}{4}} + +%\def\knoten#1#2#3{ +% \fill[color=white] \punkt{#1}{#2} circle[radius=0.3]; +% \node at \punkt{#1}{#2} {$#3$\strut}; +% \draw \punkt{#1}{#2} circle[radius=0.3]; +%} +\def\dreieck#1#2#3{ + \fill[color=gray] \punkt{#1}{#2} -- \punkt{#1+1}{#2} + -- \punkt{#1}{(#2)+1} -- cycle; +% \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1}{#2} -- \punkt{#1+1}{#2}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1+1}{#2} -- \punkt{#1}{#2+1}; +% \draw[->,line width=1pt,shorten >= 0.3cm,shorten <= 0.3cm] +% \punkt{#1}{#2+1} -- \punkt{#1}{#2}; +} + +%\def\Dreieck#1#2#3{ +% \fill[color=gray!50] \punkt{#1}{#2} -- \punkt{#1+1}{#2} +% -- \punkt{#1+1}{(#2)-1} -- cycle; +% \node at \punkt{#1+0.3333}{#2+0.3333} {$#3$\strut}; +%} + +%\def\kante#1#2#3{ +% \fill[color=white,opacity=0.8] \punkt{#1}{#2} circle[radius=0.15]; +% \node at \punkt{#1}{#2} {$\scriptstyle #3$}; +%} + +\def\gebiet{ + \dreieck{0}{0}{1} + \dreieck{1}{0}{2} + \dreieck{2}{0}{3} + \dreieck{3}{0}{4} + \dreieck{0}{1}{5} + \dreieck{2}{1}{6} + \dreieck{0}{2}{7} + \dreieck{1}{2}{8} + \dreieck{0}{3}{9} +} + +\begin{scope} +\gebiet +\draw[->,color=red] \A -- \B -- \F -- cycle; +\draw[->,color=red] \A -- \B; +\node[color=red] at ({2*\s},-0.5) {$z_1$}; +\end{scope} + +\begin{scope}[xshift=2cm] +\gebiet +\draw[color=red] \B -- \C -- \G -- cycle; +\draw[->,color=red] \B -- \C; +\node[color=red] at ({2*\s},-0.5) {$z_2$}; +\end{scope} + +\begin{scope}[xshift=4cm] +\gebiet +\draw[color=red] \C -- \D -- \H -- cycle; +\draw[->,color=red] \C -- \D; +\node[color=red] at ({2*\s},-0.5) {$z_3$}; +\end{scope} + +\begin{scope}[xshift=6cm] +\gebiet +\draw[color=red] \D -- \E -- \I -- cycle; +\draw[->,color=red] \D -- \E; +\node[color=red] at ({2*\s},-0.5) {$z_4$}; +\end{scope} + +\begin{scope}[xshift=8cm] +\gebiet +\draw[color=red] \A -- \B -- \G -- \F -- cycle; +\draw[<-,color=red] \A -- \B; +\node[color=red] at ({2*\s},-0.5) {$z_5$}; +\end{scope} + +\begin{scope}[xshift=10cm] +\gebiet +\draw[color=red] \C -- \D -- \I -- \H -- cycle; +\draw[<-,color=red] \C -- \D; +\node[color=red] at ({2*\s},-0.5) {$z_6$}; +\end{scope} + +\begin{scope}[xshift=12cm] +\gebiet +\draw[color=red] \A -- \B -- \G -- \J -- \F -- cycle; +\draw[->,color=red] \A -- \B; +\node[color=red] at ({2*\s},-0.5) {$z_7$}; +\end{scope} + +\begin{scope}[xshift=0cm,yshift=-3cm] +\gebiet +\draw[color=red] \C -- \D -- \I -- \L -- \H -- cycle; +\draw[->,color=red] \C -- \D; +\node[color=red] at ({2*\s},-0.5) {$z_8$}; +\end{scope} + +\begin{scope}[xshift=2cm,yshift=-3cm] +\gebiet +\draw[color=red] \A -- \B -- \C -- \H -- \L -- \K -- \J -- \F -- cycle; +\draw[<-,color=red] \A -- \B; +\node[color=red] at ({2*\s},-0.5) {$z_9$}; +\end{scope} + +\begin{scope}[xshift=4cm,yshift=-3cm] +\gebiet +\draw[color=red] \J -- \K -- \M -- cycle; +\draw[->,color=red] \J -- \K; +\node[color=red] at ({2*\s},-0.5) {$z_{10}$}; +\end{scope} + +\begin{scope}[xshift=6cm,yshift=-3cm] +\gebiet +\draw[color=red] \A -- \B -- \C -- \H -- \L -- \N -- \K -- \J -- \F -- cycle; +\draw[->,color=red] \A -- \B; +\node[color=red] at ({2*\s},-0.5) {$z_{11}$}; +\end{scope} + +\begin{scope}[xshift=8cm,yshift=-3cm] +\gebiet +\draw[color=red] \J -- \K -- \N -- \M -- cycle; +\draw[<-,color=red] \J -- \K; +\node[color=red] at ({2*\s},-0.5) {$z_{12}$}; +\end{scope} + +\begin{scope}[xshift=10cm,yshift=-3cm] +\gebiet +\draw[color=red] \J -- \K -- \N -- \O -- \M -- cycle; +\draw[->,color=red] \J -- \K; +\node[color=red] at ({2*\s},-0.5) {$z_{13}$}; +\end{scope} + + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/polyeder.pdf b/buch/chapters/95-homologie/images/polyeder.pdf Binary files differnew file mode 100644 index 0000000..3a8ba60 --- /dev/null +++ b/buch/chapters/95-homologie/images/polyeder.pdf diff --git a/buch/chapters/95-homologie/images/polyeder.tex b/buch/chapters/95-homologie/images/polyeder.tex new file mode 100644 index 0000000..9a900cc --- /dev/null +++ b/buch/chapters/95-homologie/images/polyeder.tex @@ -0,0 +1,109 @@ +% +% tikztemplate.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math,calc} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +% add image content here +\begin{scope}[xshift=-3.5cm,scale=0.5] +\coordinate (A) at (0,0); +\coordinate (B) at (4,0); +\coordinate (C) at (5,-2); +\coordinate (D) at (8,-1); +\coordinate (E) at (7,1); +\coordinate (F) at (7,3); +\coordinate (G) at (1,3); +\coordinate (H) at (5,4); +\coordinate (I) at (9,5); +\coordinate (J) at (4,7); +\coordinate (K) at (-1,9); +\coordinate (L) at (7,11); +\coordinate (M) at (6,-0.5); + +\fill[color=gray,opacity=0.5] (A)--(B)--(H)--(G)--cycle; +\fill[color=gray,opacity=0.5] (G)--(I)--(K)--cycle; +\fill[color=gray,opacity=0.5] (G)--(L)--(K)--cycle; + +\draw (K)--(G)--(A)--(B)--(D); +\draw (C)--(E); +\draw (G)--(I)--(K); +\draw (G)--(L)--(K); +\draw (B)--(H); +\draw (B)--(F); + +\fill (A) circle[radius=0.1]; +\fill (B) circle[radius=0.1]; +\fill (C) circle[radius=0.1]; +\fill (D) circle[radius=0.1]; +\fill (E) circle[radius=0.1]; +\fill (F) circle[radius=0.1]; +\fill (G) circle[radius=0.1]; +\fill (H) circle[radius=0.1]; +\fill (I) circle[radius=0.1]; +%\fill (J) circle[radius=0.1]; +\fill (K) circle[radius=0.1]; +\fill (L) circle[radius=0.1]; +%\fill (M) circle[radius=0.1]; + +\draw[color=red] (H) circle[radius=0.5]; +\draw[color=red] (J) circle[radius=0.5]; +\draw[color=red] (M) circle[radius=0.5]; +\draw[color=red] ($0.25*(A)+0.25*(B)+0.25*(G)+0.25*(H)$) circle[radius=0.5]; + +\end{scope} + +\begin{scope}[xshift=3.5cm,scale=0.5] +\coordinate (A) at (0,0); +\coordinate (B) at (4,0); +\coordinate (C) at (5,-2); +\coordinate (D) at (8,-1); +\coordinate (E) at (7,1); +\coordinate (F) at (7,3); +\coordinate (G) at (1,3); +\coordinate (H) at (5,4); +\coordinate (I) at (9,5); +\coordinate (J) at (4,7); +\coordinate (K) at (-1,9); +\coordinate (L) at (7,11); +\coordinate (M) at (6,-0.5); + +\fill[color=gray!50] (A)--(B)--(H)--(I)--(J)--(L)--(K)--(G)--cycle; + +\draw (K)--(G)--(A)--(B)--(D); +\draw (C)--(E); +\draw (G)--(I)--(K); +\draw (G)--(L)--(K); +\draw (B)--(H); +\draw (B)--(F); +\draw (H)--(J); +\draw (A)--(H); + +\fill (A) circle[radius=0.1]; +\fill (B) circle[radius=0.1]; +\fill (C) circle[radius=0.1]; +\fill (D) circle[radius=0.1]; +\fill (E) circle[radius=0.1]; +\fill (F) circle[radius=0.1]; +\fill (G) circle[radius=0.1]; +\fill (H) circle[radius=0.1]; +\fill (I) circle[radius=0.1]; +\fill (J) circle[radius=0.1]; +\fill (K) circle[radius=0.1]; +\fill (L) circle[radius=0.1]; +\fill (M) circle[radius=0.1]; + +\end{scope} + +\end{tikzpicture} +\end{document} + diff --git a/buch/chapters/95-homologie/images/tetraeder.jpg b/buch/chapters/95-homologie/images/tetraeder.jpg Binary files differnew file mode 100644 index 0000000..0ec168b --- /dev/null +++ b/buch/chapters/95-homologie/images/tetraeder.jpg diff --git a/buch/chapters/95-homologie/images/tetraeder.pdf b/buch/chapters/95-homologie/images/tetraeder.pdf Binary files differnew file mode 100644 index 0000000..0a57e95 --- /dev/null +++ b/buch/chapters/95-homologie/images/tetraeder.pdf diff --git a/buch/chapters/95-homologie/images/tetraeder.pov b/buch/chapters/95-homologie/images/tetraeder.pov new file mode 100644 index 0000000..b110f96 --- /dev/null +++ b/buch/chapters/95-homologie/images/tetraeder.pov @@ -0,0 +1,116 @@ +// +// tetraeder.pov +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#version 3.7; +#include "colors.inc" + +global_settings { + assumed_gamma 1 +} + +#declare imagescale = 0.169; +#declare O = <0, 0, 0>; +#declare at = 0.02; + +camera { + location <-2, 3, -10> + look_at <0, 0.18, 0> + right 16/9 * x * imagescale + up y * imagescale +} + +//light_source { +// <-14, 20, -50> color White +// area_light <1,0,0> <0,0,1>, 10, 10 +// adaptive 1 +// jitter +//} + +light_source { + <-41, 20, -20> color White + area_light <1,0,0> <0,0,1>, 10, 10 + adaptive 1 + jitter +} + +sky_sphere { + pigment { + color rgb<1,1,1> + } +} + +#declare v1 = <1,1,1>; +#declare v2 = <-1,1,-1>; +#declare farbe = rgbf<0.8,0.8,1.0,0.5>; + +#declare tetraederwinkel = acos(vdot(v1,v2)/(vlength(v1)*vlength(v2))); + +#declare O = < 0, 0, 0 >; +#declare A = < 0, 1, 0 >; +#declare B = < sin(tetraederwinkel), cos(tetraederwinkel), 0>; +#declare C = < sin(tetraederwinkel)*cos(2*pi/3), cos(tetraederwinkel), sin(2*pi/3)>; +#declare D = < sin(tetraederwinkel)*cos(2*pi/3), cos(tetraederwinkel), -sin(2*pi/3)>; + +#macro arrow(from, to, arrowthickness, c) +#declare arrowdirection = vnormalize(to - from); +#declare arrowlength = vlength(to - from); +union { + sphere { + from, 1.0 * arrowthickness + } + cylinder { + from, + from + (arrowlength - 8 * arrowthickness) * arrowdirection, + arrowthickness + } + cone { + from + (arrowlength - 8 * arrowthickness) * arrowdirection, + 2 * arrowthickness, + to - 3 * arrowthickness * arrowdirection, + 0 + } + pigment { + color c + } + finish { + specular 0.9 + metallic + } +} +#end + +union { + arrow(B, C, at, White) + arrow(D, C, at, White) + arrow(D, B, at, White) + arrow(B, A, at, White) + arrow(C, A, at, White) + arrow(D, A, at, White) + sphere { A, 4 * at } + sphere { B, 4 * at } + sphere { C, 4 * at } + sphere { D, 4 * at } + pigment { + color White + } + finish { + specular 0.9 + metallic + } +} + +mesh { + triangle { A, B, C } + triangle { A, C, D } + triangle { A, D, B } + triangle { B, C, D } + pigment { + color farbe + } +// finish { +// specular 0.9 +// metallic +// } +} diff --git a/buch/chapters/95-homologie/images/tetraeder.tex b/buch/chapters/95-homologie/images/tetraeder.tex new file mode 100644 index 0000000..e62770f --- /dev/null +++ b/buch/chapters/95-homologie/images/tetraeder.tex @@ -0,0 +1,97 @@ +% +% tetraeder.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math,calc} +\usepackage{ifthen} +\begin{document} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{7} +\def\hoehe{4} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\node at (0,0) {\includegraphics[width=8cm]{tetraeder.jpg}}; + +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} + +\def\knoten#1#2{ + %\fill[color=white,opacity=0.5] #1 circle[radius=0.2]; + \node at #1 {$#2$}; +} + +\knoten{(-2.2,-3.6)}{0}; +\knoten{( 3.3,-1.9)}{1}; +\knoten{(-3.4,-1.2)}{2}; +\knoten{(-0.75,3.6)}{3}; + +\def\s{0.2} + +\def\kante#1#2{ + %\fill[color=white,opacity=0.5] #1 circle[radius=0.2]; + \fill[color=white,opacity=0.5] + ($#1+(-\s,-\s)$) -- + ($#1+(+\s,-\s)$) -- + ($#1+(+\s,+\s)$) -- + ($#1+(-\s,+\s)$) -- cycle; + \node at #1 {$#2$}; +} + +\kante{(0.5,-2.8)}{k_0} +\kante{(-2.8,-2.3)}{k_1} +\kante{(-1.4,0)}{k_2} +\kante{(-0.4,-1.55)}{k_3} +\kante{(1.25,0.95)}{k_4} +\kante{(-2.08,1.1)}{k_5} + +\def\r{0.33} + +\def\flaeche#1#2{ + \fill[color=white,opacity=0.5] + ($#1+({-\r*cos(30)},{-\r*sin(30)})$) -- + ($#1+({\r*cos(30)},{-\r*sin(30)})$) -- + ($#1+(0,{\r})$) -- cycle; + \node at #1 {$#2$}; +} + +\flaeche{(-0.7,-5)}{f_0} +\draw (-0.7,-4.7) -- (-0.7,-3.25); +\draw[->,color=black!70] (-0.7,-3.06) -- (-0.7,-2.5); +\flaeche{(0.2,-0.5)}{f_1} +\flaeche{(-2.3,-0.7)}{f_2} +\coordinate (A) at (1,2.6); +\coordinate (B) at (0,1); + +\flaeche{($1.2*(A)-0.2*(B)$)}{f_3} + +\def\t{0.58} +\pgfmathparse{1-\t} +\xdef\T{\pgfmathresult} +\draw (A) -- ($\t*(A)+\T*(B)$); + +\def\t{0.48} +\pgfmathparse{1-\t} +\xdef\T{\pgfmathresult} +\draw[->,color=black!70] ($\t*(A)+\T*(B)$) -- (B); + + +\end{tikzpicture} + +\end{document} + diff --git a/buch/chapters/95-homologie/induzierteabb.tex b/buch/chapters/95-homologie/induzierteabb.tex new file mode 100644 index 0000000..13591d7 --- /dev/null +++ b/buch/chapters/95-homologie/induzierteabb.tex @@ -0,0 +1,204 @@ +\subsection{Induzierte Abbildung +\label{buch:subsection:induzierte-abbildung}} +Früher haben wurde eine Abbildung $f_*$ zwischen Kettenkomplexen $C_*$ und +$D_*$ so definiert, +dass sie mit den Randoperatoren verträglich sein muss. +Diese Forderung bewirkt, dass sich auch eine lineare Abbildung +\[ +H_k(f) \colon H_k(C) \to H_k(D) +\] +zwischen den Homologiegruppen ergibt, wie wir nun zeigen wollen. + +\subsubsection{Definition der induzierten Abbildung} +Um eine Abbildung von $H_k(C)$ nach $H_k(D)$ zu definieren, müssen wir +zu einem Element von $H_k(C)$ ein Bildelement konstruieren. +Ein Element in $H_k(C)$ ist eine Menge von Zyklen in $Z^C_k$, die sich +nur um einen Rand in $B_k$ unterscheiden. +Wir wählen also einen Zyklus $z\in Z_k$ und bilden ihn auf $f_k(z)$ ab. +Wegen $\partial^D_kf(z)=f\partial^C_kz = f(0) =0 $ ist auch $f_k(z)$ +ein Zyklus. +Wir müssen jetzt aber noch zeigen, dass eine andere Wahl des Zyklus +das gleiche Element in $H_k(D)$ ergibt. +Dazu genügt es zu sehen, dass sich $f(z)$ höchstens um einen Rand +ändert, wenn man $z$ um einen Rand ändert. +Sei also $b\in B^C_k$ ein Rand, es gibt also ein $w\in C_{k+1}$ mit +$\partial^C_{k+1}w=b$. +Dann gilt aber auch +\[ +f_k(z+b) += +f_k(z) + f_k(b) += +f_k(z) + f_k(\partial^C_{k+1}w) += +f_k(z) + \partial^D_{k+1}(f_k(w)). +\] +Der letzte Term ist ein Rand in $D_k$, somit ändert sich $f_k(z)$ nur +um diesen Rand, wenn man $z$ um einen Rand ändert. +$f_k(z)$ und $f_k(z+b)$ führen auf die selbe Homologieklasse. + +\subsubsection{Matrixdarstellung} +In Abschnitt~\ref{buch:subsection:basiswahl} wurde gezeigt, wie man +für die Vektorräume der Zyklen eine Basis derart finden kann, +dass die Ränder von einer Teilmenge der Basis aufgespannt werden. +Eine solche Basis kann man immer erweitern zu einer Basis von $C_k$. +Für das Folgende bezeichnen wir die Vektoren einer solche Basis von $C_k$ +mit +\[ +\{ +b_1,\dots, b_r, +z_1,\dots,z_l, +c_1,\dots,c_s +\}. +\] +wobei die Vektoren die folgende Bedeutung haben: +\begin{center} +\begin{tabular}{|l|l|} +\hline +Vektoren&Bedeutung\\ +\hline +$b_1,\dots,b_r$ & Basis für $B_k(C)$ \\ +$z_1,\dots,z_l$ & zusätzliche Vektoren für eine Basis von $Z_k(C)$ \\ +$c_1,\dots,c_s$ & zusätzliche Vektoren für eine Basis von $C_k$ \\ +\hline +\end{tabular} +\end{center} + +Wählt man eine Basis dieser Art sowohl in $C_*$ wie auch in $D_*$, +dann kann man die induzierte Abbildung als $3\times 3$-Blockmatrix +schreiben. +Man verwendet dabei, dass $f_k$ die Unterräume $B_k(C)$ und +$Z_k(C)$ in die entsprechenden Unterräume $B_k(D)$ und $Z_k(D)$ +abbildet, also +\[ +f_k(B_k(C)) \subset B_k(D) +\qquad\text{und}\qquad +f_k(Z_k(C)) \subset Z_k(D). +\] +In der Matrixdarstellung äussert sich das darin, dass die Blöcke +links unten zu Null werden. +Die Matrixdarstellung von $f_k$ hat daher die Form +\[ +f_k += +\begin{pmatrix} +f_{k,B} & * & * \\ + 0 & f_{k,Z} & * \\ + 0 & 0 & f_{k,*} +\end{pmatrix}. +\] +Genauso kann man natürlich auch die Randoperatoren in dieser Basis +ausdrücken. +Sie bilden die Zyklen auf $0$ ab und aus den Vektoren $c_1,\dots,c_s$ +werden Ränder. +Die Matrix hat daher die Form +\[ +\partial_k += +\begin{pmatrix} +0& 0 & \Delta_k \\ +0& 0 & 0 \\ +0& 0 & 0 +\end{pmatrix} +\] +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/complexbasis.pdf} +\caption{Basiswahl für den Kettenkomplex $C_k$. +Der Randoperator $\partial_k$ bildet $Z_k$ auf $0$ ab, der blaue +Unterraum, aufgespannt von den Vektoren $c_i$, wird bijektiv auf $B_{k-1}$ +abgebildet. +Eine Basis kann immer so gefunden werden, dass die Vektoren $c_i$ +von $\partial_k$ auf die Basisvektoren von $B_{k-1}$ abgebildet werden. +In dieser Basis ist $\Delta_k$ eine Einheitsmatrix. +\label{buch:homologie:fig:komplexbasis}} +\end{figure}% +Die Bedingung \eqref{buch:komplex:abbildung} für die Komplexabbildung +bekommt jetzt die Matrixform +\begin{equation} +\left. +\begin{aligned} +\partial_k^{D}\circ f_k +&= +\begin{pmatrix} +0&0&\Delta_k^{(D)}\\ +0&0&0\\ +0&0&0 +\end{pmatrix} +\begin{pmatrix} +f_{k,B} & * & * \\ + 0 & f_{k,Z} & * \\ + 0 & 0 & f_{k,*} +\end{pmatrix} += +\begin{pmatrix} +0&0&\Delta_k^{(D)}f_{k,*}\\ +0&0&0\\ +0&0&0 +\end{pmatrix} +\\ +f_{k-1}\circ \partial_k^C +&= +\begin{pmatrix} +f_{k-1,B}& * & * \\ + 0 &f_{k-1,Z}& * \\ + 0 & 0 &f_{k-1,*} +\end{pmatrix} +\begin{pmatrix} +0&0&\Delta_k^{(C)}\\ +0&0&0\\ +0&0&0 +\end{pmatrix} += +\begin{pmatrix} +0&0&f_{k-1,B}\Delta_k^{(C)}\\ +0&0&0\\ +0&0&0 +\end{pmatrix} +\end{aligned} +\right\} +\Rightarrow +\Delta_k^{(D)}f_{k,*} += +f_{k-1,B}\Delta_k^{(C)}. +\label{buch:homologie:matrixform} +\end{equation} +Für die induzierte Abbildung in Homologie ist ausschliesslich der +Block $f_{k,Z}$ notwendig, die Matrix von $H_k(f)$ in der gewählten +Basis von $H_k(C)$ bzw.~$H_k(D)$ ist also genau die Matrix $f_{k,Z}$. + + +Wie Abbildung~\ref{buch:homologie:fig:komplexbasis} können die +Basisvektoren $c_*$ in $C_k$ so gewählt werden, dass sie vom Randoperator +$\partial_k$ auf die Basisvektoren von $Z_{k-1}$ abgebildet werden. +Bei dieser Wahl wird die Matrix $\Delta_k$ eine Einheitsmatrix. + +\subsubsection{Spur} +Wir betrachten jetzt den Fall einer Selbstabbildung $f_*\colon C_*\to C_*$. +Die Basis soll so gewählt werden, dass $\Delta_k$ eine Einheitsmatrix ist. +Aus~\eqref{buch:homologie:matrixform} kann man ablesen, dass für diese +Basiswahl $f_{k,*}=f_{k-1,B}$ gilt. +Die Matrizen von $f_k$ haben daher die Form +\[ +f_k += +\begin{pmatrix} +f_{k,B} & * & * \\ + 0 & f_{k,Z} & * \\ + 0 & 0 & f_{k-1,B} +\end{pmatrix}. +\] +Entsprechend ist die Spur +\begin{equation} +\operatorname{Spur} f_k += +\operatorname{Spur} f_{k,B} ++ +\operatorname{Spur} f_{k,Z} ++ +\operatorname{Spur} f_{k-1,B}. +\label{buch:homologie:eqn:spur} +\end{equation} + + + diff --git a/buch/chapters/95-homologie/komplex.tex b/buch/chapters/95-homologie/komplex.tex index 6dd8efb..9787bb2 100644 --- a/buch/chapters/95-homologie/komplex.tex +++ b/buch/chapters/95-homologie/komplex.tex @@ -6,9 +6,107 @@ \section{Kettenkomplexe \label{buch:section:komplex}} \rhead{Kettenkomplexe} +Die algebraische Struktur, die in Abschnitt~\ref{buch:subsection:triangulation} +konstruiert wurde, kann noch etwas abstrakter konstruiert werden. +Es ergibt sich das Konzept eines Kettenkomplexes. +Die Triangulation gibt also Anlass zu einem Kettenkomplex. +So lässt sich zu einem geometrischen Objekt ein algebraisches +Vergleichsobjekt konstruieren. +Im Idealfall lassens ich anschliessend geometrische Eigenschaften mit +algebraischen Rechnungen zum Beispiel in Vektorräumen mit Matrizen +beantworten. -\subsection{Randoperator von Simplexen -\label{buch:subsection:randoperator-von-simplexen}} +\subsection{Definition +\label{buch:subsection:kettenkomplex-definition}} +Die Operation $\partial$, die für Simplizes konstruiert worden ist, +war linear und hat die Eigenschaft $\partial^2$ gehabt. +Diese Eigenschaften reichen bereits für Definition eines Kettenkomplexes. + +\begin{definition} +Eine Folge $C_0,C_1,C_2,\dots$ von Vektorräumen über dem Körper $\Bbbk$ +mit einer Folge von linearen Abbildungen +$\partial_k\colon C_k \to C_{k-1}$, dem {\em Randoperator}, +heisst ein Kettenkomplex, wenn $\partial_{k-1}\partial_k=0$ gilt +für alle $k>0$. +\end{definition} + +Die aus den Triangulationen konstruieren Vektorräme von +Abschnitt~\ref{buch:subsection:triangulation} bilden einen +Kettenkomplex. + +XXX nachrechnen: $\partial^2 = 0$ ? + +\subsection{Abbildungen +\label{buch:subsection:abbildungen}} +Wenn man verschiedene geometrische Objekte mit Hilfe von Triangulationen +vergleichen will, dann muss man auch das Konzept der Abbildungen zwischen +den geometrischen Objekten in die Kettenkomplexe transportieren. + +Eine Abbildung zwischen Kettenkomplexen muss einerseits eine lineare +Abbildung der Vektorräume $C_k$ sein, andererseits muss sich eine +solche Abbildung mit dem Randoperator vertragen. +Wir definieren daher + +\begin{definition} +Eine Abbildung $f_*$ zwischen zwei Kettenkomplexe $(C_*,\partial^C_*)$ und +$(D_*,\partial^D_*)$ heisst eine Abbildung von Kettenkomplexen, wenn +für jedes $k$ +\begin{equation} +\partial^D_k +\circ +f_{k} += +f_{k-1} +\circ +\partial^C_k +\label{buch:komplex:abbildung} +\end{equation} +gilt. +\end{definition} + +Die Beziehung~\eqref{buch:komplex:abbildung} kann übersichtlich als +kommutatives Diagramm dargestellt werden. +\begin{equation} +\begin{tikzcd} +0 + & C_0 \arrow[l, "\partial_0^C" above] + \arrow[d, "f_0"] + & C_1 \arrow[l,"\partial_1^C" above] + \arrow[d, "f_1"] + & C_2 \arrow[l,"\partial_2^C" above] + \arrow[d, "f_2"] + & \dots \arrow[l] + \arrow[l, "\partial_{3}^C" above] + & C_{k-1} + \arrow[l, "\partial_{k-1}^C" above] + \arrow[d, "f_{k-1}"] + & C_{k}\arrow[l, "\partial_{k}^C" above] + \arrow[d, "f_{k}"] + & \dots + \arrow[l,"\partial_{k+1}^C" above] +\\ +0 + & D_0 \arrow[l, "\partial_0^D" above] + & D_1 \arrow[l,"\partial_1^D" above] + & D_2 \arrow[l,"\partial_2^D" above] + & \dots \arrow[l] + \arrow[l, "\partial_{3}^D" above] + & D_{k-1} + \arrow[l, "\partial_{k-1}^D" above] + & D_{k}\arrow[l, "\partial_{k}^D" above] + & \dots + \arrow[l,"\partial_{k+1}^D" above] +\end{tikzcd} +\label{buch:komplex:abbcd} +\end{equation} +Die Relation~\eqref{buch:komplex:abbildung} drückt aus, dass man jeden +den Pfeilen im Diagram~\eqref{buch:komplex:abbcd} folgen kann und +dabei zwischen zwei Vektorräumen unabhängig vom Weg die gleiche Abbildung +resultiert. + +Die Verfeinerung einer Triangulation erzeugt eine solche Abbildung von +Komplexen. + + +% XXX simpliziale Approximation -\subsection{Kettenkomplexe und Morphismen -\label{buch:subsection:kettenkomplex}} diff --git a/buch/chapters/95-homologie/mayervietoris.tex b/buch/chapters/95-homologie/mayervietoris.tex deleted file mode 100644 index 57105f8..0000000 --- a/buch/chapters/95-homologie/mayervietoris.tex +++ /dev/null @@ -1,28 +0,0 @@ -% -% mayervietoris.tex -% -% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -% -\section{Exaktheit und die Mayer-Vietoris-Folge -\label{buch:section:mayervietoris}} -\rhead{Exaktheit und die Mayer-Vietoris-Folge} -Die Berechnung der Homologie-Gruppen ist zwar im Wesentlichen ein -kombinatorisches Problem, trotzdem ist eher aufwändig. -Oft weiss man, wie sich toplogische Räume aus einfacheren Räumen -zusammensetzen lassen. -Eine Mannigkfaltigkeit zum Beispiel wird durch die Karten -definiert, also zusammenziehbare Teilmengen von $\mathbb{R}^n$, -die die Mannigkfaltigkeit überdecken. -Das Ziel dieses Abschnittes ist, Regeln zusammenzustellen, mit denen -man die Homologie eines solchen zusammengesetzten Raumes aus der -Homologie der einzelnen Teile und aus den ``Verklebungsabbildungen'', -die die Teile verbinden, zu berechnen. - -\subsection{Kurze exakte Folgen von Kettenkomplexen -\label{buch:subsection:exaktefolgen}} - -\subsection{Schlangenlemma und lange exakte Folgen -\label{buch:subsection:schlangenlemma}} - -\subsection{Mayer-Vietoris-Folge -\label{buch:subsection:mayervietoris}} diff --git a/buch/chapters/95-homologie/simplex.tex b/buch/chapters/95-homologie/simplex.tex index 5ca2ca8..3bf1004 100644 --- a/buch/chapters/95-homologie/simplex.tex +++ b/buch/chapters/95-homologie/simplex.tex @@ -1,17 +1,17 @@ % -% simplex.tex -- simplizes und simpliziale Komplexe +% simplex.tex -- simplizes und Polyeder % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % -\section{Simplexe und simpliziale Komplexe +\section{Simplizes \label{buch:section:simplexe}} -\rhead{Simplexe und simpliziale Komplexe} +\rhead{Simplizes} Die Idee, das Dreieck und seinen Rand zu unterscheiden verlangt, dass wir zunächst Dreiecke und deren höherdimensionale Verallgemeinerungen, die sogenannten Simplizes entwickeln müssen. -\subsection{Simplexe und Rand -\label{buch:subsection:simplexe}} +\subsection{Simplizes und Rand +\label{buch:subsection:simplices}} \subsubsection{Rand eines Dreiecks} Die Inzidenz-Matrix eines Graphen hat einer Kante die beiden Endpunkte @@ -231,8 +231,127 @@ Vorzeichen zu, die Matrix ist \] \end{definition} +\subsection{Polyeder} +\begin{figure} +\centering +\includegraphics{chapters/95-homologie/images/polyeder.pdf} +\caption{Aufbau eines zweidimensionalen Polyeders aus +verschiedenen Simplizes. +Die Schnittmenge zweier Simplizes muss ein Untersimplex beider Simplizes +sein. +Die roten Kreise im linken Bild weisen auf verschiedene Situationen +hin, wo das diese Bedingung nicht erfüllt ist. +In rechten Bild sind zusätzliche Simlizes hinzugefügt worden, um +die Bedingungen eines Polyeders zu erfüllen. +\label{buch:homologie:figure:polyeder}} +\end{figure} +Aus einzelnen Simplizes können jetzt kompliziertere geometrische +Objekte gebaut werden. +Ein Graph ist ein Beispiel für ein geometrisches Objekt, welches +als Vereinigung von 1-Simplizes entsteht. +Die Vereinigung ist aber nicht beliebig, vielmehr ist die Schnittmenge +zweier beliebiger 1-Simplizes immer entweder leer, eine Menge +mit nur einem Vertex oder ein ganzes 1-Simplex. + +Dies reicht aber nicht, wie Abbildung~\ref{buch:homologie:polyeder} +zeigt. +In einem Graphen dürfen sich Kanten nicht in einem inneren Punkt treffen, +sondern nur in Endpunkten. +Verallgemeinert auf höherdimensionale Simplizes kann man dies als die +Bedingung formulieren, dass die Schnittmenge zweier beliebiger +Simplizes immer Untersimplizes beider Simplizes sein müssen. +Wir fassen dies zusammen in der folgenden Definition. + +\begin{definition} +\index{Polyeder}% +\index{Dimension eines Polyeders}% +\index{Polyeder, Dimension eines}% +Ein {\em Polyeder} ist eine Vereingung von endlich vielen Simplizes derart, +dass die Schnittmenge zweier beliebiger Simplizes immer ein Untersimplex +beider Simplizes ist. +Die {\em Dimension} des Polyeders ist die grösste Dimension der darin +enthaltenen Simplizes. +\end{definition} + +Ein Graph ist nach dieser Definition ein eindimensionales Polyeder. +Die Mengen in der Abbildung~\ref{buch:homologie:figure:polyeder} +ist kein Polyeder, kann aber leicht zu einem Polyeder gemacht werden, +indem man einzelne Kanten mit zusätzlichen Punkten unterteilt. +Auch müssen die zweidimensionalen Simplizes aufgeteilt werden. + +Die Abbildung~\ref{buch:homologie:figure:polyeder} zeigt auch, dass +die Darstellung einer Punktmenge als Polyeder nicht eindeutig ist. +Man kann die Kanten und Flächen jederzeit weiter unterteilen, ohne +dass sich die Gestalt der gesamten Menge dadurch ändert. \subsection{Triangulation -\label{buch:subsection:}} +\label{buch:subsection:triangulation}} +Unser Ziel ist, geometrische Objekte besser verstehen zu können. +Dabei sind uns Deformationen ja sogar Knicke egal, es interessiert uns +nur die ``Gestalt'' des Objekts. +Entfernungen zwischen Punkten sind ebenfalls von untergeordneter +Bedeutung, da sie bei Deformation nicht erhalten bleiben. +Der Begriff des ``topologischen Raumes'' fasst diese Ideen mathematisch +präzise ein, eine genaue Definition würde aber an dieser Stelle zu weit +führen. +Stattdessen beschränken wir uns auf eine Klasse von Punktmengen, die man +mit Simplizes beschreiben kann. + +Ein topologischer Raum zeichnet sich durch einen Nachbarschaftsbegriff +von Punkte aus, der erlaubt zu definieren, was eine stetige Abbildung ist. +Ein stetige Abbildungen bildet nahe beeinander liegende Punkte wieder +auf nahe beeinander liegende Punkte ab. +Dass nahe liegende Punkte nicht plötzlich auf weit auseinander liegende +Punkte abgebildet werden gibt die Intuition wieder, dass Deformationen +möglich sein sollen, dass der Raum dabei aber nicht ``reissen'' darf. +Zwei topologische Räume $X$ und $Y$ können daher als ``gleichgestaltig'' +betrachtet werden, wenn es zwei stetige Abbildungen $f\colon X\to Y$ +und $g\colon Y\to X$ gibt, die zu einander invers sein. +Oder wenn sich $X$ stetig auf $Y$ abbilden lässt, so dass auch die +Umkehrabbildung stetig ist. +Eine solche Abbildung heisst ein {\em Homöomorphismus}, die beiden Räume +$X$ und $Y$ heissen {\em homomorph}. + +Eine Kugel ist natürlich kein Polyeder, aber sie kann leicht homöomorph +auf ein dreidimensionales Simplex abgebildet werden. + +\begin{beispiel} +Sei $T$ ein reguläres Tetraeder mit den Ecken auf der dreidimensionalen +Einheitskugel $B^3$. +Für jeden Richtungsvektor $x\ne 0$ sei $l(x)$ Entfernung vom Mittelpunkt des +Tetraeders bis zum Durchstosspunkt einer Geraden durch den Mittelpunkt +mit Richtungsvektor $x$ durch die Oberfläche des Tetraeders. +Dann sind die Abbildungen +\[ +f\colon +T\to B^3 +: +x \mapsto\begin{cases} +\displaystyle +\frac{x}{l(x)}&\quad\text{für $x\ne 0$}\\ +0&\quad\text{für $x=0$} +\end{cases} +\qquad\text{und}\qquad +g\colon +B^3\to T +: +x \mapsto\begin{cases} +l(x) x&\quad\text{für $x\ne 0$}\\ +0&\quad\text{für $x=0$} +\end{cases} +\] +zueinander inverse stetige Abbildungen oder Homöomorphismen. +\end{beispiel} + +Im Folgenden sollen daher nur solche topologischen Räume untersucht werden, +die homöomorph sind zu einem Polyeder. +Man nennt die homöomorphe Abbildung eines Polyeders auf so einen Raum +auch eine Triangulation. +Durch Unterteilung der Simplizes in kleiner Simplizes kann eine solche +Triangulation beliebig verfeinert werden. + + + + diff --git a/buch/chapters/references.bib b/buch/chapters/references.bib index a5d0201..977bf81 100644 --- a/buch/chapters/references.bib +++ b/buch/chapters/references.bib @@ -20,6 +20,12 @@ keywords = "World Wide Web, Search engines, Information retrieval, PageRank, Goo abstract = "In this paper, we present Google, a prototype of a large-scale search engine which makes heavy use of the structure present in hypertext. Google is designed to crawl and index the Web efficiently and produce much more satisfying search results than existing systems. The prototype with a full text and hyperlink database of at least 24 million pages is available at http://google.stanford.edu/ To engineer a search engine is a challenging task. Search engines index tens to hundreds of millions of Web pages involving a comparable number of distinct terms. They answer tens of millions of queries every day. Despite the importance of large-scale search engines on the Web, very little academic research has been done on them. Furthermore, due to rapid advance in technology and Web proliferation, creating a Web search engine today is very different from three years ago. This paper provides an in-depth description of our large-scale Web search engine — the first such detailed public description we know of to date. Apart from the problems of scaling traditional search techniques to data of this magnitude, there are new technical challenges involved with using the additional information present in hypertext to produce better search results. This paper addresses this question of how to build a practical large-scale system which can exploit the additional information present in hypertext. Also we look at the problem of how to effectively deal with uncontrolled hypertext collections where anyone can publish anything they want." } +@book{buch:linalg, + title = {Lineare Algebra}, + author = {Andreas M"uller}, + url = {https://github.com/AndreasFMueller/LinAlg.git}, + year = {2010} +} @book{buch:mathsem-wavelets, title = {Mathematisches Seminar Wavelets}, @@ -33,6 +39,13 @@ abstract = "In this paper, we present Google, a prototype of a large-scale searc year = {2016}, } +@online{buch:rfc2409, + title = {The Internet Key Exchange (IKE)}, + author = { D. Harkins and D. Carrel}, + url = {https://datatracker.ietf.org/doc/html/rfc2409}, + year = {1998} +} + @online{buch:fftw, title = {Fastest Fourier Transform in the West}, url = {http://www.fftw.org/}, diff --git a/buch/common/macros.tex b/buch/common/macros.tex index 2c6eea2..5783122 100644 --- a/buch/common/macros.tex +++ b/buch/common/macros.tex @@ -108,6 +108,7 @@ \newtheorem{forderung}{Forderung}[chapter] \newtheorem{konsequenz}[satz]{Konsequenz} \newtheorem{algorithmus}[satz]{Algorithmus} +\newtheorem{ziel}[satz]{Ziel} \renewcommand{\floatpagefraction}{0.7} \definecolor{darkgreen}{rgb}{0,0.6,0} diff --git a/buch/papers/clifford/0_ElevatorPitch.tex b/buch/papers/clifford/0_ElevatorPitch.tex index 0db5617..ad9bcc2 100644 --- a/buch/papers/clifford/0_ElevatorPitch.tex +++ b/buch/papers/clifford/0_ElevatorPitch.tex @@ -1,2 +1,6 @@ -TODO... -GA [Geometric Algebra i.a.W. Clifford Algebra] provides a unified language for the whole of physics and for much of mathematics and its applications that is conceptually and computationally superior to alternative mathematical systems in many application domains.
\ No newline at end of file + +Der Nutzen, welche die Clifford Algebra hat, lässt sich am besten mit den Worten des modernen Begründers dieser erläutern. + +"GA [Geometric Algebra i.a.W. Clifford Algebra] provides a unified language for the whole of physics and for much of mathematics and its applications that is conceptually and computationally superior to alternative mathematical systems in many application domains." \cite{clifford:hestenes_GA} + +Im folgenden hoffen wir den Leser von der Nützlichkeit und der geometrischen Schönheit der Clifford Algebra zu überzeugen.
\ No newline at end of file diff --git a/buch/papers/clifford/10_Quaternionen.tex b/buch/papers/clifford/10_Quaternionen.tex index 375c6e7..d04ea38 100644 --- a/buch/papers/clifford/10_Quaternionen.tex +++ b/buch/papers/clifford/10_Quaternionen.tex @@ -6,222 +6,172 @@ \section{Quaternionen} \rhead{Quaternionen} -Wie die komplexen Zahlen eine Erweiterung der reellen Zahlen sind, sind die Quaternionen eine Erweiterung der komplexen Zahlen für den dreidimensionalen Raum. Sie haben, wie die komplexen Zahlen, eine dreh-streckende Eigenschaft. -Sie finden beispielsweise in der Computergraphik und in der Robotik Anwendung. +Wie die komplexen Zahlen eine Erweiterung der reellen Zahlen sind, sind die Quaternionen eine Erweiterung der komplexen Zahlen für den dreidimensionalen Raum. Sie haben, wie die komplexen Zahlen, eine drehstreckende Eigenschaft. +Sie finden beispielsweise in der Computergrafik und Robotik Anwendung. Die Quaternionen \begin{align} - q = w + xi + yj + zk \quad w,x,y,z \in \mathbb{R}\enspace q \in \mathbb{H} +q = w + xi + yj + zk \quad w,x,y,z \in \mathbb{R}\enspace q \in \mathbb{H} \end{align} -können dabei eine Drehstreckung mit dieser Formel erreichen +können dabei eine Drehstreckung mit \begin{align} \label{QuatRot} - \begin{split} - &v'' = qvq^{-1};\quad q,v,q^{-1} \in \mathbb{H}\\ - &\operatorname{Re}(q) = \operatorname{Re}(q^{-1})\quad \operatorname{Im}(q) = -\operatorname{Im}(q^{-1}) - \end{split} +\begin{split} +v \mapsto v'' = qvq^{-1} +\end{split} \end{align} -Auffallend ist hier schon die Ähnlichkeit zu dem Kapitel Rotation. Man könnte sich nun fragen wieso es drei imaginäre Einheiten $i,j,k$ gibt und nicht zwei, was doch näherliegender wäre. Der Grund liegt darin, weil es in der dritten Dimension drei Drehachsen gibt, anstatt nur eine. Wie im Kapitel Rotation beschrieben können wir auch hier die drei Drehungen durch Linearkombinationen von drei Bivektoren beschreiben. In der geometrischen Algebra ist es leicht herauszufinden wie viele Imaginärteile für jede weitere Dimension existieren. Dabei muss man nur die Anzahl der unabhängigen Bivektoren ermitteln. In der vierten Dimension würden es beispielsweise durch alle Vektorkombinationen von $\mathbf{e}_1, \mathbf{e}_2,\mathbf{e}_3, \mathbf{e}_4$ insgesamt 8 Bivektoren existieren (Nicht 16, da $\mathbf{e}_{ij} = -\mathbf{e}_{ji}$ nicht unabhängig voneinander sind). +erreichen, falls $q,v,q^{-1} \in \mathbb{H}$ und die Zusammenhänge +\begin{align} +\operatorname{Re}(q) = \operatorname{Re}(q^{-1})\quad\text{und}\quad \operatorname{Im}(q) = -\operatorname{Im}(q^{-1}) +\end{align} +gelten. Auffallend ist bei der abbildenden Funktion \eqref{QuatRot} schon die Ähnlichkeit zur Funktion \eqref{rotGA} im Abschnitt Drehung. Man könnte sich nun fragen wieso es drei imaginäre Einheiten $i,j,k$ gibt und nicht zwei, was doch näherliegender wäre. Der Grund liegt darin, weil es in drei Dimensionen drei Drehachsen gibt, anstatt nur eine. Wie im Abschnitt Drehung beschrieben, können wir auch hier die drei Drehungen durch Linearkombinationen von drei Bivektoren beschreiben. In der geometrischen Algebra ist es leicht herauszufinden, wie viele Imaginärteile für jede weitere Dimension existieren. Dabei muss man nur die Anzahl der unabhängigen Bivektoren ermitteln. In vier Dimensionen würden es beispielsweise durch alle Vektorkombinationen von $\mathbf{e}_1, \mathbf{e}_2,\mathbf{e}_3, \mathbf{e}_4$ insgesamt 8 Bivektoren existieren (Nicht 16, da $\mathbf{e}_{ij} = -\mathbf{e}_{ji}$ nicht unabhängig voneinander sind). -Ohne die geometrische Algebra, haben wir jetzt aber leider ein kleines Problem. Für die Darstellung der Quaternionen bräuchten wir insgesamt vier Achsen. Drei für die imaginären Einheiten und eine für die reelle Einheit. Ein weiterer Nachteil in visueller Hinsicht entsteht beim Anwenden eines Quaternion auf einen Vektor. Sie befinden sich nicht im gleichen Raum und müssen zuerst ineinander umgewandelt werden, um damit zu rechnen, wie man bei $v \in \mathbb{H}$ in der Formel (\ref{QuatRot}) sieht. +Ohne die geometrische Algebra, haben wir jetzt aber leider ein kleines Problem. Für die Darstellung der Quaternionen bräuchten wir insgesamt vier Achsen. Drei für die imaginären Einheiten und eine für die reelle Einheit. Ein weiterer Nachteil in visueller Hinsicht entsteht beim Anwenden einer Quaternion auf einen Vektor. Sie befinden sich nicht im gleichen Raum und müssen zuerst durch +\begin{align} +\mathbf{v} = x\mathbf{\hat{x}} + y\mathbf{\hat{y}} + z \mathbf{\hat{z}} \in \mathbb{R}^3 \enspace\mapsto\enspace v = 0 + xi + yj + zk \in \mathbb{H} +\end{align} +ineinander umgewandelt werden, um damit zu rechnen. \subsection{Geometrische Algebra} -Die geometrische Algebra besitzt die Fähigkeit beide Probleme zu lösen. Die Quaternionen können, wie schon im 2 dimensionalen Fall durch die gerade Grade $G_3^+(\mathbb{R}) \cong \mathbb{H}$ dargestellt werden. Da wir uns jetzt aber in $G_3(\mathbb{R})$ befinden haben wir drei Basisvektoren $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ und können somit drei Bivektoren bilden $\mathbf{e}_{12}, \mathbf{e}_{23}, \mathbf{e}_{31}$. +Die geometrische Algebra kann beide Probleme beheben. Die Quaternionen können, wie schon im zweidimensionalen Fall durch die gerade Grade $G_3^+(\mathbb{R}) \cong \mathbb{H}$ dargestellt werden. Da wir uns jetzt aber in $G_3(\mathbb{R})$ befinden haben wir drei Basisvektoren $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ und können somit drei Bivektoren $\mathbf{e}_{12}, \mathbf{e}_{23}, \mathbf{e}_{31}$ bilden. \begin{definition} - Multivektoren mit Drehstreckenden Eigenschaften in $G_3(\mathbb{R})$ (gleichbedeutend zu Quaternionen) + Die Multivektoren mit drehstreckenden Eigenschaften in $G_3(\mathbb{R})$ sind \begin{align} - \mathbf{q} = w + x\mathbf{e}_{12} + y\mathbf{e_{23}} + z\mathbf{e_{31}}; \quad w,x,y,z \in \mathbb{R};\enspace \mathbf{q} \in \mathbb{G}_3^+ + \mathbf{q} = w + x\mathbf{e}_{12} + y\mathbf{e}_{23} + z\mathbf{e}_{31} \quad w,x,y,z \in \mathbb{R}\enspace \mathbf{q} \in \mathbb{G}_3^+. \end{align} \end{definition} -Die Probleme werden dadurch gelöst, da wir die Bivektoren im Raum nicht durch einzelne Achsen darstellen müssen, sondern sie als eine orientiere Fläche darstellen können. Anstatt die Vektoren in Quaternionen umzurechnen, können wir jetzt die Vektoren separat im gleichen Raum darstellen. +Die Probleme werden dadurch gelöst, da wir die Bivektoren im Raum nicht durch einzelne Achsen darstellen müssen, sondern sie als eine orientiere Fläche darstellen können. Anstatt die Vektoren in Quaternionen umzurechnen, können wir jetzt die Vektoren separat im gleichen Raum, wie in Abbildung \ref{BildQuaternionen} gezeigt, darstellen. \begin{figure} \centering - \begin{tikzpicture} - % Koordinatensystem - \draw[thin,gray!40] (-3,-2) grid (3,3); - \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; - \draw[<->] (0,-2)--(0,3) node[above]{$a_2$}; - \draw[<->] (3,3)--(-2,-2) node[left]{$a_3$}; - - % v Vektor - \draw[line width=2pt,black,-stealth](0,0)--(2,-1) node[anchor=north]{$\boldsymbol{v}$}; - - % q Quaternion - \draw[line width=0,fill=blue!40] (0,0)--(0.75,0)--(0.75,0.75)--(0,0.75) - node[xshift=0.375cm, yshift=-0.5cm, blue]{$x\boldsymbol{e_{12}}$}; - \draw[->] (0.7,0.55) arc (0:310:0.15); - - \draw[line width=0,fill=blue!40] (0,0)--(-1,-1)--(-1,0.71)--(0,1.71) - node[xshift=-0.5cm, yshift=-1.5cm, blue]{$y\boldsymbol{e_{23}}$}; - \draw[->] (-0.1,1.1) arc (0:310:0.15); - - \draw[line width=0,fill=blue!40] (0,0)--(-0.71,-0.71)--(0.29,-0.71)--(1,0) - node[xshift=-0.7cm, yshift=-0.2cm, blue]{$z\boldsymbol{e_{31}}$}; - \draw[->] (0,-0.5) arc (0:310:0.15); - - % Basisvektoren - \draw[line width=1.5pt,gray,-stealth](0,0)--(1,0) node[anchor=south west]{$\boldsymbol{e_1}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(0,1) node[anchor=north west, yshift=0.2cm]{$\boldsymbol{e_2}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(-0.71,-0.71) node[anchor=south, yshift=0.2cm]{$\boldsymbol{e_3}$}; - \end{tikzpicture} + \includegraphics{papers/clifford/3d/dq.pdf} \caption{Darstellung eines Quaternion $\mathbf{q}$ und eines Vektors $\mathbf{v}$ im selben Raum} \label{BildQuaternionen} \end{figure} -Wie schon im 2 dimensionalen Fall beschreibt ein Bivektor, um wie viel der um 90 grad gedrehte orginale Vektor gestreckt wird. Dabei dreht jeder Bivektor den Vektor um eine andere Achse. -\\BILD?\\ -In der Computergraphik und Robotik macht eine Drehstreckung aber nicht viel Sinn. Wieso sollte ein Objekt bei einer Drehung zusätzlich noch grösser werden? Darum verwendet man sogenannte Einheitsquaternionen, welche den Betrag $|q|=1$ haben. Sie rotieren die Objekte bzw. Vektoren lediglich. + +Betrachten wir nun das Produkt +\begin{align} +\mathbf{qv} &= (w + x\mathbf{e}_{12} + y\mathbf{e}_{23} + z\mathbf{e}_{31})(a\mathbf{e}_1+b\mathbf{e}_2+c\mathbf{e}_3)\\ +&= \underbrace{w(a\mathbf{e}_1+b\mathbf{e}_2+c\mathbf{e}_3)}_{\displaystyle{w\mathbf{v}}} + \underbrace{x(-a\mathbf{e}_2+b\mathbf{e}_1}_{\displaystyle{x\mathbf{v}_{\angle 90^\circ, \parallel \mathbf{e}_{12}}}}+c\mathbf{e}_{123}) + \underbrace{y(-b\mathbf{e}_3+c\mathbf{e}_2}_{\displaystyle{y\mathbf{v}_{\angle 90^\circ, \parallel \mathbf{e}_{23}}}}+a\mathbf{e}_{123}) + \underbrace{z(a\mathbf{e}_3-c\mathbf{e}_1}_{\displaystyle{z\mathbf{v}_{\angle 90^\circ, \parallel \mathbf{e}_{31}}}}-b\mathbf{e}_{123}). +\end{align} +Wie schon im zweidimensionalen Fall \eqref{GAdrehstreck}, beschreibt im dreidimensionalen Fall mit drei Bivektoren jeder Bivektoranteil um wie viel der um 90° gedrehte zu der Ebene parallele Teil des Vektors gestreckt wird. Dabei dreht jeder Bivektor den Vektor um eine andere Achse und man sieht die drehstreckende Eigenschaft ähnlich zu den komplexen Zahlen. Der störende Trivektoranteil $(xc+ya+zb)\mathbf{e}_{123}$ bekommt man aber nur weg, indem man, wie in der Drehungsgleichung \eqref{QuatRot}, mit der Inversen Quaternion $\mathbf{q}^{-1}$ multipliziert, wobei die drehgestreckten parallelen Anteile nochmals drehgestreckt werden. Da nur so der Trivektoranteil wegfällt, sieht man, dass die Drehungsformel, der einzige vernünftige Weg ist, mit Quaternionen zu arbeiten. + +In der Computergraphik und Robotik macht eine Drehstreckung aber nicht viel Sinn. Wieso sollte ein Objekt bei einer Drehung zusätzlich noch grösser werden? Darum verwendet man sogenannte Einheitsquaternionen, welche den Betrag $|\mathbf{q}|=1$ haben und somit drehen sie die Objekte bzw. Vektoren lediglich. \begin{definition} - Einheitsquaternionen + Die Einheitsquaternionen sind definiert als \begin{align} - \mathbf{q} = \cos(\alpha) + sin(\alpha)(\tilde{x}\mathbf{e}_{12} + \tilde{y}\mathbf{e}_{23} + \tilde{z}\mathbf{e}_{31}) + \mathbf{q} = \cos(\alpha) + \sin(\alpha)(\tilde{x}\mathbf{e}_{12} + \tilde{y}\mathbf{e}_{23} + \tilde{z}\mathbf{e}_{31}) \end{align} \end{definition} -Dabei ist definiert, dass $\tilde{x}^2+\tilde{y}^2+\tilde{z}^2=1$. Somit beträgt der Betrag von $\mathbf{q}$ immer 1. +Zudem setzten wir $\tilde{x}^2+\tilde{y}^2+\tilde{z}^2=1$, damit \begin{align} - |\mathbf{q}| = \sqrt{cos(\alpha)^2 + sin(\alpha)^2(\tilde{x}^2+\tilde{y}^2+\tilde{z}^2) } = \sqrt{cos(\alpha)^2 + sin(\alpha)^2} = 1 +|\mathbf{q}| = \sqrt{\cos(\alpha)^2 + \sin(\alpha)^2(\tilde{x}^2+\tilde{y}^2+\tilde{z}^2) } = \sqrt{\cos(\alpha)^2 + \sin(\alpha)^2} = 1. \end{align} -Der Winkel $\alpha$ beschreibt dabei, wie im Bild (...) gezeigt den halben Winkel, um welchen der parallelen Anteil $\mathbf{v_{\perp}}$ des Vektors $\mathbf{v}$ zur kombinierten Bivektorebene $sin(\alpha)^2(\tilde{x}^2+\tilde{y}^2+\tilde{z}^2)$ gedreht wird. +Der Winkel $\alpha$ beschreibt dabei, wie im Bild \ref{BildQuaternionBeispiel2} gezeigt, den halben Winkel, um welchen der parallelen Anteil $\mathbf{v_{\parallel}}$ des Vektors $\mathbf{v}$ zur kombinierten Bivektorebene $sin(\alpha)(\tilde{x}\mathbf{e}_{12} + \tilde{y}\mathbf{e}_{23} + \tilde{z}\mathbf{e}_{31})$ gedreht wird. -Um einen Vektor zu drehen, verwendet man wieder die gleiche Formel, wie auch schon im zweidimensionalen Fall. +Um einen Vektor zu drehen, verwendet man die in Abschnitt 18.4 hergeleitete Formel \begin{align} \label{QuatRotGA} - \begin{split} - &\mathbf{v}'' = \mathbf{qvq}^{-1}\\ - &\operatorname{Re}(\mathbf{q}) = \operatorname{Re}(\mathbf{q}^{-1});\enspace \operatorname{Im}(\mathbf{q}) = -\operatorname{Im}(\mathbf{q}^-1) - \end{split} +\begin{split} +\mathbf{v}'' = \mathbf{qvq}^{-1}, +\end{split} +\end{align} +wobei wie auch schon bei den Quaternionen gelten muss, dass +\begin{align} \label{GAReIm} +\operatorname{Re}(\mathbf{q}) = \operatorname{Re}(\mathbf{q}^{-1}) \quad\text{und}\quad \operatorname{Im}(\mathbf{q}) = -\operatorname{Im}(\mathbf{q}^{-1}). +\end{align} +Der Grund für die Zusammenhänge \eqref{GAReIm} kann man durch die hergeleitete vereinfachte Drehungsgleichung \eqref{GAvereinfRot} sehen, weil durch den negierten Winkel $\theta$ der Reelle bzw. Grad 0 Anteil +\begin{align} +\operatorname{Re}(e^{-\theta \mathbf{e}_{12}}) = \operatorname{Re}(e^{\theta \mathbf{e}_{12}}) \end{align} -Es ist wichtig bei Quaternionen für eine reine Drehstreckung mit $q$ und $q^{-1}$ beidseitig zu multiplizieren, sonst werden die senkrechten Anteile zu den Bivektorebenen ebenfalls beeinflusst, wie man im Kapitel Rotation bei der Formel (\ref{RotAufPerpPar}) sehen kann. +und der imaginäre bzw. Grad 2 Anteil +\begin{align} +\operatorname{Im}(e^{-\theta \mathbf{e}_{12}}) = -\operatorname{Im}(e^{\theta \mathbf{e}_{12}}) +\end{align} +ist. Durch die geometrische Algebra sieht man nun, wieso es wichtig ist, bei Quaternionen für eine reine Drehstreckung mit $\mathbf{q}$ und $\mathbf{q}^{-1}$ beidseitig zu multiplizieren, sonst werden die senkrechten Anteile zu den Bivektorebenen ebenfalls beeinflusst, wie man im Abschnitt Drehung bei der Formel \eqref{RotAufPerpPar} sehen kann. \begin{beispiel} - Eine Drehung eines Vektors $\mathbf{v}= 1\mathbf{e}_2$ um 90 Grad um die $\mathbf{e}_1$-Achse und danach 90 Grad um die $\mathbf{e}_2$-Achse. Dafür nehmen wir zuerst einen Einheitsquaternion welcher um die Orientierte Ebene $\mathbf{e}_{23}$ um 90 Grad dreht + Eine Drehung eines Vektors $\mathbf{v}= 1\mathbf{e}_2$ um 90 Grad um die $\mathbf{e}_1$-Achse und danach 90 Grad um die $\mathbf{e}_2$-Achse. Dafür nehmen wir zuerst die Einheitsquaternion \begin{align} - \mathbf{q}_{23} &= \cos(\pi/4) + sin(\pi/4)(1\mathbf{e}_{23}) = e^{(\pi/4)\mathbf{e}_{23}} &= 0.71 + 0.71\mathbf{e}_{23}\\ - \mathbf{q}_{23}^{-1} &&= 0.71 - 0.71\mathbf{e}_{23} + \mathbf{q}_{23} &= \cos(\pi/4) + \sin(\pi/4)(1\mathbf{e}_{23}) = e^{(\pi/4)\mathbf{e}_{23}} &= \textstyle{\frac{\sqrt{2}}{2}}(1 + \mathbf{e}_{23})\\ + \mathbf{q}_{23}^{-1} &&= \textstyle{\frac{\sqrt{2}}{2}} (1- \mathbf{e}_{23}) \end{align} - und danach Einheitsquaternion welcher um die Orientierte Ebene $\mathbf{e}_{31}$ um 90 Grad dreht + welche um die $\mathbf{e}_{2}$-$\mathbf{e}_{3}$-Ebene um 90 Grad dreht und danach die Einheitsquaternion \begin{align} - \mathbf{q}_{31} &= \cos(\pi/4) + sin(\pi/4)(1\mathbf{e}_{31}) = e^{(\pi/4)\mathbf{e}_{31}} &= 0.71 + 0.71\mathbf{e}_{31}\\ - \mathbf{q}_{31}^{-1} &&= 0.71 - 0.71\mathbf{e}_{31} + \mathbf{q}_{31} &= \cos(\pi/4) + \sin(\pi/4)(1\mathbf{e}_{31}) = e^{(\pi/4)\mathbf{e}_{31}} &= \textstyle{\frac{\sqrt{2}}{2}}(1 + \mathbf{e}_{31})\\ + \mathbf{q}_{31}^{-1} &&= \textstyle{\frac{\sqrt{2}}{2}}(1 - \mathbf{e}_{31}), \end{align} - Um die vollständige Rotation zu beschreiben können die Einheitsquaternion multipliziert werden, wobei die Reihenfolge der Ausführung beachtet werden muss + welche um die $\mathbf{e}_{3}$-$\mathbf{e}_{1}$-Ebene um 90 Grad dreht. Um die vollständige Drehung zu beschreiben, können die Einheitsquaternionen multipliziert werden, wobei die Reihenfolge der Ausführung beachtet werden muss. Somit ist \begin{align} \label{FormelBeispielQuaternion} - \mathbf{q} &= \mathbf{q}_{31}\mathbf{q}_{23} = (0.71 + 0.71\mathbf{e}_{31})(0.71 + 0.71\mathbf{e}_{23}) &= 0.5 + 0.5\mathbf{e}_{31} + 0.5 \mathbf{e}_{23} + 0.5 \mathbf{e}_{12}\\ - \mathbf{q}^{-1} &= \mathbf{q}_{23}^{-1}\mathbf{q}_{31}^{-1} = (0.71 - 0.71\mathbf{e}_{23})(0.71 - 0.71\mathbf{e}_{31}) &= 0.5 - 0.5\mathbf{e}_{31} - 0.5 \mathbf{e}_{23} - 0.5 \mathbf{e}_{12} + \mathbf{q} &= \mathbf{q}_{31}\mathbf{q}_{23} = \textstyle{\frac{\sqrt{2}}{2}}(1 + \mathbf{e}_{31})\textstyle{\frac{\sqrt{2}}{2}}(1 + \mathbf{e}_{23}) &= \textstyle{\frac{1}{2}}(1 + \mathbf{e}_{31} + \mathbf{e}_{23} + \mathbf{e}_{12})\\ + \mathbf{q}^{-1} &= \mathbf{q}_{23}^{-1}\mathbf{q}_{31}^{-1} = \textstyle{\frac{\sqrt{2}}{2}} (1- \mathbf{e}_{23})\textstyle{\frac{\sqrt{2}}{2}}(1 -\mathbf{e}_{31}) &= \textstyle{\frac{1}{2}}(1 - \mathbf{e}_{31} - \mathbf{e}_{23} - \mathbf{e}_{12}). \end{align} - Wenn wir nun den Quaternion $\mathbf{q}$ auf den Vektor $\mathbf{v}$ anwenden + Wenn wir nun die Quaternion $\mathbf{q}$ auf den Vektor $\mathbf{v}$ anwenden, erhalten wir \begin{align} - \mathbf{v}'' = \mathbf{qvq}^{-1} &= (0.5 + 0.5\mathbf{e}_{31} + 0.5 \mathbf{e}_{23} + 0.5 \mathbf{e}_{12})(1\mathbf{e}_2)(0.5 - 0.5\mathbf{e}_{31} - 0.5 \mathbf{e}_{23} - 0.5 \mathbf{e}_{12})\\ - &= (0.5\mathbf{e}_2 + 0.5 \mathbf{e}_{123} - 0.5 \mathbf{e}_3 + 0.5 \mathbf{e}_1)(0.5 - 0.5\mathbf{e}_{31} - 0.5 \mathbf{e}_{23} - 0.5 \mathbf{e}_{12})\\ - &= (0.25 + 0.25 + 0.25 + 0.25)\mathbf{e}_1 + (0.25 + 0.25 - 0.25 - 0.25)\mathbf{e}_2 +\\ &(-0.25 + 0.25 - 0.25 + 0.25)\mathbf{e}_3 + (0.25 - 0.25 - 0.25 + 0.25)\mathbf{e}_{123}\\ - &= 1e_1 + \mathbf{v}'' = \mathbf{qvq}^{-1} &= \textstyle{\frac{1}{2}}(1 + \mathbf{e}_{31} + \mathbf{e}_{23} + \mathbf{e}_{12})(1\mathbf{e}_2)\textstyle{\frac{1}{2}}(1 - \mathbf{e}_{31} - \mathbf{e}_{23} - \mathbf{e}_{12})\\ + &= \textstyle{\frac{1}{4}}(\mathbf{e}_2 + \mathbf{e}_{123} - \mathbf{e}_3 + \mathbf{e}_1)(1 - \mathbf{e}_{31} - \mathbf{e}_{23} - \mathbf{e}_{12})\\ + &= (\textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}})\mathbf{e}_1 + (\textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}} - \textstyle{\frac{1}{4}} - \textstyle{\frac{1}{4}})\mathbf{e}_2 +\\ &\qquad(-\textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}} - \textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}})\mathbf{e}_3 + (\textstyle{\frac{1}{4}} - \textstyle{\frac{1}{4}} - \textstyle{\frac{1}{4}} + \textstyle{\frac{1}{4}})\mathbf{e}_{123}\\ + &= 1e_1. \end{align} Anders betrachtet könnte man von der Formel \eqref{FormelBeispielQuaternion} sehen, dass der Drehwinkel \begin{align} - \alpha = \arccos(w) = \arccos(0.5) = 60° + \alpha = \arccos(w) = \arccos(\textstyle{\frac{1}{2}}) = 60^\circ \end{align} und die Ebene der kombinierten Bivektoren wie in Abbildung \ref{BildQuaternionBeispiel2} aussieht. - Somit kann man sich ebenfalls Vorstellen, wie der parallele Anteil zur Ebene insgesamt um 120° rotiert wird während der senkrechte Anteil unverändert bleibt + Somit kann man sich ebenfalls vorstellen, wie der parallele Anteil zur Ebene insgesamt um 120° gedreht wird, während der senkrechte Anteil unverändert bleibt. \end{beispiel} \begin{figure} \centering - \begin{tikzpicture} - % Koordinatensystem - \draw[thin,gray!40] (-3,-2) grid (3,3); - \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; - \draw[<->] (0,-2)--(0,3) node[above]{$a_2$}; - \draw[<->] (3,3)--(-2,-2) node[left]{$a_3$}; - - % q Quaternion - \draw[line width=0,fill=blue!40] (0,0)--(1.41,0)--(1.41,1.41)--(0,1.41) - node[xshift=0.375cm, yshift=-0.5cm, blue]{$x\boldsymbol{e_{12}}$}; - \draw[->] (1.35, 1.2) arc (0:310:0.15); - - \draw[line width=0,fill=blue!40] (0,0)--(-1,-1)--(-1,0.41)--(0,1.41) - node[xshift=-0.5cm, yshift=-1.5cm, blue]{$y\boldsymbol{e_{23}}$}; - \draw[->] (-0.65,-0.5) arc (0:310:0.15); - - \draw[line width=0,fill=blue!40] (0,0)--(-1,-1)--(0.41,-1)--(1.41,0) - node[xshift=-0.7cm, yshift=-0.2cm, blue]{$z\boldsymbol{e_{31}}$}; - \draw[->] (0.4,-0.8) arc (0:310:0.15); - - % Basisvektoren - \draw[line width=1.5pt,gray,-stealth](0,0)--(2,0) node[anchor=south west]{$\boldsymbol{e_1}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(0,2) node[anchor=north west, yshift=0.2cm]{$\boldsymbol{e_2}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(-1.41,-1.41) node[anchor=south, yshift=0.2cm]{$\boldsymbol{e_3}$}; - - % v Vektor - \draw[line width=2pt,black,-stealth](-0.05,0)--(-0.05,2) node[anchor=east]{$\boldsymbol{v}$}; - % v'' Vektor - \draw[line width=2pt,black,-stealth](0,0.05)--(2,0.05) node[anchor=north]{$\boldsymbol{v}''$}; - \end{tikzpicture} + \includegraphics{papers/clifford/3d/qq.pdf} + \caption{Beispiel für Drehung um 90 Grad je um die $\mathbf{e}_1$- und $\mathbf{e}_2$-Achse.} \label{BildQuaternionBeispiel} \end{figure} \begin{figure} \centering - \begin{tikzpicture} - % q Quaternion - \draw[line width=0,fill=blue!40] (-0.75,-1)--(1.5,-0.5)--(0.55,1.35)--(-1.5,1) - node[xshift=0.375cm, yshift=-0.5cm, blue]{$\boldsymbol{q}$}; - \draw[->] (-0.7, 0.5) arc (310:0:0.15); - - % Koordinatensystem - \draw[thin,gray!40] (-3,-2) grid (3,3); - \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; - \draw[<->] (0,-2)--(0,3) node[above]{$a_2$}; - \draw[<->] (3,3)--(-2,-2) node[left]{$a_3$}; - - % Basisvektoren - \draw[line width=1.5pt,gray,-stealth](0,0)--(2,0) node[anchor=south west]{$\boldsymbol{e_1}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(0,2) node[anchor=north west, yshift=0.2cm]{$\boldsymbol{e_2}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(-1.41,-1.41) node[anchor=south, yshift=0.2cm]{$\boldsymbol{e_3}$}; - - % v Vektor - \draw[line width=2pt,black,-stealth](-0.05,0)--(-0.05,2) node[anchor=east]{$\boldsymbol{v}$}; - % vpar Vektor - \draw[line width=2pt,red,-stealth](0,0)--(-0.33,1.25) node[anchor=east]{$\boldsymbol{v_{\parallel}}$}; - % vperp Vektor - \draw[line width=2pt,green,-stealth](-0.33,1.25)--(0,2) node[anchor=east, xshift = -0.05, yshift = -0.3cm]{$\boldsymbol{v_{\perp}}$}; - % v'' Vektor - \draw[line width=2pt,black,-stealth](0,0.05)--(2,0.05) node[anchor=north, xshift = 0.25cm]{$\boldsymbol{v}''$}; - % vpar'' Vektor - \draw[line width=2pt,red,-stealth](0,0)--(1.66,-0.75) node[anchor=east, yshift = -0.2cm, xshift = -0.1cm]{$\boldsymbol{v_{\parallel}''}$}; - % vperp'' Vektor - \draw[line width=2pt,green,-stealth](1.66,-0.75)--(2,0) node[anchor=east, xshift = 0.5cm, yshift = -0.65cm]{$\boldsymbol{v_{\perp}''}$}; - - \coordinate (A) at (0,0); - \coordinate (B) at (-0.33,1.25); - \coordinate (C) at (1.66,-0.75); - \tikzset{anglestyle/.style={angle eccentricity=2, draw, thick, angle radius=0.75cm, purple}} - \draw pic ["120° $=2\alpha$", anglestyle] {angle = C--A--B}; - \end{tikzpicture} + \includegraphics{papers/clifford/3d/drehung.pdf} \caption{Beim Beispiel wird der parallele Anteil um 120° gedreht während der senkrechte Anteil zur kombinierten Ebene (Bivektoraddition) gleich bleibt} \label{BildQuaternionBeispiel2} \end{figure} \subsection{Interpolation} -In der Computergrafik wird Interpolation verwendet, um eine flüssige Drehbewegung zu erreichen. Dabei wird die gewünschte Drehbewegungen des Objektes in kleinere aufgeteilt. Man kann dabei mit zwei verschiedenen Systemen arbeiten. +In der Computergrafik wird Interpolation verwendet, um eine flüssige Drehbewegung zu erreichen. Dabei wird die ganze gewünschte Drehbewegungen des Objektes in kleinere Drehbewegungen aufgeteilt, wobei diese zeitlich nacheinander auf das Objekt angewendet werden. Als Vergleich könnte man sagen, dass ein Film auch nur Bilder sind, welche zeitlich nacheinander gezeigt werden. Man kann dabei mit zwei verschiedenen Systemen arbeiten. \begin{itemize} - \item Mit den Eulerschen Winkeln, welche für die Meisten zwar intuitiver sind, aber dafür Nachteile haben, worauf ich in diesem Abschnitt eingehen werde. Dabei kann eine ganze Drehbewegung $\mathbf{v}'' = R\mathbf{v}$ durch die Drehmatrix $R$ dargestellt werden. - \begin{align} - \begin{split} - &R = R_z(\gamma) R_y(\beta) R_x(\alpha)\\ - &R = - \begin{pmatrix} - \cos(\gamma) & -\sin(\gamma) & 0\\ \sin(\gamma) & \cos(\gamma) & 0 \\ 0 & 0 & 1 - \end{pmatrix} - \begin{pmatrix} - \cos(\beta) & 0 & \sin(\beta)\\ 0 & 1 & 0 \\ -\sin(\beta) & 0 & \cos(\beta) - \end{pmatrix} - \begin{pmatrix} - 1 & 0 & 0 \\ 0 & \cos(\alpha) & -\sin(\alpha)\\ 0 & \sin(\alpha) & \cos(\alpha) - \end{pmatrix} - \end{split} + \item Mit den Eulerschen Winkeln, welche für die Meisten zwar intuitiver sind, aber dafür Nachteile haben, worauf ich in diesem Abschnitt eingehen werde. Dabei kann eine ganze Drehbewegung $\mathbf{v}'' = R\mathbf{v}$ durch die Drehmatrix + \begin{align} \label{GADrehmatrix} + R = + \underbrace{ + \begin{pmatrix} + \cos(\gamma) & -\sin(\gamma) & 0\\ \sin(\gamma) & \cos(\gamma) & 0 \\ 0 & 0 & 1 + \end{pmatrix} + }_{\displaystyle{R_z(\gamma)}} + \underbrace{ + \begin{pmatrix} + \cos(\beta) & 0 & \sin(\beta)\\ 0 & 1 & 0 \\ -\sin(\beta) & 0 & \cos(\beta) + \end{pmatrix} + }_{\displaystyle{R_y(\beta)}} + \underbrace{ + \begin{pmatrix} + 1 & 0 & 0 \\ 0 & \cos(\alpha) & -\sin(\alpha)\\ 0 & \sin(\alpha) & \cos(\alpha) + \end{pmatrix} + }_{\displaystyle{R_x(\alpha)}} \end{align} - Wichtig dabei zu sehen ist, dass die Drehbewegungen durch die einzelnen Matrizen nacheinander ausgeführt werden. Das bedeutet, wenn man die Reihenfolge vertauscht, bekommt man eine völlig andere Drehung. Man kann die Auswirkungen der Reihenfolge gut bei einem Gimbal (REF zu BILD) sehen. Die Matrix ganz links ist die, welche als letztes Angewendet wird. Somit bildet sie die Drehung des äusseren Rings, welche auch die zwei inneren Ringe und das Objekt mitdreht. Die Matrix ganz rechts hingegen bildet nur die Drehung des inneren Rings, welche nur das Objekt selber dreht. Man kann dabei erkennen, dass vorgehen dabei sehr intuitiv ist, aber es kompliziert sein kann eine gewünschte Drehbewegung auszuführen, da sich beim Drehen der äusseren Achse, sich auch die Inneren drehen. Das bedeutet, wenn man sich eine Drehbewegung um die anfängliche x Achse mit $R_x(\alpha_2)$ wünscht, und vorher eine beliebige Drehung $R = R_z(\gamma_1) R_y(\beta_1) R_x(\alpha_1)$ ausgeführt hat, bekommt man nicht das richtige Ergebnis, da die anfängliche x-Achse durch die Drehmatrizen $R_z(\gamma_1)$ und $R_y(\beta_1)$ zu einer neuen, lokalen x-Achse wurde. - \item Andererseits mit den Quaternionen, welche die besondere Eigenschaft haben, dass eine Drehung immer um die globale Achsen ausgeführt wird, egal in welcher Rotationsposition sich das Objekt befindet. + dargestellt werden. Wichtig dabei zu sehen ist, dass die Drehbewegungen durch die einzelnen Matrizen nacheinander ausgeführt werden. Das bedeutet, wenn man die Reihenfolge vertauscht, bekommt man eine völlig andere Drehung. Man kann die Auswirkungen der Reihenfolge gut bei einem Gimbal im Bild \ref{BildReihenfolgeGimbal} sehen. Die Matrix ganz links in der Gleichung \eqref{GADrehmatrix} ist die, welche als letztes Angewendet wird. Somit bildet sie die Drehung des äusseren Rings, welche auch die zwei inneren Ringe und das Objekt mitdreht. Die Matrix ganz rechts hingegen bildet nur die Drehung des inneren Rings, welche nur das Objekt mitdreht. Man kann dabei erkennen, dass Vorgehen dabei sehr intuitiv ist, aber es kompliziert sein kann, eine gewünschte Drehbewegung auszuführen, da sich beim Drehen der äusseren Achse, sich auch die inneren drehen. Das bedeutet, wenn man sich eine Drehbewegung um die anfängliche x Achse mit $R_x(\alpha_2)$ wünscht, und vorher eine beliebige Drehung $R = R_z(\gamma_1) R_y(\beta_1) R_x(\alpha_1)$ ausgeführt hat, bekommt man nicht das richtige Ergebnis, da die anfängliche $x$-Achse durch die Drehmatrizen $R_z(\gamma_1)$ und $R_y(\beta_1)$ zu einer neuen, lokalen $x$-Achse wurde. + \item Mit den Quaternionen, welche die besondere Eigenschaft haben, dass eine Drehung immer um die globale Achsen ausgeführt wird, egal in welcher Drehungsposition sich das Objekt befindet. \end{itemize} Für Spielentwickler ist es darum meist sinnvoller Quaternionen für Drehbewegungen anzuwenden, als sich mit komplizierten Berechnungen mit Eulerschen Winkeln herumzuschlagen. + +\begin{figure} + \centering + \includegraphics[width=10cm]{papers/clifford/Bilder/ReihenfolgeGimbal.png} + \caption{Das Gimbal Lock tritt ein, wenn zwei Drehachsen in der gleichen Ebene liegen. Dies ist im rechten Bild bei der grünen und blauen Achse der Fall. Der rote Kreis würde sich an der oberen Halterung genau in die gleiche Richtung drehen, wie der grüne Kreis an der unteren Halterung. Man verliert somit eine Drehrichtung.} + \label{BildReihenfolgeGimbal} +\end{figure} + \subsection{Gimbal-Lock} -Ein weiterer Nachteil der Eulerschen Winkel ist das Gimbal-Lock. Es entsteht dann, wenn der äussere Ring Deckungsgleich über denn Inneren gedreht wird. Dabei verliert das Gimbal eine Drehrichtung, da der äussere und Innere Ring nun die gleiche Drehrichtung besitzen. Dies kann beispielsweise Probleme bei Spielen bei der Berechnung der Interpolation führen. Man hat das bei älteren Spielen dann gesehen, wenn plötzlich Gliedmassen bei den Spielermodellen in unnatürlichen Richtungen gesprungen sind. -\subsection{Fazit} -andere Darstellungsweise. Besser für Verständnis => komplexe Zahlen erscheinen ähnlicher zu Quaternionen? Eine Sprache für alle Geometrische Probleme
\ No newline at end of file +Ein weiterer Nachteil der Eulerschen Winkel ist das Gimbal-Lock. Es entsteht dann, wenn zwei Ringe Deckungsgleich übereinander gedreht werden, wie man im Bild \eqref{BildReihenfolgeGimbal} sieht. Dabei verliert das Gimbal eine Drehrichtung, da der äussere und Innere Ring nun die gleiche Drehrichtung besitzen. Dies kann beispielsweise Probleme bei Spielen bei der Berechnung der Interpolation führen. Man hat dies bei älteren Spielen wie im Bild \ref{BildGimbalLock} dann gesehen, wenn plötzlich Gliedmassen bei den Spielermodellen in unnatürliche Richtungen gesprungen sind. + +\begin{figure} + \centering + \includegraphics[width=10cm]{papers/clifford/Bilder/GimbalLock.png} + \caption{Interpolationsfehler durch Gimbal-Lock} + \label{BildGimbalLock} +\end{figure}
\ No newline at end of file diff --git a/buch/papers/clifford/11_Fazit.tex b/buch/papers/clifford/11_Fazit.tex new file mode 100644 index 0000000..79a683d --- /dev/null +++ b/buch/papers/clifford/11_Fazit.tex @@ -0,0 +1,9 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Fazit} +\rhead{Fazit} + +Die geometrische Algebra ist dafür ausgelegt, geometrische Operationen, wie die Spiegelung oder Drehung, einfach zu beschreiben. Dadurch kann sie als gute Alternative zu der linearen Algebra angewendet werden, um grafische Probleme zu lösen. Sie kann zudem zum Verständnis der drehenden Eigenschaften der komplexen Zahlen und Quaternionen beitragen und die Zusammenhänge zwischen den komplexen Zahlen und den Quaternionen zeigen.
\ No newline at end of file diff --git a/buch/papers/clifford/1_Vektordarstellung.tex b/buch/papers/clifford/1_Vektordarstellung.tex index 88a5789..ac00a33 100644 --- a/buch/papers/clifford/1_Vektordarstellung.tex +++ b/buch/papers/clifford/1_Vektordarstellung.tex @@ -1,7 +1,7 @@ \section{Vektoroperationen\label{clifford:section:Vektoroperationen}} \rhead{Vektoroperationen} \subsection{Vektordarstellung\label{clifford:section:Vektordarstellung}} -Vektoren können neben der üblichen Darstellung, auch als Linearkombination aus Basisvektoren dargestellt werden +Vektoren können neben der üblichen Spaltendarstellung, auch als Linearkombination aus Basisvektoren \begin{equation} \begin{split} \textbf{a} @@ -31,12 +31,14 @@ Vektoren können neben der üblichen Darstellung, auch als Linearkombination aus \sum_{i=1}^{n} a_i \textbf{e}_i \qquad a_i \in \mathbb{R} - , \textbf{e}_i \in \mathbb{R}^n. + , \textbf{e}_i \in \mathbb{R}^n \end{split} \end{equation} -Diese Basisvektoren sollen orthonormal sein und um die Darstellung zu vereinfachen werden sie durch $\textbf{e}_1 , \textbf{e}_2, ...$ ersetzt. +dargestellt werden. +Diese Basisvektoren werden so gewählt, dass sie orthonormal sind. +Um die Darstellung zu vereinfachen werden sie durch $\textbf{e}_1 , \textbf{e}_2, \dots$ ersetzt. \begin{beispiel} -Linearkombination von Basisvektoren in $\mathbb{R}^4$ +Eine Linearkombination von Basisvektoren in $\mathbb{R}^4$ könnte wie folgt aussehen \begin{equation} \begin{pmatrix} 42 \\ 2 \\ 1291 \\ 4 @@ -65,7 +67,7 @@ Linearkombination von Basisvektoren in $\mathbb{R}^4$ + 1291\textbf{e}_3 + - 4\textbf{e}_4 + 4\textbf{e}_4. \end{equation} +Dieses Beispiel ist für einen vier dimensionalen Vektor, dies kann selbstverständlich für beliebig viele Dimensionen nach demselben Schema erweitert werden. \end{beispiel} -Wobei Beispiel für einen vier dimensionalen Vektor ist, dies kann selbstverständlich für beliebig viele Dimensionen nach demselben Schema erweitert werden.
\ No newline at end of file diff --git a/buch/papers/clifford/2_QuadratVektoren.tex b/buch/papers/clifford/2_QuadratVektoren.tex index cfb05d6..6c6fb7d 100644 --- a/buch/papers/clifford/2_QuadratVektoren.tex +++ b/buch/papers/clifford/2_QuadratVektoren.tex @@ -1,54 +1,71 @@ \subsection{Quadrat von Vektoren} -Was eine Addition von Vektoren bedeutet ist sehr intuitiv und auch leicht geometrisch darzustellen, was allerdings das Produkt von Vektoren ergibt mag anfänglich unintuitiv wirken. +\subsubsection{Ziel der Multiplikation} +Was eine Addition von Vektoren bedeutet ist sehr intuitiv und auch leicht geometrisch darzustellen wie in Abbildung \ref{figure:addition}, was allerdings das Produkt von Vektoren ergibt mag anfänglich unintuitiv wirken. +\begin{figure}[htb] + \centering + \begin{tikzpicture} + \draw[thin,gray!40] (0,0) grid (4,4); + \draw[blue,thick,->] (0,0)--(3.5,2) node[midway,above,sloped] {$\textbf{a}$}; + \draw[red,thick,->] (3.5,2)--(1.5,3.8) node[midway,above,sloped] {$\textbf{b}$}; + \draw[black,thick,->] (0,0)--(1.5,3.8)node[midway,above,sloped] {$\textbf{a} +\textbf{b} = \textbf{c} $}; + \end{tikzpicture} + \caption{Addition von zwei Vektoren\label{figure:addition}} +\end{figure} Was soll es schon heissen zwei Vektoren miteinander zu multiplizieren? -\newline Im Folgenden werden wir versuchen diese Operation ähnlich intuitiv darzustellen. -\newline -Um sinnvoll eine neue Operation zwischen zwei Elementen einer Algebra, in diesem Fall Vektoren, zu definieren, muss man überlegen, was das Ziel dieser Operation ist. -Als grundsätzliches Ziel wird definiert, dass das Quadrat eines Vektor dessen Länge im Quadrat ergibt, da dies auch in vielen anderen Bereichen der Mathematik,zum Beispiel bei komplexen Zahlen, auch so definiert ist. -\newline -Zusätzlich wollen wir auch das Assoziativgesetz und das Kommutativgesetz für Skalare beibehalten. Wobei das Kommutativgesetz leider, oder wie man sehen wird zum Glück, in der geometrischen Algebra im generellen nicht mehr gilt. Das heisst wir dürfen ausklammern \ref{eq:assoziativ} und die Position von Skalaren im Produkt ändern \ref{eq:kommSkalar}, allerdings nicht die Position der Vektoren \ref{eq:kommVector}. + +Um sinnvoll eine neue Operation zwischen zwei Elementen einer Algebra, in diesem Fall sind diese Elemente Vektoren, zu definieren, muss man überlegen, was das Ziel dieser Operation sein soll. + +Als grundsätzliches Ziel wird definiert, dass das Quadrat eines Vektor dessen Länge im Quadrat ergibt, da dies auch in vielen anderen Bereichen der Mathematik,zum Beispiel bei komplexen Zahlen,so definiert ist. + +Zusätzlich soll auch das Assoziativgesetz für die Multiplikation von Vektoren gelten, dass heisst wir dürfen ausklammern \begin{equation} \label{eq:assoziativ} \textbf{e}_i(\textbf{e}_j + \textbf{e}_k) = - \textbf{e}_i\textbf{e}_j + \textbf{e}_i\textbf{e}_k + \textbf{e}_i\textbf{e}_j + \textbf{e}_i\textbf{e}_k. \end{equation} +Allerdings gilt das Kommutativgesetz leider, oder wie man sehen wird zum Glück, nur für skalare Elemente \begin{equation} \label{eq:kommSkalar} a\textbf{e}_ib\textbf{e}_j = - ab\textbf{e}_i\textbf{e}_j + ab\textbf{e}_i\textbf{e}_j \qquad a,b \in \mathbb{R} \end{equation} +und nicht für Vektoren \begin{equation} \label{eq:kommVector} \textbf{e}_i\textbf{e}_j \neq - \textbf{e}_j\textbf{e}_i + \textbf{e}_j\textbf{e}_i. +\end{equation} +\subsubsection{Quadrieren eines Vektors} +Betrachten wir nun mit diesen Regeln das Quadrat eines Vektors. Zuerst werden die Vektoren als Linearkombinationen geschrieben +\begin{equation} + \textbf{a}^2 = + \left ( + \sum_{i=1}^{n} a_i \textbf{e}_i + \right ) + \left ( + \sum_{i=1}^{n} a_i \textbf{e}_i + \right ) + \label{eq:quad_a_1}. +\end{equation} +Das Quadrat kann nun in zwei Summen aufgeteilt werden +\begin{equation} + \textbf{a}^2 = + \textcolor{red}{\sum_{i=1}^{n} a_i^2\textbf{e}_i^2} + + + \textcolor{blue}{\sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j } + \label{eq:quad_a_2}, +\end{equation} +wobei die roten Summe die quadrierten Terme und die blaue Summe die Mischterme beinhaltet. Da $\textbf{e}_i^2 = 1$ gilt, weil das zuvor definierte Ziel des Quadrates eines Vektors dessen Länge ergibt und die Basisvektoren Länge 1 haben, wird dies nun eingesetzt +\begin{equation} + \textbf{a}^2 = \textcolor{cyan}{\sum_{i=1}^{n} a_i^2} + \textcolor{orange}{\sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j}. + \label{eq:quad_a_3} \end{equation} -Betrachten wir nun mit diesen Regeln das Quadrat eines Vektors. -\begin{align} - \textbf{a}^2 &= - \left ( - \sum_{i=1}^{n} a_i \textbf{e}_i - \right ) - \left ( - \sum_{i=1}^{n} a_i \textbf{e}_i - \right ) - \label{eq:quad_a_1} - \\ - &= - \textcolor{red}{\sum_{i=1}^{n} a_i^2\textbf{e}_i^2} - + - \textcolor{blue}{\sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j } - \label{eq:quad_a_2} - \\ - &= \textcolor{cyan}{\sum_{i=1}^{n} a_i^2} + \textcolor{orange}{\sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j}. - \label{eq:quad_a_3} -\end{align} - \begin{beispiel} -Quadrat eines Vektors in $\mathbb{R}^2$ +Das Quadrat des Vektor $a$ in $\mathbb{R}^2$ ist \begin{equation} \begin{split} \textbf{a}^2 @@ -56,22 +73,17 @@ Quadrat eines Vektors in $\mathbb{R}^2$ &= \textcolor{red}{a_1^2\textbf{e}_1^2 + a_2^2\textbf{e}_2^2} + \textcolor{blue}{a_1\textbf{e}_1a_2\textbf{e}_2 + a_2\textbf{e}_2a_1\textbf{e}_2} \\\ & = \textcolor{cyan}{a_1^2 + a_2^2} + \textcolor{orange}{a_1b\textbf{e}_1a_2\textbf{e}_2 + a_2\textbf{e}_2a_1\textbf{e}_2} - \end{split} + \end{split}. \end{equation} - \end{beispiel} -Der Vektor wird in \ref{eq:quad_a_1} als Linearkombination geschrieben. -Das Quadrat kann, wie in \ref{eq:quad_a_2} gezeigt, in zwei Summen aufteilen werden , wobei die roten Summe die quadrierten Terme und die blaue Summe die Mischterme beinhaltet. -\newline -Da $\textbf{e}_i^2 = 1$ gilt, da zuvor vorausgesetzt wurde, dass man mit orthonormalen Einheitsvektoren arbeitet, wird dies nun eingesetzt ergibt sich \ref{eq:quad_a_3} -\newline -Die hellblaue Teil ist nun bereits Länge im Quadrat eines Vektors, also das Ziel der Multiplikation. -Daher muss der restliche Teil dieser Gleichung null ergeben. -Aus dieser Erkenntnis leiten wir in \ref{eq:Mischterme_Null} weitere Eigenschaften für die Multiplikation her. + +Die hellblaue Teil ist nun bereits die Länge im Quadrat, also das zuvor definierte Ziel der Multiplikation. +Daraus lässt sich schliessen, dass der restliche Teil dieser Gleichung null ergeben muss \begin{equation} \label{eq:Mischterme_Null} - \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j = \textcolor{blue}{a_1a_2(\textbf{e}_1\textbf{e}_2 + \textbf{e}_2\textbf{e}_1)} + a_1a_3(\textbf{e}_1\textbf{e}_3 + \textbf{e}_3\textbf{e}_1) + \dots = 0 + \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n a_ia_j\textbf{e}_i\textbf{e}_j = \textcolor{blue}{a_1a_2(\textbf{e}_1\textbf{e}_2 + \textbf{e}_2\textbf{e}_1)} + a_1a_3(\textbf{e}_1\textbf{e}_3 + \textbf{e}_3\textbf{e}_1) + \dots = 0. \end{equation} +Aus dieser Erkenntnis können weitere Eigenschaften für die Multiplikation hergeleitet werden. Da dies für beliebige $a_i$ gelten muss werden alle Terme bis auf $a_1$ und $a_2$ gleich null gesetzt. Somit fallen alle Terme bis auf den blauen weg. Wird dies weiter vereinfacht ergibt sich \begin{equation} \begin{split} @@ -81,15 +93,13 @@ Da dies für beliebige $a_i$ gelten muss werden alle Terme bis auf $a_1$ und $a_ \end{split} \end{equation} \begin{satz} - Die Multiplikation von Vektoren ist antikommutativ, wenn die multiplizierten Vektoren orthogonal sind. + Die Multiplikation von Vektoren ist antikommutativ, wenn die multiplizierten Vektoren orthogonal sind, es gilt also \begin{equation} - \textbf{e}_i\textbf{e}_j = -\textbf{e}_j\textbf{e}_i \qquad \textbf{e}_i \perp \textbf{e}_j + \textbf{e}_i\textbf{e}_j = -\textbf{e}_j\textbf{e}_i \quad \textrm{für} \quad \textbf{e}_i \perp \textbf{e}_j. \end{equation} \end{satz} -Dieses Wissen reicht nun bereits um alle Produkte der Basisvektoren zu berechnen, was in \ref{tab:multip_vec} gemacht wurde. +Dieses Wissen reicht nun bereits um alle Produkte der Basisvektoren zu berechnen, was in Tabelle \ref{tab:multip_vec} gemacht wurde. \begin{table} -\caption{Multiplikationstabelle für Vektoren} -\label{tab:multip_vec} \begin{center} \begin{tabular}{ |c|c|c|c|c|c| } \hline @@ -107,4 +117,6 @@ Dieses Wissen reicht nun bereits um alle Produkte der Basisvektoren zu berechnen \hline \end{tabular} \end{center} +\caption{Multiplikationstabelle für Vektoren} +\label{tab:multip_vec} \end{table}
\ No newline at end of file diff --git a/buch/papers/clifford/3_MultiplikationVektoren.tex b/buch/papers/clifford/3_MultiplikationVektoren.tex index 841dde4..0969b89 100644 --- a/buch/papers/clifford/3_MultiplikationVektoren.tex +++ b/buch/papers/clifford/3_MultiplikationVektoren.tex @@ -1,11 +1,14 @@ \subsection{Multiplikation von Vektoren} -Was geschieht nun wenn zwei beliebige Vektoren,$u$ und $v$, miteinander multipliziert werden? +Was geschieht nun wenn zwei beliebige Vektoren, $u$ und $v$ \begin{equation} \textbf{u} = \sum_{i=1}^{n} u_i \textbf{e}_i \qquad \textbf{v} = \sum_{i=1}^{n} v_i \textbf{e}_i \end{equation} + miteinander multipliziert werden? + + Wieder werden die Vektoren zuerst als Linearkombinationen darstellen und danach in zwei Summen aufgeteilt, eine Summe mit quadrierten Termen und eine Summe mit Mischtermen \begin{equation} \begin{split} \textbf{u}\textbf{v} @@ -18,12 +21,12 @@ Was geschieht nun wenn zwei beliebige Vektoren,$u$ und $v$, miteinander multipli \right) = \sum_{i=1}^n u_iv_i\underbrace{\textbf{e}_i^2}_{1} - + \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j + + \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j, \end{split} \end{equation} +wobei die Summe der quadrierten Termen bereits bekannt vorkommen könnte, es ist nämlich das Skalarprodukt von $u$ und $v$. Die Summe der Mischterme bilden etwas neues, dass wir das äussere Produkt von $u$ und $v$ nennen. \begin{beispiel} Multiplikation von Vektoren in $\mathbb{R}^2$ -\end{beispiel} \begin{equation} \begin{split} \textbf{u}\textbf{v} @@ -44,7 +47,7 @@ Was geschieht nun wenn zwei beliebige Vektoren,$u$ und $v$, miteinander multipli \underbrace{(u_1v_2 - u_2v_1)\textbf{e}_1\textbf{e}_2}_{\text{Äusseres Produkt}} \end{split} \end{equation} -Der linke Teil dieser Multiplikation ergibt das Skalarprodukt der zwei Vektoren, der rechte Term ergibt etwas neues das sich das äussere Produkt der zwei Vektoren nennt. +\end{beispiel} \subsubsection{Äusseres Produkt} Das äussere Produkt von zwei Vektoren wird mit einem $\wedge$ dargestellt \begin{equation} @@ -53,123 +56,118 @@ Das äussere Produkt von zwei Vektoren wird mit einem $\wedge$ dargestellt \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j \end{equation} \begin{beispiel} -Äusseres Produkt von zwei Vektoren in $\mathbb{R}^3$ -\end{beispiel} +Das äusseres Produkt von zwei Vektoren in $\mathbb{R}^3$ ist \begin{equation} - \begin{split} - u \wedge v - &= - u_1v_2\textbf{e}_1\textbf{e}_2 - + - u_1v_3\textbf{e}_1\textbf{e}_3 - + - u_2v_2\textbf{e}_2\textbf{e}_3 - + - u_2v_1\textbf{e}_2\textbf{e}_1 - + - u_3v_1\textbf{e}_3\textbf{e}_1 - + - u_3v_2\textbf{e}_3\textbf{e}_2 \\\ - &= - (u_1v_2 - u_2v_1)\textbf{e}_1\textbf{e}_2 - + - (u_1v_3 - v_3u_1)\textbf{e}_1\textbf{e}_3 - + - (u_2v_3 - u_3v_2)\textbf{e}_2\textbf{e}_3 - \end{split} + \begin{split} + u \wedge v + &= + u_1v_2\textbf{e}_1\textbf{e}_2 + + + u_1v_3\textbf{e}_1\textbf{e}_3 + + + u_2v_2\textbf{e}_2\textbf{e}_3 + + + u_2v_1\textbf{e}_2\textbf{e}_1 + + + u_3v_1\textbf{e}_3\textbf{e}_1 + + + u_3v_2\textbf{e}_3\textbf{e}_2 \\\ + &= + (u_1v_2 - u_2v_1)\textbf{e}_1\textbf{e}_2 + + + (u_1v_3 - v_3u_1)\textbf{e}_1\textbf{e}_3 + + + (u_2v_3 - u_3v_2)\textbf{e}_2\textbf{e}_3. + \end{split} \end{equation} -Im letzten Schritt des Beispiels wurden nun, mit Hilfe der antikommutativität des Produkts, die Vektorprodukte, welche die gleichen Einheitsvektoren beinhalten, zusammengefasst. Dieses Vorgehen kann man auch allgemein anwenden, wie in den Gleichungen \ref{eq:u_wedge_v}-\ref{eq:u_wedge_v_5} hergeleitet. +\end{beispiel} + +Im letzten Schritt des Beispiels wurden nun, mit Hilfe der antikommutativität des Produkts, die Vektorprodukte, welche die gleichen Einheitsvektoren beinhalten, zusammengefasst. Dieses Vorgehen kann man auch allgemein anwenden, wie in den Gleichungen \eqref{eq:u_wedge_v}-\eqref{eq:u_wedge_v_5} hergeleitet. Die Summe, \begin{align} \textbf{u}\wedge \textbf{v} &= \sum_{\begin{subarray}{l}i,j=1\\i \neq j\end{subarray}}^n - u_iv_j\textbf{e}_i\textbf{e}_j + u_iv_j\textbf{e}_i\textbf{e}_j, \label{eq:u_wedge_v} - \\ + \intertext{wird in zwei verschiedene Summen aufgeteilt. + Wobei die linke Summe jeweils den Basisvektor mit dem höheren Index an erster Stelle und die rechte Summe diesen jeweils an zweiter Stelle hat} \label{eq:u_wedge_v_1} &= \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j + - \sum_{\begin{subarray}{l}i,j=1\\j < i\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j - \\ + \sum_{\begin{subarray}{l}i,j=1\\j < i\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j. + \intertext{Nun werden die Indexe der zweiten Summe vertauscht} \label{eq:u_wedge_v_2} &= \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j + - \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_jv_i\textbf{e}_j\textbf{e}_i - \\ - \label{eq:u_wedge_v_3} + \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_jv_i\textbf{e}_j\textbf{e}_i, + \intertext{und diese wird nun mit Hilfe der Antikommutativität umgeformt zu} &= \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_iv_j\textbf{e}_i\textbf{e}_j - - \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_jv_i\textbf{e}_i\textbf{e}_j - \\ + \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n u_jv_i\textbf{e}_i\textbf{e}_j. + \intertext{Nun können die zwei Summen wieder zusammengefasst werden} \label{eq:u_wedge_v_4} &= - \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n (u_iv_j -u_jv_i)\textbf{e}_i\textbf{e}_j - \\ - \label{eq:u_wedge_v_5} + \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n (u_iv_j -u_jv_i)\textbf{e}_i\textbf{e}_j. + \intertext{Der Term in der Summe könnte einem bereits bekannt vorkommen, es ist nämlich die Determinante einer Matrix mit $u$ und $v$ als ihre Spalten} &= + \label{eq:u_wedge_v_5} \sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n \begin{vmatrix} u_i & v_i \\ u_j & v_j - \end{vmatrix}\textbf{e}_i\textbf{e}_j + \end{vmatrix}\textbf{e}_i\textbf{e}_j. \end{align} -Die Summe aus \ref{eq:u_wedge_v_1} wird in \ref{eq:u_wedge_v} in zwei verschiedene Summen aufgeteilt. -Wobei die linke Summe jeweils den Basisvektor mit dem höheren Index an erster Stelle und die rechte Summe diesen jeweils an zweiter Stelle hat. -\newline -Bei \ref{eq:u_wedge_v_2} werden die Indexe der zweiten Summe vertauscht, damit man nun bei beiden Teilen die gleiche Summe hat. -Danach werden in \ref{eq:u_wedge_v_3}, mit Hilfe der Antikommutativität, die Einheitsvektoren der zweiten Summe vertauscht. -\newline -Nun können die Summen, wie in \ref{eq:u_wedge_v_4} wieder in eine Summe zusammengefasst werden. -\newline -Der Term in der Klammer in \ref{eq:u_wedge_v_4} kann auch als Determinante einer 2x2 Matrix dargestellt werden, was in \ref{eq:u_wedge_v_5} gemacht wird. -\newline -Die Determinante einer Matrix beschreibt welche von den Spaltenvektoren aufgespannt wird, wie in Abbildung \ref{figure:det} dargestellt. -\begin{figure} -\centering -\begin{tikzpicture} - \draw[thin,gray!40] (0,0) grid (4,4); - \draw[<->] (0,0)--(4,0) ; - \draw[<->] (0,0)--(0,4) ; - \draw[line width=0,fill=gray!40] (0,0)--(3,1)--(4,3)--(1,2); - \draw[line width=2pt,blue,-stealth](0,0)--(3,1) node[anchor=north - west]{$\boldsymbol{u}$}; - \draw[line width=2pt,red,-stealth](0,0)--(1,2) node[anchor=south east]{$\boldsymbol{v}$}; - \draw[black] (2,1.5)--(-0.5,2.5) node[anchor = east]{$\begin{vmatrix} - u_i & v_i \\ - u_j & v_j - \end{vmatrix} = u_iv_j - v_iu_j$}; -\end{tikzpicture} -\caption{Geometrische Interpretation der Determinante einer 2x2 Matrix\label{figure:det}} +Die Determinante einer Matrix beschreibt die Fläche, welche von den Spaltenvektoren aufgespannt wird, wie in Abbildung \ref{figure:det} dargestellt. +\begin{figure}[htb] + \centering + \begin{minipage}[t]{.45\linewidth} + \centering + \begin{tikzpicture} + \draw[thin,gray!40] (0,0) grid (4,4); + \draw[<->] (0,0)--(4,0) ; + \draw[<->] (0,0)--(0,4) ; + \draw[line width=0,fill=gray!40] (0,0)--(3,1)--(4,3)--(1,2); + \draw[line width=2pt,blue,-stealth](0,0)--(3,1) node[anchor=north + west]{$\boldsymbol{u}$}; + \draw[line width=2pt,red,-stealth](0,0)--(1,2) node[anchor=south east]{$\boldsymbol{v}$}; + \draw[black] (2,1.5)--(1.8,3.2) node[anchor = south]{$\begin{vmatrix} + u_i & v_i \\ + u_j & v_j + \end{vmatrix} = u_iv_j - v_iu_j$}; + \end{tikzpicture} + \caption{Geometrische Interpretation der Determinante einer $2 \times 2$ Matrix\label{figure:det}} + \end{minipage}% + \hfill% + \begin{minipage}[t]{.45\linewidth} + \centering + \begin{tikzpicture} + \draw[thin,gray!40] (0,0) grid (4,4); + \draw[<->] (0,0)--(4,0) node[right]{$x$}; + \draw[<->] (0,0)--(0,4) node[above]{$y$}; + \draw[line width=0,fill=gray!40] (0,0)--(3,1)--(4,3)--(1,2); + \draw[line width=2pt,blue,-stealth](0,0)--(3,1) node[anchor=north + west]{$\boldsymbol{u}$}; + \draw[line width=2pt,red,-stealth](0,0)--(1,2) node[anchor=south east]{$\boldsymbol{v}$}; + \draw[->] (2.15,1.5) arc (0:310:0.3); + \draw[black] (2,1.5)--(2.5,3.2) node[anchor = south]{$u\wedge v = \begin{vmatrix} + u_i & v_i \\ + u_j & v_j + \end{vmatrix} e_1e_2 = (u_iv_j - v_iu_j)\textbf{e}_1\textbf{e}_2$}; + \end{tikzpicture} + \caption{Geometrische Interpretation des äusseren Produktes \label{figure:wedge}} + \end{minipage} \end{figure} -\newline Das äussere Produkt besteht nun also aus der Summe - $\sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n$ + \(\sum_{\begin{subarray}{l}i,j=1\\i < j\end{subarray}}^n\) von Flächen - $\begin{vmatrix} - u_i & v_i \\ - u_j & v_j - \end{vmatrix}$, welche in $\textbf{e}_i\textbf{e}_j$ aufgespannt sind, wie man in \ref{eq:u_wedge_v_5} sieht. + \(\begin{vmatrix} + u_i & v_i \\ + u_j & v_j + \end{vmatrix}\) +, welche in $\textbf{e}_i\textbf{e}_j$ aufgespannt sind, wie man in \ref{eq:u_wedge_v_5} sieht. Dieses Produkt $\textbf{e}_i\textbf{e}_j$ der Basisvektoren interpretiert man als Umlaufrichtung. Wobei die gebildete Fläche in Richtung des ersten Vektors umschritten wird. -Dies ist in \ref{figure:wedge} dargestellt, wobei bei diesem Beispiel die Umlaufrichtung im Gegenuhrzeigersinn ist, da die Fläche in Richtung u umschritten wird. +Dies ist in Abbildung \ref{figure:wedge} dargestellt, wobei bei diesem Beispiel die Umlaufrichtung im Gegenuhrzeigersinn ist, da die Fläche in Richtung u umschritten wird. Diese Fläche mit einer Richtung nennt man in der geometrischen Algebra einen Bivektor, da er eine Art zwei dimensionaler Vektor ist. -\begin{figure} -\centering -\begin{tikzpicture} - \draw[thin,gray!40] (0,0) grid (4,4); - \draw[<->] (0,0)--(4,0) node[right]{$x$}; - \draw[<->] (0,0)--(0,4) node[above]{$y$}; - \draw[line width=0,fill=gray!40] (0,0)--(3,1)--(4,3)--(1,2); - \draw[line width=2pt,blue,-stealth](0,0)--(3,1) node[anchor=north - west]{$\boldsymbol{u}$}; - \draw[line width=2pt,red,-stealth](0,0)--(1,2) node[anchor=south east]{$\boldsymbol{v}$}; - \draw[->] (2.15,1.5) arc (0:310:0.3); - \draw[black] (2,1.5)--(-0.5,2.5) node[anchor = east]{$u\wedge v = \begin{vmatrix} - u_i & v_i \\ - u_j & v_j - \end{vmatrix} e_1e_2 = (u_iv_j - v_iu_j)\textbf{e}_1\textbf{e}_2$}; -\end{tikzpicture} -\caption{Geometrische Interpretation des äusseren Produkt in $\mathbb{R}^2$\label{figure:wedge}} -\end{figure}
\ No newline at end of file diff --git a/buch/papers/clifford/3d/Makefile b/buch/papers/clifford/3d/Makefile new file mode 100644 index 0000000..147ca81 --- /dev/null +++ b/buch/papers/clifford/3d/Makefile @@ -0,0 +1,38 @@ +# +# Makefile +# +# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +all: dq.jpg q23.jpg q31.jpg drehung.jpg dq.pdf qq.pdf drehung.pdf + +size="+W3840 +H2160" + +dq.png: dq.pov common.inc + povray +A0.1 $(size) -Odq.png dq.pov +dq.jpg: dq.png Makefile + convert -extract 1600x1400+1500+60 dq.png -density 300 -units PixelsPerInch dq.jpg +dq.pdf: dq.jpg dq.tex + pdflatex dq.tex + +extract="1200x1200+1450+350" + +q23.png: q23.pov common.inc + povray +A0.1 $(size) -Oq23.png q23.pov +q23.jpg: q23.png Makefile + convert -extract $(extract) q23.png -density 300 -units PixelsPerInch q23.jpg + +q31.png: q31.pov common.inc + povray +A0.1 $(size) -Oq31.png q31.pov +q31.jpg: q31.png Makefile + convert -extract $(extract) q31.png -density 300 -units PixelsPerInch q31.jpg + +qq.pdf: qq.tex q31.jpg q23.jpg + pdflatex qq.tex + +drehung.png: drehung.pov common.inc + povray +A0.1 $(size) -Odrehung.png drehung.pov +drehung.jpg: drehung.png Makefile + convert -extract 1600x1450+1400+50 drehung.png -density 300 -units PixelsPerInch drehung.jpg +drehung.pdf: drehung.tex drehung.jpg + pdflatex drehung.tex + diff --git a/buch/papers/clifford/3d/common.inc b/buch/papers/clifford/3d/common.inc new file mode 100644 index 0000000..55bf6e1 --- /dev/null +++ b/buch/papers/clifford/3d/common.inc @@ -0,0 +1,271 @@ +// +// common.inc +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#version 3.7; +#include "colors.inc" + +global_settings { + assumed_gamma 1 +} + +#declare imagescale = 0.14; +#declare r = 0.02; +#declare thick = 0.040; + +camera { + location <40, 12, 15> + look_at <0, 0, 0> + right 16/9 * x * imagescale + up y * imagescale +} + +light_source { + <40, 20, 20> color White + area_light <1,0,0> <0,0,1>, 10, 10 + adaptive 1 + jitter +} + +sky_sphere { + pigment { + color rgb<1,1,1> + } +} + +// +// draw an arrow from <from> to <to> with thickness <arrowthickness> with +// color <c> +// +#macro arrow(from, to, arrowthickness, c) +#declare arrowdirection = vnormalize(to - from); +#declare arrowlength = vlength(to - from); +union { + sphere { + from, 1.1 * arrowthickness + } + cylinder { + from, + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + arrowthickness + } + cone { + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + 2 * arrowthickness, + to, + 0 + } + pigment { + color c + } + finish { + specular 0.9 + metallic + } +} +#end + + +arrow(< -3, 0, 0 >, < 3, 0, 0 >, r, White) +arrow(< 0, -3, 0 >, < 0, 3, 0 >, r, White) +arrow(< 0, 0, -3 >, < 0, 0, 3 >, r, White) + +#macro circlearrow0(e1, e2, e3, r1, r2, h, winkel) + +mesh { + #declare N = 100; + #declare phi = 0; + #declare phimax = winkel - pi / 12; + #declare phistep = (phimax - phi) / N; + #while (phi < phimax - phistep/2) + triangle { + center + r1 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) - h * e3 + } + triangle { + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) - h * e3, + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r2 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) - h * e3 + } + triangle { + center + r1 * (cos(phi ) * e1 + sin(phi ) * e2) + h * e3, + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) + h * e3, + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3 + } + triangle { + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3, + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) + h * e3, + center + r2 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3 + } + triangle { + center + r1 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) - h * e3, + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3 + } + triangle { + center + r1 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r1 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3, + center + r1 * (cos(phi ) * e1 + sin(phi ) * e2) + h * e3 + } + triangle { + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r2 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) - h * e3, + center + r2 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3 + } + triangle { + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) - h * e3, + center + r2 * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + h * e3, + center + r2 * (cos(phi ) * e1 + sin(phi ) * e2) + h * e3 + } + #declare phi = phi + phistep; + #end + + triangle { + center + r1 * e1 - h * e3, + center + r1 * e1 + h * e3, + center + r2 * e1 + h * e3 + } + triangle { + center + r2 * e1 - h * e3, + center + r2 * e1 + h * e3, + center + r1 * e1 - h * e3 + } + triangle { + center + r1 * cos(phi) * e1 + r1 * sin(phi) * e2 - h * e3, + center + r2 * cos(phi) * e1 + r2 * sin(phi) * e2 - h * e3, + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) - h * e3 + } + triangle { + center + r1 * cos(phi) * e1 + r1 * sin(phi) * e2 + h * e3, + center + r2 * cos(phi) * e1 + r2 * sin(phi) * e2 + h * e3, + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) + h * e3 + } + triangle { + center + r1 * cos(phi) * e1 + r1 * sin(phi) * e2 - h * e3, + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) - h * e3 + center + r1 * cos(phi) * e1 + r1 * sin(phi) * e2 + h * e3 + } + triangle { + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) - h * e3 + center + r1 * cos(phi) * e1 + r1 * sin(phi) * e2 + h * e3, + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) + h * e3 + } + triangle { + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) - h * e3, + center + r2 * cos(phi) * e1 + r2 * sin(phi) * e2 - h * e3, + center + r2 * cos(phi) * e1 + r2 * sin(phi) * e2 + h * e3 + } + triangle { + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) - h * e3, + center + r2 * cos(phi) * e1 + r2 * sin(phi) * e2 + h * e3, + center + 0.5*(r1+r2) * (cos(phi + pi/12) * e1 + sin(phi + pi/12) * e2) + h * e3 + } + + pigment { + color rgb<1, 0.4, 0.4> + } +} + +#end + + +#macro circlearrow(fromdirection, axis, center, r, h, winkel, anzahl) + +#declare e1 = vnormalize(fromdirection); +#declare e2 = -vnormalize(vcross(axis, fromdirection)); +#declare e3 = vnormalize(axis); + +#declare r1 = 0.4 * r; +#declare r2 = r; + +#declare w = 0; +#while (w < anzahl) + #declare a = 2 * w * pi / anzahl; + circlearrow0(e1 * cos(a) - e2 * sin(a), e1 * sin(a) + e2 * cos(a), e3, r1, r2, 1.2 * h, winkel) + #declare w = w + 1; +#end + +mesh { + #declare vlu = center - r * e1 - r * e2 - h * e3; + #declare vlo = center - r * e1 - r * e2 + h * e3; + #declare vru = center - r * e1 + r * e2 - h * e3; + #declare vro = center - r * e1 + r * e2 + h * e3; + #declare hlu = center + r * e1 - r * e2 - h * e3; + #declare hlo = center + r * e1 - r * e2 + h * e3; + #declare hru = center + r * e1 + r * e2 - h * e3; + #declare hro = center + r * e1 + r * e2 + h * e3; + triangle { vlu, vru, vro } + triangle { vlu, vro, vlo } + + triangle { vru, hru, hro } + triangle { vru, hro, vro } + + triangle { hru, hlu, hlo } + triangle { hru, hlo, hro } + + triangle { hlu, vlu, vlo } + triangle { hlu, vlo, hlo } + + triangle { vlu, vru, hru } + triangle { vlu, hru, hlu } + + triangle { vlo, vro, hro } + triangle { vlo, hro, hlo } + + pigment { + color rgb<0.6,0.6,1> + } + finish { + specular 0.96 + metallic + } +} + +#if (vlength(axis) > 0.1) +cone { + center + 1.19 * h * e3, r, center + 2 * r * e3, 0 + pigment { + color rgbt<0.6,0.6,1,0.8> + } +} +#end + +cylinder { + center, center + 2 * r * e3, 0.04*0.2 + pigment { + color rgb<1.0,0.6,0.6> + } + finish { + specular 0.96 + metallic + } +} + +#end + +#macro bogen(v1, v2, center, winkelbogen, farbe) + +union { + #declare phi = 0; + #declare phimax = winkelbogen; + #declare phistep = (phimax - phi) / N; + #while (phi < phimax - phistep/2) + cylinder { + cos(phi ) * v1 + sin(phi ) * v2 + center, + cos(phi+phistep) * v1 + sin(phi+phistep) * v2 + center, + 0.01 + } + sphere { + cos(phi ) * v1 + sin(phi ) * v2 + center, + 0.01 + } + #declare phi = phi + phistep; + #end + pigment { + color farbe + } +} + +#end diff --git a/buch/papers/clifford/3d/dq.jpg b/buch/papers/clifford/3d/dq.jpg Binary files differnew file mode 100644 index 0000000..690cfdc --- /dev/null +++ b/buch/papers/clifford/3d/dq.jpg diff --git a/buch/papers/clifford/3d/dq.pdf b/buch/papers/clifford/3d/dq.pdf Binary files differnew file mode 100644 index 0000000..797a558 --- /dev/null +++ b/buch/papers/clifford/3d/dq.pdf diff --git a/buch/papers/clifford/3d/dq.pov b/buch/papers/clifford/3d/dq.pov new file mode 100644 index 0000000..762eee2 --- /dev/null +++ b/buch/papers/clifford/3d/dq.pov @@ -0,0 +1,30 @@ +// +// dq.pov -- Drehung und Quaternion +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#include "common.inc" + +arrow(<0,0,0>, <1, sqrt(2), 2>, r, Red) + +#declare r = 0.2 * r; + +#declare drehwinkel = 0.95 * 2*pi/3 * 3; +#declare drehwinkel23 = drehwinkel; +#declare drehwinkel12 = drehwinkel / sqrt(2); +#declare drehwinkel13 = drehwinkel / 2; + +circlearrow(<1,0,0>, <0,0,1>, <1, sqrt(2), 0>, 1, thick, drehwinkel23, 1) +circlearrow(<1,0,0>, <0,1,0>, <1, 0, 2>, sqrt(2)/2, thick, drehwinkel12, 1) +circlearrow(<0,0,1>, <1,0,0>, <0, sqrt(2), 2>, 0.5, thick, drehwinkel13, 1) + +#declare l = 2.8; +#declare h = 0.0001; +union { + box { <-l,-l,-h>, <l,l,-h> } + box { <-l,-h,-l>, <l,-h,l> } + box { <-h,-l,-l>, <-h,l,l> } + pigment { + color rgbt<0.6,0.6,0.6,0.0> + } +} diff --git a/buch/papers/clifford/3d/dq.tex b/buch/papers/clifford/3d/dq.tex new file mode 100644 index 0000000..6b28452 --- /dev/null +++ b/buch/papers/clifford/3d/dq.tex @@ -0,0 +1,51 @@ +% +% dq.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math} +\usepackage{ifthen} +\begin{document} + +\definecolor{darkred}{rgb}{0.7,0,0} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{6} +\def\hoehe{6} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\node at (0,0) {\includegraphics[width=12cm]{dq.jpg}}; + +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} + +\node at (-2.8,-2.7) {$O$}; +\node at (4.7,-3.4) {$a_1$}; +\node at (-2.6,5.2) {$a_2$}; +\fill[color=white,opacity=0.7] ({-5.7-0.25},{-4.8-0.15}) rectangle ({-5.7+0.25},{-4.8+0.2}); +\node at (-5.7,-4.8) {$a_3$}; + +\node[color=blue] at (-3.6,0.8) {$y\mathbf{e}_{23}$}; +\node[color=blue] at (2.1,0.9) {$x\mathbf{e}_{12}$}; +\node[color=blue] at (1.3,-3.7) {$z\mathbf{e}_{13}$}; + +\node[color=darkred] at (1.3,0.4) {$\vec{q}$}; + +\end{tikzpicture} + +\end{document} + diff --git a/buch/papers/clifford/3d/drehung.jpg b/buch/papers/clifford/3d/drehung.jpg Binary files differnew file mode 100644 index 0000000..2347296 --- /dev/null +++ b/buch/papers/clifford/3d/drehung.jpg diff --git a/buch/papers/clifford/3d/drehung.pdf b/buch/papers/clifford/3d/drehung.pdf Binary files differnew file mode 100644 index 0000000..bc8036e --- /dev/null +++ b/buch/papers/clifford/3d/drehung.pdf diff --git a/buch/papers/clifford/3d/drehung.pov b/buch/papers/clifford/3d/drehung.pov new file mode 100644 index 0000000..b86a2c5 --- /dev/null +++ b/buch/papers/clifford/3d/drehung.pov @@ -0,0 +1,87 @@ +// +// drehung.pov -- Drehung um (1,1,1) +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#include "common.inc" + +#declare n = vnormalize(<1,1,1>); +#declare V = <0,2.6,0>; +#declare W = <0,0,2.6>; + +#declare Vparallel = vdot(n, V) * n; +#declare Vperp = V - Vparallel; +#declare Wparallel = vdot(n, W) * n; +#declare Wperp = W - Wparallel; + +arrow(<0,0,0>, 2*n, thick, Red) + +arrow(<0,0,0>, V, thick, rgb<0.0,1.0,1.0>) +arrow(<0,0,0>, W, thick, rgb<0.0,1.0,1.0>) + +circlearrow(vnormalize(vcross(<-1,0,1>,n)), -0.01 * <1,1,1>, <0,0,0>, 1, 0.8*thick, 1.98*pi/3, 3) + +arrow(<0,0,0>, Vperp, 0.99*thick, Blue) +arrow(<0,0,0>, Wperp, 0.99*thick, Blue) + +arrow(Vperp, V, thick, Green) +arrow(Wperp, W, thick, Green) + +#declare l = 2.4; +intersection { + box { <-l,-l,-l>, <l,l,l> } + //cylinder { -n, n, 3 } + plane { n, 0.01 } + plane { -n, 0.01 } + pigment { + color rgbt<0.6,0.6,1.0,0.8> + } +} + +#declare e1 = vnormalize(Vperp); +#declare e3 = n; +#declare e2 = vnormalize(vcross(e3, e1)); +#declare r = vlength(Vperp); + +mesh { + #declare phi = 0; + #declare phimax = 2*pi/3; + #declare phistep = (phimax - phi) / N; + #while (phi < phimax - phistep/2) + triangle { + <0,0,0>, + r * (cos(phi ) * e1 + sin(phi ) * e2), + r * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + } + #declare phi = phi + phistep; + #end + pigment { + color rgbt<0.2,0.2,1.0,0.4> + } +} + +mesh { + #declare phi = 0; + #declare phimax = 2*pi/3; + #declare phistep = (phimax - phi) / N; + #while (phi < phimax - phistep/2) + triangle { + r * (cos(phi ) * e1 + sin(phi ) * e2), + r * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2), + r * (cos(phi ) * e1 + sin(phi ) * e2) + Vparallel + } + triangle { + r * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2), + r * (cos(phi ) * e1 + sin(phi ) * e2) + Vparallel, + r * (cos(phi+phistep) * e1 + sin(phi+phistep) * e2) + Vparallel + } + #declare phi = phi + phistep; + #end + pigment { + color rgbt<0.2,1,0.2,0.4> + } +} + +bogen(r * e1, r * e2, <0,0,0>, 2*pi/3, Blue) +bogen(r * e1, r * e2, Vparallel, 2*pi/3, Green) + diff --git a/buch/papers/clifford/3d/drehung.tex b/buch/papers/clifford/3d/drehung.tex new file mode 100644 index 0000000..2ed6789 --- /dev/null +++ b/buch/papers/clifford/3d/drehung.tex @@ -0,0 +1,56 @@ +% +% drehung.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math} +\usepackage{ifthen} +\begin{document} + +\definecolor{darkgreen}{rgb}{0,0.6,0} +\definecolor{darkred}{rgb}{0.6,0,0} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{7} +\def\hoehe{6} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\node at (0,0) {\includegraphics[width=13cm]{drehung.jpg}}; + +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} + +\node at (6.1,-3.3) {$a_1$}; +\node at (-2.0,5.7) {$a_2$}; +\node at (-5.7,-4.9) {$a_3$}; + +\node[color=white] at (-1.9,4.4) {$\boldsymbol{v}$}; +\node[color=white] at (4.5,-2.7) {$\boldsymbol{v}''$}; + +\node[color=darkgreen] at (-3.3,4.4) {$\boldsymbol{v}_{\perp}$}; +\node[color=darkgreen] at (4.2,-4.3) {$\boldsymbol{v}''_{\perp}$}; + +\node[color=blue] at (-3.7,1.5) {$\boldsymbol{v}_{\|}$}; +\node[color=blue] at (1.9,-4.7) {$\boldsymbol{v}''_{\|}$}; + +\node[color=darkred] at (-1.6,-4.2) {$2\alpha=120^\circ$}; +\node[color=darkred] at (-4.9,-0.6) {$\boldsymbol{q}$}; + +\end{tikzpicture} + +\end{document} + diff --git a/buch/papers/clifford/3d/q23.jpg b/buch/papers/clifford/3d/q23.jpg Binary files differnew file mode 100644 index 0000000..929ef90 --- /dev/null +++ b/buch/papers/clifford/3d/q23.jpg diff --git a/buch/papers/clifford/3d/q23.pov b/buch/papers/clifford/3d/q23.pov new file mode 100644 index 0000000..2e55c96 --- /dev/null +++ b/buch/papers/clifford/3d/q23.pov @@ -0,0 +1,14 @@ +// +// q23.pov -- Drehung und Quaternion +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#include "common.inc" + +circlearrow(<1,0,0>, 0.01*<0,0,-1>, <0, 0, 0>, 1.0, thick, 0.98*pi/2, 4) + +bogen( <0,1.7,0>, <-1.7, 0, 0>, <0,0,0>, pi/2, Blue) + +arrow( <0,0,0>, <-2.0,0,0>, 0.99*thick, Blue) +arrow( <0,0,0>, <0,2.0,0>, 0.99*thick, Blue) +arrow( <0,0,0>, <0,0,2.0>, 0.99*thick, Red) diff --git a/buch/papers/clifford/3d/q31.jpg b/buch/papers/clifford/3d/q31.jpg Binary files differnew file mode 100644 index 0000000..c240b4f --- /dev/null +++ b/buch/papers/clifford/3d/q31.jpg diff --git a/buch/papers/clifford/3d/q31.pov b/buch/papers/clifford/3d/q31.pov new file mode 100644 index 0000000..4abe1ed --- /dev/null +++ b/buch/papers/clifford/3d/q31.pov @@ -0,0 +1,15 @@ +// +// q31.pov -- Drehung und Quaternion +// +// (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#include "common.inc" + +circlearrow(<1,0,0>, 0.01*<0,-1,0>, <0, 0, 0>, 1.0, thick, 0.98*pi/2, 4) + +arrow( <0,0,0>, <-2.0,0,0>, 0.99*thick, Blue) +arrow( <0,0,0>, <0,2.0,0>, 0.99*thick, Red) +arrow( <0,0,0>, <0,0,2.0>, 0.99*thick, Blue) + +bogen( <0,0,1.7>, <-1.7, 0, 0>, <0,0,0>, pi/2, Blue) + diff --git a/buch/papers/clifford/3d/qq.pdf b/buch/papers/clifford/3d/qq.pdf Binary files differnew file mode 100644 index 0000000..fd7dbfa --- /dev/null +++ b/buch/papers/clifford/3d/qq.pdf diff --git a/buch/papers/clifford/3d/qq.tex b/buch/papers/clifford/3d/qq.tex new file mode 100644 index 0000000..9baa8bb --- /dev/null +++ b/buch/papers/clifford/3d/qq.tex @@ -0,0 +1,68 @@ +% +% qq.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math} +\usepackage{ifthen} +\begin{document} + +\definecolor{darkred}{rgb}{0.7,0,0} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{4} +\def\hoehe{4} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\begin{scope}[xshift=-3.3cm] +\node at (0,0) {\includegraphics[width=6.3cm]{q23.jpg}}; +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} +\fill[color=white,opacity=0.5] ({-0.6-0.3},{-0.2-0.2}) rectangle ({-0.6+0.3},{-0.2+0.2}); +\node[color=darkred] at (-0.6,-0.2) {$\boldsymbol{q}_{23}$}; +\node[color=blue] at (-0.4,2.7) {$\boldsymbol{v}$}; +\node[color=blue] at (0.7,0.4) {$\boldsymbol{v}''_{23}$}; +\node at (3.1,-1.4) {$a_1$}; +\node at (-2.7,-2.4) {$a_3$}; +\node at (-0.7,3.4) {$a_2$}; +\end{scope} + +\setboolean{showgrid}{false} + +\begin{scope}[xshift=3.3cm] +\node at (0,0) {\includegraphics[width=6.3cm]{q31.jpg}}; +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} +\fill[color=white,opacity=0.5] ({-0.7-0.3},{-0.9-0.2}) rectangle ({-0.7+0.3},{-0.9+0.2}); +\node[color=darkred] at (-0.7,-0.9) {$\boldsymbol{q}_{13}$}; +\node[color=blue] at (0.7,0.4) {$\boldsymbol{v}''_{23}$}; +\node[color=blue] at (2.7,-0.7) {$\boldsymbol{v}''$}; +\node at (3.1,-1.4) {$a_1$}; +\node at (-2.7,-2.4) {$a_3$}; +\node at (-0.7,3.4) {$a_2$}; +\end{scope} + + +\end{tikzpicture} + +\end{document} + diff --git a/buch/papers/clifford/4_GeometrischesProdukt.tex b/buch/papers/clifford/4_GeometrischesProdukt.tex index a19e983..f18b90d 100644 --- a/buch/papers/clifford/4_GeometrischesProdukt.tex +++ b/buch/papers/clifford/4_GeometrischesProdukt.tex @@ -12,9 +12,9 @@ Ein Multivektor besteht aus den verschiedenen Bauteilen, wie zum Beispiel Vektor M = \sum \left ( \prod a_i\textbf{e}_j \right) \end{equation} \end{definition} -Besteht eine Clifford Algebra aus n Basisvektoren so hat sie n Dimensionen, dies wird nicht wie in der linearen Algebra mit $\mathbb{R}^n$ sondern mit $\mathbb{G}^n$ beschrieben. +Besteht eine Clifford Algebra aus n Basisvektoren so hat sie n Dimensionen, dies wird nicht wie in der linearen Algebra mit $\mathbb{R}^n$ sondern mit $G_n(\mathbb{R})$ beschrieben. Dies wird so geschrieben da man eine neue Algebrastruktur um die Vektoren einführt. \begin{beispiel} -Allgemeiner Multivektor in $\mathbb{G}^3$ +Allgemeiner Multivektor in $G_3(\mathbb{R})$ \begin{equation} M = a + @@ -26,34 +26,30 @@ Allgemeiner Multivektor in $\mathbb{G}^3$ \end{equation} \end{beispiel} \begin{definition} -Um das Produkt von Basisvektoren in Zukunft darzustellen wird folgende Notation definiert +Für das Produkt von Basisvektoren wird folgende Notation definiert \begin{equation} - e_ie_j = e_{ij} + e_ie_j = e_{ij}. \end{equation} \end{definition} -Nun da das geometrische Produkt vollständig definiert wurde können Multiplikationstabellen für verschiedene Dimensionen $\mathbb{G}^n$ erstellt werden. In \ref{tab:multip} ist dies für $\mathbb{G}^3$ gemacht. +Nun da das geometrische Produkt vollständig definiert wurde können Multiplikationstabellen für verschiedene Dimensionen $G_n(\mathbb{R})$ erstellt werden. In Tabelle \ref{tab:multip} ist dies für $G_3(\mathbb{R})$ gemacht. \begin{table} - \caption{Multiplikationstabelle für $\mathbb{G^3}$} \label{tab:multip} \begin{center} - \begin{tabular}{ |c|c|c|c|c|c|c|c| } + \begin{tabular}{ |c|ccc|ccc|c| } \hline 1 & $\textbf{e}_1$ & $\textbf{e}_2$ &$\textbf{e}_3$ & $\textbf{e}_{12}$ & $\textbf{e}_{13}$ & $\textbf{e}_{23}$ & $\textbf{e}_{123}$\\ \hline $\textbf{e}_1$ & 1 & $\textbf{e}_{12}$ & $\textbf{e}_{12}$ & $\textbf{e}_2$ & $\textbf{e}_3$ & $\textbf{e}_{123}$ & $\textbf{e}_{23}$\\ - \hline $\textbf{e}_2$ & $-\textbf{e}_{12}$ & 1 & $\textbf{e}_{23}$ & $-\textbf{e}_1$ & $-\textbf{e}_{123}$ & $\textbf{e}_3$ & $-\textbf{e}_{13}$\\ - \hline $\textbf{e}_3$ & $-\textbf{e}_{13}$ & $-\textbf{e}_{23}$ & 1 & $\textbf{e}_{123}$ & $-\textbf{e}_1$ & $-\textbf{e}_2$ & $\textbf{e}_{12}$\\ \hline $\textbf{e}_{12}$ & -$\textbf{e}_2$ & $\textbf{e}_1$& $\textbf{e}_{123}$ & -1 & $-\textbf{e}_{23}$ & $\textbf{e}_{13}$ & $-\textbf{e}_{3}$\\ - \hline $\textbf{e}_{13}$ & $-\textbf{e}_{3}$ & $-\textbf{e}_{123}$ & $\textbf{e}_{1}$ & $\textbf{e}_{23}$ & -1 & $-\textbf{e}_{12}$ & $\textbf{e}_{2}$\\ - \hline $\textbf{e}_{23}$ & $\textbf{e}_{123}$ & $-\textbf{e}_{3}$ & $\textbf{e}_{2}$ & $-\textbf{e}_{13}$ & $\textbf{e}_{12}$ & -1 & $-\textbf{e}_{1}$ \\ \hline $\textbf{e}_{123}$ & $\textbf{e}_{23}$ & $-\textbf{e}_{13}$ & $\textbf{e}_{12}$ & $-\textbf{e}_{3}$& $\textbf{e}_{2}$ & $-\textbf{e}_{1}$ & -1 \\ \hline \end{tabular} \end{center} + \caption{Multiplikationstabelle für $G_3(\mathbb{R})$} \end{table} diff --git a/buch/papers/clifford/6_PauliMatrizen.tex b/buch/papers/clifford/6_PauliMatrizen.tex index e41275a..4e82f28 100644 --- a/buch/papers/clifford/6_PauliMatrizen.tex +++ b/buch/papers/clifford/6_PauliMatrizen.tex @@ -10,33 +10,33 @@ Was ist der beste Weg um einen Computeralgorithmus für die Rechenoperationen in \begin{beispiel} Der Algorithmus weiss, dass er $a\mathbf{e}_1\cdot b\mathbf{e}_1$ zu $ab\cdot1$ vereinfachen kann. Dies ermöglicht zum Beispiel die Vereinfachung \begin{align} - 3\mathbf{e}_1 \cdot 2\mathbf{e}_1 + 3\mathbf{e}_2 \Rightarrow 6 + 3\mathbf{e}_2 + 3\mathbf{e}_1 \cdot 2\mathbf{e}_1 + 3\mathbf{e}_2 \Rightarrow 6 + 3\mathbf{e}_2 \end{align} \end{beispiel} Ein textueller Algorithmus ist aber sehr ineffizient. Die Pauli-Matrizen bilden eine elegante und schnellere Alternative, welche für die dreidimensionale Clifford-Algebra verwendet werden können und alle Operationen aus der Clifford-Algebra gleich wie die Matrixoperationen ausführen lassen. \begin{definition} \label{def:defPauli} Die Matrizen \begin{align} \label{Pauli} - \mathbf{e}_0 = E = - \begin{pmatrix} - 1 & 0 \\ - 0 & 1 - \end{pmatrix},\quad - \mathbf{e}_1 = - \begin{pmatrix} - 0 & 1 \\ - 1 & 0 - \end{pmatrix},\quad - \mathbf{e}_2 = - \begin{pmatrix} - 0 & -j \\ - j & 0 - \end{pmatrix},\quad - \mathbf{e}_3 = - \begin{pmatrix} - 1 & 0 \\ - 0 & -1 - \end{pmatrix} + \mathbf{e}_0 = E = + \begin{pmatrix} + 1 & 0 \\ + 0 & 1 + \end{pmatrix},\quad + \mathbf{e}_1 = + \begin{pmatrix} + 0 & 1 \\ + 1 & 0 + \end{pmatrix},\quad + \mathbf{e}_2 = + \begin{pmatrix} + 0 & -j \\ + j & 0 + \end{pmatrix},\quad + \mathbf{e}_3 = + \begin{pmatrix} + 1 & 0 \\ + 0 & -1 + \end{pmatrix} \end{align} heissen Pauli-Matrizen ($\mathbf{e}_0$ = Skalare) \end{definition} @@ -44,85 +44,85 @@ Die Matrix-Multiplikationen der Pauli-Matrizen führt auf die gleichen algebrais \begin{definition} \label{def:defPauli2} Die Bivektoren und Trivektoren hergeleitet aus den Pauli-Matrizen sind \begin{align} \label{Pauli2} - \mathbf{e}_{12} = - \begin{pmatrix} - j & 0 \\ - 0 & -j - \end{pmatrix}\quad - \mathbf{e}_{23} = - \begin{pmatrix} - 0 & j \\ - j & 0 - \end{pmatrix}\quad - \mathbf{e}_{31} = - \begin{pmatrix} - 0 & 1 \\ - -1 & 0 - \end{pmatrix}\enspace\text{und}\enspace - \mathbf{e}_{123} = - \begin{pmatrix} - j & 0 \\ - 0 & j - \end{pmatrix}. - \end{align} -\end{definition} -Dabei ist wichtig, dass sich die Matrizen gleich verhalten, wie es die Clifford-Algebra für die Basiselemente definiert hat. Zum Beispiel gilt in der Clifford-Algebra $\mathbf{e}_1^2=\mathbf{e}_0$ und $\mathbf{e}_{12}^2=-\mathbf{e}_0$, genau die selbe Relation gilt auch für die zugehörigen Matrizen, wie man durch die Matrizenrechnungen -\begin{align} - \mathbf{e}_1^2 &= + \mathbf{e}_{12} = \begin{pmatrix} - 0 & 1 \\ - 1 & 0 - \end{pmatrix}^2 = + j & 0 \\ + 0 & -j + \end{pmatrix}\quad + \mathbf{e}_{23} = \begin{pmatrix} - 1 & 0 \\ - 0 & 1 - \end{pmatrix}= \mathbf{e}_0 \quad\text{und}\\ - \mathbf{e}_{12}^2 &= + 0 & j \\ + j & 0 + \end{pmatrix}\quad + \mathbf{e}_{31} = \begin{pmatrix} - j & 0 \\ - 0 & -j - \end{pmatrix}^2 = + 0 & 1 \\ + -1 & 0 + \end{pmatrix}\enspace\text{und}\enspace + \mathbf{e}_{123} = \begin{pmatrix} - -1 & 0 \\ - 0 & -1 - \end{pmatrix} = -\mathbf{e}_0 + j & 0 \\ + 0 & j + \end{pmatrix}. + \end{align} +\end{definition} +Dabei ist wichtig, dass sich die Matrizen gleich verhalten, wie es die Clifford-Algebra für die Basiselemente definiert hat. Zum Beispiel gilt in der Clifford-Algebra $\mathbf{e}_1^2=\mathbf{e}_0$ und $\mathbf{e}_{12}^2=-\mathbf{e}_0$, genau die selbe Relation gilt auch für die zugehörigen Matrizen, wie man durch die Matrizenrechnungen +\begin{align} +\mathbf{e}_1^2 &= +\begin{pmatrix} +0 & 1 \\ +1 & 0 +\end{pmatrix}^2 = +\begin{pmatrix} +1 & 0 \\ +0 & 1 +\end{pmatrix}= \mathbf{e}_0 \quad\text{und}\\ +\mathbf{e}_{12}^2 &= +\begin{pmatrix} +j & 0 \\ +0 & -j +\end{pmatrix}^2 = +\begin{pmatrix} +-1 & 0 \\ +0 & -1 +\end{pmatrix} = -\mathbf{e}_0 \end{align} -bestätigt. Man kann bei den Definitionen \ref{def:defPauli} und \ref{def:defPauli2} sehen, dass alle Matrizen linear unabhängig voneinander sind. Das bedeutet, dass wenn man die Matrizen der Basiselemente normal addiert und zu einer Matrix zusammenfasst, kann man anschliessend die einzelnen Anteile der Basiselemente wieder herausgelesen. +bestätigt. Man kann bei den Definitionen \ref{def:defPauli} und \ref{def:defPauli2} sehen, dass alle Matrizen linear unabhängig voneinander sind. Das bedeutet, dass wenn man die Matrizen der Basiselemente normal addiert und zu einer Matrix zusammenfasst, kann man anschliessend die einzelnen Anteile der Basiselemente wieder herauslesen. \begin{hilfssatz} Ein beliebiger Multivektor \begin{align} \label{MultiVektorAllg} - M = a_0\mathbf{e}_0 + a_1\mathbf{e}_1 + a_2\mathbf{e}_3 + a_{12}\mathbf{e}_{12} + a_{23}\mathbf{e}_{23} + a_{31}\mathbf{e}_{31} + a_{123}\mathbf{e}_{123}\\ + M = a_0\mathbf{e}_0 + a_1\mathbf{e}_1 + a_2\mathbf{e}_3 + a_{12}\mathbf{e}_{12} + a_{23}\mathbf{e}_{23} + a_{31}\mathbf{e}_{31} + a_{123}\mathbf{e}_{123} \end{align} erhält durch das einsetzten der Formel Matrizen \eqref{Pauli} und \eqref{Pauli2} die Form \begin{align} - M = - \begin{pmatrix} - (a_0+a_3) + (a_{12}+a_{123})j & (a_1+a_{31})+(-a_2+a_{23})j \\ - (a_1-a_{31})+(a_2+a_{23})j & (a_0-a_3)+(-a_{12}+a_{123})j - \end{pmatrix}.\label{MultivektorMatirx} + M = + \begin{pmatrix} + (a_0+a_3) + (a_{12}+a_{123})j & (a_1+a_{31})+(-a_2+a_{23})j \\ + (a_1-a_{31})+(a_2+a_{23})j & (a_0-a_3)+(-a_{12}+a_{123})j + \end{pmatrix}.\label{MultivektorMatirx} \end{align} \end{hilfssatz} Die Anteile treten zudem immer paarweise auf und können somit immer je durch zwei Gleichungen bestimmt werden. \begin{beispiel} Die Matrix \begin{align} - M &= - \begin{pmatrix} - 1 & 0 \\ - 0 & -1j - \end{pmatrix} + M &= + \begin{pmatrix} + 1 & 0 \\ + 0 & -1j + \end{pmatrix} \end{align} soll als Multivektor in der Form \eqref{MultiVektorAllg} geschrieben werden. Dafür entnehmen wir aus \eqref{MultivektorMatirx} die Gleichungen \begin{align} - a_0 + a_3 = 1,\quad a_0 - a_3 = 0,\quad a_{12}+a_{123} = 0\enspace\text{und}\enspace -a_{12}+a_{123}=-1 + a_0 + a_3 = 1,\quad a_0 - a_3 = 0,\quad a_{12}+a_{123} = 0\enspace\text{und}\enspace -a_{12}+a_{123}=-1, \end{align} aus denen man auf \begin{align} - a_0 = \dfrac{1}{2},\quad a_3 = \dfrac{1}{2},\quad a_{12}=\dfrac{1}{2}\enspace\text{und}\enspace a_{123}=-\dfrac{1}{2} + a_0 = \dfrac{1}{2},\quad a_3 = \dfrac{1}{2},\quad a_{12}=\dfrac{1}{2}\enspace\text{und}\enspace a_{123}=-\dfrac{1}{2} \end{align} schliessen kann. Da die restlichen Realteile und Imaginärteile 0 sind, werden die anderen Anteile ebenfalls 0 sein. Daher ist \begin{align} - M = \dfrac{1}{2} \mathbf{e}_0+ \dfrac{1}{2} \mathbf{e}_3 + \dfrac{1}{2} \mathbf{e}_{12} - \dfrac{1}{2} \mathbf{e}_{123}. + M = \dfrac{1}{2} \mathbf{e}_0+ \dfrac{1}{2} \mathbf{e}_3 + \dfrac{1}{2} \mathbf{e}_{12} - \dfrac{1}{2} \mathbf{e}_{123}. \end{align} \end{beispiel} Die Clifford-Algebra ist bei der Darstellung durch Matrizen kein Ausnahmefall. Es lässt sich theoretisch jede algebraische Struktur durch Matrizen darstellen.
\ No newline at end of file diff --git a/buch/papers/clifford/7_Reflektion.tex b/buch/papers/clifford/7_Reflektion.tex index bdfb4e8..549848c 100644 --- a/buch/papers/clifford/7_Reflektion.tex +++ b/buch/papers/clifford/7_Reflektion.tex @@ -6,15 +6,15 @@ \section{Spiegelung} \rhead{Spiegelung} -Die Spiegelung ist eine grundlegende, geometrische Operation, aus welcher man weitere, wie beispielsweise die später beschriebene Rotation, ableiten kann. Da die geometrische Algebra für geometrische Anwendungen ausgelegt ist, sollte die Spiegelung auch eine einfache, praktische Formulierung besitzen. +Die Spiegelung ist eine grundlegende, geometrische Operation, aus welcher man weitere Operationen, wie beispielsweise die später beschriebene Rotation, ableiten kann. Da die geometrische Algebra für geometrische Anwendungen ausgelegt ist, sollte die Spiegelung auch eine einfache, praktische Formulierung besitzen. \begin{figure} \centering \begin{tikzpicture} \draw[thin,gray!40] (-3,-1) grid (3,3); \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; \draw[<->] (0,-1)--(0,3) node[above]{$a_2$}; + \draw[blue, line width=1.0pt] (0,3)--(0,-1) node[anchor=south east]{$\sigma_u$}; \draw[line width=2pt,black,-stealth](0,0)--(2,2) node[anchor=south east]{$\boldsymbol{v}$}; - \draw[line width=1.5pt,blue,-stealth](0,0)--(0,2.5) node[anchor=south east]{$\boldsymbol{u}$}; \draw[line width=2pt,black,-stealth](0,0)--(-2,2) node[anchor=south east]{$\boldsymbol{v'}$}; \draw[line width=1.5pt,gray,-stealth](0,0)--(1,0) node[anchor=north]{$\boldsymbol{e_1}$}; \draw[line width=1.5pt,gray,-stealth](0,0)--(0,1) node[anchor=north east]{$\boldsymbol{e_2}$}; @@ -22,62 +22,74 @@ Die Spiegelung ist eine grundlegende, geometrische Operation, aus welcher man we 0.25cm]{$\boldsymbol{v_{\perp u}}$}; \draw[line width=1.5pt,red,-stealth](-2,2)--(0,2) node[xshift=-1cm, yshift= 0.25cm]{$\boldsymbol{v_{\perp u}}$}; - \draw[line width=1.5pt,purple,-stealth](0,1.5)--(1,1.5) node[xshift=-0.5cm, yshift=-0.25cm]{$\boldsymbol{\hat{n}}$}; + \draw[line width=1.5pt,blue,-stealth](0,0.05)--(1,0.05) node[xshift=-0.5cm, yshift=-0.25cm]{$\boldsymbol{\hat{u}}$}; \end{tikzpicture} - \caption{Spiegelung des Vektors \textbf{v} an Spiegelachse bzw. Vektor \textbf{u}} + \caption{Spiegelung des Vektors $\mathbf{v}$ an der Spiegelebene $\sigma_u$ mit dem Normalenvektor $\mathbf{\hat{u}}$} \label{BildSpiegelung} \end{figure} \subsection{Linearen Algebra} Aus der linearen Algebra ist bekannt, dass man eine Spiegelung an einer Ebene wie folgt beschreiben kann. \begin{definition} - Die Spiegelungsgleichung in der linearen Algebra mit dem Normalenvektor $\mathbf{\hat{n}}$ zur Spiegelebene ist + Die Abbildung der Spiegelung in der linearen Algebra mit dem Normalenvektor $\mathbf{\hat{u}}$ zur Spiegelebene ist \begin{equation} \label{RefLinAlg} - \mathbf{v^{'}} = \mathbf{v} - 2 \cdot \mathbf{v_{\parallel \hat{n}}} = \mathbf{v} - 2 \cdot \mathbf{v_{\perp u}}. + \mathbf{v} = \mathbf{v_{\perp u}} + \mathbf{v_{\parallel u}} \enspace\mapsto\enspace \mathbf{v'} = \mathbf{v_{\perp u}} - \mathbf{v_{\parallel u}} = \mathbf{v} - 2 \cdot \mathbf{v_{\parallel u}}. \end{equation} - Per Definition sind $\mathbf{v_{\parallel \hat{n}}} = \mathbf{v_{\perp u}}$. In der geometrischen Algebra verwenden wir aber in den Formeln Vektoren, welche Spiegelachsen, nicht Spiegelebenen, repräsentieren. \end{definition} -Es scheint für diese Formel aber umständlich zu sein, weitere Spiegelungen mit weiteren Spiegelebenen anzufügen. Man kann diese Abbildung aber auch als Matrix schreiben. Sei $\mathbf{\hat{n}}$ ein Normalenvektor auf die Spiegelungs-Achse bzw. -Ebene, also $\mathbf{\hat{n}}\perp \mathbf{u}$, und sei ausserdem normiert $|\mathbf{\hat{n}}| = 1$, dann kann man die Spiegelung durch die Matrix +Es scheint für diese Formel \eqref{RefLinAlg} aber umständlich zu sein, weitere Spiegelungen mit weiteren Spiegelebenen anzufügen. Weil man $\mathbf{v_{\parallel u}}$ auch als Skalarprodukt $\mathbf{v_{\parallel u}} = \mathbf{\hat{u}} \cdot \mathbf{v}$ schreiben kann, ist es leicht diese Abbildung auch als Matrix darzustellen. Sei $\mathbf{\hat{u}}$ ein Normalenvektor auf die Spiegelungsebene, also $\mathbf{\hat{u}}\perp \sigma_u$, und sei ausserdem normiert $|\mathbf{\hat{u}}| = 1$, dann kann man die Spiegelung durch die Matrix \begin{align} - S = E - 2\dfrac{1}{|\mathbf{n}|^2}\mathbf{nn}^t + S = E - 2\mathbf{\hat{u}\hat{u}}^t \end{align} beschrieben werden. In der zweiten und dritten Dimension ergibt die Berechnung \begin{align} \label{Spiegelmatrizen} S_2 = \begin{pmatrix} - 1-2n_1^2 & -2n_1n_2 \\ - -2n_1n_2 & 1-2n_2^2 - \end{pmatrix} \quad + 1-2u_1^2 & -2u_1u_2 \\ + -2u_1u_2 & 1-2u_2^2 + \end{pmatrix}\enspace\text{und}\enspace S_3 = \begin{pmatrix} - 1-2n_1^2 & -2n_1n_2 & -2n_1n_3\\ - -2n_1n_2 & 1-2n_2^2 & -2n_2n_3\\ - -2n_1n_3 & -2n_2n_3 & 1-2n_3^2\\ + 1-2u_1^2 & -2u_1u_2 & -2u_1u_3\\ + -2u_1u_2 & 1-2u_2^2 & -2u_2u_3\\ + -2u_1u_3 & -2u_2u_3 & 1-2u_3^2\\ \end{pmatrix}. \end{align} -Diese Spiegelmatrizen gehören der orthogonalen Matrizengruppe $S\in \text{O}(n)$ an. Die Matrizengruppe $\text{O}(n)$ haben die Eigenschaft $S^t S = E$, was bedeutet, dass die Länge und Winkel bei der Abbildung beibehalten bleiben. Zusätzlich sind die Spiegelmatrizen symmetrisch, es gilt $S^t = S$. Somit liefert zweimal dieselbe Spiegelung wieder die identische Abbildung, wie man aus +Diese Spiegelmatrizen gehören der orthogonalen Matrizengruppe $S_n\in \text{O}(n)$ an. Die Matrizengruppe $\text{O}(n)$ haben die Eigenschaft $S_n^t S_n = E$, was bedeutet, dass die Länge und Winkel bei der Abbildung beibehalten bleiben. Zusätzlich sind die Spiegelmatrizen symmetrisch, es gilt $S_n^t = S_n$. Somit liefert zweimal dieselbe Spiegelung wieder die identische Abbildung, wie man aus \begin{align} - S^t S = S^2 = E + S_n^t S_n = S_n^2 = E \end{align} schliessen kann. \subsection{Geometrische Algebra} -Um die folgenden Formeln zu verstehen, definieren wir zuerst die Inverse eines Vektors, welche in dieser Form nicht in der linearen Algebra nicht existiert. +Wir definieren zuerst die Inverse eines Vektors, welche in dieser Form nicht in der linearen Algebra nicht existiert. \begin{definition} Die Inverse eines Vektors wird definiert als - \begin{align} - \mathbf{u}^{-1} = \dfrac{\mathbf{u}}{|\mathbf{u}|^2} \Rightarrow \mathbf{uu}^{-1} = \dfrac{\mathbf{u}^2}{|\mathbf{u}|^2} = 1. + \begin{align} \label{InverseGA} + \mathbf{u}^{-1} = \dfrac{\mathbf{u}}{|\mathbf{u}|^2}. \end{align} - Wie schon aus anderen algebraischen Strukturen bekannt, ergibt ein Element, hier $\mathbf{u}$, multipliziert mit dessen Inversen, hier $\mathbf{u}^{-1}$, das neutrale Element der Struktur, hier 1. \end{definition} +Diese Definition ist sinnvoll, da wegen $\mathbf{u}^2 = |\mathbf{u}|^2$ folgt +\begin{align} + \mathbf{uu}^{-1} = \mathbf{u} \frac{\mathbf{u}}{|\mathbf{u}|^2} = \frac{\mathbf{u}^2}{|\mathbf{u}|^2} = \frac{|\mathbf{u}|^2}{|\mathbf{u}|^2} = 1. +\end{align} +Der Vektor $\mathbf{u}^{-1}$ in \eqref{InverseGA} ist also tatsächlich das inverse Element im Sinne des Produktes in der geometrischen Algebra. Die geometrische Algebra leitet aus der obigen Formel \eqref{RefLinAlg} für eine Spiegelung eine einfache und intuitive Form her, welche auch für weitere Operationen erweitert werden kann. \begin{definition} - Die Spiegelungsgleichung in der geometrischen Algebra mit der Spiegelachse $\mathbf{u}$ ist definiert als + Die Abbildung der Spiegelung in der geometrischen Algebra mit dem senkrechten Vektor $\mathbf{u}$ zur Spiegelungsebene $\sigma_u$ ist \begin{align}\label{RefGA} - \mathbf{v}' = \mathbf{uvu}^{-1} + \mathbf{v} \enspace\mapsto\enspace \mathbf{v}' = -\mathbf{uvu}^{-1} \end{align} \end{definition} +Diese Abbildung muss stimmen, weil man durch die Schlussfolgerungen \eqref{uperpv} und \eqref{uparallelv} die Zusammenhänge +\begin{align} + \mathbf{uv_{\perp u}} = -\mathbf{v_{\perp u}u} \enspace\text{und}\enspace \mathbf{uv_{\parallel u}}=\mathbf{v_{\parallel u}u} +\end{align} +der geometrischen Produkte findet und somit die Abbildung aus der geometrischen Algebra \eqref{RefGA} wegen +\begin{align} + \mathbf{v}' = -\mathbf{uvu}^{-1} = -\mathbf{uv_{\perp u}u}^{-1} - \mathbf{uv_{\parallel u}u}^{-1} = -(-\mathbf{v_{\perp u}}\underbrace{\mathbf{u})\mathbf{u}^{-1}}_{1} -(\mathbf{v_{\parallel u}}\underbrace{\mathbf{u})\mathbf{u}^{-1}}_{1} = \mathbf{v_{\perp u}} - \mathbf{v_{\parallel u}} +\end{align} +gleichbedeutend zu der Definition \eqref{RefLinAlg} der Spiegelung ist. -verwendet man für $\mathbf{u}$ nur einen Einheitsvektor $\mathbf{\hat{u}}$, welcher die Länge 1 besitzt, wird die Gleichung zu +Verwendet man für $\mathbf{u}$ nur einen Einheitsvektor $\mathbf{\hat{u}}$, welcher die Länge 1 besitzt, wird die Gleichung \eqref{RefGA} zu \begin{align} - \mathbf{v'} = \mathbf{\hat{u}v\hat{u}} + \mathbf{v'} = -\mathbf{\hat{u}v\hat{u}} \end{align} vereinfacht. Im Gegensatz zu den Abbildungen in der linearen Algebra, welche in jeder anderen Dimension, durch andere Matrizen \eqref{Spiegelmatrizen} beschrieben werden müssen, ist es in der geometrischen Algebra immer der gleiche Vorgehensweise. Zudem ist diese kompakte Schreibweise in der linearen Algebra nicht möglich, da bis auf das Vektorprodukt in der dritten Dimension keine Multiplikation von Vektoren definiert ist.
\ No newline at end of file diff --git a/buch/papers/clifford/7_Spiegelung.tex b/buch/papers/clifford/7_Spiegelung.tex new file mode 100644 index 0000000..c79d908 --- /dev/null +++ b/buch/papers/clifford/7_Spiegelung.tex @@ -0,0 +1,100 @@ +% +% teil1.tex -- Beispiel-File für das Paper +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Spiegelung} +\rhead{Spiegelung} + +Die Spiegelung ist eine grundlegende, geometrische Operation, aus welcher man weitere Operationen, wie beispielsweise die später beschriebene Drehung, ableiten kann. Da die geometrische Algebra für geometrische Anwendungen ausgelegt ist, sollte die Spiegelung auch eine einfache, praktische Formulierung besitzen. +\begin{figure} + \centering + \begin{tikzpicture} + \draw[thin,gray!40] (-3,-1) grid (3,3); + \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; + \draw[<->] (0,-1)--(0,3) node[above]{$a_2$}; + \draw[blue, line width=1.0pt] (0,3)--(0,-1) node[anchor=south east]{$\sigma_u$}; + \draw[line width=2pt,black,-stealth](0,0)--(2,2) node[anchor=south east]{$\boldsymbol{v}$}; + \draw[line width=2pt,black,-stealth](0,0)--(-2,2) node[anchor=south east]{$\boldsymbol{v'}$}; + \draw[line width=1.5pt,gray,-stealth](0,0)--(1,0) node[anchor=north]{$\boldsymbol{e_1}$}; + \draw[line width=1.5pt,gray,-stealth](0,0)--(0,1) node[anchor=north east]{$\boldsymbol{e_2}$}; + \draw[line width=1.5pt,red,-stealth](0,2)--(2,2) node[xshift=-1cm, yshift= + 0.25cm]{$\boldsymbol{v_{\parallel u}}$}; + \draw[line width=1.5pt,red,-stealth](-2,2)--(0,2) node[xshift=-1cm, yshift= + 0.25cm]{$\boldsymbol{v_{\parallel u}}$}; + \draw[line width=1.5pt,blue,-stealth](0,0.05)--(1,0.05) node[xshift=-0.5cm, yshift=-0.25cm]{$\boldsymbol{\hat{u}}$}; + \end{tikzpicture} + \caption{Spiegelung des Vektors $\mathbf{v}$ an der Spiegelebene $\sigma_u$ mit dem Normalenvektor $\mathbf{\hat{u}}$} + \label{BildSpiegelung} +\end{figure} + +\subsection{Linearen Algebra} +Aus der linearen Algebra ist bekannt, dass man eine Spiegelung an einer Ebene, wie in Abbildung \ref{BildSpiegelung} gezeigt, wie folgt beschreiben kann. +\begin{definition} + Die Abbildung der Spiegelung in der linearen Algebra mit dem Normalenvektor $\mathbf{\hat{u}}$ zur Spiegelebene ist + \begin{equation} \label{RefLinAlg} + \mathbf{v} = \mathbf{v_{\perp u}} + \mathbf{v_{\parallel u}} \enspace\mapsto\enspace \mathbf{v'} = \mathbf{v_{\perp u}} - \mathbf{v_{\parallel u}} = \mathbf{v} - 2 \cdot \mathbf{v_{\parallel u}}. + \end{equation} +\end{definition} +Es scheint für diese Formel \eqref{RefLinAlg} aber umständlich zu sein, weitere Spiegelungen mit weiteren Spiegelebenen anzufügen. Weil man $\mathbf{v_{\parallel u}}$ auch als Skalarprodukt $\mathbf{v_{\parallel u}} = \mathbf{\hat{u}} \cdot \mathbf{v}$ schreiben kann, ist es leicht diese Abbildung auch als Matrix darzustellen. Sei $\mathbf{\hat{u}}$ ein Normalenvektor auf die Spiegelungsebene, also $\mathbf{\hat{u}}\perp \sigma_u$, und sei ausserdem normiert $|\mathbf{\hat{u}}| = 1$, dann kann man die Spiegelung durch die Matrix +\begin{align} +S = E - 2\mathbf{\hat{u}\hat{u}}^t +\end{align} +beschrieben werden. In zwei und drei Dimensionen ergibt die Berechnung +\begin{align} \label{Spiegelmatrizen} +S_2 = \begin{pmatrix} +1-2u_1^2 & -2u_1u_2 \\ +-2u_1u_2 & 1-2u_2^2 +\end{pmatrix}\quad\text{und}\quad +S_3 = \begin{pmatrix} +1-2u_1^2 & -2u_1u_2 & -2u_1u_3\\ +-2u_1u_2 & 1-2u_2^2 & -2u_2u_3\\ +-2u_1u_3 & -2u_2u_3 & 1-2u_3^2\\ +\end{pmatrix}. +\end{align} +Diese Spiegelmatrizen gehören der orthogonalen Matrizengruppe $S_n\in \text{O}(n)$ an. Die Matrizengruppe $\text{O}(n)$ hat die Eigenschaft $S_n^t S_n = E$, was bedeutet, dass die Länge und Winkel bei der Abbildung beibehalten bleiben. Zusätzlich sind die Spiegelmatrizen symmetrisch, es gilt $S_n^t = S_n$. Somit liefert zweimal dieselbe Spiegelung wieder die identische Abbildung, wie man aus +\begin{align} +S_n^t S_n = S_n^2 = E +\end{align} +schliessen kann. + +\subsection{Geometrische Algebra} +Wir definieren zuerst die Inverse eines Vektors, welche in dieser Form nicht in der linearen Algebra nicht existiert. +\begin{definition} + Die Inverse eines Vektors wird definiert als + \begin{align} \label{InverseGA} + \mathbf{u}^{-1} = \dfrac{\mathbf{u}}{|\mathbf{u}|^2}. + \end{align} +\end{definition} +Diese Definition ist sinnvoll, da wegen $\mathbf{u}^2 = |\mathbf{u}|^2$ folgt +\begin{align} +\mathbf{uu}^{-1} = \mathbf{u} \frac{\mathbf{u}}{|\mathbf{u}|^2} = \frac{\mathbf{u}^2}{|\mathbf{u}|^2} = \frac{|\mathbf{u}|^2}{|\mathbf{u}|^2} = 1. +\end{align} +Der Vektor $\mathbf{u}^{-1}$ in \eqref{InverseGA} ist also tatsächlich das inverse Element im Sinne des Produktes in der geometrischen Algebra. +Die geometrische Algebra leitet aus der obigen Formel \eqref{RefLinAlg} für eine Spiegelung eine einfache und intuitive Form her, welche auch für weitere Operationen erweitert werden kann. +\begin{definition} + Die Abbildung der Spiegelung in der geometrischen Algebra mit dem senkrechten Vektor $\mathbf{u}$ zur Spiegelungsebene $\sigma_u$ ist + \begin{align}\label{RefGA} + \mathbf{v} \enspace\mapsto\enspace \mathbf{v}' = -\mathbf{uvu}^{-1} + \end{align} +\end{definition} +Um zu überprüfen, ob die Spiegelungsgleichung \eqref{RefGA} wirklich eine Spiegelung ist, setzen wir zuerst in diese Gleichung $\mathbf{v} = \mathbf{v_{\perp u}} + \mathbf{v_{\parallel u}}$ ein. Wir bekommen somit +\begin{align} +\mathbf{v}' = -\mathbf{uv_{\perp u}u}^{-1} - \mathbf{uv_{\parallel u}u}^{-1}. +\end{align} +Danach können wir mit Hilfe der aus der Schlussfolgerung \eqref{uperpv} und \eqref{uparallelv} hergeleiteten Zusammenhänge +\begin{align} +\mathbf{uv_{\perp u}} = -\mathbf{v_{\perp u}u} \quad\text{und}\quad \mathbf{uv_{\parallel u}}=\mathbf{v_{\parallel u}u}, +\end{align} +die Gleichung weiter umformen zu +\begin{align} +\mathbf{v}' = -(-\mathbf{v_{\perp u}}\underbrace{\mathbf{u})\mathbf{u}^{-1}}_{1} -(\mathbf{v_{\parallel u}}\underbrace{\mathbf{u})\mathbf{u}^{-1}}_{1} = \mathbf{v_{\perp u}} - \mathbf{v_{\parallel u}}. +\end{align} +Man sieht, dass das Resultat $\mathbf{v}' = \mathbf{v_{\perp u}} - \mathbf{v_{\parallel u}}$ +gleichbedeutend zu der Definition \eqref{RefLinAlg} der Spiegelung ist. + +Verwendet man für $\mathbf{u}$ nur einen Einheitsvektor $\mathbf{\hat{u}}$, welcher die Länge 1 besitzt, wird die Gleichung \eqref{RefGA} zu +\begin{align} +\mathbf{v'} = -\mathbf{\hat{u}v\hat{u}} +\end{align} +vereinfacht. Im Gegensatz zu den Abbildungen in der linearen Algebra, welche in jeder anderen Dimension, durch andere Matrizen \eqref{Spiegelmatrizen} beschrieben werden müssen, ist es in der geometrischen Algebra immer der gleiche Vorgehensweise. Zudem ist diese kompakte Schreibweise in der linearen Algebra nicht möglich, da bis auf das Vektorprodukt in drei Dimensionen keine Multiplikation von Vektoren definiert ist.
\ No newline at end of file diff --git a/buch/papers/clifford/8_Rotation.tex b/buch/papers/clifford/8_Rotation.tex index 6a3251a..43d8f8a 100644 --- a/buch/papers/clifford/8_Rotation.tex +++ b/buch/papers/clifford/8_Rotation.tex @@ -3,168 +3,196 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Rotation} -\rhead{Rotation} +\section{Drehung} +\rhead{Drehung} -Eine Rotation kann man aus zwei aufeinanderfolgenden Spiegelungen bilden. Das wird für einige zuerst eine verwirrende Aussage sein, da man aus den vorherig gezeigten Formeln annehmen könnte, dass die Spiegelung schon für eine Drehung ausreicht. Obwohl sich die Längen, Winkel und Volumen sich bei einer Spiegelung, wie bei einer Rotation, nicht ändert, sind sie doch verschieden, da die Orientierung bei der Spiegelung invertiert wird. Stellt man sich beispielsweise ein Objekt im Dreidimensionalen vor und spiegelt dieses an einer Fläche, dann ist es unmöglich nur durch eine Rotation (egal an welchem Punkt) das ursprüngliche Objekt deckungsgleich auf das Gespiegelte zu drehen. Hingegen ist es wiederum möglich ein zweifach gespiegeltes Objekt durch eine Drehung zu erreichen. Das liegt daran, da die Orientierung zweimal invertiert wurde. -\\(Hier wird noch ein Bild für das Verständnis eingefügt) +Eine Drehung kann man aus zwei aufeinanderfolgenden Spiegelungen bilden. Das kann vielleicht zuerst eine verwirrende Aussage sein, da man aus den vorherig gezeigten Formeln annehmen könnte, dass die Spiegelung schon für eine Drehung ausreicht. Obwohl sich die Längen, Winkel und Volumen sich bei einer Spiegelung, wie bei einer Drehung, nicht ändert, sind sie doch verschieden, da die Orientierung bei der Spiegelung invertiert wird. Stellt man sich, wie im Bild \ref{BildSpiegRot} dargestellt, beispielsweise ein Objekt vor und spiegelt dieses an einer Ebene, dann ist es unmöglich, nur durch eine Drehung (egal an welchem Punkt) das ursprüngliche Objekt deckungsgleich auf das Gespiegelte zu drehen. Hingegen ist es wiederum möglich ein zweifach gespiegeltes Objekt durch eine Drehung zu erreichen. Das liegt daran, da die Orientierung zweimal invertiert wurde. + +\begin{figure} + \centering + \includegraphics[width=10cm]{papers/clifford/images/spiegelung.pdf} + \caption{Der wesentliche Unterschied zwischen Spiegelung und Drehung ist die Umkehrung der Orientierung} + \label{BildSpiegRot} +\end{figure} \begin{figure} \centering \begin{tikzpicture} - \draw[thin,gray!40] (-3,-1) grid (3,3); - \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; - \draw[<->] (0,-1)--(0,3) node[above]{$a_2$}; - \draw[line width=2pt,black,-stealth](0,0)--(2,2) node[anchor=south east]{$\boldsymbol{v}$}; - \draw[line width=1.5pt,blue,-stealth](0,0)--(0,2.5) node[anchor=south east]{$\boldsymbol{u}$}; - \draw[line width=2pt,black,-stealth](0,0)--(-2,2) node[anchor=south east]{$\boldsymbol{v'}$}; - \draw[line width=1.5pt,red,-stealth](0,0)--(-2.31, 0.957) node[anchor=south east]{$\boldsymbol{w}$}; - \draw[line width=2pt,black,-stealth](0,0)--(-2.828,0) node[anchor=south east]{$\boldsymbol{v''}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(1,0) node[anchor=north]{$\boldsymbol{e_1}$}; - \draw[line width=1.5pt,gray,-stealth](0,0)--(0,1) node[anchor=north west]{$\boldsymbol{e_2}$}; - - \coordinate (A) at (0,0); - \coordinate (B) at (0,2.5); - \coordinate (C) at (-2.31, 0.957); - \tikzset{anglestyle/.style={angle eccentricity=1.25, draw, thick, angle radius=1.25cm}} - \draw pic ["$\theta$", anglestyle] {angle = B--A--C}; + \draw[thin,gray!40] (-3,-1) grid (3,3); + \draw[<->] (-3,0)--(3,0) node[right]{$a_1$}; + \draw[<->] (0,-1)--(0,3) node[above]{$a_2$}; + \draw[line width=1.0pt,green,-stealth](2,2)--(-2,2) node[anchor=south west]{$\boldsymbol{-2v_{\parallel u}}$}; + \draw[line width=1.0pt,green,-stealth](-2,2)--(-2.828,0) node[anchor=north west]{$\boldsymbol{-2v'_{\parallel w}}$}; + \draw[blue, line width=1.0pt] (0,3)--(0,-1) node[anchor=south east]{$\sigma_u$}; + \draw[red, line width=1.0pt] (-3,1.24)--(2.21,-1) node[anchor=south]{$\sigma_w$}; + \draw[line width=2pt,black,-stealth](0,0)--(2,2) node[anchor=south east]{$\boldsymbol{v}$}; + \draw[line width=1.5pt,blue,-stealth](0,0)--(2.5, 0) node[anchor=south east]{$\boldsymbol{u}$}; + \draw[line width=2pt,black,-stealth](0,0)--(-2,2) node[anchor=south east]{$\boldsymbol{v'}$}; + \draw[line width=1.5pt,red,-stealth](0,0)--(0.957, 2.31) node[anchor=south east]{$\boldsymbol{w}$}; + \draw[line width=2pt,black,-stealth](0,0)--(-2.828,0) node[anchor=south east]{$\boldsymbol{v''}$}; + \draw[line width=1.5pt,gray,-stealth](0,0)--(1,0) node[anchor=north]{$\boldsymbol{e_1}$}; + \draw[line width=1.5pt,gray,-stealth](0,0)--(0,1) node[anchor=north east]{$\boldsymbol{e_2}$}; + + \coordinate (A) at (0,0); + \coordinate (B) at (2.5,0); + \coordinate (C) at (0.957, 2.31); + \tikzset{anglestyle/.style={angle eccentricity=1.25, purple, draw, thick, angle radius=1cm}} + \draw pic ["$\theta$", anglestyle] {angle = B--A--C}; + \coordinate (D) at (0,0); + \coordinate (E) at (1,1); + \coordinate (F) at (-1, 0); + \tikzset{anglestyle/.style={angle eccentricity=1.25, purple, draw, thick, angle radius=1.25cm}} + \draw pic ["$2\theta$", anglestyle] {angle = E--D--F}; \end{tikzpicture} - \caption{Rotation des Vektors $\textbf{v}$ um $2\theta$} - \label{BildRotation} + \caption{Drehung des Vektors $\textbf{v}$ um $2\theta$} + \label{BildDrehung} \end{figure} \subsection{Linearen Algebra} In der linearen Algebra haben wir Drehungen durch die Matrizen der Gruppe $\text{SO}(n)$ beschrieben. Beispielsweise besteht $\text{SO}(2)$ aus den Matrizen \begin{align} - D = - \begin{pmatrix} - \cos(\alpha) & \sin(\alpha) \\ - -\sin(\alpha) & \cos(\alpha) - \end{pmatrix},\quad - \alpha \in [0, 2\pi). -\end{align} -Diese Drehmatrizen gehören der speziellen orthogonalen Matrizengruppe $D\in \text{SO}(n) = \text{SL}_n(\mathbb{R})\enspace \cap \enspace \text{O}(n)$ an. $\text{SL}_n(\mathbb{R})$ beinhaltet die Matrizen mit scherenden Eigenschaften. Diese Drehmatrizen haben die Eigenschaft $D^t D = E \enspace \land \enspace \det(D)=1$. Da $\det(D) = 1$ und nicht $-1$ sein kann fallen alle Spiegelungen aus der Menge heraus. $\det(D) = -1$ bedeutet, dass eine Orientierungsinversion stattfindet. -\\(BILD Mengen Spezieller Matrizen von Herrn Müller Präsentation) +D = +\begin{pmatrix} +\cos(\alpha) & \sin(\alpha) \\ +-\sin(\alpha) & \cos(\alpha) +\end{pmatrix},\quad +\alpha \in [0, 2\pi). +\end{align} +Diese Drehmatrizen gehören der speziellen orthogonalen Matrizengruppe $D\in \text{SO}(n) = \text{SL}_n(\mathbb{R})\enspace \cap \enspace \text{O}(n)$ an. $\text{SL}_n(\mathbb{R})$ beinhaltet die Matrizen mit scherenden Eigenschaften. Die Drehmatrizen haben die Eigenschaft $D^t D = E \enspace \land \enspace \det(D)=1$. Da $\det(D) = 1$ und nicht $-1$ sein kann fallen alle Spiegelungen aus der Menge heraus. $\det(D) = -1$ bedeutet, dass eine Orientierungsinversion stattfindet. Eine übersichtliche Darstellung der beschriebenen Matrizengruppen sieht man in der Abbildung \ref{BildMatrizenGruppen} + +\begin{figure} + \centering + \includegraphics[width=10cm]{papers/clifford/Bilder/MatrizenGruppen.png} + \caption{Matrizengruppen} + \label{BildMatrizenGruppen} +\end{figure} \subsection{Geometrische Algebra} -Da wir jetzt aus der Geometrie wissen, dass eine Rotation durch zwei Spiegelungen gebildet werden kann, können wir die Rotation mit der Formel \eqref{RefGA} einfach herleiten. +Da wir jetzt aus der Geometrie wissen, dass eine Drehung durch zwei Spiegelungen gebildet werden kann, können wir die Drehung mit der Formel \eqref{RefGA} einfach herleiten. \begin{satz} - Eine Rotation + Durch zwei nacheinander auf einen Vektor $\mathbf{v}$ angewendete Spiegelungen lässt sich eine Drehung \begin{align} \label{rotGA} - \mathbf{v}'' = \mathbf{wv}'\mathbf{w}^{-1} = \mathbf{w}(\mathbf{uvu}^{-1})\mathbf{w}^{-1} = (\mathbf{wu})\mathbf{v}(\mathbf{u}^{-1}\mathbf{w}^{-1}) + \mathbf{v}'' = -\mathbf{wv}'\mathbf{w}^{-1} = -\mathbf{w}(-\mathbf{uvu}^{-1})\mathbf{w}^{-1} = (\mathbf{wu})\mathbf{v}(\mathbf{u}^{-1}\mathbf{w}^{-1}) \end{align} - lässt sich durch zwei nacheinander auf einen Vektor $\mathbf{v}$ angewendete Spiegelungen beschreiben. + beschreiben. \end{satz} Die Vektoren $\mathbf{w}$ und $\mathbf{u}$ bilden hier wiederum die Spiegelachsen. Diese Formel versuchen wir jetzt noch durch Umstrukturierung zu verbessern. \subsubsection{Exponentialform} -Dazu leiten wir zuerst die Exponentialform eines Vektors her. Es wird dabei zur Vereinfachung davon ausgegangen, dass alle Vektoren $\mathbf{w}, \mathbf{u}, \mathbf{v}$ in der $\mathbf{e}_{12}$ Ebene liegen. Weitere Drehungen können in höheren Dimensionen durch Linearkombinationen von Drehungen in den $\mathbf{e}_{ij}, i\not=j$ Ebenen erreicht werden. Für die Herleitung erweitern wir nun als erstes die Polarform +Dazu leiten wir zuerst die Exponentialform eines Vektors her. Es wird dabei zur Vereinfachung davon ausgegangen, dass alle Vektoren $\mathbf{w}, \mathbf{u}, \mathbf{v}$ in der $\mathbf{e}_{1}$-$\mathbf{e}_{2}$-Ebene liegen. Weitere Drehungen können in höheren Dimensionen durch Linearkombinationen von Drehungen in den $\mathbf{e}_{i}$-$\mathbf{e}_{j}$-Ebenen $(i\not=j)$ erreicht werden. Für die Herleitung ersetzen wir als erstes in der Polarform \begin{align} - \mathbf{w} = |\mathbf{w}| \left(\cos(\theta_w) \mathbf{e}_1 + \sin(\theta_w) \mathbf{e}_2\right) +\mathbf{w} = |\mathbf{w}| \left(\cos(\theta_w) \mathbf{e}_1 + \sin(\theta_w) \mathbf{e}_2\right) \end{align} -eines Vektors mit $\mathbf{e}_1^2 = 1$ beim Sinus +eines Vektors einen Faktor 1 durch $1=\mathbf{e}_1^2$ und erhalten beim Sinus \begin{align}\label{e1ausklammern} - \mathbf{w} &= |\mathbf{w}| \left(\cos(\theta_w) \mathbf{e}_1 + \sin(\theta_w) \mathbf{e}_1\mathbf{e}_1\mathbf{e}_2\right), +\mathbf{w} &= |\mathbf{w}| \left(\cos(\theta_w) \mathbf{e}_1 + \sin(\theta_w) \mathbf{e}_1\mathbf{e}_1\mathbf{e}_2\right). \end{align} -um dann $\mathbf{e}_1$ +In einem zweiten Schritt klammern wir $\mathbf{e}_1$ aus, dies ergibt \begin{align} - \mathbf{w} = |\mathbf{w}|\mathbf{e}_1\left(\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12}\right) \label{ExponentialGA} +\mathbf{w} = |\mathbf{w}|\mathbf{e}_1\left(\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12}\right). \label{ExponentialGA} \end{align} -ausklammern zu können. Die Ähnlichkeit des Klammerausdrucks zu der Eulerschen Formel bei den Komplexen Zahlen ist nun schon gut erkennbar. Versuchen wir nun mithilfe der Reihenentwicklungen +Die Ähnlichkeit des Klammerausdrucks in der Formel \eqref{ExponentialGA} zu der Eulerschen Formel bei den komplexen Zahlen ist nun schon gut erkennbar. Versuchen wir nun mithilfe der Reihenentwicklungen \begin{align} - \sin(\theta_w)\mathbf{e}_{12}&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta_w^{2n+1}}{(2n+1)!}}\mathbf{e}_{12} =\theta_w\mathbf{e}_{12}-{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}-\cdots \\ - \cos(\theta_w)&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta_w^{2n}}{(2n)!}} =1-{\frac {\theta_w^{2}}{2!}}+{\frac {\theta_w^{4}}{4!}}-\cdots +\sin(\theta_w)\mathbf{e}_{12}&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta_w^{2n+1}}{(2n+1)!}}\mathbf{e}_{12} =\theta_w\mathbf{e}_{12}-{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}-\cdots \\ +\cos(\theta_w)&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta_w^{2n}}{(2n)!}} =1-{\frac {\theta_w^{2}}{2!}}+{\frac {\theta_w^{4}}{4!}}-\cdots \end{align} -den Zusammenhang auch hier herzustellen. Verwenden wir jetzt noch die Eigenschaft, dass $\mathbf{e}_{12}^2=-1, \enspace\mathbf{e}_{12}^3=-\mathbf{e}_{12}, \dots$, bei dem Klammerausdruck in Formel \eqref{ExponentialGA} +diesen Zusammenhang auch hier herzustellen. Setzt man diese beiden Reihenentwicklungen in \eqref{ExponentialGA} ein, erhält man \begin{align} - \cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12} &= 1+\theta_w\mathbf{e}_{12}-{\frac {\theta_w^{2}}{2!}}-{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}+{\frac {\theta_w^{4}}{4!}}+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}-\cdots\\ - &= 1 \mathbf{e}_{12}^0+\theta_w\mathbf{e}_{12}^1+{\frac {\theta_w^{2}}{2!}}\mathbf{e}_{12}^2+{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}^3+{\frac {\theta_w^{4}}{4!}}\mathbf{e}_{12}^4+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}^5+\cdots - \label{ExponentialGA2} +\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12} &= 1+\theta_w\mathbf{e}_{12}-{\frac {\theta_w^{2}}{2!}}-{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}+{\frac {\theta_w^{4}}{4!}}+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}-\cdots \end{align} -dann sieht man die Übereinstimmung mit der Reihenentwicklung der Exponentialfunktion +Dies sieht noch nicht wie eine Exponentialreihe aus, da $\mathbf{e}_{12}$ nur in jedem zweiten Term auftritt. Da aber $\mathbf{e}_{12}=-1$ gibt, erhält man für \begin{align} - &e^{\theta_w\mathbf{e}_{12}}=\sum _{n=0}^{\infty }{\frac {(\theta_w\mathbf{e}_{12})^{n}}{n!}}={\frac {(\theta_w\mathbf{e}_{12})^{0}}{0!}}+{\frac {(\theta_w\mathbf{e}_{12})^{1}}{1!}}+{\frac {(\theta_w\mathbf{e}_{12})^{2}}{2!}}+{\frac {(\theta_w\mathbf{e}_{12})^{3}}{3!}}+\cdots\\ - &\Rightarrow \mathbf{w} = |w|\mathbf{e}_1 e^{\theta_w \mathbf{e}_{12}} = |w|\mathbf{e}_1\left(\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12}\right). +e^{\theta_w\mathbf{e}_{12}} = 1 \mathbf{e}_{12}^0+\theta_w\mathbf{e}_{12}^1+{\frac {\theta_w^{2}}{2!}}\mathbf{e}_{12}^2+{\frac {\theta_w^{3}}{3!}}\mathbf{e}_{12}^3+{\frac {\theta_w^{4}}{4!}}\mathbf{e}_{12}^4+{\frac {\theta_w^{5}}{5!}}\mathbf{e}_{12}^5+\cdots +\label{ExponentialGA2} +\end{align} +Man sieht, dass die beiden Reihen übereinstimmen. Es folgt somit +\begin{align}\label{EulerGA} +e^{\theta_w \mathbf{e}_{12}} = \cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_{12}, +\end{align} +was zeigt, dass es eine Euler-Formel mit $\mathbf{e}_{12}$ anstelle der imaginären Einheit $j$ gibt. + +Wenn man jetzt den Vektor \eqref{ExponentialGA} durch die eulersche Schreibweise +\begin{align} +\mathbf{w} = |\mathbf{w}|\mathbf{e}_1e^{\theta_w\mathbf{e}_{12}} \end{align} -Man kann die Exponentialform des Vektors ähnlich wie die der komplexen Zahlen interpretieren. Der Einheitsvektor $\mathbf{e}_1$ wird um die Länge $|\mathbf{w}|$ gestreckt und um $\theta_w$ gedreht. -Bei den komplexen Zahlen würden man vom Punkt 1 anstatt $\mathbf{e}_1$ ausgehen. +ersetzt, kann die Exponentialform des Vektors ähnlich wie die der komplexen Zahlen interpretieren. Der Einheitsvektor $\mathbf{e}_1$ wird um die Länge $|\mathbf{w}|$ gestreckt und um $\theta_w$ gedreht. \subsubsection{Vektormultiplikation} -Nun werden wir das Produkt von zwei Vektoren $\mathbf{wu}$ -\begin{align} - \mathbf{wu} = |\mathbf{w}|\mathbf{e}_1 e^{\theta_w \mathbf{e}_{12}}|\mathbf{u}|\mathbf{e}_1 e^{\theta_u \mathbf{e}_{12}} +Nun werden wir das Vektorprodukt +\begin{align} \label{VektorproduktformelGA} +\mathbf{wu} = |\mathbf{w}|\mathbf{e}_1 e^{\theta_w \mathbf{e}_{12}}|\mathbf{u}|\mathbf{e}_1 e^{\theta_u \mathbf{e}_{12}} \end{align} -so umformen, dass wir eine bessere Darstellung erhalten. Wir tauschen dafür zuerst beim Vektor $\mathbf{w}$ die Reihenfolge von -$\mathbf{e}_1$ mit dem Exponentialterm $e^{\theta_w \mathbf{e}_{12}}$, indem wir bei der Gleichung \eqref{e1ausklammern}, anstatt mit $\mathbf{e}_1\mathbf{e}_1\mathbf{e}_2$ mit $\mathbf{e}_2\mathbf{e}_1\mathbf{e}_1$ erweitern +so umformen, dass wir die Drehung nur durch Exponentialterme beschreiben können. Wir tauschen dafür zuerst beim Vektor $\mathbf{w}$ die Reihenfolge von +$\mathbf{e}_1$ mit dem Exponentialterm $e^{\theta_w \mathbf{e}_{12}}$, indem wir bei der Gleichung \eqref{e1ausklammern} $1=\mathbf{e}_1^2$ an einer anderen Position einsetzten. Wir erhalten \begin{align} - \mathbf{w} &= |\mathbf{w}|\left(\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_2\mathbf{e}_1\right)\mathbf{e}_1\\ - &= |\mathbf{w}|e^{\theta_w \mathbf{e}_{21}}\mathbf{e}_1\\ - &= |\mathbf{w}|e^{-\theta_w \mathbf{e}_{12}}\mathbf{e}_1 +\mathbf{w} &= |\mathbf{w}|\left(\cos(\theta_w)+ \sin(\theta_w) \mathbf{e}_2\mathbf{e}_1\right)\mathbf{e}_1. \end{align} -und umstrukturiert wieder in die Vektorproduktformel einsetzen +Mithilfe der Formel \eqref{EulerGA} und dem Wissen, dass $\mathbf{e}_{21}= -\mathbf{e}_{12}$ können wir die Umformung \begin{align} - \mathbf{wu} = |\mathbf{w}||\mathbf{u}|e^{-\theta_w \mathbf{e}_{12}}\mathbf{e}_1\mathbf{e}_1 e^{\theta_u \mathbf{e}_{12}}\\ - \mathbf{wu} = |\mathbf{w}||\mathbf{u}|e^{(\theta_u-\theta_w) \mathbf{e}_{12}}. +|\mathbf{w}|e^{-\theta_w \mathbf{e}_{12}}\mathbf{e}_1 \end{align} -Der Term $\mathbf{u}^{-1}\mathbf{w}^{-1}$ +ausführen. Diese wichtige Umstrukturierung können wir wieder in die Vektorproduktformel \eqref{VektorproduktformelGA} einsetzen un erhalten \begin{align} - \mathbf{u}^{-1}\mathbf{w}^{-1} = \dfrac{1}{|\mathbf{w}||\mathbf{u}|}e^{(\theta_w-\theta_u) \mathbf{e}_{12}} +\mathbf{wu} &= |\mathbf{w}|\,|\mathbf{u}|e^{-\theta_w \mathbf{e}_{12}}\mathbf{e}_1\mathbf{e}_1 e^{\theta_u \mathbf{e}_{12}}\\ +&= |\mathbf{w}|\,|\mathbf{u}|e^{(\theta_u-\theta_w) \mathbf{e}_{12}}. \end{align} -kann durch die selbe Methode zusammengefasst werden. -Wenn wir den Winkel zwischen den Vektoren $\mathbf{w}$ und $\mathbf{u}$ als $\theta = \theta_w - \theta_u$ definieren erhalten wir +Das inverse Vektorprodukt +\begin{align} +\mathbf{u}^{-1}\mathbf{w}^{-1} = \dfrac{1}{|\mathbf{w}|\,|\mathbf{u}|}e^{(\theta_w-\theta_u) \mathbf{e}_{12}} +\end{align} +kann durch die selbe Methode vereinfacht werden. +Wenn wir den Winkel zwischen den Vektoren $\mathbf{w}$ und $\mathbf{u}$ als $\theta = \theta_w - \theta_u$ definieren erhalten wir als endgültige Form der Vektorprodukte \begin{align}\label{wuExpo} - \mathbf{wu} = |\mathbf{w}||\mathbf{u}|e^{-\theta \mathbf{e}_{12}}\\ - \mathbf{u}^{-1}\mathbf{w}^{-1} = \dfrac{1}{|\mathbf{w}||\mathbf{u}|}e^{\theta \mathbf{e}_{12}} \label{wuExpoInv} +\mathbf{wu} &= |\mathbf{w}|\,|\mathbf{u}|e^{-\theta \mathbf{e}_{12}}\enspace\text{und}\\ +\mathbf{u}^{-1}\mathbf{w}^{-1} &= \dfrac{1}{|\mathbf{w}|\,|\mathbf{u}|}e^{\theta \mathbf{e}_{12}} \label{wuExpoInv}. \end{align} -die finale Form der Vektorprodukte. \subsubsection{Umstrukturierte Drehungsgleichung} -Setzten wir nun unsere neuen Erkenntnisse in die Gleichung \eqref{rotGA} ein +Setzten wir nun unsere neuen Erkenntnisse in die Gleichung \eqref{rotGA} ein, erhalten wir \begin{align} - \mathbf{v''} = (|\mathbf{w}||\mathbf{u}|e^{-\theta \mathbf{e}_{12}}) \mathbf{v}( \dfrac{1}{|\mathbf{w}||\mathbf{u}|}e^{\theta \mathbf{e}_{12}}), +\mathbf{v''} = (|\mathbf{w}|\,|\mathbf{u}|e^{-\theta \mathbf{e}_{12}})\mathbf{v}\biggl(\dfrac{1}{|\mathbf{w}|\,|\mathbf{u}|}e^{\theta \mathbf{e}_{12}}\biggr) \end{align} -erhalten wir durch die Kürzungen der Längen die vereinfachte Drehungsgleichung -\begin{align} - \mathbf{v''} = e^{-\theta \mathbf{e}_{12}} v e^{\theta \mathbf{e}_{12}}. +und können durch die Kürzungen der Längen die vereinfachte Drehungsgleichung +\begin{align} \label{GAvereinfRot} +\mathbf{v''} = e^{-\theta \mathbf{e}_{12}} v e^{\theta \mathbf{e}_{12}} \end{align} - -Wir wissen nun, dass das diese beidseitige Multiplikation die Länge von $\mathbf{v}$ nicht verändert, da sich die Längen von $\mathbf{w}$ und $\mathbf{u}$ kürzen. Betrachten wir nun den Effekt der Exponentialterme auf $\mathbf{v}$. Dabei Teilen wir den Vektor $\mathbf{v}$ auf in einen Anteil $\mathbf{v_\parallel}$, welcher auf der Ebene $\mathbf{e}_{12}$ liegt, und einen Anteil $\mathbf{v_\perp}$, welcher senkrecht zu der Ebene steht. Wir bekommen durch Einsetzten nun diese Form +bilden. Wir wissen nun, dass das diese beidseitige Multiplikation die Länge von $\mathbf{v}$ nicht verändert, da sich die Längen von $\mathbf{w}$ und $\mathbf{u}$ kürzen. Betrachten wir nun den Effekt der Exponentialterme auf $\mathbf{v}$. Dabei teilen wir den Vektor $\mathbf{v}$ auf in einen Anteil $\mathbf{v_\parallel}$, welcher auf der Ebene $\mathbf{e}_{12}$ liegt, und einen Anteil $\mathbf{v_\perp}$, welcher senkrecht zu der Ebene steht. Wir bekommen durch Einsetzten nun diese Form \begin{align} \label{RotAufPerpPar} - \mathbf{v}'' = e^{-\theta \mathbf{e}_{12}} (\mathbf{v_\perp + v_\parallel}) e^{\theta \mathbf{e}_{12}} = e^{-\theta \mathbf{e}_{12}} \mathbf{v_\perp} e^{\theta \mathbf{e}_{12}} + e^{-\theta \mathbf{e}_{12}} \mathbf{v_\parallel} e^{\theta \mathbf{e}_{12}}. +\mathbf{v}'' = e^{-\theta \mathbf{e}_{12}} (\mathbf{v_\perp + v_\parallel}) e^{\theta \mathbf{e}_{12}} = e^{-\theta \mathbf{e}_{12}} \mathbf{v_\perp} e^{\theta \mathbf{e}_{12}} + e^{-\theta \mathbf{e}_{12}} \mathbf{v_\parallel} e^{\theta \mathbf{e}_{12}}. \end{align} -Auf eine allgemeine Herleitung wird hier zwar verzichtet, aber man kann zeigen, dass die Reihenfolge so umstrukturiert werden kann +Auf eine allgemeine Herleitung wird hier zwar verzichtet, aber man kann zeigen, dass man die Reihenfolge der Vektoranteile $\mathbf{v_\perp}$ und $\mathbf{v_\parallel}$ mit dem Exponentialterm $e^{-\theta \mathbf{e}_{12}}$ so vertauschen kann, dass sich \begin{align} - \mathbf{v}'' = \mathbf{v_\perp} e^{-\theta \mathbf{e}_{12}} e^{\theta \mathbf{e}_{12}} + \mathbf{v_\parallel} e^{-(-\theta) \mathbf{e}_{12}} e^{\theta \mathbf{e}_{12}}, +\mathbf{v}'' = \mathbf{v_\perp} e^{-\theta \mathbf{e}_{12}} e^{\theta \mathbf{e}_{12}} + \mathbf{v_\parallel} e^{-(-\theta) \mathbf{e}_{12}} e^{\theta \mathbf{e}_{12}} \end{align} -dass der Winkel beim parallelen Anteil negiert wird. An der Zusammengefassten Gleichung +ergibt. Der Winkel wird beim parallelen Anteil negiert. An der Zusammengefassten Gleichung \begin{align}\label{RotParPerp} - \mathbf{v}'' = \mathbf{v_\perp} + \mathbf{v_\parallel} e^{2\theta \mathbf{e}_{12}} +\mathbf{v}'' = \mathbf{v_\perp} + \mathbf{v_\parallel} e^{2\theta \mathbf{e}_{12}} \end{align} kann man sehen, dass nur der parallele Anteil $\mathbf{v_\parallel}$ des Vektors $\mathbf{v}$ auf der Ebene $\mathbf{e}_{12}$ um $2\theta$ gedreht wird. Der senkrechte Anteil $\mathbf{v_\perp}$ bleibt gleich. Wichtig dabei zu sehen ist, dass nur der Winkel zwischen den Vektoren $\mathbf{w}$ und $\mathbf{u}$ von Bedeutung ist. Die Länge und Richtung der einzelnen Vektoren spielt keine Rolle. Zeigen wir nun diese Eigenschaften an einem Beispiel \begin{beispiel} - Gegeben sei ein Vektor $\mathbf{v} = 1\mathbf{e}_1 + 2\mathbf{e}_2 + 3\mathbf{e}_3$ mit zur $\mathbf{e}_{12}$-Ebene parallelen Anteil $\mathbf{v_\parallel} = 1\mathbf{e}_1 + 2\mathbf{e}_2$ und senkrechten Anteil $\mathbf{v_\perp} = 3\mathbf{e}_3$. Zusätzlich sind die Spiegelachsen $\mathbf{u} = \mathbf{e}_1$ und $\mathbf{w} = 2\mathbf{e}_2$ gegeben. Gesucht ist der rotierte Vektor $\mathbf{v}''$. Bestimmen wir als erstes das Vektorprodukt $\mathbf{wu}$ + Gegeben sei ein Vektor $\mathbf{v} = 1\mathbf{e}_1 + 2\mathbf{e}_2 + 3\mathbf{e}_3$ mit zur $\mathbf{e}_{12}$-Ebene parallelen Anteil $\mathbf{v_\parallel} = 1\mathbf{e}_1 + 2\mathbf{e}_2$ und senkrechten Anteil $\mathbf{v_\perp} = 3\mathbf{e}_3$. Zusätzlich sind die Spiegelachsen $\mathbf{u} = \mathbf{e}_1$ und $\mathbf{w} = 2\mathbf{e}_2$ gegeben. Gesucht ist der rotierte Vektor $\mathbf{v}''$. Bestimmen wir als erstes das Vektorprodukt \begin{align} - \mathbf{wu} = (2\mathbf{e}_2)(\mathbf{e}_1) = -2\mathbf{e}_{12} + \mathbf{wu} = (2\mathbf{e}_2)(\mathbf{e}_1) = -2\mathbf{e}_{12} \end{align} - und das Produkt der Inversen $\mathbf{u}^{-1}\mathbf{w}^{-1}$ + und das Produkt der Inversen \begin{align} - \mathbf{u}^{-1}\mathbf{w}^{-1} = (\dfrac{\mathbf{e}_1}{1^2})(\dfrac{2\mathbf{e}_2}{2^2}) = \dfrac{1}{2}\mathbf{e}_{12}. + \mathbf{u}^{-1}\mathbf{w}^{-1} = \biggl(\dfrac{\mathbf{e}_1}{1^2}\biggr) \left(\dfrac{2\mathbf{e}_2}{2^2}\right) = \dfrac{1}{2}\mathbf{e}_{12}. \end{align} - Der rotierte Vektor $\mathbf{v}''$ können wir nun durch das einsetzten und auflösen der Produkte in die Gleichung \eqref{rotGA} + Den gedrehten Vektor $\mathbf{v}''$ können wir nun durch Einsetzen und Auflösen der Produkte in die Gleichung \eqref{rotGA} bestimmen. Der Rechnenvorgang ist \begin{align} - \mathbf{v}'' = (\mathbf{wu})\mathbf{v}(\mathbf{u}^{-1}\mathbf{w}^{-1}) &= (-2e_{12})(1\mathbf{e}_1 + \mathbf{e}_2 + 1\mathbf{e}_3)(\dfrac{1}{2}\mathbf{e}_{12})\\ - &= (2\mathbf{e}_2-2\mathbf{e}_1-2\mathbf{e}_{123})(\dfrac{1}{2}\mathbf{e}_{12})\\ - &= -1\mathbf{e}_1 - 1\mathbf{e}_2 + 1\mathbf{e}_3 + \mathbf{v}'' = (\mathbf{wu})\mathbf{v}(\mathbf{u}^{-1}\mathbf{w}^{-1}) &= (-2e_{12})(1\mathbf{e}_1 + \mathbf{e}_2 + 1\mathbf{e}_3)(\textstyle{\frac{1}{2}}\mathbf{e}_{12})\\ + &= (2\mathbf{e}_2-2\mathbf{e}_1-2\mathbf{e}_{123})(\textstyle{\frac{1}{2}}\mathbf{e}_{12})\\ + &= -1\mathbf{e}_1 - 1\mathbf{e}_2 + 1\mathbf{e}_3. \end{align} - finden. Aus dem Resultat $\mathbf{v}''= -1\mathbf{e}_1 + 1\mathbf{e}_2 + 1\mathbf{e}_3$ können wir bestätigen, dass + Aus dem Resultat $\mathbf{v}''= -1\mathbf{e}_1 + 1\mathbf{e}_2 + 1\mathbf{e}_3$ können wir bestätigen, dass \begin{itemize} \item die Länge $|\mathbf{v}| = \sqrt{3}$ zur Länge $|\mathbf{v}''|=\sqrt{3}$ gleich blieb. \item sich der parallele Anteil $\mathbf{v_\parallel}'' = -1\mathbf{e}_1 - 1\mathbf{e}_2$ gedreht hat und der senkrechte Anteil $\mathbf{v_\perp}'' = 1\mathbf{e}_3$ unverändert blieb. \item der parallele Teil sich genau um $2\theta=180$° gedreht hat. $\theta$ kann übrigens durch die Umformung des Produkt $\mathbf{wu}$ in die Exponentialschreibweise \begin{align} - &\mathbf{wu} = -2\mathbf{e}_{12} = 2(0-1\mathbf{e}_{12})=2(\cos(\dfrac{-\pi}{2} + \sin(\dfrac{-\pi}{2})\mathbf{e}_{12})) = 2e^{(-\pi/2)\mathbf{e}_{12}} + &\mathbf{wu} = -2\mathbf{e}_{12} = 2(0-1\mathbf{e}_{12})=2(\cos\biggl(\dfrac{-\pi}{2}\biggr) + \sin\biggl(\dfrac{-\pi}{2}\biggr)\mathbf{e}_{12}) = 2e^{(-\pi/2)\mathbf{e}_{12}} \end{align} durch einen Vergleich mir der Formel \eqref{wuExpo} \begin{align} - \theta = -(\dfrac{-\pi}{2}) = \dfrac{\pi}{2} + \theta = -\biggl(\dfrac{-\pi}{2}\biggr) = \dfrac{\pi}{2} \end{align} - ausgelesen werden. + ausgelesen werden. \qedhere \end{itemize} -\end{beispiel}
\ No newline at end of file +\end{beispiel}
\ No newline at end of file diff --git a/buch/papers/clifford/9_KomplexeZahlen.tex b/buch/papers/clifford/9_KomplexeZahlen.tex index 70107da..12fa546 100644 --- a/buch/papers/clifford/9_KomplexeZahlen.tex +++ b/buch/papers/clifford/9_KomplexeZahlen.tex @@ -6,23 +6,34 @@ \section{Komplexe Zahlen} \rhead{Komplexe Zahlen} -Die komplexen Zahlen finden eine Vielzahl von Anwendungsgebiete in den Ingenieurwissenschaften. Das liegt daran, weil die komplexen Zahlen Rotationen und Schwingungen gut beschreiben können. Nach dem vorherigen Kapitel überrascht es wahrscheinlich nicht viele, dass es möglich ist komplexe Zahlen in der geometrischen Algebra darzustellen. Sie können durch die geraden Grade der 2 Dimensionalen geometrischen Algebra vollständig beschrieben werden: $\mathbf{g}_n \in G_2^+(\mathbb{R}) \cong \mathbb{C}$. Das bedeutet eine komplexe Zahl kann durch ein Skalar (Grad 0) und einem Bivektor (Grad 2) dargestellt werden +Die komplexen Zahlen finden eine Vielzahl von Anwendungsgebiete in den Ingenieurwissenschaften. Das liegt daran, weil die komplexen Zahlen Drehungen und Schwingungen gut beschreiben können. Nach dem vorherigen Abschnitt ist es nicht überraschend, dass es möglich ist, komplexe Zahlen in der geometrischen Algebra darzustellen. Sie können durch die geraden Grade der zweidimensionalen geometrischen Algebra vollständig beschrieben werden: $\mathbf{g}_n \in G_2^+(\mathbb{R}) \cong \mathbb{C}$. Das bedeutet eine komplexe Zahl \begin{align} - a_0 + a_1 j \cong a_0 + a_1 \mathbf{e}_{12} = \mathbf{g}_n\quad a_0, a_1 \in \mathbb{R}\\ - |r|e^{\theta j} \cong |r|e^{\theta \mathbf{e}_{12}} = \mathbf{g}_n; \quad r, \theta \in \mathbb{R} +a_0 + a_1 j \cong a_0 + a_1 \mathbf{e}_{12} = \mathbf{g}_n\quad a_0, a_1 \in \mathbb{R}\\ +|r|e^{\theta j} \cong |r|e^{\theta \mathbf{e}_{12}} = \mathbf{g}_n; \quad r, \theta \in \mathbb{R} \end{align} -weil $j$ und $\mathbf{e}_{12}$ beide die Eigenschaft besitzen quadriert $-1$ zu ergeben +kann durch ein Skalar (Grad 0) und einem Bivektor (Grad 2) dargestellt werden, weil $j$ und $\mathbf{e}_{12}$ beide die Eigenschaft \begin{align} - j^2 = -1\quad \mathbf{e}_{12}^2 = -1 +j^2 = -1\quad\text{und}\quad\mathbf{e}_{12}^2 = -1 \end{align} -Man beachte, dass wenn wir, wie bei den komplexen Zahlen, Elemente von $G_2^+(\mathbb{R})$ miteinander Multiplizieren, ist es nicht, wie im Kapitel Rotation bei der Formel (\ref{rotGA})beschrieben, eine Multiplikation von zwei $g_n$ mit einem Vektor. Im zweidimensionalen bewirken beide Multiplikationen grundsätzlich das Gleiche (eine Drehstreckung), aber die Multiplikation von mehreren $g_n$ ist kommutativ, wie wir es von den komplexen Zahlen kennen. +besitzen. Die Kommutativität \begin{align} - \mathbf{g}_1\mathbf{g}_2 = \mathbf{g}_2\mathbf{g}_1 \quad&\Leftrightarrow\quad (a + b \mathbf{e}_{12})(f + g \mathbf{e}_{12}) = (f + g \mathbf{e}_{12})(a + b \mathbf{e}_{12})\\ - \mathbf{g}_1\mathbf{v}\not= \mathbf{v}\mathbf{g}_1 \quad&\Leftrightarrow\quad(a + b \mathbf{e}_{12})(x\mathbf{e}_1+y\mathbf{e}_2)\not= (x\mathbf{e}_1+y\mathbf{e}_2)(a + b \mathbf{e}_{12}) +\begin{split} +\mathbf{g}_1\mathbf{g}_2 = \mathbf{g}_2\mathbf{g}_1 \enspace&\Leftrightarrow\enspace (a + b \mathbf{e}_{12})(f + g \mathbf{e}_{12}) = (f + g \mathbf{e}_{12})(a + b \mathbf{e}_{12})\\ &\Leftrightarrow\enspace |\mathbf{g}_1|\,|\mathbf{g}_2|e^{(\theta_{g_1} + \theta_{g_2})\mathbf{e}_{12}} = |\mathbf{g}_2|\,|\mathbf{g}_1|e^{(\theta_{g_2} + \theta_{g_1})\mathbf{e}_{12}}, +\end{split} \end{align} -Um später die Auswirkung der Quaternionen besser zu verstehen, möchte ich kurz darauf eingehen, was ein $g_n$ für eine Auswirkung auf einen Vektor hat. -Wir kennen diesen Effekt schon von den komplexen Zahlen. Wenn eine komplexe Zahl $c_1=a+bj$ mit einer zweiten $c_2=f+gj$ multipliziert wird, dann kann man diese so aufteilen. +welche wir schon von den komplexen Zahlen her kennen, ist dabei eine in der geometrischen Algebra nur selten anzutreffende Eigenschaft. Beispielsweise ist das geometrische Produkt von \begin{align} - c = c_1\cdot c_2 = (a + bj)(d + ej) = f\cdot(a+bj) + gj\cdot(a+bj) +\mathbf{g}_1\mathbf{v}\not= \mathbf{v}\mathbf{g}_1 \quad\Leftrightarrow\quad(a + b \mathbf{e}_{12})(x\mathbf{e}_1+y\mathbf{e}_2)\not= (x\mathbf{e}_1+y\mathbf{e}_2)(a + b \mathbf{e}_{12}) \end{align} -Dabei ist $f\cdot(a+bj)$ die jetzige komplexe Zahl $c_1$ um den Faktor $f$ steckt und $gj\cdot(a+bj)$ die um 90° im Gegenuhrzeigersinn gedrehte Zahl $c_2$ um den Faktor $g$ streckt. Diese Anteile addiert ergeben, dann den um $c_2$ dreh-gestreckten Vektor $c_1$. Die wirklichen Vorteile der geometrischen Algebra werden sich aber erst bei den Quaternionen zeigen.
\ No newline at end of file +und auch die im folgenden Kapitel behandelten Quaternionen sind nicht kommutativ. + +Um später die Auswirkung der Quaternionen auf Vektoren besser zu verstehen, möchten wir kurz darauf eingehen, was ein $\mathbf{g}_n$ für eine Auswirkung auf einen Vektor hat. +Wir kennen diesen Effekt schon von den komplexen Zahlen. Wenn eine komplexe Zahl $c_1=a+bj$ mit einer zweiten $c_2=f+gj$ multipliziert wird, dann kann man +\begin{align} +c = c_1\cdot c_2 = (a + bj)(d + ej) = \underbrace{a\cdot(d+ej)}_{\displaystyle{a\cdot c_2}} + \underbrace{bj\cdot(d+ej)}_{\displaystyle{b\cdot c_2 \cdot (1\angle 90^\circ)}} +\end{align} +so aufteilen. Dabei ist $a\cdot(d+ej)$ die komplexe Zahl $c_2$ um den Faktor $a$ steckt und $bj\cdot(d+ej)$ die um 90° im Gegenuhrzeigersinn gedrehte Zahl $c_2$ um den Faktor $b$ streckt. Diese Anteile addiert ergeben dann den um $c_1$ drehgestreckten Vektor $c_2$. Den gleichen Effekt hat +\begin{align}\label{GAdrehstreck} +\mathbf{v}' = \mathbf{g}\mathbf{v} = (a + b\mathbf{e}_{12})(d\mathbf{e}_{1} + e\mathbf{e}_{2}) = a(d\mathbf{e}_{1} + e\mathbf{e}_{2}) + b\mathbf{e}_{12}(d\mathbf{e}_{1} + e\mathbf{e}_{2}) +\end{align} +in der zweidimensionalen geometrischen Algebra. Im Falle der komplexen Zahlen macht es jetzt noch nicht wirklich Sinn in die geometrische Algebra zu wechseln. Die potenziellen Vorteile der geometrischen Algebra werden sich aber erst bei den Quaternionen zeigen.
\ No newline at end of file diff --git a/buch/papers/clifford/Bilder/ReihenfolgeGimbal.png b/buch/papers/clifford/Bilder/ReihenfolgeGimbal.png Binary files differnew file mode 100644 index 0000000..625757d --- /dev/null +++ b/buch/papers/clifford/Bilder/ReihenfolgeGimbal.png diff --git a/buch/papers/clifford/Bilder/test.png b/buch/papers/clifford/Bilder/test.png Binary files differdeleted file mode 100644 index 1633a2e..0000000 --- a/buch/papers/clifford/Bilder/test.png +++ /dev/null diff --git a/buch/papers/clifford/Makefile.inc b/buch/papers/clifford/Makefile.inc index e168ae8..fe32eba 100644 --- a/buch/papers/clifford/Makefile.inc +++ b/buch/papers/clifford/Makefile.inc @@ -14,7 +14,9 @@ dependencies-clifford = \ papers/clifford/4_GeometrischesProdukt.tex \ papers/clifford/5_PolareDarstellung.tex \ papers/clifford/6_PauliMatrizen.tex \ - papers/clifford/7_Reflektion.tex \ + papers/clifford/7_Spiegelung.tex \ papers/clifford/8_Rotation.tex \ papers/clifford/9_KomplexeZahlen.tex \ - papers/clifford/10_Quaternionen.tex + papers/clifford/10_Quaternionen.tex \ + papers/clifford/11_Fazit.tex + diff --git a/buch/papers/clifford/images/Makefile b/buch/papers/clifford/images/Makefile new file mode 100644 index 0000000..cc621fb --- /dev/null +++ b/buch/papers/clifford/images/Makefile @@ -0,0 +1,13 @@ +# +# Makefile +# +# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +all: spiegelung.pdf + +spiegelung.pdf: spiegelung.tex punkte.tex + pdflatex spiegelung.tex + +punkte.tex: spiegelung.m + octave spiegelung.m + diff --git a/buch/papers/clifford/images/punkte.tex b/buch/papers/clifford/images/punkte.tex new file mode 100644 index 0000000..41d2247 --- /dev/null +++ b/buch/papers/clifford/images/punkte.tex @@ -0,0 +1,21 @@ +\coordinate (A) at (2.300,1.700); +\coordinate (B) at (4.300,2.500); +\coordinate (C) at (2.800,2.700); +\coordinate (S) at (3.133,2.300); +\coordinate (G1) at (0.489,0.873); +\coordinate (G1oben) at (4.886,8.725); +\coordinate (G1unten) at (-4.886,-8.725); +\coordinate (G2) at (0.336,-0.942); +\coordinate (G2oben) at (3.363,-9.417); +\coordinate (G2unten) at (-3.363,9.417); +\coordinate (A1) at (0.248,2.849); +\coordinate (B1) at (-0.115,4.973); +\coordinate (C1) at (0.839,3.798); +\coordinate (S1) at (0.324,3.873); +\coordinate (A2) at (-1.997,2.048); +\coordinate (B2) at (-3.061,3.921); +\coordinate (C2) at (-3.055,2.407); +\coordinate (S2) at (-2.704,2.792); +\def\winkela{60.7512} +\def\winkelb{48.9027} +\coordinate (G) at (0.489,0.873); diff --git a/buch/papers/clifford/images/spiegelung.m b/buch/papers/clifford/images/spiegelung.m new file mode 100644 index 0000000..a086cb5 --- /dev/null +++ b/buch/papers/clifford/images/spiegelung.m @@ -0,0 +1,66 @@ +# +# spiegelung.m +# +# +fn = fopen("punkte.tex", "w"); + + +a = [ 2.3; 1.7 ]; +b = [ 4.3; 2.5 ]; +c = [ 2.8; 2.7 ]; +s = (a + b + c)/3; + +fprintf(fn, "\\coordinate (A) at (%.3f,%.3f);\n", a(1, 1), a(2, 1)); +fprintf(fn, "\\coordinate (B) at (%.3f,%.3f);\n", b(1, 1), b(2, 1)); +fprintf(fn, "\\coordinate (C) at (%.3f,%.3f);\n", c(1, 1), c(2, 1)); +fprintf(fn, "\\coordinate (S) at (%.3f,%.3f);\n", s(1, 1), s(2, 1)); + +n1 = [ -2.5; 1.4 ]; +n1 = n1 / norm(n1); +S1 = eye(2) - 2 * (n1 * n1'); +g1 = [ n1(2,1); -n1(1,1) ]; + +fprintf(fn, "\\coordinate (G1) at (%.3f,%.3f);\n", g1(1,1), g1(2,1)); +fprintf(fn, "\\coordinate (G1oben) at (%.3f,%.3f);\n", 10*g1(1,1), 10*g1(2,1)); +fprintf(fn, "\\coordinate (G1unten) at (%.3f,%.3f);\n", -10*g1(1,1), -10*g1(2,1)); + +n2 = [ 1.4; 0.5 ]; +n2 = n2 / norm(n2); +S2 = eye(2) - 2 * (n2 * n2'); +g2 = [ n2(2,1); -n2(1,1) ]; + +fprintf(fn, "\\coordinate (G2) at (%.3f,%.3f);\n", g2(1,1), g2(2,1)); +fprintf(fn, "\\coordinate (G2oben) at (%.3f,%.3f);\n", 10*g2(1,1), 10*g2(2,1)); +fprintf(fn, "\\coordinate (G2unten) at (%.3f,%.3f);\n", -10*g2(1,1), -10*g2(2,1)); + +D = S2 * S1; + +a1 = S1 * a; +b1 = S1 * b; +c1 = S1 * c; +s1 = S1 * s; + +fprintf(fn, "\\coordinate (A1) at (%.3f,%.3f);\n", a1(1, 1), a1(2, 1)); +fprintf(fn, "\\coordinate (B1) at (%.3f,%.3f);\n", b1(1, 1), b1(2, 1)); +fprintf(fn, "\\coordinate (C1) at (%.3f,%.3f);\n", c1(1, 1), c1(2, 1)); +fprintf(fn, "\\coordinate (S1) at (%.3f,%.3f);\n", s1(1, 1), s1(2, 1)); + +a2 = D * a; +b2 = D * b; +c2 = D * c; +s2 = D * s; + +fprintf(fn, "\\coordinate (A2) at (%.3f,%.3f);\n", a2(1, 1), a2(2, 1)); +fprintf(fn, "\\coordinate (B2) at (%.3f,%.3f);\n", b2(1, 1), b2(2, 1)); +fprintf(fn, "\\coordinate (C2) at (%.3f,%.3f);\n", c2(1, 1), c2(2, 1)); +fprintf(fn, "\\coordinate (S2) at (%.3f,%.3f);\n", s2(1, 1), s2(2, 1)); + +winkel1 = atan2(g1(2,1), g1(1,1)) * (180 / pi); +winkel2 = acosd(g1' * g2); + +fprintf(fn, "\\def\\winkela{%.4f}\n", winkel1); +fprintf(fn, "\\def\\winkelb{%.4f}\n", 180 - winkel2); + +fprintf(fn, "\\coordinate (G) at (%.3f,%.3f);\n", g1(1,1), g1(2,1)); + +fclose(fn); diff --git a/buch/papers/clifford/images/spiegelung.pdf b/buch/papers/clifford/images/spiegelung.pdf Binary files differnew file mode 100644 index 0000000..a17d369 --- /dev/null +++ b/buch/papers/clifford/images/spiegelung.pdf diff --git a/buch/papers/clifford/images/spiegelung.tex b/buch/papers/clifford/images/spiegelung.tex new file mode 100644 index 0000000..0960456 --- /dev/null +++ b/buch/papers/clifford/images/spiegelung.tex @@ -0,0 +1,85 @@ +% +% spiegelung.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math,calc} +\begin{document} +\def\skala{1.1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\definecolor{darkgreen}{rgb}{0,0.6,0} + +\def\punkt#1{ +\fill #1 circle[radius=0.06]; +} + +\coordinate (M) at (0,0); + +\fill[color=blue] (M) circle[radius=0.06]; +\node[color=blue] at (M) [left] {$M$}; + +\input{punkte.tex} + +\fill[color=red!30] (A) -- (B) -- (C) -- cycle; +\draw[color=red] (A) -- (B) -- (C) -- cycle; +\node at (A) [below] {$A$}; +\node at (B) [above right] {$B$}; +\node at (C) [above] {$C$}; +\node at (S) {$\circlearrowleft$}; + +\fill[color=red!30] (A1) -- (B1) -- (C1) -- cycle; +\draw[color=red] (A1) -- (B1) -- (C1) -- cycle; +\node at (A1) [below] {$A'$}; +\node at (B1) [above] {$B'$}; +\node at (C1) [above right] {$C'$}; +\node at (S1) {$\circlearrowright$}; + +\fill[color=red!30] (A2) -- (B2) -- (C2) -- cycle; +\draw[color=red] (A2) -- (B2) -- (C2) -- cycle; +\node at (A2) [below] {$A''$}; +\node at (B2) [above] {$B''$}; +\node at (C2) [left] {$C''$}; +\node at (S2) {$\circlearrowleft$}; + +\draw[color=gray,dotted] (A) -- (A1); +\draw[color=gray,dotted] (B) -- (B1); +\draw[color=gray,dotted] (C) -- (C1); + +\draw[color=gray,dotted] (A1) -- (A2); +\draw[color=gray,dotted] (B1) -- (B2); +\draw[color=gray,dotted] (C1) -- (C2); + +\punkt{(A)} +\punkt{(B)} +\punkt{(C)} +\punkt{(A1)} +\punkt{(B1)} +\punkt{(C1)} +\punkt{(A2)} +\punkt{(B2)} +\punkt{(C2)} + +\fill[color=darkgreen!30] (M) -- (G1) arc ({\winkela}:{\winkela+\winkelb}:1) -- cycle; +\draw[color=darkgreen] (G1) arc ({\winkela}:{\winkela+\winkelb}:1); +\node[color=darkgreen] at ({\winkela+0.5*\winkelb}:0.7) {$\alpha$}; + +\node at ($6*(G1)$) [right] {$g\mathstrut$}; +\node at ($-5.6*(G2)$) [left] {$h\mathstrut$}; + +\clip (-3,-0.2) rectangle (4.5,5.5); + +\draw[line width=1pt] (G1oben) -- (G1unten); +\draw[line width=1pt] (G2oben) -- (G2unten); + +\fill[color=blue] (M) circle[radius=0.06]; + +\end{tikzpicture} +\end{document} + diff --git a/buch/papers/clifford/main.tex b/buch/papers/clifford/main.tex index ec44963..3649b20 100644 --- a/buch/papers/clifford/main.tex +++ b/buch/papers/clifford/main.tex @@ -16,10 +16,11 @@ \input{papers/clifford/4_GeometrischesProdukt.tex} \input{papers/clifford/5_PolareDarstellung.tex} \input{papers/clifford/6_PauliMatrizen.tex} -\input{papers/clifford/7_Reflektion.tex} +\input{papers/clifford/7_Spiegelung.tex} \input{papers/clifford/8_Rotation.tex} \input{papers/clifford/9_KomplexeZahlen.tex} \input{papers/clifford/10_Quaternionen.tex} +\input{papers/clifford/11_Fazit.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/clifford/references.bib b/buch/papers/clifford/references.bib index ff829d6..9090005 100644 --- a/buch/papers/clifford/references.bib +++ b/buch/papers/clifford/references.bib @@ -4,32 +4,13 @@ % (c) 2020 Autor, Hochschule Rapperswil % -@online{clifford:bibtex, - title = {BibTeX}, - url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} -} - -@book{clifford:numerical-analysis, - title = {Numerical Analysis}, - author = {David Kincaid and Ward Cheney}, - publisher = {American Mathematical Society}, - year = {2002}, - isbn = {978-8-8218-4788-6}, - inseries = {Pure and applied undegraduate texts}, - volume = {2} -} - -@article{clifford:mendezmueller, - author = { Tabea Méndez and Andreas Müller }, - title = { Noncommutative harmonic analysis and image registration }, - journal = { Appl. Comput. Harmon. Anal.}, - year = 2019, - volume = 47, - pages = {607--627}, - url = {https://doi.org/10.1016/j.acha.2017.11.004} +@article{clifford:hestenes_GA, + author = { David Hestenes, Garret Eugene Sobczyk and James S. Marsh }, + title = { Clifford Algebra to Geometric Calculus. A Unified Language for Mathematics and Physics }, + journal = { American Journal of Physics }, + year = 1985, + volume = 53, + pages = {24}, + url = {https://www.researchgate.net/publication/258944244_Clifford_Algebra_to_Geometric_Calculus_A_Unified_Language_for_Mathematics_and_Physics} } diff --git a/buch/papers/erdbeben/teil0.tex b/buch/papers/erdbeben/teil0.tex index afa1244..d32b316 100644 --- a/buch/papers/erdbeben/teil0.tex +++ b/buch/papers/erdbeben/teil0.tex @@ -77,7 +77,7 @@ Deshalb nehmen wir $f$ als dritte Grösse in den Zustandsvektor auf und definier \[ x = (s_1, s_2, f)^T. - \] +\] Für die Standard-Form $\dot x = Ax$ brauchen wir als nächstes die Ableitungen aller Elemente von $x$. Für $\dot s_1$ und $\dot s_2$ folgen diese direkt aus Gleichung (20.1), aber über $\dot f$ wissen wir nichts. Wir müssen also eine Annahme treffen: $\dot f = 0$. Diese Annahme ist im Allgemeinen falsch, aber etwas Besseres haben wir zurzeit nicht zur Verfügung. diff --git a/buch/papers/ifs/images/FIC.pdf b/buch/papers/ifs/images/FIC.pdf index 1c76dfe..525a857 100644 --- a/buch/papers/ifs/images/FIC.pdf +++ b/buch/papers/ifs/images/FIC.pdf @@ -1,7 +1,7 @@ %PDF-1.6
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diff --git a/buch/papers/ifs/images/Makefile b/buch/papers/ifs/images/Makefile new file mode 100644 index 0000000..c6d3fb5 --- /dev/null +++ b/buch/papers/ifs/images/Makefile @@ -0,0 +1,9 @@ +# +# Makefile +# +# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +chaosspiel.pdf: chaosspiel.tex \ + farnnotweight-eps-converted-to.pdf \ + farnrightwight-eps-converted-to.pdf + pdflatex chaosspiel.tex diff --git a/buch/papers/ifs/images/chaosspiel.pdf b/buch/papers/ifs/images/chaosspiel.pdf Binary files differnew file mode 100644 index 0000000..23f0dd2 --- /dev/null +++ b/buch/papers/ifs/images/chaosspiel.pdf diff --git a/buch/papers/ifs/images/chaosspiel.tex b/buch/papers/ifs/images/chaosspiel.tex new file mode 100644 index 0000000..7c69ad3 --- /dev/null +++ b/buch/papers/ifs/images/chaosspiel.tex @@ -0,0 +1,37 @@ +% +% tikztemplate.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +% add image content here + +\begin{scope}[xshift=-3.6cm] +%\clip (-3.3,-3) rectangle (3.3,3); +\node at (0,0) { +\includegraphics[width=6.8cm]{farnnotweight-eps-converted-to.pdf} +}; +\node at (0.2,-5.7) {(a)}; +\end{scope} + +\begin{scope}[xshift=3.6cm] +%\clip (-3.3,-3) rectangle (3.3,3); +\node at (0,0) { +\includegraphics[width=6.8cm]{farnrightwight-eps-converted-to.pdf} +}; +\node at (0.2,-5.7) {(b)}; +\end{scope} + +\end{tikzpicture} +\end{document} + diff --git a/buch/papers/ifs/images/farnnotweight-eps-converted-to.pdf b/buch/papers/ifs/images/farnnotweight-eps-converted-to.pdf Binary files differindex 35bff32..f5e4093 100644 --- a/buch/papers/ifs/images/farnnotweight-eps-converted-to.pdf +++ b/buch/papers/ifs/images/farnnotweight-eps-converted-to.pdf diff --git a/buch/papers/ifs/images/farnrightwight-eps-converted-to.pdf b/buch/papers/ifs/images/farnrightwight-eps-converted-to.pdf Binary files differindex 3652e8f..fa69d77 100644 --- a/buch/papers/ifs/images/farnrightwight-eps-converted-to.pdf +++ b/buch/papers/ifs/images/farnrightwight-eps-converted-to.pdf diff --git a/buch/papers/ifs/teil0.tex b/buch/papers/ifs/teil0.tex index 833748c..af2105e 100644 --- a/buch/papers/ifs/teil0.tex +++ b/buch/papers/ifs/teil0.tex @@ -5,7 +5,7 @@ % \section{Einleitung \label{ifs:section:teil0}} \rhead{Was ist ein Iteriertes Funktionsschema} -Mit der Hilfe von Iterierten Funktionsschemata (IFS) kann mit nur wenigen affinen Funktionen, komplexe Bilder beschreiben werden. +Mit der Hilfe von Iterierten Funktionsschemata (IFS) können mit nur wenigen affinen Funktionen komplexe Bilder beschrieben werden. In der Regel sind diese Bilder Fraktale. Wie es dazu kommt, und wie man mit IFS auch Bilder komprimieren kann, wollen wir in diesem Kapitel untersuchen. diff --git a/buch/papers/ifs/teil1.tex b/buch/papers/ifs/teil1.tex index a75b529..caba120 100644 --- a/buch/papers/ifs/teil1.tex +++ b/buch/papers/ifs/teil1.tex @@ -7,29 +7,27 @@ \label{ifs:section:teil1}} \rhead{Problemstellung} Bevor wir die IFS ansehen, schauen wir uns Fraktale genauer an. - - Über die genaue Definition von Fraktalen sind sich die Mathematiker nicht einig. -In diesem Kapitel orientieren wir uns an den Eigenschaften welche Kenneth Falconer in seinem Buch Fractal Geometry \cite{ifs:fractal-geometry} beschreibt. +In diesem Kapitel orientieren wir uns an den Eigenschaften, welche Kenneth Falconer in seinem Buch {\em Fractal Geometry} \cite{ifs:fractal-geometry} beschreibt. Von einem Fraktal $F$ können wir folgende Eigenschaften erwarten: \begin{enumerate} \item $F$ hat eine unendlich feine Struktur \item $F$ kann nicht mit der klassischen Geometrie beschrieben werden. \item Oftmals hat $F$ eine Form von Selbstähnlichkeit. - \item Die 'fraktale Dimension' ist grösser als die topologische Dimension + Man spricht von einer selbstähnlichen Menge, wenn sich diese Menge überdecken lässt mit echten Teilmengen, die zur ganzen Menge ähnlich sind. + \item Die `fraktale Dimension' ist grösser als die topologische Dimension. \item Viele Fraktale lassen sich auf eine simple Art definieren. Es genügen zum Beispiel nur wenige Funktionen, welche rekursiv ausgeführt werden, um ein Fraktal zu definieren. \end{enumerate} \subsection{Koch Kurve \label{ifs:subsection:lilkoch}} Diese Eigenschaften möchten wir nun am Beispiel der Koch Kurve näher anschauen. -In Abbildung \ref{ifs:kochkurve8} sehen wir die Koch Kurve. Sie besteht aus lauter kleineren Kopien von sich selber. -Den Konstruktionsvorgang ist in Abbildung \ref{ifs:kochconst} dargestellt. +In Abbildung \ref{ifs:kochkurve8} sehen wir die Koch Kurve. Sie besteht aus lauter kleineren Kopien von sich selbst. +Der Konstruktionsvorgang ist in Abbildung \ref{ifs:kochconst} dargestellt. Gestartet wird mit einer einzelnen Strecke der Länge $a$. Diese wird in ersten Schritt durch vier gleich langen Streckenabschnitte der Länge $\frac{a}{3}$ ersetzt. In \ref{ifs:kochconstb} ist die Anordnung dieser vier Streckenabschnitte ersichtlich. Dieser Schritt wird nun für jeden der resultierten Streckenabschnitten wiederholt. Die Kurve besteht also aus vier kleineren Kopien der ganzen Kurve, was auch unter Selbstähnlichkeit bekannt ist. -Man spricht von einer selbstähnlichen Menge, wenn sich diese Menge überdecken lässt mit echten Teilmengen, die zur ganzen Menge ähnlich sind. \begin{figure} @@ -66,7 +64,7 @@ berechnen. In jedem Schritt wird die Länge um den Faktor $\frac{4}{3}$ verlängert. Daraus resultiert, dass die Länge gegen $\infty$ divergiert. -Die Fläche unter der Kurve lässt sich folgendermassen berechnen +Die Fläche zwischen der Strecke von $O$ nach $(1,0)$ und der Kurve lässt sich folgendermassen berechnen \begin{align*} A_0 &= 0 \\ A_1 &= \left( \frac{a}{3}\right)^2 \frac{\sqrt{3}}{4} = a^2 \frac{\sqrt{3}}{36}\\ @@ -88,22 +86,22 @@ Wie wir sehen ist die Koch-Kurve ein Objekt mit endlicher Fläche, aber unendlic Zu guter Letzt bestimmen wir die Dimension der Kurve. Es gibt viele verschiedene Methoden die Dimension zu definieren. Diese können dann auch unterschiedliche Resultate liefern. Vor allem im Zusammenhang mit Fraktalen findet man in der Literatur unterschiedliche Arten. -In diesem Beispiel werden wir die Ähnlichkeits-Dimension \cite{ifs:fractal-geometry}. +Da die Kochsche Kurve selbstähnlich ist, ist die Ähnlichkeits-Dimension \cite{ifs:fractal-geometry} die angemessene Messzahl für die Dimension. Die Ähnlichkeits-Dimension $D$ ist das Verhältnis der Logarithmen der Anzahl Kopien $N$ des Originales und deren Skalierungsfaktor $\epsilon$ \begin{align*} D = - \frac{\log N}{\log \epsilon }. \end{align*} -Mit ihr kann man einfach die Dimension selbstähnlicher Mengen bestimmen. -Als Beispiel nehmen wir ein gleichseitiges Dreieck. Dieses besteht aus $N = 4$ Kopien mit halber ($\epsilon = 1/2$) Kantenlänge $l$, Abbildung \ref{ifs:trinagle}. +Die Ähnlichkeits-Dimension stimmt für viele gewöhnliche Geometrische Objekte mit der intuitiven Vorstellung von Dimension überein. +Zum Beispiel besteht ein Dreieck aus $N = 4$ Kopien mit halber ($\epsilon = 1/2$) Kantenlänge $l$, Abbildung \ref{ifs:trinagle}. Somit hat das Dreieck die Dimension $D = 2$. Die Koch Kurve besteht aus $N = 4$ Kopien mit Kantenlänge $\epsilon =l \cdot 1/3$. Ihre Ähnlichkeits-Dimension ist somit \begin{align*} D = - \frac{\log N }{\log \epsilon } = - \frac{\log 4 }{\log 1/3 } \approx 1.2619. \end{align*} -Wie wir nun sehen besitzt die Koch-Kurve alle oben beschriebenen Eigenschaften von Fraktalen. -Dies muss jedoch nicht bei allen Fraktalen der Fall. Sonst wäre die Frage nach einer 'richtigen' Definition einfach zu beantworten. +Wie wir nun sehen, besitzt die Koch-Kurve alle oben beschriebenen Eigenschaften von Fraktalen. +Dies muss jedoch nicht bei allen Fraktalen der Fall sein. Sonst wäre die Frage nach einer `richtigen' Definition einfach zu beantworten. \begin{figure} \centering \begin{tikzpicture} diff --git a/buch/papers/ifs/teil2.tex b/buch/papers/ifs/teil2.tex index fd10634..d0110ed 100644 --- a/buch/papers/ifs/teil2.tex +++ b/buch/papers/ifs/teil2.tex @@ -6,8 +6,9 @@ \section{Fraktale mit IFS \label{ifs:section:teil2}} \rhead{Teil 2} -Wollen wir nun eine bestimmte Art anschauen, wie man Fraktale machen kann. -Zur Veranschaulichung dieser Methode nehmen wir das Sierpinski Dreieck. +Wollen wir nun eine bestimmte Art anschauen, wie man Fraktale erzeugen kann. +Im Beispiel auf Seite \pageref{ifs:trinagle} haben wir ein Dreieck aus 4 skalierten Kopien zusammengefügt. +Lässt man die Kopie im Zentrum des Dreiecks weg, entsteht die Grundlage des sogenannten Sierpinski-Dreieck in Abbildung \ref{ifs:sierpinski10}. \begin{figure} \centering \includegraphics[width=0.5\textwidth]{papers/ifs/images/sierpinski} @@ -92,21 +93,22 @@ Man kann sogar noch einen Schritt weiter gehen, und sagen: Wenn wir die Funktion \label{ifs:sierpconst} \end{figure} Im Beispiel der Abbildung \ref{ifs:sierpconst} sehen wir, wie das Bild nach jeder Iteration dem Sierpinski-Dreieck ähnlicher wird. -Der Abstand zum Original wird immer kleiner, und konvergiert gegen null. +Der `Abstand' zum Original wird immer kleiner, und konvergiert gegen null. \subsection{Iterierte Funktionensysteme \label{ifs:subsection:IteratedFunktionensysteme}} In diesem Abschnitt wollen wir die Erkenntnis, wie wir aus einer beliebigen Menge ein Sierpinski-Dreieck generieren können, verallgemeinern. -$S_1,\dots,S_n$ sind Kontraktionen auf die Menge $D \subset \mathbb{R}^n$. Es gilt +$S_1,\dots,S_n$ sind Kontraktionen auf einer Menge $D \subset \mathbb{R}^n$. Es gilt \begin{align} |S_i(x) - S_i(y)| \leq c_i|x - y| \end{align} für jedes i mit einem $c_i < 1$. -Der Banachsche Fixpunktsatz besagt, dass für solche Kontraktionen ein Eindeutiges $A$ existiert, für das $S(A) = A$ gilt. +Man kann zeigen, dass für solche Kontraktionen ein eindeutiges $A$ existiert, für das $S_i(A) = A$ gilt. Den Beweis kann man in \cite{ifs:Rousseau2012} nachlesen. -Hat man nicht nur eine sondern mehrere Kontraktionen, dann existiert eine eindeutige kompakte Menge $F$ für die gilt + +Hat man nicht nur eine sondern mehrere Kontraktionen, dann existiert eine eindeutige kompakte Menge $F$, für die gilt \begin{equation} F = \bigcup\limits_{i = 1}^{m} S_i(F). \end{equation} @@ -115,17 +117,17 @@ Weiter definieren wir die Transformation S auf kompakte Mengen $E$ ohne die leer S(E) = \bigcup\limits_{i = 1}^m S_i(E). \label{ifs:transformation} \end{equation} -Wird diese Transformation Iterativ ausgeführt, das heisst $S^0(E) = E, S^k(E) = S(S^{k-1}(E))$, gilt +Wird diese Transformation iterativ ausgeführt, das heisst $S^0(E) = E, S^k(E) = S(S^{k-1}(E))$, gilt \begin{equation} F = \bigcap\limits_{k = 1}^{\infty} S^k(E). \label{ifs:ifsForm} \end{equation} -In Worte gefasst bedeutet das, dass jede Gruppe von Kontraktionen iterativ ausgeführt, gegen eine eindeutige Menge konvergiert. +In Worte gefasst bedeutet das, dass jede Gruppe von Kontraktionen iterativ ausgeführt gegen eine eindeutige Menge konvergiert. Diese Menge ist auch als Attraktor eines IFS bekannt. Der Beweis für die Existenz eines eindeutigen Attraktors ist in \cite{ifs:fractal-geometry} beschrieben. \subsection{Beispiel: Barnsley-Farn} -Der Barnsley-Farn, Abbildung \ref{ifs:farn}, ist ein Beispiel eines Fraktal, welches mit einem IFS generiert werden kann. +Der Barnsley-Farn, Abbildung \ref{ifs:farn}, ist ein Beispiel eines Fraktals, welches mit einem IFS generiert werden kann. Wie man schnell erkennen kann, besteht der Farn aus Blättern, welche eine grosse Ähnlichkeit zum ganzen Farn haben. Die vier affinen Transformationen \begin{align} @@ -153,7 +155,7 @@ Die vier affinen Transformationen \begin{pmatrix} 0 \\ 1.6 - \end{pmatrix}\\ + \end{pmatrix},\\ & {S_3(x,y)} = \begin{pmatrix} @@ -183,25 +185,25 @@ Die vier affinen Transformationen \begin{pmatrix} 0 \\ 0.44 - \end{pmatrix}\\ + \end{pmatrix},\\ \label{ifs:farnFormel} \end{align} -, welche für die konstruktion des Farns benötigt werden sind in der Abbildung \ref{ifs:farncolor} farblich dargestellt. +welche für die Konstruktion des Farns benötigt werden, sind in der Abbildung \ref{ifs:farncolor} farblich dargestellt. Das gesamte Farnblatt ist in der schwarzen Box. -Auf diese werden die Transformationen angewendet +Auf diese werden die Transformationen angewendet. $S_1$ erstellt den Stiel des Farnblattes (rot). -Die Transformation bildet das Gesamte Blatt auf die Y-Achse ab. +Die Transformation bildet das gesamte Blatt auf die $y$-Achse ab. $S_2$ (grün) erstellt den Hauptteil des Farnes. Sie verkleinert und dreht das gesamte Bild und stellt es auf das Ende des Stiels aus $S_1$. -$S_3$ bildet das gesamte Blatt auf das blaue Teilblatt unten Links ab. +$S_3$ bildet das gesamte Blatt auf das blaue Teilblatt unten links ab. $S_4$ spiegelt das Blatt und bildet es auf das magentafarbene Teilblatt ab. \subsection{Erzeugung eines Bildes zu einem IFS} -Es gibt zwei verschiedene Methoden um das Bild zu einem IFS zu erzeugen. +Es gibt zwei verschiedene Methoden, um das Bild zu einem IFS zu erzeugen. Die erste Methode ist wahrscheinlich die intuitivste. -Wir beginnen mit einm Startbild, zum Beispiel ein Schwarzes Quadrat, und bilden dieses mit den affinen Transformationen des IFS ab. -Das neue Bild, dass entsteht, ist die nächste Iterierte. +Wir beginnen mit einem Startbild, zum Beispiel ein schwarzes Quadrat, und bilden dieses mit den affinen Transformationen des IFS ab. +Das neue Bild, das entsteht, ist die nächste Iterierte. Dieses wird wieder mit den Transformationen abgebildet. -Wir wiederholen den letzten schritt, bis wir zufrieden mit der neusten Iterierten sind. +Wir wiederholen den letzten Schritt, bis wir zufrieden mit der neusten Iterierten sind. Diesen Vorgang haben wir beim Sierpinski-Dreieck in Abbildung \ref{ifs:sierpconst} gebraucht. In Abbildung \ref{ifs:sierpinski10} ist die zehnte Iterierte zu sehen. @@ -213,11 +215,12 @@ Bis jetzt wurde immer davon gesprochen, die Transformationen auf die gesamte Men Bei komplizierteren IFS welche viele Iterationen brauchen, bis man den Attraktor erkennen kann, ist die erste Methode ziemlich rechenintensiv. Beim Chaosspiel werden die Transformationen nicht auf die Menge angewendet, sondern nur auf einen einzelnen Punkt. Der Startpunkt kann dabei ein beliebiger Punkt in $E$ sein. -Es wird bei jedem Iterationsschritt nur eine Transformation, welche zufällig gewählt wurde, angewendet. +Es wird bei jedem Iterationsschritt nur eine Transformation $S_i$, welche zufällig gewählt wurde, angewendet. + Da, wie wir beim Barnsley-Farn gut sehen, nicht jede Transformation gleich viel des Bildes ausmacht, werden diese beim Chaosspiel gewichtet. -Je mehr eine Transformation kontrahiert, desto weniger Punkte braucht es um die resultierende Teilabbildung darzustellen. -Im Fall des Barnsley-Fern wird $S_1$ in $1\%$, $S_2$ in $85\%$ und $S_3 \& S_4$ in $7\%$ der Iterationen ausgeführt. -Wir sehen auch in Abbildung \ref{ifs:farncolor} gut, dass der rote Stiel, $S_1$, einiges weniger Punkte braucht als der grüne Hauptteil des Blattes, $S_2$. +Je mehr eine Transformation kontrahiert, desto weniger Punkte braucht es, um die resultierende Teilabbildung darzustellen. +Im Fall des Barnsley-Farns wird $S_1$ in $1\%$, $S_2$ in $85\%$ und $S_3$ und $S_4$ in $7\%$ der Iterationen ausgeführt. +Wir sehen auch in Abbildung \ref{ifs:farncolor} gut, dass der rote Stiel, $S_1$, viel weniger Punkte braucht als der grüne Hauptteil des Blattes, $S_2$. In Abbildung \ref{ifs:farnNoWeight} wurden die vier gleich stark gewichtet. Man sieht, dass trotzt gleich vieler Iterationen wie in Abbildung \ref{ifs:farn}, der Farn nicht so gut abgebildet wird. @@ -245,12 +248,13 @@ In jeder Kopie des ganzen Farns fehlen die Punkte für dieses rechte untere Teil \begin{figure} \centering - \subfigure[]{ - \label{ifs:farnNoWeight} - \includegraphics[width=0.45\textwidth]{papers/ifs/images/farnnotweight}} - \subfigure[]{ - \label{ifs:farnrightWeight} - \includegraphics[width=0.45\textwidth]{papers/ifs/images/farnrightwight}} + \includegraphics{papers/ifs/images/chaosspiel.pdf} + %\subfigure[]{ + % \label{ifs:farnNoWeight} + % \includegraphics[width=0.45\textwidth]{papers/ifs/images/farnnotweight}} + %\subfigure[]{ + % \label{ifs:farnrightWeight} + % \includegraphics[width=0.45\textwidth]{papers/ifs/images/farnrightwight}} \caption{(a) Chaosspiel ohne Gewichtung (b) $S_4$ zu wenig gewichtet} \label{ifs:farnweight} \end{figure} diff --git a/buch/papers/ifs/teil3.tex b/buch/papers/ifs/teil3.tex index 78fb935..cebb664 100644 --- a/buch/papers/ifs/teil3.tex +++ b/buch/papers/ifs/teil3.tex @@ -6,32 +6,31 @@ \section{Fraktale Bildkomprimierung \label{ifs:section:teil3}} \rhead{Fraktale Bildkomprimierung} -Mit dem Prinzip dieser IFS ist es auch möglich Bilder zu Komprimieren. -Diese Idee hatte der Mathematiker Michael Barnsley, welcher mit seinem Buch Fractals Everywhere einen wichtigen Beitrag zum Verständnis von Fraktalen geliefert hat. -Das Ziel ist es ein IFS zu finden, welches das Bild als Attraktor hat. +Mit dem Prinzip dieser IFS ist es auch möglich, Bilder zu komprimieren. +Diese Idee hatte der Mathematiker Michael Barnsley, welcher mit seinem Buch {\em Fractals Everywhere} einen wichtigen Beitrag zum Verständnis von Fraktalen geliefert hat. +Das Ziel ist, ein IFS zu finden, welches das Bild als Attraktor hat. In diesem Unterkapitel wollen wir eine Methode dafür anschauen, wie sie in \cite{ifs:Rousseau2012} beschrieben ist. Es ist wohl nicht falsch zu sagen, dass Ähnlichkeiten zur gesamten Menge, wie wir sie zum Beispiel beim Barnsley Farn gesehen haben, bei Bilder aus dem Alltag eher selten anzutreffen sind. Ein IFS, wie wir es in \ref{ifs:subsection:IteratedFunktionensysteme} definiert haben, wird uns also nicht weiter helfen. -Die Lösung dazu sind Partitionierte IFS (PIFS) \cite{ifs:pifs}. +Anders sieht es mit partitionierten IFS (PIFS) \cite{ifs:pifs} aus. + In \ref{ifs:transformation} wurde definiert, dass die Kontraktionen $S_i$ bei IFS auf die gesamte Menge $E$ angewendet werden. Bei einem PIFS wird der Attraktor in disjunkte Teilmengen aufgeteilt. Für jede dieser Teilmengen $R_i$ braucht es dann eine grössere Teilmenge, welche mit einer affinen Transformation eine zu $R_i$ ähnliche Menge bildet. -Wir müssen nicht mehr Ähnlichkeiten zum ganzen Bild finden, sondern zwischen Teilen des Bildes. +Wir müssen nicht mehr Ähnlichkeiten zum ganzen Bild finden, sondern nur zwischen Teilen des Bildes. Doch wie finden wir das PIFS, welches das Bild als Attraktor hat? -\subsection{das Kompressionsverfahren +\subsection{Das Kompressionsverfahren \label{ifs:subsection:malorum}} Wir beschränken das Verfahren für Graustufenbilder. Wie das Verfahren für Farbbilder verwendet werden kann, wird später erläutert. -Ein Graustufenbild kann man als Pixelraster mit einer x und y Achse verstehen. +Ein Graustufenbild kann man als Pixelraster mit einer $x$ und $y$ Achse verstehen. Jedem dieser Pixel wird ein Grauwert zugeordnet. -Ein Bild ist also eine Funktion, die jedem Pixel einen Grauwert $z$ zuweist -\begin{align*} - z = f(x,y). -\end{align*} +Ein Bild ist also eine Funktion, die jedem Pixel einen Grauwert \(z = f(x,y)\) zuweist. + +Wir suchen ein PIFS, welches das zu komprimierende Bild als Attraktor hat. +In einem ersten Schritt teilen wir das Bild in disjunkte benachbarte $b \times b$ Pixel-Quadrate auf. Diese Blöcke nennen wir Range-Blöcke der Menge $R=\{R_0,R_1,...R_m\}$. Diese sind als Raster im rechten Bild der Abbildung \ref{ifs:FIC} dargestellt. -Wir suchen ein PIFS welches das zu komprimierende Bild als Attraktor hat. -In einem ersten Schritt teilen wir das Bild in disjunkte benachbarte $b \times b$ Pixel-Quadrate auf. Diese Blöcke nennen wir Range-Blöcke der Menge $R=\{R_0,R_1,...R_m\}$ Im nächsten Schritt teilen wir das Bild in alle möglichen $2b \times 2b$ Pixel-Quadrate auf. Diese sind die Domain-Blöcke der Menge $D = \{D_0,D_1,...D_n\}$. Im dritten und letzten Schritt wird für jeden Range-Block $R_i$ ein Domain-Block $D_j$ gesucht, welcher ihm am ähnlichsten ist. Zwei Beispiele wie solche Domain-, und Range-Block Paare aussehen können, sehen wir in Abbildung \ref{ifs:FIC} @@ -57,8 +56,10 @@ Zuerst brauchen wir die Transformation g_i \end{pmatrix} \end{align*} -um ein Element aus $D$ auf ein Element von $R$ Abzubilden. -Wenn wir die Grauwerte ausser acht lassen, haben wir die affine Abbildung +um ein Element aus $D$ auf ein Element von $R$ abzubilden. +Das bestimmen der besten Transformation kann man in drei Schritte aufteilen. + +\textbf{Schritt 1: }Wenn wir die Grauwerte ausser acht lassen, haben wir die affine Abbildung \begin{align} t_i(x,y) = \begin{pmatrix} @@ -83,39 +84,47 @@ Wir sind auf folgende acht Abbildungen beschränkt: \item Drehung um 90, 180 oder 270 Grad. \item Spiegelung an der vertikalen, horizontalen und den Diagonalachsen. \end{itemize} -Da wir ein $2b \times 2b$ Feld auf ein $b \times b$ Feld abbilden möchten, müssen wir zuerst $G_j$ um $1/2$ skalieren. -Dies erreichen wir, indem wir alle disjunkten $2 \times 2$ px Blöcke mit einem Pixel des Grautones deren Mittelwertes ersetzen. +Da wir ein $2b \times 2b$ Feld auf ein $b \times b$ Feld abbilden möchten, müssen wir zuerst $D_j$ um $1/2$ skalieren. +Dies erreichen wir, indem wir alle disjunkten $2 \times 2$ Pixel Blöcke mit einem Pixel des Grautones deren Mittelwertes ersetzen. - -Die Parameter $s_i$ und $g_i$ beschreiben die Änderung des Grautones. $s$ verändert den Kontrast und $g$ verschiebt die Grautöne auf die richtige Helligkeit, sie bilden die lineare Funktion +\textbf{Schritt 2: }Es muss nicht nur eine geometrische Abbildung, sondern auch eine Abbildung für die Grautöne gewählt werden. Letztere lässt sich mit den Parametern $s_i$ und $g_i$ beschrieben. +Wir suchen einen linearen Zusammenhang zwischen den Grautönen des Domain-, und Range-Block. $s_i$ verändert den Kontrast und $g_i$ verschiebt die Grautöne auf die richtige Helligkeit, sie bilden die lineare Funktion \begin{align*} z' = s_i z + g_i. \end{align*} Für die Bestimmung dieser Parameter führen wir zuerst die Bildfunktionen $f_{R_i}$ und $\tilde{f_{R_i}}$ ein. -$f_{R_i}$ ist die Bildfunktion des Range-Blockes $R_i$ und $\tilde{f_{R_i}}$ ist die Bildfunktion des zuerst Skalierten und dann mit \ref{ifs:affTrans} transformierten Domain-Blocks $D_j$. +$f_{R_i}$ ist die Bildfunktion des Range-Blockes $R_i$ und $\tilde{f_{R_i}}$ ist die Bildfunktion des zuerst skalierten und dann mit \eqref{ifs:affTrans} transformierten Domain-Blocks $D_j$. -Wir suchen $s_i$ und $g_i$ so das +Wir suchen $s_i$ und $g_i$ so das der quadratische Abstand zwischen \begin{align*} - f_{R_i} = s_i \tilde{f_{R_i}} + g_i = \bar{f_{R_i}}. + \bar{f_{R_i}} = s_i \tilde{f_{R_i}} + g_i \end{align*} -Die Parameter lassen sich mit +und $f_{R_i}$ am kleinsten ist. +Dies ist ein klassisches Problem der linearen Regression. Die Parameter lassen sich mit \begin{align*} - s = \frac{\operatorname{cov}(f_{R_i}), f(\tilde{f_{R_i}}))}{\operatorname{var}(\tilde{f_{R_i}})} \\ - g = E(f_{R_i}) - s E(f(\tilde{f_{R_i}})) + s_i = \frac{\operatorname{cov}(f_{R_i}, \tilde{f_{R_i}})}{\operatorname{var}(\tilde{f_{R_i}})} \\ + g_i = E(f_{R_i}) - s E(\tilde{f_{R_i}}) \end{align*} berechnen. +Die Varianz und Kovarianz erstrecken sich über die Grauwerte der Pixel der Blöcke. Mit diesen Parametern haben wir nun die Transformation vollständig bestimmt. -Um zu beurteilen wie ähnlich der Domain-Block $D_j$ mit der gefundenen Transformation $T$ dem Range-Block ist, berechnet man den quadratischen Abstand + +Um zu beurteilen wie ähnlich der Domain-Block $D_j$ mit der gefundenen Transformation $T$ dem Range-Block ist, berechnet man den quadratischen Fehler \begin{align*} e = d(f_{R_i}, \bar{f_{R_i}}). \end{align*} -Dieser Abstand sollte so klein wie möglich sein. +$e$ sollte so klein wie möglich sein. + +\textbf{Schritt 3: } +Somit haben wir die zwei Schritte um eine Transformation $T_i$ zu finden. +Wir führen den zweiten Schritt für jede der acht möglichen affinen Abbildungen vom ersten Schritt aus, und bestimmen den jeweilig resultierenden Fehler $e$. +Es resultieren acht $T_j$ mit ihren jeweiligen Fehlern. -Wir bestimmen die Parameter $s$ und $g$ für jede der acht möglichen affinen Abbildungen und das mit jedem Domain-Block. -Die Kombination von $D_j$ und $T_i$, welche den kleinsten Abstand $e$ hat, ist die beste. +Um den besten Domain-Block zu finden, führen wir die drei Schritte für jeden Domain-Block aus. +Der Domain-Block $D_j$, welcher die Transformation $T_j$ mit dem kleinsten Fehler $e$ hat, ist der ähnlichste. -Diese Schritte führen wir für jeden Range-Block $R_i$ aus. -Am Ende des Algorithmus haben wir für jeden Range-Block den zugehörigen Domain-Block und Transformation gefunden. +Wir suchen nun für jeden Range-Block $R_i$ den ähnlichsten Domain-Block. +Am Ende des Algorithmus haben wir für jeden Range-Block den zugehörigen Domain-Block und die dazugehörige Transformation gefunden. \begin{figure} \centering @@ -128,7 +137,7 @@ Am Ende des Algorithmus haben wir für jeden Range-Block den zugehörigen Domain Mit den gefundenen Abbildungen lässt sich das Bild generieren. Wir beginnen wie schon im letzten Kapitel mit einer beliebigen Startmenge. In unserem Fall ist dieses ein Bild $f_0$ derselben Grösse. -Nun ersetzen wir jedes $R_i$ mit der Transformierten des zugehörigen Domain-Blocks $T(G_j)$. +Nun ersetzen wir jedes $R_i$ mit der Transformierten des zugehörigen Domain-Blocks $T(D_j)$. Dies wird verkürzt als Operator $W$ geschrieben. So erhalten wir ein neues Bild $f_1 = W(f_0)$. Dieses Vorgehen führen wir iteriert aus bis wir von $f_n = W(f_{n-1})$ zu $f_{n-1}$ kaum mehr einen Unterschied feststellen. Die Iteration hat nun ihren Attraktor, das Bild, erreicht. @@ -140,22 +149,21 @@ Teilt man ein Bild in die drei Farbkanäle auf, das heisst, es wird nur noch ein Nun wendet man auf jeden dieser Farbkanalbilder den Algorithmus an, und fügt nach der Rekonstruktion die Kanäle wieder zusammen. \subsubsection{Performance des Verfahren} -Dieser Grundalgorithmus der fraktalen Bildkompression ist recht langsam und skaliert auch schlecht für grössere Bilder. -Dies resultiert aus eigenen Experimenten. +Experimentelle Beobachtungen haben gezeigt, dass dieser Grundalgorithmus der fraktalen Bildkompression recht langsam ist und auch schlecht für grössere Bilder skaliert. Man kann die Laufzeit zwar verbessern indem man die Domain-Blöcke auch disjunkt macht, und für weniger detailreiche Bilder ein grösseres $b$ wählt, jedoch wird er auch so nicht so schnell wie zum Beispiel das JPEG-Verfahren. Es wurden bessere Algorithmen der fraktalen Bildkompression entwickelt, doch auch diese können, vor allem in der Laufzeit, noch nicht mit herkömmlichen Komprimierungsverfahren mithalten. \subsection{Beispiel} -Wir Verwenden dafür den oben beschriebenen Algorithmus, welcher uns für jeden Range-Block die benötigten Parameter liefert. +Wir verwenden dafür den oben beschriebenen Algorithmus, welcher uns für jeden Range-Block die benötigten Parameter liefert. Mit diesen lässt sich das Bild im Anschluss wieder Rekonstruieren. -Die Range-Blöcke wurden $4\times4$ gewählt und die Dommain dementsprechend $8\times8$. +Die Range-Blöcke wurden $4\times4$ gewählt und die Domain dementsprechend $8\times8$. Um etwas Zeit bei der Komprimierung zu ersparen, wurden nur disjunkte Domain-Blöcke gebraucht. -Als erstes Beispiel wählen wir das 360x360px Bild von Rapperswil in Abbildung \ref{ifs:original}. -Das Startbild ist ein mittelgraues 360x360px Bild, Abbildung \ref{ifs:bild0}. -Es kann jedoch ein beliebiges Startbild +Als erstes Beispiel wählen wir das 360$\times$360 Pixel Bild von Rapperswil in Abbildung \ref{ifs:original}. +Das Startbild ist ein mittelgraues 360$\times$360 Pixel Bild, Abbildung \ref{ifs:bild0}. +Es kann jedoch ein beliebiges Startbild sein. Nun lassen wir das PIFS laufen. Wie wir in Abbildung \ref{ifs:rappirecoa} sehen, ist schon nach der ersten Iteration das Bild schon erkennbar. -Nach der fünften Iteration , Abbildung \ref{ifs:rappirecoc} gibt es fast keinen Unterschied mehr zur letzten Iteration, wir können die Rekonstruktion beenden. +Nach der fünften Iteration, Abbildung \ref{ifs:rappirecoc} gibt es fast keinen Unterschied mehr zur letzten Iteration, wir können die Rekonstruktion beenden. \begin{figure} \centering \includegraphics[width=0.4\textwidth]{papers/ifs/images/original} diff --git a/buch/papers/mceliece/Makefile.inc b/buch/papers/mceliece/Makefile.inc index ed1affa..53ecf7a 100644 --- a/buch/papers/mceliece/Makefile.inc +++ b/buch/papers/mceliece/Makefile.inc @@ -7,8 +7,8 @@ dependencies-mceliece = \ papers/mceliece/packages.tex \ papers/mceliece/main.tex \ papers/mceliece/references.bib \ - papers/mceliece/teil0.tex \ - papers/mceliece/teil1.tex \ - papers/mceliece/teil2.tex \ - papers/mceliece/teil3.tex + papers/mceliece/einleitung.tex \ + papers/mceliece/aufbau.tex \ + papers/mceliece/funktionsweise.tex \ + papers/mceliece/fazit.tex diff --git a/buch/papers/mceliece/aufbau.tex b/buch/papers/mceliece/aufbau.tex new file mode 100644 index 0000000..200cb7b --- /dev/null +++ b/buch/papers/mceliece/aufbau.tex @@ -0,0 +1,161 @@ +% +% einleitung.tex -- Beispiel-File für die Einleitung +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Aufbau\label{mceliece:section:Aufbau}} +\rhead{Aufbau} +Das McEliece-Kryptosystem besteht aus folgenden Elementen: + +\subsection{Datenvektor $d_k$ +\label{mceliece:subsection:d_k}} +In diesem Vektor der Länge $k$ sind die zu verschlüsselnden Daten enthalten. + +Beispiel: +\[d_4= +\begin{pmatrix} + 1\\ + 1\\ + 1\\ + 0 +\end{pmatrix} +\] + +\subsection{Binäre Zufallsmatrix $S_k$ +\label{mceliece:subsection:s_k}} +$S_k$ ist eine Binäre Zufallsmatrix der Grösse $k \times k$. +Auch muss diese Matrix in $\mathbb{F}_2$ invertierbar sein. +Für kleine Matrizen kann durchaus jedes Matrizenelement zufällig generiert werden, +wobei danach mithilfe des Gauss-Algorithmus deren Inverse bestimmt werden kann. +Da eine solche Matrix möglicherweise singulär ist, muss in diesem Fall eine neue Zufallsmatrix erzeugt werden. +Für grössere Matrizen existieren bessere Methoden, auf welche hier nicht weiter eingegangen wird \cite{mceliece:GenerationRandMatrix}. + +Beispiel: +\[S_4= + \begin{pmatrix} + 0 & 0 & 1 & 1\\ + 0 & 0 & 0 & 1\\ + 0 & 1 & 0 & 1\\ + 1 & 0 & 0 & 1 + \end{pmatrix} +\] + +\[ + S_4^{-1}= + \begin{pmatrix} + 0 & 1 & 0 & 1\\ + 0 & 1 & 1 & 0\\ + 1 & 1 & 0 & 0\\ + 0 & 1 & 0 & 0\\ + \end{pmatrix} +\] + +\subsection{Linear-Code-Generatormatrix $G_{n,k}$ +\label{mceliece:subsection:g_nk}} +Das wichtigste Element des McEliece-Systems ist ein fehlerkorrigierender Code, +der in der Lage ist, $t$ Fehler zu korrigieren. +Im Zusammenhang mit McEliece werden dabei meist binäre Goppa-Codes \cite{mceliece:goppa} verwendet, +es können prinzipiell auch andere Codes wie beispielsweise Reed-Solomon verwendet werden, +jedoch besitzen einige (unter anderem auch Reed-Solomon) Codes Schwachstellen \cite{mceliece:lorenz}. +Das Codieren mit diesem linearen Code kann mithilfe dessen Generatormatrix $G_{n,k}$ erfolgen. +Da es sich um einen fehlerkorrigierenden Code handelt, +wird das Codewort länger als das Datenwort, +es wird also Redundanz hinzugefügt, +um die Fehlerkorrektur möglich zu machen. + +Beispiel +\[ + G_{7,4}= + \begin{pmatrix} + 1 & 0 & 0 & 0\\ + 1 & 1 & 0 & 0\\ + 0 & 1 & 1 & 0\\ + 1 & 0 & 1 & 1\\ + 0 & 1 & 0 & 1\\ + 0 & 0 & 1 & 0\\ + 0 & 0 & 0 & 1 + \end{pmatrix} +\] + +\subsection{Permutations-Matrix $P_n$ +\label{mceliece:subsection:p_n}} +Mit der zufällig generierten Permutationsmatrix $P_n$ wird die Reihenfolge der Bits geändert. +Mit der Inversen $P_n^{-1}$ kann die Bitvertauschung rückgängig gemacht werden. + +Beispiel +\[ + P_7= + \begin{pmatrix} + 0 & 1 & 0 & 0 & 0 & 0 & 0\\ + 0 & 0 & 0 & 0 & 0 & 0 & 1\\ + 0 & 0 & 0 & 0 & 0 & 1 & 0\\ + 0 & 0 & 1 & 0 & 0 & 0 & 0\\ + 0 & 0 & 0 & 1 & 0 & 0 & 0\\ + 1 & 0 & 0 & 0 & 0 & 0 & 0\\ + 0 & 0 & 0 & 0 & 1 & 0 & 0 + \end{pmatrix} +\] +, +\[ + P_7^{-1}=P_7^t= + \begin{pmatrix} + 0 & 0 & 0 & 0 & 0 & 1 & 0\\ + 1 & 0 & 0 & 0 & 0 & 0 & 0\\ + 0 & 0 & 0 & 1 & 0 & 0 & 0\\ + 0 & 0 & 0 & 0 & 1 & 0 & 0\\ + 0 & 0 & 0 & 0 & 0 & 0 & 1\\ + 0 & 0 & 1 & 0 & 0 & 0 & 0\\ + 0 & 1 & 0 & 0 & 0 & 0 & 0 + \end{pmatrix} +\] + +\subsection{Public-Key $K_{n,k}$ +\label{mceliece:subsection:k_nk}} +Der öffentliche Schlüssel, welcher zum Verschlüsseln verwendet wird, +berechnet sich aus den bereits bekannten Matrizen wiefolgt: +\[ + K_{n,k}=P_{n}\cdot G_{n,k}\cdot S_{k}\,. +\] + +Beispiel +\[ + K_{7,4}= + \begin{pmatrix} + 0 & 0 & 1 & 0\\ + 1 & 0 & 0 & 1\\ + 0 & 0 & 1 & 1\\ + 1 & 1 & 1 & 1\\ + 0 & 1 & 0 & 1\\ + 0 & 1 & 0 & 0\\ + 1 & 0 & 0 & 0 + \end{pmatrix} +\] + +\subsection{Fehler-Vektor $e_n$ +\label{mceliece:subsection:e_n}} +Dieser Vektor der Länge $n$ besteht aus $t$ Einsen, welche zufällig innerhalb des Vektors angeordnet sind, +alle anderen Einträge sind Null. +Dieser Fehlervektor besitzt also gleich viele Einer, +wie die Anzahl Fehler, die der Linearcode der Generatormatrix $G_{n,k}$ zu korrigieren vermag. + +Beispiel +\[ + E_7= + \begin{pmatrix} + 0\\ + 0\\ + 1\\ + 0\\ + 0\\ + 0\\ + 0 + \end{pmatrix} +\] + +\subsection{Daten-Vektor $d_k$ +\label{mceliece:subsection:d_k}} +In diesem Vektor der länge $k$ ist die Nachricht (oder einen Teil davon) enthalten. + +\subsection{Code-Vektor $c_n$ +\label{mceliece:subsection:c_n}} +In diesem Vektor der länge $n$ ist die verschlüsselte Nachricht (oder einen Teil davon) enthalten.
\ No newline at end of file diff --git a/buch/papers/mceliece/einleitung.tex b/buch/papers/mceliece/einleitung.tex new file mode 100644 index 0000000..cebb8ed --- /dev/null +++ b/buch/papers/mceliece/einleitung.tex @@ -0,0 +1,16 @@ +% +% teil1.tex -- Beispiel-File für das Paper +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Einleitung +\label{mceliece:section:einleitung}} +\rhead{Einleitung} +Beim McEliece-Kryptosystem handelt es sich um ein asymetrisches Verschlüsselungsverfahren, welches erlaubt, +Daten verschlüsselt über ein Netzwerk zu übermitteln, ohne dass vorab ein gemeinsamer, +geheimer Schlüssel unter den Teilnehmern ausgetauscht werden müsste. +Eine andere, bereits erläuterte Variante einer asymetrischen Verschlüsselung ist das Diffie-Hellman-Verfahren \ref{buch:subsection:diffie-hellman}. +Im Gegensatz zu Diffie-Hellman gilt das McEliece-System als Quantencomputerresistent +und das Verschlüsseln/Entschlüsseln von Nachrichten wird hauptsächlich mit Matrizenoperationen durchgeführt. + + diff --git a/buch/papers/mceliece/fazit.tex b/buch/papers/mceliece/fazit.tex new file mode 100644 index 0000000..186708b --- /dev/null +++ b/buch/papers/mceliece/fazit.tex @@ -0,0 +1,57 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Fazit +\label{mceliece:section:fazit}} +\rhead{Fazit} +Ein kurzer Vergleich des McEliece-Systems +mit dem oft verwendeten RSA-System soll zeigen, wo dessen Vor- und Nachteile liegen. + +\subsection{Resourcen} +Eine Eigenheit des McEliece-Systems ist das hinzufügen von Rauschen (mit Fehlervektor $e_n$). +Damit diese mit dem Lienarcode-Decoder wieder entfernt werden können, +wird Redundanz benötigt, +weshalb dessen Kanalefizienz (Nutzbits/Übertragungsbits) sinkt. +Die Schlüsselgrösse des McEliece-Systems ist deshalb so riesig, weil es sich um eine zweidimensionale Matrix handelt, währenddem RSA mit nur zwei Skalaren auskommt. +Das McEliece-System benötigt dafür weniger Rechenaufwand beim Verschlüsseln/Entschlüsseln, da die meisten Operationen mit Matrixmultiplikationen ausgeführt werden können (Aufwand ist in binären Operationen pro Informationsbit)\cite{mceliece:CodeBasedCrypto}. +Beim Rechenaufwand sei noch erwähnt, +dass asymetrische Verschlüsselungen meist nur dazu verwendet werden, +um einen Schlüssel für eine symetrische Verschlüsselung auszutauschen. +\begin{center} +\begin{tabular}{c|c|c} + &McEliece (n=2048, k=1718, t = 30) &RSA (2048, e = 216 + 1)\\ + \hline + Schlüssegrösse: (Public) &429.5 KByte &0.5 KByte \\ + Kanaleffizienz: &83.9 \% &100 \% \\ + Verschlüsselungsaufwand: &1025 &40555 \\ + Entschlüsselungsaufwand: &2311 &6557176, 5 +\end{tabular} +\end{center} + +\subsection{Sicherheit} +Grosse unterschiede zwischen den beiden Kryptosystemen gibt es jedoch bei der Sicherheit. +Der Kern der RSA-Verschlüsselung beruht auf dem Problem, eine grosse Zahl in ihre beiden Primfaktoren zu zerlegen. +Bei genügend grossen Zahlen ist diese Zerlegung auch mit den heute besten verfügbaren Computern kaum innerhalb vernünftiger Zeit zu lösen. +Weiter ist aber bekannt, +dass mithilfe des sogenannten Shor-Algorithmus \cite{mceliece:shor} und einem Quantencomputer auch diese Zerlegung zügig realisiert werden könnte, +was zur Folge hätte, dass die Verschlüsselung von RSA unwirksam würde. +Zurzeit sind die Quantencomputer jedoch noch bei weitem nicht in der Lage, grosse Zahlen mithilfe dieses Algorithmuses zu zerlegen. +Das McEliece-System hingegen beruht auf dem Problem des ``Syndrome decoding'' (Korrektur von Bitfehlern eines Codewortes, das mit einem entsprechenden Linearcode codiert wurde). +Für das ``Syndrome decoding'' sind bis heute keine Methoden bekannt, +welche nennenswerte Vorteile gegenüber dem Durchprobieren (brute-force) bringen, +auch nicht mithilfe eines Quantencomputers. +\begin{center} +\begin{tabular}{c|c|c} + &McEliece &RSA \\ +\hline + Grundlage Verschlüsselung &Syndrome decoding &Integer factoring\\ + Aufwand (gewöhnliche CPU) &exponential &< exponential \\ + Aufwand (Quantencomputer) &> polynomial &$\mathcal{O}(\log(N)^3)$ +\end{tabular} +\end{center} +Die Verbreitung des McEliece-Kryptosystems ist zurzeit äusserst gering. +Das liegt einerseits an der immensen Grösse des öffentlichen Schlüssels, +andererseits wird aber auch in naher Zukunft nicht mit einem genügend starken Quantencomputer gerechnet, +welcher andere asymetrische Verschlüsselungen gefährden würde. diff --git a/buch/papers/mceliece/funktionsweise.tex b/buch/papers/mceliece/funktionsweise.tex new file mode 100644 index 0000000..7c69b13 --- /dev/null +++ b/buch/papers/mceliece/funktionsweise.tex @@ -0,0 +1,83 @@ +% +% teil2.tex -- Beispiel-File für teil2 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Funktionsweise +\label{mceliece:section:funktionsweise}} +\rhead{Funktionsweise} +Um den Ablauf des Datenaustausches mittels McEliece-Verschlüsselung zu erläutern, +wird ein Szenario verwendet, +bei dem Bob an Alice eine verschlüsselte Nachticht über ein öffentliches Netzwerk zukommen lässt. + +\subsection{Vorbereitung +\label{mceliece:section:vorbereitung}} +Damit der Nachrichtenaustausch stattfinden kann, muss Alice (Empfängerin) +zuerst ein Schlüsselpaar definieren. +Dazu erstellt sie die einzelnen Matrizen $S_k$, $G_{n,k}$ und $P_n$. +Diese drei einzelnen Matrizen bilden den privaten Schlüssel von Alice +und sollen geheim bleiben. +Der öffentliche Schlüssel $K_{n,k}$ hingegen berechnet sich +aus der Multiplikation der privaten Matrizen (Abschnitt \ref{mceliece:subsection:k_nk}) +und wird anschliessend Bob zugestellt. + +\subsection{Verschlüsselung +\label{mceliece:section:verschl}} +Bob berechnet nun die verschlüsselte Nachricht $c_n$, indem er seine Daten $d_k$ +mit dem öffentlichen Schlüssel $K_{n,k}$ von Alice multipliziert +und anschliessend durch eine Addition mit einem Fehlervektor $e_n$ einige Bitfehler hinzufügt. +\[ + c_n\,=\,K_{n,k}\cdot d_k + e_n\,. +\] +Dabei wird für jede Nachricht (oder für jedes Nachrichtenfragment) +einen neuen, zufälligen Fehlervektor generiert. +Die verschlüsselte Nachricht $c_n$ wird anschliessend Alice zugestellt. + +\subsection{Entschlüsselung +\label{mceliece:section:entschl}} +Alice entschlüsselt die erhaltene Nachricht in mehreren einzelnen Schritten. +Um etwas Transparenz in diese Prozedur zu bringen, wird der öffentliche Schlüssel $K_{n,k}$ mit seinen Ursprungsmatrizen dargestellt. +\begin{align*} + c_n\,&=\,K_{n,k}\cdot d_k + e_n \\ + &= P_{n}\cdot G_{n,k}\cdot S_{k}\cdot d_k + e_n +\end{align*} +Zuerst wird der Effekt der Permutationsmatrix rückgängig gemacht, +indem das Codewort mit dessen Inversen $P_n^{-1}$ multipliziert wird. +\begin{align*} + c_{n}''\,=\,P_n^{-1}\cdot c_n\,&= P_n^{-1}\cdot P_{n}\cdot G_{n,k}\cdot S_{k}\cdot d_k + P_n^{-1}\cdot e_n \\ + &= G_{n,k}\cdot S_{k}\cdot d_k + P_n^{-1}\cdot e_n \\ +\end{align*} +Eine weitere Vereinfachung ist nun möglich, +weil $P_n^{-1}$ einerseits auch eine gewöhnliche Permutationsmatrix ist +und andererseits ein zufälliger Fehlervektor $e_n$ multipliziert mit einer Permutationsmatrix +wiederum einen gleichwertigen, zufälligen Fehlervektor $e_n'$ ergibt. +\begin{align*} + c_{n}''\,&=\,G_{n,k}\cdot S_{k}\cdot d_k + P_n^{-1}\cdot e_n \\ + &=\,G_{n,k}\cdot S_{k}\cdot d_k + e'_n\quad \quad \quad | \, + e'_n\,=\,P_n^{-1}\cdot e_n +\end{align*} +Dank des fehlerkorrigierenden Codes, der durch die implizite Multiplikation mittels $G_{n,k}$ auf die Daten angewendet wurde, +können nun die Bitfehler, verursacht durch den Fehlervektor $e'_n$, +entfernt werden. +Da es sich bei diesem Schritt nicht um eine einfache Matrixmultiplikation handelt, +wird die Operation durch eine Funktion dargestellt. +Wie dieser Decoder genau aufgebaut ist, +hängt vom verwendeten Linearcode ab. +\begin{align*} + c_{k}'\,&=\text{Linear-Code-Decoder($c''_n$)}\\ + &=\text{Linear-Code-Decoder($G_{n,k}\cdot S_{k}\cdot d_k + e'_n$)}\\ + &=S_{k}\cdot d_k +\end{align*} +Zum Schluss wird das inzwischen fast entschlüsselte Codewort $c'_k$ mit der inversen der zufälligen Binärmatrix $S^{-1}$ multipliziert, +womit der Inhalt der ursprünglichen Nachricht nun wiederhergestellt wurde. +\begin{align*} + c_{k}'\,&=S_{k}\cdot d_k \quad | \cdot S_k^{-1}\\ + d'_{k}\,=\,S_{k}^{-1} \cdot c'_k&=S_{k}^{-1} \cdot S_{k}\cdot d_k\\ + &=d_k +\end{align*} + +\subsection{Beispiel} + +TODO: +-alle Beispielmatrizen- und Vektoren hierhin zügeln, numerisches Beispiel kreieren\\ +-erläutern des 7/4-codes (ja/nein)?
\ No newline at end of file diff --git a/buch/papers/mceliece/main.tex b/buch/papers/mceliece/main.tex index dbbaaac..352a6be 100644 --- a/buch/papers/mceliece/main.tex +++ b/buch/papers/mceliece/main.tex @@ -8,29 +8,10 @@ \begin{refsection} \chapterauthor{Reto Fritsche} -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} - -\input{papers/mceliece/teil0.tex} -\input{papers/mceliece/teil1.tex} -\input{papers/mceliece/teil2.tex} -\input{papers/mceliece/teil3.tex} +\input{papers/mceliece/einleitung.tex} +\input{papers/mceliece/aufbau.tex} +\input{papers/mceliece/funktionsweise.tex} +\input{papers/mceliece/fazit.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/mceliece/references.bib b/buch/papers/mceliece/references.bib index 47798d3..0388ff4 100644 --- a/buch/papers/mceliece/references.bib +++ b/buch/papers/mceliece/references.bib @@ -4,32 +4,45 @@ % (c) 2020 Autor, Hochschule Rapperswil % -@online{mceliece:bibtex, - title = {BibTeX}, - url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} +@online{mceliece:GenerationRandMatrix, + title = {Efficient Generation of Random Nonsingular Matrices}, + url = {https://www.researchgate.net/publication/2729950_Efficient_Generation_of_Random_Nonsingular_Matrices}, + date = {Januar 1993}, + year = {2021}, + month = {7}, + day = {29} } -@book{mceliece:numerical-analysis, - title = {Numerical Analysis}, - author = {David Kincaid and Ward Cheney}, - publisher = {American Mathematical Society}, - year = {2002}, - isbn = {978-8-8218-4788-6}, - inseries = {Pure and applied undegraduate texts}, - volume = {2} +@online{mceliece:lorenz, + title = {Cryptography based on error correcting codes}, + url = {https://algo.epfl.ch/_media/en/projects/lorenz_thesis.pdf}, + date = {2007-07-27}, + year = {2021}, + month = {7}, + day = {29} } -@article{mceliece:mendezmueller, - author = { Tabea Méndez and Andreas Müller }, - title = { Noncommutative harmonic analysis and image registration }, - journal = { Appl. Comput. Harmon. Anal.}, - year = 2019, - volume = 47, - pages = {607--627}, - url = {https://doi.org/10.1016/j.acha.2017.11.004} +@online{mceliece:shor, + title = {Shor's algorithm}, + url = {https://en.wikipedia.org/wiki/Shor%27s_algorithm}, + year = {2021}, + month = {8}, + day = {9} } +@online{mceliece:CodeBasedCrypto, + title = {Code based cryptography and steganography}, + url = {https://www.researchgate.net/publication/268009418_Code_Based_Cryptography_and_Steganography}, + date = {2013-05-30}, + year = {2021}, + month = {8}, + day = {9} +} + +@online{mceliece:goppa, + title = {Binary Goppa code}, + url = {https://en.m.wikipedia.org/wiki/Binary_Goppa_code}, + year = {2021}, + month = {8}, + day = {10} +}
\ No newline at end of file diff --git a/buch/papers/mceliece/teil0.tex b/buch/papers/mceliece/teil0.tex deleted file mode 100644 index b98f8be..0000000 --- a/buch/papers/mceliece/teil0.tex +++ /dev/null @@ -1,22 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 0\label{mceliece:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{mceliece:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - - diff --git a/buch/papers/mceliece/teil1.tex b/buch/papers/mceliece/teil1.tex deleted file mode 100644 index 06035a6..0000000 --- a/buch/papers/mceliece/teil1.tex +++ /dev/null @@ -1,55 +0,0 @@ -% -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{mceliece:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{mceliece:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{mceliece:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{mceliece:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{mceliece:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. - - diff --git a/buch/papers/mceliece/teil2.tex b/buch/papers/mceliece/teil2.tex deleted file mode 100644 index fd247c7..0000000 --- a/buch/papers/mceliece/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{mceliece:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{mceliece:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/mceliece/teil3.tex b/buch/papers/mceliece/teil3.tex deleted file mode 100644 index 421b331..0000000 --- a/buch/papers/mceliece/teil3.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil3.tex -- Beispiel-File für Teil 3 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 3 -\label{mceliece:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{mceliece:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/multiplikation/Makefile b/buch/papers/multiplikation/Makefile index 8f04c2c..8f04c2c 100644..100755 --- a/buch/papers/multiplikation/Makefile +++ b/buch/papers/multiplikation/Makefile diff --git a/buch/papers/multiplikation/Makefile.inc b/buch/papers/multiplikation/Makefile.inc index b78d67e..074020f 100644..100755 --- a/buch/papers/multiplikation/Makefile.inc +++ b/buch/papers/multiplikation/Makefile.inc @@ -7,8 +7,7 @@ dependencies-multiplikation = \ papers/multiplikation/packages.tex \ papers/multiplikation/main.tex \ papers/multiplikation/references.bib \ - papers/multiplikation/teil0.tex \ - papers/multiplikation/teil1.tex \ - papers/multiplikation/teil2.tex \ - papers/multiplikation/teil3.tex + papers/multiplikation/einlteung.tex \ + papers/multiplikation/loesungsmethoden.tex \ + papers/multiplikation/problemstellung.tex diff --git a/buch/papers/multiplikation/code/Figure_1.png b/buch/papers/multiplikation/code/Figure_1.png Binary files differnew file mode 100755 index 0000000..9def15a --- /dev/null +++ b/buch/papers/multiplikation/code/Figure_1.png diff --git a/buch/papers/multiplikation/code/MM.c b/buch/papers/multiplikation/code/MM.c new file mode 100755 index 0000000..2588262 --- /dev/null +++ b/buch/papers/multiplikation/code/MM.c @@ -0,0 +1,466 @@ +#include <stdio.h>
+#include <stdint.h>
+#include <stdlib.h>
+#include <time.h>
+#include <omp.h>
+#include "c_matrix.h"
+#include <gsl/gsl_cblas.h>
+#include <string.h>
+
+void MM(int *A, int *B, int *C, int n);
+void openMP_MM(int *A, int *B, int *C, int n);
+void winograd(int *A, int *B, int *C, int n);
+int winograd_inner(int *a, int *b, int n);
+void run_algo(void (*algo)(), char alog_name[], int print);
+void run_algo_cblas(int print);
+void MM_dc(int *A, int *B, int *C, int n);
+void strassen(int *A, int *B, int *C, int n);
+void printMatrix(int *C, int n);
+void printMatrix_double(double *C, int n);
+void split(int *in, int *out, int n, int col, int row);
+void join(int *in, int *out, int n, int col, int row);
+void add(int *A, int *B, int *C, int n);
+void sub(int *A, int *B, int *C, int n);
+void multiply(int *A, int *B, int *C, int n);
+
+int main() {
+ // omp_set_dynamic(0);
+ // omp_set_num_threads(4);
+// run_algo(openMP_MM, "openMP_MM",0);
+ run_algo(MM_dc, "MM_dc",0);
+
+ run_algo(strassen, "strassen",0);
+
+ run_algo(MM, "MM", 0);
+ run_algo(winograd, "winograd", 0);
+ run_algo_cblas(0);
+
+ return 0;
+}
+
+void MM(int *A, int *B, int *C, int n) {
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ int sum = 0;
+ for (int k = 0; k < n; ++k) {
+ sum += (*((A + i * n) + k)) * (*((B + k * n) + j));
+ }
+ *((C + i * n) + j) = sum;
+ }
+ }
+}
+
+int winograd_inner(int *a, int *b, int n){
+ int ab = 0;
+ if(n%2==0)
+ {
+ int xi = 0;
+ int etha = 0;
+ for(int i = 0; i<n/2;++i)
+ {
+ xi += a[2*i]*a[2*i+1];
+ etha += b[2*i]*b[2*i+1];
+ ab += (a[2*i]+b[2*i+1])*(a[2*i+1]+b[2*i]);
+ }
+ ab = ab-etha-xi;
+ }
+ return ab;
+ }
+
+ void winograd(int *A, int *B, int *C, int n) {
+
+ int xi_array[n];
+ int etha_array[n];
+ int xi = 0;
+ int etha = 0;
+ int ab = 0;
+
+ for (int i = 0; i < n; ++i) {
+ xi = 0;
+ etha = 0;
+ for(int k = 0;k<n/2;++k)
+ {
+ xi += (*((A + i * n) + 2*k))*(*((A + i * n) + (2*k+1)));
+ etha += (*((B + 2*k * n) + i))*(*((B + (2*k+1) * n) + i));
+ }
+ xi_array[i] = xi;
+ etha_array[i] = etha;
+ }
+
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ ab = 0;
+ for(int k = 0;k<n/2;++k)
+ {
+ ab += ((*((A + i * n) + 2*k))+(*((B + (2*k+1) * n) + j)))*((*((A + i * n) + (2*k+1)))+(*((B + 2*k * n) + j)));
+ }
+ *((C + i * n) + j) = ab-etha_array[j]-xi_array[i];
+ }
+ }
+
+
+
+
+ // for (int i = 0; i < n; ++i) {
+ // int *a = (int*) malloc(n * sizeof(int));
+ // for(int k = 0; k<n; ++k)
+ // {
+ // a[k] = (*((A + i * n) + k));
+ // }
+ //
+ // for (int j = 0; j < n; ++j) {
+ // int *b = (int*) malloc(n * sizeof(int));
+ // for(int k = 0; k<n; ++k)
+ // {
+ // b[k] =(*((B + k * n) + j));
+ // }
+ // *((C + i * n) + j) = winograd_inner(a,b,n);
+ // }
+ // }
+ }
+
+
+void openMP_MM(int *A, int *B, int *C, int n) {
+
+ #pragma omp parallel for
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ int sum = 0;
+ for (int k = 0; k < n; ++k) {
+ sum += (*((A + i * n) + k)) * (*((B + k * n) + j));
+ }
+ *((C + i * n) + j) = sum;
+ }
+ }
+}
+
+void MM_dc(int *A, int *B, int *C, int n) {
+ if (n <= 2) {
+ MM((int*) A, (int*) B, (int*) C, n);
+ } else {
+ int *A11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ split((int*) A, (int*) A11, n / 2, 0, 0);
+ split((int*) A, (int*) A12, n / 2, 0, n / 2);
+ split((int*) A, (int*) A21, n / 2, n / 2, 0);
+ split((int*) A, (int*) A22, n / 2, n / 2, n / 2);
+ split((int*) B, (int*) B11, n / 2, 0, 0);
+ split((int*) B, (int*) B12, n / 2, 0, n / 2);
+ split((int*) B, (int*) B21, n / 2, n / 2, 0);
+ split((int*) B, (int*) B22, n / 2, n / 2, n / 2);
+
+ int *tmp1 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp2 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp3 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp4 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp5 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp6 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp7 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *tmp8 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ MM_dc((int*) A11, (int*) B11, (int*) tmp1, n / 2);
+ MM_dc((int*) A12, (int*) B21, (int*) tmp2, n / 2);
+ MM_dc((int*) A11, (int*) B12, (int*) tmp3, n / 2);
+ MM_dc((int*) A12, (int*) B22, (int*) tmp4, n / 2);
+ MM_dc((int*) A21, (int*) B11, (int*) tmp5, n / 2);
+ MM_dc((int*) A22, (int*) B21, (int*) tmp6, n / 2);
+ MM_dc((int*) A21, (int*) B12, (int*) tmp7, n / 2);
+ MM_dc((int*) A22, (int*) B22, (int*) tmp8, n / 2);
+
+ free(A11);
+ free(A12);
+ free(A21);
+ free(A22);
+ free(B11);
+ free(B12);
+ free(B21);
+ free(B22);
+
+ int *C11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ add((int*) tmp1, (int*) tmp2, (int*) C11, n / 2);
+ add((int*) tmp3, (int*) tmp4, (int*) C12, n / 2);
+ add((int*) tmp5, (int*) tmp6, (int*) C21, n / 2);
+ add((int*) tmp7, (int*) tmp8, (int*) C22, n / 2);
+
+ free(tmp1);
+ free(tmp2);
+ free(tmp3);
+ free(tmp4);
+ free(tmp5);
+ free(tmp6);
+ free(tmp7);
+ free(tmp8);
+
+ join((int*) C11, (int*) C, n / 2, 0, 0);
+ join((int*) C12, (int*) C, n / 2, 0, n / 2);
+ join((int*) C21, (int*) C, n / 2, n / 2, 0);
+ join((int*) C22, (int*) C, n / 2, n / 2, n / 2);
+
+ free(C11);
+ free(C12);
+ free(C21);
+ free(C22);
+
+ }
+}
+
+void strassen(int *A, int *B, int *C, int n) {
+ if (n <= 2) {
+
+ int P, Q, R, S, T, U, V;
+ P = ((*((A + 0 * n) + 0)) + (*((A + 1 * n) + 1)))
+ * ((*((B + 0 * n) + 0)) + (*((B + 1 * n) + 1)));
+ Q = ((*((A + 1 * n) + 0)) + (*((A + 1 * n) + 1)))
+ * ((*((B + 0 * n) + 0)));
+ R = ((*((A + 0 * n) + 0)))
+ * ((*((B + 0 * n) + 1)) - (*((B + 1 * n) + 1)));
+ S = ((*((A + 1 * n) + 1)))
+ * ((*((B + 1 * n) + 0)) - (*((B + 0 * n) + 0)));
+ T = ((*((A + 0 * n) + 0)) + (*((A + 0 * n) + 1)))
+ * ((*((B + 1 * n) + 1)));
+ U = ((*((A + 1 * n) + 0)) - (*((A + 0 * n) + 0)))
+ * ((*((B + 0 * n) + 0)) + (*((B + 0 * n) + 1)));
+ V = ((*((A + 0 * n) + 1)) - (*((A + 1 * n) + 1)))
+ * ((*((B + 1 * n) + 0)) + (*((B + 1 * n) + 1)));
+ (*((C + 0 * n) + 0)) = P + S - T + V;
+ (*((C + 0 * n) + 1)) = R + T;
+ (*((C + 1 * n) + 0)) = Q + S;
+ (*((C + 1 * n) + 1)) = P + R - Q + U;
+
+ } else {
+ int *A11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *A22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *B22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ split((int*) A, (int*) A11, n / 2, 0, 0);
+ split((int*) A, (int*) A12, n / 2, 0, n / 2);
+ split((int*) A, (int*) A21, n / 2, n / 2, 0);
+ split((int*) A, (int*) A22, n / 2, n / 2, n / 2);
+ split((int*) B, (int*) B11, n / 2, 0, 0);
+ split((int*) B, (int*) B12, n / 2, 0, n / 2);
+ split((int*) B, (int*) B21, n / 2, n / 2, 0);
+ split((int*) B, (int*) B22, n / 2, n / 2, n / 2);
+
+ int *P = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *Q = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *R = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *S = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *T = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *U = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *V = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ int *addA = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *addB = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ add((int*) A11, (int*) A22, (int*) addA, n / 2);
+ add((int*) B11, (int*) B22, (int*) addB, n / 2);
+ strassen((int*) addA, (int*) addB, (int*) P, n / 2);
+
+ add((int*) A21, (int*) A22, (int*) addA, n / 2);
+ strassen((int*) addA, (int*) B11, (int*) Q, n / 2);
+
+ sub((int*) B12, (int*) B22, (int*) addB, n / 2);
+ strassen((int*) A11, (int*) addB, (int*) R, n / 2);
+
+ sub((int*) B21, (int*) B11, (int*) addB, n / 2);
+ strassen((int*) A22, (int*) addB, (int*) S, n / 2);
+
+ add((int*) A11, (int*) A12, (int*) addA, n / 2);
+ strassen((int*) addA, (int*) B22, (int*) T, n / 2);
+
+ sub((int*) A21, (int*) A11, (int*) addA, n / 2);
+ add((int*) B11, (int*) B12, (int*) addB, n / 2);
+ strassen((int*) addA, (int*) addB, (int*) U, n / 2);
+
+ sub((int*) A12, (int*) A22, (int*) addA, n / 2);
+ add((int*) B21, (int*) B22, (int*) addB, n / 2);
+ strassen((int*) addA, (int*) addB, (int*) V, n / 2);
+
+ free(A11);
+ free(A12);
+ free(A21);
+ free(A22);
+ free(B11);
+ free(B12);
+ free(B21);
+ free(B22);
+ free(addA);
+ free(addB);
+
+ int *C11 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C12 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C21 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *C22 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ int *resAdd1 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+ int *resAdd2 = (int*) malloc(n / 2 * n / 2 * sizeof(int));
+
+ add((int*) R, (int*) T, (int*) C12, n / 2);
+ add((int*) Q, (int*) S, (int*) C21, n / 2);
+
+ add((int*) P, (int*) S, (int*) resAdd1, n / 2);
+ add((int*) resAdd1, (int*) V, (int*) resAdd2, n / 2);
+ sub((int*) resAdd2, (int*) T, (int*) C11, n / 2);
+
+ add((int*) P, (int*) R, (int*) resAdd1, n / 2);
+ add((int*) resAdd1, (int*) U, (int*) resAdd2, n / 2);
+ sub((int*) resAdd2, (int*) Q, (int*) C22, n / 2);
+
+ free(P);
+ free(Q);
+ free(R);
+ free(S);
+ free(T);
+ free(U);
+ free(V);
+ free(resAdd1);
+ free(resAdd2);
+
+ join((int*) C11, (int*) C, n / 2, 0, 0);
+ join((int*) C12, (int*) C, n / 2, 0, n / 2);
+ join((int*) C21, (int*) C, n / 2, n / 2, 0);
+ join((int*) C22, (int*) C, n / 2, n / 2, n / 2);
+
+ free(C11);
+ free(C12);
+ free(C21);
+ free(C22);
+ }
+}
+
+void add(int *A, int *B, int *C, int n) {
+ for (int i = 0; i < n; i++) {
+ for (int j = 0; j < n; j++) {
+ *((C + i * n) + j) = *((A + i * n) + j) + *((B + i * n) + j);
+ }
+ }
+}
+
+void sub(int *A, int *B, int *C, int n) {
+ for (int i = 0; i < n; i++) {
+ for (int j = 0; j < n; j++) {
+ *((C + i * n) + j) = *((A + i * n) + j) - *((B + i * n) + j);
+ }
+ }
+}
+
+void multiply(int *A, int *B, int *C, int n) {
+ int mul;
+
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ mul = (*((A + i * n) + j)) * (*((B + i * n) + j));
+ *((C + i * n) + j) = mul;
+ }
+ }
+}
+
+void split(int *in, int *out, int n, int col, int row) {
+ for (int i1 = 0, i2 = col; i1 < n; i1++, i2++)
+ for (int j1 = 0, j2 = row; j1 < n; j1++, j2++) {
+ *((out + i1 * n) + j1) = *((in + i2 * n * 2) + j2);
+
+ }
+}
+
+void join(int *in, int *out, int n, int col, int row) {
+ for (int i1 = 0, i2 = col; i1 < n; i1++, i2++)
+ for (int j1 = 0, j2 = row; j1 < n; j1++, j2++)
+ *((out + i2 * n * 2) + j2) = *((in + i1 * n) + j1);
+}
+
+void printMatrix(int *C, int n) {
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ printf("%d ", *((C + i * n) + j));
+ }
+ printf("\n");
+ }
+}
+
+void printMatrix_double(double *C, int n) {
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ printf("%.0f ", *((C + i * n) + j));
+ }
+ printf("\n");
+ }
+}
+
+void run_algo(void (*algo)(), char alog_name[], int print)
+{
+ FILE *fptr;
+
+ char fileName[40] = "meas/";
+ strcat(fileName, alog_name);
+ strcat(fileName, ".txt");
+ fptr = fopen(fileName, "w");
+
+
+ for(int i=0; i<n_arrays; ++i)
+ {
+ for(int j = 0; j<10; ++j)
+ {
+ int *C = (int*) malloc(n[i] * n[i] * sizeof(int));
+ double dtime = omp_get_wtime();
+ algo(Ap[i], Bp[i], (int*) C, n[i]);
+ dtime = omp_get_wtime() - dtime;
+ // printf("The %s program took %f seconds to execute \n", alog_name, dtime);
+ fprintf(fptr, "%f,%d\n", dtime, n[i]);
+
+ if(print==1)
+ {
+ printMatrix((int*)C, n[i]);
+ }
+ free(C);
+ }
+ }
+ fclose(fptr);
+
+}
+
+void run_algo_cblas(int print)
+{
+
+ FILE *fptr;
+
+ fptr = fopen("meas/blas.txt", "w");
+ for(int i=0; i<n_arrays; ++i)
+ {
+ for(int j = 0; j<10; ++j)
+ {
+ double *dC = (double*) malloc(n[i] * n[i] * sizeof(double));
+ double dtime = omp_get_wtime();
+ cblas_dgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans, n[i], n[i], n[i], 1.0, dAp[i], n[i],
+ dBp[i], n[i], 0.0, dC, n[i]);
+ dtime = omp_get_wtime() - dtime;
+ // printf("The cblas program took %f seconds to execute \n", dtime);
+ fprintf(fptr, "%f,%d\n",dtime, n[i]);
+
+ if(print==1)
+ {
+ printMatrix_double( (double*)dC, n[i]);
+ }
+
+ free(dC);
+ }
+ }
+ fclose(fptr);
+
+}
diff --git a/buch/papers/multiplikation/code/MM.py b/buch/papers/multiplikation/code/MM.py new file mode 100644 index 0000000..8057850 --- /dev/null +++ b/buch/papers/multiplikation/code/MM.py @@ -0,0 +1,339 @@ +#!/usr/bin/env python3 +# -*- coding: utf-8 -*- +""" +Created on Fri Mar 19 07:31:29 2021 + +@author: nunigan +""" +import scipy.stats +import numpy as np +import time +import matplotlib.pyplot as plt +from scipy.optimize import curve_fit +import tikzplotlib +def MM(A, B): + n = np.shape(A)[0] + C = np.zeros((n, n)) + for i in range(n): + for j in range(n): + C[i, j] = 0 + for k in range(n): + C[i, j] += A[i, k]*B[k, j] + return C + + +def MM_dc(A, B): + n = np.shape(A)[0] + if(n <= 2): + C = np.zeros((n, n)) + C[0, 0] = A[0, 0]*B[0, 0]+A[0, 1]*B[1, 0] + C[0, 1] = A[0, 0]*B[0, 1]+A[0, 1]*B[1, 1] + C[1, 0] = A[1, 0]*B[0, 0]+A[1, 1]*B[1, 0] + C[1, 1] = A[1, 0]*B[0, 1]+A[1, 1]*B[1, 1] + return C + else: + A11, A12, A21, A22 = A[:n//2, :n//2], A[:n//2, n//2:], A[n//2:, :n//2], A[n//2:, n//2:] + B11, B12, B21, B22 = B[:n//2, :n//2], B[:n//2, n//2:], B[n//2:, :n//2], B[n//2:, n//2:] + C11 = MM_dc(A11, B11) + MM_dc(A12, B21) + C12 = MM_dc(A11, B12) + MM_dc(A12, B22) + C21 = MM_dc(A21, B11) + MM_dc(A22, B21) + C22 = MM_dc(A21, B12) + MM_dc(A22, B22) + C = np.vstack((np.hstack((C11, C12)), np.hstack((C21, C22)))) + return C + + +def strassen(A, B): + n = np.shape(A)[0] + if(n <= 2): + C = np.zeros((n, n)) + P = (A[0, 0]+A[1, 1])*(B[0, 0]+B[1, 1]) + Q = (A[1, 0]+A[1, 1])*B[0, 0] + R = A[0, 0]*(B[0, 1]-B[1, 1]) + S = A[1, 1]*(B[1, 0]-B[0, 0]) + T = (A[0, 0]+A[0, 1])*B[1, 1] + U = (A[1, 0]-A[0, 0])*(B[0, 0]+B[0, 1]) + V = (A[0, 1]-A[1, 1])*(B[1, 0]+B[1, 1]) + C[0, 0] = P+S-T+V + C[0, 1] = R+T + C[1, 0] = Q+S + C[1, 1] = P+R-Q+U + return C + else: + m = n//2 + A11, A12, A21, A22 = A[:m, :m], A[:m, m:], A[m:, :m], A[m:, m:] + B11, B12, B21, B22 = B[:m, :m], B[:m, m:], B[m:, :m], B[m:, m:] + P = strassen((A11+A22),(B11+B22)) + Q = strassen((A21+A22),B11) + R = strassen(A11,(B12-B22)) + S = strassen(A22,(B21-B11)) + T = strassen((A11+A12),B22) + U = strassen((A21-A11),(B11+B12)) + V = strassen((A12-A22),(B21+B22)) + + C11 = P+S-T+V + C12 = R+T + C21 = Q+S + C22 = P+R-Q+U + + C = np.vstack((np.hstack((C11, C12)), np.hstack((C21, C22)))) + return C + +def winograd_inner(a, b): + n = np.shape(a)[0] + if n%2 == 0: + xi = np.sum(a[::2]*a[1::2]) + etha = np.sum(b[::2]*b[1::2]) + # print("xi = {}, etha = {}".format(xi, etha)) + ab = np.sum((a[::2]+b[1::2])*(a[1::2]+b[::2]))-xi-etha + else: + xi = np.sum(a[0:-1:2]*a[1::2]) + etha = np.sum(b[0:-1:2]*b[1::2]) + ab = np.sum((a[0:-1:2]+b[1::2])*(a[1::2]+b[0:-1:2]))-xi-etha+a[-1]*b[-1] + return ab + +def winograd(A, B): + m,n = np.shape(A) + n2,p = np.shape(B) + C = np.zeros((m,p)) + for i in range(np.shape(A)[0]): + for j in range(np.shape(B)[1]): + C[i,j] = winograd_inner(A[i,:], B[:,j]) + return C + +def winograd2(A, B): + m,n = np.shape(A) + n2,p = np.shape(B) + C = np.zeros((m,p)) + xi = np.zeros((m)) + eta = np.zeros((p)) + ab = 0 + for i in range(m): + for j in range(n//2): + xi[i] += A[i,2*j]*A[i,2*j+1] + + for i in range(p): + for j in range(n//2): + eta[i] += B[2*j,i]*B[2*j+1,i] + + if n%2==0: + for i in range(m): + for j in range(p): + ab = 0 + for k in range(n//2): + ab += (A[i,2*k]+B[2*k+1,j])*(A[i,2*k+1]+B[2*k,j]) + C[i,j] = ab-eta[j]-xi[i] + else: + for i in range(m): + for j in range(p): + ab = 0 + for k in range(n//2): + ab += (A[i,2*k]+B[2*k+1,j])*(A[i,2*k+1]+B[2*k,j]) + C[i,j] = ab-eta[j]-xi[i]+A[i,-1]*B[-1,j] + + return C + +def test_perfomance(n): + + t_mm = [] + t_mm_dc = [] + t_mm_strassen = [] + t_wino = [] + t_np = [] + + for i in n: + A = np.random.randn(i, i) + B = np.random.randn(i, i) + # A = np.random.randint(-100, 100,(i, i)) + # B = np.random.randint(-100, 100,(i, i)) + + start = time.time() + C3 = strassen(A, B) + t_mm_strassen.append(time.time() - start) + + start = time.time() + C1 = MM(A, B) + t_mm.append(time.time() - start) + + start = time.time() + C2 = MM_dc(A, B) + t_mm_dc.append(time.time() - start) + + start = time.time() + C4 = winograd2(A, B) + t_wino.append(time.time() - start) + + start = time.time() + C = A@B + t_np.append(time.time() - start) + + plt.figure(figsize=(13,8)) + plt.rcParams['font.family'] = 'STIXGeneral' + plt.rc('axes', labelsize=23) + plt.rc('xtick', labelsize=23) + plt.rc('ytick', labelsize=23) + plt.plot(n, t_mm, label='Standard', lw=5) + plt.plot(n, t_mm_dc, label='Divide and conquer', lw=5) + plt.plot(n, t_mm_strassen, label='Strassen', lw=5) + plt.plot(n, t_wino, label='Winograd', lw=5) + plt.plot(n, t_np, label='NumPy A@B', lw=5) + # plt.xscale('log', base=2) + plt.legend() + plt.xlabel("n") + plt.ylabel("time (s)") + plt.grid(True, which="both", ls="-") + plt.tight_layout() + # plt.yscale('log') + plt.legend(fontsize=19) + plt.savefig('meas_' + str(max(n))+ '.pdf') + arr = np.array([n, t_mm, t_mm_dc, t_mm_strassen, t_wino, t_np]) + np.savetxt('meas_' + str(max(n))+ '.txt',arr) + return t_np + + +def plot(num): + arr = np.loadtxt('meas_{}.txt'.format(num)) + n, t_mm, t_mm_dc, t_mm_strassen, t_wino, t_np = arr + plt.figure(figsize=(13,8)) + plt.rcParams['font.family'] = 'STIXGeneral' + plt.rc('axes', labelsize=23) + plt.rc('xtick', labelsize=23) + plt.rc('ytick', labelsize=23) + plt.plot(n, t_mm, label='3 For Loops', lw=5) + plt.plot(n, t_mm_dc, label='Divide and Conquer', lw=5) + plt.plot(n, t_mm_strassen, label='Strassen', lw=5) + plt.plot(n, t_wino, label='Winograd', lw=5) + plt.plot(n, t_np, label='NumPy A@B', lw=5) + plt.legend() + plt.xlabel("n") + # plt.yscale('log', base=10) + plt.ylabel("time (s)") + plt.grid(True) + plt.tight_layout() + # plt.yscale('log') + plt.legend(fontsize=19) + plt.savefig('meas_' + str(num)+ '.pdf') + return arr + +def plot_c_res(ave, num): + + MM = np.loadtxt("meas/MM.txt", delimiter=',') + winograd = np.loadtxt("meas/winograd.txt", delimiter=',') + blas = np.loadtxt("meas/blas.txt", delimiter=',') + MM_dc = np.loadtxt("meas/MM_dc.txt", delimiter=',') + strassen = np.loadtxt("meas/strassen.txt", delimiter=',') + + MM_t = MM[:,0] + MM_n = MM[:,1] + # MM_t = np.mean(MM_t.reshape(-1,ave),axis=1) + # MM_n = np.mean(MM_n.reshape(-1,ave),axis=1) + + MM_dc_t = MM_dc[:,0] + MM_dc_n = MM_dc[:,1] + # MM_dc_t = np.mean(MM_dc_t.reshape(-1,ave),axis=1) + # MM_dc_n = np.mean(MM_dc_n.reshape(-1,ave),axis=1) + + strassen_t = strassen[:,0] + strassen_n = strassen[:,1] + # strassen_t = np.mean(strassen_t.reshape(-1,ave),axis=1) + # strassen_n = np.mean(strassen_n.reshape(-1,ave),axis=1) + + winograd_t = winograd[:,0] + winograd_n = winograd[:,1] + # winograd_t = np.mean(winograd_t.reshape(-1,ave),axis=1) + # winograd_n = np.mean(winograd_n.reshape(-1,ave),axis=1) + + blas_t = blas[:,0] + blas_n = blas[:,1] + # blas_t = np.mean(blas_t.reshape(-1,ave),axis=1) + # blas_n = np.mean(blas_n.reshape(-1,ave),axis=1) + + + + def func(x, a,b): + return b*x**a + + # popt, pcov = curve_fit(func, blas_n, blas_t) + # popt1, pcov2 = curve_fit(func, blas_n, winograd_t) + # popt2, pcov2 = curve_fit(func, blas_n, MM_t) + + plt.figure(figsize=(13,8)) + plt.rcParams['font.family'] = 'STIXGeneral' + plt.rc('axes', labelsize=23) + plt.rc('xtick', labelsize=23) + plt.rc('ytick', labelsize=23) + plt.loglog(MM_n, MM_t, '.', label='3 For Loops', lw=5) + plt.loglog(winograd_n, winograd_t, '.', label='Winograd MM', lw=5) + plt.loglog(blas_n, blas_t, '.', label='Blas', lw=5) + plt.loglog(strassen_n, strassen_t, '.', label='Strassen', lw=5) + plt.loglog(MM_dc_n, MM_dc_t, '.', label='Divide and Conquer', lw=5) + plt.xlabel("n") + # plt.yscale('log', base=10) + # plt.xscale('log', base=2) + plt.ylabel("time (s)") + plt.grid(True, which="both", ls="-") + plt.tight_layout() + plt.legend(fontsize=19) + plt.savefig('c_meas_' + str(num)+ '.pdf') + + # plt.plot(blas_n, func(blas_n, *popt), 'r-', label='fit blas: a=%5.5f, b=%5.10f' % tuple(popt)) + # plt.plot(blas_n, func(blas_n, *popt1), 'r-', label='fit winograd: a=%5.5f, b=%5.10f' % tuple(popt1)) + # plt.plot(blas_n, func(blas_n, *popt2), 'r-', label='fit MM: a=%5.5f, b=%5.10f' % tuple(popt2)) + + plt.legend() + # return [MM_n,winograd_n,blas_n,strassen_n,MM_dc_n] + + + return [MM_t,winograd_t,blas_t,strassen_t,MM_dc_t] + + +def mean_confidence_interval(data, confidence=0.95): + a = 1.0 * np.array(data) + n = len(a) + m, se = np.mean(a), scipy.stats.sem(a) + h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1) + return m, h + +# test%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +if __name__ == '__main__': + # A = plot_c_res(10, 4096) + # name = ['MM', 'Wino', 'blas', 'strassen', 'dc'] + # for i in range(5): + # ci_inner = [] + # print(name[i]) + # for j in range(11): + # m,h=mean_confidence_interval(A[i][j*10:(j+1)*10]) + # print("({},{})".format(2**(j+1),m)) + # np.savetxt('meas/ci/' + name[i]+'.txt',ci_inner) + + arr = plot(4096) + # n = np.logspace(1,12,12,base=2,dtype=(np.int)) + # n=[2048,4096] + # n = np.arange(1,50,2) + # A = np.random.randint(-10, 6, (5,3)) + # B = np.random.randint(-10, 6, (3,5)) + + # C = winograd2(A, B) + # C_test = A@B + # print(C) + # print(C_test) + # print(np.equal(C, C_test)) + + # t_np = test_perfomance(n) + # C = strassen(A, B) + # C_test = A@B + + + # plot_c_res() + # def func(x, a): + # return x**a + + # popt, pcov = curve_fit(func, n, t_np, bounds=(2, 3)) + + + # plt.figure() + # plt.plot(n, t_np, 'b-', label='data') + # plt.plot(n, func(n, *popt), 'r-', label='fit: a=%5.3f' % tuple(popt)) + # plt.xlabel('x') + # plt.ylabel('y') + # plt.legend() +
\ No newline at end of file diff --git a/buch/papers/multiplikation/code/__pycache__/MM.cpython-38.pyc b/buch/papers/multiplikation/code/__pycache__/MM.cpython-38.pyc Binary files differnew file mode 100644 index 0000000..7768772 --- /dev/null +++ b/buch/papers/multiplikation/code/__pycache__/MM.cpython-38.pyc diff --git a/buch/papers/multiplikation/code/c_matrix.h b/buch/papers/multiplikation/code/c_matrix.h new file mode 100644 index 0000000..63d5390 --- /dev/null +++ b/buch/papers/multiplikation/code/c_matrix.h @@ -0,0 +1,177 @@ +/* Seminar Matrizen, autogenerated File, Michael Schmid, 10/08/2021, 05:46:32 */ + +#include <stdint.h> +const int A0[][2] = + { + {60,-84}, + {-66,-1} + }; +const int B0[][2] = + { + {-45,87}, + {-38,-73} + }; +const double dB0[][2] = + { + {-45,87}, + {-38,-73} + }; +const double dA0[][2] = + { + {60,-84}, + {-66,-1} + }; +const int A1[][4] = + { + {-72,-19,-91,62}, + {-36,-74,-44,-47}, + {-39,-31,50,-93}, + {-81,2,-17,-86} + }; +const int B1[][4] = + { + {-66,39,-23,52}, + {-88,-13,13,-13}, + {-45,-70,28,-20}, + {96,5,88,96} + }; +const double dB1[][4] = + { + {-66,39,-23,52}, + {-88,-13,13,-13}, + {-45,-70,28,-20}, + {96,5,88,96} + }; +const double dA1[][4] = + { + {-72,-19,-91,62}, + {-36,-74,-44,-47}, + {-39,-31,50,-93}, + {-81,2,-17,-86} + }; +const int A2[][8] = + { + {-36,-2,-58,-32,34,-89,49,-55}, + {-68,-73,52,-3,-51,-37,-31,70}, + {73,-90,-21,-79,-15,96,-99,12}, + {68,-25,38,-73,-60,35,-99,72}, + {-43,-87,48,-84,-100,37,80,53}, + {-27,88,-5,-82,-57,-27,20,10}, + {-91,-47,54,-90,-99,-76,50,-18}, + {69,-36,76,5,-67,-38,-95,91} + }; +const int B2[][8] = + { + {-84,22,-13,-66,-42,51,66,0}, + {37,-65,66,-85,-10,-23,77,5}, + {1,41,-79,0,63,-37,-10,29}, + {72,66,-99,92,-28,65,25,-40}, + {69,-49,65,-18,64,-97,-47,30}, + {36,86,66,-12,-17,89,1,-37}, + {-100,11,27,23,-75,-23,96,-9}, + {68,90,-87,-99,-70,-28,98,-76} + }; +const double dB2[][8] = + { + {-84,22,-13,-66,-42,51,66,0}, + {37,-65,66,-85,-10,-23,77,5}, + {1,41,-79,0,63,-37,-10,29}, + {72,66,-99,92,-28,65,25,-40}, + {69,-49,65,-18,64,-97,-47,30}, + {36,86,66,-12,-17,89,1,-37}, + {-100,11,27,23,-75,-23,96,-9}, + {68,90,-87,-99,-70,-28,98,-76} + }; +const double dA2[][8] = + { + {-36,-2,-58,-32,34,-89,49,-55}, + {-68,-73,52,-3,-51,-37,-31,70}, + {73,-90,-21,-79,-15,96,-99,12}, + {68,-25,38,-73,-60,35,-99,72}, + {-43,-87,48,-84,-100,37,80,53}, + {-27,88,-5,-82,-57,-27,20,10}, + {-91,-47,54,-90,-99,-76,50,-18}, + {69,-36,76,5,-67,-38,-95,91} + }; +const int A3[][16] = + { + {-24,65,21,19,94,70,-90,-81,53,-41,-23,-1,58,-80,-54,59}, + {-42,76,-19,98,29,-56,92,14,45,11,82,83,48,-13,81,66}, + {43,-57,-67,95,5,72,11,0,-47,55,-24,36,84,54,-31,-54}, + {-39,-40,19,97,-82,-56,27,95,81,-21,-50,-74,-35,-87,-28,-26}, + {-74,-98,79,92,-24,-48,99,94,55,-83,70,98,-24,18,-67,14}, + {20,76,11,-23,-56,21,0,42,64,86,-74,44,93,-76,-30,97}, + {13,20,-73,-11,-30,80,53,-8,60,21,17,-42,82,-72,-6,-80}, + {36,-93,-64,-21,20,-85,15,24,99,81,-52,64,71,-56,52,63}, + {32,9,-2,-85,17,62,-98,-35,75,-58,-44,-20,-47,89,-95,52}, + {93,-43,86,68,-6,-25,90,57,60,-10,65,-97,43,46,-60,-41}, + {43,-33,0,50,-100,26,-60,95,39,-70,-61,-81,9,-23,-99,-4}, + {20,61,15,43,-96,93,-55,38,-29,-1,-10,26,-87,18,64,6}, + {-98,-84,51,16,-14,86,52,59,44,-39,-2,10,82,-66,54,19}, + {89,-49,-37,-6,-53,40,-11,46,-51,-56,86,34,11,13,-20,-49}, + {-90,14,28,-45,-25,-56,-51,-61,28,-8,51,91,95,-10,-85,58}, + {8,-44,88,-71,-27,11,89,37,86,-78,-44,-56,-87,0,-42,-61} + }; +const int B3[][16] = + { + {62,-30,62,92,29,-93,-95,44,-33,-88,-29,9,-88,-42,-90,-70}, + {60,37,-44,-93,-87,6,-53,2,-29,53,-49,59,6,83,-15,50}, + {-19,85,-49,-14,84,-4,12,88,-83,-81,-24,-16,-12,-42,-63,-71}, + {-42,-78,-58,-61,-29,67,-28,-46,64,7,6,-13,88,-42,95,-24}, + {-90,-56,8,-30,-89,70,37,-29,24,-8,-10,-2,-25,-63,-95,-91}, + {10,-81,42,-28,-13,-68,-72,-20,-22,5,-79,-50,-88,62,57,69}, + {-67,24,-71,-43,11,48,33,-93,-82,-65,-4,5,-15,25,-54,-45}, + {-49,19,-29,90,-97,-87,78,-39,-75,-85,-79,-35,54,3,-73,7}, + {-7,39,70,-42,32,-100,56,4,-24,-57,38,-49,-50,-44,79,-42}, + {37,-65,-55,22,-97,-42,-76,95,97,-27,38,11,0,-81,-23,35}, + {26,-70,10,-29,47,-70,-52,29,-13,-18,5,34,18,32,87,91}, + {-84,41,-19,96,-51,-19,81,75,81,92,2,-40,-42,-69,-10,-61}, + {-30,98,71,-51,91,-59,58,86,86,-22,-84,7,66,-55,-52,23}, + {-71,-44,-9,90,26,18,26,-10,-85,64,-47,3,72,81,74,-8}, + {52,-59,-91,22,8,-63,84,9,-11,-54,-78,-71,-98,42,96,57}, + {18,-39,34,-50,-62,-96,-2,-78,52,94,-33,2,-19,-9,-86,-75} + }; +const double dB3[][16] = + { + {62,-30,62,92,29,-93,-95,44,-33,-88,-29,9,-88,-42,-90,-70}, + {60,37,-44,-93,-87,6,-53,2,-29,53,-49,59,6,83,-15,50}, + {-19,85,-49,-14,84,-4,12,88,-83,-81,-24,-16,-12,-42,-63,-71}, + {-42,-78,-58,-61,-29,67,-28,-46,64,7,6,-13,88,-42,95,-24}, + {-90,-56,8,-30,-89,70,37,-29,24,-8,-10,-2,-25,-63,-95,-91}, + {10,-81,42,-28,-13,-68,-72,-20,-22,5,-79,-50,-88,62,57,69}, + {-67,24,-71,-43,11,48,33,-93,-82,-65,-4,5,-15,25,-54,-45}, + {-49,19,-29,90,-97,-87,78,-39,-75,-85,-79,-35,54,3,-73,7}, + {-7,39,70,-42,32,-100,56,4,-24,-57,38,-49,-50,-44,79,-42}, + {37,-65,-55,22,-97,-42,-76,95,97,-27,38,11,0,-81,-23,35}, + {26,-70,10,-29,47,-70,-52,29,-13,-18,5,34,18,32,87,91}, + {-84,41,-19,96,-51,-19,81,75,81,92,2,-40,-42,-69,-10,-61}, + {-30,98,71,-51,91,-59,58,86,86,-22,-84,7,66,-55,-52,23}, + {-71,-44,-9,90,26,18,26,-10,-85,64,-47,3,72,81,74,-8}, + {52,-59,-91,22,8,-63,84,9,-11,-54,-78,-71,-98,42,96,57}, + {18,-39,34,-50,-62,-96,-2,-78,52,94,-33,2,-19,-9,-86,-75} + }; +const double dA3[][16] = + { + {-24,65,21,19,94,70,-90,-81,53,-41,-23,-1,58,-80,-54,59}, + {-42,76,-19,98,29,-56,92,14,45,11,82,83,48,-13,81,66}, + {43,-57,-67,95,5,72,11,0,-47,55,-24,36,84,54,-31,-54}, + {-39,-40,19,97,-82,-56,27,95,81,-21,-50,-74,-35,-87,-28,-26}, + {-74,-98,79,92,-24,-48,99,94,55,-83,70,98,-24,18,-67,14}, + {20,76,11,-23,-56,21,0,42,64,86,-74,44,93,-76,-30,97}, + {13,20,-73,-11,-30,80,53,-8,60,21,17,-42,82,-72,-6,-80}, + {36,-93,-64,-21,20,-85,15,24,99,81,-52,64,71,-56,52,63}, + {32,9,-2,-85,17,62,-98,-35,75,-58,-44,-20,-47,89,-95,52}, + {93,-43,86,68,-6,-25,90,57,60,-10,65,-97,43,46,-60,-41}, + {43,-33,0,50,-100,26,-60,95,39,-70,-61,-81,9,-23,-99,-4}, + {20,61,15,43,-96,93,-55,38,-29,-1,-10,26,-87,18,64,6}, + {-98,-84,51,16,-14,86,52,59,44,-39,-2,10,82,-66,54,19}, + {89,-49,-37,-6,-53,40,-11,46,-51,-56,86,34,11,13,-20,-49}, + {-90,14,28,-45,-25,-56,-51,-61,28,-8,51,91,95,-10,-85,58}, + {8,-44,88,-71,-27,11,89,37,86,-78,-44,-56,-87,0,-42,-61} + }; +const int *Ap[4] = {(int*) A0,(int*) A1,(int*) A2,(int*) A3}; +const int *Bp[4] = {(int*) B0,(int*) B1,(int*) B2,(int*) B3}; +const double *dAp[4] = {(double*) dA0,(double*) dA1,(double*) dA2,(double*) dA3}; +const double *dBp[4] = {(double*) dB0,(double*) dB1,(double*) dB2,(double*) dB3}; +int n[4] = {2,4,8,16}; +int n_arrays = 4; diff --git a/buch/papers/multiplikation/code/c_meas_1024.pdf b/buch/papers/multiplikation/code/c_meas_1024.pdf Binary files differnew file mode 100644 index 0000000..95b68b5 --- /dev/null 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b/buch/papers/multiplikation/code/c_meas_64.pdf diff --git a/buch/papers/multiplikation/code/c_meas_8.pdf b/buch/papers/multiplikation/code/c_meas_8.pdf Binary files differnew file mode 100644 index 0000000..9682aca --- /dev/null +++ b/buch/papers/multiplikation/code/c_meas_8.pdf diff --git a/buch/papers/multiplikation/code/ci.txt b/buch/papers/multiplikation/code/ci.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/ci.txt diff --git a/buch/papers/multiplikation/code/helper_class.py b/buch/papers/multiplikation/code/helper_class.py new file mode 100755 index 0000000..3b74f67 --- /dev/null +++ b/buch/papers/multiplikation/code/helper_class.py @@ -0,0 +1,106 @@ +#!/usr/bin/env python3 +# -*- coding: utf-8 -*- +""" +Created on Fri Mar 12 09:02:48 2021 + +@author: nunigan +""" + +from datetime import datetime +import numpy as np + +class Helper(): + def __init__(self): + pass + + def write_c_matrix(self, n_array): + + with open('c_matrix.h', 'w') as file: + file.writelines('/* Seminar Matrizen, autogenerated File, Michael Schmid, {} */ \n \n'.format(datetime.now().strftime("%d/%m/%Y, %H:%M:%S"))) + + file.writelines('#include <stdint.h> \n') + + + + for k, n in enumerate(n_array): + A = np.random.randint(-100,100,(n,n)) + B = np.random.randint(-100,100,(n,n)) + file.writelines('const int A{}[][{}] = \n'.format(k, n)) + file.writelines(' {\n') + for i in range(n): + file.writelines(' {') + for j in range(n): + if j == n-1: + file.writelines('{}'.format(A[i,j])) + else: + file.writelines('{},'.format(A[i,j])) + if i == n-1: + file.writelines('}\n') + else: + file.writelines('},\n') + + file.writelines(' };\n') + + file.writelines('const int B{}[][{}] = \n'.format(k,n)) + file.writelines(' {\n') + for i in range(n): + file.writelines(' {') + for j in range(n): + if j == n-1: + file.writelines('{}'.format(B[i,j])) + else: + file.writelines('{},'.format(B[i,j])) + if i == n-1: + file.writelines('}\n') + else: + file.writelines('},\n') + + file.writelines(' };\n') + + file.writelines('const double dB{}[][{}] = \n'.format(k,n)) + file.writelines(' {\n') + for i in range(n): + file.writelines(' {') + for j in range(n): + if j == n-1: + file.writelines('{}'.format(B[i,j])) + else: + file.writelines('{},'.format(B[i,j])) + if i == n-1: + file.writelines('}\n') + else: + file.writelines('},\n') + + file.writelines(' };\n') + + file.writelines('const double dA{}[][{}] = \n'.format(k,n)) + file.writelines(' {\n') + for i in range(n): + file.writelines(' {') + for j in range(n): + if j == n-1: + file.writelines('{}'.format(A[i,j])) + else: + file.writelines('{},'.format(A[i,j])) + if i == n-1: + file.writelines('}\n') + else: + file.writelines('},\n') + + file.writelines(' };\n') + + file.writelines('const int *Ap[{}] = {{{}}}; \n'.format(len(n_array),",".join(['(int*) A'+str(element) for element in np.arange(len(n_array))]))) + file.writelines('const int *Bp[{}] = {{{}}}; \n'.format(len(n_array),",".join(['(int*) B'+str(element) for element in np.arange(len(n_array))]))) + file.writelines('const double *dAp[{}] = {{{}}}; \n'.format(len(n_array),",".join(['(double*) dA'+str(element) for element in np.arange(len(n_array))]))) + file.writelines('const double *dBp[{}] = {{{}}}; \n'.format(len(n_array),",".join(['(double*) dB'+str(element) for element in np.arange(len(n_array))]))) + file.writelines('int n[{}] = {{{}}}; \n'.format(len(n_array),",".join([str(element) for element in n_array]))) + file.writelines('int n_arrays = {};\n'.format(len(n_array))) + +# test%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +if __name__ == '__main__': + + helper = Helper() + # n = np.arange(2,10) + n = np.logspace(1,11,11,base=2,dtype=(np.int)) + # n=[8192] + # C = helper.write_c_matrix(n) diff --git a/buch/papers/multiplikation/code/meas/MM.txt b/buch/papers/multiplikation/code/meas/MM.txt new file mode 100644 index 0000000..7bffb6e --- /dev/null +++ b/buch/papers/multiplikation/code/meas/MM.txt @@ -0,0 +1,110 @@ +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000001,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000001,4 +0.000001,4 +0.000002,8 +0.000002,8 +0.000002,8 +0.000002,8 +0.000002,8 +0.000002,8 +0.000002,8 +0.000002,8 +0.000001,8 +0.000001,8 +0.000011,16 +0.000011,16 +0.000011,16 +0.000011,16 +0.000011,16 +0.000021,16 +0.000011,16 +0.000011,16 +0.000011,16 +0.000011,16 +0.000090,32 +0.000093,32 +0.000083,32 +0.000082,32 +0.000090,32 +0.000080,32 +0.000080,32 +0.000080,32 +0.000089,32 +0.000126,32 +0.000771,64 +0.000651,64 +0.000651,64 +0.000651,64 +0.000731,64 +0.000673,64 +0.000745,64 +0.000672,64 +0.000671,64 +0.000707,64 +0.005642,128 +0.005579,128 +0.005768,128 +0.005745,128 +0.005518,128 +0.005877,128 +0.005513,128 +0.005850,128 +0.005769,128 +0.005581,128 +0.052188,256 +0.051988,256 +0.051888,256 +0.051518,256 +0.051709,256 +0.051543,256 +0.051707,256 +0.051845,256 +0.051495,256 +0.051834,256 +0.507020,512 +0.504111,512 +0.502049,512 +0.529743,512 +0.501028,512 +0.502097,512 +0.503490,512 +0.502079,512 +0.506688,512 +0.504163,512 +4.538722,1024 +4.291473,1024 +4.516302,1024 +4.374630,1024 +4.719557,1024 +4.438999,1024 +4.641680,1024 +4.407959,1024 +4.441451,1024 +4.677313,1024 +129.433279,2048 +129.277802,2048 +129.284817,2048 +129.086884,2048 +129.197444,2048 +129.350999,2048 +129.264250,2048 +129.295723,2048 +129.402601,2048 +129.300820,2048 diff --git a/buch/papers/multiplikation/code/meas/MM_dc.txt b/buch/papers/multiplikation/code/meas/MM_dc.txt new file mode 100644 index 0000000..b78b925 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/MM_dc.txt @@ -0,0 +1,110 @@ +0.000003,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000002,4 +0.000001,4 +0.000001,4 +0.000001,4 +0.000001,4 +0.000001,4 +0.000001,4 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+17.639024,1024 +17.625402,1024 +17.722286,1024 +17.611777,1024 +17.653120,1024 +17.748270,1024 +17.691817,1024 +17.614448,1024 +141.943689,2048 +141.580812,2048 +141.882050,2048 +141.516253,2048 +141.351237,2048 +141.641167,2048 +141.596407,2048 +141.607048,2048 +141.469723,2048 +141.515550,2048 diff --git a/buch/papers/multiplikation/code/meas/blas.txt b/buch/papers/multiplikation/code/meas/blas.txt new file mode 100644 index 0000000..9414d8f --- /dev/null +++ b/buch/papers/multiplikation/code/meas/blas.txt @@ -0,0 +1,110 @@ +0.000001,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,2 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000000,4 +0.000001,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000000,8 +0.000003,16 +0.000003,16 +0.000003,16 +0.000003,16 +0.000003,16 +0.000003,16 +0.000012,16 +0.000003,16 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+7.619210,2048 diff --git a/buch/papers/multiplikation/code/meas/ci/MM.txt b/buch/papers/multiplikation/code/meas/ci/MM.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/ci/MM.txt diff --git a/buch/papers/multiplikation/code/meas/ci/Wino.txt b/buch/papers/multiplikation/code/meas/ci/Wino.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/ci/Wino.txt diff --git a/buch/papers/multiplikation/code/meas/ci/blas.txt b/buch/papers/multiplikation/code/meas/ci/blas.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/ci/blas.txt diff --git a/buch/papers/multiplikation/code/meas/ci/dc.txt b/buch/papers/multiplikation/code/meas/ci/dc.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/ci/dc.txt diff --git a/buch/papers/multiplikation/code/meas/ci/strassen.txt b/buch/papers/multiplikation/code/meas/ci/strassen.txt new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/ci/strassen.txt diff --git a/buch/papers/multiplikation/code/meas/old/8196/MM.txt b/buch/papers/multiplikation/code/meas/old/8196/MM.txt new file mode 100644 index 0000000..0edf9f6 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/old/8196/MM.txt @@ -0,0 +1 @@ +9376.173434,8192 diff --git a/buch/papers/multiplikation/code/meas/old/8196/MM_dc.txt b/buch/papers/multiplikation/code/meas/old/8196/MM_dc.txt new file mode 100644 index 0000000..36f6ff0 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/old/8196/MM_dc.txt @@ -0,0 +1 @@ +9606.402522,8192 diff --git a/buch/papers/multiplikation/code/meas/old/8196/blas.txt b/buch/papers/multiplikation/code/meas/old/8196/blas.txt new file mode 100644 index 0000000..b5989fb --- /dev/null +++ b/buch/papers/multiplikation/code/meas/old/8196/blas.txt @@ -0,0 +1 @@ +478.429957,8192 diff 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a/buch/papers/multiplikation/code/meas/old/MM_dc.txt b/buch/papers/multiplikation/code/meas/old/MM_dc.txt new file mode 100644 index 0000000..f6be928 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/old/MM_dc.txt @@ -0,0 +1,12 @@ +0.000003,2 +0.000002,4 +0.000010,8 +0.000068,16 +0.000594,32 +0.004264,64 +0.036289,128 +0.324645,256 +2.612010,512 +19.928951,1024 +159.333884,2048 +1147.106865,4096 diff --git a/buch/papers/multiplikation/code/meas/old/blas.txt b/buch/papers/multiplikation/code/meas/old/blas.txt new file mode 100644 index 0000000..92a61b9 --- /dev/null +++ b/buch/papers/multiplikation/code/meas/old/blas.txt @@ -0,0 +1,12 @@ +0.000001,2 +0.000001,4 +0.000001,8 +0.000003,16 +0.000022,32 +0.000179,64 +0.001278,128 +0.010165,256 +0.074739,512 +0.704748,1024 +6.845095,2048 +55.845038,4096 diff --git a/buch/papers/multiplikation/code/meas/old/strassen.txt b/buch/papers/multiplikation/code/meas/old/strassen.txt new file mode 100644 index 0000000..fdfbf2b --- /dev/null +++ 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+\definecolor{color0}{rgb}{0.886274509803922,0.290196078431373,0.2} +\definecolor{color1}{rgb}{0.203921568627451,0.541176470588235,0.741176470588235} +\definecolor{color2}{rgb}{0.596078431372549,0.556862745098039,0.835294117647059} +\definecolor{color3}{rgb}{0.984313725490196,0.756862745098039,0.368627450980392} + +\begin{axis}[ +axis background/.style={fill=white!89.8039215686275!black}, +axis line style={white}, +legend cell align={left}, +legend style={ + fill opacity=0.8, + draw opacity=1, + text opacity=1, + at={(0.03,0.97)}, + anchor=north west, + draw=white!80!black, + fill=white!89.8039215686275!black +}, +tick align=outside, +tick pos=left, +x grid style={white}, +xlabel={n}, +xmajorgrids, +xmin=-4.3, xmax=134.3, +xtick style={color=white!33.3333333333333!black}, +y grid style={white}, +ylabel={time (s)}, +ymajorgrids, +ymin=-0.0834965705871582, ymax=1.75356960296631, +ytick style={color=white!33.3333333333333!black} +] +\addplot [line width=2pt, color0] +table {% +2 1.57356262207031e-05 +4 5.96046447753906e-05 +8 0.000428915023803711 +16 0.00276041030883789 +32 0.0217020511627197 +64 0.160412073135376 +128 1.3419406414032 +}; +\addlegendentry{Standard MM} +\addplot [line width=2pt, color1] +table {% +2 6.43730163574219e-06 +4 6.69956207275391e-05 +8 0.00048065185546875 +16 0.00336766242980957 +32 0.0257236957550049 +64 0.231612205505371 +128 1.67006659507751 +}; +\addlegendentry{Divide and conquer MM} +\addplot [line width=2pt, color2] +table {% +2 2.90870666503906e-05 +4 0.000133275985717773 +8 0.000703096389770508 +16 0.00453472137451172 +32 0.0282893180847168 +64 0.181003332138062 +128 1.40816903114319 +}; +\addlegendentry{Strassen MM} +\addplot [line width=2pt, white!46.6666666666667!black] +table {% +2 2.19345092773438e-05 +4 9.01222229003906e-05 +8 0.000406503677368164 +16 0.00258469581604004 +32 0.0171687602996826 +64 0.126588344573975 +128 1.02698183059692 +}; +\addlegendentry{Winograd MM} +\addplot [line width=2pt, color3] +table {% +2 1.45435333251953e-05 +4 1.1444091796875e-05 +8 7.39097595214844e-06 +16 1.28746032714844e-05 +32 2.83718109130859e-05 +64 0.000111103057861328 +128 0.000159025192260742 +}; +\addlegendentry{np MM} +\end{axis} + +\end{tikzpicture} diff --git a/buch/papers/multiplikation/einlteung.tex b/buch/papers/multiplikation/einlteung.tex new file mode 100755 index 0000000..9b03a4e --- /dev/null +++ b/buch/papers/multiplikation/einlteung.tex @@ -0,0 +1,51 @@ +% +% einleitung.tex -- Beispiel-File für die Einleitung +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Matrizenmultiplikation \label{multiplikation:section:einleitung}} +\rhead{Matrizenmultiplikation} + +Die Multiplikation zweier Matrizen ist eine wichtige Operation, die in verschiedensten Teilen der Mathematik Anwendung findet. +Die Beschreibung der Multiplikation aus der Definition 2.10: +Eine $m\times n$-Matrix $\mathbf{A}\in M_{m\times n}(\Bbbk)$ und eine +$n\times p$-Matrix $\mathbf{B}\in M_{n\times p}(\Bbbk)$ haben als Produkt +eine $m\times p$-Matrix $\mathbf{C}=\mathbf{AB}\in M_{m\times p}(\Bbbk)$ mit den +Koeffizienten +\begin{equation} +C_{ij} = \sum_{k=1}^n A_{ik} B_{kj}. +\label{multiplikation:eq:MM} +\end{equation} +Grafisch kann die Matrizenmultiplikation $\mathbf{AB}=\mathbf{C}$ wie in Abbildung \ref{multiplikation:fig:mm_viz} visualisiert werden. +Im Fall einer Matrizengr\"osse von $2\times 2$ kann die Matrixgleichung +\begin{equation} + \begin{bmatrix} +A_{11} & A_{12}\\ +A_{21} & A_{22} +\end{bmatrix} +\begin{bmatrix} +B_{11} & B_{12}\\ +B_{21} & B_{22} +\end{bmatrix} += +\begin{bmatrix} +C_{11} & C_{12}\\ +C_{21} & C_{22} +\end{bmatrix} +\end{equation} +explizt als Gleichungen +\begin{equation} \label{multiplikation:eq:MM_exp} +\begin{split} +C_{11} &= A_{11} \cdot B_{11} + A_{12} \cdot B_{21}\\ +C_{12} &= A_{11} \cdot B_{12} + A_{12} \cdot B_{22}\\ +C_{21} &= A_{21} \cdot B_{11} + A_{22} \cdot B_{21}\\ +C_{22} &= A_{21} \cdot B_{12} + A_{22} \cdot B_{22} +\end{split} +\end{equation} +der einzelnen Terme geschrieben werden. +\begin{figure} + \center + \includegraphics[]{papers/multiplikation/images/mm_visualisation} + \caption{Grafische Illustration der Matrizenmultiplikation} + \label{multiplikation:fig:mm_viz} +\end{figure} diff --git a/buch/papers/multiplikation/images/algo_tab.pdf b/buch/papers/multiplikation/images/algo_tab.pdf Binary files differnew file mode 100644 index 0000000..7f2bb4f --- /dev/null +++ b/buch/papers/multiplikation/images/algo_tab.pdf diff --git a/buch/papers/multiplikation/images/algo_tab.tex b/buch/papers/multiplikation/images/algo_tab.tex new file mode 100644 index 0000000..50ce392 --- /dev/null +++ b/buch/papers/multiplikation/images/algo_tab.tex @@ -0,0 +1,122 @@ +\documentclass{article} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{algorithm} +\usepackage{algpseudocode} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{pgfplots} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{multicol} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usetikzlibrary{decorations.text} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + + + +\begin{table}[t] + \begin{tabular}{ll} + \begin{minipage}{0.4\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:b1} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{B1}{$a, b$} + \State \textbf{return} $a+b$ + \EndFunction + \State + \State + \end{algorithmic} + \end{algorithm} + \end{minipage} + & + \begin{minipage}{0.4\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:b2} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{B2}{$a, b$} + \State $ x \gets a+b $ + \State $ y \gets a \cdot b $ + \State \textbf{return} $x+y$ + \EndFunction + \end{algorithmic} +\end{algorithm} + + \end{minipage} + \end{tabular} +\end{table} + +\begin{table} + \begin{tabular}[t]{ll} + \begin{minipage}{0.4\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \label{multiplikation:alg:linear} + \Function{L}{$\mathbf{a}, \mathbf{b}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[i] $ + \EndFor + + \State \textbf{return} $sum$ + + \EndFunction + \State + \State + \end{algorithmic} + \end{algorithm} + \end{minipage} + & + \begin{minipage}{0.4\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:q1} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{Q}{$\mathbf{A}, \mathbf{B}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \For{$j = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[j] $ + \EndFor + \EndFor + \State \textbf{return} $sum$ + \EndFunction + \end{algorithmic} + \end{algorithm} + \end{minipage} + \end{tabular} +\end{table} + +dhdfh +\end{document} diff --git a/buch/papers/multiplikation/images/bigo.pdf b/buch/papers/multiplikation/images/bigo.pdf Binary files differnew file mode 100644 index 0000000..2519553 --- /dev/null +++ b/buch/papers/multiplikation/images/bigo.pdf diff --git a/buch/papers/multiplikation/images/bigo.tex b/buch/papers/multiplikation/images/bigo.tex new file mode 100644 index 0000000..63fd0fd --- /dev/null +++ b/buch/papers/multiplikation/images/bigo.tex @@ -0,0 +1,112 @@ +\documentclass[border=10pt,varwidth]{standalone} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{pgfplots} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usetikzlibrary{decorations.text} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +\begin{tikzpicture} + +\begin{axis}[ + xmode=log, ymode=log, + xmin=1e-0, xmax=5000, + ymin=10e-1, ymax=1e7, + grid=both, + major grid style={black!50}, + xlabel = data input size, + ylabel = {time}, + legend pos=north west, + very thick, + yticklabels=\empty, + xticklabels=\empty, + scale only axis=true, + width=12cm, height=8cm, + legend cell align={left} + ] +\addplot [ + domain= 1:5000, + samples=100, + color=red, +] +{1}; +\addlegendentry{$\mathcal{O}(1)$} +\addplot [ + domain= 1:5000, + samples=100, + color=green, +] +{x}; +\addlegendentry{$\mathcal{O}(n)$} +\addplot [ + domain= 1:50000, + samples=100, + color=blue, +] +{x^2}; +\addlegendentry{$\mathcal{O}\left(n^2\right)$} +\addplot [ + domain= 1:500, + samples=100, + color=purple, +] +{x^3}; +\addlegendentry{$\mathcal{O}\left(n^3\right)$} +\addplot [ + domain= 1:500, + samples=100, + color=black, +] +{exp(x) - 1.7}; +\addlegendentry{$\mathcal{O}\left(e^n\right)$} +\addplot [ + domain= 1:5000, + samples=100, + color=orange, +] +{log2(x)+1}; +\addlegendentry{$\mathcal{O}(\log n)$} + +\addplot [ + domain= 1:5000, + samples=100, + color=gray, +] +{x*log2(x)+1}; +\addlegendentry{$\mathcal{O}(n \log n)$} +\end{axis} +\end{tikzpicture} + +\end{document} diff --git a/buch/papers/multiplikation/images/c_meas_4096.pdf b/buch/papers/multiplikation/images/c_meas_4096.pdf Binary files differnew file mode 100644 index 0000000..304015a --- /dev/null +++ b/buch/papers/multiplikation/images/c_meas_4096.pdf diff --git a/buch/papers/multiplikation/images/meas_1024.pdf b/buch/papers/multiplikation/images/meas_1024.pdf Binary files differnew file mode 100644 index 0000000..70c7ec1 --- /dev/null +++ b/buch/papers/multiplikation/images/meas_1024.pdf diff --git a/buch/papers/multiplikation/images/meas_c.pdf b/buch/papers/multiplikation/images/meas_c.pdf Binary files differnew file mode 100644 index 0000000..521151e --- /dev/null +++ b/buch/papers/multiplikation/images/meas_c.pdf diff --git a/buch/papers/multiplikation/images/meas_c.tex b/buch/papers/multiplikation/images/meas_c.tex new file mode 100644 index 0000000..12d3527 --- /dev/null +++ b/buch/papers/multiplikation/images/meas_c.tex @@ -0,0 +1,150 @@ + +\documentclass[border=10pt,varwidth]{standalone} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{pgfplots} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usepackage{pgfplotstable} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usetikzlibrary{decorations.text} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +\begin{tikzpicture} +\begin{axis}[ +xmode=log, ymode=log, +xmin=30, xmax=10000, +ymin=1e-5, ymax=2e4, +grid=both, +major grid style={black!50}, +xlabel = data input ($n$), +ylabel = {time ($s$)}, +legend pos=north west, +very thick, +scale only axis=true, +width=12cm, height=8cm, + log basis x={10}, + legend cell align={left} +] +\addlegendentry{Winograd} +\addplot[ color=blue, + error bars/.cd, y dir=both, y explicit, +] coordinates { +%(2,1e-07) +%(4,5e-07) +%(8,2.0000000000000003e-06) +%(16,1.1999999999999999e-05) +(32,8.329999999999999e-05) +(64,0.0006479) +(128,0.0052873) +(256,0.052674599999999995) +(512,0.5249752000000001) +(1024,4.671161) +(2048,136.6769777) +(4096,1179.261048) +(8192,10071.512655) +}; +\addlegendentry{Strassen} +\addplot [ color=black, +]coordinates { +%(2,1e-07) +%(4,2.1e-06) +%(8,1.13e-05) +%(16,7.07e-05) +(32,0.0005041) +(64,0.003596) +(128,0.0254481) +(256,0.1781817) +(512,1.2555) +(1024,8.8302371) +(2048,61.9018691) +(4096,414.648901) +(8192,3014.235467) +}; + +\addlegendentry{MM div and conq} +\addplot[ color=green, +] coordinates { +%(2,3e-07) +%(4,1.1e-06) +%(8,8.6e-06) +%(16,7.819999999999999e-05) +(32,0.0005940000000000001) +(64,0.0044339) +(128,0.0348443) +(256,0.29484730000000003) +(512,2.2228507) +(1024,17.659234500000004) +(2048,141.6103936) +(4096,1147.106865) +(8192,9606.402522) +}; + +\addlegendentry{MM} +\addplot [ color=red, +]coordinates { +%(2,0.0) +%(4,3e-07) +%(8,1.8000000000000001e-06) +%(16,1.1999999999999999e-05) +(32,8.93e-05) +(64,0.0006923) +(128,0.0056842) +(256,0.051771500000000005) +(512,0.5062468000000001) +(1024,4.5048086) +(2048,129.2894619) +(4096,1111.312696) +(8192,9376.173434) +}; +\addlegendentry{BLAS} +\addplot[ color=purple, +] coordinates { +%(2,1e-07) +%(4,0.0) +%(8,1e-07) +%(16,3.9e-06) +(32,2.1000000000000002e-05) +(64,0.00018580000000000002) +(128,0.0012649) +(256,0.0096489) +(512,0.0773765) +(1024,0.7643868) +(2048,7.6320993999999995) +(4096,55.845038) +(8192,478.429957) +}; +\end{axis} +\end{tikzpicture} + +\end{document} diff --git a/buch/papers/multiplikation/images/meas_python.pdf b/buch/papers/multiplikation/images/meas_python.pdf Binary files differnew file mode 100644 index 0000000..fe89773 --- /dev/null +++ b/buch/papers/multiplikation/images/meas_python.pdf diff --git a/buch/papers/multiplikation/images/meas_python.tex b/buch/papers/multiplikation/images/meas_python.tex new file mode 100644 index 0000000..ad43cf6 --- /dev/null +++ b/buch/papers/multiplikation/images/meas_python.tex @@ -0,0 +1,145 @@ + +\documentclass[border=10pt,varwidth]{standalone} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{pgfplots} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usepackage{pgfplotstable} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usetikzlibrary{decorations.text} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +\begin{tikzpicture} +\begin{axis}[ +xmode=log, ymode=log, +xmin=30, xmax=4200, +ymin=0.01, ymax=70000, +grid=both, +major grid style={black!50}, +xlabel = data input ($n$), +ylabel = {time ($s$)}, +legend pos=north west, +very thick, +scale only axis=true, +width=12cm, height=8cm, + log basis x={10}, + legend cell align={left} +] +\addlegendentry{Winograd} +\addplot[ color=blue, +] coordinates { +% (2, 2.7895e-05 ) +% (4, 0.000104904) +% (8, 0.000552893) +% (16, 0.0045557 ) +(32, 0.0187144 ) +(64, 0.153069 ) +(128, 1.19476 ) +(256, 8.29899 ) +(512, 68.3699 ) +(1024,537.374 ) +(2046,4884.61) +(4096,43597.1) +}; +\addlegendentry{Strassen} +\addplot [ color=black, +]coordinates { + % (2,2.09808e-05 ) + % (4,0.000174284 ) + % (8,0.000943899 ) + % (16,0.00475407 ) + (32,0.0485256 ) + (64,0.220414 ) + (128,1.44718 ) + (256,9.93866 ) + (512,63.961 ) +(1024,461.494 ) +(2046,3860.57) +(4096,22904.3) +}; + +\addlegendentry{MM div and conq} +\addplot[ color=green, +] coordinates { + % (2,8.10623e-06 ) + % (4,9.01222e-05 ) + % (8,0.000729084 ) + % (16,0.00497079 ) + (32,0.02719 ) + (64,0.26528 ) + (128,1.77787 ) + (256,13.27 ) + (512,105.397 ) +(1024,847.321 ) +(2046,7375.93) +(4096,58466) +}; + +\addlegendentry{MM} +\addplot [ color=red, +]coordinates { + % (2,1.85966e-05) + % (4,8.29697e-05 ) + % (8,0.000547171) + % (16,0.00305367 ) + (32, 0.0240743 ) + (64, 0.186895 ) + (128, 1.56369 ) + (256, 11.0062 ) + (512, 85.4768) +(1024,750.757 ) +(2046,6154.18) +(4096,46813.3) +}; +% \addlegendentry{NumPy} +% \addplot[ color=blue, +% ] coordinates { +% % (2,1.83582e-05 ) +% % (4,7.86781e-06) +% % (8,1.00136e-05) +% % (16,5.4121e-05 ) +% (32,4.26769e-05) +% (64,0.000118494) +% (128,0.000244141 ) +% (256,0.000695705 ) +% (512,0.00221705 ) +% (1024,0.0188088 ) +% (2046,0.215739) +% (4096,1.49159) +% }; + +\end{axis} +\end{tikzpicture} + +\end{document} diff --git a/buch/papers/multiplikation/images/mm_visualisation.pdf b/buch/papers/multiplikation/images/mm_visualisation.pdf Binary files differnew file mode 100644 index 0000000..9309df1 --- /dev/null +++ b/buch/papers/multiplikation/images/mm_visualisation.pdf diff --git a/buch/papers/multiplikation/images/mm_visualisation.tex b/buch/papers/multiplikation/images/mm_visualisation.tex new file mode 100644 index 0000000..6e8f789 --- /dev/null +++ b/buch/papers/multiplikation/images/mm_visualisation.tex @@ -0,0 +1,45 @@ + + \begin{tikzpicture}[ampersand replacement=\&] + + \matrix (A)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (0,0) + { + A_{1,1} \& \cdots \& A_{1,k} \& \cdots \& A_{1,n} \\ + \vdots \& \& \vdots \& \& \vdots \\ + A_{i,1} \& \cdots \& A_{i,k} \& \cdots \& A_{i,n} \\ + \vdots \& \& \vdots \& \& \vdots \\ + A_{m,1} \& \cdots \& A_{m,k} \& \cdots \& A_{m,n} \\ + }; + + \node [right=0.1 of A] (mul) {$\cdot$}; + + + \matrix (B)[right=0.1 of mul, matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] + { + B_{1,1} \& \cdots \& B_{1,j} \& \cdots \& B_{1,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + B_{k,1} \& \cdots \& B_{k,j} \& \cdots \& B_{k,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + B_{n,1} \& \cdots \& B_{n,j} \& \cdots \& B_{n,p} \\ + }; + + \node [right=0.1 of B] (eq) {$=$}; + + \matrix (C)[right=0.1 of eq, matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] + { + C_{1,1} \& \cdots \& C_{1,j} \& \cdots \& C_{1,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + C_{i,1} \& \cdots \& C_{i,j} \& \cdots \& C_{i,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + C_{m,1} \& \cdots \& C_{m,j} \& \cdots \& C_{m,p} \\ + }; + + + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=green, fit=(A-3-1)(A-3-5)] {}; + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=blue, fit=(B-1-3)(B-5-3)] {}; + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=red, fit=(C-3-3)] {}; + + + \end{tikzpicture} + +\end{document} + diff --git a/buch/papers/multiplikation/images/strassen.pdf b/buch/papers/multiplikation/images/strassen.pdf Binary files differnew file mode 100644 index 0000000..d150125 --- /dev/null +++ b/buch/papers/multiplikation/images/strassen.pdf diff --git a/buch/papers/multiplikation/images/strassen.tex b/buch/papers/multiplikation/images/strassen.tex new file mode 100644 index 0000000..b51a9d5 --- /dev/null +++ b/buch/papers/multiplikation/images/strassen.tex @@ -0,0 +1,231 @@ +\documentclass[border=10pt]{standalone} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{pgfplots} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usetikzlibrary{decorations.text} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +\begin{tikzpicture}[ampersand replacement=\&] + +\foreach \i in {1,...,4} +{ + \small{ + \matrix (X\i)[matrix of math nodes,nodes in empty cells, + nodes = {draw, minimum size=10mm, + anchor=center, + inner sep=0pt, outer sep=0pt}, + column sep=-\pgflinewidth, + row sep=-\pgflinewidth, + ] at (0,-\i*5) + { + A_{11}B_{11} \& A_{12}B_{11} \& A_{21}B_{11} \& A_{22}B_{11} \\ + A_{11}B_{21} \& A_{12}B_{21} \& A_{21}B_{21} \& A_{22}B_{21} \\ + A_{11}B_{11} \& A_{12}B_{12} \& A_{21}B_{12} \& A_{22}B_{12} \\ + A_{11}B_{22} \& A_{12}B_{22} \& A_{21}B_{22} \& A_{22}B_{22} \\ + };} + + \foreach \j in {1,...,7} + { + \matrix(M\i\j)[matrix of math nodes,nodes in empty cells, + nodes = {draw, minimum size=10mm, + anchor=center, + inner sep=0pt, outer sep=0pt}, + column sep=-\pgflinewidth, + row sep=-\pgflinewidth, + ] at (\j*5,-\i*5) + { + \& \& \& \\ + \& \& \& \\ + \& \& \& \\ + \& \& \& \\ + }; + } +} + +\huge{ + \node at (-3,-20) {$\mathbf{C}_{22}=$}; + \node at (-3,-15) {$\mathbf{C}_{21}=$} ; + \node at (-3,-10) {$\mathbf{C}_{12}=$} ; + \node at (-3,-5) {$\mathbf{C}_{11}=$} ; + + \node at (5,-2) {$\mathbf{P}$}; + \node at (10,-2) {$\mathbf{Q}$}; + \node at (15,-2) {$\mathbf{R}$}; + \node at (20,-2) {$\mathbf{S}$}; + \node at (25,-2) {$\mathbf{T}$}; + \node at (30,-2) {$\mathbf{U}$}; + \node at (35,-2) {$\mathbf{V}$}; +} + + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X1-1-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X1-2-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X2-3-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X2-4-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X3-1-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X3-2-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X4-3-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X4-4-4)] {}; + +% P +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-1-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M21-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M21-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M21-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M21-1-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M31-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M31-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M31-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M31-1-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-1-1)] {}; + +% Q +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M12-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M12-1-3)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M22-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M22-1-3)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M32-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M32-1-3)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M42-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M42-1-3)] {}; + +% R + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M13-3-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M13-4-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M23-3-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M23-4-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M33-3-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M33-4-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M43-3-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M43-4-1)] {}; + +% S + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M14-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M14-2-4)] {}; + + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M24-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M24-2-4)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M34-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M34-2-4)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M44-1-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M44-2-4)] {}; + +%T + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M15-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M15-4-2)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M25-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M25-4-2)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M35-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M35-4-2)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M45-4-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M45-4-2)] {}; + +% U + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M16-1-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M16-1-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M16-3-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M16-3-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M26-1-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M26-1-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M26-3-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M26-3-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M36-1-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M36-1-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M36-3-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M36-3-1)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M46-1-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M46-1-1)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M46-3-3)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M46-3-1)] {}; + +%V + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M17-2-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M17-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M17-2-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M17-4-2)] {}; + + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M27-2-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M27-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M27-2-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M27-4-2)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M37-2-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M37-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M37-2-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M37-4-2)] {}; + +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M47-2-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M47-4-4)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M47-2-2)] {}; +\node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=gray, fit=(M47-4-2)] {}; + + + + + +\end{tikzpicture} + +\end{document} diff --git a/buch/papers/multiplikation/loesungsmethoden.tex b/buch/papers/multiplikation/loesungsmethoden.tex new file mode 100755 index 0000000..8d0c0a8 --- /dev/null +++ b/buch/papers/multiplikation/loesungsmethoden.tex @@ -0,0 +1,506 @@ +% +% teil2.tex -- Beispiel-File für teil2 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% + +\section{Algorithmen} +\rhead{Algorithmen} + +In diesem Abschnitt werden mehrere Algorithmen zur Berechnung der Matrizenmultiplikation vorgestellt, auch werden Bibliotheken zur unkomplizierten Verwendung von vordefinierten Algorithmen gezeigt. + +\subsection{Standardalgorithmus} + +Die Standardmethode ist im Algorithmus \ref{multiplikation:alg:smm} implementiert. +Hierf\"ur wurde die Gleichung \eqref{multiplikation:eq:MM} direkt umgesetzt. +Die \texttt{for i} Schleife iteriert \"uber alle Zeilen der $\mathbf{A}$ Matrix, die \texttt{for j} Schleife iteriert \"uber alle Spalten der $\mathbf{B}$ Matrix und die \texttt{for k} Schleife iteriert \"uber alle Eintr\"age dieser Zeilen bzw. Spalten. +\begin{algorithm}\footnotesize\caption{Matrizenmultiplikation} + \label{multiplikation:alg:smm} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{MM}{$\textbf{A}, \textbf{B}$} + \State $sum \gets 0$ + \State $n \gets columns(\textbf{A}) == rows(\textbf{B})$ + \State $m \gets rows(\textbf{A})$ + \State $p \gets columns(\textbf{B})$ + \State $\textbf{C} \gets zeros(m,p)$ + \For{$i = 0,1,2 \dots,m-1$} + \For{$j = 0,1,2 \dots,p-1$} + \State $sum \gets 0$ + \For{$k = 0,1,2 \dots,n-1$} + \State $sum \gets sum + \textbf{A}[i][k] \cdot \textbf{B}[k][j]$ + \EndFor + \State $\textbf{C}[i][j] \gets sum $ + \EndFor + \EndFor + \State \textbf{return} $\textbf{C}$ + \EndFunction + \end{algorithmic} +\end{algorithm} +Die Laufzeit dieser Struktur mit drei \texttt{for} Schleifen ist $\mathcal{O} (n^3)$. + +\subsubsection{Divide and Conquer Methode} + +F\"ur gewisse Algorithmen f\"uhren \textit{Divide and Conquer} Ans\"atze \cite{multiplikation:DAC} zu markant besseren Laufzeiten. +Die Grundidee ist, dass ein Problem in mehrere, meist simplere und kleinere Teilprobleme aufgeteilt wird. +Das bekannteste Beispiel ist wohl die \textit{Fast Fourier Transform} wobei die Laufzeit von $\mathcal{O} (n^2)$ zu $\mathcal{O}(n \log n)$ verbessert werden kann. + +Die Matrizenmultiplikation kann ebenfalls mit solch einem Ansatz berechnet werden. +Zur vereinfachten Veranschaulichung kann die Situation mit $\mathbf{A}$ und $\mathbf{B}$ der Gr\"osse $2^n \times 2^n$ verwendet werden. +Die Matrizen $\mathbf{A}$ und $\mathbf{B}$ werden in jeweils vier Blockmatrizen der Gr\"osse $2^{n-1} \times 2^{n-1}$ aufgeteilt. +Das Matrizenprodukt +\begin{equation} +\mathbf{A}\mathbf{B}= +\begin{bmatrix} +\mathbf{A}_{11} & \mathbf{A}_{12}\\ +\mathbf{A}_{21} & \mathbf{A}_{22} +\end{bmatrix} +\begin{bmatrix} +\mathbf{B}_{11} & \mathbf{B}_{12}\\ +\mathbf{B}_{21} & \mathbf{B}_{22} +\end{bmatrix} += +\begin{bmatrix} +\mathbf{C}_{11} & \mathbf{C}_{12}\\ +\mathbf{C}_{21} & \mathbf{C}_{22} +\end{bmatrix} +\end{equation} +mit \begin{equation} +\mathbf{C}_{ij} = \sum_{k=1}^{2n} \mathbf{A}_{ik} \mathbf{B}_{kj}, +\label{multiplikation:eq:MM_block} +\end{equation} +ist identisch zu der Gleichung \eqref{multiplikation:eq:MM}, f\"ur die Multiplikation der Untermatrizen $\mathbf{A}_{ik}$ und $\mathbf{B}_{kj}$ wird die Matrizenmultiplikation verwendet. + +Der Algorithmus \ref{multiplikation:alg:devide_mm} zeigt den \textit{Divide and Conquer} Ansatz, +Die Grundstruktur dieser Methode besteht aus dem rekursiven Aufruf der Funktion mit den erzeugten Blockmatrizen. +Der rekursive Aufruf wird bis zu der Gr\"osse der Matrizen von $N = 2 \times 2$ durchgef\"uhrt. +\begin{algorithm}\footnotesize\caption{Divide and Conquer Matrizenmultiplikation} + \setlength{\lineskip}{7pt} + \label{multiplikation:alg:devide_mm} + \begin{algorithmic} + \Function{MM}{$\textbf{A}, \textbf{B}, n$} + \If{$n = 2$} + \State $ \mathbf{C} \gets zeros(n, n)$ + \State $C[0, 0] \gets A[0][0]\cdot B[0][0]+A[0][1]\cdot B[1][0]$ + \State $C[0, 1] \gets A[0][0]\cdot B[0][1]+A[0][1]\cdot B[1][1]$ + \State $C[1, 0] \gets A[1][0]\cdot B[0][0]+A[1][1]\cdot B[1][0]$ + \State $C[1, 1] \gets A[1][0]\cdot B[0][1]+A[1][1]\cdot B[1][1]$ + \Else + \State $ m \gets n/2$ + \State $\mathbf{A11}, \mathbf{A12}, \mathbf{A21}, \mathbf{A22} \gets \mathbf{A}[:m][:m], \mathbf{A}[:m][m:], \mathbf{A}[m:][:m], \mathbf{A}[m:][m:]$ + \State $\mathbf{B11}, \mathbf{B12}, \mathbf{B21}, \mathbf{B22} \gets \mathbf{B}[:m][:m], \mathbf{B}[:m][m:], \mathbf{B}[m:][:m], \mathbf{B}[m:][m:]$ + + \State $\mathbf{C11} \gets \text{MM}(\mathbf{A11}, \mathbf{B11},n) + \text{MM}(\mathbf{A12}, \mathbf{B21},n)$ + \State $\mathbf{C12} \gets \text{MM}(\mathbf{A11},\mathbf{B12},n) + \text{MM}(\mathbf{A12}, \mathbf{B22},n)$ + \State $\mathbf{C21} \gets \text{MM}(\mathbf{A21}, \mathbf{B11},n) + \text{MM}(\mathbf{A22}, \mathbf{B21},n)$ + \State $\mathbf{C22} \gets \text{MM}(\mathbf{A21}, \mathbf{B12},n) + \text{MM}(\mathbf{A22}, \mathbf{B22},n)$ + \State $ C \gets vstack(hstack(C11, C12), hstack(C21, C22))$ + + \EndIf + \State \textbf{return} $\textbf{C}$ + + \EndFunction + \end{algorithmic} +\end{algorithm} + +Die Laufzeit dieser rekursiven Funktion kann mit dem \textit{Master Theorem} \cite{multiplikation:master_theorem} berechnet werden. Das \textit{Master Theorem} bestimmt die Zeitkomplexit\"at von rekursiven Algorithmen. +Ohne auf dieses vertieft einzugehen, bestimmt die Anzahl rekursiver Aufrufe $\mathcal{T} $ der Funktion die Laufzeit. +In diesem Fall wird die Funktion pro Durchlauf acht mal rekursiv aufgerufen, dies f\"uhrt zu +\begin{equation} \label{multiplikation:eq:laufzeitdac} + \mathcal{T}(n) = 8 \cdot \mathcal{T} \left(\frac{n}{2}\right ) + n^2 = \mathcal{O}(n^{\log_2 8}) = \mathcal{O} (n^{3} ), +\end{equation} +also einer kubischen Laufzeit. +Die Addition zweier Matrizen $\mathbf{A} + \mathbf{B} = \mathbf{C}$ hat eine Laufzeit von $\mathcal{O}(n^{2})$ und kann neben dem dominierendem Anteil von $\mathcal{O}(n^{3})$ ignoriert werden. +In diesem Fall hat der \textit{Divide and Conquer} Ansatz zu keiner Verbesserung gef\"uhrt. + + +\subsection{Strassens Algorithmus} + +Strassens Algorithmus \cite{multiplikation:strassen_1969} beschreibt die Matrizenmultiplikation mit einer Vielzahl von Additionen, Subtraktionen und Multiplikationen von Blockmatrizen. +Die sieben grundlegenden Terme +\begin{equation} \label{multiplikation:eq:strassen} +\begin{split} +\text{\textbf{P}} &= \left(\mathbf{A}_{11} + \mathbf{A}_{22}\right ) \cdot \left(\mathbf{B}_{11} + \mathbf{B}_{22}\right ) \\ +\text{\textbf{Q}} &= \left(\mathbf{A}_{21} + \mathbf{A}_{22}\right ) \cdot \mathbf{B}_{11} \\ +\text{\textbf{R}} &= \mathbf{A}_{11} \cdot \left(\mathbf{B}_{12}-\mathbf{B}_{22}\right ) \\ +\text{\textbf{S}} &= \mathbf{A}_{22} \cdot \left(-\mathbf{B}_{11}+\mathbf{B}_{21}\right ) \\ +\text{\textbf{T}} &= \left(\mathbf{A}_{11} + \mathbf{A}_{12}\right ) \cdot \mathbf{B}_{22} \\ +\text{\textbf{U}} &= \left(-\mathbf{A}_{11} + \mathbf{A}_{21}\right ) \cdot \left(\mathbf{B}_{11} + \mathbf{B}_{12}\right ) \\ +\text{\textbf{V}} &= \left(\mathbf{A}_{12} - \mathbf{A}_{22}\right ) \cdot \left(\mathbf{B}_{21} + \mathbf{B}_{22}\right ) +\end{split} +\end{equation} +aus $\mathbf{A}$ und $\mathbf{B}$ werden f\"ur die Berechnung der Bl\"ocke +\begin{equation} \label{multiplikation:eq:strassen2} +\begin{split} +\mathbf{C}_{11} &= \text{\textbf{P}} + \text{\textbf{S}} - \text{\textbf{T}} + \text{\textbf{V}} \\ +\mathbf{C}_{21} &= \text{\textbf{R}} + \text{\textbf{T}} \\ +\mathbf{C}_{12} &= \text{\textbf{Q}} + \text{\textbf{S}}\\ +\mathbf{C}_{22} &= \text{\textbf{P}} + \text{\textbf{R}} - \text{\textbf{Q}} + \text{\textbf{U}} +\end{split} +\end{equation} +der Matrix $\mathbf{C}$ gebraucht. +\begin{algorithm}\footnotesize\caption{Strassen Matrizenmultiplikation} + \label{multiplikation:alg:strassen} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{strassen}{$\textbf{A}, \textbf{B}, n$} + \If{$n = 2$} + \State $ \mathbf{C} \gets zeros((n, n))$ + \State $P \gets (A[0][0]+A[1][1])\cdot( B[0][0]+B[1][1])$ + \State $Q \gets (A[1][0]+A[1][1])\cdot B[0][0]$ + \State $R \gets A[0][0]\cdot (B[0][1]-B[1][1])$ + \State $S \gets A[1][1]\cdot (B[1][0]-B[0][0])$ + \State $T \gets (A[0][0]+A[0][1])\cdot B[1][1]$ + \State $U \gets (A[1][0]-A[0][0])\cdot (B[0][0]+B[0][1])$ + \State $V \gets (A[0][1]-A[1][1])\cdot (B[1][0]+B[1][1])$ + \State $C[0][0] \gets P+S-T+V$ + \State $C[0][1] \gets R+T$ + \State $C[1][0] \gets Q+S$ + \State $C[1][1] \gets P+R-Q+U$ + \Else + \State $ m \gets n/2$ + \State $\mathbf{A11}, \mathbf{A12}, \mathbf{A21}, \mathbf{A22} \gets \mathbf{A}[:m][:m], \mathbf{A}[:m][m:], \mathbf{A}[m:][:m], \mathbf{A}[m:][m:]$ + \State $\mathbf{B11}, \mathbf{B12}, \mathbf{B21}, \mathbf{B22} \gets \mathbf{B}[:m][:m], \mathbf{B}[:m][m:], \mathbf{B}[m:][:m], \mathbf{B}[m:][m:]$ + + \State $ \mathbf{P} \gets \text{strassen}((\mathbf{A11}+ \mathbf{A22}),(\mathbf{B11}+\mathbf{B22}), m)$ + \State $ \mathbf{Q} \gets \text{strassen}((\mathbf{A21}+ \mathbf{A22}), \mathbf{B11},m)$ + \State $ \mathbf{R} \gets \text{strassen}( \mathbf{A11},(\mathbf{B12}- \mathbf{B22}),m)$ + \State $ \mathbf{S} \gets \text{strassen}( \mathbf{A22},(\mathbf{B21}- \mathbf{B11}),m)$ + \State $ \mathbf{T} \gets \text{strassen}((\mathbf{A11}+ \mathbf{A12}), \mathbf{B22},m)$ + \State $ \mathbf{U} \gets \text{strassen}((\mathbf{A21}- \mathbf{A11}),(\mathbf{B11}+\mathbf{B12}),m)$ + \State $ \mathbf{V} \gets \text{strassen}((\mathbf{A12}- \mathbf{A22}),(\mathbf{B21}+\mathbf{B22}),m)$ + + + + \State $\mathbf{C11} \gets \mathbf{P+S-T+V}$ + \State $\mathbf{C12} \gets \mathbf{R+T}$ + \State $\mathbf{C21} \gets \mathbf{Q+S}$ + \State $\mathbf{C22} \gets \mathbf{P+R-Q+U}$ + \State $ C \gets vstack(hstack(C11, C12), hstack(C21, C22))$ + + \EndIf + \State \textbf{return} $\textbf{C}$ + + \EndFunction + \end{algorithmic} +\end{algorithm} +Strassens Methode wird in der Abbildung \ref{multiplikation:fig:strassen} grafisch dargestellt. +Jedes Feld steht f\"ur eine Multiplikation zweier Matrizenelementen von $\mathbf{A}$ oder $\mathbf{B}$ . +Die gr\"unen Felder auf der linken Seite, zeigen die Addition, welche f\"ur den dazugeh\"origen Term ben\"otigt wird. +Die sieben Spalten beschreiben die Matrizen $\mathbf{P,Q,R, \ldots, V}$. +Rote Felder stehen f\"ur eine Subtraktion und die gr\"unen f\"ur eine Addition. +Graue Felder bedeuten, dass die dazugehörige Spalte nicht für die Berechnung benötigt wird. +\begin{figure} + \center + \includegraphics[width=\linewidth]{papers/multiplikation/images/strassen.pdf} + \caption{Der Algorithmus von Strassen verwendet Multiplikationen zur Berechnung der sieben Blockmatrizen $\mathbf{P}$ bis $\mathbf{V}$ aus $\mathbf{A}$ und $\mathbf{B}$, aus denen sich die Blöcke es Produktes $\mathbf{C}=\mathbf{AB}$ ausschliesslich durch Addition und Subtraktion bilden lassen. Die einzelnen Felder in den Quadraten stellen alle möglichen Produkte von Matrizen $\mathbf{A}_{ik}$ und $\mathbf{B}_{jl}$ dar. In den grossen Quadraten am linken Rand sind diejenigen Produkte grün markiert, welche zusammen die entsprechenden Blöcke $\mathbf{C}_{il}$ von $\mathbf{C}$ ergeben. In den Spalten $\mathbf{P}$ bis $\mathbf{V}$ sind die Produkte farblich hervorgehoben, die in der Definition der entsprechenden Matrix vorkommen. Grün und rot symbolisieren die Vorzeichen, mit denen die Produkte kombiniert werden müssen. Graue Felder werden für die Berechnung von $\mathbf{C}_{il}$ nicht benötigt.} + \label{multiplikation:fig:strassen} +\end{figure} + +Die Funktion wird sieben mal rekursiv aufgerufen. +Dies f\"uhrt nach dem \textit{Master Theorem} zu einer Laufzeit von +\begin{equation} \label{multiplikation:eq:laufzeitstrassen} +\mathcal{T}(n) = +7 \cdot \mathcal{T}\left(\frac{n}{2}\right) + n^2 = \mathcal{O}(n^{\log_2 7} ) = \mathcal{O}(n^{2.8074} ) +\end{equation} +und ist somit schneller als die Standardmethode. +Man beachte, dass die Anzahl von Additionen und Subtraktionen gr\"osser und die Anzahl der Multiplikationen kleiner wurde. + +\subsection{Winograds Algorithmus} + +Einen weiteren Ansatz lieferte Shmuel Winograd im Jahre 1968 \cite{multiplikation:winograd_1968}. +Er beschrieb einen neuen Algorithmus f\"ur das Skalarprodukt +\begin{equation} \label{multiplikation:eq:skalar} + \langle x,y \rangle = \sum_{i=1}^{n}x_i y_i. +\end{equation} +F\"ur jeden Vektor berechne +\begin{equation} + \xi = \sum_{j=1}^{ \lfloor n/2 \rfloor} x_{2j-1} \cdot x_{2j} +\end{equation} +und +\begin{equation} + \eta = \sum_{j=1}^{ \lfloor n/2 \rfloor} y_{2j-1} \cdot y_{2j}, +\end{equation} +die jeweils nur von $x$ und $y$ abhängen. +Dazu werden $2 \cdot \lfloor n/2 \rfloor \leq n$ Multiplikationen benötigt. +Das Skalarprodukt ist nun geben mit +\begin{equation} + \langle x,y \rangle = + \begin{cases} + \displaystyle \quad \sum_{j=1}^{ \lfloor n/2 \rfloor} (x_{2j-1} + y_{2j})(x_{2j}+y_{2j-1})-\xi - \eta & \text{wenn $n$ gerade}\\ + \displaystyle \quad \sum_{j=1}^{ \lfloor n/2 \rfloor} (x_{2j-1} + y_{2j})(x_{2j}+y_{2j-1})-\xi - \eta + x_n y_n & \text{wenn $n$ ungerade}. + \end{cases} +\end{equation} +Das Skalarprodukt kann also mit $ \lfloor \frac{n+1}{2} \rfloor$ weiteren Multiplikationen berechnet werden. +Angenommen man hat $N$ Vektoren, mit welchen man $T$ Skalarprodukte berechnen m\"ochte. +Daf\"ur werden $N\lfloor n/2 \rfloor + T\lfloor (n+1)/2 \rfloor $ Multiplikationen ben\"otigt. +Die Summen f\"ur $\xi$ und $\eta$ m\"ussen nur einmal berechnet werden. +Für die ursprüngliche Gleichung \eqref{multiplikation:eq:skalar} für das Skalarprodukt benötigt man $Tn$ Multiplikationen. +Damit können wir die Laufzeit der Methode von Winograd mit der Laufzeit der Standardmethode vergleichen. Sie ist kleiner als die Laufzeit für die Standardmethode, wenn gilt +\begin{equation}\label{multiplikation:eq:eff} +\begin{array}{crcl} + & N\lfloor n/2\rfloor + T\lfloor(n+1)/2\rfloor \approx Nn/2 + Tn/2 & \le & Tn \\ +\Leftrightarrow & Nn/2 & \le & Tn/2 \\ +\Leftrightarrow & N & \le & T. +\end{array} +\end{equation} +Eine Matrizenmultiplikation mit $\mathbf{A}$ einer $m \times n$ und $\mathbf{B}$ einer $n \times p$ Matrix, entspricht $N=m+p$ Vektoren mit welchen man $T=mp$ Skalarprodukte berechnet. +Dies f\"uhrt zu +\begin{equation} + (m+p) \left \lfloor \frac{n}{2} \right \rfloor + mp \left \lfloor \frac{n+1}{2} \right \rfloor = \frac{mn}{2} + \frac{pn}{2} + \frac{mpn}{2} + \frac{mp}{2} +\end{equation} +Multiplikationen. +Wenn $m,p,n$ gross werden, dominiert der Term $\frac{mpn}{2}$ und es werden $\frac{mpn}{2}$ Multiplikationen ben\"otigt, was im Vergleich zu den $mpn$ Multiplikation der Standardmethode nur die H\"alfte ist. +Mit dem gleichen Ansatz wie in der Gleichung \eqref{multiplikation:eq:eff} aber mit quadratischen Matrizen, muss +\begin{align} + \begin{split} +N=2n, &\quad T = n^2 \\ + 2n &\leq n^2 \\ + 2 &\leq n +\end{split} +\end{align} +sein, damit man etwas einspart. +Die Implementation kann Algorithmus \ref{multiplikation:alg:winograd} entnommen werden. +Falls $m=n=p$, werden $\frac{n^3}{2}$ Multiplikationen benötigt. +Im Abschnitt \ref{muliplikation:sec:bigo} wurde bereits erläutert: falls $n \rightarrow \infty$ können Konstanten vernachlässigt werden und + somit entsteht für diesen Algorithmus wieder die ursprüngliche Laufzeit von $\mathcal{O}(n^3 )$. +\begin{algorithm}\footnotesize\caption{Winograds Matrizenmultiplikation} + \setlength{\lineskip}{7pt} + \label{multiplikation:alg:winograd} + \begin{algorithmic} + \Function{Winograd}{$\textbf{A}, \textbf{B}, n$} + \State $ m \gets rows(\mathbf{A})$ + \State $ n \gets columns(\mathbf{A}) == rows(\mathbf{B})$ + \State $ p \gets columns(\mathbf{B})$ + \State $ \mathbf{\xi} \gets zeros(m)$ + \State $ \mathbf{\eta} \gets zeros(p)$ + + + \For{$i = 0,1,2 \dots,m-1$} + \For{$j = 0,1,2 \dots,\lfloor n/2 \rfloor-1$} + \State $\xi[i] \gets \xi[i]+A[i,2 j]A[i,2 j+1]$ + \EndFor + \EndFor + + \For{$i = 0,1,2 \dots,p-1$} + \For{$j = 0,1,2 \dots,\lfloor n/2 \rfloor-1$} + \State $\eta[i] \gets \eta[i]+B[2 j,i]B[2 j+1,i]$ + \EndFor + \EndFor + + \If{$n \% 2 == 0$} + \For{$i = 0,1,2 \dots,m-1$} + \For{$j = 0,1,2 \dots,p-1$} + \State $ab \gets 0$ + \For{$k = 0,1,2 \dots,\lfloor n/2 \rfloor-1$} + \State $ab \gets ab + (A[i,2k]+B[2k+1,j])(A[i,2k+1]+B[2k,j])$ + \EndFor + \State $C[i,j] \gets ab-\eta[j]-\xi[i]$ + \EndFor + \EndFor + \Else + \For{$i = 0,1,2 \dots,n-1$} + \For{$j = 0,1,2 \dots,n-1$} + \State $ab \gets 0$ + \For{$k = 0,1,2 \dots,\lfloor n/2 \rfloor-1$} + \State $ab \gets ab + (A[i,2k]+B[2k+1,j])(A[i,2k+1]+B[2k,j])$ + \EndFor + \State $C[i,j] \gets ab-\eta[j]-\xi[i]+A[i,-1]B[-1,j]$ + \EndFor + \EndFor + \EndIf + \State \textbf{return} $\textbf{C}$ + + \EndFunction + \end{algorithmic} +\end{algorithm} + + +\subsection{Basic Linear Algebra Subprograms (BLAS)} + +Die gebräuchliche Methode f\"ur die Anwendung einer optimierten Matrizenmultiplikation ist die Verwendung einer Subroutine aus den \textit{Basic Linear Algebra Subprograms (BLAS)} \cite{multiplikation:BLAS}. +Die meisten numerischen Bibliotheken von high-level Skriptsprachen wie \texttt{Matlab}, \texttt{NumPy (Python)}, \texttt{GNU Octave} oder \texttt{Mathematica} ben\"utzen eine Form von \textit{BLAS}. + +\textit{BLAS} sind dabei in drei unterschiedliche Levels aufgeteilt. + +\begin{itemize} + \item Level 1 + \begin{itemize} + \item Operationen der Art: $\mathbf{y} \leftarrow \alpha \mathbf{x}+\mathbf{y}$ + \item Dieses Level hat $\mathcal{O}(n)$ Charakteristik + \end{itemize} + \item Level 2 + \begin{itemize} + \item Operationen der Art: $\mathbf{y} \leftarrow \alpha \mathbf{A}\mathbf{x}+\beta \mathbf{y}$ + \item Dieses Level hat $\mathcal{O}(n^2)$ Charakteristik + \end{itemize} + \item Level 3 + \begin{itemize} + \item Operationen der Art: $\mathbf{C} \leftarrow \alpha \mathbf{A}\mathbf{B}+\beta\mathbf{C}$ + \item Dieses Level hat $\mathcal{O}(n^3)$ Charakteristik + \end{itemize} +\end{itemize} + +Die \textit{BLAS} sind auf die modernen Computerprozessoren optimiert und k\"onnen dank einer ausgeklügelter Verwendung der Speicherarchitektur zu erheblichen Leistungsoptimierungen f\"uhren. + + +%\subsubsection{General Matrix Multiplication (GEMM)} +% +%Die \textit{Double-GEMM} \cite{multiplikation:DGEMM} ist definiert als: +% +%\textit{DGEMM performs one of the matrix-matrix operations} +%$$ +% C := \alpha \cdot op( A )\cdot op( B ) + \beta \cdot C, +% $$ +% \textit{where op( X ) is one of} +%$$ +%op( X ) = X \quad \text{ or } \quad op( X ) = X^T, +%$$ +% \textit{alpha and beta are scalars, and A, B and C are matrices, with op( A ) +% an m by k matrix, op( B ) a k by n matrix and C an m by n matrix. +% } + +%Die Implementation von $\alpha\mathbf{A}\mathbf{B} + \beta \mathbf{C} = \mathbf{C}$, wobei $\alpha = 1.0$ und $\beta = 0.0$ in der \texttt{C}-Version von \textit{BLAS}, ist als +%\begin{lstlisting}[style=multiplikationC] +%cblas_dgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans, +% m, n, k, 1, A, m , B, k, 0, C, m); +%\end{lstlisting} +%definiert. + + + +\section{Implementation}\label{multiplikation:section:Implementation} +\rhead{Implementation} + +Folgende Algorithmen wurden jeweils in \texttt{C} und \texttt{Python} implementiert. +\begin{itemize} + \item Standard Matrizenmultiplikation + \item \textit{Divide and Conquer} Matrizenmultiplikation + \item Strassens Matrizenmultiplikation + \item Winograds Matrizenmultiplikation + \item \texttt{BLAS} Matrizenmultiplikation in \texttt{C} + \item \texttt{Numpy} Matrizenmultiplikation in \texttt{Python} +\end{itemize} + +Der Code kann im zum Buch gehörigem \textit{GitHub} \footnote{\url{https://github.com/AndreasFMueller/SeminarMatrizen.git}} Repository gefunden werden. +Anzumerken ist, dass die Matrizenmultiplikation von \texttt{NumPy} als einzige Implementation Multiprocessing und Multithreading verwendet, dies f\"uhrt zu den tiefen Messzeiten. +In Abbildung \ref{multiplikation:fig:python} und Abbildung \ref{multiplikation:fig:c_meas_4096} sind de Messresultate grafisch dargestellt. Die selben Messresultate sind tabellarisch in Tabelle \ref{multiplikation:tab:messung_Python} und Tabelle \ref{multiplikation:tab:messung_C} ersichtlich. + +Die gezeigten Algorithmen haben alle eine Laufzeit der Form $\mathcal{O}(n^k) $. +Bei einer doppelt logarithmischen Darstellung unterscheiden sich diese in Geraden mit unterschiedlichen Steigungen. +Bei den grafisch gezeigten Messresultate, können diese Steigungen gut erkannt werden, wobei die tiefere Laufzeit des Strassen Algorithmus eindrücklich zu sehen ist. +Der benötigte Overhead der Algorithmen zeigt sich in unterschiedlichen $y$-Achsenschnittpunkte. + +In der Messung mit der Programmiersprache \texttt{C} kann ein typischer Cache-Effekt beobachtet wer- +den. +Bei den Algorithmen von Winograd und der Standardmethode hat bei einer Matrizengrösse von $n = 2048$ wohl eine Zeile der Matrix nicht an einer Cache Speicherstelle Platz. +Diese beiden Algorithmen sind die Einzigen, welche \texttt{for}-Schleifen über die ganze Breite der Matrizen verwenden. +Dies führt dazu, dass ganze Zeilen zwischengespeichert werden müssen. +Bei den anderen Algorithmen ist dies nicht der Fall. + +Die Hardwareinformationen des verwendeten Computers sind in der Tabelle \ref{multiplikation:tab:pc_config} aufgelistet. + + +\begin{table} + \begin{center} + \begin{tabular}{r l l l l l} + \hline + \hline + \textbf{n} & \textbf{MM (\textit{s})} & \textbf{MM DC (\textit{s})} & \textbf{Strassen (\textit{s})} & \textbf{Winograd (\textit{s})} & \textbf{BLAS (\textit{s})} \\ + \hline + \multicolumn{6}{c}{} \\ + \textbf{32} & \phantom{000}0.000089 & \phantom{000}0.000594 & \phantom{000}0.0005 & \phantom{0000}0.00008 & \phantom{00}0.000021 \\ + \textbf{64} & \phantom{000}0.00069 & \phantom{000}0.0044 & \phantom{000}0.0036 & \phantom{0000}0.00064 & \phantom{00}0.00018 \\ + \textbf{128} & \phantom{000}0.0057 & \phantom{000}0.035 & \phantom{000}0.025 & \phantom{0000}0.0052 & \phantom{00}0.0012 \\ + \textbf{256} & \phantom{000}0.052 & \phantom{000}0.29 & \phantom{000}0.178 & \phantom{0000}0.053 & \phantom{00}0.0096 \\ + \textbf{512} & \phantom{000}0.51 & \phantom{000}2.22 & \phantom{000}1.25 & \phantom{0000}0.55 & \phantom{00}0.077 \\ + \textbf{1024} & \phantom{000}4.50 & \phantom{00}17.65 & \phantom{000}8.83 & \phantom{0000}4.67 & \phantom{00}0.764 \\ + \textbf{2048} & \phantom{0}129.28 & \phantom{0}141.61 & \phantom{00}61.901 & \phantom{00}136.67 & \phantom{00}7.63 \\ + \textbf{4096} & 1111.31 & 1147.10 & \phantom{0}414.64 & \phantom{0}1179.26 & \phantom{0}55.84 \\ + \textbf{8192} & 9376.17 & 9606.40 & 3014.23 & 10071.51 & 478.42 \\ + \multicolumn{6}{c}{} \\ + \hline + \hline + \end{tabular} + \end{center} + \caption{Laufzeiten der verschieden Algorithmen in der Programmiersprache \texttt{C}} + \label{multiplikation:tab:messung_C} + \end{table} + + + + \begin{table} + \begin{center} + \begin{tabular}{r l l l l l} + \hline + \hline + \textbf{n} & \textbf{MM (\textit{s})} & \textbf{MM DC (\textit{s})} & \textbf{Strassen (\textit{s})} & \textbf{Winograd (\textit{s})} & \textbf{NumPy(\textit{s})} \\ + \hline + \multicolumn{6}{c}{} \\ + \textbf{32} &\phantom{0000}0.0240 & \phantom{0000}0.0271& \phantom{0000}0.04852 & \phantom{0000}0.01871 & 0.0000426 \\ + \textbf{64} &\phantom{0000}0.186 & \phantom{0000}0.265 & \phantom{0000}0.2204 & \phantom{0000}0.1530& 0.000118 \\ + \textbf{128} &\phantom{0000}1.563 & \phantom{0000}1.777 & \phantom{0000}1.447 & \phantom{0000}1.1947 & 0.000244 \\ + \textbf{256} &\phantom{000}11.006 & \phantom{000}13.27 & \phantom{0000}9.938 & \phantom{0000}8.298& 0.000695 \\ + \textbf{512} &\phantom{000}85.476 & \phantom{00}105.397 & \phantom{000}63.961 & \phantom{000}68.360 & 0.00221\\ + \textbf{1024} &\phantom{00}750.757 & \phantom{00}847.321 & \phantom{00}461.494 & \phantom{00}537.374 & 0.0188 \\ + \textbf{2048} &\phantom{0}6154.18 & \phantom{0}7375.93 & \phantom{0}3860.57 & \phantom{0}4884.61 & 0.215 \\ + \textbf{4096} & 46813.30 & 58466.00 & 22904.30 & 43597.10 & 1.49 \\ + \multicolumn{6}{c}{} \\ + \hline + \hline + \end{tabular} + \end{center} + \caption{Laufzeiten der verschieden Algorithmen in der Skriptsprache \texttt{Python}} + \label{multiplikation:tab:messung_Python} + \end{table} + + \begin{table} + \begin{center} + \begin{tabular}{c c c c} + \hline + \hline + \textbf{CPU} & \textbf{OS} & \textbf{GPU } & \textbf{Memory } \\ + \hline + \multicolumn{4}{c}{} \\ + Intel® Core™ i7-4770K CPU & Ubuntu 20.04.2 LTS & Radeon RX 570 & 32 GB 1600 MHz \\ + @ 3.50GHz × 8 & 64-bit & & \\ + \multicolumn{4}{c}{} \\ + \hline + \hline + \end{tabular} + \end{center} + \caption{Messsystem} + \label{multiplikation:tab:pc_config} + \end{table} + +\begin{figure} + \center + \includegraphics[width=\linewidth]{papers/multiplikation/images/meas_c} + \caption{Doppelt logarithmisch dargestellte Laufzeiten, der verschieden Algorithmen, in der Programmiersprache \texttt{C}. + Die Steigung der Messreihe mit Strassens Algorithmus ist deutlich kleiner als deren der anderen Algorithmen. + Die Messung von Winograd ist beinahe gleich wie die Messung mit der Standardmethode, deshalb ist sie nicht gut sichtbar.} + \label{multiplikation:fig:c_meas_4096} +\end{figure} + + +\begin{figure} + \center + \includegraphics[width=\linewidth]{papers/multiplikation/images/meas_python} + \caption{Doppelt logarithmisch dargestellte Laufzeiten, der verschieden Algorithmen, in der Skriptsprache \texttt{Python}. + Die Steigung der Messreihe mit Strassens Algorithmus ist deutlich kleiner als deren der anderen Algorithmen. +} + \label{multiplikation:fig:python} +\end{figure} + +\section{Fazit} +\rhead{Fazit} + +Wie man im Abschnitt \ref{multiplikation:section:Implementation} sehen kann, sind die gezeigten Algorithmen trotz der theoretisch geringeren Zeitkomplexitäten den Implementationen der numerischen Bibliotheken klar unterlegen. +Ein optimierter Speicherzugriff hat einen weitaus grösseren Einfluss auf die Laufzeit als die Zeitkomplexität des Algorithmus. + +Doch haben Entdeckungen wie jene von Strassen und Winograd ihre Daseinsberechtigung. +Nicht auf jeden Computersystemen können die \textit{BLAS} angewandt werden. +Denke man an sehr kleine Mikrocontroller ohne Floatingpoint Recheneinheiten oder auch an \textit{Field Programmable Gate Arrays (FPGA's)}. +Der Overhead der gezeigten Algorithmen ist in allen Fällen grösser als bei der Standardmethode (z.B. sieben rekursive Aufrufe gegenüber drei \texttt{for}-Schleifen). +Um diesem entgegenzuwirken muss der Laufzeitunterschied zwischen Addition und Multiplikation gross genug sein. +Wenn dies gegeben ist und dazu noch grosse Matritzen multipliziert werden, kann die Verwendung der Algorithmen von Strassen oder Winograd zu einer Senkung der Laufzeit führen. diff --git a/buch/papers/multiplikation/main.tex b/buch/papers/multiplikation/main.tex index 42f2768..4a23109 100644..100755 --- a/buch/papers/multiplikation/main.tex +++ b/buch/papers/multiplikation/main.tex @@ -1,36 +1,40 @@ +% !TEX root = ../../buch.tex % % main.tex -- Paper zum Thema <multiplikation> % -% (c) 2020 Hochschule Rapperswil +% (c) 2021 Hochschule Rapperswil % -\chapter{Thema\label{chapter:multiplikation}} -\lhead{Thema} +\definecolor{mygreen}{RGB}{28,172,0} % color values Red, Green, Blue +\definecolor{mylilas}{RGB}{170,55,241} +\definecolor{backcolour}{rgb}{0.95,0.95,0.92} +\lstdefinestyle{multiplikationC}{ + numbers=left, + belowcaptionskip=1\baselineskip, + breaklines=true, + frame=l, + framerule=0pt, + framesep=-1pt, + xleftmargin=1em, + language=C, + showstringspaces=false, + basicstyle=\ttfamily, + keywordstyle=\bfseries\color{green!40!black}, + commentstyle=\itshape\color{purple!40!black}, + identifierstyle=\color{blue}, + stringstyle=\color{red}, + numberstyle=\ttfamily\tiny, + backgroundcolor=\color{backcolour} +} + +\chapter{Schnelle Matrizenmultiplikation\label{chapter:multiplikation}} +\lhead{Schnelle Matrizenmultiplikation} \begin{refsection} -\chapterauthor{Hans Muster} +\chapterauthor{Michael Schmid} -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} -\input{papers/multiplikation/teil0.tex} -\input{papers/multiplikation/teil1.tex} -\input{papers/multiplikation/teil2.tex} -\input{papers/multiplikation/teil3.tex} +\input{papers/multiplikation/einlteung.tex} +\input{papers/multiplikation/problemstellung.tex} +\input{papers/multiplikation/loesungsmethoden.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/multiplikation/packages.tex b/buch/papers/multiplikation/packages.tex index e4173c0..e4173c0 100644..100755 --- a/buch/papers/multiplikation/packages.tex +++ b/buch/papers/multiplikation/packages.tex diff --git a/buch/papers/multiplikation/papers/Strassen_GPU.pdf b/buch/papers/multiplikation/papers/Strassen_GPU.pdf Binary files differnew file mode 100755 index 0000000..4ce7625 --- /dev/null +++ b/buch/papers/multiplikation/papers/Strassen_GPU.pdf diff --git a/buch/papers/multiplikation/papers/Strassen_original_1969.pdf b/buch/papers/multiplikation/papers/Strassen_original_1969.pdf Binary files differnew file mode 100755 index 0000000..b647fc0 --- /dev/null +++ b/buch/papers/multiplikation/papers/Strassen_original_1969.pdf diff --git a/buch/papers/multiplikation/papers/assay_fast_MM.pdf 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+% +% common.tex -- gemeinsame Definitionen +% +% (c) 2021 Michael Schmid, OST Campus Rapperswil +% +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{epic} +\usepackage{color} +\usepackage{array} +\usepackage{algorithm} +\usepackage{ifthen} +\usepackage{adjustbox} +\usepackage[noend]{algpseudocode} +\usepackage{neuralnetwork} +\usepackage{amsmath} +\usepackage{lmodern} +\usepackage{tikz} +\usetikzlibrary{decorations.text} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usepackage{pgfplots} +\usepackage{listings} +\usepackage{svg} + +\definecolor{codegreen}{rgb}{0,0.6,0} +\definecolor{codegray}{rgb}{0.5,0.5,0.5} +\definecolor{codepurple}{rgb}{0.58,0,0.82} +\definecolor{backcolour}{rgb}{0.95,0.95,0.92} +\definecolor{ost}{rgb}{164,0,136} + +\lstdefinestyle{mystyle}{ + backgroundcolor=\color{backcolour}, + commentstyle=\color{codegreen}, + keywordstyle=\color{magenta}, + numberstyle=\tiny\color{codegray}, + stringstyle=\color{codepurple}, + basicstyle=\footnotesize, + breakatwhitespace=false, + breaklines=true, + captionpos=b, + keepspaces=true, + numbers=left, + numbersep=2pt, + showspaces=false, + showstringspaces=false, + showtabs=false, + tabsize=2 +} + +\usetikzlibrary{fit} +\tikzset{% + highlight/.style={rectangle,rounded corners,fill=red!15,draw,fill opacity=0.5,inner sep=0pt} +} +\newcommand{\tikzmark}[2]{\tikz[overlay,remember picture,baseline=(#1.base)] \node (#1) {#2};} +% +\newcommand{\Highlight}[1][submatrix]{% + \tikz[overlay,remember picture]{ + \node[highlight,fit=(left.north west) (right.south east)] (#1) {};} +} + + +\lstset{style=mystyle} +\lstdefinestyle{mystyle}{ + morekeywords={cwt,contourf,datetick} +} + + +\usetikzlibrary{shapes.geometric} +\mode<beamer>{% +\usetheme[]{Frankfurt}} +\beamertemplatenavigationsymbolsempty +\title[]{Fast Matrix Multiplication} +\author[]{Michael Schmid} +\usecolortheme[named=ost]{structure} + +\date[]{31.05.2021} +\newboolean{presentation} diff --git a/buch/papers/multiplikation/presentation/presentation.nav b/buch/papers/multiplikation/presentation/presentation.nav new file mode 100644 index 0000000..2a01568 --- /dev/null +++ b/buch/papers/multiplikation/presentation/presentation.nav @@ -0,0 +1,59 @@ +\headcommand {\slideentry {0}{0}{1}{1/1}{}{0}} +\headcommand {\beamer@framepages {1}{1}} +\headcommand {\beamer@sectionpages {1}{1}} +\headcommand {\beamer@subsectionpages {1}{1}} +\headcommand {\sectionentry {1}{Big $\mathcal {O}$}{2}{Big $\mathcal {O}$}{0}} +\headcommand {\slideentry {1}{0}{1}{2/4}{}{0}} +\headcommand {\beamer@framepages {2}{4}} +\headcommand {\slideentry {1}{0}{2}{5/6}{}{0}} +\headcommand {\beamer@framepages {5}{6}} +\headcommand {\slideentry {1}{0}{3}{7/8}{}{0}} +\headcommand {\beamer@framepages {7}{8}} +\headcommand {\slideentry {1}{0}{4}{9/10}{}{0}} +\headcommand {\beamer@framepages {9}{10}} +\headcommand {\slideentry {1}{0}{5}{11/12}{}{0}} 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{\beamer@partpages {1}{50}} +\headcommand {\beamer@subsectionpages {50}{50}} +\headcommand {\beamer@sectionpages {50}{50}} +\headcommand {\beamer@documentpages {50}} +\headcommand {\gdef \inserttotalframenumber {21}} diff --git a/buch/papers/multiplikation/presentation/presentation.pdf b/buch/papers/multiplikation/presentation/presentation.pdf Binary files differnew file mode 100644 index 0000000..842e68c --- /dev/null +++ b/buch/papers/multiplikation/presentation/presentation.pdf diff --git a/buch/papers/multiplikation/presentation/presentation.snm b/buch/papers/multiplikation/presentation/presentation.snm new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/presentation/presentation.snm diff --git a/buch/papers/multiplikation/presentation/presentation.tex b/buch/papers/multiplikation/presentation/presentation.tex new file mode 100644 index 0000000..2a4af45 --- /dev/null +++ b/buch/papers/multiplikation/presentation/presentation.tex @@ -0,0 +1,12 @@ +% +% MathSem-yyy-xxx.tex -- Präsentation +% +% (c) 2021 Michael Schmid, OST campus Rapperswil +% + +\documentclass[aspectratio=169]{beamer} +\input{common.tex} +%\setboolean{presentation}{true} +\begin{document} +\input{slides/slides.tex} +\end{document} diff --git a/buch/papers/multiplikation/presentation/slides/algo.tex b/buch/papers/multiplikation/presentation/slides/algo.tex new file mode 100644 index 0000000..0c3d130 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/algo.tex @@ -0,0 +1,111 @@ +\begin{frame} + \frametitle{Algorithm} + \begin{columns} + \begin{column}{0.6\textwidth} + \begin{algorithm}[H]\caption{Square Matrix Multiplication} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{MM}{$\textbf{A}, \textbf{B}, \textbf{C}$} + \State $sum \gets 0$ + \State $n \gets columns(\textbf{A}) == rows(\textbf{B})$ + \State $m \gets rows(\textbf{A})$ + \State $p \gets columns(\textbf{B})$ + + \For{$i = 0,1,2 \dots,m-1$} + \For{$j = 0,1,2 \dots,p-1$} + \State $sum \gets 0$ + \For{$k = 0,1,2 \dots,n-1$} + \State $sum \gets sum + \textbf{A}[i][k] \cdot \textbf{B}[k][j]$ + \EndFor + \State $\textbf{C}[i][j] \gets sum $ + \EndFor + \EndFor + \State \textbf{return} $\textbf{C}$ + \EndFunction + \end{algorithmic} + \end{algorithm} +\end{column} +\begin{column}{0.4\textwidth} + \scalebox{0.6}{\parbox{\linewidth}{ + + \begin{tikzpicture}[ampersand replacement=\&,remember picture,overlay] + + \matrix (A)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (2,-2.8) + { + A_{1,1} \& \cdots \& A_{1,k} \& \cdots \& A_{1,n} \\ + \vdots \& \& \vdots \& \& \vdots \\ + A_{i,1} \& \cdots \& A_{i,k} \& \cdots \& A_{i,n} \\ + \vdots \& \& \vdots \& \& \vdots \\ + A_{m,1} \& \cdots \& A_{m,k} \& \cdots \& A_{m,n} \\ + }; + + \matrix (B)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (7.5,1.2) + { + B_{1,1} \& \cdots \& B_{1,j} \& \cdots \& B_{1,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + B_{k,1} \& \cdots \& B_{k,j} \& \cdots \& B_{k,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + B_{n,1} \& \cdots \& B_{n,j} \& \cdots \& B_{n,p} \\ + }; + + \matrix (C)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (7.5,-2.8) + { + C_{1,1} \& \cdots \& C_{1,j} \& \cdots \& C_{1,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + C_{i,1} \& \cdots \& C_{i,j} \& \cdots \& C_{i,p} \\ + \vdots \& \& \vdots \& \& \vdots \\ + C_{m,1} \& \cdots \& C_{m,j} \& \cdots \& C_{m,p} \\ + }; + + + \begin{scope}[on background layer] + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=green, fit=(A-3-1)(A-3-5)] {}; + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=blue, fit=(B-1-3)(B-5-3)] {}; + \node[opacity=0.5, rounded corners=2pt, inner sep=-1pt, fill=red, fit=(C-3-3)] {}; + + \end{scope} + + + + + \end{tikzpicture} + }} + \end{column} +\end{columns} +\end{frame} + + +\begin{frame} + \frametitle{Algorithm} + +\begin{columns} + \begin{column}{0.6\textwidth} +\begin{algorithm}[H]\caption{Square Matrix Multiplication} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{MM}{$\textbf{A}, \textbf{B}, \textbf{C}$} + \State $sum \gets 0$ + \State $n \gets columns(\textbf{A}) == rows(\textbf{B})$ + \State $m \gets rows(\textbf{A})$ + \State $p \gets columns(\textbf{B})$ + + \For{$i = 0,1,2 \dots,m-1$} + \For{$j = 0,1,2 \dots,p-1$} + \State $sum \gets 0$ + \For{$k = 0,1,2 \dots,n-1$} + \State $sum \gets sum + \textbf{A}[i][k] \cdot \textbf{B}[k][j]$ + \EndFor + \State $\textbf{C}[i][j] \gets sum $ + \EndFor + \EndFor + \State \textbf{return} $\textbf{C}$ + \EndFunction + \end{algorithmic} +\end{algorithm} +\end{column} +\begin{column}{0.4\textwidth} +\Huge$\mathcal{O}(n^3)$ +\end{column} +\end{columns} + +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/bigO.tex b/buch/papers/multiplikation/presentation/slides/bigO.tex new file mode 100644 index 0000000..d425da8 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/bigO.tex @@ -0,0 +1,251 @@ + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} +\begin{itemize} + \item <1-> Time complexity of an algorithm + \item <2-> How many multiplications in a function + \item <3-> Drop Constants +\end{itemize} +\end{frame} + + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} + \onslide<1->{ + + \begin{algorithm}[H]\caption{Foo 1} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{foo}{$a, b$} + \State \textbf{return} $a+b$ + \EndFunction + \end{algorithmic} + \end{algorithm} +} +\onslide<2->{ +$\mathcal{O}(1)$ + } +\end{frame} + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} + \onslide<1->{ + + \begin{algorithm}[H]\caption{Foo 2} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{foo}{$a, b$} + \State $ x \gets a+b $ + \State $ y \gets a \cdot b $ + \State \textbf{return} $x+y$ + \EndFunction + \end{algorithmic} + \end{algorithm} +} +\onslide<2->{ +$\mathcal{O}(1) + \mathcal{O}(1) = 2\mathcal{O}(1) = \mathcal{O}(1) $ + } +\end{frame} + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} + \onslide<1->{ + + \begin{algorithm}[H]\caption{Foo 3} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{foo}{$\mathbf{A}, \mathbf{B}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[i] $ + \EndFor + + \State \textbf{return} $sum$ + + \EndFunction + \end{algorithmic} + \end{algorithm} +} +\onslide<2->{ +$\mathcal{O}(n)$ + } +\end{frame} + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} + \onslide<1->{ + + \begin{algorithm}[H]\caption{Foo 4} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{foo}{$\mathbf{A}, \mathbf{B}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \For{$j = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[j] $ + \EndFor + \EndFor + \State \textbf{return} $sum$ + \EndFunction + \end{algorithmic} + \end{algorithm} +} +\onslide<2->{ +$\mathcal{O}(n^2)$ + } +\end{frame} + +% \begin{frame} +% \frametitle{Big $\mathcal{O}$ notation} +% \onslide<1->{ +% +% \begin{algorithm}[H]\caption{Fibonacci} +% \setlength{\lineskip}{7pt} +% \begin{algorithmic}[1] +% \Function{fib}{$n$} +% \If{$n <= 1$} +% \State \textbf{return} $1$ +% \Else +% \State \textbf{return} fib($n-1$) + fib($n-2$) +% \EndIf +% +% \EndFunction +% \end{algorithmic} +% \end{algorithm} +% } +% \onslide<2->{ +% \[ +% \langle x,y \rangle = +% \begin{cases} +% \displaystyle $\mathcal{O}(1)$ & \text{if $n \leq 2$}\\ +% \displaystyle $ 2 \mathcal{T}(\frac{n}{2})$ & \text{if $n > 2$} +% \end{cases} +% \] } +% \end{frame} + + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} +\begin{tikzpicture} +\begin{axis}[ + axis lines = left, + xlabel = $n$ (Data Input), + ylabel = {$t$ (time)}, + legend pos=north east, + very thick, + ymax = 20, + yticklabels=\empty, + xticklabels=\empty, + scale only axis=true, + width=12cm, height=6cm, + ] +%Below the red parabola is defined +\addplot [ + domain= 1:6, + samples=100, + color=red, +] +{1}; +\addlegendentry{$\mathcal{O}(1)$} +%Here the blue parabloa is defined +\addplot [ + domain= 1:6, + samples=100, + color=green, +] +{x}; +\addlegendentry{$\mathcal{O}(n)$} +\addplot [ + domain= 1:6, + samples=100, + color=blue, +] +{x^2}; +\addlegendentry{$\mathcal{O}(n^2)$} +\addplot [ + domain= 1:6, + samples=100, + color=purple, +] +{x^3}; +\addlegendentry{$\mathcal{O}(n^3)$} +\addplot [ + domain= 1:3, + samples=100, + color=black, +] +{exp(x)}; +\addlegendentry{$\mathcal{O}(e^n)$} +\addplot [ + domain= 1:6, + samples=100, + color=orange, +] +{log2(x)}; +\addlegendentry{$\mathcal{O}(\log n)$} +\end{axis} +\end{tikzpicture} + +\end{frame} + +\begin{frame} + \frametitle{Big $\mathcal{O}$ notation} +\begin{tikzpicture} +\begin{axis}[ + axis lines = left, + xlabel = $n$ (Data Input), + ylabel = {$t$ (time)}, + legend pos=north east, + very thick, + ymax = 500, + yticklabels=\empty, + xticklabels=\empty, + scale only axis=true, + width=12cm, height=6cm, + ] +\addplot [ + domain= 1:20, + samples=100, + color=red, +] +{1}; +\addlegendentry{$\mathcal{O}(1)$} +\addplot [ + domain= 1:20, + samples=100, + color=green, +] +{x}; +\addlegendentry{$\mathcal{O}(n)$} +\addplot [ + domain= 1:20, + samples=100, + color=blue, +] +{x^2}; +\addlegendentry{$\mathcal{O}(n^2)$} +\addplot [ + domain= 1:10, + samples=100, + color=purple, +] +{x^3}; +\addlegendentry{$\mathcal{O}(n^3)$} +\addplot [ + domain= 1:10, + samples=100, + color=black, +] +{exp(x)}; +\addlegendentry{$\mathcal{O}(e^n)$} +\addplot [ + domain= 1:20, + samples=100, + color=orange, +] +{log2(x)}; +\addlegendentry{$\mathcal{O}(\log n)$} +\end{axis} +\end{tikzpicture} + +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/blas.tex b/buch/papers/multiplikation/presentation/slides/blas.tex new file mode 100644 index 0000000..ed498a3 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/blas.tex @@ -0,0 +1,18 @@ +\begin{frame} +\frametitle{BLAS, LAPACK} +\begin{itemize} + \item Basic Linear Algebra Subprograms + \begin{itemize} + \item $\mathbf{y} = \alpha \mathbf{x}+\mathbf{y}$ + \item $\mathbf{y} = \alpha \mathbf{A}\mathbf{x}+ \beta \mathbf{y}$ + \item $\mathbf{C} = \alpha \mathbf{A}\mathbf{B}+ \beta \mathbf{C}$ + + \end{itemize} + \item Linear Algebra Package + \begin{itemize} + \item QR decomposition + \item Singular value decomposition + \item Eigenvalues + \end{itemize} +\end{itemize} +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/conclusuion.tex b/buch/papers/multiplikation/presentation/slides/conclusuion.tex new file mode 100644 index 0000000..e69de29 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/conclusuion.tex diff --git a/buch/papers/multiplikation/presentation/slides/logo.pdf b/buch/papers/multiplikation/presentation/slides/logo.pdf Binary files differnew file mode 100644 index 0000000..d78ca88 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/logo.pdf diff --git a/buch/papers/multiplikation/presentation/slides/meas.tex b/buch/papers/multiplikation/presentation/slides/meas.tex new file mode 100644 index 0000000..489c010 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/meas.tex @@ -0,0 +1,42 @@ +\begin{frame} + \frametitle{Measurements Python} + \only<1>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_8.pdf}} + \only<2>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_16.pdf}} + \only<3>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_32.pdf}} + \only<4>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_64.pdf}} + \only<5>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_128.pdf}} + \only<6>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_256.pdf}} + \only<7>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_512.pdf}} + \only<8>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/meas_1024.pdf}} +\end{frame} + + +\begin{frame} + \frametitle{Measurements C} + \only<1>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_8.pdf}} + \only<2>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_16.pdf}} + \only<3>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_32.pdf}} + \only<4>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_64.pdf}} + \only<5>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_128.pdf}} + \only<6>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_256.pdf}} + \only<7>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_512.pdf}} + \only<8>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_1024.pdf}} + \only<9>{ + \includegraphics[width=\textwidth,height=0.9\textheight,keepaspectratio]{../code/c_meas_2048.pdf}} +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/nn.tex b/buch/papers/multiplikation/presentation/slides/nn.tex new file mode 100644 index 0000000..e74e970 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/nn.tex @@ -0,0 +1,97 @@ + +\begin{frame} + \frametitle{Neural Network} + \centering +\newcommand{\inputnum}{4} + +% Hidden layer neurons'number +\newcommand{\hiddennumA}{5} +\newcommand{\hiddennumB}{6} + +% Output layer neurons'number +\newcommand{\outputnum}{4} + +\begin{tikzpicture} + + +% Input Layer +\foreach \i in {1,...,\inputnum} +{ + \node[circle, + minimum size = 6mm, + fill=blue!30] (Input-\i) at (0,-\i) {}; +} + +% Hidden Layer1 +\foreach \i in {1,...,\hiddennumA} +{ + \node[circle, + minimum size = 6mm, + fill=red!50, + yshift=(\hiddennumA-\inputnum)*5 mm + ] (Hidden1-\i) at (2.5,-\i) {}; +} + +% Hidden Layer2 +\foreach \i in {1,...,\hiddennumB} +{ + \node[circle, + minimum size = 6mm, + fill=red!50, + yshift=(\hiddennumB-\inputnum)*5 mm + ] (Hidden2-\i) at (5,-\i) {}; +} + +% Output Layer +\foreach \i in {1,...,\outputnum} +{ + \node[circle, + minimum size = 6mm, + fill=green!50, + yshift=(\outputnum-\inputnum)*5 mm + ] (Output-\i) at (7.5,-\i) {}; +} + +% Connect neurons In-Hidden +\foreach \i in {1,...,\inputnum} +{ + \foreach \j in {1,...,\hiddennumA} + { + \draw[->, shorten >=1pt] (Input-\i) -- (Hidden1-\j); + } +} + +% Connect neurons In-Hidden +\foreach \i in {1,...,\hiddennumA} +{ + \foreach \j in {1,...,\hiddennumB} + { + \draw[->, shorten >=1pt] (Hidden1-\i) -- (Hidden2-\j); + } +} + +% Connect neurons Hidden-Out +\foreach \i in {1,...,\hiddennumB} +{ + \foreach \j in {1,...,\outputnum} + { + \draw[->, shorten >=1pt] (Hidden2-\i) -- (Output-\j); + } +} + +% Inputs +\foreach \i in {1,...,\inputnum} +{ + \draw[<-, shorten <=1pt] (Input-\i) -- ++(-1,0) + node[left]{\LARGE{$x_{\i}$}}; +} + +% Outputs +\foreach \i in {1,...,\outputnum} +{ + \draw[->, shorten <=1pt] (Output-\i) -- ++(1,0) + node[right]{\LARGE{$y_{\i}$}}; +} + +\end{tikzpicture} +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/parcomp.tex b/buch/papers/multiplikation/presentation/slides/parcomp.tex new file mode 100644 index 0000000..1ba39ee --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/parcomp.tex @@ -0,0 +1,66 @@ +% !TEX root = presentation.tex + +\begin{frame} + \frametitle{Vector-Matrix Multiplication} +\center{ + \begin{tikzpicture}[ampersand replacement=\&] + + \matrix (A)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] + { + A_{1,1} \& A_{1,2} \& A_{1,3} \& A_{1,4} \\ + }; + + \matrix (B)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (5,-0.95) + { + B_{1,1} \& B_{1,2} \& B_{1,3} \& B_{1,4} \& B_{1,5} \\ + B_{2,1} \& B_{2,2} \& B_{2,3} \& B_{2,4} \& B_{2,5} \\ + B_{3,1} \& B_{3,2} \& B_{3,3} \& B_{3,4} \& B_{3,5} \\ + B_{4,1} \& B_{4,2} \& B_{4,3} \& B_{4,4} \& B_{4,5} \\ + }; + + \matrix (C)[matrix of math nodes, label skeleton, left delimiter=[,right delimiter={]}] at (5,-3) + { + C_{1,1} \& C_{1,2} \& C_{1,3} \& C_{1,4} \& C_{1,5}\\ + }; + + \foreach \i in {1,...,4} + { + \pgfmathtruncatemacro{\ii}{\i+1} + \onslide<\ii>{ + + \foreach \j in {1,...,5} + { + \draw[thick] (A-1-\i.south) to [out=-90,in=135]node[visible on=<\i->, anchor=north]{} (B-\i-\j.center); + + } + } + } + + + \end{tikzpicture} +} +\end{frame} + + +\begin{frame} + \frametitle{DSP Architecture} +\scalebox{2}{ + \begin{tikzpicture} + \node (mul) at (0,0) [circle,draw=black,inner sep=0pt,minimum size=0.5cm] {X}; + \node (mac) at (2,0) [circle,draw=black,inner sep=0pt,minimum size=0.5cm] {\textbf{+}}; + + \node at (-2,0.3) {$A[n]$}; + \node at (0.4,2) {$B[n]$}; + \node at (4,0.3) {$C[n]$}; + + \draw[thick, ->] (-2,0) --++ (mul); + \draw[thick, ->] (0,2) --++ (mul); + \draw[thick, ->] (mul) -- (mac); + \draw[thick] (mac) --++ (1,0) node (i) {}; + \draw[thick, ->] (i.center) --++ (0,1) --++ (-1,0) -- (mac); + \draw[thick, ->] (i.center) --++ (1,0); + + + \end{tikzpicture} + } +\end{frame} diff --git a/buch/papers/multiplikation/presentation/slides/slides.tex b/buch/papers/multiplikation/presentation/slides/slides.tex new file mode 100644 index 0000000..64edb86 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/slides.tex @@ -0,0 +1,15 @@ +% !TEX root = presentation.tex +\begin{frame} +\titlepage +\end{frame} +% +\section{Big $\mathcal{O}$} +\input{slides/BigO.tex} +\section{Strassen's Algorithm} +\input{slides/strassen.tex} +% \input{slides/nn.tex} +\section{Measurements} +\input{slides/meas.tex} +% \input{slides/parcomp.tex} +\section{How To Matrix Multiply} +\input{slides/blas.tex} diff --git a/buch/papers/multiplikation/presentation/slides/strassen.tex b/buch/papers/multiplikation/presentation/slides/strassen.tex new file mode 100644 index 0000000..c3398d5 --- /dev/null +++ b/buch/papers/multiplikation/presentation/slides/strassen.tex @@ -0,0 +1,429 @@ +\begin{frame} + \frametitle{Strassen's Algorithm} + \includegraphics[page=1,width=\textwidth,height=0.8\textheight,keepaspectratio]{../papers/Strassen_original_1969.pdf} + \includegraphics[page=2,width=\textwidth,height=0.8\textheight,keepaspectratio]{../papers/Strassen_original_1969.pdf} \includegraphics[page=3,width=\textwidth,height=0.8\textheight,keepaspectratio]{../papers/Strassen_original_1969.pdf} + \end{frame} + +\begin{frame} + \frametitle{Strassen's Algorithm} + \centering + \large +\onslide<1->{ + $ + \mathbf{A B = C} + $ +} + +\onslide<2->{ + + +\medskip + $ + \begin{bmatrix} + A_{11} & A_{12}\\ + A_{21} & A_{22} + \end{bmatrix} + \begin{bmatrix} + B_{11} & B_{12}\\ + B_{21} & B_{22} + \end{bmatrix} + = + \begin{bmatrix} + C_{11} & C_{12}\\ + C_{21} & C_{22} + \end{bmatrix} + $ + } + + + \onslide<3->{ + +\medskip +$ +C_{11} = A_{11} \cdot B_{11} + A_{12} \cdot B_{21} +$ + +$ +C_{12} = A_{11} \cdot B_{12} + A_{12} \cdot B_{22} +$ + +$ +C_{21} = A_{21} \cdot B_{11} + A_{22} \cdot B_{21} +$ + +$ +C_{22} = A_{21} \cdot B_{12} + A_{22} \cdot B_{22} +$ +} +\end{frame} + +\input{slides/algo.tex} + + + +\begin{frame} + \frametitle{Strassen's Algorithm} + \begin{columns} + \begin{column}{0.5\textwidth} + \onslide<1->{ + \large + \begin{math} + \begin{aligned} + \text{I} &= (A_{11} + A_{22}) \cdot (B_{11} + B_{22}) \\ + \text{II} &= (A_{21} + A_{22}) \cdot B_{11} \\ + \text{III} &= A_{11} \cdot (B_{12}-B_{22}) \\ + \text{IV} &= A_{22} \cdot (-B_{11}+B_{21}) \\ + \text{V} &= (A_{11} + A_{12}) \cdot B_{22} \\ + \text{VI} &= (-A_{11} + A_{21}) \cdot (B_{11} + B_{12}) \\ + \text{VII} &= (A_{12} - A_{22}) \cdot (B_{21} + B_{22}) \\ + \end{aligned} + \end{math} + } + \end{column} + + \begin{column}{0.5\textwidth} + \onslide<2->{ + \large + \begin{math} + \begin{aligned} + C_{11} &= \text{I} + \text{IV} - \text{V} + \text{VII} \\ + C_{21} &= \text{II} + \text{IV} \\ + C_{12} &= \text{III} + \text{V}\\ + C_{22} &= \text{I} + \text{III} - \text{II} + \text{VI} \\ + \end{aligned} + \end{math} + } + \end{column} +\end{columns} + +\onslide<3->{ + +\bigskip +\centering +\tiny +\begin{math} +\begin{aligned} + C_{11} &= (A_{11} + A_{22}) \cdot (B_{11} + B_{22}) + A_{22} \cdot (-B_{11}+B_{21}) - (A_{11} + A_{12}) \cdot B_{22} + (A_{12} - A_{22}) \cdot (B_{21} + B_{22}) \\ + C_{11} &= A_{11}B_{11} + A_{11}B_{22} + A_{22}B_{11} + A_{22}B_{22} -A_{22}B_{11}+A_{22}B_{21} - A_{11}B_{22} - A_{12}B_{22}+ A_{12}B_{21} + A_{12}B_{22} - A_{22}B_{21} - A_{22}B_{22} \\ + C_{11} &= A_{11}B_{11} + A_{12}B_{21} +\end{aligned} +\end{math} +} + +\end{frame} + + +\begin{frame} +\begin{adjustbox}{width=\textwidth} +\begin{tikzpicture}[ampersand replacement=\&] + + \foreach \i in {1,...,4} + { + \small{ + \matrix (X\i)[matrix of math nodes,nodes in empty cells, + nodes = {draw, minimum size=10mm, + anchor=center, + inner sep=0pt, outer sep=0pt}, + column sep=-\pgflinewidth, + row sep=-\pgflinewidth, + ] at (0,-\i*5) + { + A_{11}B_{11} \& A_{12}B_{11} \& A_{21}B_{11} \& A_{22}B_{11} \\ + A_{11}B_{21} \& A_{12}B_{21} \& A_{21}B_{21} \& A_{22}B_{21} \\ + A_{11}B_{11} \& A_{12}B_{12} \& A_{21}B_{12} \& A_{22}B_{12} \\ + A_{11}B_{22} \& A_{12}B_{22} \& A_{21}B_{22} \& A_{22}B_{22} \\ + };} + + \foreach \j in {1,...,7} + { + \matrix(M\i\j)[matrix of math nodes,nodes in empty cells, + nodes = {draw, minimum size=10mm, + anchor=center, + inner sep=0pt, outer sep=0pt}, + column sep=-\pgflinewidth, + row sep=-\pgflinewidth, + ] at (\j*5,-\i*5) + { + \& \& \& \\ + \& \& \& \\ + \& \& \& \\ + \& \& \& \\ + }; + } + } + +\huge{ + \node at (-3,-20) {$C_{22}=$}; + \node at (-3,-15) {$C_{21}=$} ; + \node at (-3,-10) {$C_{12}=$} ; + \node at (-3,-5) {$C_{11}=$} ; + + \node at (5,-2) {I}; + \node at (10,-2) {II}; + \node at (15,-2) {III}; + \node at (20,-2) {IV}; + \node at (25,-2) {V}; + \node at (30,-2) {VI}; + \node at (35,-2) {VII}; + } + + + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X1-1-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X1-2-2)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X2-3-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X2-4-2)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X3-1-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X3-2-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X4-3-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(X4-4-4)] {}; + + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-4-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M11-1-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M14-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M14-2-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M15-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M15-4-2)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M17-2-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M17-4-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M17-2-2)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M17-4-2)] {}; + + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M23-3-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M23-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M25-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M25-4-2)] {}; + + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M32-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M32-1-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M34-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M34-2-4)] {}; + + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-4-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M41-1-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M42-1-4)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M42-1-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M43-3-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M43-4-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M46-1-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M46-1-1)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=green, fit=(M46-3-3)] {}; + \node[opacity=0.5, rounded corners=0pt, inner sep=-1pt, fill=red, fit=(M46-3-1)] {}; +\end{tikzpicture} +\end{adjustbox} +\end{frame} + + +\begin{frame} + \frametitle{Strassen's Algorithm} + \begin{columns} + \begin{column}{0.5\textwidth} + \large + \begin{math} + \begin{aligned} + \text{I} &= (A_{11} + A_{22}) \cdot (B_{11} + B_{22}) \\ + \text{II} &= (A_{21} + A_{22}) \cdot B_{11} \\ + \text{III} &= A_{11} \cdot (B_{12}-B_{22}) \\ + \text{IV} &= A_{22} \cdot (-B_{11}+B_{21}) \\ + \text{V} &= (A_{11} + A_{12}) \cdot B_{22} \\ + \text{VI} &= (-A_{11} + A_{21}) \cdot (B_{11} + B_{12}) \\ + \text{VII} &= (A_{12} - A_{22}) \cdot (B_{21} + B_{22}) \\ + \end{aligned} + \end{math} + + \end{column} + + \begin{column}{0.5\textwidth} + \large + \begin{math} + \begin{aligned} + C_{11} &= \text{I} + \text{IV} - \text{V} + \text{VII} \\ + C_{21} &= \text{II} + \text{IV} \\ + C_{12} &= \text{III} + \text{V}\\ + C_{22} &= \text{I} + \text{III} - \text{II} + \text{VI} \\ + \end{aligned} + \end{math} + + \end{column} +\end{columns} +\end{frame} + + + +\begin{frame} + \frametitle{Strassen's Algorithm} + +\begin{columns} + \begin{column}{0.5\textwidth} +\large +\begin{math} +\begin{aligned} +\text{\textbf{I}} &= (\mathbf{A_{11}} + \mathbf{A_{22}}) \cdot (\mathbf{B_{11}} + \mathbf{B_{22}}) \\ +\text{\textbf{II}} &= (\mathbf{A_{21}} + \mathbf{A_{22}}) \cdot \mathbf{B_{11}} \\ +\text{\textbf{III}} &= \mathbf{A_{11}} \cdot (\mathbf{B_{12}}-\mathbf{B_{22}}) \\ +\text{\textbf{IV}} &= \mathbf{A_{22}} \cdot (-\mathbf{B_{11}}+\mathbf{B_{21}}) \\ +\text{\textbf{V}} &= (\mathbf{A_{11}} + \mathbf{A_{12}}) \cdot \mathbf{B_{22}} \\ +\text{\textbf{VI}} &= (-\mathbf{A_{11}} + \mathbf{A_{21}}) \cdot (\mathbf{B_{11}} + \mathbf{B_{12}}) \\ +\text{\textbf{VII}} &= (\mathbf{A_{12}} - \mathbf{A_{22}}) \cdot (\mathbf{B_{21}} + \mathbf{B_{22}}) \\ +\end{aligned} +\end{math} + +\end{column} + +\begin{column}{0.5\textwidth} + \large + \begin{math} + \begin{aligned} + \mathbf{C_{11}} &= \text{\textbf{I}} + \text{\textbf{IV}} - \text{\textbf{V}} + \text{\textbf{VII}} \\ + \mathbf{C_{21}} &= \text{\textbf{II}} + \text{\textbf{IV}} \\ + \mathbf{C_{12}} &= \text{\textbf{III}} + \text{\textbf{V}}\\ + \mathbf{C_{22}} &= \text{\textbf{I}} + \text{\textbf{III}} - \text{\textbf{II}} + \text{\textbf{VI}} \\ + \end{aligned} + \end{math} + +\end{column} +\end{columns} + +\end{frame} + +\begin{frame} + \frametitle{Algorithm} + \onslide<1->{ + + \scalebox{0.45}{\parbox{\linewidth}{ + \begin{algorithm}[H]\caption{Strassen Matrix Multiplication} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{strassen}{$\textbf{A}, \textbf{B}, n$} + \If{$n = 2$} + \State $ \mathbf{C} \gets zeros((n, n))$ + \State $P \gets (A[0][0]+A[1][1])\cdot( B[0][0]+B[1][1])$ + \State $Q \gets (A[1][0]+A[1][1])\cdot B[0][0]$ + \State $R \gets A[0][0]\cdot (B[0][1]-B[1][1])$ + \State $S \gets A[1][1]\cdot (B[1][0]-B[0][0])$ + \State $T \gets (A[0][0]+A[0][1])\cdot B[1][1]$ + \State $U \gets (A[1][0]-A[0][0])\cdot (B[0][0]+B[0][1])$ + \State $V \gets (A[0][1]-A[1][1])\cdot (B[1][0]+B[1][1])$ + \State $C[0][0] \gets P+S-T+V$ + \State $C[0][1] \gets R+T$ + \State $C[1][0] \gets Q+S$ + \State $C[1][1] \gets P+R-Q+U$ + \Else + \State $ m \gets n/2$ + \State $\mathbf{A11}, \mathbf{A12}, \mathbf{A21}, \mathbf{A22} \gets \mathbf{A}[:m][:m], \mathbf{A}[:m][m:], \mathbf{A}[m:][:m], \mathbf{A}[m:][m:]$ + \State $\mathbf{B11}, \mathbf{B12}, \mathbf{B21}, \mathbf{B22} \gets \mathbf{B}[:m][:m], \mathbf{B}[:m][m:], \mathbf{B}[m:][:m], \mathbf{B}[m:][m:]$ + + \State $ \mathbf{P} \gets \text{strassen}((\mathbf{A11}+ \mathbf{A22}),(\mathbf{B11}+\mathbf{B22}), m)$ + \State $ \mathbf{Q} \gets \text{strassen}((\mathbf{A21}+ \mathbf{A22}), \mathbf{B11},m)$ + \State $ \mathbf{R} \gets \text{strassen}( \mathbf{A11},(\mathbf{B12}- \mathbf{B22}),m)$ + \State $ \mathbf{S} \gets \text{strassen}( \mathbf{A22},(\mathbf{B21}- \mathbf{B11}),m)$ + \State $ \mathbf{T} \gets \text{strassen}((\mathbf{A11}+ \mathbf{A12}), \mathbf{B22},m)$ + \State $ \mathbf{U} \gets \text{strassen}((\mathbf{A21}- \mathbf{A11}),(\mathbf{B11}+\mathbf{B12}),m)$ + \State $ \mathbf{V} \gets \text{strassen}((\mathbf{A12}- \mathbf{A22}),(\mathbf{B21}+\mathbf{B22}),m)$ + + + + \State $\mathbf{C11} \gets \mathbf{P+S-T+V}$ + \State $\mathbf{C12} \gets \mathbf{R+T}$ + \State $\mathbf{C21} \gets \mathbf{Q+S}$ + \State $\mathbf{C22} \gets \mathbf{P+R-Q+U}$ + \State $ C \gets vstack((hstack((C11, C12)), hstack((C21, C22))))$ + + \EndIf + \State \textbf{return} $\textbf{C}$ + + \EndFunction + \end{algorithmic} + \end{algorithm} + }}} +% \[ +% \mathcal{T}(n) = \left\{\begin{array}{lr} +% 1, & \text{if} n \leq 2\\ +% 7 \mathcal{T}(\frac{n}{2}) + n^2, & \text{if} n > 2\\ +% \end{array}\right\} +% \] +\only<2>{ + $ + \mathcal{T}(n) = + \begin{cases} + 1 & \text{if } n \leq 2\\ + 7 \cdot \mathcal{T}(\frac{n}{2}) + n^2 & \text{if } n > 2 + \end{cases} = \mathcal{O}(n^{\log_2 7})$ + +} +\only<3>{ + $ + \mathcal{T}(n) = + \begin{cases} + 1 & \text{if } n \leq 2\\ + 7 \cdot \mathcal{T}(\frac{n}{2}) + n^2 & \text{if } n > 2 + \end{cases} = \mathcal{O}(n^{2.81})$ + +} + +\end{frame} + +\begin{frame} + \frametitle{Algorithm} + \onslide<1->{ + + \scalebox{0.45}{\parbox{\linewidth}{ + \begin{algorithm}[H]\caption{Strassen Matrix Multiplication} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{MM}{$\textbf{A}, \textbf{B}, n$} + \If{$n = 2$} + \State $ \mathbf{C} \gets zeros((n, n))$ + \State $C[0, 0] \gets A[0][0]*B[0][0]+A[0][1]*B[1][0]$ + \State $C[0, 1] \gets A[0][0]*B[0][1]+A[0][1]*B[1][1]$ + \State $C[1, 0] \gets A[1][0]*B[0][0]+A[1][1]*B[1][0]$ + \State $C[1, 1] \gets A[1][0]*B[0][1]+A[1][1]*B[1][1]$ + \Else + \State $ m \gets n/2$ + \State $\mathbf{A11}, \mathbf{A12}, \mathbf{A21}, \mathbf{A22} \gets \mathbf{A}[:m][:m], \mathbf{A}[:m][m:], \mathbf{A}[m:][:m], \mathbf{A}[m:][m:]$ + \State $\mathbf{B11}, \mathbf{B12}, \mathbf{B21}, \mathbf{B22} \gets \mathbf{B}[:m][:m], \mathbf{B}[:m][m:], \mathbf{B}[m:][:m], \mathbf{B}[m:][m:]$ + + \State $\mathbf{C11} \gets \text{MM}(\mathbf{A11}, \mathbf{B11}) + \text{MM}(\mathbf{A12}, \mathbf{B21})$ + \State $\mathbf{C12} \gets \text{MM}(\mathbf{A11},\mathbf{B12}) + \text{MM}(\mathbf{A12},\mathbf{B22})$ + \State $\mathbf{C21} \gets \text{MM}(\mathbf{A21}, \mathbf{B11}) + \text{MM}(\mathbf{A22}, \mathbf{B21})$ + \State $\mathbf{C22} \gets \text{MM}(\mathbf{A21}, \mathbf{B12}) + \text{MM}(\mathbf{A22}, \mathbf{B22})$ + \State $ C \gets vstack((hstack((C11, C12)), hstack((C21, C22))))$ + + \EndIf + \State \textbf{return} $\textbf{C}$ + + \EndFunction + \end{algorithmic} + \end{algorithm} + \bigskip + \bigskip + \bigskip + \bigskip + \bigskip + }}} + +\only<2>{ + + + $ + \mathcal{T}(n) = + \begin{cases} + 1 & \text{if } n \leq 2\\ + 8 \cdot \mathcal{T}(\frac{n}{2}) + n^2 & \text{if } n > 2 + \end{cases} = \mathcal{O}(n^{\log_2 8})$ + +} +\only<3>{ + $ + \mathcal{T}(n) = + \begin{cases} + 1 & \text{if } n \leq 2\\ + 8 \cdot \mathcal{T}(\frac{n}{2}) + n^2 & \text{if } n > 2 + \end{cases} = \mathcal{O}(n^{3})$ + +} + +\end{frame} diff --git a/buch/papers/multiplikation/presentation/tikz/algo.pdf b/buch/papers/multiplikation/presentation/tikz/algo.pdf Binary files differnew file mode 100644 index 0000000..752f42e --- /dev/null +++ b/buch/papers/multiplikation/presentation/tikz/algo.pdf diff --git a/buch/papers/multiplikation/presentation/tikz/algo.tex b/buch/papers/multiplikation/presentation/tikz/algo.tex new file mode 100644 index 0000000..0b2c567 --- /dev/null +++ b/buch/papers/multiplikation/presentation/tikz/algo.tex @@ -0,0 +1,52 @@ +\documentclass[border=10pt]{article} +\usepackage[left=25mm,right=25mm,top=25mm,bottom=25mm]{geometry} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{times} +\usepackage{geometry} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{algorithm} +\usepackage[noend]{algpseudocode} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +\begin{algorithm}[H]\caption{Square Matrix Multiplication} + \setlength{\lineskip}{7pt} + \begin{algorithmic}[1] + \Function{MM}{$\textbf{A}, \textbf{B}, \textbf{C}, n$} + \State $sum \gets 0$ + \For{$i = 0,1,2 \dots,n-1$} + \For{$j = 0,1,2 \dots,n-1$} + \State $sum \gets 0$ + \For{$k = 0,1,2 \dots,n-1$} + \State $sum \gets sum + \textbf{A}[i][k] \cdot \textbf{B}[k][j]$ + \EndFor + \State $\textbf{C}[i][j] \gets sum $ + \EndFor + \EndFor + \EndFunction + \end{algorithmic} +\end{algorithm} +\end{document} diff --git a/buch/papers/multiplikation/problemstellung.tex b/buch/papers/multiplikation/problemstellung.tex new file mode 100755 index 0000000..879b210 --- /dev/null +++ b/buch/papers/multiplikation/problemstellung.tex @@ -0,0 +1,137 @@ +% +% teil1.tex -- Beispiel-File für das Paper +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Laufzeiten von Algorithmen} +\rhead{Laufzeiten von Algorithmen} +Wegen der breiten Anwendung der Matrizenmultiplikation ist eine effiziente Ausführung dieser Operation von grosser Bedeutung. +Das Ziel dieses Papers ist, verschiedenen Algorithmen der Matrizenmultiplikation vorzustellen. +Gezielt wird auf Algorithmen eingegangen, welche das Problem schneller als der Standardalgorithmus l\"osen. + +\label{muliplikation:sec:bigo} +Die Big $\mathcal{O}$ Notation beschreibt die Laufzeitkomplexit\"at eines Algorithmus in Relation zur Inputgrösse \cite{multiplikation:bigo}. +$f(x) \in \mathcal{O}(g(x))$ besagt, dass die Funktion $f$ nicht wesentlich schneller w\"achst als $g$, wenn $x \rightarrow \infty$. +Dies ist gegeben, falls es für $f \in \mathcal{O}(n^k)$ eine Konstante $C$ gibt, mit $f(n) \leq Cn^k$. +% Es gibt eine Konstante $K$ derart, dass $f(x) \le K g(x)$ für $x\to\infty$. +Vereinfacht werden f\"ur Algorithmen die folgenden Sprechweisen verwendet: +\begin{itemize} + \item $f \in \mathcal{O}(1) \rightarrow f$ ist beschr\"ankt + \item $f \in \mathcal{O}(n) \rightarrow f$ w\"achst linear + \item $f \in \mathcal{O} (n^2 ) \rightarrow f$ w\"achst quadratisch + \item $f \in \mathcal{O}(\log n) \rightarrow f$ w\"achst logarithmisch + \item $f \in \mathcal{O}(n \log n) \rightarrow f$ hat super-lineares Wachstum + \item $f \in \mathcal{O} (e^n ) \rightarrow f$ w\"achst exponentiell + \item usw. +\end{itemize} + +Konstanten werden nicht beachtet, eine Laufzeit von $4n^2$ führt, für $n \rightarrow \infty$ zu $\mathcal{O}(n^2)$. +In der Abbildung \ref{multiplikation:fig:bigo} k\"onnen die verschiedenen Laufzeiten miteinander verglichen werden. +Bei einer doppelt logarithmischen Darstellung werden Polynome der Form $f(x) = x^k$ als Gerade und Exponentialfunktionen der Form $f(x) = a^x$ als nach oben gekr\"ummte Kurven abgebildet. + + + +\subsubsection{Beispielalgorithmen} + +Es folgen einige Beispiele von Algorithmen, welche zu einer bestimmten Zeitkomplexit\"atsklasse zugeteilt werden k\"onnen. + + +\begin{table}[t] + \begin{tabular}{ll} + \begin{minipage}{0.48\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:b1} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{B1}{$a, b$} + \State \textbf{return} $a+b$ + \EndFunction + \State + \State + \end{algorithmic} + \end{algorithm} + \end{minipage} + & + \begin{minipage}{0.48\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:b2} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{B2}{$a, b$} + \State $ x \gets a+b $ + \State $ y \gets a \cdot b $ + \State \textbf{return} $x+y$ + \EndFunction + \end{algorithmic} + \end{algorithm} + \end{minipage} \\ + \begin{minipage}{0.48\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \label{multiplikation:alg:linear} + \Function{L}{$\mathbf{a}, \mathbf{b}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[i] $ + \EndFor + + \State \textbf{return} $sum$ + + \EndFunction + \State + \State + \end{algorithmic} + \end{algorithm} + \end{minipage} + & + \begin{minipage}{0.48\textwidth} + \begin{algorithm}[H]\footnotesize\caption{} + \label{multiplikation:alg:q1} + \setlength{\lineskip}{7pt} + \begin{algorithmic} + \Function{Q}{$\mathbf{A}, \mathbf{B}$,n} + \State $ sum \gets 0$ + \For{$i = 0,1,2 \dots,n$} + \For{$j = 0,1,2 \dots,n$} + \State $ sum \gets sum + A[i] \cdot B[j] $ + \EndFor + \EndFor + \State \textbf{return} $sum$ + \EndFunction + \end{algorithmic} + \end{algorithm} + \end{minipage} + \end{tabular} +\end{table} + +%\begin{table} +% \begin{tabular}[t]{ll} + +% \end{tabular} +%\end{table} + +\paragraph{Beschr\"ankter Algorithmus} +Algorithmus \ref{multiplikation:alg:b1} ist ein Beispiel mit beschränkter Laufzeit $\mathcal{O}(1)$ +Da $a$ und $b$ Skalare sind, hat keine Gr\"osse $n$ einen Einfluss auf die Laufzeit. + +Wie erwähnt werden Konstanten nicht beachtet, der Algorithmus \ref{multiplikation:alg:b2} f\"uhrt ebenso zu $\mathcal{O}(1)$ und nicht zu $\mathcal{O}(2)$. + + +\paragraph{Linearer Algorithmus} + +Der Algorithmus \ref{multiplikation:alg:linear} hat ein lineares Verhalten. +Die \texttt{for}-Schleife wird $n$-mal durchlaufen und f\"uhrt deshalb zu $\mathcal{O}(n)$. + +\paragraph{Quadratischer Algorithmus} + +Der Algorithmus \ref{multiplikation:alg:q1} hat ein quadratisches Verhalten. +Die beiden \texttt{for}-Schleifen werden jeweils $n$-mal durchlaufen und f\"uhrt deshalb zu $\mathcal{O} (n^2 )$. + + +\begin{figure} + \center + \includegraphics[]{papers/multiplikation/images/bigo} + \caption{Laufzeiten von verschiedensten Zeitkomplexitäten. Bei einer doppelt logarithmischen Darstellung werden Polynome der Form $f(x) = x^k$ als Gerade und Exponentialfunktionen der Form $f(x) = a^x$ als nach oben gekr\"ummte Kurven dargestellt.} + \label{multiplikation:fig:bigo} +\end{figure} diff --git a/buch/papers/multiplikation/references.bib b/buch/papers/multiplikation/references.bib index 7149fb1..8815386 100644..100755 --- a/buch/papers/multiplikation/references.bib +++ b/buch/papers/multiplikation/references.bib @@ -33,3 +33,70 @@ url = {https://doi.org/10.1016/j.acha.2017.11.004} } +@article{multiplikation:winograd_1968, + title={A New Algorithm for Inner Product}, + volume={C-17}, + DOI={10.1109/tc.1968.227420}, + number={7}, + journal={IEEE Transactions on Computers}, + author={Winograd, S.}, + year={1968}, + pages={693–694} +} + +@article{multiplikation:strassen_1969, + title={Gaussian elimination is not optimal}, + volume={13}, + DOI={10.1007/bf02165411}, + number={4}, + journal={Numerische Mathematik}, + author={Strassen, Volker}, + year={1969}, + pages={354–356} +} + +@online{multiplikation:bigo, + title = {Big O notation}, + url = {https://en.wikipedia.org/wiki/Big_O_notation}, + date = {2021-07-27}, + year = {2021}, + month = {7}, + day = {27} +} + +@online{multiplikation:master_theorem, + title = {Master theorem (analysis of algorithms)}, + url = {https://en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms)}, + date = {2021-07-28}, + year = {2021}, + month = {7}, + day = {28} +} + + +@online{multiplikation:DAC, + title = {Divide-and-conquer algorithm}, + url = {https://en.wikipedia.org/wiki/Divide-and-conquer_algorithm}, + date = {2021-07-28}, + year = {2021}, + month = {7}, + day = {28} +} + +@online{multiplikation:BLAS, + title = {BLAS (Basic Linear Algebra Subprograms)}, + url = {http://www.netlib.org/blas/}, + date = {2021-08-01}, + year = {2021}, + month = {8}, + day = {01} +} + +@online{multiplikation:DGEMM, + title = {DGEMM}, + url = {http://www.netlib.org/lapack/explore-html/d1/d54/group__double__blas__level3_gaeda3cbd99c8fb834a60a6412878226e1.html#gaeda3cbd99c8fb834a60a6412878226e1}, + date = {2021-08-01}, + year = {2021}, + month = {8}, + day = {01} +} diff --git a/buch/papers/multiplikation/teil0.tex b/buch/papers/multiplikation/teil0.tex deleted file mode 100644 index 082b7f5..0000000 --- a/buch/papers/multiplikation/teil0.tex +++ /dev/null @@ -1,22 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 0\label{multiplikation:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{multiplikation:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - - diff --git a/buch/papers/multiplikation/teil1.tex b/buch/papers/multiplikation/teil1.tex deleted file mode 100644 index 0a6903a..0000000 --- a/buch/papers/multiplikation/teil1.tex +++ /dev/null @@ -1,55 +0,0 @@ -% -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{multiplikation:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{multiplikation:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{multiplikation:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{multiplikation:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{multiplikation:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. - - diff --git a/buch/papers/multiplikation/teil2.tex b/buch/papers/multiplikation/teil2.tex deleted file mode 100644 index efbf31a..0000000 --- a/buch/papers/multiplikation/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{multiplikation:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{multiplikation:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/multiplikation/teil3.tex b/buch/papers/multiplikation/teil3.tex deleted file mode 100644 index f58508b..0000000 --- a/buch/papers/multiplikation/teil3.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil3.tex -- Beispiel-File für Teil 3 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 3 -\label{multiplikation:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{multiplikation:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. 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+\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsthm} +\usepackage{lipsum} +\usepackage{amscd} +\usepackage{graphicx} +\usepackage{fancyhdr} +\usepackage{textcomp} +\usepackage{txfonts} +\usepackage[all]{xy} +\usepackage{paralist} +\usepackage[colorlinks=true]{hyperref} +\usepackage{array} +\usepackage{tikz} +\usepackage{slashed} +\usepackage{pdfpages} +\usepackage{cite} +\usepackage{url} +\usepackage{amsmath,amsfonts,amssymb} +\usepackage{tikz} +\usetikzlibrary{arrows,matrix,positioning} +\usetikzlibrary{overlay-beamer-styles} +\usetikzlibrary{matrix.skeleton} +\usetikzlibrary{automata,positioning} +\usepackage{listings} +\usepackage{multirow} +\usepackage{color} + +\begin{document} + +$ +A= +\begin{bmatrix} +A_{11} & A_{12}\\ +A_{21} & A_{22} +\end{bmatrix}, +B= +\begin{bmatrix} +B_{11} & B_{12}\\ +B_{21} & B_{22} +\end{bmatrix}, +C= +\begin{bmatrix} +C_{11} & C_{12}\\ +C_{21} & C_{22} +\end{bmatrix} +$ + +\medskip +$ +A \cdot B = C +$ + +\medskip +$ +C_{11} = A_{11} \cdot B_{11} + A_{12} \cdot B_{21}\\ +C_{12} = A_{11} \cdot B_{12} + A_{12} \cdot B_{22}\\ +C_{21} = A_{21} \cdot B_{11} + A_{22} \cdot B_{21}\\ +C_{22} = A_{21} \cdot B_{12} + A_{22} \cdot B_{22} +$ + +\medskip +\begin{math} +\begin{aligned} +\text{I} &= (A_{11} + A_{22}) \cdot (B_{11} + B_{22}) \\ +\text{II} &= (A_{21} + A_{22}) \cdot B_{11} \\ +\text{III} &= A_{11} \cdot (B_{12}-B_{22}) \\ +\text{IV} &= A_{22} \cdot (-B_{11}+B_{21}) \\ +\text{V} &= (A_{11} + A_{12}) \cdot B_{22} \\ +\text{VI} &= (-A_{11} + A_{21}) \cdot (B_{11} + B_{12})) \\ +\text{VII} &= (A_{12} - A_{22}) \cdot (B_{21} + B_{22}) \\ +\end{aligned} +\end{math} + + +\medskip +\begin{math} +\begin{aligned} +C_{11} &= \text{I} + \text{IV} - \text{V} + \text{VII} \\ +C_{21} &= \text{II} + \text{IV} \\ +C_{12} &= \text{III} + \text{V}\\ +C_{22} &= \text{I} + \text{III} - \text{II} + \text{VI} \\ +\end{aligned} +\end{math} + + +\medskip +\begin{math} +\begin{aligned} +C_{11} &= \text{II} + \text{IV} \\ +C_{11} &= (A_{11} + A_{22}) \cdot (B_{11} + B_{22}) + A_{22} \cdot (-B_{11}+B_{21}) - (A_{11} + A_{12}) \cdot B_{22} + (A_{12} - A_{22}) \cdot (B_{21} + B_{22})C_{21} \\ +C_{11} &= A_{11}B_{11} + A_{11}B_{22} + A_{22}B_{11} + A_{22}B_{22} -A_{22}B_{11}+A_{22}B_{21} - A_{11}B_{22} - A_{12}B_{22}+ A_{12}B_{21} + A_{12}B_{22} - A_{22}B_{21} - A_{22}B_{22} \\ +C_{11} &= A_{11}B_{11} + A_{12}B_{21} +\end{aligned} +\end{math} + +\section{Winograd} + +$ +x_1 y_1 + x_2 y_2 = (x_1 +y_2)(y_1 + x_2)-x_1 x_2 - y_1 y_2 +$ + +$ +x = (x_1, \cdots, x_n), y=(y_1, \cdots, y_n) +$ + +\[ +\xi = \sum_{j=1}^{ \lfloor n/2 \rfloor} x_{2j-1} \cdot x_{2j} +\] + +\[ +\eta = \sum_{j=1}^{ \lfloor n/2 \rfloor} y_{2j-1} \cdot y_{2j} +\] + +\[ +\langle x,y \rangle = +\begin{cases} + \displaystyle \sum_{j=1}^{ \lfloor n/2 \rfloor} (x_{2j-1} + y_{2j})(x_{2j}+y_{2j-1})-\xi - \eta & \text{if $n$ is even}\\ +\displaystyle \sum_{j=1}^{ \lfloor n/2 \rfloor} (x_{2j-1} + y_{2j})(x_{2j}+y_{2j-1})-\xi - \eta + x_n y_n & \text{if $n$ is odd} +\end{cases} +\] + +\end{document} diff --git a/buch/papers/multiplikation/tikz_formulas/algo_graph.fdb_latexmk b/buch/papers/multiplikation/tikz_formulas/algo_graph.fdb_latexmk new file mode 100644 index 0000000..ddfa880 --- /dev/null +++ b/buch/papers/multiplikation/tikz_formulas/algo_graph.fdb_latexmk @@ -0,0 +1,245 @@ +# Fdb version 3 +["pdflatex"] 1621585121 "algo_graph.tex" "algo_graph.pdf" "algo_graph" 1621585184 + "/dev/null" 1621583990 0 d41d8cd98f00b204e9800998ecf8427e "" + "/etc/texmf/web2c/texmf.cnf" 1619433543 475 c0e671620eb5563b2130f56340a5fde8 "" + "/usr/share/texlive/texmf-dist/fonts/enc/dvips/base/8r.enc" 1165713224 4850 80dc9bab7f31fb78a000ccfed0e27cab "" + "/usr/share/texlive/texmf-dist/fonts/map/fontname/texfonts.map" 1577235249 3524 cb3e574dea2d1052e39280babc910dc8 "" + "/usr/share/texlive/texmf-dist/fonts/tfm/jknappen/ec/ecrm1000.tfm" 1136768653 3584 adb004a0c8e7c46ee66cad73671f37b4 "" + 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file mode 100644 index 0000000..5689825 --- /dev/null +++ b/buch/papers/munkres/figures/ganzzahlige_punkte.png diff --git a/buch/papers/munkres/main.tex b/buch/papers/munkres/main.tex index 4dd20fa..201e70b 100644 --- a/buch/papers/munkres/main.tex +++ b/buch/papers/munkres/main.tex @@ -3,29 +3,11 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:munkres}} -\lhead{Thema} +\chapter{Das Zuordnungsproblem und der Munkres-Algorithmus\label{chapter:munkres}} +\lhead{Das Zuordnungsproblem und der Munkres-Algorithmus} \begin{refsection} -\chapterauthor{Hans Muster} +\chapterauthor{Marc Kühne} -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} \input{papers/munkres/teil0.tex} \input{papers/munkres/teil1.tex} diff --git a/buch/papers/munkres/teil0.tex b/buch/papers/munkres/teil0.tex index de522c7..0578429 100644 --- a/buch/papers/munkres/teil0.tex +++ b/buch/papers/munkres/teil0.tex @@ -3,20 +3,8 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 0\label{munkres:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{munkres:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - +\section{Einleitung\label{munkres:section:teil0}} +\rhead{Einleitung} +Im Bereich der Unternehmensplanung (Operations Research) gibt es verschiedene Fragestellungen. Eine davon ist das sogenannte Transportproblem. Zum Transport einheitlicher Objekte von mehreren Angebots- zu mehreren Nachfrageorten ist ein optimaler, d. h. kostenminimaler Plan zu finden, wobei die vorhandenen und zu liefernden Mengen an den einzelnen Standorten gegeben sowie die jeweiligen Transportkosten pro Einheit zwischen allen Standorten bekannt sind. +Nun gibt es im Bereich des klassischen Transportproblems Sonderfälle. Ein Sonderfall ist z.B. das Zuordnungsproblem. diff --git a/buch/papers/munkres/teil1.tex b/buch/papers/munkres/teil1.tex index f4f5e39..aad45cc 100644 --- a/buch/papers/munkres/teil1.tex +++ b/buch/papers/munkres/teil1.tex @@ -3,53 +3,85 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 1 +\section{Beschrieb des Zuordnungsproblems \label{munkres:section:teil1}} \rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt + +Das Spezielle an einem Zuordnungsproblem ist, dass es an jedem Ort nur eine Einheit angeboten bzw. nachgefragt wird. Es werden hier nicht Mengen möglichst kostenminimal von einem zum anderen +Ort transportiert, sondern es geht um die kostenminimale Zuordnung von z.B. Personen oder Bau-Maschinen auf bestimmte Orte, Stellen oder Aufgaben. +Um dieses Problem in einer einfachen, händischen Art und Weise zu lösen wurde der Munkres-Algorithmus, auch die Ungarische Methode genannt, entwickelt. Diese Methode ist ein weiteres Hauptthema dieses Kapitels. + +\subsection{Zuordnungsproblem an einem konkreten Beispiel +\label{munkres:subsection:bonorum}} +Als Beispiel betrachten wir den Fall, wo ein Bauunternehmer einen Bauingenieur beauftragt, eine optimale Transportroute für die Umplatzierung seiner Kräne zu eruieren. Das heisst, die Transportstrecke für die Umplatzierung seine Kräne +soll möglichst klein werden. +Die Frage lautet: Wie sind die Kräne umzusetzen, damit deren Transportstrecke minimal wird? Bei der normalen Optimierung dürfen normalerweise beliebige reelle Werte $\mathbb{R}$ angenommen werden. +Beim Beispiel mit den Kräne gibt es aber ein Problem. Bei der Suche nach der optimalen Lösung darf nur die Methode der ganzzahligen Optimierung gewählt werden. Materialien kann man aufteilen, jedoch Maschinen nicht. Die Bauarbeiter auf der neuen Baustelle benötigen einen ganzen Kran und nicht nur einen halben Kran. Es muss immer ein ganzer Kran (Anzahl 1) von A nach B oder gar kein Kran (Anzahl 0) verschoben werden. +Für solche Optimierungsprobleme für reelle Variablen sind verschiedene Verfahren entwickelt worden, die im Allgemeinen auch sehr effizient sind. Das reelle Problem ist also in einer einfachen Art und Weise lösbar. Doch das Problem bleibt, wie in der Illustration oben ersichtlich. Es kann mit ganzzahligen Punkten kein Optimum erzielt werden. Das Ziel ist es an das Optimum so nah wie möglich heranzukommen und dies ist eine vergleichsweise träge und langsame Angelegenheit. + +\begin{figure} +\centering +\includegraphics[width=8cm]{papers/munkres/figures/ganzzahlige_punkte} +\caption{Problem der Ganzzahligkeit.} +\label{munkres:Vr2} +\end{figure} + + +\subsection{Zuordnungsproblem abstrakt +\label{munkres:subsection:bonorum}} + +In einem Zuordnungsproblem sind alle Angebots- und Bedarfsmengen gleich 1 \begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{munkres:equation1} +a_{i}=b_{j}=1 \end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{munkres:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{munkres:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{munkres:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. +Das Ziel ist es die Gesamtkosten zu minimieren. Mit Hilfe einer $n\times n$ Matrix +\[ +A += +\begin{pmatrix} +a_{11}&a_{12}&\dots &a_{1n}\\ +a_{21}&a_{22}&\dots &a_{2n}\\ +\vdots&\vdots&\ddots&\vdots\\ +a_{n1}&a_{n2}&\dots &a_{nn} +\end{pmatrix} +\in \mathbb{R}^{n,n} +\] +kann der Faktor Kosten mit in die Rechnung eingebracht werden. +In den Zellen dieser Matrix sind die Zahlen $a_{i,j}$ dargestellt, welche den Weg in z.B. Kilometer beschreiben. +Sie entstehen, wenn man z.B. einem Kran $i$ dem Einsatzort $j$ zuordnet. + +\subsection{Alternative Darstellungen des Zuordnungsproblems +\label{munkres:subsection:bonorum}} +\subsubsection{Netzwerk} +Ein (Fluss- oder Transport-) Netzwerk (engl. network) ist ein zusammenhängender Graph, bei dem jede Kante einen Fluss aufnehmen kann und jede Kante eine Kapazität für den Fluss hat. Die Menge des Flusses auf einer Kante kann die Kapazität der Kante nicht überschreiten. Ein Fluss muss die Einschränkung erfüllen, dass die Menge des Flusses in einen Knoten gleich der Menge des Flusses aus ihm heraus ist. Ein Fluss-Netzwerk (engl. flow network) ist ein Netzwerk, dessen Kanten zusätzlich Kosten pro Mengeneinheit des Flusses zugeordnet sind. Typischerweise will man einen Fluss durch die Kanten bestimmen, der den Einschränkungen des Netzwerks genügt und dessen Gesamtkosten minimal sind. Im Bild 21.2 dargestellt sind in den eckigen Klammern links die externen Flüsse $[1]$ für jeden Kran und in den eckigen Klammern rechts eine $[-1]$ für jeden Baustellenort. Die Kosten sind entlang der Kanten als Zahlen in Klammern dargestellt. +\subsubsection{Matrix} +Im Bild 21.3 ist eine typische $4\times 4$ Matrix dargestellt. Die Zeilen A1 bis A4 betreffen z.B. vier bestehende Maschinenlager eines Unternehmers. In den Spalten B1 bis B4 sind vier neue Baustellenorte zugewiesen. Die Zahlen in der Matrix bedeuten z.B. die Distanz in Kilometer von dem jeweiligen Lager zur jeweiligen Baustelle. +\subsubsection{Bitpartiter Graph} +Ein bipartiter Graph ist ein mathematisches Modell für Beziehungen +zwischen den Elementen zweier Mengen. Es eignet sich sehr gut zur Untersuchung von Zuordnungsproblemen. Zwischen zwei Gruppen von Objekten wird hierbei eine eindeutige Zuordnung hergestellt. Der Graph ist in Abbildung 21.4 ersichtlich. +\begin{itemize} +\item 3 = Anzahl der Knoten aus Menge A. +\item 3 = Anzahl der Knoten aus Menge B. +\end{itemize} + + +\begin{figure} +\centering +\includegraphics[width=5cm]{papers/munkres/figures/Netzwerkdarstellung} +\caption{Typische Netzwerkdarstellung eines Zuordnungsproblems.} +\label{munkres:Vr2} +\end{figure} +\begin{figure} +\centering +\includegraphics[width=5cm]{papers/munkres/figures/Matrixdarstellung} +\caption{Typische 4x4 Matrixdarstellung eines Zuordnungsproblems.} +\label{munkres:Vr2} +\end{figure} +\begin{figure} +\centering +\includegraphics[width=5cm]{papers/munkres/figures/bipartiter_graph} +\caption{$K_{3,3}$ vollständig bipartiter Graph mit 3 Knoten pro Teilmenge.} +\label{munkres:Vr2} +\end{figure} diff --git a/buch/papers/munkres/teil2.tex b/buch/papers/munkres/teil2.tex index 23536b9..2fe24f8 100644 --- a/buch/papers/munkres/teil2.tex +++ b/buch/papers/munkres/teil2.tex @@ -3,38 +3,10 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 2 +\section{Schwierigkeit der Lösung (Permutationen) \label{munkres:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{munkres:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. +\rhead{Schwierigkeit der Lösung (Permutationen)} +Eine Permutation ist eine Anordnung von Objekten in einer bestimmten Reihenfolge oder eine Umordnung von Objekten aus einer vorgegebenen Reihung. Ist eine optimale Zuordnung gefunden, so steht in jeder Zeile und jeder Spalte der Matrix genau ein Element, das zur optimalen Lösung gehört, eine solche Gruppe von Positionen wird auch als Transversale der Matrix bezeichnet. +Die Problemstellung kann auch so formuliert werden, dass man die Zeilen- oder die Spaltenvektoren so umordnet soll, dass die Summe der Elemente in der Hauptdiagonale maximal wird. Hieraus wird sofort ersichtlich, dass es in einer $n$×$n$-Matrix genau so viele Möglichkeiten gibt, die Zeilen- bzw. Spaltenvektoren zu ordnen, wie es Permutationen von $n$ Elementen gibt, also $n!$. Außer bei kleinen Matrizen ist es nahezu aussichtslos, die optimale Lösung durch Berechnung aller Möglichkeiten zu finden. Schon bei einer 10×10-Matrix gibt es nahezu 3,63 Millionen (3.628.800) zu berücksichtigende Permutationen. diff --git a/buch/papers/munkres/teil3.tex b/buch/papers/munkres/teil3.tex index b67ad74..fd25a74 100644 --- a/buch/papers/munkres/teil3.tex +++ b/buch/papers/munkres/teil3.tex @@ -3,38 +3,86 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 3 +\section{Der Munkres-Algorithmus (Ungarische Methode) \label{munkres:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum +\rhead{Der Munkres-Algorithmus (Ungarische Methode)} + +Mit der ungarischen Methode können also Optimierungsprobleme gelöst +werden, die bei gewichteten Zuordnungen in bipartiten Graphen entstehen. +Mit ihr kann die eindeutige Zuordnung von Objekten aus zwei Gruppen so +optimiert werden, dass die Gesamtkosten minimiert werden bzw.~der +Gesamtgewinn maximiert werden kann. + +\subsection{Geschichte +\label{munkres:subsection:malorum}} +Die Ungarische Methode wurde 1955 von Harold Kuhn entwickelt und veröffentlicht. +Der Name ``Ungarische Methode'' ergab sich, weil der Algorithmus +weitestgehend auf den früheren Arbeiten zweier ungarischer Mathematiker +basierte: Dénes Kőnig und Jenő Egerváry. +James Munkres überprüfte den Algorithmus im Jahr 1957 und stellte fest, +dass der Algorithmus (stark) polynomiell ist. +Seitdem ist der Algorithmus auch als Kuhn-Munkres oder +Munkres-Zuordnungsalgorithmus bekannt. +Die Zeitkomplexität des ursprünglichen Algorithmus war $O(n^4)$, +später wurde zudem festgestellt, dass er modifiziert werden kann, +um eine $O(n^3)$-Laufzeit zu erreichen. + +\subsection{Besondere Leistung der Ungarischen Methode +\label{munkres:subsection:malorum}} +Die Ungarische Methode ist ein kombinatorischer Optimierungsalgorithmus, der das Zuordnungsproblem +in polynomieller Zeit löst. +Der Begriff polynomielle Laufzeit bedeutet, dass die Laufzeit des Programms +wie $n^2$, $n^3$, $n^4$, etc.~wächst und vernünftig skaliert. $n$ ist hierbei die ''Grösse'' des Problems. + +\subsection{Unterschiedliche Anzahl von Quellen und Zielen +\label{munkres:subsection:malorum}} +Es gibt Fälle, in welchen das Ausgangsproblem keine quadratische Form besitzt. Das ist z. B. dann der Fall, wenn drei Mitarbeiter vier verschiedene Eignungstests absolvieren müssen. In diesem Fall wird in der Ungarischen Methode die Matrix künstlich mittels einer Dummy Position zu einem Quadrat ergänzt. Dummy-Positionen werden dann mit der größten vorhandenen Zahl aus der Matrix besetzt. Beispielsweise wird eine $3\times 4$ zu einer $4\times 4$-Matrix. + +\subsection{Beispiel eines händischen Verfahrens +\label{munkres:subsection:malorum}} + +Die ungarische Methode kann in einem einfachen händischen Beispiel erläutert werden. Wir gehen von der Kostenmatrix $A$ aus. Diese Matrix wird in mehreren Schritten immer weiter reduziert. Anschliessend erfolgen mehrere Zuordnungen. Hierbei ist zu beachten, dass jede Zeile und jede Spalte immer genau eine eindeutige Zuordnung ergibt. Es gibt Situationen, in denen man nichts mehr tun muss, um eine optimale Zuordnung zu finden. Eine optimale Zuordnung ohne zusätzliche Kosten ist eine Auswahl genau eines Feldes in jeder Zeile und Spalte, welches 0 enthält. Das Ziel des Algorithmus ist also, die Matrix so zu ändern, dass genügend Nullen in der Matrix vorkommen. Es ist zudem wichtig, dass man nach jeder Modifikation der Matrix testet, ob man bereits eine Zuordnung machen kann, also genügend Nullen hat. +Das Vorgehen wird in den nachfolgenden Schritten 1-6 beschrieben und auch in der Abbildung 21.5 dargestellt. + +\begin{enumerate} +\item Man beginnt mit der Zeilen-Reduktion. Pro Zeile eruiert man die kleinste Zahl. Diese kleinste Zahl, jeweils in rot markiert, wird bei allen anderen Ziffern in der jeweiligen Zeile subtrahiert. Mit dieser Subtraktion zieht man die unvermeidbaren Kosten ab, die man hat, um eine Baustelle zu erreichen. Man erkennt, dass die Nullen mit zwei Linien abdeckbar sind. Das heisst es gibt zwei Spalten bei denen noch keine Zuordnungen möglich sind. + +\item Auch im zweiten Schritt werden mittels der Spalten-Reduktion die unvermeidbaren Weg-Kosten abgezogen. Man zieht die kleinste Zahl, wiederum in rot markiert, in jeder Spalte von allen Zahlen in der Spalte ab. +Die Nullen können somit mit drei Linien abgedeckt werden. Im Idealfall hat die Matrix in jeder Zeile und Spalte bereits genügend viele Nullen, so dass man bereits eine Zuordnung ohne Mehrkosten machen kann. Dies ist jedoch noch nicht der Fall. Es sollen weitere Nullen in die Matrix hineingebracht werden. + +\item Es bleiben jetzt einige Felder übrig, für die noch keine Zuordnung möglich ist. Die kleinste Ziffer wird dabei aus den noch nicht mit blau markierten Zahlen ausgewählt werden. Im Beispiel ist es die Zahl 1. Das Feld mit dem kleinsten Eintrag beinhaltet die Kosten, die unvermeidlich sind, wenn man für diese Felder auch noch eine Zuordnung machen will. Um neue Nullen zu bekommen, lagert man jetzt die Kosten auf die anderen Zeilen und Spalten um. Dies tut man, indem man in allen nicht abgedeckten Feldern die minimalen Kosten subtrahiert und in den blau markierten Kreuzungspunkten dazu addiert. +Dieser Schritt 3 muss so oft wiederholt werden, bis genügend viele Nullen in der Matrix vorhanden sind. + +\item In Schritt 4 sollen jetzt möglichst viele Nullen markiert werden, welche freistehend sind. +Freistehend bedeutet, dass sowohl in der jeweiligen Zeile und Spalte keine andere markierte Null vorhanden ist. + +\item Alle markierten Nullen werden jetzt in eine 1 umgewandelt. Die restlichen Ziffern in der Matrix, exklusiv die einsen, sollen jetzt ignoriert und durch eine Null ersetzt werden. + +\item Zu guter Letzt werden überall wo eine 1 steht, die Zahlen aus der Ausgangsmatrix eingefügt. Nach Einsetzen der Zahlen können die in rot markierten Zahlen aufsummiert werden. Man erhält den minimalsten Transportweg von total 13 Kilometer. +\end{enumerate} + +\begin{figure} +\centering +\includegraphics[width=8cm]{papers/munkres/figures/Ungarische_Methode_Beispiel.png} +\caption{Händisches Beispiel des Munkres Algorithmus, minimalster Transportweg.} +\label{munkres:Vr2} +\end{figure} + +\subsection{Zuordnung der Kräne \label{munkres:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. +Als Resultat des Munkres-Algorithmus werden in Abbildung 21.6 nebst dem minimalsten Transportweg auch die optimalste Zuweisung der Kräne auf die neuen Standorte ersichtlich. +Es können die folgenden Zuordnungen aus der Matrix abgelesen werden: +\begin{itemize} +\item Der Kran von Baustelle A1 soll zur Baustelle B2. +\item Der Kran von Baustelle A2 soll zur Baustelle B3. +\item Der Kran von Baustelle A3 soll zur Baustelle B4. +\item Der Kran von Baustelle A4 soll zur Baustelle B1. +\end{itemize} +\begin{figure} +\centering +\includegraphics[width=3cm]{papers/munkres/figures/Ungarische_Methode_Beispiel_Zuw.png} +\caption{Händisches Beispiel des Munkres Algorithmus, Zuweisung der Kräne } +\label{munkres:Vr2} +\end{figure}
\ No newline at end of file diff --git a/buch/papers/munkres/teil4.tex b/buch/papers/munkres/teil4.tex new file mode 100644 index 0000000..9a27227 --- /dev/null +++ b/buch/papers/munkres/teil4.tex @@ -0,0 +1,9 @@ +% +% teil4.tex -- Beispiel-File für Teil 4 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{- +\label{munkres:section:teil4}} +\rhead{-} + diff --git a/buch/papers/munkres/teil5.tex b/buch/papers/munkres/teil5.tex new file mode 100644 index 0000000..b938c50 --- /dev/null +++ b/buch/papers/munkres/teil5.tex @@ -0,0 +1,8 @@ +% +% teil5.tex -- Beispiel-File für Teil 5 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{- +\label{munkres:section:teil5}} +\rhead{-} diff --git a/buch/papers/punktgruppen/Makefile b/buch/papers/punktgruppen/Makefile index f92dc95..03ad15a 100644 --- a/buch/papers/punktgruppen/Makefile +++ b/buch/papers/punktgruppen/Makefile @@ -11,11 +11,16 @@ SOURCES := \ symmetry.tex TIKZFIGURES := \ + tikz/atoms-grid-still.tex \ + tikz/atoms-grid-force.tex \ + tikz/atoms-piezo-still.tex \ + tikz/atoms-piezo-force-vertical.tex \ + tikz/atoms-piezo-force-horizontal.tex \ tikz/combine-symmetries.tex \ tikz/lattice.tex \ - tikz/piezo-atoms.tex \ tikz/piezo.tex \ tikz/projections.tex \ + tikz/stereographic-projections.tex \ tikz/symmetric-shapes.tex FIGURES := $(patsubst tikz/%.tex, figures/%.pdf, $(TIKZFIGURES)) @@ -28,7 +33,7 @@ figures/%.pdf: tikz/%.tex pdflatex --output-directory=figures $< .PHONY: standalone -standalone: standalone.tex $(SOURCES) +standalone: standalone.tex $(SOURCES) $(FIGURES) mkdir -p standalone cd ../..; \ pdflatex \ diff --git a/buch/papers/punktgruppen/Makefile.inc b/buch/papers/punktgruppen/Makefile.inc index 8cde9d7..fbb073e 100644 --- a/buch/papers/punktgruppen/Makefile.inc +++ b/buch/papers/punktgruppen/Makefile.inc @@ -11,8 +11,15 @@ dependencies-punktgruppen = \ papers/punktgruppen/crystals.tex \ papers/punktgruppen/piezo.tex \ papers/punktgruppen/references.bib \ - papers/punktgruppen/tikz/combine-symmetries.tex \ + papers/punktgruppen/tikz/atoms-grid-force.tex \ + papers/punktgruppen/tikz/atoms-grid-still.tex \ + papers/punktgruppen/tikz/atoms-piezo-force-horizontal.tex \ + papers/punktgruppen/tikz/atoms-piezo-force-vertical.tex \ + papers/punktgruppen/tikz/atoms-piezo-still.tex \ + papers/punktgruppen/tikz/combine-symmetries.tex \ papers/punktgruppen/tikz/lattice.tex \ papers/punktgruppen/tikz/piezo-atoms.tex \ papers/punktgruppen/tikz/piezo.tex \ - papers/punktgruppen/tikz/projections.tex + papers/punktgruppen/tikz/projections.tex \ + papers/punktgruppen/tikz/stereographic-projections.tex \ + papers/punktgruppen/tikz/symmetric-shapes.tex diff --git a/buch/papers/punktgruppen/crystals.tex b/buch/papers/punktgruppen/crystals.tex index 1aec16f..0a9d3b6 100644 --- a/buch/papers/punktgruppen/crystals.tex +++ b/buch/papers/punktgruppen/crystals.tex @@ -1,8 +1,7 @@ \section{Kristalle} -%einleitung sollte noch an das ende von der Symmetrie angepasst werden -Unter dem Begriff Kristall sollte sich jeder ein Bild machen können. -Wir werden uns aber nicht auf sein Äusseres fokussieren, sondern was ihn im Inneren ausmacht. -Die Innereien eines Kristalles sind glücklicherweise relativ einfach definiert. +Eine nicht allzu häufig gestellte Frage ist, wie ein Kristall definiert ist. +Um zu klären, was ein Kristall mit Symmetrien zu tun hat, ist jedoch genau diese Frage äusserst relevant. +Glücklicherweise ist das Innere eines Kristalles relativ einfach definiert. \begin{definition}[Kristall] Ein Kristall besteht aus Atomen, welche sich in einem Muster arrangieren, welches sich in drei Dimensionen periodisch wiederholt. \end{definition} @@ -12,117 +11,160 @@ Die Innereien eines Kristalles sind glücklicherweise relativ einfach definiert. \includegraphics[]{papers/punktgruppen/figures/lattice} \caption{ Zweidimensionales Kristallgitter. - \texttt{TODO: make wider and shorter} \label{fig:punktgruppen:lattice} } \end{figure} \subsection{Kristallgitter} Ein zweidimensionales Beispiel eines solchen Muster ist Abbildung \ref{fig:punktgruppen:lattice}. -Für die Überschaubarkeit haben wir ein simples Motiv eines einzelnen grauen Punktes gewählt und betrachten dies nur in Zwei Dimensionen. -Die eingezeichneten Vektoren $\vec{a}$ und $\vec{b}$ sind die kleinstmöglichen Schritte im Raum bis sich das Kristallgitter wiederholt. -Wird ein beliebiger grauer Gitterpunkt in \ref{fig:punktgruppen:lattice} gewählt -und um eine ganzzahlige Linearkombination von $\vec{a}$ und $\vec{b}$ verschoben, -endet er zwangsweise auf einem Gitterpunkt, wenn nicht wieder am selben Ort. -Im Dreidimensionalen-Raum können alle Gitterpunkte mit derselben Idee und einem zusätzlichen Vektor $\vec{c}$ also +Für die Überschaubarkeit haben wir ein simples Motiv eines einzelnen grauen Punktes dargestellt und betrachten dies nur in zwei Dimensionen. +Die eingezeichneten Vektoren \(\vec{a}_1\) und \(\vec{a}_2\) sind die kleinstmöglichen Schritte im Raum bis sich das Kristallgitter wiederholt. +Wird ein beliebiger grauer Gitterpunkt in Abbildung \ref{fig:punktgruppen:lattice} gewählt und um eine ganzzahlige Linearkombination von \(\vec{a}_1\) und \(\vec{a}_2\) verschoben, endet er zwangsweise auf einem Gitterpunkt, wenn nicht wieder am selben Ort. +Im dreidimensionalen Raum können alle Gitterpunkte mit derselben Idee und einem zusätzlichen Vektor \(\vec{a}_3\) also \[ - \vec{r} = n_1 \vec{a} + n_2 \vec{b} + n_3 \vec{c} + \vec{r} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3 = \sum_i n_i \vec{a}_i \] -erreicht werden sofern $\{n_1,n_2,n_3\} \in \mathbb{Z}$ sind. -Sind die Vektoren $\vec{a}$ , $\vec{b}$ , $\vec{c}$ gegeben , -ist ein Kristallgitter eindeutig beschrieben, weswegen sie auch als Grundvektoren bekannt sind. +erreicht werden sofern \(n_1,n_2,n_3 \in \mathbb{Z}\) sind. +Sind die Vektoren \(\vec{a}_1\), \(\vec{a}_2\), \(\vec{a}_3\) gegeben, ist ein Kristallgitter eindeutig beschrieben, weswegen sie auch als Grundvektoren bekannt sind. -\subsection{Translationssymmetrie} +\subsection{Translationssymmetrie} Da sich das ganze Kristallgitter wiederholt, wiederholen sich auch dessen Eigenschaften periodisch mit den Grundvektoren. -Sollte man sich auf einem Gitterpunkt in einem Kristall aufhalten, ist es unmöglich zu wissen, auf welchem Gitterpunkt man sich befindet, -da die Umgebungen aller Punkte Identisch sind. -Mit anderen worten: Jedes Kristallgitter $ G $ ist \emph{Translationssymmetrisch} in der Translation +Sollte man sich auf einem Gitterpunkt in einem Kristall aufhalten, ist es unmöglich zu wissen, auf welchem Gitterpunkt man sich befindet, da die Umgebungen aller Punkte identisch sind. +Mit anderen Worten: Jedes Kristallgitter $ G $ ist \emph{translationssymmetrisch} in der Translation \[ - Q_i(G) = G + \vec{a_i} -\] wobei der Vektor $a_i$ ein Grundvektor sein muss. + \vec{Q}_i(G) = G + \vec{a}_i, +\] +wobei der Vektor $\vec{a}_i$ ein Grundvektor sein muss. Da die Translationssymmetrie beliebig oft mit allen Grundvektoren angewendet werden kann, können wir auch sagen, dass alle Verschiebungen um eine Linearkombination -der Vektoren $\vec{a}$ , $\vec{b}$ und $\vec{c}$ erlaubt sind oder kurz, um $\vec{r}$. -Verschiebungen um $\vec{r}$ bewirken demnach keine Veränderungen, -solange wir ein unendlich grosses Kristallgitter verschieben. +der Vektoren $\vec{a}_1$ , $\vec{a}_2$ und $\vec{a}_3$ erlaubt sind. +Dabei sollte erwähnt werden, dass eine Translationssymmetrie nur in unendlich grossen Kristallgittern besteht. -\subsection{Limitierte Kristallsymmetrien} +\subsection{Einschränkungen durch Kristallsymmetrien} \label{sec:punktgruppen:Translationssymmetrie} Die Translationssymmetrie ist wohl keine grosse Überraschung, wenn man die Abbildung \ref{fig:punktgruppen:lattice} betrachtet. - Was nicht direkt ersichtlich ist, ist das auch wenn die Grundvektoren frei gewählt werden können, - können nur Rotationssymmetrische Kristalle bestimmter Rotationswinkel erzeugt werden. - + Was nicht direkt ersichtlich ist, ist dass bei beliebigen Grundvektoren nicht beliebige Symmetrien erstellt werden können. + Dies weil die Translationssymmetrie eines Kristalles weitere Symmetrien deutlich einschränkt. + \begin{figure} \centering \includegraphics[]{papers/punktgruppen/figures/combine-symmetries} \caption{ Translations und Rotationssymmetrisches Kristallgitter - \texttt{TODO: make wider and change color (yellow)} } \label{fig:punktgruppen:rot-geometry} \end{figure} - \subsubsection{Translationssymmetrie $Q$ in Kombination mit Rotationssymmetrie $C_\alpha$} % Müssen uns auf eine schreibweise für Symmetrie Operationen einigen oder sicher am Ende überprüfen - In Abbildung \ref{fig:punktgruppen:rot-geometry} Sehen wir Gitterpunkte und deren Zusammenhänge. +\begin{satz} \label{thm:punktgruppen:crystal-restriction} + Die Rotationssymmetrien eines Kristalls sind auf 2-fach, 3-fach, 4-fach und 6-fach beschränkt. + Mit anderen Worten: Es sind nur Drehwinkel von + 0\(^{\circ}\), + 60\(^{\circ}\), + 90\(^{\circ}\), + 120\(^{\circ}\) und + 180\(^{\circ}\) + m\"oglich. +\end{satz} + +\begin{proof} + In Abbildung \ref{fig:punktgruppen:rot-geometry} sehen wir Gitterpunkte und deren Zusammenhänge. \begin{itemize} - \item $A$ ist unser erster Gitterpunkt. - - \item $A'$ ist gegeben, weil wir $A$ mit der Translation $Q$ um einen Grundvektor verschieben und wir wissen, - dass nach einer Translation wieder ein Gitterpunkt an der Verschobenen Stelle sein muss. - \item $B$ entsteht, weil wir die Rotationssymmetrie $C_\alpha$ auf den Punkt $A$ anwenden. - Dadurch dreht sich das ganze Gitter um den Winkel $\alpha$. - Für uns bedeutet dies lediglich, dass unser zweiter Punkt $A'$ abgedreht wird. - An der neuen Position von $A'$ muss also auch ein Punkt sein, um die Rotationssymmetrie zu erfüllen. - \item $B$ ist unser Name für diesen neuen Punkt. - Da auch die Eigenschaften des Kristallgittes periodisch mit dem Gitter sein müssen, dürfen wir $C_\alpha$ auch auf $A'$ anwenden. - Also wenden wir $C_\alpha$ invertiert - \footnote{Eine Rotationssymmetrie muss auch in die inverse Richtung funktionieren. - Genauere Überlegungen hierzu werden dem Leser überlassen, da sich die Autoren nicht explizit mit dieser Frage Auseinander gesetzt haben.} - auch auf $A'$ an. - Dies dreht $A$ auf einen neuen Punkt. - \item $B'$ ist kein zufälliger Name für diesen neuen Punkt, denn wir wissen, dass zwischen allen Punkten eine Translationssymmetrie bestehen muss. - Die Translationssymmetrie zwischen $B$ und $B'$ ist hier als $Q'$ bezeichnet. + \item \(A\) ist unser erster Gitterpunkt. + + \item \(A'\) ist gegeben, weil wir \(A\) mit der Translation \(\vec{Q}\) um einen Grundvektor verschieben und wir wissen, + dass nach einer Translation wieder ein Gitterpunkt an der verschobenen Stelle sein muss. + \item \(B\) entsteht, weil wir die Rotationssymmetrie \(C_n\) auf den Punkt \(A\) anwenden. + Dadurch dreht sich das ganze Gitter um den Winkel \(360^\circ/n\). + Für uns bedeutet dies lediglich, dass unser zweiter Punkt \(A'\) abgedreht wird. + An der neuen Position \(B\) von \(A'\) muss also auch ein Punkt des Gitters sein, um die Rotationssymmetrie zu erfüllen. + \item \(B\) ist unser Name für diesen neuen Punkt. + Da auch die Eigenschaften des Kristallgitters periodisch mit dem Gitter sein müssen, dürfen wir \(C_n\) auch auf \(A'\) anwenden. + Also wenden wir \(C_n^{-1}\) auch auf \(A'\) an. + Dies dreht \(A\) auf einen neuen Punkt. + \item \(B'\) ist kein zufälliger Name für diesen neuen Punkt, denn wir wissen, dass zwischen allen Punkten eine Translationssymmetrie bestehen muss. + Die Translationssymmetrie zwischen \(B\) und \(B'\) ist hier als \(\vec{Q}'\) bezeichnet. \end{itemize} Mit den gegebenen Punkten lassen sich geometrische Folgerungen ziehen. - Wir beginnen, indem wir die Länge der Translation $Q$ mit jener von $Q'$ vergleichen. - Aus Abbildung \ref{fig:punktgruppen:rot-geometry} ist ersichtlich, dass $|Q| = |Q'|+ 2x$. - Ist $Q$ ein Grundvektor so muss $|Q'|$ ein ganzes vielfaches von $|Q|$ sein. Also + Wir beginnen, indem wir die Länge der Verschiebung \(|\vec{Q}| = Q\) setzen und \(|\vec{Q}'| = Q'\). + Aus Abbildung \ref{fig:punktgruppen:rot-geometry} ist ersichtlich, dass \(Q' = Q + 2x\). + Da \(\vec{Q}\) eine Translation um ein Grundvektor ist , muss \(\vec{Q}'\) ein ganzes Vielfaches von \(\vec{Q}\) sein. + Demnach ist auch die Länge \[ - |Q'| = n|Q| = |Q| + 2x + Q' = nQ = Q + 2x . \] - Die Strecke $x$ lässt sich auch mit hilfe der Trigonometrie und dem angenommenen Rotationswinkel $\alpha$ ausdrücken: + Die Strecke \(x\) lässt sich auch mit Hilfe der Trigonometrie und dem angenommenen Rotationswinkel \(\alpha\) ausdrücken: \[ - n|Q| = |Q| + 2|Q|\sin(\alpha - \pi/2) + nQ = Q + 2Q\sin(\alpha - \pi/2) . \] - Wir können mit $|Q|$ dividieren um unabhängig von der Läge des Grundvektors zu werden, - was auch Sinn macht, da eine Skalierung eines Kristalles seine Symmetrieeigenschaften nicht tangieren soll. - Zusätzlich können wir den Sinusterm vereinfachen. + Wir können durch \(Q\), dividieren um unabhängig von der Läge des Grundvektors zu werden, was auch Sinn macht, + da eine Skalierung eines Kristalles seine Symmetrieeigenschaften nicht tangiert. + Zusätzlich können wir den Sinusterm vereinfachen. Somit wird \[ - n = 1 - 2\cos\alpha - \alpha = \cos^{-1}\left(\frac{1-n}{2}\right) + n = 1 - 2\cos\alpha \quad\text{oder}\quad + \alpha = \cos^{-1}\left(\frac{1-n}{2}\right). \] Dies schränkt die möglichen Rotationssymmetrien auf - \[ + \( \alpha \in \left\{ 0^\circ, 60^\circ, 90^\circ, 120^\circ, 180^\circ\right\} - \] + \) ein. +\end{proof} \begin{figure} \centering - \includegraphics[]{papers/punktgruppen/figures/projections} - \caption{Kristallklassen mit zugehöriger Schönfliesnotation} - \label{fig:punktgruppen:Kristallkassen} + \includegraphics[height=6cm]{papers/punktgruppen/figures/stereographic-projections} + \caption{ + Stereografische Projektion einer \(C_{i}\) Symmetrie. Es wird eine Linie vom magentafarbenen Punkt auf der oberen Hälfte der Kugel zum Südpol gezogen. + Wo die Linie die Ebene schneidet (\(z = 0\)), ist die Projektion des Punktes. + Die Koordinaten der Projektionen sind einfach zu berechnen: ein Punkt auf eine Kugel mit Radius \(r\) mit den Koordinaten \(x, y, z,\) wird auf \(xr/(r + z), yr/(r + z)\) projiziert. + Für den orangefarbenen Punkt unterhalb des Äquators wird die Linie zum Nordpol gezogen und die Projektionsformel hat stattdessen einen Nenner von \(r - z\). + } + \label{fig:punktgruppen:stereographic-projections} \end{figure} \subsection{Kristallklassen} -Vorgehend wurde gezeigt, dass in einem zweidimensionalen Kristallgitter nicht alle Symmetrien möglich sind. -Mit weiteren ähnlichen überlegungen gezeigt werden kann, dass Kristalle im dreidimensionalen Raum -\footnote{Alle $17$ möglichen zweidimensionalen Symmetrien sind als Wandmustergruppen bekannt} -nur auf genau $32$ Arten punktsymmetrisch sein können. -Diese $32$ möglichen Punktsymmetrien scheinen durchaus relevant zu sein, denn sie werden unter anderem als Kristallklassen bezeichnet. -Eine mögliche Art, die Klassen zu benennen ist nacht dem Mathematiker Arthur Moritz Schönflies, -welcher sich mit der Klasifizierung dieser Symmetrien auseinandergesetzt hat. -Auf der Abbildung \ref{fig:punktgruppen:Kristallkassen} sind die möglichen Punktsymmetrien mit deren Schönfliesnotation aufgelistet. -Als Darstellungsmethode wurde die stereographische Projektion gewählt, wobei $5$ Klassen aus Gründen der Überschaubarkeit nicht gezeichnet wurden. + +Im vorausgegangenen Abschnitt wurde gezeigt, dass in einem zweidimensionalen Kristallgitter nicht alle Symmetrien möglich sind. + Mit weiteren ähnlichen Überlegungen kann gezeigt werden, dass Kristalle im dreidimensionalen Raum nur auf genau 32 Arten rein punktsymmetrische Symmetriegruppen bilden können. + Diese 32 möglichen Symmetriegruppen scheinen durchaus relevant zu sein, denn sie werden unter anderem als Kristallklassen bezeichnet. + Die 32 möglichen Kristallklassen sind auf Abbildung \ref{fig:punktgruppen:kristallklassen} zu sehen. + Die Darstellung von dreidimensionalen Punktsymmetrien wurde mit der stereographischen Projektion ermöglicht (siehe Abbildung \ref{fig:punktgruppen:stereographic-projections}), wobei die gestrichelten Klassen aus Gründen der Überschaubarkeit nicht im Detail gezeichnet wurden. + + +\begin{figure} + \centering + \includegraphics[]{papers/punktgruppen/figures/projections} + \caption{Kristallklassen mit zugehörigem Schönflies-Symbol} + \label{fig:punktgruppen:kristallklassen} +\end{figure} + +\subsubsection{Schönflies-Symbolik} + +Jede der 32 Kristallklassen auf der Abbildung \ref{fig:punktgruppen:kristallklassen} ist mit ihrem zugehörigen Schönflies-Symbol bezeichnet. + Die Schönflies-Symbolik stammt von dem Mathematiker Arthur Moritz Schönflies, welcher sich unter anderem mit der Klasifizierung der Punktgruppen auseinandergesetzt hat. + Er hat Untergruppen gebildet, welche als Grossbuchstaben in Abbildung \ref{fig:punktgruppen:kristallklassen} zu sehen sind. + \begin{itemize} + \item In Kristallen ist nur die Drehgruppe \(C\), Diedergruppe \(D\), Drehspiegelgruppe \(S\), Tetraedergruppe \(T\) und die Oktaedergruppe \(O\) zu finden. + Es gäbe auch die Ikosaedergruppe \(I\) und die Kugelgruppe \(K\), diese sind aber nach Satz \ref{thm:punktgruppen:crystal-restriction} nicht kompatibel mit der Translationssymmetrie eines Kristalles und daher in der Kristallographie nicht relevant. + \item Dank Abschnitt \ref{sec:punktgruppen:Translationssymmetrie} wissen wir, wieso in Abbildung \ref{fig:punktgruppen:kristallklassen} auf \(C\) nur ganz bestimmte Subskripte folgen. + Ist im Subskript eine Zahl \(n\) zu finden, steht dies für eine \(n\)-fache Symmetrie. + Daher darf \(C_5\) auf der Abbildung \ref{fig:punktgruppen:kristallklassen} nicht vorkommen, da \(360^\circ/5 = 72^\circ\) was nach Satz \ref{thm:punktgruppen:crystal-restriction} keine mögliche Rotationssymmetrie eines Kristalles ist. + \item Sind im Subskript Buchstaben, definieren diese weitere Symmetrieeigenschaften der Klasse. + Für die folgenden Betrachtungen müssen wir uns Abbildung \ref{fig:punktgruppen:kristallklassen} genauer ansehen. + Dabei ist mit horizontal flach auf dem Papier gemeint. + \begin{itemize} + \item[\(h\)] bezeichnet eine horizontale Spiegelebene und + \item[\(v\)] eine Symmetrieebene, was eine Spiegelebene ist, die sich mit der Symmetrie mitdreht. + Zum Beispiel hat \(C_{3v}\) eine vertikale Spiegelebene, die durch die 3-fache Drehsymmetrie als 3 Spiegelebenen erscheinen. + \item[\(s\)] ist ein spezielles Subskript um die beiden Symmetriegruppen \(C_{1v}\) und \(C_{1h}\) zu beschreiben, weil \(C_{1v} = C_{1h}\). + \item[\(d\)] symbolisiert eine diagonale Symmetrieebene. + Es wird ersichtlich wie diagonal gemeint ist, wenn man \(D_2\) zu \(D_{2d}\) vergleicht. + \item[\(i\)] steht für ein Inversionszentrum. Hat eine Symmetriegruppe ein Inversionszentrum, bedeutet dies dass sie im Ursprung punktsymmetrisch ist. + \end{itemize} + \end{itemize} +Zu beachten ist jedoch, dass manche Symmetriegruppen mit mehreren Schönflies-Symbolen beschieben werden können. + \(C_{3i}\) beschreibt genau das selbe wie \(S_6\), da eine dreifache Rotationssymmetrie mit einem Inversionszentrum einer sechsfachen Drehspiegelsymmetrie entspricht. + +%% vim:spell spelllang=de showbreak=.. breakindent linebreak: diff --git a/buch/papers/punktgruppen/figures/atoms-grid-force.pdf b/buch/papers/punktgruppen/figures/atoms-grid-force.pdf Binary files differnew file mode 100644 index 0000000..b3e6215 --- /dev/null +++ b/buch/papers/punktgruppen/figures/atoms-grid-force.pdf diff --git a/buch/papers/punktgruppen/figures/atoms-grid-still.pdf b/buch/papers/punktgruppen/figures/atoms-grid-still.pdf Binary files differnew file mode 100644 index 0000000..752014d --- /dev/null +++ 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b/buch/papers/punktgruppen/intro.tex @@ -1,14 +1,16 @@ \section{Einleitung} -Es gibt viele Möglichkeiten sich in Kristallen zu verlieren. -Auch wen man nur die mathematischen Betrachtunngsweisen berüksichtigt, hat man noch viel zu viele Optionen sich mit Kristallen zu beschäftigen. -In diesem Kapitel ist daher der Fokus ``nur'' auf die Symmetrie gelegt. -Zu beginn werden wir zeigen was eine Symmetrie ausmacht und dass sie noch weit mehr in sich verbirgt als nur schön auszusehen. -Die vorgestellten Symmetrien sind äusserst gut geeignet um die Grundeigenschaften eines Kristalles zu Beschreiben. -Mit etwas kiffligen geometrischen Überlegungen kann man zeigen wass in der Welt der Kristallographie alles möglich ist oder nicht. -Die Einschränkungen sind durchaus wilkommen, dank ihnen halten sich die möglichen Kristallgitter in Grenzen und Lassen sich Kategorisieren. -Kategorien sind nicht nur für einen besseren Überblich nützlich, sondern kann man aus ihnen auch auf Physikalische Eigenschaften schliessen, als spannendes Beispiel: Die Piezoelektrizität. -Die Piezoelektrizität ist vielleicht noch nicht jedem bekannt, sie versteckt sich aber in diversen Altagsgegenständen zum Beispiel sorgen sie in den meisten Feuerzeugen für die Zündung. -Ein Funken Interesse ist hoffentlich geweckt um sich mit dem scheinbar trivialen thema der Symmetrie auseinander zu setzten. - +Es gibt viele Möglichkeiten sich in Kristallen zu verlieren. +Auch wenn man nur die mathematischen Betrachtungsweisen berücksichtigt, hat man noch viel zu viele Optionen, sich mit Kristallen zu beschäftigen. +In diesem Kapitel wird daher der Fokus ``nur'' auf die Symmetrie gelegt. +Zu Beginn werden wir zeigen, was eine Symmetrie ausmacht und dass sie noch weit mehr in sich verbirgt als nur schön auszusehen. +Die vorgestellten Symmetrien sind äusserst gut geeignet, um die Grundeigenschaften eines Kristalles zu beschreiben. +Mit etwas kniffligen geometrischen Überlegungen kann man zeigen, was in der Welt der Kristallographie alles möglich ist oder nicht. +Diese erlauben alle möglichen Kristalle nach ihren Symmetrien in erstaunlich wenige Klassen zu kategorisieren. +Kategorien sind nicht nur für einen besseren Überblick nützlich, sondern kann man aus ihnen auch auf physikalische Eigenschaften schliessen. +Als spannendes Beispiel: Die Piezoelektrizität. +Piezoelektrizität beschreibt einen Effekt, ohne welchen diverse Altagsgegenständen nicht besonders nützlich wären. +Zum Beispiel sorgt er in den allermeisten Feuerzeugen für die Zündung. +Hiermit ist hoffentlich ein Funken Interesse geweckt um sich mit dem scheinbar trivialen Thema der Symmetrie auseinander zu setzten. +%% vim:linebreak breakindent showbreak=.. spell spelllang=de: diff --git a/buch/papers/punktgruppen/main.tex b/buch/papers/punktgruppen/main.tex index a6e246c..556fc2b 100644 --- a/buch/papers/punktgruppen/main.tex +++ b/buch/papers/punktgruppen/main.tex @@ -18,6 +18,8 @@ \nocite{punktgruppen:pinter-algebra} \nocite{punktgruppen:sands-crystal} \nocite{punktgruppen:lang-elt2} +\nocite{punktgruppen:ouchem} +\nocite{punktgruppen:restriction} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/punktgruppen/piezo.tex b/buch/papers/punktgruppen/piezo.tex index e6b595a..1cf9b98 100644 --- a/buch/papers/punktgruppen/piezo.tex +++ b/buch/papers/punktgruppen/piezo.tex @@ -1,6 +1,5 @@ \section{Piezoelektrizität} -Die Piezoelektrizität ist per Definition spannend. -Sie beschreibt die Eigenschaft, dass gewisse Kristalle eine elektrische Spannung erzeugen, wenn machanischer Druck auf sie ausgeübt wird. +Die Piezoelektrizität ist die spannende Eigenschaft, dass gewisse Kristalle eine elektrische Spannung erzeugen, wenn mechanischer Druck auf sie ausgeübt wird. \begin{figure} \centering @@ -10,65 +9,69 @@ Sie beschreibt die Eigenschaft, dass gewisse Kristalle eine elektrische Spannung \end{figure} \subsection{Polarisierung} -Piezoelektrizität basiert darauf, dass zwischen den Oberflächen des Kristalles ein Ladungsungleichgewicht entsteht siehe Abbildung\ref{fig:punktgruppen:basicPiezo}. -Dieses Ungleichgewicht resultiert, -weil durch den mechanischen Druck auf der einen Oberfläche des Kristalles positiv Ione näher an die Oberfläche gelangen, -wärend auf der gegenüberliegenden Oberfläche sich mehr negative Ionen Sammeln. -Das sich die atomare Struktur eines Kristalles unter Druck genau so verformt ist nicht bei jedem Kristall gegeben. + +Piezoelektrizität basiert darauf, dass zwischen den Oberflächen des Kristalles ein Ladungsungleichgewicht entsteht (siehe Abbildung\ref{fig:punktgruppen:basicPiezo}). +Dieses Ungleichgewicht resultiert, weil durch den mechanischen Druck auf der einen Oberfläche des Kristalles positive Ionen näher an die Oberfläche gelangen, wärend auf der gegenüberliegenden Seite dasselbe mit negativen Ionen passiert. +Es besitzt jedoch nicht jeder Kristall eine atomare Struktur, welche sich unter Druck genau so verformt. Der Aufbau und somit auch die Symmetrie des Kristalles sind daher relevant für die Entstehung dieses Effektes. + \begin{figure} \centering - \includegraphics[]{papers/punktgruppen/figures/piezo-atoms} + \begin{tabular}{c |c} + \subfigure[][\label{fig:punktgruppen:atoms-piezo}]{\includegraphics{papers/punktgruppen/figures/atoms-piezo-still}} & + \subfigure[][\label{fig:punktgruppen:atoms-grid}]{\includegraphics{papers/punktgruppen/figures/atoms-grid-still}} \\ + \subfigure[][\label{fig:punktgruppen:atoms-piezo-fv}]{\includegraphics{papers/punktgruppen/figures/atoms-piezo-force-vertical}} + \hspace{2mm} + \subfigure[][\label{fig:punktgruppen:atoms-piezo-fh}]{\includegraphics{papers/punktgruppen/figures/atoms-piezo-force-horizontal}} + \hspace{3mm} & \hspace{3mm} + \subfigure[][\label{fig:punktgruppen:atoms-grid-f}]{\includegraphics{papers/punktgruppen/figures/atoms-grid-force}} \\ + \end{tabular} \caption{ Kristallstrukturen mit und ohne piezoelektrischer Eigenschaft. - \texttt{TODO: adapt figure for paper with subfigure markers.} } \label{fig:punktgruppen:atomPiezo} \end{figure} \subsection{Atomarer Aufbau} -Die Polarisation resultiert über eine gesamte Oberfläche eines Kristalles, entscheidend ist aber der atomare Aufbau. + +Die Polarisation entsteht an der Oberfläche eines Kristalles, die Erklärung dazu finden wir jedoch im atomaren Aufbau. Wir wollen dazu die verschiedenen Kristallstrukturen auf Abbildung \ref{fig:punktgruppen:atomPiezo} diskutieren. -In Abbildung \ref{fig:punktgruppen:atomPiezo} gilt für alle Strukturen, dass rote Kreise Positive Ionen und blaue negative Ionen repräsentieren. -%liste oder anderes format?.. -Struktur$(a)$ zeigt ein piezoelektrisches Material in Ruhe. Struktur $(b)$ ist dasselbe Kristallgitter, jedoch wird es senkrecht belastet. -Eingezeichnet ist auch das elektrische Feld, welches entsteht, weil mitlleren Ladungsträger weiter auseinander gerdrückt werden. -Als hilfe zur Vorstellung kann man $(b)$ zwischen zwei leitende Platten setzen, -so wird ersichtlich, dass mit wachsendem Druck eine negative Ladung an die rechte Platte gedrückt wird, -während sich die positiven Ionen weiter entfernen. -$(d)$ ist nicht piezoelektrisch. -Dies wird ersichtlich, wenn man $(d)$ unterdruck setzt und sich die Struktur zu $(e)$ verformt. -Setzt man $(e)$ gedanklich auch zwischen zwei leitende Platten scheint es als würden rechts mehr Positive Ionen in die Platte gedrückt werden -und links umgekehrt. -Dies ist aber nicht mehr der Fall, wenn der Kristall nach oben und periodisch wiederholt. -Struktur $(c)$ zeigt $(a)$ in unter horizontaler Belastung. -Was in zwischen $(b)$ und $(c)$ zu beobachten ist, ist dass das entstandene Ladungsdifferenz orthogonal zu der angelegten Kraft entsteht, -im Gegensatz zu $(b)$. -Daraus kann man schlissen, dass $(a)$ keine Rotationssymmetrie von $90^\circ$ besitzen kann, weil die Eigenschaften ändern bei einer $90^\circ$ Drehung. -Das Fehlen dieser Rotationssymmetrie kann mit betrachten von $(a)$ bestätigt werden. - -\subsection{Punktsymmetrie}\footnote{In der Literatur wird ein Punktsymmetrisches Kristallgitter oft als Kristallgitter mit Inversionszentrum bezeichnet.} -Piezoelektrische Kristalle können nicht Punktsymmetrisch sein. +In Abbildung \ref{fig:punktgruppen:atomPiezo} gilt für alle Strukturen, dass rote Kreise positive Ionen und blaue negative Ionen repräsentieren. +Struktur \subref{fig:punktgruppen:atoms-piezo} zeigt ein piezoelektrisches Material in Ruhe. +Struktur \subref{fig:punktgruppen:atoms-piezo-fv} ist dasselbe Kristallgitter, jedoch wird es senkrecht belastet. +Eingezeichnet ist auch das elektrische Feld, welches entsteht, weil die Ladungsträger ganz links und rechts weiter auseinander gedrückt werden. +Als Hilfe zur Vorstellung kann man \subref{fig:punktgruppen:atoms-piezo-fv} zwischen zwei leitende Platten setzen, so wird ersichtlich, dass mit wachsendem Druck eine negative Ladung an die rechte Platte gedrückt wird, während sich die positiven Ionen weiter entfernen. + + +Die Struktur \subref{fig:punktgruppen:atoms-grid} ist nicht piezoelektrisch. +Dies wird ersichtlich, wenn man \subref{fig:punktgruppen:atoms-grid} unter Druck setzt und sich die Struktur zu \subref{fig:punktgruppen:atoms-grid-f} verformt. +Setzt man \subref{fig:punktgruppen:atoms-grid-f} gedanklich auch zwischen zwei leitende Platten, scheint es, als würden rechts mehr positive Ionen in die Platte gedrückt werden und links umgekehrt. +Dies ist aber nicht mehr der Fall, wenn sich die Struktur nach oben und unten periodisch wiederholt. + + +Struktur \subref{fig:punktgruppen:atoms-piezo-fh} zeigt \subref{fig:punktgruppen:atoms-piezo} in unter horizontaler Belastung. +Was zwischen \subref{fig:punktgruppen:atoms-piezo-fv} und \subref{fig:punktgruppen:atoms-piezo-fh} zu beobachten ist, dass die entstandene Ladungsdifferenz orthogonal zu der angelegten Kraft entsteht, im Gegensatz zu \subref{fig:punktgruppen:atoms-piezo-fh}. +Daraus kann man schliessen, dass \subref{fig:punktgruppen:atoms-piezo} keine Rotationssymmetrie von \(90^\circ\) besitzen kann, weil die Eigenschaften der Struktur sich bei einer \(90^\circ\) Drehung ändern. +Das Fehlen dieser Rotationssymmetrie bestätigt sich auch wenn \subref{fig:punktgruppen:atoms-piezo} als Hexagon betrachtet wird. + + +\subsection{Punktsymmetrie} + +Piezoelektrische Kristalle können nicht punktsymmetrisch sein. Kristallgitter, bei welchen eine Punktspiegelung eine symmetrische Operation ist, können keine piezoelektrische Kristalle bilden. -Auf Abbildung \ref{fig:punktgruppen:atomPiezo} ist bewusst $(a)$ ein nicht Punktsymmetrischer Kristall mit einem Punktsymmetrischen $(d)$ verglichen worden. -Als vereinfachte Erklärung kann mann sich wieder das Bild vor augen führen, eines Kristalles, -welcher unter Druck auf der einen Seite negative und der anderen Seite positive Ionen an seine Oberfläche verdrängt. -Spiegelt man nun den Kristall um den Gitterpunkt in der mitte des Kristalles, so würden die negativen Ionen auf den Positiven auf der anderen seite landen, -was der Definition einer Symmetrie deutlich widerspricht. +Auf Abbildung \ref{fig:punktgruppen:atomPiezo} ist bewusst \subref{fig:punktgruppen:atoms-piezo} ein nicht punktsymmetrischer Kristall mit einem punktsymmetrischen \subref{fig:punktgruppen:atoms-grid} verglichen worden. +Als vereinfachte Erklärung kann man sich wieder das Bild eines Kristalles wie \subref{fig:punktgruppen:atoms-piezo} vor Augen führen, welcher unter Druck auf der einen Seite negative und der anderen Seite positive Ionen an seine Oberfläche verdrängt. +Spiegelt man nun den Kristall um den Gitterpunkt in der Mitte des Kristalles, so würden die negativen Ionen auf den positiven auf der anderen Seite landen, was der Definition einer Symmetrie deutlich widerspricht. + \subsection{Vom Kristall zum Feuer} -Piezoelektrizität hat durchaus nutzen im Alltag. -Feuerzeuge welche nicht auf dem Prinzip beruhen einen Zündstein abzuschleifen, -sonder ohne Verschleiss auf Knopfdruck einen Zündfunken erzeugen, basieren auf dem Prinzip der Piezoelektrizität. -Drückt der Nutzende auf den Zündknopf spannt sich eine Feder bis zu einer Konfigurierten Spannung. -Wird vom Nutzenden weiter gedrückt entspannt sich die Feder schlagartig und beschleunigt mit der gespeicherten Energie ein Hammer, -welcher auf das Piezoelement aufschlägt. -Der augenblicklich hohe Druck sorgt an den Piezokontakten für eine eben so Kurze aber hohe elekrische Spannung. + +Piezoelektrizität hat durchaus Nutzen im Alltag. +Feuerzeuge welche nicht auf dem Prinzip beruhen einen Zündstein abzuschleifen, sondern ohne Verschleiss auf Knopfdruck einen Zündfunken erzeugen, basieren auf dem Prinzip der Piezoelektrizität. +Drückt der Nutzende auf den Zündknopf, spannt sich eine Feder bis zu einer konfigurierten Spannung. +Drückt der Nutzende stärker zu, entspannt sich die Feder schlagartig und beschleunigt mit der gespeicherten Energie ein Hammer, welcher auf das Piezoelement aufschlägt. +Der augenblicklich hohe Druck sorgt an den Piezokontakten für eine eben so kurze aber hohe elektrische Spannung. Die Spannung reicht aus, um eine Funkenstrecke zu überwinden und so eine entflammbares Gas zu entzünden. -Sollten Sie also eines Tages in die Situation geraten, in welcher Sie zwei verschiedene Kristalle vor sich haben -und ein piezoelektrisches Feuerzeug bauen müssen, -wobei Sie aber wissen, dass einer eine Punktsymmetrie aufweist, -versuche sie es mit dem anderen. -Ich muss aber anmerken, dass aus den $21$ möglichen Kristallsymmetrien ohne Punktsymmetrie einer nicht piezoelektrisch ist. -ein wenig glück brauchen Sie also immer noch. +Sollte der Leser eines Tages in die Situation geraten, in welcher er zwei verschiedene Kristalle vor sich hat und ein piezoelektrisches Feuerzeug bauen musst, wobei bekannt ist, dass der eine eine Punktsymmetrie aufweist, empfiehlt es sich, sich mit dem anderen zu versuchen. + diff --git a/buch/papers/punktgruppen/references.bib b/buch/papers/punktgruppen/references.bib index 9edb8bd..7928b22 100644 --- a/buch/papers/punktgruppen/references.bib +++ b/buch/papers/punktgruppen/references.bib @@ -26,10 +26,29 @@ @book{punktgruppen:lang-elt2, title = {Elektrotechnik 2}, - author = {Hans-Dieter Lang}, + author = {Hans-Dieter Lang Ph.D}, publisher = {Fachhochschule Ostschweiz Rapperswil}, year = {2020}, month = {2}, inseries = {Vorlesungsskript zum Modul ELT}, } +@online{punktgruppen:ouchem, + title = {Symmetry in Crystallography}, + author = {Dept. of Chemistry \& Biochemistry{,} Chemical Crystallography Laboratory{,} University of Oklahoma}, + year = {2019}, + month = {11}, + day = {17}, + url = {http://archive.today/2021.07.22-083802/http://xrayweb.chem.ou.edu/notes/symmetry.html}, + urldate = {2021-07-22}, +} + +@online{punktgruppen:restriction, + title = {Structure of Materials: Allowed Rotational Symmetry in Crystals}, + author = {Silvija Gradecak-Garaj{,} Massachusetts Institute of Technology (MIT)}, + year = {2020}, + month = {4}, + day = {9}, + url = {https://www.youtube.com/watch?v=Ia2eHF1ZKoI}, + urldate = {2021-07-30}, +} diff --git a/buch/papers/punktgruppen/symmetry.tex b/buch/papers/punktgruppen/symmetry.tex index 1dc6f98..4a8d911 100644 --- a/buch/papers/punktgruppen/symmetry.tex +++ b/buch/papers/punktgruppen/symmetry.tex @@ -1,175 +1,137 @@ \section{Symmetrie} Das Wort Symmetrie ist sehr alt und hat sich seltsamerweise von seinem -ursprünglichen griechischen Wort -\(\mathrm{\Sigma\nu\mu\mu\varepsilon\tau\rho\iota\alpha}\) -\footnote{\emph{Symmetr\'ia}: ein gemeinsames Mass habend, gleichmässig, -verhältnismässig} fast nicht verändert. In der Alltagssprache mag es ein -locker definierter Begriff sein, aber in der Mathematik hat Symmetrie eine sehr -präzise Bedeutung. +ursprünglichen griechischen Wort \(\mathrm{\Sigma\upsilon\mu\mu\varepsilon\tau\rho\iota\alpha}\)\footnote{\emph{Symmetr\'ia}: ein gemeinsames Mass habend, gleichmässig,verhältnismässig} fast nicht verändert. +In der Alltagssprache mag es ein locker definierter Begriff sein, in der Mathematik hat Symmetrie jedoch eine sehr präzise Bedeutung. \begin{definition}[Symmetrie] - Ein mathematisches Objekt wird als symmetrisch bezeichnet, wenn es unter einer - bestimmten Operation invariant ist. + Ein mathematisches Objekt wird als symmetrisch bezeichnet, wenn es unter einer bestimmten Operation invariant ist. \end{definition} -Die intuitivsten Beispiele kommen aus der Geometrie, daher werden wir mit -einigen geometrischen Beispielen beginnen. Wie wir jedoch später sehen werden, -ist das Konzept der Symmetrie eigentlich viel allgemeiner. +Die intuitivsten Beispiele kommen aus der Geometrie, daher werden wir mit einigen geometrischen Beispielen beginnen. +Wie wir jedoch später sehen werden, ist das Konzept der Symmetrie eigentlich viel allgemeiner. \begin{figure} - \centering - \includegraphics{papers/punktgruppen/figures/symmetric-shapes} - \caption{ - Beispiele für geometrisch symmetrische Formen. - \label{fig:punktgruppen:geometry-example} - } + \centering + \includegraphics{papers/punktgruppen/figures/symmetric-shapes} + \caption{ + Beispiele für geometrisch symmetrische Formen. + \label{fig:punktgruppen:geometry-example} + } \end{figure} \subsection{Geometrische Symmetrien} -In Abbildung \ref{fig:punktgruppen:geometry-example} haben wir einige Formen, -die offensichtlich symmetrisch sind. Zum Beispiel hat das Quadrat eine Gerade, an -deren es gespiegelt werden kann, ohne sein Aussehen zu verändern. Regelmässige -Polygone mit \(n\) Seiten sind auch gute Beispiele, um eine diskrete -Rotationssymmetrie zu veranschaulichen, was bedeutet, dass eine Drehung um -einen Punkt um einen bestimmten Winkel \(360^\circ/n\) die Figur unverändert -lässt. Das letzte Beispiel auf der rechten Seite ist eine unendliche -Rotationssymmetrie. Sie wird so genannt, weil es unendlich viele Werte für -\(\alpha \in \mathbb{R}\) gibt, die die Form unverändert lassen. Dies ist -hoffentlich ausreichend, um die Bedeutung hinter der Notation zu verstehen, die -nun eingeführt wird. - -% Vieleicht eine kurze Einführung in für die Definition, ich habe das gefühl, dass in der Definition die Symmetrie-Operation und die Gruppe auf einmal erklährt wird -\subsubsection{Symetriegruppe} -\texttt{TODO: review this paragraph, explain what is \(\mathds{1}\).} +In Abbildung \ref{fig:punktgruppen:geometry-example} haben wir einige Formen, die offensichtlich symmetrisch sind. +Zum Beispiel hat das Quadrat eine Gerade, an der es gespiegelt werden kann, ohne sein Aussehen zu verändern. +Regelmässige Polygone mit \(n\) Seiten sind auch gute Beispiele, um eine diskrete Rotationssymmetrie zu veranschaulichen, was bedeutet, dass eine Drehung um einen Punkt um einen bestimmten Winkel \(360^\circ/n\) die Figur unverändert lässt. +Das letzte Beispiel auf der rechts ist eine unendliche Rotationssymmetrie. Sie wird so genannt, weil es unendlich viele Werte für den Drehwinkel \(\alpha \in \mathbb{R}\) gibt, die die Form unverändert lassen. Ein Objekt kann mehr als nur eine Symmetrie aufweisen. -Als Beispiel, kann das Quadrat in Abbildung \ref{fig:punktgruppen:geometry-example} -nicht nur um $\sigma$ sondern auch Diagonal gespiegelt werden oder um $90^\circ$ gedreht werden. -Fässt man die möglichen Symmetrien zusammen, entsteht eine Symmetriegruppe. +Zum Beispiel kann das Quadrat in Abbildung \ref{fig:punktgruppen:geometry-example} nicht nur um \(\sigma\) sondern auch diagonal gespiegelt werden oder um \(90^\circ\) gedreht werden. +Fasst man die möglichen Symmetrien zusammen, entsteht eine Symmetriegruppe. \begin{definition}[Symmetriegruppe] - Sei \(g\) eine Operation, die ein mathematisches Objekt unverändert lässt. - Bei einer anderen Operation \(h\) definieren wir die Komposition \(h\circ g\) - als die Anwendung der Operationen nacheinander. Alle Operationen bilden unter - Komposition eine Gruppe, die Symmetriegruppe genannt wird. -\end{definition} % ich lese diese Definition ein wenig holprig, vieleicht können wir sie zusammen anschauen - -% Nach meinem Geschmack könne es hier auch eine einleitung wie mein Beispiel geben dammit man den Text flüssiger lesen kann -\begin{definition}[Zyklische Untergruppe, Erzeuger] - Sei \(g\) ein Element einer Symmetriegruppe \(G\). Alle möglichen - Kompositionen von \(g\) und \(g^{-1}\) bilden eine sogenannte zyklische - Untergruppe von \(G\), und \(g\) wird ihr Erzeuger genannt. Die erzeugte - Untergruppe \(\langle g \rangle\) wird mit spitzen Klammern um den Erzeuger - bezeichnet. + Seien \(g\) und \(h\) umkehrbare Operationen, sogenannte Symmetrieoperationen, die ein mathematisches Objekt unverändert lassen. + Die Komposition \(h\circ g\) definieren wir als die Anwendung der Operationen nacheinander. + Alle möglichen Symmetrieoperationen bilden unter Komposition eine Gruppe, die Symmetriegruppe genannt wird. \end{definition} -Mit dem oben Gesagten können wir das \(n\)-Gon Beispiel formalisieren. -Bezeichnen wir mit \(r\) eine Drehung im Gegenuhrzeigersinn von \(360^\circ/n\) -um einen Punkt. Diese Definition reicht aus, um die gesamte Symmetriegruppe -\[ - C_n = \langle r \rangle - = \left\{\mathds{1}, r, r^2, \ldots, r^{n-1}\right\} -\] -der Drehungen eines \(n\)-Gons zu definieren. Das liegt daran, -dass wir durch die mehrfache Verwendung von \(r\) jeden Winkel erzeugen, der -die Rotationssymmetrie bewahrt. Hier die Potenzen von \(r\) sind als -wiederholte Komposition gemeint, dass heisst \(r^n = r\circ r \circ \cdots -r\circ r\). Wenn wir diese Idee nun erweitern, können wir mit einem -Erzeugendensystemen komplexere Strukturen aufbauen. +Eine Gruppe benötigt ausserdem auch zwingend ein neutrales Element, welches wir mit \(\mathds{1}\) bezeichnen. +Die Anwendung der neutralen Operation ist gleichbedeutend damit, alles unverändert zu lassen. +Weiterhin muss in einer Gruppe für jede Operation \(g\) auch eine inverse Operation \(g^{-1}\) vorkommen, die rückgängig macht, was \(g\) getan hat. +Somit ist \(\mathds{1}\) auch äquivalent dazu, eine Operation und dann ihre Inverse anzuwenden. + Die Definition der Symmetriegruppe ist mit der Kompositionsoperation gegeben, sie wird aber auch oft als Multiplikation geschrieben. +Das liegt daran, dass in manchen Fällen die Zusammensetzung algebraisch durch eine Multiplikation berechnet wird. +Die Verwendung einer multiplikativen Schreibweise ermöglicht es, einige Ausdrücke kompakter zu schreiben, z.B. +durch Verwendung von Potenzen \(r^n = r\circ r \circ \cdots r\circ r\) für eine wiederholte Komposition. -\begin{definition}[Erzeugendensysteme] - % please fix this unreadable mess - Jede Gruppe kann durch eines oder mehrere ihrer Elemente generiert werden. - Wir lassen \(g_1, g_2, \ldots, g_n\) erzeugenden Elemente einer - Symmetriegruppe sein. Da es mehrere Erzeuger gibt, müssen auch die - sogenannte Definitionsgleichungen gegeben werden, die die - Multiplikationstabelle vollständig definieren. Die Gleichungen sind ebenfalls - in den Klammern angegeben. Die erzeugende Elementen zusammen mit der - Definitionsgleichungen bauen ein Erzeugendensysteme. +\begin{definition}[Zyklische Untergruppe, Erzeuger] + Sei \(g\) ein Element einer Symmetriegruppe \(G\). + Alle möglichen Kompositionen von \(g\) und \(g^{-1}\) bilden eine sogenannte zyklische Untergruppe von \(G\), wobei \(g\) Erzeuger der Untergruppe genannt wird. + Die von \(g\) erzeugte Untergruppe \(\langle g \rangle = \{ g^k : k \in \mathbb{Z} \}\) wird mit spitzen Klammern bezeichnet. \end{definition} +\begin{beispiel} + Um die Syntax zu verstehen, betrachten wir eine durch \(a\) erzeugte Gruppe \(G = \langle a \rangle\). + Das bedeutet, dass \(G\) die Elemente \(a, aa, aaa, \ldots\) sowie \(a^{-1}, a^{-1}a^{-1}, \ldots\) und ein neutrales Element \(\mathds{1} = aa^{-1}\) enthält. +\end{beispiel} +\begin{beispiel} + Als anschaulicheres Beispiel können wir eine zyklische Untergruppe des \(n\)-Gon formalisieren. + Wir bezeichnen mit \(r\) eine Drehung im Gegenuhrzeigersinn von \(360^\circ/n\) um einen Punkt. + Diese Definition reicht aus, um die gesamte Symmetriegruppe + \[ + C_n = \langle r \rangle + = \{\mathds{1}, r, r^2, \ldots, r^{n-1}\} + \] + der Drehungen eines \(n\)-Gons zu erzeugen. + Das liegt daran, dass wir durch die mehrfache Verwendung von \(r\) jeden Winkel erzeugen k\"onnen, der die Rotationssymmetrie bewahrt. + In ähnlicher Weise, aber weniger interessant, enthält die Reflexionssymmetriegruppe \(\langle\sigma\rangle\) nur \(\left\{\mathds{1}, \sigma\right\}\), weil \(\sigma^2 = \mathds{1}\). +\end{beispiel} -\texttt{TODO: should put examples for generators?} \\ +Wenn wir diese Idee nun erweitern, können wir mit einem Erzeugendensystem +komplexere Strukturen aufbauen. -Die Reflexionssymmetriegruppe ist nicht so interessant, da sie nur -\(\left\{\mathds{1}, \sigma\right\}\) enthält. Kombiniert man sie jedoch mit -der Rotation, erhält man die so genannte Diedergruppe -\[ - D_n = \langle r, \sigma : r^{n-1} = \sigma^2 = (\sigma r)^2 = \mathds{1} \rangle - = \left\{ - \mathds{1}, r, \ldots, r^{n-1}, \sigma, \sigma r, \ldots, \sigma r^{n-1} - \right\}. -\] +%@Naoki Are you ok with my grammar fixes I'm not 101% shore how to use the word Erzeugendensystem? +\begin{definition}[Erzeugendensystem] + Jede diskrete Gruppe kann durch eines oder mehrere ihrer Elemente generiert werden. + Wir lassen \(g_1, g_2, g_3, \ldots\) erzeugenden Elemente einer Symmetriegruppe sein. + Da es mehrere Erzeuger gibt, müssen auch die sogenannten Definitionsgleichungen gegeben werden, die die Multiplikationstabelle vollständig definieren. + Die Gleichungen sind ebenfalls in den Klammern angegeben. + Die erzeugenden Elementen bauen zusammen mit den Definitionsgleichungen ein Erzeugendensystem. +\end{definition} +\begin{beispiel} + Wir werden nun alle Symmetrien eines \(n\)-Gons beschreiben, was bedeutet, dass wir die Operationen \(r\) und \(\sigma\) kombinieren. + Die Definitionsgleichungen sind \(r^n = \mathds{1}\), \(\sigma^2 = \mathds{1}\) und \((\sigma r)^2 = \mathds{1}\). + Die ersten beiden sind ziemlich offensichtlich. + Die letzte wird oft auch als Inversion bezeichnet, weil die Anwendung von \(\sigma r\) dasselbe ist wie das Ziehen einer Linie von einem Punkt, die durch den Ursprung geht, und das Verschieben des Punktes auf die andere Seite des Nullpunkts. + Wenn man dies zweimal macht, geht man zurück zum Anfangspunkt. + Daraus ergibt sich die so genannte Diedergruppe + \begin{align*} + D_n &= \langle r, \sigma : r^n = \sigma^2 = (\sigma r)^2 = \mathds{1} \rangle \\ + &= \{ + \mathds{1}, r, \ldots, r^{n-1}, \sigma, \sigma r, \ldots, \sigma r^{n-1} + \}. \qedhere + \end{align*} +\end{beispiel} -Die Symmetrieoperationen, die wir bis jetzt besprochen haben, haben immer -mindestens einen Punkt gehabt, der wieder auf sich selbst abgebildet wird. Im -Fall der Rotation war es der Drehpunkt, bei der Spiegelung die Punkte der -Spiegelachse. Dies ist jedoch keine Voraussetzung für eine Symmetrie, da es -Symmetrien gibt, die jeden Punkt zu einem anderen Punkt verschieben können. -Diesen Spezialfall, bei dem mindestens ein Punkt unverändert bleibt, nennt man -Punktsymmetrie. +Die Symmetrieoperationen, die wir bis jetzt besprochen haben, haben immer mindestens einen Punkt gehabt, der wieder auf sich selbst abgebildet wird. +Im Fall der Rotation war es der Drehpunkt, bei der Spiegelung die Punkte der Spiegelachse. +Dies ist jedoch keine Voraussetzung für eine Symmetrie, da es Symmetrien gibt, die jeden Punkt zu einem anderen Punkt verschieben können. + Diesen Spezialfall, bei dem immer mindestens ein Punkt unverändert bleibt, nennt man Punktsymmetrie. \begin{definition}[Punktgruppe] - Wenn jede Operation in einer Symmetriegruppe die Eigenschaft hat, mindestens - einen Punkt unverändert zu lassen, sagt man, dass die Symmetriegruppe eine - Punktgruppe ist. + Wenn es einen Punkt gibt, der von jeder Gruppenoperation unverändert gelassen wird, ist die Symmetriegruppe eine Punktgruppe. \end{definition} \subsection{Algebraische Symmetrien} -Wir haben nun unseren Operationen Symbole gegeben, mit denen es tatsächlich -möglich ist, Gleichungen zu schreiben. Die naheliegende Frage ist dann, könnte -es sein, dass wir bereits etwas haben, das dasselbe tut? Natürlich, ja. +Wir haben nun unseren Operationen Symbole gegeben, mit denen es tatsächlich möglich ist, Gleichungen zu schreiben. +Die anschliessende Frage ist dann, ob wir bereits mathematische Objekte haben, mit denen wir Gleichungen schreiben, die sich auf die gleiche Weise verhalten. +Die Antwort lautet natürlich ja. Um es formaler zu beschreiben, werden wir einige Begriffe einführen. \begin{definition}[Gruppenhomomorphismus] - Seien \(G\) und \(H\) Gruppe mit unterschiedlicher Operation \(\diamond\) - bzw. \(\star\). Ein Homomorphismus\footnote{ Für eine ausführlichere - Diskussion siehe \S\ref{buch:grundlagen:subsection:gruppen} im Buch.} ist - eine Funktion \(f: G \to H\), so dass für jedes \(a, b \in G\) gilt - \(f(a\diamond b) = f(a) \star f(b)\). Man sagt, dass der Homomorphismus - \(f\) \(G\) in \(H\) transformiert. + \(G\) und \(H\) seien Gruppen mit unterschiedlichen Operationen \(\diamond\) bzw. + \(\star\). + Ein Homomorphismus\footnote{ Für eine ausführlichere Diskussion siehe \S\ref{buch:grundlagen:subsection:gruppen} im Buch.} ist eine Funktion \(f: G \to H\), so dass für jedes \(a, b \in G\) gilt \(f(a\diamond b) = f(a) \star f(b)\). + Man sagt, dass der Homomorphismus \(f\) \(G\) in \(H\) transformiert. \end{definition} \begin{beispiel} - Die Rotationssymmetrie des Kreises \(C_\infty\), mit einem unendlichen - Kontinuum von Werten \(\alpha \in \mathbb{R}\), entspricht perfekt dem - komplexen Einheitskreis. Der Homomorphismus \(\phi: C_\infty \to \mathbb{C}\) - ist durch die Eulersche Formel \(\phi(r) = e^{i\alpha}\) gegeben. + Die Rotationssymmetrie des Kreises \(C_\infty\), mit einem unendlichen Kontinuum von Werten \(\alpha \in \mathbb{R}\), entspricht genau dem komplexen Einheitskreis. + Der Homomorphismus \(\phi: C_\infty \to \mathbb{C}\) ist durch die Eulersche Formel \(\phi(r) = e^{i\alpha}\) gegeben. \end{beispiel} \begin{definition}[Darstellung einer Gruppe] - Die Darstellung einer Gruppe ist ein Homomorphismus, der eine Symmetriegruppe - auf eine Menge von Matrizen abbildet. - \[ - \Phi: G \to \operatorname{GL}_n(\mathbb{R}). - \] - Äquivalent kann man sagen, dass ein Element aus der Symmetriegruppe auf einen - Vektorraum \(V\) wirkt, indem man definiert \(\Phi : G \times V \to V\). + Die Darstellung einer Gruppe ist ein Homomorphismus + \[ + \Phi: G \to \operatorname{GL}_n(\mathbb{R}), + \] + der eine Symmetriegruppe auf eine Menge von Matrizen abbildet. + Äquivalent kann man sagen, dass ein Element aus der Symmetriegruppe auf einen Vektorraum \(V\) wirkt, indem man \(\Phi : G \times V \to V\) definiert. \end{definition} \begin{beispiel} - Die Elemente \(r^k \in C_n\), wobei \(0 < k < n\), stellen abstrakt eine - Drehung von \(2\pi k/n\) um den Ursprung dar. Die mit der Matrix - \[ - \Phi(r^k) = \begin{pmatrix} - \cos(2\pi k/n) & -\sin(2\pi k/n) \\ - \sin(2\pi k/n) & \cos(2\pi k/n) - \end{pmatrix} - \] - definierte Funktion von \(C_n\) nach \(O(2)\) ist eine Darstellung von - \(C_n\). In diesem Fall ist die erste Gruppenoperation die Komposition und - die zweite die Matrixmultiplikation. Man kann überprüfen, dass \(\Phi(r^2 - \circ r) = \Phi(r^2)\Phi(r)\). + Die Elemente \(r^k \in C_n\), wobei \(0 < k < n\), stellen abstrakt eine Drehung von \(2\pi k/n\) um den Ursprung dar. + Die mit der Matrix + \[ + \Phi(r^k) = \begin{pmatrix} + \cos(2\pi k/n) & -\sin(2\pi k/n) \\ + \sin(2\pi k/n) & \cos(2\pi k/n) + \end{pmatrix} + \] + definierte Funktion von \(C_n\) nach \(O(2)\) ist eine Darstellung von \(C_n\). + In diesem Fall ist die erste Gruppenoperation die Komposition und die zweite die Matrixmultiplikation. + Man kann überprüfen, dass \(\Phi(r^2 \circ r) = \Phi(r^2)\Phi(r)\). \end{beispiel} - -\texttt{TODO: rewrite section on translational symmetry.} -%% TODO: title / fix continuity -% Um das Konzept zu illustrieren, werden wir den umgekehrten Fall diskutieren: -% eine Symmetrie, die keine Punktsymmetrie ist, die aber in der Physik sehr -% nützlich ist, nämlich die Translationssymmetrie. Von einem mathematischen -% Objekt \(U\) wird gesagt, dass es eine Translationssymmetrie \(Q(x) = x + a\) -% hat, wenn es die Gleichung -% \[ -% U(x) = U(Q(x)) = U(x + a), -% \] -% für ein gewisses \(a\), erfüllt. Zum Beispiel besagt das erste Newtonsche -% Gesetz, dass ein Objekt, auf das keine Kraft einwirkt, eine -% zeitranslationsinvariante Geschwindigkeit hat, d.h. wenn \(\vec{F} = \vec{0}\) -% dann \(\vec{v}(t) = \vec{v}(t + \tau)\). - -% \subsection{Sch\"onflies notation} - -% vim:ts=2 sw=2 spell spelllang=de: diff --git a/buch/papers/punktgruppen/tikz/atoms-grid-force.tex b/buch/papers/punktgruppen/tikz/atoms-grid-force.tex new file mode 100644 index 0000000..05742cf --- /dev/null +++ b/buch/papers/punktgruppen/tikz/atoms-grid-force.tex @@ -0,0 +1,42 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + \begin{tikzpicture}[ + >=latex, + node distance = 2mm, + charge/.style = { + circle, draw = black, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \matrix[nodes = { charge }, row sep = 5mm, column sep = 1cm] { + \node[positive] (NW) {}; & \node[negative] (N) {}; & \node [positive] (NE) {}; \\ + \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\ + \node[positive] (SW) {}; & \node[negative] (S) {}; & \node [positive] (SE) {}; \\ + }; + + \foreach \d in {NW, N, NE} { + \draw[orange, very thick, <-] (\d) to ++(0,.7); + } + + \foreach \d in {SW, S, SE} { + \draw[orange, very thick, <-] (\d) to ++(0,-.7); + } + + \draw[gray, dashed] (W) to (N) to (E) to (S) to (W); + \end{tikzpicture} +\end{document} diff --git a/buch/papers/punktgruppen/tikz/atoms-grid-still.tex b/buch/papers/punktgruppen/tikz/atoms-grid-still.tex new file mode 100644 index 0000000..4e43856 --- /dev/null +++ b/buch/papers/punktgruppen/tikz/atoms-grid-still.tex @@ -0,0 +1,33 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + \begin{tikzpicture}[ + >=latex, + node distance = 2mm, + charge/.style = { + circle, draw = black, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \matrix[nodes = { charge }, row sep = 8mm, column sep = 8mm] { + \node[positive] {}; & \node[negative] (N) {}; & \node [positive] {}; \\ + \node[negative] (W) {}; & \node[positive] {}; & \node [negative] (E) {}; \\ + \node[positive] {}; & \node[negative] (S) {}; & \node [positive] {}; \\ + }; + \draw[gray, dashed] (W) to (N) to (E) to (S) to (W); + \end{tikzpicture} +\end{document} diff --git a/buch/papers/punktgruppen/tikz/atoms-piezo-force-horizontal.tex b/buch/papers/punktgruppen/tikz/atoms-piezo-force-horizontal.tex new file mode 100644 index 0000000..e4c3f93 --- /dev/null +++ b/buch/papers/punktgruppen/tikz/atoms-piezo-force-horizontal.tex @@ -0,0 +1,47 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + \begin{tikzpicture}[ + >=latex, + node distance = 2mm, + charge/.style = { + circle, draw = black, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \node[charge, positive, yshift= 2.5mm] (C1) at ( 60:1.5cm) {}; + \node[charge, negative, yshift= 2.5mm] (C2) at (120:1.5cm) {}; + \node[charge, positive, xshift= 2.5mm] (C3) at (180:1.5cm) {}; + \node[charge, negative, yshift=-2.5mm] (C4) at (240:1.5cm) {}; + \node[charge, positive, yshift=-2.5mm] (C5) at (300:1.5cm) {}; + \node[charge, negative, xshift=-2.5mm] (C6) at (360:1.5cm) {}; + + \draw[black] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + + \draw[orange, very thick, <-] (C6) to ++(.7,0); + \draw[orange, very thick, <-] (C3) to ++(-.7,0); + + \node[black] (E) {\(\vec{E}_p\)}; + \begin{scope}[node distance = .5mm] + \node[blue!50, right = of E] {\(-\)}; + \node[red!50, left = of E] {\(+\)}; + \end{scope} + % \draw[gray, thick, dotted] (E) to ++(0,2); + % \draw[gray, thick, dotted] (E) to ++(0,-2); + \end{tikzpicture} +\end{document} diff --git a/buch/papers/punktgruppen/tikz/atoms-piezo-force-vertical.tex b/buch/papers/punktgruppen/tikz/atoms-piezo-force-vertical.tex new file mode 100644 index 0000000..892ab42 --- /dev/null +++ b/buch/papers/punktgruppen/tikz/atoms-piezo-force-vertical.tex @@ -0,0 +1,52 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + \begin{tikzpicture}[ + >=latex, + node distance = 2mm, + charge/.style = { + circle, draw = black, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \node[charge, positive, yshift=-2.5mm] (C1) at ( 60:1.5cm) {}; + \node[charge, negative, yshift=-2.5mm] (C2) at (120:1.5cm) {}; + \node[charge, positive, xshift=-2.5mm] (C3) at (180:1.5cm) {}; + \node[charge, negative, yshift= 2.5mm] (C4) at (240:1.5cm) {}; + \node[charge, positive, yshift= 2.5mm] (C5) at (300:1.5cm) {}; + \node[charge, negative, xshift= 2.5mm] (C6) at (360:1.5cm) {}; + + \draw[black] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + + \foreach \d in {C1, C2} { + \draw[orange, very thick, <-] (\d) to ++(0,.7); + } + + \foreach \d in {C4, C5} { + \draw[orange, very thick, <-] (\d) to ++(0,-.7); + } + + \node[black] (E) {\(\vec{E}_p\)}; + \begin{scope}[node distance = .5mm] + \node[red!50, right = of E] {\(+\)}; + \node[blue!50, left = of E] {\(-\)}; + \end{scope} + % \draw[gray, thick, dotted] (E) to ++(0,2); + % \draw[gray, thick, dotted] (E) to ++(0,-2); + \end{tikzpicture} +\end{document} diff --git a/buch/papers/punktgruppen/tikz/atoms-piezo-still.tex b/buch/papers/punktgruppen/tikz/atoms-piezo-still.tex new file mode 100644 index 0000000..2eb78ba --- /dev/null +++ b/buch/papers/punktgruppen/tikz/atoms-piezo-still.tex @@ -0,0 +1,34 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + \begin{tikzpicture}[ + >=latex, + node distance = 2mm, + charge/.style = { + circle, draw = black, thick, + minimum size = 5mm + }, + positive/.style = { fill = red!50 }, + negative/.style = { fill = blue!50 }, + ] + + \foreach \x/\t [count=\i] in {60/positive, 120/negative, 180/positive, 240/negative, 300/positive, 360/negative} { + \node[charge, \t] (C\i) at (\x:1.5cm) {}; + } + + \draw[black] (C1) to (C2) to (C3) to (C4) to (C5) to (C6) to (C1); + \node[circle, draw=gray, fill=gray, outer sep = 0, inner sep = 0, minimum size = 3mm] {}; + % \draw[gray, dashed] (C2) to (C4) to (C6) to (C2); + \end{tikzpicture} +\end{document} diff --git a/buch/papers/punktgruppen/tikz/combine-symmetries.tex b/buch/papers/punktgruppen/tikz/combine-symmetries.tex index 84e0a76..fa669ae 100644 --- a/buch/papers/punktgruppen/tikz/combine-symmetries.tex +++ b/buch/papers/punktgruppen/tikz/combine-symmetries.tex @@ -13,6 +13,7 @@ \begin{document} \begin{tikzpicture}[ + >=latex, dot/.style = { draw, circle, thick, black, fill = gray!40!white, minimum size = 2mm, @@ -45,7 +46,7 @@ (A2) ++(-.5,0) arc (180:60:.5); \draw[red!80!black, dashed, thick, ->] (A2) to (B2); - \draw[yellow!50!orange, thick, ->] + \draw[cyan!40!blue, thick, ->] (B1) to node[above, midway] {\(\vec{Q}'\)} (B2); \draw[gray, dashed, thick] (A1) to (A1 |- B1) node (Xl) {}; diff --git a/buch/papers/punktgruppen/tikz/lattice.tex b/buch/papers/punktgruppen/tikz/lattice.tex index 9c05af3..a6b1876 100644 --- a/buch/papers/punktgruppen/tikz/lattice.tex +++ b/buch/papers/punktgruppen/tikz/lattice.tex @@ -13,23 +13,24 @@ \begin{document} \begin{tikzpicture}[ - dot/.style = { - draw, circle, thick, black, fill = gray!40!white, - minimum size = 2mm, - inner sep = 0pt, - outer sep = 1mm, - }, + >=latex, + dot/.style = { + draw, circle, thick, black, fill = gray!40!white, + minimum size = 2mm, + inner sep = 0pt, + outer sep = 1mm, + }, ] \begin{scope} - \clip (-2,-2) rectangle (3,4); + \clip (-2,-2) rectangle (7,2); \foreach \y in {-7,-6,...,7} { \foreach \x in {-7,-6,...,7} { \node[dot, xshift=3mm*\y] (N\x\y) at (\x, \y) {}; } } \end{scope} - \draw[black, thick] (-2, -2) rectangle (3,4); + \draw[black, thick] (-2, -2) rectangle (7,2); \draw[red!80!black, thick, ->] (N00) to node[midway, below] {\(\vec{a}_1\)} (N10); diff --git a/buch/papers/punktgruppen/tikz/piezo-atoms.tex b/buch/papers/punktgruppen/tikz/piezo-atoms.tex index 82a2710..1811392 100644 --- a/buch/papers/punktgruppen/tikz/piezo-atoms.tex +++ b/buch/papers/punktgruppen/tikz/piezo-atoms.tex @@ -13,6 +13,7 @@ \begin{document} \begin{tikzpicture}[ + >=latex, node distance = 2mm, charge/.style = { circle, draw = black, thick, diff --git a/buch/papers/punktgruppen/tikz/piezo.tex b/buch/papers/punktgruppen/tikz/piezo.tex index 1d16ab7..6542f26 100644 --- a/buch/papers/punktgruppen/tikz/piezo.tex +++ b/buch/papers/punktgruppen/tikz/piezo.tex @@ -12,12 +12,14 @@ \usetikzlibrary{calc} \begin{document} -\begin{tikzpicture} +\begin{tikzpicture}[ + >=latex, + ] \begin{scope}[ node distance = 0cm ] \node[ - rectangle, fill = gray!60!white, + rectangle, fill = gray!20!white, minimum width = 3cm, minimum height = 2cm, ] (body) {\(\vec{E}_p = \vec{0}\)}; @@ -43,9 +45,9 @@ xshift = 7cm ] \node[ - rectangle, fill = gray!40!white, + rectangle, fill = gray!20!white, minimum width = 3cm, minimum height = 1.5cm, - ] (body) {\(\vec{E}_p = \vec{0}\)}; + ] (body) {\(\vec{E}_p \neq \vec{0}\)}; \node[ draw, rectangle, thick, black, fill = red!50, diff --git a/buch/papers/punktgruppen/tikz/projections.tex b/buch/papers/punktgruppen/tikz/projections.tex index a763e77..e8a4a2e 100644 --- a/buch/papers/punktgruppen/tikz/projections.tex +++ b/buch/papers/punktgruppen/tikz/projections.tex @@ -13,6 +13,7 @@ \begin{document} \begin{tikzpicture}[ + >=latex, classcirc/.style = { draw = gray, thick, circle, minimum size = 12mm, @@ -43,7 +44,7 @@ \node[classcirc] (C2h) {} node[classlabel] {\(C_{2h}\)}; & \node[classcirc] (D2) {} node[classlabel] {\(D_{2}\)}; \\ - \node[classcirc] (D3d) {} node[classlabel] {\(D_{3d}\)}; & + \node[classcirc] (D3d) {} node[classlabel] {\(C_{3v}\)}; & \node[classcirc] (C2v) {} node[classlabel] {\(C_{2v}\)}; & \node[classcirc] (D2h) {} node[classlabel] {\(D_{2h}\)}; & \node[classcirc] (D3) {} node[classlabel] {\(D_{3}\)}; & diff --git a/buch/papers/punktgruppen/tikz/stereographic-projections.tex b/buch/papers/punktgruppen/tikz/stereographic-projections.tex new file mode 100644 index 0000000..7d612fb --- /dev/null +++ b/buch/papers/punktgruppen/tikz/stereographic-projections.tex @@ -0,0 +1,108 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{tikz-3dplot} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + +\tdplotsetmaincoords{60}{130} +\pgfmathsetmacro{\l}{2} + +\begin{tikzpicture}[ + >=latex, + tdplot_main_coords, + dot/.style = { + black, fill = black, circle, + outer sep = 0, inner sep = 0, + minimum size = 1mm + }, + round/.style = { + draw = orange, thick, circle, + minimum size = 1mm, + inner sep = 0pt, outer sep = 0pt, + }, + cross/.style = { + cross out, draw = magenta, thick, + minimum size = 1mm, + inner sep = 0pt, outer sep = 0pt + }, + ] + + % origin + \coordinate (O) at (0,0,0); + + % poles + \coordinate (NP) at (0,0,\l); + \coordinate (SP) at (0,0,-\l); + + % axis + % \draw[->] (O) -- ++(1.5*\l,0,0); + % \draw[->] (O) -- ++(0,1.5*\l,0); + % \draw[->] (O) -- ++(0,0,1.5*\l); + + % gray unit circle + \tdplotdrawarc[gray, thick]{(O)}{\l}{0}{360}{}{}; + \draw[gray, dotted] (-\l, 0, 0) to (\l, 0, 0); + \draw[gray, dotted] (0, -\l, 0) to (0, \l, 0); + + % meridians + \foreach \phi in {0, 30, 60, ..., 150}{ + \tdplotsetrotatedcoords{\phi}{90}{0}; + \tdplotdrawarc[lightgray, dashed, tdplot_rotated_coords]{(O)}{\l}{0}{360}{}{}; + } + + % dot above and its projection + \pgfmathsetmacro{\phi}{120} + \pgfmathsetmacro{\theta}{60} + + \pgfmathsetmacro{\px}{cos(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\py}{sin(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\pz}{cos(\theta)*\l}) + + \coordinate (A) at (\px,\py,\pz); + \coordinate (Aproj) at ({\px * \l / (\l + \pz)}, {\py * \l / (\l + \pz)}, 0); + + % dot below and its projection + \pgfmathsetmacro{\phi}{-60} + \pgfmathsetmacro{\theta}{120} + + \pgfmathsetmacro{\px}{cos(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\py}{sin(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\pz}{cos(\theta)*\l}) + + \coordinate (B) at (\px,\py,\pz); + \coordinate (Bproj) at ({\px * \l / (\l - \pz)}, {\py * \l / (\l - \pz)}, 0); + + % projection lines + \draw[gray] (A) to (SP); + \draw[gray] (SP) to (O) to (Aproj); + + \draw[gray] (B) to (NP); + \draw[gray] (NP) to (O) to (Bproj); + + % dots + \draw (O) node[dot] {}; + \draw (SP) node[dot] {}; + \draw (NP) node[dot] {}; + \draw (A) node[dot, fill = magenta, minimum size = 1.5mm] {}; + \draw (B) node[dot, fill = orange, minimum size = 1.5mm] {}; + + % projection markers + \draw[very thick, magenta] + (Aproj) ++(.15,0) to ($(Aproj)+(-.15, 0)$) + (Aproj) ++(0,.15) to ($(Aproj) +(0, -.15)$); + + \tdplotdrawarc[orange, very thick]{(Bproj)}{.1}{0}{360}{}{}; + +\end{tikzpicture} +\end{document} +% vim:ts=2 sw=2 et: diff --git a/buch/papers/punktgruppen/tikz/symmetric-shapes.tex b/buch/papers/punktgruppen/tikz/symmetric-shapes.tex index b2c051f..688fb61 100644 --- a/buch/papers/punktgruppen/tikz/symmetric-shapes.tex +++ b/buch/papers/punktgruppen/tikz/symmetric-shapes.tex @@ -14,6 +14,7 @@ \begin{document} \begin{tikzpicture}[ + >=latex, node distance = 2cm, shapetheme/.style = { very thick, draw = black, fill = magenta!20!white, diff --git a/buch/papers/reedsolomon/Makefile b/buch/papers/reedsolomon/Makefile index 9c96e88..4be963e 100644 --- a/buch/papers/reedsolomon/Makefile +++ b/buch/papers/reedsolomon/Makefile @@ -4,6 +4,52 @@ # (c) 2020 Prof Dr Andreas Mueller # -images: - @echo "no images to be created in reedsolomon" +SOURCES := \ + anwendungen.tex \ + codebsp.tex \ + decmitfehler.tex \ + decohnefehler.tex \ + dtf.tex \ + einleitung.tex \ + endlichekoerper.tex \ + hilfstabellen.tex \ + idee.tex \ + main.tex \ + packages.tex \ + rekonstruktion.tex \ + restetabelle1.tex \ + restetabelle2.tex \ + standalone.tex \ + zusammenfassung.tex + +TIKZFIGURES := \ + tikz/polynom2.tex \ + tikz/fourier.tex + +FIGURES := $(patsubst tikz/%.tex, figures/%.pdf, $(TIKZFIGURES)) + + +all: images standalone + + +.PHONY: images +images: $(FIGURES) + +figures/%.pdf: tikz/%.tex + mkdir -p figures + pdflatex --output-directory=figures $< + +.PHONY: standalone +standalone: standalone.tex $(SOURCES) $(FIGURES) + mkdir -p standalone + cd ../..; \ + pdflatex \ + --halt-on-error \ + --shell-escape \ + --output-directory=papers/reedsolomon/standalone \ + papers/reedsolomon/standalone.tex; + cd standalone; \ + bibtex standalone; \ + makeindex standalone; + diff --git a/buch/papers/reedsolomon/Makefile.inc b/buch/papers/reedsolomon/Makefile.inc index 6a676f8..ea51f7a 100644 --- a/buch/papers/reedsolomon/Makefile.inc +++ b/buch/papers/reedsolomon/Makefile.inc @@ -6,9 +6,17 @@ dependencies-reedsolomon = \ papers/reedsolomon/packages.tex \ papers/reedsolomon/main.tex \ - papers/reedsolomon/references.bib \ - papers/reedsolomon/teil0.tex \ - papers/reedsolomon/teil1.tex \ - papers/reedsolomon/teil2.tex \ - papers/reedsolomon/teil3.tex + papers/reedsolomon/einleitung.tex \ + papers/reedsolomon/idee.tex \ + papers/reedsolomon/dtf.tex \ + papers/reedsolomon/endlichekoerper.tex \ + papers/reedsolomon/codebsp.tex \ + papers/reedsolomon/decohnefehler.tex \ + papers/reedsolomon/decmitfehler.tex \ + papers/reedsolomon/rekonstruktion.tex \ + papers/reedsolomon/zusammenfassung.tex \ + papers/reedsolomon/anwendungen.tex \ + papers/reedsolomon/hilfstabellen.tex \ + papers/reedsolomon/references.bib + diff --git a/buch/papers/reedsolomon/RS presentation/images/polynom1 - Kopie.tex b/buch/papers/reedsolomon/RS presentation/images/polynom1 - Kopie.tex new file mode 100644 index 0000000..038e93e --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/images/polynom1 - Kopie.tex @@ -0,0 +1,33 @@ +% polynome1 +%------------------- +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\newcommand{\teiler}{40} +\begin{document} + + +\begin{tikzpicture}[>=latex,thick] + + \begin{axis}[ + axis lines = left, + xlabel = \(x\), + ylabel = {\(f(x)\)}, + ] + %Below the red parabola is defined + \addplot[ + color=blue, + ] + coordinates { + (0,23.1)(10,27.5)(20,32)(30,37.8)(40,44.6)(60,61.8)(80,83.8)(100,114) + }; + %Here the blue parabola is defined + + \end{axis} +\end{tikzpicture} +\end{document} + diff --git a/buch/papers/reedsolomon/anwendungen.tex b/buch/papers/reedsolomon/anwendungen.tex index 83e0f94..b9b1d69 100644 --- a/buch/papers/reedsolomon/anwendungen.tex +++ b/buch/papers/reedsolomon/anwendungen.tex @@ -6,14 +6,38 @@ \section{Anwendungen des Reed-Solomon-Codes \label{reedsolomon:section:anwendung}} \rhead{Anwendungen} -\textcolor{red}{Platzierung der Bilder? Quellenangabe der Bilder?} -In den vorherigen Abschnitten haben wir betrachtet, wie Reed-Solomon-Codes in der Theorie Funktionieren. +In den vorherigen Abschnitten haben wir betrachtet, wie Reed-Solomon-Codes in der Theorie funktionieren. In diesem Abschnitt werden wir einige Anwendungen vorstellen, bei denen ein Reed-Solomon-Code zum Einsatz kommt. + +Dabei teilen all diese Anwendungen das gleiche Problem: Die Daten können nur durch einen (höchst Wahrscheinlichen) fehlerbehafteten Kanal empfangen werden. Es gibt keine andere Methode, an diese Daten zu kommen, als über diesen Kanal. + +In der Netzwerktechnik zum Beispiel ist es üblich, dass bei Paketverluste oder beschädigt empfangene Datenpaketen diese einfach noch einmal innert wenigen Millisekunden angefordert werden können. +In der Raumfahrt ist dies nicht möglich, da aufgrund der beschränkten Speichermöglichkeit die gesammelten Daten so rasch wie möglich zur Erde gesendet werden. +Diese Daten wiederum brauchen aufgrund der grossen Distanz Stunden bis die Daten beim Empfänger ankommen. +Fehlerhafte Daten kann also auf Grund der Zeitverzögerung nicht mehr angefordert werden. + +Bei CDs oder DVDs gibt es zwar kein zeitliches Problem, jedoch erschweren Kratzer, Verschmutzungen oder Produktionsfehler das Lesen einer solchen Disk. +Da vor allem Produktionsfehler und Kratzer irreversibel sind und die Disk nicht nach jedem Kratzer ersetzt werden muss, so wird die korrekte Ausgabe der gespeicherten Information durch die Fehlerkorrektur sichergestellt. + +Einen ähnlichen Ansatz verfolgen QR-Codes, wobei die Information auch dann noch gelesen werden kann wenn der Code nicht mehr vollständig vorhanden ist. + +%Wie man sieht, eignen sich Reed-Solomon-Codes vor allem für Anwendungen, bei der die Informationen nicht auf einen Anderen Weg beschafft werden kann. +% +% +%, bei denen die Wahrscheinlichkeit hoch ist, dass während der Übertragung +% +%Es ist deshalb umso wichtiger die Daten Codiert zu lesen um so gleich die Lesefehler zu korrigieren. +% +% da aufgrund der grossen Distanz Stunden vergehen können bis gesendete Daten auf der Erde empfangen werden kann. +% + Obwohl alle diese Codes nach dem gleichen Prinzip arbeiten gibt es starke Unterschiede in deren Funktionsweise. Dies kommt vor allem daher, da die Codes nur Ressourcen zur Verfügung haben, die von der Hardware bereitstellt wird, auf denen die Codes implementiert wurden. Diese Codes bedienen sich daher verschiedener Tricks und Optimierungen um möglichst effizient zu arbeiten. -% + +Um die Fähigkeit eines verwendeten Reed-Solomon-Codes zu beschreiben verwendet man die Notation ($n$,$k$), wobei $n$ die Grösse des Nachrichtenblocks angibt und $k$ die Anzahl der Stellen, die für Nutzdaten gebraucht werden können. + %Dies kommt vor allem daher, da diese Codes an ihre Hardware gebunden sind, auf denen sie implementiert worden sind. %Deshalb wurden diese Codes stark optimiert damit sie möglichst Effizient arbeiten können. % @@ -45,8 +69,23 @@ Diese Codes bedienen sich daher verschiedener Tricks und Optimierungen um mögli %In den letzten abschnitten haben wir uns ausführlich die Funktionsweise des Reed-Solomon-Codes angeschaut. In diesem Abschnitt möchten wir dem Leser ein paar bekannte beispiele vorstellen, in denen Reed-Solomon-Codes zum einsatz kommen. Es sei jedoch angemerkt, dass diese Anwendungen in der Umsetzung oft ein wenig anderst funktionieren als hier vorgestellt. Dies wurde vor allem wegen technischen optimierungen realisiert. (technische tricks und finessen), von der logik jedoch sehr stark an unserem Beispiel orientieren \subsection{Raumfahrt} -Obwohl Reed-Solomon-Codes bereits in den 1960er entwickelt wurden fanden sie erstmals Anwendung in der Voyager Raumsonde der NASA. Die Daten der zwei im Jahre 1977 gestarteten Sonden werden mit einem RS(255,233)-Code \textcolor{red}{benötigt das weitere erklärungen, wie z.b. 255: grösse nachrichtenblock, 233: anzahl der nutzbaren daten ?} zusammen mit einem konventionellen Faltungscode übertragen. +Obwohl Reed-Solomon-Codes bereits in den 1960er entwickelt wurden fanden sie erstmals Anwendung in der Voyager Raumsonde der NASA. Die Daten der zwei im Jahre 1977 gestarteten Sonden (siehe Abbildung \ref{fig:voyager}) werden mit einem ($255$,$233$)-Code +Codiert. +Der Nachrichtenblock hat somit eine Länge von $255$ Zahlen, wovon $233$ als Nutzlast zur Verfügung stehen. +Damit ist es möglich bis zu $11$ Fehler im Nachrichtenblock zu korrigieren. +Der Codierte Nachrichtenblock wird in kleinere Blöcke aufgeteilt, mit einem Faltungscode erneut Codiert und anschliessend gesendet. +Ein Faltungscode ist wie ein Reed-Solomon-Code in der Lage Fehler zu korrigieren, +Codiert seine Information aber auf eine andere weise. Aus jedem unterteilten Block wird vor dem Versenden ein Paritätsbit erzeugt und dem Block angehängt. Anhand diesem Paritätsbit überprüft der Empfänger, ob bei der Übertragung der Block beschädigt wurde. Ist dies der Fall, wird der Block bei der Decodierung nicht beachtet. Diese so entstandenen ``Lücken'' im Datenstrom werden wiederum vom Reed-Solomon-Code korrigiert. Dieses Zusammenspiel beider Codes garantiert so eine hohe Robustheit gegenüber Übertragungsfeher. +% +% Funktioniert aber nach einem ganz anderen Prinzip. +% +%Durch diese doppelte Codierung wird eine äusserst hohe Übertragungssicherheit garantiert. +% +%Dabei steht die Zahl 255 für grösse des Nachrichtenblocks, der die Anzahl 233 +% +% +% \textcolor{red}{benötigt das weitere Erklärungen, wie z.b. 255: grösse Nachrichtenblock, 233: anzahl der nutzbaren daten ?} zusammen mit einem konventionellen Faltungscode übertragen. Eine von der Sonde gesendete Nachricht hat eine Blockgrösse von 255 Zeichen, wovon 233 für die Nutzdaten gebraucht werden können. Dieser Code ist somit in der Lage 11 Fehler in einem Nachrichtenblock zu korrigieren. % % Die zwei im Jahre 1977 gestarteten Sonden senden Daten mit der Hilfe eines RS(255,233)-Code für die digitalen Bilder sowie einem konventionellen Faltungscode. % @@ -56,14 +95,14 @@ Obwohl Reed-Solomon-Codes bereits in den 1960er entwickelt wurden fanden sie ers \begin{figure} \centering \includegraphics[width=0.5\textwidth]{papers/reedsolomon/images/Voyager_Sonde} - \caption{Voyager Raumsonde} + \caption{Mit einer Entfernung von über 22.8 Milliarden Kilometer ist die Voyager 1 Raumsonde das am weitesten entfernte, von Menschen erschaffene Objekt. Obwohl ihre Schwestersonde Voyager 2 zuerst ins All gestartet wurde befindet Sie sich ``nur'' 19 Milliarden Kilometer weit weg von der Erde. Aufgrund abnehmender Batterieleistung werden die beiden Sonden ihre wissenschaftlichen Aktivitäten etwa 2025 einstellen, bleiben aber bis in die 2030er mit uns in Kontakt.} \label{fig:voyager} \end{figure} \subsection{CD/DVD} Compact discs verwenden sogar zwei ineinander verschachtelte Reed-Solomon-Codes, einen (32,28)-Code und einen (28,24)-Code. -Beide Codes sind in der Lage, Fehler aus dem jeweils anderen gelesenen Block zu korrigieren. Dieses spezielle zusammenspielen dieser beiden Codes werden auch Cross-interleaved Reed-Solomon-Codes (CIRC) genannt. -Diese Vorgehensweise erzielt eine hohe Robustheit gegenüber Produktionsfehler oder Verschmutzung auf der Disc. Bei CD's sind diese in der Lage bis zu 4000 fehlerhafte Bits am Stück (ca. $2.5mm$) zu erkennen und zu korrigieren. +Beide Codes sind in der Lage, Fehler aus dem jeweils anderen gelesenen Block zu korrigieren. Dieses spezielle Zusammenspielen dieser beiden Codes werden auch Cross-interleaved Reed-Solomon-Codes (CIRC) genannt. +Diese Vorgehensweise erzielt eine hohe Robustheit gegenüber Produktionsfehlern oder Verschmutzung auf der Disc. Bei CDs sind diese in der Lage, bis zu 4000 fehlerhafte Bits am Stück (ca. $2.5mm$) zu erkennen und zu korrigieren. Die Digital Video Disc funktioniert nach dem selben Konzept mit grösseren Codeblöcken. Die DVD verwendet einen (208,192)-Code und einen (182,172)-Code. @@ -72,13 +111,25 @@ Die Digital Video Disc funktioniert nach dem selben Konzept mit grösseren Codeb \begin{figure} \centering - \includegraphics[width=0.5\textwidth]{papers/reedsolomon/images/Compact_Disc} - \caption{Compact Disc} + \subfigure[]{ + \includegraphics[width=0.45\textwidth]{papers/reedsolomon/images/Compact_Disc} + } + \subfigure[]{ + \includegraphics[width=0.45\textwidth]{papers/reedsolomon/images/Compact_Disc_zoomed_in} + } + \caption{CDs kamen 1982 auf den Markt. Sie funktioniert durch das Einpressen oder Einbrennen von Punkten und Strichen, die die Daten repräsentieren. Gelesen werden diese wiederum durch die Reflektion eines Lasers an diesen Punkten und Strichen.} \label{fig:cd} \end{figure} \subsection{QR-Codes} -Quick Response Codes funktionieren nach einem sehr ähnlichen Prinzip wie in unserem Beispiel, nur dass QR-Codes in einem $\mathbb{F}_{256}$ Körper arbeiten. Je nach grösse der Codierung ist der QR-Code im Endeffekt robuster gegen Beschädigungen. Bei Low Level Codes können 7\% der Daten Wiederhergestellt werden, beim High Level Code sind das sogar 30\%. +Quick Response Codes oder auch QR-Codes funktionieren nach einem sehr ähnlichen Prinzip wie in unserem Beispiel der Abschnitte \ref{reedsolomon:section:codebsp} - \ref{reedsolomon:section:rekonstruktion} nur das QR-Codes in einem $\mathbb{F}_{256}$ Körper arbeiten. Die physische Grösse eines Codes ist stark abhängig von der Menge an codierten Daten sowie dem verwendeten Fehlerkorrektur-Level. Es ist so auf dem ersten Blick nicht ersichtlich, wie viel Nutzinformationen ein Qr-Code enthält. Die QR-Codes in Abbildung \ref{fig:qr} zeigen jeweils die Gleiche Information mit unterschiedlichem Fehlerkorrektur-Level. Codes mit einem höheren Korrektur-Level können auch für Designer-Codes Zweckentfremdet werden. Dabei wird z.B. das Firmenlogo oder einen Schriftzug über den Qr-Code gelegt, ohne das die Funktion des Codes beeinträchtigt wird. Ein Beispiel dazu ist unter Abbildung \ref{fig:designqr} zu finden. + +% + +%So kann auf den ersten Blick nicht +% +% +% funktionieren nach einem sehr ähnlichen Prinzip wie in unserem Beispiel, nur dass QR-Codes in einem $\mathbb{F}_{256}$ Körper arbeiten. Je nach grösse der Codierung ist der QR-Code im Endeffekt robuster gegen Beschädigungen. Bei Low Level Codes können 7\% der Daten Wiederhergestellt werden, beim High Level Code sind das sogar 30\%. \begin{figure} \centering @@ -88,6 +139,30 @@ Quick Response Codes funktionieren nach einem sehr ähnlichen Prinzip wie in uns \subfigure[]{ \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/qrcode_l} } - \caption{(a) High Level Code, (b) Low Level Code} +% \subfigure[]{ +% \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/designer_qrcode_ohnelogo} +% } +% \subfigure[]{ +% \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/designer_qrcode} +% } + \caption{Anhand der grösse würde man darauf schliessen, dass bei (a) mehr Informationen Codiert sind als bei (b). Tatsächlich aber beinhalten beide Codes die gleiche Information. Das liegt daran, da die Fehlerkorrekturfähigkeit von QR-Codes sich in insgesamt vier Levels aufteilen lassen. Der höchste Fehlerkorrektur-Level, der bei (a) angewendet wurde, ist in der Lage, bis zu 30\% der Daten wiederherzustellen. Der kleinste Level schafft etwa 7\%, der in (b) veranschaulicht wird. Da die Grösse also nichts über die Menge an Daten aussagt, könnte es sich bei (a) auch um einen Code mit viel Nutzdaten und kleinem Fehlerkorrektur-Level handeln. Der Unterschied ist von Auge nicht sichtbar.} \label{fig:qr} \end{figure} + +\begin{figure} + \centering +% \subfigure[]{ +% \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/qrcode_h} +% } +% \subfigure[]{ +% \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/qrcode_l} +% } + \subfigure[]{ + \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/designer_qrcode_ohnelogo} + } + \subfigure[]{ + \includegraphics[width=0.4\textwidth]{papers/reedsolomon/images/designer_qrcode} + } + \caption{Während (a) noch einen unveränderten QR-Code repräsentiert, handelt es sich bei (b) nun um einen Designer-QR-Code. Beide Codes verfügen über einen mittleren Fehlerkorrektur-Level von theoretisch 15\%. Da bei (b) jetzt einen Teil des Codes durch ein Logo verdeckt wird, schränkt sich die Fehlerkorrekturfähigkeit je nach Grösse des verdeckten Teils mehr oder weniger stark ein. Unser Designer-Code in (b) ist nur noch in der Lage etwa 9\% des Codes zu rekonstruieren.} + \label{fig:designqr} +\end{figure}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/codebsp.tex b/buch/papers/reedsolomon/codebsp.tex index 8430ebd..eb4e82f 100644 --- a/buch/papers/reedsolomon/codebsp.tex +++ b/buch/papers/reedsolomon/codebsp.tex @@ -76,7 +76,7 @@ dar. \subsection{Der Ansatz der diskreten Fouriertransformation \label{reedsolomon:subsection:diskFT}} -In einem vorherigen Abschnitt \textcolor{red}{(???)} haben wir schon einmal die diskrete Fouriertransformation zum Codieren einer Nachricht verwendet. In den endlichen Körpern wird dies jedoch nicht gelingen, da die Eulerische Zahl $e$ in endlichen Körpern nicht existiert. +Im vorherigen Abschnitt \ref{reedsolomon:section:dtf} haben wir schon einmal die diskrete Fouriertransformation zum Codieren einer Nachricht verwendet. In den endlichen Körpern wird dies jedoch nicht gelingen, da die Eulerische Zahl $e$ in endlichen Körpern nicht existiert. Wir wählen deshalb eine Zahl $a$, die die gleichen Aufgaben haben soll wie $e^{\frac{j}{2 \pi}}$ in der diskreten Fouriertransformation, nur mit dem Unterschied, dass $a$ in $\mathbb{F}_{11}$ ist. Dazu soll die Potenz von $a$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken. Dazu ändern wir die Darstellung von \[ diff --git a/buch/papers/reedsolomon/dtf.tex b/buch/papers/reedsolomon/dtf.tex index 025be3a..9647775 100644 --- a/buch/papers/reedsolomon/dtf.tex +++ b/buch/papers/reedsolomon/dtf.tex @@ -1,30 +1,124 @@ % -% teil3.tex -- Beispiel-File für Teil 3 +% dtf.tex -- Idee mit DFT % -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Diskrete Fourien Transformation +\section{Übertragung mit Hilfe der diskrten Fourier-Transformation \label{reedsolomon:section:dtf}} \rhead{Umwandlung mit DTF} -Um die Polynominterpolation zu umgehen, gehen wir nun über in die Fourientransformation. -Dies wird weder eine erklärung der Forientransorfmation noch ein genauer gebrauch -für den Reed-Solomon-Code. Dieser Abschnitt zeigt nur wie die Fourientransformation auf Fehler reagiert. -wobei sie dann bei späteren Berchnungen ganz nütlich ist. +Die Grundidee eines fehlerkorrigierenden Code ist, dass Informationen eines Datenpunktes +durch die Codierung auf viele übertragene Werte verteilt werden. +Die Decodierung ist in der Lage, den ursprünglichen Datenwert zu rekonstruieren, +sogar wenn einzelne wenige übertragene Werte beschädigt worden sind. +\par +Die Fourier-Transformation transformiert einen einzelnen Wert, +eine Dirac-Funktion, auf ein Spektrum, welches sich über die ganze Frequenzachse erstreckt. +Aus der Filtertheorie ist bekannt, dass der ursprüngliche Impuls mehr oder weniger rekonstruierbar ist, + vorausgesetzt, es gehen nicht zu viele Frequenzen bei der Übertragung verloren. +\par +Es liegt daher nahe zu versuchen, die Fourier-Transformation +für Codierung und Decodierung zu verwenden. -\subsection{Übertragungsabfolge -\label{reedsolomon:subsection:Übertragungsabfolge}} -Das Signal.... sind die Daten, Zahlen welche übertragen werden sollen. -Das speziell ist das wir 100 Punkte übertragen und von 64 bis 100, -werden nur Null Punkte übertragen, dies weiss auch unser Empfänger. -Nun wird das Signal in Abbildung... codiert... -Somit wird die Information jedes Punktes auf das ganze spektrum von 0 bis 100 übertragen. -Kommen nuun drei Fehler... hinzu zu diesem codierten Signal sind diese nicht zu erkennen. -Nach dem Empfangen... und decodieren ... erkennt man die fehlerhafte information in den Punkten 64 bis 100. -Filtert man nur diese Punkte heraus und Transformiert sie mit Fourier erhält man die stellen an denen die Fehler sich eingeschlichen haben. +\subsection{Beispiel mit Fehlerkorrektur mit Fourier-Transformation +\label{reedsolomon:subsection:sendbsp}} +Das folgende Beispiel soll zeigen, wie die Idee der Fehlerkorrektur umgesetzt wurde. +Die Fehlererkennung des Reed-Solomon-Codes funktioniert nach einem sehr Ähnlichen Prinzip. -\subsection{Diskrete Fourientransformation Zusamenhang -\label{reedsolomon:subsection:dtfzusamenhang}} -Die Diskrete Fourientransformation ist definiert als -.... +%Das folgende Beispiel soll zeigen, wie Fehlerkorrektur möglich ist. +%Dieses auf eine Art, die der Funktionsweise des Reed-Solomon-Codes, +%der später erklärt wird, analog ist. +\par +Der Auftrag besteht darin, 64 Datenwerte zu übertragen, 32 Fehler erkennen können und bis zu 16 Fehler zu rekonstruieren. +Mit Hilfe der Fourier-Transformation werden die \textcolor{blue}{blauen Datenpunkte} transformiert, +zu den \textcolor{darkgreen}{grünen Übertragungspunkten}. +Durch eine Rücktransformation können die \textcolor{blue}{blauen Datenpunkte} wieder rekonstruiert werden. +\begin{figure}%[!ht] + \centering + \resizebox{\textwidth}{!}{ + \includegraphics[width=\textwidth]{papers/reedsolomon/figures/fourier} + %\input{papers/reedsolomon/tikz/plotfftraw.tex} + } + \caption{Übertragungsabfolge \ref{reedsolomon:subsection:sendbsp}} + \label{fig:sendorder} +\end{figure} +In der Abbildung \ref{fig:sendorder} wird eine Übertragung Schritt für Schritt illustriert. +In der folgenden Aufzählung werden diese einzelne Schritte erklärt und erläutert: +\begin{enumerate}[(1)] + \item Das Signal besteht aus 64 zufälligen, ganzzahligen Datenwerten zwischen 0 und 10. + Für die Rekonstruktion werden zusätzliche Datenwerte benötigt, wir fügen deshalb 32 Werte hinzu. + Diese setzen wir willkürlich alle auf Null und nennen sie Fehlerkorrekturstellen. + Wir erhalten so einen erweiterten Signalvektor der Länge $N =96$. + \item Mit der Fourier-Transformation wird der ganze Signalvektor codiert. + Dadurch wird jede Informationseinheit auf alle Punkte des Spektrums verteilt. + \item Wir dürfen annehmen, dass bei der Übertragung, nur einzelne übertragene + Werte durch Fehler verändert werden. + \par + Im Beispiel sind dies die Werte an den Stellen 6, 20 und 74 (\textcolor{red}{rote Kurve}), + die um einen Betrag verändert werden. + Dieser ist bis zu 150-mal kleiner als die ursprünglich codierten Werte. + Der Empfänger erkennt daher im allgemeinen nicht, ob und wo Übertragungsfehler aufgetreten sind. + \item Ohne Übertragungsfehler kann der Signalvektor durch die inverse Fourier-Transformation vollständig + wiederhergestellt werden. + Dazu gehören auch die Nullen an den Fehlerkorrekturstellen 64 - 96. + \par + Sind Übertragungsfehler aufgetreten, werden an diesen Stellen die Werte von Null abweichen. + Somit haben wir bereits Fehler erkannt. + \item Die Werte an den Fehlerkorrekturstellen 64 - 96, die nicht mehr Null sind, nennen wir das Syndrom. + Im Syndrom steckt nur Information über die Fehler, sie werden durch die inverse Fourier-Transformation erzeugt. + \item Um die Fehler zu rekonstruieren, kann man versuchen, die Information im Syndrom mit Fourier-Transformation zu transformieren. + Da das Syndrom nur ein Teil der Fehlerinformation ist, liefert die Fourier-Transformation eine Approximation der Fehler. + Diese Approximation der Fehler ist genau genug, um die Fehlerstellen zu lokalisieren. +\end{enumerate} +Im Beispiel haben wir mit dem Syndrom nur etwa ein Drittel der Fehlerinformation, es ist daher zu erwarten, +dass die Fehlerwerte auch nur ein Drittel so gross sind. +\par +Damit können die Fehler korrigiert und die Originaldaten wiederhergestellt werden. +Der Rekonstruktionsauftrag ist damit erfolgreich ausgeführt. +\subsection{Fourier-Transformation und Polynome\label{reedsolomon:subsection:ftandpolynom}} +Im Abschnitt \externaldocument{papers/reedsolomon/idee}\ref{reedsolomon:section:polynomansatz} +wurden Werte eines Polynoms zur Codierung verwendet. +Die 7 Übertragungspunkte könnten ein Polynom +\begin{equation} + \textcolor{darkgreen}{p(x)} + = + \textcolor{blue}{a_0} + \textcolor{blue}{a_1}x + \textcolor{blue}{a_2}x^2 + + \textcolor{gray}{a_3}x^3 + \textcolor{gray}{a_4}x^4 + \textcolor{gray}{a_5}x^5 + + \textcolor{gray}{a_6}x^6 +\label{reedsolomon:equationpoly} +\end{equation} +sechsten Grades bestimmen. +Durch die Wahl von $\textcolor{gray}{a_3=0}$, $\textcolor{gray}{a_4=0}$, $\textcolor{gray}{a_5=0}$, $\textcolor{gray}{a_6=0}$ +erzeugen wir die für die Fehlerkorrektur nötige Redundanz, ganz analog zum Schritt (1) im Beispiel. +\par +Die Analogie geht aber noch weiter. + Schreibt man + \( w = + e^{-\frac{2\pi j}{N} k}\) + \label{reedsolomon:DFT_summand}, damit wird aus der Formel + \begin{equation} + \hat{c}_{k} + = \frac{1}{N} \sum_{n=0}^{N-1} + {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} + ,\label{reedsolomon:DFT} + \end{equation} + für die diskrete-Fourier-Transformation das Polynom + \begin{equation} + q(w)= + \frac{{f}_0}{N} + \frac{{f}_1}{N} w^1 + \frac{{f}_2}{N} w^2 + \dots + \frac{{f}_{N-1}}{N} w^{N-1}. + \label{reedsolomon:DFT_polynom} + \end{equation} + Im Beispiel werden aber Werte des Polynoms + \begin{equation} + \textcolor{darkgreen}{q(w)}= + \frac{\textcolor{blue}{{f}_0}}{N} + \frac{\textcolor{blue}{{f}_1}}{N} w^1 + \frac{\textcolor{blue}{{f}_2}}{N} w^2 + \dots + + \frac{\textcolor{blue}{{f}_{63}}}{N} w^{63} + \frac{\textcolor{gray}{{f}_{64}}}{N} w^{64} + \textcolor{gray}{\dots} + \frac{\textcolor{gray}{{f}_{N-1}}}{N} w^{N-1} + \label{reedsolomon:DFT_polynom2} + \end{equation} + für verschiedene \( w = e^{-\frac{2\pi j}{N} k}, k=1, \dots ,N-1\) übermittelt. +Das Syndrom entstand durch die Wahl ${f_{64}}=0$ bis ${f}_{N-1}=0$ (graue Koeffizenten). +\par +Die Polynominterpolation und die Fourier-Transformation rechnen beide mit reellen Zahlen. +Wenn die Approximation nicht mehr genügend gut ist um die Fehler zu erkennen und rekonstruieren, +dann brauchen wir andere Varianten. +Um dieser Approximation zu entkommen, verlassen wir die reellen Zahlen und gehen zum endlichen Körpern, oder auch Galois-Körper genannt. +Dieser bietet uns einige Vorteile.
\ No newline at end of file diff --git a/buch/papers/reedsolomon/einleitung.tex b/buch/papers/reedsolomon/einleitung.tex index 3d40db1..f99ad82 100644 --- a/buch/papers/reedsolomon/einleitung.tex +++ b/buch/papers/reedsolomon/einleitung.tex @@ -6,14 +6,12 @@ \section{Einleitung \label{reedsolomon:section:einleitung}} \rhead{Einleitung} -Der Reed-Solomon-Code ist entstaden im ... vom .. um, -das Problem der Daten Übertragung zu lösen. -In deiesem Abschnitt wird möglichst verständlich die mathematische Abfolge, Funktion oder Algorithmus erklärt. -Es wird jedoch nicht auf die technische Umsetzung oder Implementierung eingegangen. -Um beim Daten Übertragen fehler zu erkennen könnte man die Daten jeweils doppelt senden, -und so jeweilige Fehler zu erkennen. -Doch dies braucht schnell unmengen an Daten, wenn man nach vielen Fehler absichern möchte. -Der Reed-Solomon-Code macht dies auf eine andere, clevere Weise. +Der Reed-Solomon-Code wurde von den beiden Mathematiker Irving S. Reed und Gustave Solomon im Jahre 1960 entwickelt. +Dabei haben sie das Problem der Fehlerhaften Datenübertragung gelöst. +In diesem Abschnitt wird möglichst verständlich die mathematische Abfolge und +Funktionsweise des Reed-Solomon-Code erklärt. +Es wird jedoch nicht auf die technische Umsetzung oder Implementierung eingegangen, jedoch wird im Abschnitt \ref{reedsolomon:section:anwendung} einige Anwendungen des Reed-Solomon-Codes vorgestellt. + diff --git a/buch/papers/reedsolomon/endlichekoerper.tex b/buch/papers/reedsolomon/endlichekoerper.tex index 1d196fd..3019dd7 100644 --- a/buch/papers/reedsolomon/endlichekoerper.tex +++ b/buch/papers/reedsolomon/endlichekoerper.tex @@ -3,21 +3,63 @@ % % (c) 2021 Michael Steiner, Hochschule Rapperswil % -\section{Reed-Solomon in Endlichen Körpern +\section{Reed-Solomon in endlichen Körpern \label{reedsolomon:section:endlichekoerper}} \rhead{Reed-Solomon in endlichen Körpern} -\[ -\textcolor{red}{\text{TODO: (warten auf den 1. Teil)}} -\] -Das Rechnen in endlichen Körpern bietet einige Vorteile: +Im vorherigen Abschnitt haben wir gesehen, dass wir die Fehler mittels Approximation suchen und somit nur ungefähre Angaben haben, wo sich Fehler aufhalten. +Um dies zu ändern wechseln wir vom komplexen Zahlenraum in endliche Körper. +In endlichen Körpern gibt es keine Approximationen wie bei den rationalen und reellen Zahlen. +Alle Zahlen sind richtig oder falsch, ``fast richtig'' gibt es nicht. +Zudem beschränken sich die arithmetischen Rechenoperationen auf das Addieren und Multiplizieren. +Wir können also nur ganze Zahlen als Resultat erhalten. +Dies erleichtert auch die Umsetzung auf ein digitales System, da Computer in der Regel lieber mit ganzen als mit gebrochenen oder komplexen Zahlen arbeiten. -\begin{itemize} - \item Konkrete Zahlen: In endlichen Körpern gibt es weder rationale noch komplexe Zahlen. Zudem beschränken sich die möglichen Rechenoperationen auf das Addieren und Multiplizieren. Somit können wir nur ganze Zahlen als Resultat erhalten. - - \item Digitale Fehlerkorrektur: lässt sich nur in endlichen Körpern umsetzen. - -\end{itemize} +Um jetzt eine Nachricht in einem endlichen Körpern zu konstruieren gehen, wir im Grunde gleich vor wie im Beispiel aus dem Abschnitt \ref{reedsolomon:subsection:sendbsp}. +Eine Nachricht besteht aus einem Nutzdatenteil und einem Fehlerkorrekturteil. +Diese Nachricht wird codiert, übertragen und beim Empfänger wieder decodiert. +In endlichen Körpern können wir jedoch nicht mehr die Fouriertransformation zur Hilfe nehmen. +Wir müssen also eine Alternative finden, welche die gleichen Eigenschaften wie die Fouriertransformation aufweist, aber im endlichen Körper verwendet werden kann. +Auch beim Decodieren müssen wir uns etwas einfallen lassen, wenn die Vorgehensweise mit dem Lokator auch in endlichen Körpern funktionieren soll. Die folgenden Abschnitte widmen sich deshalb der genaueren Betrachtung eines Reed-Solomon-Codes und wie er in endlichen Körpern funktioniert. -Um jetzt eine Nachricht in den endlichen Körpern zu konstruieren legen wir fest, dass diese Nachricht aus einem Nutzdatenteil und einem Fehlerkorrekturteil bestehen muss. Somit ist die zu übertragende Nachricht immer grösser als die Daten, die wir übertragen wollen. Zudem müssen wir einen Weg finden, den Fehlerkorrekturteil so aus den Nutzdaten zu berechnen, dass wir die Nutzdaten auf der Empfängerseite wieder rekonstruieren können, sollte es zu einer fehlerhaften Übertragung kommen. - -Nun stellt sich die Frage, wie wir eine fehlerhafte Nachricht korrigieren können, ohne ihren ursprünglichen Inhalt zu kennen. Der Reed-Solomon-Code erzielt dies, indem aus dem Fehlerkorrekturteil ein sogenanntes ``Lokatorpolynom'' generiert werden kann. Dieses Polynom gibt dem Emfänger an, welche Stellen in der Nachricht feherhaft sind. +% +%Damit all diese Probleme möglichst verständlich +% +% +%Um all diese Probleme und möglichst +% +% +%um Fehler zu erkennen und mittels Lokatorpolynom +% +% +% ein Lokatorpolynom zu finden. +% +% +% +% Eine Nachricht besteht aus einem Nutzdatenanteil und einem Fehlerkorrekturteil, +% +% +% +%In diesem Zahlenraum gibt es nur Natürliche Zahlen und es darf nur Addiert oder Multipliziert werden. +%Der grosse Vorteil an endlichen Körper ist, dass dich der einfacher Digital umsetzen lässt. +% +% +%Dieser Zahlenraum bringt eine Menge von neuen Regeln mit sich. +%So gibt es dort nur Natürliche Zahlen und die Arithmetischen Rechenoperationen sind beschränkt auf die Addition und Multiplikation. +% +% +% +%\[ +%\textcolor{red}{\text{TODO: (warten auf den 1. Teil)}} +%\] +%Das Rechnen in endlichen Körpern bietet einige Vorteile: +% +%\begin{itemize} +% \item Konkrete Zahlen: In endlichen Körpern gibt es weder rationale noch komplexe Zahlen. Zudem beschränken sich die möglichen Rechenoperationen auf das Addieren und Multiplizieren. Somit können wir nur ganze Zahlen als Resultat erhalten. +% +% \item Digitale Fehlerkorrektur: lässt sich nur in endlichen Körpern umsetzen. +% +%\end{itemize} +% +%Um jetzt eine Nachricht in den endlichen Körpern zu konstruieren legen wir fest, dass diese Nachricht aus einem Nutzdatenteil und einem Fehlerkorrekturteil bestehen muss. Somit ist die zu übertragende Nachricht immer grösser als die Daten, die wir übertragen wollen. Zudem müssen wir einen Weg finden, den Fehlerkorrekturteil so aus den Nutzdaten zu berechnen, dass wir die Nutzdaten auf der Empfängerseite wieder rekonstruieren können, sollte es zu einer fehlerhaften Übertragung kommen. +% +%Nun stellt sich die Frage, wie wir eine fehlerhafte Nachricht korrigieren können, ohne ihren ursprünglichen Inhalt zu kennen. Der Reed-Solomon-Code erzielt dies, indem aus dem Fehlerkorrekturteil ein sogenanntes ``Lokatorpolynom'' generiert werden kann. Dieses Polynom gibt dem Emfänger an, welche Stellen in der Nachricht feherhaft sind. diff --git a/buch/papers/reedsolomon/experiments/f.m b/buch/papers/reedsolomon/experiments/f.m index 6bdc741..bf2587c 100644 --- a/buch/papers/reedsolomon/experiments/f.m +++ b/buch/papers/reedsolomon/experiments/f.m @@ -1,8 +1,8 @@ -# -# f.m -- Reed-Solomon-Visualisierung mit FFT -# -# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -# +% +% f.m -- Reed-Solomon-Visualisierung mit FFT +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule + N = 64; b = 32; l = N + b; @@ -51,6 +51,7 @@ syndrom(1:N,1) = zeros(N,1) plot(abs(syndrom)); xlim([1, l]); title("Syndrom"); + pause() locator = abs(fft(syndrom)) @@ -59,3 +60,13 @@ plot(locator); xlim([1, l]); title("Locator"); pause() + + +writematrix([transpose(counter), abs(signal)], 'signal.txt') +writematrix([transpose(counter), abs(codiert)], 'codiert.txt') +writematrix([transpose(counter), fehler], 'fehler.txt') +writematrix([transpose(counter), abs(empfangen)], 'empfangen.txt') +writematrix([transpose(counter), abs(decodiert)], 'decodiert.txt') +writematrix([transpose(counter), abs(syndrom)], 'syndrom.txt') +writematrix([transpose(counter), locator], 'locator.txt') + diff --git a/buch/papers/reedsolomon/experiments/plot.tex b/buch/papers/reedsolomon/experiments/plot.tex new file mode 100644 index 0000000..4b156bb --- /dev/null +++ b/buch/papers/reedsolomon/experiments/plot.tex @@ -0,0 +1,103 @@ +% polynome1 +%------------------- +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usepackage{pgfplotstable} +\usepackage{filecontents} +\usetikzlibrary{arrows,intersections,math} +\newcommand{\x}{10} +\newcommand{\y}{-8} +\begin{document} + +\begin{tikzpicture}[] + + %--------------------------------------------------------------- + %Knote + \matrix[draw = none, column sep=20mm, row sep=4mm]{ + \node(signal) [] { + \begin{tikzpicture} + \begin{axis}[ + title = {\Large {Signal}}, + xlabel={Anzahl Übertragene Zahlen}, + xtick={0,20,40,64,80,98},] + \addplot[blue] table[col sep=comma] {signal.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(codiert) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Codiert}}] + \addplot[] table[col sep=comma] {codiert.txt}; + \end{axis} + \end{tikzpicture}}; \\ + + &\node(fehler) [] { + \begin{tikzpicture} + \begin{axis}[scale=0.6, title = {\Large {Fehler}}] + \addplot[red] table[col sep=comma] {fehler.txt}; + \end{axis} + \end{tikzpicture}};\\ + + \node(decodiert) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Decodiert}}] + \addplot[blue] table[col sep=comma] {decodiert.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(empfangen) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Empfangen}}] + \addplot[] table[col sep=comma] {empfangen.txt}; + \end{axis} + \end{tikzpicture}};\\ + + \node(syndrom) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Syndrom}}] + \addplot[blue] table[col sep=comma] {syndrom.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(locator) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Locator}}] + \addplot[] table[col sep=comma] {locator.txt}; + \end{axis} + \end{tikzpicture}};\\ + }; + %------------------------------------------------------------- + %FFT & IFFT deskription + + \draw[thin,gray,dashed] (0,12) to (0,-12); + \node(IFFT) [scale=0.7] at (0,12.3) {IFFT}; + \draw[<-](IFFT.south west)--(IFFT.south east); + \node(FFT) [scale=0.7, above of=IFFT] {FFT}; + \draw[->](FFT.north west)--(FFT.north east); + + \draw[thick, ->,] (fehler.west)++(-1,0) +(0.05,0.5) -- +(-0.1,-0.1) -- +(0.1,0.1) -- +(0,-0.5); + %Arrows + \draw[ultra thick, ->] (signal.east) to (codiert.west); + \draw[ultra thick, ->] (codiert.south) to (fehler.north); + \draw[ultra thick, ->] (fehler.south) to (empfangen.north); + \draw[ultra thick, ->] (empfangen.west) to (decodiert.east); + \draw[ultra thick, ->] (syndrom.east) to (locator.west); + \draw(decodiert.south east)++(-1.8,1) ellipse (1.3cm and 0.8cm) ++(-1.3,0) coordinate(zoom) ; + \draw[ultra thick, ->] (zoom) to[out=180, in=90] (syndrom.north); + + %item + \node[circle, draw, fill =lightgray] at (signal.north west) {1}; + \node[circle, draw, fill =lightgray] at (codiert.north west) {2}; + \node[circle, draw, fill =lightgray] at (fehler.north west) {3}; + \node[circle, draw, fill =lightgray] at (empfangen.north west) {4}; + \node[circle, draw, fill =lightgray] at (decodiert.north west) {5}; + \node[circle, draw, fill =lightgray] at (syndrom.north west) {6}; + \node[circle, draw, fill =lightgray] at (locator.north west) {7}; + +\end{tikzpicture} +\end{document} + diff --git a/buch/papers/reedsolomon/figures/fourier.pdf b/buch/papers/reedsolomon/figures/fourier.pdf Binary files differnew file mode 100644 index 0000000..4995141 --- /dev/null +++ b/buch/papers/reedsolomon/figures/fourier.pdf diff --git a/buch/papers/reedsolomon/figures/plotfft.pdf b/buch/papers/reedsolomon/figures/plotfft.pdf Binary files differnew file mode 100644 index 0000000..80adafb --- /dev/null +++ b/buch/papers/reedsolomon/figures/plotfft.pdf diff --git a/buch/papers/reedsolomon/figures/polynom2.pdf b/buch/papers/reedsolomon/figures/polynom2.pdf Binary files differnew file mode 100644 index 0000000..55a50ac --- /dev/null +++ b/buch/papers/reedsolomon/figures/polynom2.pdf diff --git a/buch/papers/reedsolomon/idee.tex b/buch/papers/reedsolomon/idee.tex index 4a7716a..daa2913 100644 --- a/buch/papers/reedsolomon/idee.tex +++ b/buch/papers/reedsolomon/idee.tex @@ -1,58 +1,157 @@ % -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% idee.tex -- Polynom Idee % \section{Idee \label{reedsolomon:section:idee}} \rhead{Problemstellung} -Das Problem liegt darin Informationen, Zahlen, -zu Übertragen und Fehler zu erkennen. -Beim Reed-Solomon-Code kann man nicht nur Fehler erkenen, -man kann sogar einige Fehler korrigieren. +Um Fehler in einer Datenübertragung zu erkennen, könnte man die Daten jeweils doppelt senden, +also den gleiche Wert immer zweimal versenden. +Tritt ein Fehler ein wird sich dies in der Differenz der beiden Werten bemerkbar machen. +Aber wie erkennen wir, welcher nun der richtige ist? Die Lösung ist simpel: Wir übertragen den Wert einfach dreimal. +Wenn jetzt ein Fehler auftritt, kann durch die beiden unveränderten Werten den richtigen bestimmt werden. +Doch was machen wir, wenn bei dieser Übertragung zwei Fehler auftreten? +Oder noch schlimmer: Was wenn zweimal derselbe Fehler auftritt? Die beiden Fehlerhaften Werte überstimmen bei der Evaluierung den gesendeten Datenwert, der dann unwiderruflich verloren geht. +Wir könnten dies noch steigern mit vier, fünf oder mehr gleichen Übertragenen Werte. Dies erhöht zwar die Robustheit der gesendeten Daten, führt aber auch dazu, dass wir durch die Mehrfachübertragung nur sehr wenige Nutzdaten versenden können. +Gerade in unserer heutigen Zeit wäre dies ein enorm grosses Problem und aus diesem Grund wurden alternative Ansätze ausgearbeitet um dieses grundlegende Problem zu lösen. +% +% +%Gerade in der heutigen modernen Zeit bei dem hohen bedarf an Daten würden unsere Kommunikationssysteme bei weitem nicht ausreichen um den einen einzigen Datenwert mehrfach zu übertragen +% +% Gerade in der Heutigen modernen Zeit bei diesem enormen mass an daten die wir alle tagtäglich anfordern Währe dies wohl unmöglich, wenn wir die daten auf diese Weise +% +% +% +% +% +%Wenn es uns gelingt, Fehler nach Ihrer Übertragung zu erkennen, dann könnten wir in einem neuen Ansatz den fehlerhaft empfangenen Wert noch einmal anfordern. +%Wir stellen fest, dass für viele alltägliche Anwendungen völlig ausreichend ist. +% +%Was ist, wenn wir aber eine Datenquelle haben, von der wir nur einmalig lesen können? +% +% +% +%Beim Übertragen von drei Werten können wir maximal 2 Fehler erkennen aber nicht mehr korrigieren. +%Wenn wir noch mehr Werte +% +%Wir Übertragen Ziemlich viele Werte für so wenige Nutzdaten. Hinzu kommt, dass wir bei dieser Vorgehensweise gerade mal bestimmen können, dass überhaupt Fehler aufgetreten sind +% +% +%Wir haben also drei Werte die bestimmt einen Fehler korrigieren können, was ziemlich viele Werte um einen Fehler zu korrigieren. +% +% um so jeweils einzelne Fehler zu erkennen. +%Wenn jedoch mehr als nur ein Fehler erkannt werden und sogar noch das Original rekonstruiert werden soll, dann sollen die Daten drei oder vierfach versendet werden. +%Doch nur schon um einen Fehler zu erkennen werden überproportional viele Daten doppelt und dreifach versendet. +%Das Hauptproblem ist, dass Informationen Fehlerfrei Übertragen werden sollen. Um dies zu erreichen muss gleich nach dem Empfangen Fehler erkannt und korrigiert werden. +% +%Das Problem liegt darin, Informationen oder Zahlen beim Übertragen gleichzeitig noch +% +%Das Problem liegt darin, das Informationen oder Zahlen zu Übertragen und gleichzeitig Fehler zu erkennen +% +% +%Das Problem liegt darin Informationen, Zahlen, zu Übertragen und Fehler zu erkennen und zu korrigieren. +%Der Unterschied des Fehler Erkennens und Korrigirens, ist das beim Erkennen nur die Frage beantwortet wird: Ist die Übertragung fehlerhaft oder nicht? +%Beim Korrigieren werden Fehler erkannt und dann zusätzlich noch die Originalwerte rekonstruiert. +%Eine weitere Möglichkeit wäre, dass der Empfänger nach einer fehlerhaften Übertragung die selben Daten nochmals anfordert. +%Dies führt wieder zu unerwünschten mehrfachen Übertragung. +%In Anwendungen des Reed-Solomon-Codes Abschnitt \externaldocument{papers/reedsolomon/anwendungen} \ref{reedsolomon:section:anwendung} +% ist diese vom Empfänger gesteuerte erneute Übertragen meistens nicht sinnvoll oder sogar unmöglich. +%Der Reed-Solomon-Code macht dies Übertragung auf eine andere, clevere Weise. +\subsection{Polynom-Ansatz +\label{reedsolomon:section:polynomansatz}} \rhead{Polynom-Ansatz} -Eine Idee ist die Daten, -ein Polynom zu bilden und dieses dann mit bestimmten Punkten überträgt. -Nehmen wir als beisbiel die Zahlen \textcolor{blue}{2}, \textcolor{blue}{1}, \textcolor{blue}{5}, -welche uns dann das Polynom +Eine zentrale Idee des Reed-Solomon-Code ist, aus den Daten ein Polynom zu bilden. +Mit dieser Polynomfunktion wird dann eine Anzahl von Werten übertragen. +\begin{beispiel} Nehmen wir die Zahlen \textcolor{blue}{2}, \textcolor{blue}{1} und \textcolor{blue}{5}, welche übertragen werden sollen. Daraus bilden wir das Polynom \begin{equation} p(x) = -2x^2 + 1x + 5 +\textcolor{blue}{2}x^2 + \textcolor{blue}{1}x + \textcolor{blue}{5}. \label{reedsolomon:equation1} \end{equation} -ergeben. -Übertragen werden nun die stellen 1, 2, 3\dots 7 dieses Polynomes. -Grafisch sieht man dies dann im Abbild //TODO -Wenn ein Fehler sich in die Übertragung eingeschlichen hatt, muss der Leser/Empfänger erkennen, welches das Richtige Polynom ist. -Der Leser/Empfänger weiss, mit welchem Grad das Polynom entwickelt wurde. -\subsection{Beispiel} -Für das Beispeil aus der Gleichung \ref{reedsolomon:equation1}, -ist ein Polynome zweiten Grades durch drei Punkte eindeutig bestimmbar. -Hat es Fehler in der Übertragunge gegeben, kann man diese erkennen, -da alle Punkte, die korrekt sind, auf dem Polynom liegen müssen. -Ab wie vielen Fehler ist das Polynom nicht mehr erkennbar beim Übertragen von 7 Punkten? -Bei 2 Fehlern kann man noch eindeutig bestimmen, dass das Polynom mit 4 Punkten, -gegenüber dem mit 5 Punkten falsch liegt. -Werden es mehr Fehler kann nur erkennt werden das das Polynom nicht stimmt. -Das Orginale Polynom kann aber nicht mehr gefunden werden. -Dabei sollten mehr Übertragungspunkte gegeben werden. - -\section{Fehlerbestimmung -\label{reedsolomon:section:Fehlerbestimmmung}} -So wird ein Muster indentifiziert, welches genau vorherbestimmen kann, -wie gross das Polynom sein muss und wie viele Übertragungspunkte gegeben werden müssen. -Durch ein klein wenig Überlegung ist klar das die anzahl Zahlen (Daten, ab hier verwenden wir das Wort Nutzlast), -die dan Entschlüsselt werden sollen den Grad des Polynoms minus 1 ergeben. -Für die Anzahl an Übertragungspunkte, muss bestimmt werden wieviel Fehler erkennt und korrigiert werden sollen. -Mit Hilfe der Tabelle.... sieht man das es bei $$t$$ Fehlern und $$k$$ Nutzlast, -für das Übertragen $$k+2t$$ Punkte gegben werden müssen. - -Ein toller Nebeneffekt ist das dadurch auch $$2t$$ Fehler erkannt werden. -um zurück auf unser Beispiel zu kommen, -können von den 7 Übertragungspunkten bis zu $$2t = 2*2 = 4 $$ Punkten falsch liegen -und es wird kein eindeutiges Polynom 2ten Grades erkannt, und somit die Nutzlast Daten als fehlerhaft deklariert. - -Ein Polynom durch Punkt mit Polynom Interpolation zu rekonstruieren ist schwierig und Fehleranfällig. +\par +Ein Polynom zweiten Grades ist durch drei Punkte eindeutig bestimmbar. +Bei einer fehlerlosen Übertragung können wir mit 3 übertragenen Werten + das Polynom durch Polynominterpolation volständig rekonstruieren. +Wir brauchen Polynominterpolation als Methode, um aus den Punkten wieder ein Polynom zu bilden. +Die Koeffizente des rekonstruierten Polynoms sind dann unsere gesendeten Zahlen \textcolor{blue}{2}, \textcolor{blue}{1} und \textcolor{blue}{5}. +\par +Wie können wir nun Fehler erkennen oder sogar korrigieren? +Versuchen wir doch, mehr Werte zu übertragen, wie zum Beispiel 7 Werte. +Übertragen werden nun die \textcolor{darkgreen}{grünen Werte} + des \textcolor{blue}{blauen Polynomes} an den Stellen 1, 2, 3, \dots , 7. +In Abbildung \ref{fig:polynom} ist das zu den \textcolor{blue}{Datenpunkten} gehörige Polynom blau dargestellt, +die \textcolor{darkgreen}{übertragenen Werte} des Polynoms sind grün, wobei diese Punkte aufgrund von Übertragungsfehler jetzt eine Parabel darstellen. +Die Fehlerhaften Punkte lassen sich sehr einfach bestimmen, weil diese nicht auf der ursprünglichen Funktion liegen. +Somit können die roten Punkte auf der Parabel durch die grauen ersetzt werden und sind damit korrigiert. + +Bisher konnten wir von 7 Zahlen zwei Fehler erkennen und korrigieren. Können wir in diesem Beispiel noch mehr Fehler korrigieren? +Wir erhöhen dazu die Fehleranzahl Schritt für Schritt: +\begin{itemize} + \item[\textit{1 Fehler}:] Bei einem Fehler können konkurrenzierende, aber falsche Polynome zusammen mit zwei originalen Punkten entstehen. + Dabei können aber maximal 3 Punkte auf diesem Konkurrenzpolynom sein. + Da 6 > 3 ist haben wir unser originales Polynom gefunden. + \item[\textit{2 Fehler}:] Bei Zwei Fehlern kann ein Fehler mit zwei originalen Punkten ein konkurrenzierendes, aber falsches Polynom bilden. + Da der zweite \textcolor{red}{Fehler} frei wählbar ist, kann dieser auch auf dem \textcolor{gray}{Konkurrenzpolynom} liegen, wie in der Abbilbung \ref{fig:polynom} zu sehen ist. + Nun haben wir, ein \textcolor{blue}{originales Polynom} mit \textcolor{darkgreen}{5} übereinstimmenden und ein konkurrenzierendes mit 4 Punkten. + Da 5 noch grösser als 4 ist, können wir sagen, welches das Originalpolynom ist. + \item[\textit{3 Fehler}:] Bei Drei kann genau wie bei 1 oder 2 Fehler, ein konkurenzierendes Polynom mit einem Fehler und zwei originalen Punkten bestimmt werden. + Auch hier sind die anderen Fehler frei wählbar und liegen auf dem Konkurrenzpolynom. + Nun ist es so das 5 Punkte auf diesem konkurenzierenden Polynom und 4 Punkte auf dem originalen. + Das Originalpolynom kann nicht mehr gefunden werden. + \item[\textit{4 Fehler}:] Bei Vier kann noch erkannt werden, dass Fehler aufgetreten sind, da 3 originale Punkte das ursprüngliche Polynom ergeben. + Somit haben wir mindestens 2 verschieden Polynome, was bedeutet, dass Fehler entstanden sind. + \item[\textit{5 Fehler:}] Bei Fünf kann mit den 2 originalen Punkte das Originale Polynom nicht mehr erkannt werden und + somit kann auch keine Aussage mehr gemacht werden, ob Fehler aufgetreten sind oder nicht. +\end{itemize} + +\begin{figure}%[!ht] + \centering + %\includegraphics[width=\textwidth]{papers/reedsolomon/figures/polynom2} + \input{papers/reedsolomon/tikz/polynomraw.tex} + \caption{Polynom $p(x)$ von der Gleichung\eqref{reedsolomon:equation1}} + \label{fig:polynom} +\end{figure} +\qedhere +\end{beispiel} + +\section{Anzahl Übertragungswerte bestimmen +\label{reedsolomon:section:Fehlerkorrekturstellen}} +Um zu bestimmen, wie viele zusätzliche \textcolor{darkgreen}{Übertragungspunkte} notwendig sind um die Fehler zu korrigieren, + muss man zuerst wissen, wie viele \textcolor{blue}{Datenwerte} gesendet und wie viele \textcolor{red}{Fehler} erkannt werden sollen. +Die Anzahl Datenwerte ergeben die Anzahl Polynomkoeffizenten \textcolor{blue}{$k$} und somit den Grad $k-1$ des Polynoms. +Die Bestimmung der Anzahl der Fehler \textcolor{red}{$t$}, welche korrigiert werden können, braucht Redundanz. +Bilden wir verschieden grosse Polynome und untersuchen diese mit unterschiedlich vielen Fehlern erkennt man allmählich ein Muster. + +\begin{table}%[!ht] + \centering + \begin{tabular}{ c c | c} + \hline + Nutzlas & Fehler & Übertragen \\ + \hline + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ + 3 & 3 & 9 Werte eines Polynoms vom Grad 2 \\ + \hline + $k$ & $t$ & $k+2t$ Werte eines Polynoms vom Grad $k-1$ \\ + \hline + \end{tabular} + \caption{ Fehlerkorrekturstellen Bestimmung.} + \label{tab:fehlerkorrekturstellen} +\end{table} +\par +Es müssen mehr Punkte auf dem \textcolor{blue}{originalen Polynom} liegen, als auf dem konkurenzierenden. +Somit braucht man für die Übertragung pro \textcolor{red}{Fehler} zwei Übertragungspunkte mehr. +Wie in der Tabelle \ref{tab:fehlerkorrekturstellen} ersichtlich ist ergeben sich diese Anzahl an \textcolor{darkgreen}{Punkte} für die Übertragung. +\begin{equation} + \textcolor{darkgreen}{u}= + \textcolor{blue}{k}+2\textcolor{red}{t}. + \label{reedsolomon:equation2} +\end{equation} + +Ein Nebeneffekt ist, dass auch $2t$ Fehler erkannt werden können, die aber nicht korrigiert werden können. +Um die Polynomkoeffizenten nach der Übertragung zu rekonstruieren, haben wir jedes mal die Polynominterpolationsmethode angewendet. +Diese Polynominterpolation ist leider schwierig zu berechnen und sehr fehleranfällig. +Es wäre daher einfacher, wenn wir eine alternative Vorgehensweise finden könnten. + diff --git a/buch/papers/reedsolomon/images/Compact_Disc_zoomed_in.png b/buch/papers/reedsolomon/images/Compact_Disc_zoomed_in.png Binary files differnew file mode 100644 index 0000000..69556d0 --- /dev/null +++ b/buch/papers/reedsolomon/images/Compact_Disc_zoomed_in.png diff --git a/buch/papers/reedsolomon/images/designer_qrcode.png b/buch/papers/reedsolomon/images/designer_qrcode.png Binary files differnew file mode 100644 index 0000000..a9e0505 --- /dev/null +++ b/buch/papers/reedsolomon/images/designer_qrcode.png diff --git a/buch/papers/reedsolomon/images/designer_qrcode_ohnelogo.png b/buch/papers/reedsolomon/images/designer_qrcode_ohnelogo.png Binary files differnew file mode 100644 index 0000000..fe4251d --- /dev/null +++ b/buch/papers/reedsolomon/images/designer_qrcode_ohnelogo.png diff --git a/buch/papers/reedsolomon/main.tex b/buch/papers/reedsolomon/main.tex index 6bd04f2..017fe94 100644 --- a/buch/papers/reedsolomon/main.tex +++ b/buch/papers/reedsolomon/main.tex @@ -8,29 +8,9 @@ \begin{refsection} \chapterauthor{Joshua Bär und Michael Steiner} -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} - % Joshua \input{papers/reedsolomon/einleitung.tex} \input{papers/reedsolomon/idee.tex} -\input{papers/reedsolomon/teil2.tex} \input{papers/reedsolomon/dtf.tex} % Michael @@ -45,6 +25,13 @@ Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren \nocite{reedsolomon:weitz} \nocite{reedsolomon:informationkommunikation} +\nocite{reedsolomon:voyager_programm} +\nocite{reedsolomon:voyager} +\nocite{reedsolomon:cd_wiki} +\nocite{reedsolomon:cd} +\nocite{reedsolomon:strichepunkte} +\nocite{reedsolomon:qr_wiki} +\nocite{reedsolomon:qr} %\nocite{reedsolomon:mendezmueller} \printbibliography[heading=subbibliography] diff --git a/buch/papers/reedsolomon/packages.tex b/buch/papers/reedsolomon/packages.tex index 3643731..40c6ea3 100644 --- a/buch/papers/reedsolomon/packages.tex +++ b/buch/papers/reedsolomon/packages.tex @@ -8,3 +8,7 @@ % following example %\usepackage{packagename} +\usepackage{pgfplots} +\usepackage{filecontents} +\usepackage{xr} + diff --git a/buch/papers/reedsolomon/references.bib b/buch/papers/reedsolomon/references.bib index 731bd35..b84b5a4 100644 --- a/buch/papers/reedsolomon/references.bib +++ b/buch/papers/reedsolomon/references.bib @@ -23,3 +23,65 @@ volume = {1} } +@online{reedsolomon:voyager_programm, + title = {Information über das Voyager Programm}, + url = {https://de.wikipedia.org/wiki/Voyager-Programm}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +} + +@online{reedsolomon:voyager, + title = {Bild der Voyager Raumsonde}, + url = {https://en.wikipedia.org/wiki/Voyager_1}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +} + +@online{reedsolomon:cd_wiki, + title = {Alles über die CD}, + url = {https://de.wikipedia.org/wiki/Compact_Disc}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +} + +@online{reedsolomon:cd, + title = {Abbildung einer CD}, + url = {https://www.stickpng.com/img/electronics/compact-discs/stack-compact-disc}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +} + +@online{reedsolomon:strichepunkte, + title = {Abbildung der Striche und Punkte einer CD}, + url = {https://www.researchgate.net/figure/The-readable-area-of-a-CD-is-magnified-in-order- to-see-the-pit-and-land-sizing-The_fig7_303401629}, + date = {2021-07-26}, + year = {2021}, + month = {7}, + day = {26} +} + +@online{reedsolomon:qr_wiki, + title = {Funktionsweise des QR-Codes}, + url = {https://de.wikipedia.org/wiki/QR-Code}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +} + +@online{reedsolomon:qr, + title = {Tool zum erstellen von QR-Codes}, + url = {https://www.qrcode-generator.ch}, + date = {2021-07-19}, + year = {2021}, + month = {7}, + day = {19} +}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/standalone.tex b/buch/papers/reedsolomon/standalone.tex new file mode 100644 index 0000000..c850d1f --- /dev/null +++ b/buch/papers/reedsolomon/standalone.tex @@ -0,0 +1,30 @@ +\documentclass{book} + +\input{common/packages.tex} + +% additional packages used by the individual papers, add a line for +% each paper +\input{papers/common/addpackages.tex} + +% workaround for biblatex bug +\makeatletter +\def\blx@maxline{77} +\makeatother +\addbibresource{chapters/references.bib} + +% Bibresources for each article +\input{papers/common/addbibresources.tex} + +% make sure the last index starts on an odd page +\AtEndDocument{\clearpage\ifodd\value{page}\else\null\clearpage\fi} +\makeindex + +%\pgfplotsset{compat=1.12} +\setlength{\headheight}{15pt} % fix headheight warning +\DeclareGraphicsRule{*}{mps}{*}{} + +\begin{document} + \input{common/macros.tex} + \def\chapterauthor#1{{\large #1}\bigskip\bigskip} + \input{papers/reedsolomon/main.tex} +\end{document} diff --git a/buch/papers/reedsolomon/standalone/standalone.pdf b/buch/papers/reedsolomon/standalone/standalone.pdf Binary files differnew file mode 100644 index 0000000..dfa9eea --- /dev/null +++ b/buch/papers/reedsolomon/standalone/standalone.pdf diff --git a/buch/papers/reedsolomon/teil2.tex b/buch/papers/reedsolomon/teil2.tex deleted file mode 100644 index b2adc9f..0000000 --- a/buch/papers/reedsolomon/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{reedsolomon:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{reedsolomon:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/reedsolomon/tikz/Makefile b/buch/papers/reedsolomon/tikz/Makefile new file mode 100644 index 0000000..1753f37 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/Makefile @@ -0,0 +1,7 @@ +# +# Makefile +# +# (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +fourier.pdf: fourier.tex + pdflatex fourier.tex diff --git a/buch/papers/reedsolomon/tikz/codiert.txt b/buch/papers/reedsolomon/tikz/codiert.txt new file mode 100644 index 0000000..4a481d8 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/codiert.txt @@ -0,0 +1,96 @@ +0,284 +1,131.570790435043 +2,41.9840308053375 +3,12.1189172092243 +4,23.8408857476069 +5,69.1793197789512 +6,24.0186013379153 +7,37.3066577242559 +8,18.2010889773887 +9,12.3214904922455 +10,15.6627133315015 +11,24.5237955316204 +12,32.1114345314062 +13,44.9845039238714 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+94,0.0261327016093158 +95,0.0314909067039411 diff --git a/buch/papers/reedsolomon/tikz/empfangen.txt b/buch/papers/reedsolomon/tikz/empfangen.txt new file mode 100644 index 0000000..38c13b0 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/empfangen.txt @@ -0,0 +1,96 @@ +0,284 +1,131.570790435043 +2,41.9840308053375 +3,12.1189172092243 +4,23.8408857476069 +5,69.1793197789512 +6,23.6290258699579 +7,37.3066577242559 +8,18.2010889773887 +9,12.3214904922455 +10,15.6627133315015 +11,24.5237955316204 +12,32.1114345314062 +13,44.9845039238714 +14,13.5324640263625 +15,10.1736266929292 +16,4.58257569495584 +17,23.217268502288 +18,16.5769107917917 +19,6.89948680823017 +20,5.55320238736303 +21,10.4219666223433 +22,43.6179140616243 +23,35.9073375743642 +24,15.0332963783729 +25,21.7594021268945 +26,23.2496572716993 +27,17.9815599423852 +28,11.3577742151117 +29,38.467599433197 +30,28.3035029562577 +31,9.54321919833388 +32,21.377558326432 +33,17.6292439561917 +34,12.6951848921471 +35,20.0667752354841 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+84,32.1114345314062 +85,24.5237955316204 +86,15.6627133315015 +87,12.3214904922455 +88,18.2010889773887 +89,37.3066577242559 +90,24.0186013379153 +91,69.1793197789512 +92,23.8408857476069 +93,12.1189172092243 +94,41.9840308053375 +95,131.570790435043 diff --git a/buch/papers/reedsolomon/tikz/fehler.txt b/buch/papers/reedsolomon/tikz/fehler.txt new file mode 100644 index 0000000..23f1a83 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/fehler.txt @@ -0,0 +1,96 @@ +0,0 +1,0 +2,0 +3,0 +4,0 +5,0 +6,2 +7,0 +8,0 +9,0 +10,0 +11,0 +12,0 +13,0 +14,0 +15,0 +16,0 +17,0 +18,0 +19,0 +20,2 +21,0 +22,0 +23,0 +24,0 +25,0 +26,0 +27,0 +28,0 +29,0 +30,0 +31,0 +32,0 +33,0 +34,0 +35,0 +36,0 +37,0 +38,0 +39,0 +40,0 +41,0 +42,0 +43,0 +44,0 +45,0 +46,0 +47,0 +48,0 +49,0 +50,0 +51,0 +52,0 +53,0 +54,0 +55,0 +56,0 +57,0 +58,0 +59,0 +60,0 +61,0 +62,0 +63,0 +64,0 +65,0 +66,0 +67,0 +68,0 +69,0 +70,0 +71,0 +72,0 +73,0 +74,1 +75,0 +76,0 +77,0 +78,0 +79,0 +80,0 +81,0 +82,0 +83,0 +84,0 +85,0 +86,0 +87,0 +88,0 +89,0 +90,0 +91,0 +92,0 +93,0 +94,0 +95,0 diff --git a/buch/papers/reedsolomon/tikz/fourier.pdf b/buch/papers/reedsolomon/tikz/fourier.pdf Binary files differnew file mode 100644 index 0000000..7e0198b --- /dev/null +++ b/buch/papers/reedsolomon/tikz/fourier.pdf diff --git a/buch/papers/reedsolomon/tikz/fourier.tex b/buch/papers/reedsolomon/tikz/fourier.tex new file mode 100644 index 0000000..7b4ccea --- /dev/null +++ b/buch/papers/reedsolomon/tikz/fourier.tex @@ -0,0 +1,139 @@ +% +% Plot der Übertrangungsabfolge ins FFT und zurück mit IFFT +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{pgfplots} +\usepackage{pgfplotstable} +\usepackage{csvsimple} +\usepackage{filecontents} + +\def\plotwidth{7.5cm} +\def\plotheight{5.5cm} +\def\xverschiebung{4.5cm} +\def\yverschiebung{-7cm} +\def\yyverschiebung{-14cm} + +\def\marke#1{ + \coordinate (M) at (-0.8,4.6); + \fill[color=lightgray] (M) circle[radius=0.3]; + \draw (M) circle[radius=0.3]; + \node at (M) {#1}; +} + +\definecolor{darkgreen}{rgb}{0,0.6,0} + +\begin{document} +\begin{tikzpicture}[>=latex,thick] + +\fill[color=blue!10] (-5.7,-14.5) rectangle (2.6,5.0); +\fill[color=darkgreen!10] (2.6,-14.5) rectangle (11.1,5.0); + +\draw[dashed,line width=2pt,color=lightgray] (2.6,4.9) -- (2.6,-14.4); +\coordinate (B) at (2.6,-1.3); +\node[color=gray] at (B) [rotate=90,above] {Zeitbereich\strut}; +\node[color=gray] at (B) [rotate=90,below] {Frequenzbereich\strut}; + +\begin{scope}[xshift=-\xverschiebung,yshift=0cm] + \begin{axis} + [title = {\large Signal\strut}, + xtick={0,32,64,96}, + axis background/.style={fill=white}, + width=\plotwidth,height=\plotheight] + \addplot[blue,line width=1pt] table[col sep=comma] + {tikz/signal.txt}; + \end{axis} + \marke{1} +\end{scope} + +\begin{scope}[xshift=\xverschiebung,yshift=0cm] + \begin{axis}[title = {\large Codiert\strut}, + xtick={0,32,64,96}, + axis background/.style={fill=white}, + width=\plotwidth,height=\plotheight] + \addplot[color=black!60!green,line width=1pt] + table[col sep=comma] + {tikz/codiert.txt}; + \end{axis} + \marke{2} + \draw[->,line width=1pt] (3,-0.4) -- node[right] {Übertragung} (3,-2.2); +\end{scope} + +\definecolor{pink}{rgb}{0.6,0.2,1} + +\begin{scope}[xshift=-\xverschiebung,yshift=\yverschiebung] + %\fill[color=pink!20] (4.65,0.35) ellipse (1.1cm and 0.5cm); + \begin{axis}[title = {\large Decodiert\strut}, + xtick={0,32,64,96}, + axis background/.style={fill=white}, + width=\plotwidth,height=\plotheight] + \addplot[blue,line width=1pt] + table[col sep=comma] {tikz/decodiert.txt}; + \end{axis} + \marke{4} + \draw[color=pink] (4.65,0.35) ellipse (1.1cm and 0.5cm); + \draw[->,color=pink,line width=1pt] + (4.65,-0.15) to[out=-90,in=90] (3,-2.2); +\end{scope} + +\begin{scope}[xshift=\xverschiebung,yshift=\yverschiebung] + \begin{axis}[title = {\large Empfangen {\color{red} mit Fehlern}\strut}, + xtick={0,96}, + axis background/.style={fill=white}, + axis y line*=left, + width=\plotwidth,height=\plotheight] + \addplot[color=black!60!green,line width=1pt] + table[col sep=comma] + {tikz/empfangen.txt}; + \end{axis} + \begin{axis}[xtick={6,20,74}, axis y line*=right, + width=\plotwidth,height=\plotheight] + \addplot[red,line width=1pt] + table[col sep=comma] {tikz/fehler.txt}; + \end{axis} + \marke{3} +\end{scope} + +\begin{scope}[xshift=-\xverschiebung,yshift=\yyverschiebung] + \begin{axis}[title = {\large \color{pink}Syndrom\strut}, + xtick={0,32,64,96}, + axis background/.style={fill=white}, + width=\plotwidth,height=\plotheight] + \addplot[pink,line width=1pt] + table[col sep=comma] {tikz/syndrom.txt}; + \end{axis} + \marke{5} +\end{scope} + +\begin{scope}[xshift=\xverschiebung,yshift=\yyverschiebung] + % Beschriftung Rechts + \begin{axis}[axis x line= none, axis y line*=right, ytick={0.3}, + xtick={0,32,64,96}, + axis background/.style={fill=white}, + width=\plotwidth,height=\plotheight] + \addplot[color=black!60,line width=1pt] {0.3}; + \end{axis} + \begin{axis}[title = {\large Lokator\strut},axis y line*=left, + xtick={0,6,20,74,96}, + width=\plotwidth,height=\plotheight] + \addplot[gray,line width=1pt] + table[col sep=comma] {tikz/locator.txt}; + \end{axis} + \marke{6} +\end{scope} + +% Fourier-Transformations-Pfeile + +\draw[->,line width=1pt] (1.8,2) -- node[above] {DFT\strut} (3.8,2); + +\begin{scope}[yshift=\yverschiebung] +\draw[<-,line width=1pt] (1.8,2) -- node[above] {DFT$\mathstrut^{-1}$} (3.8,2); +\end{scope} + +\begin{scope}[yshift=\yyverschiebung] +\draw[->,line width=1pt] (1.8,2) -- node[above] {DFT\strut} (3.8,2); +\end{scope} + +\end{tikzpicture} +\end{document} diff --git a/buch/papers/reedsolomon/tikz/locator.txt b/buch/papers/reedsolomon/tikz/locator.txt new file mode 100644 index 0000000..b28988c --- /dev/null +++ b/buch/papers/reedsolomon/tikz/locator.txt @@ -0,0 +1,96 @@ +0,0.0301224340567056 +1,0.141653026854885 +2,0.138226631799377 +3,0.0339903276086929 +4,0.310585462557496 +5,0.551427312631385 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+50,0.0181978813094817 +51,0.0271098219177584 +52,0.0386836665079729 +53,0.0180518611046889 +54,0.0272138992557141 +55,0.0381891287148314 +56,0.0180809085252469 +57,0.0281418959420061 +58,0.0384596362516637 +59,0.0182864418432272 +60,0.0302250788423173 +61,0.0397874837986351 +62,0.0186786556701694 +63,0.0342489348284216 +64,0.0429932815348666 +65,0.0192777878591759 +66,0.0422808966931999 +67,0.0506815964680563 +68,0.0201167847752226 +69,0.0615048274405271 +70,0.0744953894508454 +71,0.021246054596492 +72,0.142602265816215 +73,0.273502052865436 +74,0.325309673287599 +75,0.272705389655349 +76,0.149074257381345 +77,0.0247199397628712 +78,0.0680137859566976 +79,0.075388270873485 +80,0.0273637831604903 +81,0.0407867704453274 +82,0.0632964886441949 +83,0.0309749128751093 +84,0.0315202035072035 +85,0.0627625211892184 +86,0.0360843918243497 +87,0.02794920551495 +88,0.0677921493367236 +89,0.0437167157553067 +90,0.0270640150996317 +91,0.0783380025231622 +92,0.0561293738314281 +93,0.0278742033265809 +94,0.0981443889498639 +95,0.0794543457386548 diff --git a/buch/papers/reedsolomon/tikz/plotfft.tex b/buch/papers/reedsolomon/tikz/plotfft.tex new file mode 100644 index 0000000..77c4dc3 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/plotfft.tex @@ -0,0 +1,104 @@ +% +% Plot der Übertrangungsabfolge ins FFT und zurück mit IFFT +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{pgfplots} +\usepackage{pgfplotstable} +\usepackage{csvsimple} +\usepackage{filecontents} + + + +\begin{document} +\begin{tikzpicture}[] + + %--------------------------------------------------------------- + %Knote + \matrix(m) [draw = none, column sep=25mm, row sep=2mm]{ + + \node(signal) [] { + \begin{tikzpicture} + \begin{axis} + [title = {\Large {Signal}}, + xtick={0,20,40,64,80,98}] + \addplot[blue] table[col sep=comma] {tikz/signal.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(codiert) [] { + \begin{tikzpicture}[] + % Beschriftung Rechts + \begin{axis}[axis x line= none, axis y line*=right,ytick={0}] + \addplot[color=white] {0}; + \end{axis} + + \begin{axis}[ title = {\Large {Codiert}}, axis y line*=left] + \addplot[color=black!60!green] table[col sep=comma] {tikz/codiert.txt}; + \end{axis} + \end{tikzpicture}}; \\ + + \node(decodiert) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Decodiert}}] + \addplot[blue] table[col sep=comma] {tikz/decodiert.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(empfangen) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Empfangen \space + \space Fehler}}, + xtick={0,40,60,100}, axis y line*=left] + \addplot[color=black!60!green] table[col sep=comma] {tikz/empfangen.txt}; + \end{axis} + \begin{axis}[xtick={7,21,75}, axis y line*=right] + \addplot[red] table[col sep=comma] {tikz/fehler.txt}; + \end{axis} + \end{tikzpicture}};\\ + + \node(syndrom) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Syndrom}}] + \addplot[black] table[col sep=comma] {tikz/syndrom.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(locator) [] { + \begin{tikzpicture} + % Beschriftung Rechts + \begin{axis}[axis x line= none, axis y line*=right, ytick={0.3}]; + \addplot[color=black!60] {0.3}; + \end{axis} + + \begin{axis}[title = {\Large {Locator}},axis y line*=left] + \addplot[gray] table[col sep=comma] {tikz/locator.txt}; + \end{axis} + \end{tikzpicture}};\\ + }; + %------------------------------------------------------------- + %FFT & IFFT deskription + + \draw[thin,gray,dashed] (0,9) to (0,-9); + \node(IFFT) [scale=0.9] at (0,9.3) {IFFT}; + \draw[stealth-](IFFT.south west)--(IFFT.south east); + \node(FFT) [scale=0.9, above of=IFFT] {FFT}; + \draw[-stealth](FFT.north west)--(FFT.north east); + + %Arrows + \draw[thick, ->] (signal.east) to (codiert.west); + \draw[thick, ->] (codiert.south) to (empfangen.north); + \draw[thick, ->] (empfangen.west) to (decodiert.east); + \draw[thick, ->] (syndrom.east) to (locator.west); + \draw[thick](decodiert.south east)++(-1.8,1) ellipse (1.3cm and 0.8cm) ++(-1.3,0) coordinate(zoom) ; + \draw[thick, ->] (zoom) to[out=180, in=90] (syndrom.north); + + %item + \node[circle, draw, fill =lightgray] at (signal.north west) {1}; + \node[circle, draw, fill =lightgray] at (codiert.north west) {2}; + \node[circle, draw, fill =lightgray] at (empfangen.north west) {3}; + \node[circle, draw, fill =lightgray] at (decodiert.north west) {4}; + \node[circle, draw, fill =lightgray] at (syndrom.north west) {5}; + \node[circle, draw, fill =lightgray] at (locator.north west) {6}; +\end{tikzpicture} +\end{document}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/tikz/plotfftraw.tex b/buch/papers/reedsolomon/tikz/plotfftraw.tex new file mode 100644 index 0000000..db35734 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/plotfftraw.tex @@ -0,0 +1,81 @@ + +\begin{tikzpicture}[] + + %--------------------------------------------------------------- + %Knote + \matrix(m) [draw = none, column sep=25mm, row sep=2mm]{ + + \node(signal) [] { + \begin{tikzpicture} + \begin{axis} + [title = {\Large {Signal}}, + xtick={0,20,40,64,80,98}] + \addplot[blue] table[col sep=comma] {tikz/signal.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(codiert) [] { + \begin{tikzpicture}[] + \begin{axis}[ title = {\Large {Codiert \space + \space Fehler}}, + xtick={0,40,60,100}, axis y line*=left] + \addplot[green] table[col sep=comma] {tikz/codiert.txt}; + \end{axis} + \begin{axis}[xtick={7,21,75}, axis y line*=right] + \addplot[red] table[col sep=comma] {tikz/fehler.txt}; + \end{axis} + \end{tikzpicture}}; \\ + + \node(decodiert) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Decodiert}}] + \addplot[blue] table[col sep=comma] {tikz/decodiert.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(empfangen) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Empfangen}}] + \addplot[green] table[col sep=comma] {tikz/empfangen.txt}; + \end{axis} + \end{tikzpicture}};\\ + + \node(syndrom) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Syndrom}}] + \addplot[black] table[col sep=comma] {tikz/syndrom.txt}; + \end{axis} + \end{tikzpicture}}; & + + \node(locator) [] { + \begin{tikzpicture} + \begin{axis}[title = {\Large {Locator}}] + \addplot[gray] table[col sep=comma] {tikz/locator.txt}; + \end{axis} + \end{tikzpicture}};\\ + }; + %------------------------------------------------------------- + %FFT & IFFT deskription + + \draw[thin,gray,dashed] (0,9) to (0,-9); + \node(IFFT) [scale=0.9] at (0,9.3) {IFFT}; + \draw[stealth-](IFFT.south west)--(IFFT.south east); + \node(FFT) [scale=0.9, above of=IFFT] {FFT}; + \draw[-stealth](FFT.north west)--(FFT.north east); + + \draw[thick, ->,] (codiert)++(-1,0) +(0.05,0.5) -- +(-0.1,-0.1) -- +(0.1,0.1) -- +(0,-0.5); + %Arrows + \draw[thick, ->] (signal.east) to (codiert.west); + \draw[thick, ->] (codiert.south) to (empfangen.north); + \draw[thick, ->] (empfangen.west) to (decodiert.east); + \draw[thick, ->] (syndrom.east) to (locator.west); + \draw[thick](decodiert.south east)++(-1.8,1) ellipse (1.3cm and 0.8cm) ++(-1.3,0) coordinate(zoom) ; + \draw[thick, ->] (zoom) to[out=180, in=90] (syndrom.north); + + %item + \node[circle, draw, fill =lightgray] at (signal.north west) {1}; + \node[circle, draw, fill =lightgray] at (codiert.north west) {2+3}; + \node[circle, draw, fill =lightgray] at (empfangen.north west) {4}; + \node[circle, draw, fill =lightgray] at (decodiert.north west) {5}; + \node[circle, draw, fill =lightgray] at (syndrom.north west) {6}; + \node[circle, draw, fill =lightgray] at (locator.north west) {7}; +\end{tikzpicture}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/tikz/polynom2.tex b/buch/papers/reedsolomon/tikz/polynom2.tex new file mode 100644 index 0000000..80557fb --- /dev/null +++ b/buch/papers/reedsolomon/tikz/polynom2.tex @@ -0,0 +1,60 @@ +% polynome +%------------------- + +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{pgfplots} + + +\begin{document} +% Teiler für das Skalieren der Grafik /40 +\newcommand{\teiler}{40} + + +%////////////////////////////////////// + +\begin{tikzpicture}[>=latex,thick,] + \draw[color=blue, line width=1.4pt] + plot[domain=0:8, samples=100] + ({\x},{(2*\x^2+1*\x+5)/\teiler}); + + \draw[->] (-0.2,0) -- (8,0) coordinate[label={$x$}]; + \draw[->] (0,-0.2) -- (0,150/\teiler) coordinate[label={right:$p(x)$}]; + + \def\punkt#1{ + \fill[color=green] #1 circle[radius=0.08]; + \draw #1 circle[radius=0.07]; + } + + \def\hellpunkt#1{ + \fill[color=lightgray] #1 circle[radius=0.08]; + \draw[gray] #1 circle[ radius=0.07]; + } + + \draw[color=gray,line width=1pt,dashed] + plot[domain=0.5:7, samples=100] + ({\x},{(7.832*\x^2-51.5*\x+121.668)/\teiler}); + + + \punkt{(1,8/\teiler)} + \hellpunkt{(2,15/\teiler)} + \hellpunkt{(3,26/\teiler)} + \punkt{(4,41/\teiler)} + \punkt{(5,60/\teiler)} + \punkt{(6,83/\teiler)} + \punkt{(7,110/\teiler)} + + + + \def\erpunkt#1{ + \fill[color=red] #1 circle[radius=0.08]; + \draw #1 circle[radius=0.07]; + } + \erpunkt{(2,50/\teiler)} + \erpunkt{(3,37.66/\teiler)} + + \draw(0,100/\teiler) -- (-0.1,100/\teiler) coordinate[label={left:$100$}]; + \draw(1,0) -- (1,-0.1) coordinate[label={below:$1$}]; +\end{tikzpicture} +\end{document} diff --git a/buch/papers/reedsolomon/tikz/polynomraw.tex b/buch/papers/reedsolomon/tikz/polynomraw.tex new file mode 100644 index 0000000..02968fd --- /dev/null +++ b/buch/papers/reedsolomon/tikz/polynomraw.tex @@ -0,0 +1,50 @@ +% polynomraw + +\newcommand{\teiler}{40} + + +%////////////////////////////////////// + +\begin{tikzpicture}[>=latex,thick,] + \draw[color=blue, line width=1.4pt] + plot[domain=0:8, samples=100] + ({\x},{(2*\x^2+1*\x+5)/\teiler}); + + \draw[->] (-0.2,0) -- (8,0) coordinate[label={$x$}]; + \draw[->] (0,-0.2) -- (0,150/\teiler) coordinate[label={right:$p(x)$}]; + + \def\punkt#1{ + \fill[color=green] #1 circle[radius=0.08]; + \draw #1 circle[radius=0.07]; + } + + \def\hellpunkt#1{ + \fill[color=lightgray] #1 circle[radius=0.08]; + \draw[gray] #1 circle[ radius=0.07]; + } + + \draw[color=gray,line width=1pt,dashed] + plot[domain=0.5:7, samples=100] + ({\x},{(7.832*\x^2-51.5*\x+121.668)/\teiler}); + + + \punkt{(1,8/\teiler)} + \hellpunkt{(2,15/\teiler)} + \hellpunkt{(3,26/\teiler)} + \punkt{(4,41/\teiler)} + \punkt{(5,60/\teiler)} + \punkt{(6,83/\teiler)} + \punkt{(7,110/\teiler)} + + + + \def\erpunkt#1{ + \fill[color=red] #1 circle[radius=0.08]; + \draw #1 circle[radius=0.07]; + } + \erpunkt{(2,50/\teiler)} + \erpunkt{(3,37.66/\teiler)} + + \draw(0,100/\teiler) -- (-0.1,100/\teiler) coordinate[label={left:$100$}]; + \draw(1,0) -- (1,-0.1) coordinate[label={below:$1$}]; +\end{tikzpicture}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/tikz/signal.txt b/buch/papers/reedsolomon/tikz/signal.txt new file mode 100644 index 0000000..c4fa5f8 --- /dev/null +++ b/buch/papers/reedsolomon/tikz/signal.txt @@ -0,0 +1,96 @@ +0,6 +1,6 +2,0 +3,6 +4,4 +5,0 +6,5 +7,2 +8,1 +9,2 +10,1 +11,2 +12,0 +13,6 +14,3 +15,5 +16,7 +17,5 +18,5 +19,4 +20,1 +21,5 +22,9 +23,9 +24,3 +25,2 +26,6 +27,6 +28,4 +29,2 +30,9 +31,1 +32,1 +33,1 +34,2 +35,6 +36,6 +37,1 +38,9 +39,7 +40,7 +41,1 +42,9 +43,9 +44,10 +45,9 +46,8 +47,5 +48,2 +49,4 +50,1 +51,0 +52,9 +53,3 +54,3 +55,3 +56,5 +57,6 +58,0 +59,8 +60,6 +61,9 +62,3 +63,4 +64,0 +65,0 +66,0 +67,0 +68,0 +69,0 +70,0 +71,0 +72,0 +73,0 +74,0 +75,0 +76,0 +77,0 +78,0 +79,0 +80,0 +81,0 +82,0 +83,0 +84,0 +85,0 +86,0 +87,0 +88,0 +89,0 +90,0 +91,0 +92,0 +93,0 +94,0 +95,0 diff --git a/buch/papers/reedsolomon/tikz/syndrom.txt b/buch/papers/reedsolomon/tikz/syndrom.txt new file mode 100644 index 0000000..8ca9eed --- /dev/null +++ b/buch/papers/reedsolomon/tikz/syndrom.txt @@ -0,0 +1,96 @@ +0,0 +1,0 +2,0 +3,0 +4,0 +5,0 +6,0 +7,0 +8,0 +9,0 +10,0 +11,0 +12,0 +13,0 +14,0 +15,0 +16,0 +17,0 +18,0 +19,0 +20,0 +21,0 +22,0 +23,0 +24,0 +25,0 +26,0 +27,0 +28,0 +29,0 +30,0 +31,0 +32,0 +33,0 +34,0 +35,0 +36,0 +37,0 +38,0 +39,0 +40,0 +41,0 +42,0 +43,0 +44,0 +45,0 +46,0 +47,0 +48,0 +49,0 +50,0 +51,0 +52,0 +53,0 +54,0 +55,0 +56,0 +57,0 +58,0 +59,0 +60,0 +61,0 +62,0 +63,0 +64,0.0275599094902563 +65,0.0115837187254191 +66,0.025877761014238 +67,0.0224618032819697 +68,0.04410594689944 +69,0.0474504002669341 +70,0.0227694695500626 +71,0.0271436638090525 +72,0.0104166666666667 +73,0.0271436638090523 +74,0.0227694695500608 +75,0.0474504002669343 +76,0.0441059468994397 +77,0.0224618032819701 +78,0.0258777610142379 +79,0.0115837187254183 +80,0.027559909490256 +81,0.0245124379481793 +82,0.0499782237195209 +83,0.0401432022864265 +84,0.0232923747656228 +85,0.0237974288564099 +86,0.0143895905726624 +87,0.0271745729691685 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+79,0.0115837187254183 +80,0.027559909490256 +81,0.0245124379481793 +82,0.0499782237195209 +83,0.0401432022864265 +84,0.0232923747656228 +85,0.0237974288564099 +86,0.0143895905726624 +87,0.0271745729691685 +88,0.0275599094902567 +89,0.0515501672184983 +90,0.0358255004834542 +91,0.024700508366373 +92,0.0210194725405171 +93,0.0177592928994296 +94,0.0261327016093158 +95,0.0314909067039411 diff --git a/buch/papers/spannung/Einleitung.tex b/buch/papers/spannung/Einleitung.tex index b1588ff..8e0d36d 100644 --- a/buch/papers/spannung/Einleitung.tex +++ b/buch/papers/spannung/Einleitung.tex @@ -1,17 +1,18 @@ \section{Einleitung\label{spannung:section:Einleitung}} \rhead{Einleitung} Das Hook'sche Gesetz beschreibt die Beziehung von Spannung und Dehnung von linear-elastischen Materialien im Eindimensionalen. -In diesem Kapitel geht es darum das Hook'sche Gesetz im Dreidimensionalen zu beschreiben. +In diesem Kapitel geht es darum, das Hook'sche Gesetz im Dreidimensionalen zu beschreiben. Durch variable Krafteinwirkungen entstehen in jedem Punkt des Materials eine Vielzahl an unterschiedlichen Spannungen. In jedem erdenklichen Punkt im Dreidimensionalen herrscht daher ein entsprechender individueller Spannungszustand. Um das Hook'sche Gesetz für den 3D Spannungszustand formulieren zu können, reichen Skalare nicht aus. -Darum werden Vektoren, Matrizen und Tensoren zur Hilfe gezogen. +Darum werden Vektoren, Matrizen und Tensoren zu Hilfe gezogen. Mit diesen lässt sich eine Spannungsformel für den 3D Spannungszustand bilden. Diese Spannungsformel ist Grundlage für Computerprogramme und geotechnische Versuche, wie der Oedometer-Versuch. -Um die mathematische Untersuchung vorzunehmen, beschäftigt man sich zuerst mit den spezifischen Gegebenheiten und Voraussetzungen. -Ebenfalls gilt es ein paar wichtige Begriffe und deren mathematischen Zeichen einzuführen. -In diesem Kapitel gehen wir auch auf die Zusammenhänge von Spannung, Dehnungen und Verformungen an elastischen Materialien ein, +Um die mathematischen und physikalischen Berechnungen anwenden zu können, +müssen vorerst ein paar spezifische Bedingungen vorausgesetzt und Annahmen getroffen werden. +Ebenfalls gilt es, ein paar wichtige Begriffe und deren mathematischen Zeichen einzuführen. +In diesem Kapitel gehen wir auch auf die Zusammenhänge von Spannungen, Dehnungen und Verformungen an elastischen Materialien ein, wie sie in gängigen Lehrbüchern der Mechanik oder der Geotechnik behandelt werden, z.~B.~\cite{spannung:Grundlagen-der-Geotechnik}. \section{Spannungsausbreitung\label{spannung:section:Spannungsausbreitung}} @@ -29,7 +30,7 @@ Belastet man den Boden mit einer Spannung so wird diese in den Boden geleitet und von diesem kompensiert. Im Boden entstehen unterschiedlich hohe Zusatzspannungen. Diese Zusatzspannung breitet sich räumlich im Boden aus. -Im Falle einer konstanten Flächenlast $\sigma$ siehe Abbildung~\ref{spannung:Bild4} breitet sich die Zusatzspannung zwiebelartig aus. +Im Falle einer konstanten Flächenlast $\sigma$ siehe Abbildung~\ref{fig:Bild4} breitet sich die Zusatzspannung zwiebelartig aus. \begin{figure} \centering @@ -38,11 +39,11 @@ Im Falle einer konstanten Flächenlast $\sigma$ siehe Abbildung~\ref{spannung:Bi \label{fig:Bild4} \end{figure} -Mit der Tiefe $t$ nimmt diese permanent ab (siehe Abbildung~\ref{spannung:Bild5}). -Wie diese Geometrie der Ausbreitung ist, kann durch viele Modelle und Ansätze näherungsweise beschrieben werden. +Mit der Tiefe $t$ nimmt diese permanent ab (siehe Abbildung~\ref{fig:Bild5}). +Wie diese Geometrie der Ausbreitung aussieht, kann durch viele Modelle und Ansätze näherungsweise beschrieben werden. Diese Zusatzspannung $\sigma$ ist im Wesentlichen abhängig von $(x,y,t)$. Je nach Modell werden noch andere Parameter berücksichtigt. -Das können beispielsweise jenste Bodenkennwerte oder auch der Wassergehalt sein. +Das können beispielsweise verschiedene Bodenkennwerte oder auch der Wassergehalt sein. \begin{figure} \centering @@ -72,18 +73,18 @@ berechnet werden mit: t &= \text{Tiefe [\si{\meter}]} \\ s &= \text{Setzung, Absenkung [m].} \end{align*} -Diese Zusammenhänge sind wie erwähnt unter anderem im Lehrbuch [\cite{spannung:Grundlagen-der-Geotechnik}] beschrieben. +Diese Zusammenhänge sind wie erwähnt unter anderem im Lehrbuch \cite{spannung:Grundlagen-der-Geotechnik} beschrieben. In der praktischen Geotechnik wird man allerdings weitaus schwierigere Situationen antreffen. -Ein Beispiel wäre eine Baugrube mit einem Baugrubenabschluss, wo ein Teil des Bodens abgetragen ist (siehe Abbildung~\ref{spannung:Bild3}). +Ein Beispiel wäre eine Baugrube mit einem Baugrubenabschluss, wo ein Teil des Bodens abgetragen ist (siehe Abbildung~\ref{fig:Bild3}). Die Ausbreitung der Zusatzspannung $\sigma(x,y,t)$ würde hier deutlich komplizierter ausfallen. Dies bedeutet auch eine komplexere Setzung der Bodenoberfläche infolge einer Flächenlast $\sigma$. Aus allen zusätzlichen Spannungen müssen die adäquaten Dehnungen mit Hilfe einer Spannungsgleichung berechnet werden. Diese beruht auf Annahmen nach Hooke auf einem linear-elastischen Boden. -Generell wird im Ingenieurwesen versucht Phänomene möglichst nach dem Hook'schen Gesetz abbilden zu können. +Generell wird im Bauingenieurwesen oder auch im Maschinenbau versucht, manche Phänomene möglichst nach dem Hook'schen Gesetz abbilden zu können. \begin{figure} \centering \includegraphics[width=0.45\linewidth,keepaspectratio]{papers/spannung/Grafiken/Bild3.png} - \caption{Beispiel eines Lastauftrags auf den Boden bei einer komplexeren Situation, welches kompliziertere Spannungsausbreitung zur Folge hat} + \caption{Beispiel eines Lastauftrags auf den Boden bei einer komplexeren Situation, welche kompliziertere Spannungsausbreitung zur Folge hat} \label{fig:Bild3} \end{figure} diff --git a/buch/papers/spannung/main.tex b/buch/papers/spannung/main.tex index bbdf730..d2aeda9 100644 --- a/buch/papers/spannung/main.tex +++ b/buch/papers/spannung/main.tex @@ -3,7 +3,7 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:spannung}} +\chapter{Dreidimensionaler Spannungszustand\label{chapter:spannung}} \lhead{Dreiachsiger Spannungszustand} \begin{refsection} \chapterauthor{Adrian Schuler und Thomas Reichlin} diff --git a/buch/papers/spannung/teil0.tex b/buch/papers/spannung/teil0.tex index 7647252..089c28e 100644 --- a/buch/papers/spannung/teil0.tex +++ b/buch/papers/spannung/teil0.tex @@ -1,9 +1,10 @@ \section{Der Spannungszustand\label{spannung:section:Der Spannungsustand}} \rhead{Der Spannungszustand} -Ein Spannungszustand ist durch alle Spannungen, welche in einem beliebigen Punkt im Körper wirken, definiert (siehe Abbildung~\ref{spannung:Bild2}). +Ein Spannungszustand ist durch alle Spannungen, welche in einem beliebigen Punkt im Körper wirken, definiert (siehe Abbildung~\ref{fig:Bild2}). Änderungen der äusseren Kräfte verändern die inneren Spannungszustände im Material. -Um alle Spannungen eines Punktes darstellen zu können, wird ein infinitesimales Bodenelement in Form eines Würfels modellhaft vorgestellt. -Man spricht auch von einem Elementarwürfel, da dieser elementar klein ist. +Um alle Spannungen eines Punktes darstellen zu können, +stellt man sich modellhaft ein infinitesimales Bodenelement in Form eines Würfels vor. +Man spricht auch von einem Elementarwürfel. \begin{figure} \centering @@ -15,19 +16,19 @@ Man spricht auch von einem Elementarwürfel, da dieser elementar klein ist. Es werden jeweils drei Seiten dieses Würfels betrachtet, wobei die drei gegenüberliegenden Seiten im Betrag die selben Spannungen aufweisen, sodass der Elementarwürfel im Gleichgewicht ist. Wäre dieses Gleichgewicht nicht vorhanden, käme es zu Verschiebungen und Drehungen. -Das infinitesimale Bodenteilchen hat die Koordinaten $1$, $2$, $3$. +Das infinitesimale Bodenteilchen hat die Koordinatenachsen $1$, $2$, $3$. Veränderungen der Normalspannungen können durch Schubspannungen kompensiert werden und umgekehrt. -So sind insgesamt neun verschiedene Spannungen möglich, wobei drei Normal- und sechs Schubspannungen sind. +So sind insgesamt neun verschiedene Spannungen möglich, konkret sind dies drei Normal- und sechs Schubspannungen. Normalspannungen wirken normal (mit rechtem Winkel) zur angreifenden Fläche und Schubspannungen parallel zur angreifenden Fläche. Alle Beträge dieser neun Spannungen am Elementarwürfel bilden den Spannungszustand. -Daraus können die äquivalenten Dehnungen $\varepsilon$ mit Hilfe des Hook'schen Gesetz berechnet werden. +Daraus können die äquivalenten Dehnungen $\varepsilon$ mit Hilfe des Hook'schen Gesetzes berechnet werden. Daher gibt es auch den entsprechenden Dehnungszustand. \section{Spannungszustand\label{spannung:section:Spannungsustand}} \rhead{Spannungszustand} -Im einachsigen Spannungszustand herrscht nur die Normalspannung $\sigma_{11}$ (siehe Abbildung~\ref{spannung:Bild1}). +Im einachsigen Spannungszustand herrscht nur die Normalspannung $\sigma_{11}$ (siehe Abbildung~\ref{fig:Bild1}). Das Hook'sche Gesetz beschreibt genau diesen 1D Spannungszustand. Nach Hooke gilt: \[ @@ -59,7 +60,7 @@ mit A &= \text{Fläche [\si{\meter\squared}].} \end{align*} Diese Beziehung gilt bei linear-elastischen Materialien, welche reversible Verformungen zulassen. -Es ist praktisch die relative Dehnung $\varepsilon$ anzugeben und nicht eine absolute Längenänderung $\Delta l$. +Es ist praktisch, die relative Dehnung $\varepsilon$ anzugeben und nicht eine absolute Längenänderung $\Delta l$. \begin{figure} \centering \includegraphics[width=0.35\linewidth,keepaspectratio]{papers/spannung/Grafiken/Bild1.png} @@ -73,10 +74,10 @@ Mithilfe vom Elastizitätsmodul $E$ als Proportionalitätskonstante lässt sich E\cdot\varepsilon \] beschreiben. -Im Falle, dass $E$ nicht konstant ist, kann dieser näherungsweise durch +Im Falle, dass $E$ nicht konstant ist, wird dieser durch \[ E = -\frac{\Delta\sigma}{\Delta\varepsilon} +\frac{\text{d}\sigma}{\text{d}\varepsilon} \] -ausgedrückt werden.
\ No newline at end of file +ausgedrückt.
\ No newline at end of file diff --git a/buch/papers/spannung/teil1.tex b/buch/papers/spannung/teil1.tex index 74516c1..647b452 100644 --- a/buch/papers/spannung/teil1.tex +++ b/buch/papers/spannung/teil1.tex @@ -1,8 +1,8 @@ \section{Skalare, Vektoren, Matrizen und Tensoren\label{spannung:section:Skalare,_Vektoren,_Matrizen_und_Tensoren}} \rhead{Skalare, Vektoren, Matrizen und Tensoren} -Der Begriff Tensor kann als Überbegriff, der mathematischen Objekte Skalar, Vektor und Matrix, betrachtet werden. +Der Begriff Tensor kann als Überbegriff der mathematischen Objekte Skalar, Vektor und Matrix, betrachtet werden. Allerdings sind noch höhere Stufen dieser Objekte beinhaltet. -Ein Skalar, ein Vektor oder eine Matrix ist daher auch ein Tensor. +Skalare, Vektoren oder Matrizen sind daher auch Tensoren. Ein Skalar ist ein Tensor 0. Stufe. Mit einem Vektor können mehrere Skalare auf einmal beschrieben werden. Ein Vektor hat daher die Stufe 1 und ist höherstufig als ein Skalar. @@ -14,11 +14,10 @@ Jede Stufe von Tensoren verlangt andere Rechenregeln. So zeigt sich auch der Nachteil von Tensoren mit Stufen höher als 2. Man ist also bestrebt höherstufige Tensoren mit Skalaren, Vektoren oder Matrizen zu beschreiben. -Der Begriff Tensor wurde 1840 von Rowan Hamilton in die Mathematik eingeführt. +In den 40er Jahren vom 19. Jahrhundert wurde der Begriff Tensor von Rowan Hamilton in die Mathematik eingeführt. James Clerk Maxwell hat bereits mit Tensoren operiert, ohne den Begriff Tensor gekannt zu haben. Erst Woldemar Voigt hat den Begriff in die moderne Bedeutung von Skalar, Matrix und Vektor verallgemeinert. Er hat in der Elastizitätstheorie als erstes Tensoren eingesetzt und beschrieben. Auch Albert Einstein hat solche Tensoren eingesetzt, um in der Relativitätstheorie die Änderung der 4D Raumzeit beschreiben zu können. \cite{spannung:Tensor} -\cite{spannung:Voigtsche-Notation} diff --git a/buch/papers/spannung/teil2.tex b/buch/papers/spannung/teil2.tex index 6326eab..8620afe 100644 --- a/buch/papers/spannung/teil2.tex +++ b/buch/papers/spannung/teil2.tex @@ -3,7 +3,7 @@ Durch komplexe Spannungsausbreitungen im Boden entstehen im 3D Spannungszustand unterschiedliche Normal- und Schubspannungen. \begin{figure} \centering - \includegraphics[width=0.4\linewidth,keepaspectratio]{papers/spannung/Grafiken/infinitesimalerWuerfel.png} + \includegraphics[width=0.30\linewidth,keepaspectratio]{papers/spannung/Grafiken/infinitesimalerWuerfel.png} \caption{Beispiel eines Spannungszustandes; Vergrösserung eines infinitesimalen Bodenteilchen} \label{fig:infinitesimalerWuerfel} \end{figure} @@ -49,7 +49,7 @@ Der Dehnungstensor ist ebenfalls ein Tensor 2. Stufe und kann somit auch als $3\ dargestellt werden und beschreibt den gesamten Dehnungszustand. Der Spannungs- und Dehnungstensor 2. Stufe kann je in einen Tensor 1. Stufe überführt werden, welches ein Spaltenvektor ist. -Gemäss der Hadamard-Algebra dürfen Zeile um Zeile in eine Spalte notiert werden, sodass es einen Spaltenvektor ergibt. +Man darf Zeile um Zeile in eine Spalte notieren, sodass es einen Spaltenvektor ergibt. So ergibt sich der Spannungsvektor \[ @@ -79,7 +79,7 @@ So ergibt sich der Spannungsvektor \sigma_{33} \end{pmatrix} \] -und Dehnungsvektor +und der Dehnungsvektor \[ \overline{\varepsilon} = @@ -140,14 +140,6 @@ C_{3311} & C_{3312} & C_{3313} & C_{3321} & C_{3322} & C_{3323} & C_{3331} & C_{ \end{pmatrix} \] geschrieben werden kann. -Dieser Elastizitätstensor muss für isotrope Materialien zwingend symmetrisch sein. -Folglich gilt: -\[ -\overline{\overline{C}} -= -\overline{\overline{C}}~^{T} -. -\] Die allgemeine Spannungsgleichung lautet nun: \[ \vec\sigma @@ -155,8 +147,7 @@ Die allgemeine Spannungsgleichung lautet nun: \overline{\overline{C}}\cdot\vec{\varepsilon} . \] - -Als Indexnotation +Sie kann ebenfalls als Indexnotation \[ \sigma_{ij} = @@ -164,7 +155,15 @@ Als Indexnotation \sum_{l=1}^3 C_{ijkl}\cdot\varepsilon_{kl} \] -kann dies ebenfalls geschrieben werden. +geschrieben werden. +Der Elastizitätstensor muss für isotrope Materialien zwingend symmetrisch sein. +Folglich gilt: +\[ +\overline{\overline{C}} += +\overline{\overline{C}}~^{T} +. +\] Die Konstanten $C$ werden nun nach dem Hook'schen Gesetz mit Hilfe des Elastizitätsmoduls $E$ definiert. Da dieser Modul durch die eindimensionale Betrachtung definiert ist, @@ -221,7 +220,7 @@ definiert ist. Trägt man die Konstanten in die Matrix ein, ergibt sich \end{pmatrix} . \] -Die Normalspannung $\sigma_{22}$ lässt sich exemplarisch als +Die Normalspannung $\sigma_{22}$ lässt sich zum Beispiel als \[ \sigma_{22} = @@ -229,11 +228,13 @@ Die Normalspannung $\sigma_{22}$ lässt sich exemplarisch als \] berechnen. +Reduzierte Spannungs- und Dehnungsgleichungen + Man betrachte nun die Eigenschaften des Elastizitätstensors. Dieser ist quadratisch und symmetrisch, die verschiedenen Einträge wechseln sich aber miteinander ab. Es ergeben sich keine Blöcke mit einheitlichen Einträgen. -Allerdings weiss man, dass im isotropen Boden der Spannungs-, Dehnungs- und daher auch Elastizitätstensor symmetrisch sind. +Allerdings weiss man, dass im isotropen Boden der Spannungs-, Dehnungs- und daher auch der Elastizitätstensor symmetrisch sind. Wäre dem nicht so, würde sich das Material je nach Richtung unterschiedlich elastisch verhalten. Diese Symmetrie setzt daher voraus, dass \[ @@ -399,7 +400,7 @@ Somit lässt sich die reduzierte allgemeine Spannungsgleichung mit \] beschreiben. Die Konstanten $C$ werden wieder nach dem Hook'schen Gesetz definiert. -Dies ergibt die Spannungsformel, welche weit möglichst vereinfacht ist: +Dies ergibt die Spannungsgleichung, welche weit möglichst vereinfacht ist: \begin{equation} \begin{pmatrix} \sigma_{11}\\ @@ -433,7 +434,7 @@ Dies ergibt die Spannungsformel, welche weit möglichst vereinfacht ist: Im Elastizitätstensor fallen zwei $3\times3$ Blöcke auf, welche nur Einträge mit $0$ haben. Der Tensor besagt also, dass diese jeweiligen Dehnungen keinen Einfluss auf unsere Spannung haben. -Man sieht nun auch ganz gut, dass sich im Vergleich zu der allgemeinen Spannungsgleichung, die Einträge verschoben haben. +Man sieht nun auch ganz gut, dass sich im Vergleich zu der allgemeinen Spannungsgleichung die Einträge verschoben haben. Da nach Voigt zuerst die Normalspannungen und anschliessend die Schubspannungen notiert worden sind, ergeben sich die $3\times3$ Blöcke. Man betrachte als Beispiel die Berechnung von $\sigma_{33}$. @@ -441,8 +442,8 @@ Es ist ersichtlich, dass die Schubdehnungen keinen Einfluss auf $\sigma_{33}$ ha Der Einfluss der zu $\sigma_{33}$ äquivalenten Dehnung $\varepsilon_{33}$ hat den grössten Einfluss. Die anderen Normalspannungen $\sigma_{11}$ und $\sigma_{22}$ haben einen unter anderem mit $\nu$ korrigierten Einfluss. -Von $\overline{\overline{C}}$ bildet man noch die inverse Matrix $\overline{\overline{C}}\mathstrut^{-1}$ um die Gleichung umstellen zu können. -Dadurch erhält man die Dehnungsgleichung: +Von $\overline{\overline{C}}$ bildet man die inverse Matrix $\overline{\overline{C}}\mathstrut^{-1}$, mithilfe des Gauss - Jordan Algorithmus, um die Gleichung umstellen zu können. +Durch einige Berechnungsschritte erhält man die Dehnungsgleichung: \[ \vec{\varepsilon} diff --git a/buch/papers/spannung/teil3.tex b/buch/papers/spannung/teil3.tex index 3e456c3..a9080ea 100644 --- a/buch/papers/spannung/teil3.tex +++ b/buch/papers/spannung/teil3.tex @@ -30,7 +30,7 @@ q \label{spannung:Invariante_q} . \end{equation} -Diese Zusammenhänge werden im Skript [\cite{spannung:Stoffgesetze-und-numerische-Modellierung-in-der-Geotechnik}] aufgezeigt. +Diese Zusammenhänge werden im Skript \cite{spannung:Stoffgesetze-und-numerische-Modellierung-in-der-Geotechnik} aufgezeigt. Die hydrostatische Spannung $p$ kann gemäss Gleichung \eqref{spannung:Invariante_p} als \[ p @@ -38,28 +38,28 @@ p \frac{\sigma_{11}+2\sigma_{33}}{3} \] vereinfacht werden. -Die deviatorische Spannung $q$ wird gemäss Gleichung \eqref{spannung:Invariante_q}als +Die deviatorische Spannung $q$ wird gemäss Gleichung \eqref{spannung:Invariante_q} als \[ q = \sigma_{11}-\sigma_{33} \] -vereinfacht. Man kann $p$ als Isotrop und $q$ als Schub betrachten. +vereinfacht. Man kann $p$ als Druck und $q$ als Schub betrachten. -Die Invarianten können mit der Spannungsformel \eqref{spannung:Spannungsgleichung} berechnet werden. +Die Invarianten $p$ und $q$ können mit der Spannungsgleichung \eqref{spannung:Spannungsgleichung} berechnet werden. Durch geschickte Umformung dieser Gleichung, lassen sich die Module als Faktor separieren. Dabei entstehen spezielle Faktoren mit den Dehnungskomponenten. So ergibt sich \[ -\overbrace{\frac{\sigma_{11}+2\sigma_{33}}{3}}^{p} +\overbrace{\frac{\sigma_{11}+2\sigma_{33}}{3}}^{\displaystyle{p}} = -\frac{E}{3(1-2\nu)} \overbrace{(\varepsilon_{11} - 2\varepsilon_{33})}^{\varepsilon_{v}} +\frac{E}{3(1-2\nu)} \overbrace{(\varepsilon_{11} - 2\varepsilon_{33})}^{\displaystyle{{\varepsilon_{v}}}} \] und \[ -\overbrace{\sigma_{11}-\sigma_{33}}^{q} +\overbrace{\sigma_{11}-\sigma_{33}}^{\displaystyle{q}} = -\frac{3E}{2(1+\nu)} \overbrace{\frac{2}{3}(\varepsilon_{11} - \varepsilon_{33})}^{\varepsilon_{s}} +\frac{3E}{2(1+\nu)} \overbrace{\frac{2}{3}(\varepsilon_{11} - \varepsilon_{33})}^{\displaystyle{\varepsilon_{s}}} . \] Die Faktoren mit den Dehnungskomponenten können so mit @@ -79,8 +79,8 @@ eingeführt werden, mit \varepsilon_{v} &= \text{Hydrostatische Dehnung [-]} \\ \varepsilon_{s} &= \text{Deviatorische Dehnung [-].} \end{align*} -Die hydrostatische Dehnung $\varepsilon_{v}$ kann mit einer Kompression verglichen werden. -Die deviatorische Dehnung $\varepsilon_{s}$ kann mit einer Verzerrung verglichen werden. +Die hydrostatische Dehnung $\varepsilon_{v}$ kann mit einer Kompression und +die deviatorische Dehnung $\varepsilon_{s}$ mit einer Verzerrung verglichen werden. Diese zwei Gleichungen kann man durch die Matrixschreibweise \begin{equation} @@ -90,8 +90,8 @@ Diese zwei Gleichungen kann man durch die Matrixschreibweise \end{pmatrix} = \begin{pmatrix} - \frac{3E}{2(1+\nu)} & 0 \\ - 0 & \frac{E}{3(1-2\nu)} + \displaystyle{\frac{3E}{2(1+\nu)}} & 0 \\ + 0 & \displaystyle{\frac{E}{3(1-2\nu)}} \end{pmatrix} \begin{pmatrix} \varepsilon_{s}\\ @@ -100,9 +100,11 @@ Diese zwei Gleichungen kann man durch die Matrixschreibweise \label{spannung:Matrixschreibweise} \end{equation} vereinfachen. -Man hat so eine Matrix multipliziert mit einem Vektor und erhält einen Vektor. -Änderungen des Spannungszustandes können mit dieser Gleichung vollumfänglich erfasst werden. +Änderungen des Spannungszustandes können mit diesen Gleichungen vollumfänglich erfasst werden. +Diese Spannungsgleichung mit den zwei Einträgen ($p$ und $q$) ist gleichwertig +wie die ursprüngliche Spannungsgleichung mit den neun Einträgen +($\sigma_{11}$, $\sigma_{12}$, $\sigma_{13}$, $\sigma_{21}$, $\sigma_{22}$, $\sigma_{23}$, $\sigma_{31}$, $\sigma_{32}$, $\sigma_{33}$). Mit dieser Formel \eqref{spannung:Matrixschreibweise} lassen sich verschieden Ergebnisse von Versuchen analysieren und berechnen. -Ein solcher Versuch, den oft in der Geotechnik durchgeführt wird, ist der Oedometer-Versuch. +Ein solcher Versuch, der oft in der Geotechnik durchgeführt wird, ist der Oedometer-Versuch. Im nächsten Kapitel wird die Anwendung der Matrix an diesem Versuch beschrieben. diff --git a/buch/papers/spannung/teil4.tex b/buch/papers/spannung/teil4.tex index 2f2e4ce..00b2d4f 100644 --- a/buch/papers/spannung/teil4.tex +++ b/buch/papers/spannung/teil4.tex @@ -1,6 +1,6 @@ -\section{Oedometer-Versuch\label{spannung:section:Oedometer-Versuch}} -\rhead{Oedometer-Versuch} -Mit dem Oedometer-Versuch kann der oedometrische Elastizitätsmodul $E_{OED}$ bestimmt werden. +\section{Oedometrischer Elastizitätsmodul\label{spannung:section:Oedometrischer Elastizitätsmodul}} +\rhead{Oedometrischer Elastizitätsmodul} +Mit dem Oedometer-Versuch kann der oedometrische Elastizitätsmodul $E_{\text{OED}}$ bestimmt werden. Dieser beschreibt ebenfalls das Verhältnis zwischen Spannung und Dehnung, allerdings unter anderen Bedingungen. Diese Bedingung ist das Verhindern der seitlichen Verformung, sprich der Dehnung in Richtung $1$ und $2$. Es wird ein Probeelement mit immer grösseren Gewichten belastet, welche gleichmässig auf das Material drücken. @@ -43,8 +43,8 @@ Diese lautet nun: \end{pmatrix} = \begin{pmatrix} - \frac{E_{OED}}{(1+\nu)} & 0 \\ - 0 & \frac{E_{OED}}{3(1-2\nu)} + \displaystyle{\frac{E_{\text{OED}}}{(1+\nu)}} & 0 \\ + 0 & \displaystyle{\frac{E_{\text{OED}}}{3(1-2\nu)}} \end{pmatrix} \begin{pmatrix} \varepsilon_{11}\\ @@ -52,28 +52,28 @@ Diese lautet nun: \end{pmatrix} . \] -Daraus lässt sich bei jedem Setzungsgrad der oedometrische Elastitzitätsmodul $E_{OED}$ und die seitlichen Spannungen $\sigma_{33}$ mit den 2 Gleichungen +Daraus lässt sich bei jedem Setzungsgrad der oedometrische Elastitzitätsmodul $E_{\text{OED}}$ und die seitlichen Spannungen $\sigma_{33}$ mit den zwei Gleichungen \[ \sigma_{11}-\sigma_{33} = -\frac{E_{OED}}{(1+\nu)}\cdot\varepsilon_{11} +\frac{E_{\text{OED}}}{(1+\nu)}\cdot\varepsilon_{11} \] und \[ \sigma_{11}+2\sigma_{33} = -\frac{E_{OED}}{3(1-2\nu)}\cdot\varepsilon_{11} +\frac{E_{\text{OED}}}{3(1-2\nu)}\cdot\varepsilon_{11} \] berechnen. -Mit diesen Gleichungen hat man das Gleichungssystem um $E_{OED}$ und $\sigma_{33}$ zu berechnen. +Mit diesen Gleichungen hat man das Gleichungssystem um $E_{\text{OED}}$ und $\sigma_{33}$ zu berechnen. Die Poisson-Zahl muss als Kennwert gemäss der Bodenklasse gewählt werden. -Den Versuch kann man auf einem $\sigma$-$\varepsilon$-Diagramm abtragen (siehe Abbildung~\ref{spannung:DiagrammOedometer-Versuch}). +Den Versuch kann man auf einem $\sigma$-$\varepsilon$-Diagramm abtragen (siehe Abbildung~\ref{fig:DiagrammOedometer-Versuch}). Durch die Komprimierung nimmt der Boden mehr Spannung auf, und verformt sich zugleich weniger stark. -Mit diesem ermittelten $E_{OED}$ kann man nun weitere Berechnungen für die Geotechnik durchführen. +Mit diesem ermittelten $E_{\text{OED}}$ kann man nun weitere Berechnungen für die Geotechnik durchführen. \begin{figure} \centering - \includegraphics[width=0.5\linewidth,keepaspectratio]{papers/spannung/Grafiken/DiagrammOedometer-Versuch.png} + \includegraphics[width=0.45\linewidth,keepaspectratio]{papers/spannung/Grafiken/DiagrammOedometer-Versuch.png} \caption{Diagramm Charakteristik verschiedener Elastizitätsmodule bei gleichem Material} \label{fig:DiagrammOedometer-Versuch} \end{figure}
\ No newline at end of file diff --git a/buch/papers/verkehr/main.tex b/buch/papers/verkehr/main.tex index 6348993..98d0581 100644 --- a/buch/papers/verkehr/main.tex +++ b/buch/papers/verkehr/main.tex @@ -3,8 +3,7 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:verkehr}} -\lhead{Verkehrsfluss und Verkehrsnetze} +\chapter{Verkehrsfluss und Verkehrsnetze\label{chapter:verkehr}} \begin{refsection} \chapterauthor{Pascal Andreas Schmid und Robine Luchsinger} diff --git a/buch/papers/verkehr/section1.tex b/buch/papers/verkehr/section1.tex index d96d450..8994066 100644 --- a/buch/papers/verkehr/section1.tex +++ b/buch/papers/verkehr/section1.tex @@ -1,118 +1,98 @@ -\section{Einführung} \label{section:verkehr/einfuehrung} -\subsection{Verkehrsnetze} Das Verkehrsnetz besteht aus allen Anlagen, auf oder unter der Erdoberfläche, auf denen eine räumliche Fortbewegung von Personen oder auch Gütern stattfindet. Verkehrsnetze sind ein Bestandteil der Verkehrsinfrastruktur, die auf topografischen Karten festgehalten werden. Sie umfassen den Schienenverkehr, alle Strassen und Wege, wie auch Flugplätze und alle dazugehörigen Bauwerke. Aus verkehrsgeografischer Sicht besteht das Verkehrsnetz aus Kanten, Knotenpunkten und dem Hinterland. Die Knotenpunkte werden auch hier durch die Kanten verbunden, die den Verkehrsstrom aufnehmen, wobei das Hinterland durch einzelne Knoten versorgt wird. Die Aufteilung in Kanten und Knotenpunkte ermöglicht eine Vereinfachung komplexer Verkehrsnetze, damit sie mittels der Graphentheorie untersucht werden können. Grundsätzlich können kurze Wege zwischen den Knotenpunkten das Ziel beim Aufbau eines Verkehrsnetzes sein. Es kann aber auch versucht werden, die Bau- und Unterhaltskosten des Verkehrsnetzes in einem gewissen Rahmen zu halten. Aus diesen Vorgaben ergibt sich dann, je nach dem was gewünscht wird, eine grob- oder feinmaschige Struktur des Netzes. Ziel ist aber ein möglichst wirtschaftliches und optimales Verkehrsnetz. -\subsection{Suchalgorithmen} +\section{Suchalgorithmen} +Inbesondere bei Graphen in Form von Verkehrsnetzen ist das Finden eines kürzesten Weges von Interesse. Mathematisch betrachtet handelt es sich hierbei um ein Optimierungsproblem, bei dem die Summe der Kantengewichte zwischen zwei Knoten minimiert werden soll. Zu diesem Zweck existieren verschiedene Suchalgorithmen. In den folgenden Abschnitten wird auf eine Auswahl davon eingegangen. Zuvor ist es jedoch notwendig, einige Begriffe und Eigenschaften von Suchalgorithmen zu definieren. -\subsubsection{Dijkstra-Algorithmus} -Der Algorithmus von Dijkstra ist benannt nach seinem Erfinder dem Mathematik- und Infomratikprofessor Edsger Dijkstra. Den Algorithmus hat er im Jahr 1959 erfunden. -Der Algorithmus von Dijkstra ist ein Greedy-Algorithmus (gieriger Algorithmus), der schrittweise einen Folgezustand auswählt, damit beim Zeitpunkt der Wahl der grösste Gewinn bzw. das beste Ergebnis erzielt werden kann. -Trotz der Schnelligkeit der Greedy-Algorithmen, können viele Probleme nicht optimal gelöst werden. -Vereinfacht wird beim Dijkstra-Algorithmus, ausgehend von einem Startknoten so lange dem kürzesten Pfad gefolgt, bis der Zielknoten erreicht wird. Dabei muss für jeden besuchten Knoten die Kostenfunktion als auch der Pfad dahin (vorheriger Knoten) gespeichert werden. -Dadurch wird hingegen garantiert, dass, wenn der Zielknoten erreicht wird, auch der kürzeste Pfad gefunden wurde. -Grundlegende Voraussetzung für den Dijkstra-Algorithmus ist die strikte Positivität der Kantengewichte. Andernfalls würde ein wiederholtes Ablaufen einer Kante mit negativem Gewicht zu einer stetigen Reduktion der Kostenfunktion führen, was zu einer unendlichen Schlaufe führen würde. +Einerseits wird zwischen optimalen und nicht-optimalen Algorithmen unterschieden. Ein Suchalgorithmus gilt als optimal, falls er einen günstigsten Pfad zwischen zwei Knoten findet. Es gilt zu beachten, dass im Falle des Vorhandenseins von mehrerern Pfaden mit identischer, minimaler Summe der Kantengewichte zwischen zwei Knoten, mindestens einer dieser Pfade gefunden wird. -Gegeben sei ein Netzwerk mit $n$ Knoten und dem Startknoten $a$. -Alle Kanten sind mit $k(i, j)$ bewertet. -Gesucht wird der kürzeste Pfad zwischen dem Startknoten und allen übrigen Knoten im Netz. -$D(i)$ ist die kürzeste Distanz vom Startknoten $a$ zum Knoten $i, V(i)$ ist der unmittelbare Vorgängerknoten vom Knoten $i$ auf dem kürzesten Weg vom Startknoten $a$ zum Konten $i$ und die Menge $M$ ist die Menge einer bestimmten Auswahl an Knoten. +Weiter wird zwischen informierten und uninformierten Algorithmen differenziert. Während uninformierte Suchalgorithmen den Suchraum schematisch auf Basis der Eigenschaften des Graphen absuchen, bis eine günstigste Lösung gefunden wurde, verwenden informierte Suchalgorithmen eine Heuristik zur Abschätzung der Suchrichtung. Oftmals wird bei informierten Algorithmen ein Verlust der Optimalität zugunsten einer verbesserten Rechenzeit in Kauf genommen. Es exisitieren jedoch auch Heurstiken, die eine optimale Lösung gewährleisten. -Dabei gilt -\begin{equation}M={a}\end{equation} -\begin{equation}D(a)=0\end{equation} wobei -\begin{equation}D(i)=\infty\end{equation} und -\begin{equation}i \neq a \end{equation} -Ausserdem gilt \begin{equation}V(i)=(-) \text{für alle Knoten $i$}\end{equation}\\ +Eine besondere Art von Suchalgorithmen stellen die sogenannten Greedy-Algorithmen, zu deutsch gierige Algorithmen, dar. Sie zeichnen sich dadurch aus, dass sie stets den zurzeit günstigsten Folgezustand auswählen. Dadurch sind sie in der Regel äusserst effizient, garantieren bei vielen Problemstellungen jedoch keine optimale Lösung. -%THEORIE... -Iteration +\subsection{Dijkstra-Algorithmus} +Der Algorithmus von Dijkstra ist benannt nach seinem Erfinder dem Mathematik- und Informatikprofessor Edsger Dijkstra. Er gehört zur Klasse der uninformierten Greedy-Algorithmen. Zudem ist die Optimalität bei strikt positiven Kantengewichten gewährleistet. +Vorteilhaft ist die einfache Implementierung. Abhängig von der Programmiersprache sind zwischen 30 und 40 Zeilen an Code ausreichend, damit er den kürzesten Pfad zwischen einem Startknoten $a$ und Zielknoten $b$ finden kann. -1. Auswahl eines Knotens \begin{equation} K\in M \text{mit} D(K)=D(i);i\in M\end{equation} +Die für dieses Paper verwendete programmierte Funktion (MATLAB) verwendet eine abgewandelte Form der gewichteten Adjazenz-Matrix $A$, für welche gilt: +Der Matrix-Eintrag $A_{i,j}$ enthält das Kantengewicht der Kante von Knoten $j$ nach $i$ auf. Falls keine Kante zwischen $j$ und $i$ vorhanden ist, beträgt der Eintrag $\infty$. Dies vereinfacht die Implementierung zur Bestimmung des nächst-günstigsten Pfades. +Zudem werden zwei Hilfs-Vektoren $\vec{d}$ und $\vec{b}$ der Länge $n$ eingeführt, wobei $n$ die Anzahl Knoten des Graphen ist. Im Vektoreintrag $\vec{d}(i)$ wird das kummulierte Kantengewicht zur Erreichung von Knoten $i$ vom Startknoten $a$ gespeichert. Der Eintrag $\vec{d}(a)$ beträgt somit $0$. Im Vektor $\vec{b}$ wird zudem vermerkt, falls ein Knoten bereits als Ziel eines kürzesten Pfads gefunden wurde und somit für die weitere Suche nicht mehr berücksichtigt werden muss ($\vec{b}(i)=1$, sonst $\vec{b}(i)=0$). -2. Für alle Nachfolger $N(j)$ vom Knoten $K$ gilt: -\begin{equation}D(K) + k_Kj < D(j)\end{equation} dann wird \begin{equation}D(j) = D(K) + k_Kj, V(j) = K\end{equation} gesetzt und somit wird der Knoten $j$ in die Menge $M$ aufgenommen. +Ausgehend vom Startknoten $a$ wird nun anhand der Matrix $A$ in der Spalte $a$ nach dem kleinsten Eintrag gesucht. Somit wird der Folgeknoten $c$ gefunden. Dieser Vorgang wird nun wiederholt, wobei jedoch sämtliche von Knoten $a$ und $c$ erreichbaren Knoten berücksichtigt werden, die noch nicht besucht wurden. In anderen Worten alle nicht verschwindenden Einträge $i$ der Spalten $a$ und $c$ der Matrix $A$, für welche gilt $\vec{b}(i)=0$. Ausschlaggebend für die folgende Auswahl ist die Summe der kummulierten Kantengewichte und des Kantengewichts des nächsten Knotens. Als Beispiel zur Erreichung von Knoten $k$ über Knoten $j$: +\begin{equation} +\vec{d}(k)=\vec{d}(j)+A(k,j) +\end{equation} +Diese Iteration wird solange durchgeführt, bis der Folgeknoten dem Zielknoten entspricht. -3. Der ausgewählte Knoten \begin{equation}K\in M\text{wird aus der Menge herausgelöscht}\end{equation}\\ -Diese drei Schritte werden so lange wiederholt bis gilt -\begin{equation}M=\{\}\end{equation} +\subsection{A*-Algorithmus} +Der A*-Algorithmus basiert auf dem Dijkstra-Algorithmus, verwendet jedoch eine Heuristik zur Abschätzung der günstigsten Suchrichtung. Somit handelt es sich um einen informierten Greedy-Algorithmus, der abhängig von der verwendeten Heuristik auch optimal sein kann. Er wurde von Peter Hart, Nils Nilsson und Bertram Raphael entwickelt. -\subsubsection{A*-Algorithmus} -Suchalgorithmen werden nach einfachen (uninformierte) und heuristischen (informierten) Algorithmen unterschieden. Während einfache Algorithmen den Suchraum intuitiv durchsuchen, beziehen heuristische Algorithmen Wissen über den Suchraum mit ein. -Der A*-Algorithmus geht auf seine Erfinder Peter Hart, Nils Nilsson und Bertram Raphael zurück, die den Algorithmus erstmals im Jahr 1968 beschrieben. -Der A*-Algorithmus ist ein heuristischer Suchalgorithmus, der den kürzesten Pfad zwischen zwei Knoten in einem Graphen mit positiven Kantengewichten berechnet. -Im Gegensatz zu einfachen Suchalgorithmen, wird beim A*-Algorithmus eine Schätzfunktion, die sogenannte Heuristik, verwendet. Dies ermöglicht ein zielgerichtetes Suchen und gleichzeitig wird die Laufzeit verringert. -Ausserdem findet der A*-Algorithmus immer eine optimale Lösung, sofern eine vorhanden ist. -Der A*-Algorithmus wird als Verallgemeinerung gehandhabt und gilt als Erweiterung des Dijkstra-Algorithmus. +\subsection{Anwendung A*-Algorithmus} +Wie oben erwähnt basiert der A*-Algorithmus auf dem Shortest-Path-Algorithmus von Dijkstra. Gemäss dem Algorihtmus von Dijkstra werden von einem Startknoten aus die jeweiligen Nachbarknoten, die Nachbarknoten der Nachbarknoten usw. verarbeitet. Die Kantengewichte werden dabei aufsummiert und die Priorität wird auf die Kante gelegt, die das geringste Gewicht aufweist. Mit diesem Verfahren wird sichergestellt, dass die erste gefundene Lösung auch eine optimale Lösung darstellt.\\ -\subsubsection{Anwendung A*-Algorithmus} -Wie oben erwähnt basiert der A*-Algorithmus auf dem Shortest-Path-Algorithmus von Dijkstra. Gemäss dem Algorihtmus von Dijkstra werden von einem Startknoten aus die jeweiligen Nachbarknoten, die Nachbarknoten der Nachbarknoten usw. verarbeitet. Die Kantengewichte werden dabei aufsummiert und die Priorität wird auf die Kante gelegt, die das geringste Gewicht aufweist. Mit diesem Verfahren wird sichergestellt, dass die erste gefundene Lösung auch die optimalste Lösung darstellt.\\ +Der A*-Algorithmus unterscheidet sich vom Dijkstra-Algorithmus dahingehend, dass bei der Auswahl des Folgeknotens, nicht nur die Summe der Kantengewichte $\vec{d}(j)+A(k,j)$, sondern zusätzlich die für jeden Knoten definierte Abschätzfunktion $f(k)$ hinzuaddiert wird. Dies passiert jedoch nur bei der \emph{Auswahl} des Folgeknotens. Der Wert von $f(k)$ wird nicht im Eintrag $\vec{d}(k)$ gespeichert. Somit wird gewährleistet, dass der gefundene Pfad, der Summe der Kantengewichte entspricht. Ein Beispiel dafür, wie eine Abschätzfunktion gebildet werden kann findet sich in Abschnitt \ref{sec:verkehr/euklidische} -Die Kantengewichte werden für jeden Knoten in Form einer Funktion dargestellt -\begin{equation}f(n)=g(n)\end{equation} mit -\begin{equation}g(n)=\text{Summe aller Kantengewichte vom Startknoten bis n}\end{equation}\\ -Der A*-Algorithmus erweitert die Vorgehensweise des Algorithmus von Dijkstra um die Heuristik $h(n)$, die für jeden Knoten $n$ die geschätzte Entfernung zum Zielknoten beschreibt. -Somit gilt: -\begin{equation}f(n)=g(n)+h(n)\end{equation}\\ -Wie auch der Algorithmus von Dijkstra findet der A*-Algorithmus die optimalste Lösung. +\subsection{Euklidische Heuristik} +\label{sec:verkehr/euklidische} +Bei Verkehrsnetzen ist die euklidische Distanz eine gängige und zuverlässige Heurstik. Dabei wird zu den effektiven Reisekosten zum aktuellen Knoten die euklidische Distanz bis zum Zielknoten hinzuaddiert. Dadurch wird die Kostenfunktion konsequent nie überschätzt. Dies stellt eine Voraussetzung an eine zulässige Heuristik dar. Unter Verwendung dieser Heuristik gilt der A*-Algorithmus als optimal. -\subsubsection{Floyd-Warshall-Algorithmus} -Der Floyd-Warshall-Algorithmus, auch Tripel-Algorithmus genannt, wurde erstmals im Jahr 1962 von seinen Namensgebern Robert Floyd und Stephen Warshall vorgestellt. -Der Floyd-Warshall-Algorithmus sucht kürzeste Wege innerhalb eines Graphen. Er ermittelt aber nicht nur die Distanz zwischen zwei Knoten, sondern berechnet die kürzesten Wege zwischen allen Knotenpaaren eines gewichteten Graphen. Somit werden die kürzesten , beziehungsweise die optimalsten Wege zwischen allen Paaren von Knoten berechnet. Der Floyd-Warhshall-Algrithmus kann ausserdem mit negativen Kantengewichten umgehen, sofern der Graph aber keinen negativen Kreis (Zyklus) aufweist. Ist dies der Fall, führt der Algorithmus zu einem falschen Ergebnis. -Ein Kreis (Zyklus) in einem Graphen ist ein Weg, bei dem Start- und Endpunkt den gleichen Knoten aufweisen. Dieser wird negativ, wenn die Summe der gewichteten Kanten kleiner als Null wird.\\ -Der Floyd-Warshall-Algorithmus besteht grundsätzlich aus Floyd's Berechnung der kürzesten Distanzen zwischen zwei Knoten und Warshall's Konstruktion der kürzesten Wege. Werden diese beiden Teilgebiete zusammengefügt, ergibt sich der Floyd-Warshall-Algorithmus. +Bei der euklidischen Heuristik wird die Abschätzfunktion $f(k)$ für jeden Knoten $k$ durch euklidische Distanz zum Zielknoten $b$ gebildet. +\begin{equation} +f(k)=\sqrt{(x_k-x_b)^2+(y_k-y_b)^2} +\end{equation} + +Was bei einem physischen Verkehrsnetz einfach zu bewältigen ist, da Koordinaten von Verkehrsnetzen zur Berechnung der Distanz verwendet werden können, ist bei virtuellen Netzwerken (z.B. Servernetzen) entweder nicht möglich, oder nicht relevant. Hier können hingegen andere Eigenschaften des Netzwerks verwendet werden, auf welche in diesem Paper nicht weiter eingegangen wird. -\subsubsection{Anwendung Floyd-Warshall-Algorithmus} +\subsection{Floyd-Warshall-Algorithmus} +Der Floyd-Warshall-Algorithmus, auch Tripel-Algorithmus genannt, wurde erstmals im Jahr 1962 von seinen Namensgebern Robert Floyd und Stephen Warshall vorgestellt. +Der Floyd-Warshall-Algorithmus sucht kürzeste Wege innerhalb eines Graphen. Er ermittelt aber nicht nur die Distanz zwischen zwei Knoten, sondern berechnet die kürzesten Wege zwischen allen Knotenpaaren eines gewichteten Graphen. Somit werden die günstigsten Wege zwischen allen Paaren von Knoten berechnet. Der Floyd-Warhshall-Algrithmus kann ausserdem mit negativen Kantengewichten umgehen, sofern der Graph keinen negativen Kreis (Zyklus) aufweist. Ein Kreis, sprich ein Weg mit identischem Start- und Zielknoten, ist negativ, falls die Summe der Kantengewichte des Weges kleiner als null ist. Ist dies der Fall, führt der Algorithmus zu einem falschen Ergebnis. -Wie oben erwähnt, besteht der Floyd-Warshall-Algorithmus aus dem Teil von Floyd zur Berechnung der kürzesten Pfade und dem Teil von Warshall zur Konstruktion der kürzesten Pfade. +\subsection{Anwendung Floyd-Warshall-Algorithmus} %THEORIE... -Als erstes wird eine Gewichtsmatrix $W$ mit den Matrixeinträgen $W[i, j]$ erstellt. +In einem ersten Schritt wird eine Gewichtsmatrix $W$ mit den Matrixeinträgen $W(i, j)$ erstellt. Der Algorithmus berechnet danach in einer Hauptschleife alle Knoten $k$ von 1 bis $n$. Dabei versucht er in jeder Iteration alle Wege von $i$ nach $j$ durch die Wege $(i, k)$ und $(k, j)$ zu verbessern. -Falls dieser mögliche Umweg zu einer Verbesserung führt, wird der Algorithmus aktualisiert. +Falls dieser mögliche Umweg zu einer Verbesserung führt, wird der entsprechende Eintrag aktualisiert. Die aktuelle Gewichtung der Pfade wird mit -\begin{equation}d[i, j]=min[d[i,j], d[i,k] + d[k,i]]\end{equation} +\begin{equation}d(i, j)=\min\{d(i,j), d(i,k) + d(k,i)\}\end{equation} ermittelt. -\subsubsection{Euklidische Heuristik} -Bei Verkehrsnetzen ist die euklidische Distanz eine gängige und zuverlässige Heurstik. Dabei wird zu den effektiven Reisekosten zum aktuellen Knoten die euklidische Distanz bis zum Zielknoten hinzuaddiert. Dadurch wird die Kostenfunktion konsequent nie überschätzt. Dies stellt eine Voraussetzung an eine zulässige Heuristik dar. -Was bei einem physischen Verkehrsnetz einfach zu bewältigen ist, da Koordinaten von Verkehrsnetzen zur Berechnung der Distanz verwendet werden können, ist bei virtuellen Netzwerken (z.B. Servernetzen) entweder nicht möglich, oder nicht relevant. -\subsection{PageRank-Algorithmus} -Der PageRank-Algorithmus wurde von den Gründern von Google, Larry Page und Sergey Brin im Jahr 1996 entwickelt und zum Patent angemeldet. Zwei Jahre später gründeten sie ihr Unternehmen Google Inc.. -Beim PageRank-Algorithmus handelt es sich um den Algorithmus von Google, aus dem die Google-Matrix abgeleitet wird. -Die Google-Matrix ist eine immens grosse Matrix mit Millionen Zeilen und Spalten, die für die schnelle und vor allem exakte Bestimmung der PageRanks (Gewichtung) eine grosse Bedeutung hat. -Der PageRank-Algorithmus analysiert und gewichtet beispielsweise die Verlinkungsstruktur verschiedener Websites des World Wide Web anhand ihrer Struktur. -Der PageRank wird umso höher, je mehr hochwertige Links auf eine Webseite verweisen und je höher die Gewichtung einer Webseite ist, desto grösser ist der Effekt.\\ -Dabei handelt es sich um einen iterativen Prozess. Ausgegangen wird von der Adjazenz-Matrix $A$, für welche gilt. -%THEORIE... -Grundsätzlich setzt sich der PageRank Algorithmus mit der Fragestellung auseinander, wie eine Suchmaschine wie Google Suchresultate bewertet und somit sortieren soll. Öfters aufgerufene Resultate sollen schliesslich höher gewichtet werden. Dabei wird angenommen, dass eine Website populärer ist, je mehr andere Websites darauf verweisen. +\section{PageRank-Algorithmus} +Der PageRank-Algorithmus wurde von den Gründern von Google, Larry Page und Sergey Brin im Jahr 1996 entwickelt und zum Patent angemeldet. Zwei Jahre später gründeten sie ihr Unternehmen Google Inc. +Beim PageRank-Algorithmus handelt es sich nicht um einen Suchalgorithmus, stattdessen werden Knoten aufgrund der Vernetzung des vorliegenden Graphen bewertet. +Verwendet wird er beispielsweise um die Verlinkungsstruktur verschiedener Websites des World Wide Web anhand ihrer Struktur zu bewerten und relevante Suchergebnisse zu ermittteln. Der PageRank wird umso höher, je mehr hochwertige Links auf eine Webseite verweisen und je höher die Gewichtung einer Webseite ist, desto grösser ist der Effekt. +Dabei handelt es sich um einen iterativen Prozess. Ausgegangen wird von der Adjazenz-Matrix $A$, für welche folgendes gilt: \begin{equation} -A_{i,j}=\left\{ \begin{matrix} -1 & \text{Kante von $j$ nach $i$} \\ 0 & \text{keine Kante von $j$ nach $i$} -\end{matrix} - \right. +A_{i,j} = \begin{cases} +1&\quad\text{Kante von $j$ nach $i$}\\ +0&\quad\text{keine Kante von $j$ nach $i$} +\end{cases} \label{verkehr:Adja} \end{equation} +%THEORIE... +Grundsätzlich setzt sich der PageRank Algorithmus mit der Fragestellung auseinander, wie eine Suchmaschine wie Google Suchresultate bewertet und somit sortieren soll. Öfters aufgerufene Resultate sollen schliesslich höher gewichtet werden. Dabei wird angenommen, dass eine Website populärer ist, je mehr andere Websites darauf verweisen. + + -Für ungerichtete Graphen mit $n$ Knoten gilt \begin{equation}A_{i,j}=A_{j,i}\end{equation} und weiter \begin{equation}A_{i,i}=0\quad\forall i\in \left\{1...n\right\}\end{equation} +Für ungerichtete Graphen mit $n$ Knoten gilt \begin{equation}A_{i,j}=A_{j,i}\end{equation} und weiter \begin{equation}A_{i,i}=0\quad\forall i\in \left\{1\dots n\right\}\end{equation} Beim PageRank-Algorithmus wird eine abgewandelte Form der Adjazenz-Matrix verwendet. -Dabei werden die Matrix-Einträge spaltenweise durch die jeweilige Spaltensumme geteilt. -\begin{equation} P_{i,j}=\frac{A_{i,j}}{\sum_{i=1}^{n}A_{i,j}} \end{equation} +Dabei werden die Matrix-Einträge spaltenweise durch die jeweilige Spaltensumme geteilt, so entsteht die Link-Matrix +\[ P_{i,j}=\frac{A_{i,j}}{\sum_{k=1}^{n}A_{k,j}} \] Anschliessend multipliziert man diese Matrix $P$ mit einem Spaltenvektor $\Vec{r_0}$ mit $n$ Einträgen, für welchen gilt: -\begin{equation} \Vec{r_0}(i) = \frac{1}{n} \quad\forall i\in \left\{1...n\right\} \end{equation} +\( \Vec{r_0}(i) = \frac{1}{n} \quad\forall i\in \left\{1\dots n\right\} \) Dieser Vektor stellt ein neutrales Ranking dar. Alle Knoten werden gleich gewichtet. -Dadurch erhält man wiederum einen $n$-zeiligen Spaltenvektor $\Vec{r_1}$, der das "erste" Ranking darstellt. Durch Multiplikation der ursprünglichen Matrix $P$ mit dem 1. Ranking-Vektor $\Vec{r_1}$ wird auf Basis des ersten Rankings ein zweites erstellt. -\begin{equation} \Vec{r_2} = P\cdot\Vec{r_1} = P\cdot(P\cdot\Vec{r_0}) = P^2\cdot\Vec{r_0}\end{equation} -somit -\begin{equation} \Vec{r_i} = P^i\cdot\Vec{r_0}\end{equation} -Der Vektor $\Vec{r_i}$ konvergiert zu einem Eigenvektor von $P$ und stellt das abschliessende Ranking dar. +Dadurch erhält man wiederum einen $n$-zeiligen Spaltenvektor $\Vec{r_1}$, der das ``erste'' Ranking darstellt. Durch Multiplikation der ursprünglichen Matrix $P$ mit dem 1. Ranking-Vektor $\Vec{r_1}$ wird auf Basis des ersten Rankings ein zweites erstellt: +\( \Vec{r_2} = P\cdot\Vec{r_1} = P\cdot(P\cdot\Vec{r_0}) = P^2\cdot\Vec{r_0}\) +und somit allgemein: +\( \Vec{r_i} = P^i\cdot\Vec{r_0}\) +Der Vektor $\Vec{r_i}$ konvergiert zu einem Eigenvektor von $P$ der das abschliessende Ranking darstellt. diff --git a/buch/papers/verkehr/section2.tex b/buch/papers/verkehr/section2.tex index 638d9dd..527885e 100644 --- a/buch/papers/verkehr/section2.tex +++ b/buch/papers/verkehr/section2.tex @@ -1,12 +1,12 @@ \section{Versuchsreihe} \label{section:verkehr/versuchsreihe} -Um zwei der vorgestellten Suchalgorithmen zu vergleichen, wurden zwei Versuchsreihen erstellt. Dazu wurden in einem ersten Schritt zufällige Netzwerke generiert und anschliessend der \emph{Dijkstra}-, sowie der \emph{$A^*$}-Algorithmus auf das Netzwerk angewandt. -Dieser Vorgang wurde für die zufällig generierten Netzwerke mit einer Knotenzahl von 10, 20 50, 100, 200, 500 und 1000 je zehnmal repetiert. -Die Anzahl der Knoten im abgesuchten Netzwerk wirkt sich direkt auf die Rechenzeit aus. Der \emph{Dijkstra}-Algorithmus weist eine Zeitkomplexität von $\mathcal{O}(E\log{}V)$ auf, wobei $E$ die Anzahl Kanten (engl. \emph{edges}) und $V$ die Anzahl Knoten (engl. \emph{vertices}) darstellt. -Für den \emph{A*}-Algorithmus ist die Zeitkomplexität einerseits abhängig von der verwendeten Heuristik, andererseits aber auch vom vorliegenden Netzwerk selbst. Aus diesem Grund lässt sich keine defintive Angabe zu $\mathcal{O}$ machen. +Um zwei der vorgestellten Suchalgorithmen zu vergleichen, wurden zwei Versuchsreihen erstellt. Dazu wurden in einem ersten Schritt zufällige Netzwerke generiert und anschliessend der Dijkstra- und der A*-Algorithmus auf das Netzwerk angewandt. +Dieser Vorgang wurde für die zufällig generierten Netzwerke mit einer Knotenzahl von 10, 20 50, 100, 200, 500 und 1000 je zehnmal wiederholt. +Die Anzahl der Knoten im abgesuchten Netzwerk wirkt sich direkt auf die Rechenzeit aus. Der \emph{Dijkstra}-Algorithmus weist eine Zeitkomplexität von $\mathcal{O}(|E|\log{}|V|)$ auf, wobei $E$ die Menge der Kanten (engl. \emph{edges}) und $V$ die Menge der Knoten (engl. \emph{vertices}) des Graphen $G$ darstellt. +Für den A*-Algorithmus ist die Zeitkomplexität einerseits abhängig von der verwendeten Heuristik, andererseits aber auch vom vorliegenden Netzwerk selbst. Aus diesem Grund lässt sich keine definitive Angabe zur Zeitkomplexität machen. -Die beiden Versuchsreihen unterscheiden sich zudem dahingehend, dass der Start- und Zielknoten bei der ersten Versuchsreihe im Netzwerk diametral gegenüber liegen. Dadurch gehen viele Knoten verloren, welcher \emph{Dijkstra} als uninformierter Suchalgorithmus absuchen würde. In der zweiten Veruschsreihe werden hingegen Start- un Zielpunkt zufällig im Netzwerk ausgewählt. Es wird deshalb erwwartet, dass die Unterschiede in der Rechenzeit der beiden Algorithmen in der zweiten Versuchsreihe deutlich ausgeprägter sind. +Die beiden Versuchsreihen unterscheiden sich zudem dahingehend, dass der Start- und Zielknoten bei der ersten Versuchsreihe im Netzwerk diametral gegenüber liegen. Dadurch gehen viele Knoten verloren, welcher \emph{Dijkstra} als uninformierter Suchalgorithmus absuchen würde. In der zweiten Veruschsreihe werden hingegen Start- un Zielpunkt zufällig im Netzwerk ausgewählt. Es wird deshalb erwartet, dass die Unterschiede in der Rechenzeit der beiden Algorithmen in der zweiten Versuchsreihe deutlich ausgeprägter sind. \subsection{Einfluss der Knotenzahl auf die Rechenzeit} \label{verkehr:Knotenzahl} @@ -19,9 +19,9 @@ Die beiden Versuchsreihen unterscheiden sich zudem dahingehend, dass der Start- \label{verkehr:Vr1} \end{figure} -In \ref{verkehr:Vr1} ist ersichtlich, dass der Unterschied in der Rechenzeit zwischen \emph{Dijkstra} und \emph{A*} erst aber einer Knotenzahl von ca. $n=500$ merklich ansteigt. Dieses etwas überraschende Resultat ist darauf zurückzuführen, dass bei steigender Knotenzahl die Abweichung des effektiven kürzesten Pfades von der Distanz der Luftlinie abnimmt. +In \ref{verkehr:Vr1} ist ersichtlich, dass der Unterschied in der Rechenzeit zwischen Dijkstra und A* erst ab einer Knotenzahl von ca. $n=500$ merklich ansteigt. Dieses etwas überraschende Resultat ist darauf zurückzuführen, dass bei steigender Knotenzahl die Abweichung des effektiven kürzesten Pfades von der Distanz der Luftlinie abnimmt. Die Effektivität von \emph{A*} mit euklidischer Heuristik ist wiederum grösser, wenn die Abweichung des kürzesten Pfads von der Luftlinie minimal ist. -Bei Betrachtung von \ref{verkehr:pathDifference} wird dies ersichtlich, wobei die relative Abweichung erstaunlicherweise bei einer Knotenzahl von $n=100$ maximal ist und nach $n=500$ nur noch marginal abnimmt. +Abbildung \ref{verkehr:pathDifference} illustriert dies, wobei die relative Abweichung erstaunlicherweise bei einer Knotenzahl von $n=100$ maximal ist und nach $n=500$ nur noch marginal abnimmt. \begin{figure} \centering @@ -36,13 +36,13 @@ Bei Betrachtung von \ref{verkehr:pathDifference} wird dies ersichtlich, wobei di \begin{figure} \centering -\includegraphics[width=12cm]{papers/verkehr/figures/chart_Vr2.png}\\ +\includegraphics[width=12cm]{papers/verkehr/figures/chart_Vr2.png} \caption{Gemessene Rechenzeiten der zweiten Versuchsreihe in Abhängigkeit der Knotenzahl.} \label{verkehr:Vr2} \end{figure} -Zum Vergleich der Resultate in \ref{verkehr:Knotenzahl} zeigt \ref{verkehr:Vr2} die Rechenzeiten der zweiten Versuchsreihe, in welcher die Start- und Zielknoten zufällig im Netzwerk ausgewählt wurden. Einerseits ist eine reduzierte durchschnittliche Rechenzeit festzustellen, was schlicht daran liegt, dass die zufällige Wahl der Knoten dazu führt, dass diese tendenziell weniger weit auseinander liegen.\\ -Des weiteren ist festzustellen, dass sich die Unterschiede der Rechenzeiten zwischen \emph{Dijkstra} und \emph{A*} deutlich früher abzeichnen. Dieses Phänomen lässt sich leicht durch die zielgerichtete Suche des \emph{A*}-Algorithmus erklären. +Zum Vergleich der Resultate in Abschnitt \ref{verkehr:Knotenzahl} zeigt Abbildung \ref{verkehr:Vr2} die Rechenzeiten der zweiten Versuchsreihe, in welcher die Start- und Zielknoten zufällig im Netzwerk ausgewählt wurden. Einerseits ist eine reduzierte durchschnittliche Rechenzeit festzustellen, was daran liegt, dass die zufällige Wahl der Knoten dazu führt, dass diese tendenziell weniger weit auseinander liegen. +Des weiteren ist festzustellen, dass sich die Unterschiede der Rechenzeiten zwischen Dijkstra und A* deutlich früher abzeichnen. Dieses Phänomen lässt sich leicht durch die zielgerichtete Suche des A*-Algorithmus erklären. \begin{figure} \centering @@ -52,4 +52,4 @@ Des weiteren ist festzustellen, dass sich die Unterschiede der Rechenzeiten zwis \label{verkehr:Comparison} \end{figure} -In \ref{verkehr:Comparison} ist ersichtlich, dass bei einem im Netzwerk liegenden Startknoten die zielgerichtete Suche von \emph{A*} deutlich ausgeprägter zum Zuge kommt, als wenn dieser am Rand des Netzwerks liegen würde. +In Abbildung \ref{verkehr:Comparison} ist ersichtlich, dass bei einem im Netzwerk liegenden Startknoten die zielgerichtete Suche von \emph{A*} deutlich ausgeprägter zum Zuge kommt, als wenn dieser am Rand des Netzwerks liegen würde. diff --git a/buch/papers/verkehr/section3.tex b/buch/papers/verkehr/section3.tex index 99a0d92..9aa8ae4 100644 --- a/buch/papers/verkehr/section3.tex +++ b/buch/papers/verkehr/section3.tex @@ -1,8 +1,9 @@ \section{Ausblick} \subsection{Optimierungsprobleme bei Graphen} -Das Finden eines kürzesten Pfades, sprich die Minimierung der Summe der Kantengewichte, ist nur eines der Optimierungsprobleme, die sich im Bereich von Grafen aufstellen lassen. Verschiedene, ähnliche Problemstellungen lassen sich teilweise mit denselben Algorithmen lösen.\\ -Im Bereich vom Computernetzwerken könnte zum Beispiel die Minimierung der Knotenzahl zur Datenübbertragung von Interesse sein. Dabei lässt sich dieses Problem einfach dadurch lösen, dass dem \emph{Dijkstra}, oder dem \emph{A*}-Algorithmus anstelle der Graph-Matrix (mit Kantengewichten als Einträgen) die Adjazenz-Matrix als Argument übergeben wird. Der gefundene kürzeste Pfad enstpricht der Anzahl benutzter Kanten, bzw. der Anzahl besuchter Knoten. +Das Finden eines kürzesten Pfades, sprich die Minimierung der Summe der Kantengewichte, ist nur eines der Optimierungsprobleme, die sich im Bereich von Graphen aufstellen lassen. Verschiedene, ähnliche Problemstellungen lassen sich teilweise mit denselben Algorithmen lösen. + +Im Bereich vom Computernetzwerken könnte zum Beispiel die Minimierung der Knotenzahl zur Datenübbertragung von Interesse sein. Dabei lässt sich dieses Problem einfach dadurch lösen, dass dem Dijkstra- oder dem A*-Algorithmus anstelle der gewichteten Adjazenz-Matrix (mit Kantengewichten als Einträgen) die ungewichtet Adjazenz-Matrix als Argument übergeben wird. Der gefundene kürzeste Pfad enstpricht der Anzahl benutzter Kanten, bzw. der Anzahl besuchter Knoten. \subsection{Wahl der Heuristik} -Ein grundlegendes Problem bei der Anwendung des \emph{A*} oder ähnlicher informierter Suchalgorithmen ist die Wahl der Heurstik. Bei einem physischen Verkehrsnetz kann bspw. die euklidische Distanz problems ermittelt werde. Bei einem regionalen Netzwerk ist die Annahme eines orthogonalen X-Y-Koordinatenetzes absolut ausreichend. Dies gilt z.B. auch für das Vernessungsnetz der Schweiz\footnote{Die aktuelle Schweizer Referenzsystem LV95 benutzt ein E/N-Koordinatennetz, wobei aufgrund zunehmender Abweichung vom Referenzellipsoid bei grosser Entfernung vom Nullpunkt ein Korrekturfaktor für die Höhe angebracht werden muss.} Bei überregionalen Netzwerken (Beispiel: Flugverbindungen) ist hingegen eine Berechnung im dreidimensionalen Raum, oder vereinfacht als Projektion auf das Geoid notwendig. Anonsten ist der Ablauf bei der Ausführung des Algorithmus allerdings identisch.\\ +Ein grundlegendes Problem bei der Anwendung des A* oder ähnlicher informierter Suchalgorithmen ist die Wahl der Heurstik. Bei einem physischen Verkehrsnetz kann bspw. die euklidische Distanz problems ermittelt werde. Bei einem regionalen Netzwerk ist die Annahme eines orthogonalen X-Y-Koordinatenetzes absolut ausreichend. Dies gilt z.B. auch für das Vernessungsnetz der Schweiz\footnote{Die aktuelle Schweizer Referenzsystem LV95 benutzt ein E/N-Koordinatennetz, wobei aufgrund zunehmender Abweichung vom Referenzellipsoid bei grosser Entfernung vom Nullpunkt ein Korrekturfaktor für die Höhe angebracht werden muss.} Bei überregionalen Netzwerken (Beispiel: Flugverbindungen) ist hingegen eine Berechnung im dreidimensionalen Raum, oder vereinfacht als Projektion auf das Geoid notwendig. Anonsten ist der Ablauf bei der Ausführung des Algorithmus allerdings identisch. In nicht-physischen Netzwerken stellt sich jedoch eine zweite Problematik. Da eine physische Distanz entweder nicht ermittelt werden kann, oder aber nicht ausschlaggebend ist, sind andere Netzwerk-Eigenschaften zur Beurteilung beizuziehen. Die Zuverlässigkeit ist dabei aber in den meisten Fällen nicht vergleichbar hoch, wie bei der euklidischen Heuristik. Oftmals werden deshalb bei derartigen Problem auch Algorithmen angewendet, die eine deutlich optimierte Zeitkomplexität aufweisen, dafür aber nicht mit Sicherheit den effizienstesten Pfad finden. |
