diff options
Diffstat (limited to 'buch/papers')
52 files changed, 8260 insertions, 1 deletions
diff --git a/buch/papers/reedsolomon/.gitignor b/buch/papers/reedsolomon/.gitignor new file mode 100644 index 0000000..52a02ac --- /dev/null +++ b/buch/papers/reedsolomon/.gitignor @@ -0,0 +1,24 @@ +RS.aux +RS.bbl +RS.bib +RS.blg +RS.idx +RS.ilg +RS.ind +RS.log +RS.out +RS.pdf +RS.run.xml +RS.toc +*.aux +*.lof +*.log +*.lot +*.fls +*.out +*.toc +*.fmt +*.fot +*.cb +*.cb2 +.*.lb diff --git a/buch/papers/reedsolomon/RS presentation/README.txt b/buch/papers/reedsolomon/RS presentation/README.txt new file mode 100644 index 0000000..4d0620f --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/README.txt @@ -0,0 +1 @@ +Dies ist die Presentation des Reed-Solomon-Code
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--- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.out @@ -0,0 +1,8 @@ +\BOOKMARK [2][]{Outline0.1}{Einführung}{}% 1 +\BOOKMARK [2][]{Outline0.2}{Polynom\040Ansatz}{}% 2 +\BOOKMARK [2][]{Outline0.3}{Diskrete\040Fourier\040Transformation}{}% 3 +\BOOKMARK [2][]{Outline0.4}{Reed-Solomon in Endlichen Körpern}{}% 4 +\BOOKMARK [2][]{Outline0.5}{Codierung\040eines\040Beispiels}{}% 5 +\BOOKMARK [2][]{Outline0.6}{Decodierung\040ohne\040Fehler}{}% 6 +\BOOKMARK [2][]{Outline0.7}{Decodierung\040mit\040Fehler}{}% 7 +\BOOKMARK [2][]{Outline0.8}{Nachricht\040Rekonstruieren}{}% 8 diff --git a/buch/papers/reedsolomon/RS presentation/RS.pdf b/buch/papers/reedsolomon/RS presentation/RS.pdf Binary files differnew file mode 100644 index 0000000..d9d6693 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.pdf diff --git a/buch/papers/reedsolomon/RS presentation/RS.snm b/buch/papers/reedsolomon/RS presentation/RS.snm new file mode 100644 index 0000000..86859c9 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.snm @@ -0,0 +1 @@ +\beamer@slide {ft_discrete}{21} diff --git a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz Binary files differnew file mode 100644 index 0000000..04bd239 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex new file mode 100644 index 0000000..6ee6cc7 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -0,0 +1,934 @@ +\documentclass[11pt,aspectratio=169]{beamer} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{lmodern} +\usepackage[ngerman]{babel} +\usepackage{tikz} +\usetheme{Hannover} + +\begin{document} + \author{Joshua Bär und Michael Steiner} + \title{Reed-Solomon-Code} + \subtitle{} + \logo{} + \institute{OST Ostschweizer Fachhochschule} + \date{26.04.2021} + \subject{Mathematisches Seminar} + %\setbeamercovered{transparent} + \setbeamercovered{invisible} + \setbeamertemplate{navigation symbols}{} + \begin{frame}[plain] + \maketitle + \end{frame} +%------------------------------------------------------------------------------- +\section{Einführung} + \begin{frame} + \frametitle{Reed-Solomon-Code:} + \begin{itemize} + \visible<1->{\item Für Übertragung von Daten} + \visible<2->{\item Ermöglicht Korrektur von Übertragungsfehler} + \visible<3->{\item Wird verwendet in: CD, QR-Codes, Voyager-Sonde, etc.} + \end{itemize} + \end{frame} +%------------------------------------------------------------------------------- +\section{Polynom Ansatz} + \begin{frame} + \begin{itemize} + \item Beispiel $2, 1, 5$ versenden und auf 2 Fehler absichern + \end{itemize} + \end{frame} + \begin{frame} + \frametitle{Beispiel} + Übertragen von + ${f}_2=\textcolor{blue}{2}$, ${f}_1=\textcolor{blue}{1}$, ${f}_0=\textcolor{blue}{5}$ + als $ p(w) = \textcolor{blue}{2}w^2 + \textcolor{blue}{1}w + \textcolor{blue}{5} $. + + + Versende $ (p(1),p(2),\dots,p(7))$ + \visible<2->{ = (\textcolor{green}{8},} + \only<2>{\textcolor{green}{15},} + \only<3>{\textcolor{red}{50},} + \only<2>{\textcolor{green}{26},} + \only<3>{\textcolor{red}{37},} + \visible<2->{\textcolor{green}{41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})} + \only<2>{\includegraphics[scale = 1.2]{images/polynom1.pdf}} + \only<3>{\includegraphics[scale = 1.2]{images/polynom2.pdf}} + \visible<3>{ + \newline + \textcolor{green}{7} Zahlen versenden, um \textcolor{blue}{3} Zahlen gegen \textcolor{red}{2} Fehlern abzusichern.} + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Parameter} + \begin{center} + \begin{tabular}{ c c c } + \hline + Nutzlas & Fehler & Versenden \\ + \hline + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ +\visible<1->{3}& +\visible<1->{3}& +\visible<1->{9 Werte eines Polynoms vom Grad 2} \\ + &&\\ +\visible<1->{$k$} & +\visible<1->{$t$} & +\visible<1->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ + \hline + &&\\ + &&\\ + \multicolumn{3}{l} { + \visible<1>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} + } + \end{tabular} + \end{center} + \end{frame} + +%------------------------------------------------------------------------------- + +\section{Diskrete Fourier Transformation} + \begin{frame} + \frametitle{Idee} + \begin{itemize} + \item Fourier-transformieren + \item Übertragung + \item Rücktransformieren + \end{itemize} + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \begin{figure} + \only<1>{ + \includegraphics[width=0.9\linewidth]{images/fig1.pdf} + } + \only<2>{ + \includegraphics[width=0.9\linewidth]{images/fig2.pdf} + } + \only<3>{ + \includegraphics[width=0.9\linewidth]{images/fig3.pdf} + } + \only<4>{ + \includegraphics[width=0.9\linewidth]{images/fig4.pdf} + } + \only<5>{ + \includegraphics[width=0.9\linewidth]{images/fig5.pdf} + } + \only<6>{ + \includegraphics[width=0.9\linewidth]{images/fig6.pdf} + } + \only<7>{ + \includegraphics[width=0.9\linewidth]{images/fig7.pdf} + } + \end{figure} + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Diskrete Fourier Transformation} + \begin{itemize} + \item Diskrete Fourier-Transformation gegeben durch: + \visible<1->{ + \[ + \label{ft_discrete} + \hat{c}_{k} + = \frac{1}{N} \sum_{n=0}^{N-1} + {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} + \]} + \visible<2->{ + \item Ersetzte + \[ + w = e^{-\frac{2\pi j}{N} k} + \]} + \visible<3->{ + \item Wenn $N$ konstant: + \[ + \hat{c}_{k}=\frac{1}{N}( {f}_0 w^0 + {f}_1 w^1 + {f}_2 w^2 + \dots + {f}_{N-1} w^N) + \]} + \end{itemize} + \end{frame} + +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Diskrete Fourier Transformation} + \[ + \begin{pmatrix} + \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n + \end{pmatrix} + = \frac{1}{N} + \begin{pmatrix} + w^0 & w^0 & w^0 & \dots &w^0 \\ + w^0 & w^1 &w^2 & \dots &w^{N-1} \\ + w^0 & w^2 &w^4 & \dots &w^{2(N-1)} \\ + \vdots & \vdots &\vdots &\ddots &\vdots \\ + w^0 & w^{1(N-1)}&w^{2(N-1)}& \dots &w^{(N-1)(N-1)} \\ + \end{pmatrix} + \begin{pmatrix} + \textcolor{blue}{f_0} \\ + \textcolor{blue}{f_1} \\ + \textcolor{blue}{f_2} \\ + \vdots \\ + 0 \\ + \end{pmatrix} + \] + \end{frame} +%------------------------------------------------------------------------------- + + \begin{frame} + \frametitle{Probleme und Fragen} + + Wie wird der Fehler lokalisiert? + \visible<2>{ + \newline + Indem in einem endlichen Körper gerechnet wird. + } + \end{frame} + +%------------------------------------------------------------------------------- + + +\section{Reed-Solomon in Endlichen Körpern} + + \begin{frame} + \frametitle{Reed-Solomon in Endlichen Körpern} + + \begin{itemize} + \onslide<1->{\item Warum endliche Körper?} + + \onslide<2->{\qquad konkrete Zahlen $\rightarrow$ keine Rundungsfehler} + + \onslide<3->{\qquad digitale Fehlerkorrektur} + + %\onslide<4->{\qquad bessere Laufzeit} + + \vspace{10pt} + + \onslide<4->{\item Nachricht = Nutzdaten + Fehlerkorrekturteil} + + \vspace{10pt} + + \onslide<5->{\item aus Fehlerkorrekturteil die Fehlerstellen finden} + + \onslide<6->{\qquad $\Rightarrow$ gesucht ist ein Lokatorpolynom} + +% \vspace{10pt} + +% \onslide<1->{\item Im Fehlerfall sollen wir aus der Nachricht ein Lokatorpolynom berechnen können, welches die fehlerhaften Stellen beinhaltet} + +% Wir sollten im Fehlerfall in der Lage sein, aus der Nachricht ein Lokatorpolynom zu berechnen, welches die Fehlerhaften Stellen beinhaltet + + \end{itemize} + +% TODO + +% erklärung und einführung der endlichen körper, was wollen wir erreichen? + +% wir versenden im endefekt mehr daten als unsere nachricht umfasst, damit die korrektur sichergestellt werden kann + +% sollten wir fehler bekommen, was uns die korrekturstellen mitgeteilt wird, dann ist es unsere aufgabe ein lokatorpolynom zu finden, welches uns verrät, auf welchen zeilen der Fehler aufgetreten ist + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Definition eines Beispiels} + + \begin{itemize} + + \onslide<1->{\item endlicher Körper $q = 11$} + + \onslide<2->{ist eine Primzahl} + + \onslide<3->{beinhaltet die Zahlen $\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}$} + + \vspace{10pt} + + \onslide<4->{\item Nachrichtenblock $=$ Nutzlast $+$ Fehlerkorrekturstellen} + + \onslide<5->{$n = q - 1 = 10$ Zahlen} + + \vspace{10pt} + + \onslide<6->{\item Max.~Fehler $t = 2$} + + \onslide<7->{maximale Anzahl von Fehler, die wir noch korrigieren können} + + \vspace{10pt} + + \onslide<8->{\item Nutzlast $k = n -2t = 6$ Zahlen} + + \onslide<9->{Fehlerkorrkturstellen $2t = 4$ Zahlen} + + \onslide<10->{Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$} + + \onslide<11->{als Polynom $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Codierung eines Beispiels} + \begin{frame} + \frametitle{Codierung} + + \begin{itemize} + \onslide<1->{\item Ansatz aus den komplexen Zahlen mit der diskreten Fouriertransformation} + + \vspace{10pt} + + \onslide<2->{\item Eulersche Zahl $\mathrm{e}$ existiert nicht in $\mathbb{F}_{11}$} + + \vspace{10pt} + + \onslide<3->{\item Wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken} + + \onslide<4->{$\mathbb{Z}_{11}\setminus\{0\} = \{a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9\}$} + + \vspace{10pt} + + \onslide<5->{\item Wir wählen $a = 8$} + + \onslide<6->{$\mathbb{Z}_{11}\setminus\{0\} = \{1,8,9,6,4,10,3,2,5,7\}$} + + \onslide<7->{$8$ ist eine primitive Einheitswurzel} + + \vspace{10pt} + + \onslide<8->{\item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$} + + \onslide<9->{$\Rightarrow$ \qquad können wir auch als Matrix schreiben} + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Codierung} + + \begin{itemize} + \onslide<1->{\item Übertragungsvektor $v$} + + \onslide<2->{\item $v = A \cdot m$} + + \end{itemize} + + \[ + \onslide<3->{ + v = \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^3& 8^6& 8^9& 8^{12}& 8^{15}& 8^{18}& 8^{21}& 8^{24}& 8^{27}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^8& 8^{16}& 8^{24}& 8^{32}& 8^{40}& 8^{48}& 8^{56}& 8^{64}& 8^{72}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 1 \\ 8 \\ 5 \\ 2 \\ 7 \\ 4 \\ 0 \\ 0 \\ 0 \\ 0 \\ + \end{pmatrix} + } + \] + + \begin{itemize} + \onslide<4->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Decodierung ohne Fehler} + \begin{frame} + \frametitle{Decodierung ohne Fehler} + + \begin{itemize} + \onslide<1->{\item Der Empfänger erhält den unveränderten Vektor $v = [5,3,6,5,2,10,2,7,10,4]$} + + \vspace{10pt} + + \onslide<2->{\item Wir suchen die Inverse der Matrix $A$} + + \vspace{10pt} + + \end{itemize} + + \begin{columns}[t] + \begin{column}{0.55\textwidth} + \onslide<3->{ Inverse der Fouriertransformation} + \vspace{10pt} + \onslide<4->{ + \[ + F(\omega) = \int_{-\infty}^{\infty} f(t) \mathrm{e}^{-j\omega t} dt + \] + } + \vspace{10pt} + \onslide<5->{ + \[ + \mathfrak{F}^{-1}(F(\omega)) = f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega + \] + } + \end{column} + \begin{column}{0.45\textwidth} + \onslide<6->{Inverse von $a$} + + \vspace{10pt} + + \onslide<7->{ + \[ + 8^{1} \Rightarrow 8^{-1} + \] + } + + \onslide<8->{Inverse finden wir über den Eulkidischen Algorithmus} + \vspace{10pt} + \end{column} + \end{columns} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Euklidische Algorithmus} + + \begin{columns}[t] + \begin{column}{0.50\textwidth} + + Recap aus der Vorlesung: + + Gegeben $a \in \mathbb{F}_p$, finde $b = a^{-1} \in \mathbb{F}_p$ + + \begin{tabular}{rcl} + $a b$ &$\equiv$& $1 \mod p$\\ + $a b$ &$=$& $1 + n p$\\ + $a b - n p$ &$=$& $1$\\ + &&\\ + $\operatorname{ggT}(a,p)$&$=$& $1$\\ + $sa + tp$&$=$& $1$\\ + $b$&$=$&$s$\\ + $n$&$=$&$-t$ + \end{tabular} + + \end{column} + \begin{column}{0.50\textwidth} + + \begin{center} + \onslide<1->{ + \begin{tabular}{| c | c c | c | r r |} + \hline + $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ + \hline + & & & & $1$& $0$\\ + $0$& $8$& $11$& $0$& $0$& $1$\\ + $1$& $11$& $8$& $1$& $1$& $0$\\ + $2$& $8$& $3$& $2$& $-1$& $1$\\ + $3$& $3$& $2$& $1$& $3$& $-2$\\ + $4$& $2$& $1$& $2$& \textcolor<2->{blue}{$-4$}& \textcolor<2->{red}{$3$}\\ + $5$& $1$& $0$& & $11$& $-8$\\ + \hline + \end{tabular} + } + + \vspace{10pt} + + \begin{tabular}{rcl} + \onslide<3->{$\textcolor{blue}{-4} \cdot 8 + \textcolor{red}{3} \cdot 11$ &$=$& $1$}\\ + \onslide<4->{$7 \cdot 8 + 3 \cdot 11$ &$=$& $1$}\\ + \onslide<5->{$8^{-1}$ &$=$& $7$} + + \end{tabular} + + \end{center} + + \end{column} + \end{columns} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Decodierung mit Inverser Matrix} + + \begin{itemize} + \onslide<1->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} + + \onslide<2->{\item $m = 1/10 \cdot A^{-1} \cdot v$} + + \onslide<3->{\item $m = 10 \cdot A^{-1} \cdot v$} + + \end{itemize} + \onslide<4->{ + \[ + m = 10 \cdot \begin{pmatrix} + 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ + 7^0& 7^1& 7^2& 7^3& 7^4& 7^5& 7^6& 7^7& 7^8& 7^9\\ + 7^0& 7^2& 7^4& 7^6& 7^8& 7^{10}& 7^{12}& 7^{14}& 7^{16}& 7^{18}\\ + 7^0& 7^3& 7^6& 7^9& 7^{12}& 7^{15}& 7^{18}& 7^{21}& 7^{24}& 7^{27}\\ + 7^0& 7^4& 7^8& 7^{12}& 7^{16}& 7^{20}& 7^{24}& 7^{28}& 7^{32}& 7^{36}\\ + 7^0& 7^5& 7^{10}& 7^{15}& 7^{20}& 7^{25}& 7^{30}& 7^{35}& 7^{40}& 7^{45}\\ + 7^0& 7^6& 7^{12}& 7^{18}& 7^{24}& 7^{30}& 7^{36}& 7^{42}& 7^{48}& 7^{54}\\ + 7^0& 7^7& 7^{14}& 7^{21}& 7^{28}& 7^{35}& 7^{42}& 7^{49}& 7^{56}& 7^{63}\\ + 7^0& 7^8& 7^{16}& 7^{24}& 7^{32}& 7^{40}& 7^{48}& 7^{56}& 7^{64}& 7^{72}\\ + 7^0& 7^9& 7^{18}& 7^{27}& 7^{36}& 7^{45}& 7^{54}& 7^{63}& 7^{72}& 7^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 5 \\ 2 \\ 10 \\ 2 \\ 7 \\ 10 \\ 4 \\ + \end{pmatrix} + \] + } + + \begin{itemize} + \onslide<5->{\item $m = [0,0,0,0,4,7,2,5,8,1]$} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Decodierung mit Fehler} + \begin{frame} + \frametitle{Decodierung mit Fehler - Ansatz} + + \begin{itemize} + \onslide<1->{\item Gesendet: $v = [5,3,6,5,2,10,2,7,10,4]$} + + \onslide<2->{\item Empfangen: $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} + + \onslide<3->{\item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]$} + + \end{itemize} + + \onslide<4->{Wie finden wir die Fehler?} + + \begin{itemize} + \onslide<5->{\item $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} + + \onslide<6->{\item $r(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + 5X^5 + 4X^4 + 5X^3 + 7X^2 + 6X + 7$} + + %\only<7->{\item $e(X) = r(X) - m(X)$} + + \onslide<7->{\item $e(X) = r(X) - m(X)$} + + \end{itemize} + + \begin{center} + \onslide<8->{ + \begin{tabular}{c c c c c c c c c c c} + \hline + $i$& $0$& $1$& $2$& $3$& $4$& $5$& $6$& $7$& $8$& $9$\\ + \hline + $r(a^{i})$& \onslide<9->{$5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$}\\ + $m(a^{i})$& \onslide<10->{$5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$}\\ + $e(a^{i})$& \onslide<11->{$0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$}\\ + \hline + \end{tabular} + } + \end{center} + + + \begin{itemize} + \onslide<12->{\item Alle Stellen, die nicht Null sind, sind Fehler} + \end{itemize} + + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Nullstellen des Fehlerpolynoms finden} + + \begin{itemize} + \onslide<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} + + \vspace{10pt} + + \onslide<2->{\item $f(X) = X^{10}-1 = 0$ \qquad für $X \in \{1,2,3,4,5,6,7,8,9,10\}$} + + \vspace{10pt} + + \onslide<3->{\item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$} + + \vspace{10pt} + + \onslide<4->{\item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$} + + \vspace{10pt} + + \onslide<5->{\item $\operatorname{ggT}$ gibt uns eine Liste der Nullstellen, an denen es keine Fehler gegeben hat} + + \vspace{10pt} + + \onslide<6->{$\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9)$} + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Nullstellen des Fehlerpolynoms finden} + + \begin{itemize} + + \onslide<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} + + \vspace{10pt} + + \onslide<1->{\item $f(X) = X^{10}-1 = 0$ \qquad für $X = [1,2,3,4,5,6,7,8,9,10]$} + + \vspace{10pt} + + \onslide<1->{\item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$} + + \vspace{10pt} + + \onslide<1->{\item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$} + + \vspace{10pt} + + \onslide<1->{\item $\operatorname{kgV}$ gibt uns eine Liste von aller Nullstellen, die wir in $e$ und $d$ zerlegen können} + + \vspace{10pt} + + \onslide<2->{$\operatorname{kgV}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot $ + + \qquad \qquad \qquad \qquad $(X-a^7)(X-a^8)(X-a^9) \cdot q(X)$} + + \onslide<3->{$= d(X) \cdot e(X)$} + + \vspace{10pt} + + \onslide<4->{\item Lokatorpolynom $d(X) = (X-a^3)(X-a^8)$} + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Kennen wir $e(X)$?} + + \begin{itemize} + + \onslide<1->{\item $e(X)$ ist unbekannt auf der Empfängerseite} + + \vspace{10pt} + + \onslide<2->{\item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt?} + + \vspace{10pt} + + \onslide<3->{\item $m$ ist nicht gänzlich unbekannt: $m = [0,0,0,0,?,?,?,?,?,?]$ + + In den bekannten Stellen liegt auch die Information, wo es Fehler gegeben hat} + + \vspace{10pt} + + \onslide<4->{\item Daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$} + + \vspace{10pt} + + \onslide<5->{\item $f(X) = X^{10} - 1 = X^{10} + 10$} + + \vspace{10pt} + + \onslide<6->{\item Jetzt können wir den $\operatorname{ggT}$ von $f(X)$ und $e(X)$ berechnen} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Euklidische Algorithmus (nochmal)} + + \onslide<1->{$\operatorname{ggT}(f(X),e(X))$ hat den Grad $8$} + \onslide<2->{ + \[ + \arraycolsep=1.4pt + \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + X^{10}& & & & & & &+& 10& & & & &:&5X^9&+&7X^8&+& 4X^7&+&10X^6&+&p(X)&=&9X&+&5\\ + X^{10}&+& 8X^9&+& 3X^8&+&2X^7&+& p(X)& & & & & & & & & & & & & & & & \\ \cline{1-9} + && 3X^9&+& 8X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ + && 3X^9&+& 2X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ \cline{3-9} + & & & &6X^8&+&0X^7&+&p(X)& & & & & & & & & & & & \\ + \end{array} + \] + } + \onslide<3->{ + \[ + \arraycolsep=1.4pt + \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + 5X^9&+& 7X^8&+& 4X^7&+& 10X^6&+& p(X)& & & & &:&6X^8&+&0X^7& & & & & & &=&10X&+&3\\ + 5X^9&+& 0X^8&+& p(X)& & & & & & & & & & & & & & & & & & & & \\ \cline{1-5} + && 7X^8&+& p(X)& & & & & & & & & & & & & & & & \\ + \end{array} + \] + } + \vspace{10pt} + + \onslide<4->{$\operatorname{ggT}(f(X),e(X)) = 6X^8$} + + \vspace{10pt} + + \onslide<5->{ $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen } + + \end{frame} + +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Erweiterte Euklidische Algorithmus} + + \begin{center} + + \begin{tabular}{| c | c | c c |} + \hline + $k$ & $q_i$ & $e_i$ & $f_i$\\ + \hline + & & $0$& $1$\\ + $0$& $9X + 5$& $1$& $0$\\ + $1$& $10X + 3$& $9X+5$& $1$\\ + $2$& & \textcolor<2->{blue}{$2X^2 + 0X + 5$}& $10X + 3$\\ + \hline + \end{tabular} + + \end{center} + + \vspace{10pt} + + \begin{tabular}{ll} + \onslide<3->{Somit erhalten wir den Faktor& $d(X) = 2X^2 + 5$\\} + \onslide<4->{Faktorisiert erhalten wir& $d(X) = 2(X-5)(X-6)$\\} + \onslide<5->{Lokatorpolynom& $d(X) = (X-a^i)(X-a^i)$} + \end{tabular} + + \vspace{10pt} + + \onslide<6->{ + \begin{center} + $a^i = 5 \qquad \Rightarrow \qquad i = 3$ + + $a^i = 6 \qquad \Rightarrow \qquad i = 8$ + \end{center} + } + + \onslide<7->{$d(X) = (X-a^3)(X-a^8)$} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Nachricht Rekonstruieren} + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \begin{itemize} + + \onslide<1->{\item $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} + + \onslide<2->{\item $d(X) = (X-\textcolor<4->{red}{a^3})(X-\textcolor<4->{red}{a^8})$} + + \end{itemize} + \onslide<3->{ + \[ + \textcolor{gray}{ + \begin{pmatrix} + a^0 \\ a^1 \\ a^2 \\ \textcolor<4->{red}{a^3} \\ a^4 \\ a^5 \\ a^6 \\ a^7 \\ \textcolor<4->{red}{a^8} \\ a^9 \\ + \end{pmatrix}} + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ \textcolor<4->{red}{8} \\ 2 \\ 10 \\ 2 \\ 7 \\ \textcolor<4->{red}{1} \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + \textcolor<4->{red}{8^0}& \textcolor<4->{red}{8^3}& \textcolor<4->{red}{8^6}& \textcolor<4->{red}{8^9}& \textcolor<4->{red}{8^{12}}& \textcolor<4->{red}{8^{15}}& \textcolor<4->{red}{8^{18}}& \textcolor<4->{red}{8^{21}}& \textcolor<4->{red}{8^{24}}& \textcolor<4->{red}{8^{27}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + \textcolor<4->{red}{8^0}& \textcolor<4->{red}{8^8}& \textcolor<4->{red}{8^{16}}& \textcolor<4->{red}{8^{24}}& \textcolor<4->{red}{8^{32}}& \textcolor<4->{red}{8^{40}}& \textcolor<4->{red}{8^{48}}& \textcolor<4->{red}{8^{56}}& \textcolor<4->{red}{8^{64}}& \textcolor<4->{red}{8^{72}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ + \end{pmatrix} + \] + } + + \begin{itemize} + \onslide<5->{\item Fehlerstellen entfernen} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& \textcolor<4->{green}{8^0}& \textcolor<4->{green}{8^0}& \textcolor<4->{green}{8^0}& \textcolor<4->{green}{8^0}\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& \textcolor<4->{green}{8^6}& \textcolor<4->{green}{8^7}& \textcolor<4->{green}{8^8}& \textcolor<4->{green}{8^9}\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& \textcolor<4->{green}{8^{12}}& \textcolor<4->{green}{8^{14}}& \textcolor<4->{green}{8^{16}}& \textcolor<4->{green}{8^{18}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& \textcolor<4->{green}{8^{24}}& \textcolor<4->{green}{8^{28}}& \textcolor<4->{green}{8^{32}}& \textcolor<4->{green}{8^{36}}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& \textcolor<4->{green}{8^{30}}& \textcolor<4->{green}{8^{35}}& \textcolor<4->{green}{8^{40}}& \textcolor<4->{green}{8^{45}}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& \textcolor<4->{green}{8^{36}}& \textcolor<4->{green}{8^{42}}& \textcolor<4->{green}{8^{48}}& \textcolor<4->{green}{8^{54}}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& \textcolor<4->{green}{8^{42}}& \textcolor<4->{green}{8^{49}}& \textcolor<4->{green}{8^{56}}& \textcolor<4->{green}{8^{63}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& \textcolor<4->{green}{8^{54}}& \textcolor<4->{green}{8^{63}}& \textcolor<4->{green}{8^{72}}& \textcolor<4->{green}{8^{81}}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ \textcolor<2->{green}{m_6} \\ \textcolor<2->{green}{m_7} \\ \textcolor<2->{green}{m_8} \\ \textcolor<2->{green}{m_9} \\ + \end{pmatrix} + \] + + \begin{itemize} + \onslide<3->{\item Nullstellen entfernen} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor<3->{red}{7} \\ \textcolor<3->{red}{4} \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + \textcolor<3->{red}{8^0}& \textcolor<3->{red}{8^7}& \textcolor<3->{red}{8^{14}}& \textcolor<3->{red}{8^{21}}& \textcolor<3->{red}{8^{28}}& \textcolor<3->{red}{8^{35}}\\ + \textcolor<3->{red}{8^0}& \textcolor<3->{red}{8^9}& \textcolor<3->{red}{8^{18}}& \textcolor<3->{red}{8^{27}}& \textcolor<3->{red}{8^{36}}& \textcolor<3->{red}{8^{45}}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \vspace{5pt} + + \begin{itemize} + \onslide<2->{\item Matrix in eine Quadratische Form bringen} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \vspace{5pt} + + \begin{itemize} + \onslide<2->{\item Matrix Invertieren} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + = + \begin{pmatrix} + 1& 1& 1& 1& 1& 1\\ + 1& 8& 9& 6& 4& 10\\ + 1& 9& 4& 3& 5& 1\\ + 1& 4& 5& 9& 3& 1\\ + 1& 10& 1& 10& 1& 10\\ + 1& 3& 9& 5& 4& 1\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \begin{center} + \onslide<2->{$\Downarrow$} + \end{center} + \[ + \onslide<3->{ + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + = + \begin{pmatrix} + 6& 4& 4& 6& 2& 1\\ + 2& 7& 10& 3& 4& 7\\ + 1& 8& 9& 8& 3& 4\\ + 3& 6& 6& 4& 5& 9\\ + 10& 10& 9& 8& 1& 6\\ + 1& 9& 6& 4& 7& 6\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + } + \] + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + = + \begin{pmatrix} + 6& 4& 4& 6& 2& 1\\ + 2& 7& 10& 3& 4& 7\\ + 1& 8& 9& 8& 3& 4\\ + 3& 6& 6& 4& 5& 9\\ + 10& 10& 9& 8& 1& 6\\ + 1& 9& 6& 4& 7& 6\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + \] + + \begin{itemize} + \onslide<2->{\item $m = [4,7,2,5,8,1]$} + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + +\end{document} diff --git a/buch/papers/reedsolomon/RS presentation/RS.toc b/buch/papers/reedsolomon/RS presentation/RS.toc new file mode 100644 index 0000000..095b5e6 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.toc @@ -0,0 +1,9 @@ +\babel@toc {ngerman}{} +\beamer@sectionintoc {1}{Einführung}{2}{0}{1} +\beamer@sectionintoc {2}{Polynom Ansatz}{5}{0}{2} +\beamer@sectionintoc {3}{Diskrete Fourier Transformation}{13}{0}{3} +\beamer@sectionintoc {4}{Reed-Solomon in Endlichen Körpern}{27}{0}{4} +\beamer@sectionintoc {5}{Codierung eines Beispiels}{29}{0}{5} +\beamer@sectionintoc {6}{Decodierung ohne Fehler}{31}{0}{6} +\beamer@sectionintoc {7}{Decodierung mit Fehler}{36}{0}{7} +\beamer@sectionintoc {8}{Nachricht Rekonstruieren}{43}{0}{8} diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.aux 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presentation/RS_handout.tex @@ -0,0 +1,907 @@ +\documentclass[11pt,aspectratio=169]{beamer} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{lmodern} +\usepackage[ngerman]{babel} +\usepackage{tikz} +\usetheme{Hannover} + +\begin{document} + \author{Joshua Bär und Michael Steiner} + \title{Reed-Solomon-Code} + \subtitle{} + \logo{} + \institute{OST Ostschweizer Fachhochschule} + \date{26.04.2021} + \subject{Mathematisches Seminar} + %\setbeamercovered{transparent} + \setbeamercovered{invisible} + \setbeamertemplate{navigation symbols}{} + \begin{frame}[plain] + \maketitle + \end{frame} +%------------------------------------------------------------------------------- +\section{Einführung} + \begin{frame} + \frametitle{Reed-Solomon-Code:} + \begin{itemize} + \item Für Übertragung von Daten + \item Ermöglicht Korrektur von Übertragungsfehler + \item Wird verwendet in: CD, QR-Codes, Voyager-Sonde, etc. + \end{itemize} + \end{frame} +%------------------------------------------------------------------------------- +\section{Polynom Ansatz} + \begin{frame} + \begin{itemize} + \item $2, 1, 5$ versenden und auf 2 Fehler absichern + \end{itemize} + \frametitle{Beispiel} + Übertragen von + ${f}_2=\textcolor{blue}{2}$, ${f}_1=\textcolor{blue}{1}$, ${f}_0=\textcolor{blue}{5}$ + als $ p(w) = \textcolor{blue}{2}w^2 + \textcolor{blue}{1}w + \textcolor{blue}{5} $. + \newline + Versende $ (p(1),p(2),\dots,p(7)) = (\textcolor{green}{8}, + \textcolor{red}{50}, \textcolor{red}{37}, + \textcolor{green}{41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})$ + \includegraphics[scale = 1.2]{images/polynom2.pdf} + \newline + \textcolor{green}{7} Zahlen versenden, um \textcolor{blue}{3} Zahlen gegen \textcolor{red}{2} Fehlern abzusichern. + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Parameter} + \begin{center} + \begin{tabular}{ c c c } + \hline + Nutzlas & Fehler & Versenden \\ + \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 \\ + &&\\ + $k$ & $t$ & $k+2t$ Werte eines Polynoms vom Grad $k-1$ \\ + \hline + &&\\ + &&\\ + \multicolumn{3}{l} { + Ausserdem können bis zu $2t$ Fehler erkannt werden! + } + \end{tabular} + \end{center} + \end{frame} + +%------------------------------------------------------------------------------- + +\section{Diskrete Fourier Transformation} + \begin{frame} + \frametitle{Idee} + \begin{itemize} + \item Fourier-transformieren + \item Übertragung + \item Rücktransformieren + \end{itemize} + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \begin{figure} + \only<1>{ + \includegraphics[width=0.9\linewidth]{images/fig1.pdf} + } + \only<2>{ + \includegraphics[width=0.9\linewidth]{images/fig2.pdf} + } + \only<3>{ + \includegraphics[width=0.9\linewidth]{images/fig3.pdf} + } + \only<4>{ + \includegraphics[width=0.9\linewidth]{images/fig4.pdf} + } + \only<5>{ + \includegraphics[width=0.9\linewidth]{images/fig5.pdf} + } + \only<6>{ + \includegraphics[width=0.9\linewidth]{images/fig6.pdf} + } + \only<7>{ + \includegraphics[width=0.9\linewidth]{images/fig7.pdf} + } + \end{figure} + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Diskrete Fourier Transformation} + \begin{itemize} + \item Diskrete Fourier-Transformation gegeben durch: + + \[ + \label{ft_discrete} + \hat{c}_{k} + = \frac{1}{N} \sum_{n=0}^{N-1} + {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} + \] + + \item Ersetzte + \[ + w = e^{-\frac{2\pi j}{N} k} + \] + + \item Wenn $N$ konstant: + \[ + \hat{c}_{k}=\frac{1}{N}( {f}_0 w^0 + {f}_1 w^1 + {f}_2 w^2 + \dots + {f}_{N-1} w^N) + \] + \end{itemize} + \end{frame} + +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Diskrete Fourier Transformation} + \[ + \begin{pmatrix} + \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n + \end{pmatrix} + = \frac{1}{N} + \begin{pmatrix} + w^0 & w^0 & w^0 & \dots &w^0 \\ + w^0 & w^1 &w^2 & \dots &w^{N-1} \\ + w^0 & w^2 &w^4 & \dots &w^{2(N-1)} \\ + \vdots & \vdots &\vdots &\ddots &\vdots \\ + w^0 & w^{1(N-1)}&w^{2(N-1)}& \dots &w^{(N-1)(N-1)} \\ + \end{pmatrix} + \begin{pmatrix} + \textcolor{blue}{f_0} \\ + \textcolor{blue}{f_1} \\ + \textcolor{blue}{f_2} \\ + \vdots \\ + 0 \\ + \end{pmatrix} + \] + \end{frame} +%------------------------------------------------------------------------------- + + \begin{frame} + \frametitle{Probleme und Fragen} + + Wie wird der Fehler lokalisiert? + \newline + Indem in einem endlichen Körper gerechnet wird. + + \end{frame} + +%------------------------------------------------------------------------------- + + +\section{Reed-Solomon in Endlichen Körpern} + + \begin{frame} + \frametitle{Reed-Solomon in Endlichen Körpern} + + \begin{itemize} + \item Warum endliche Körper? + + \qquad konkrete Zahlen $\rightarrow$ keine Rundungsfehler + + \qquad digitale Fehlerkorrektur + + %\onslide<4->{\qquad bessere Laufzeit} + + \vspace{10pt} + + \item Nachricht = Nutzdaten + Fehlerkorrekturteil + + \vspace{10pt} + + \item aus Fehlerkorrekturteil die Fehlerstellen finden + + \qquad $\Rightarrow$ gesucht ist ein Lokatorpolynom + +% \vspace{10pt} + +% \onslide<1->{\item Im Fehlerfall sollen wir aus der Nachricht ein Lokatorpolynom berechnen können, welches die fehlerhaften Stellen beinhaltet} + +% Wir sollten im Fehlerfall in der Lage sein, aus der Nachricht ein Lokatorpolynom zu berechnen, welches die Fehlerhaften Stellen beinhaltet + + \end{itemize} + +% TODO + +% erklärung und einführung der endlichen körper, was wollen wir erreichen? + +% wir versenden im endefekt mehr daten als unsere nachricht umfasst, damit die korrektur sichergestellt werden kann + +% sollten wir fehler bekommen, was uns die korrekturstellen mitgeteilt wird, dann ist es unsere aufgabe ein lokatorpolynom zu finden, welches uns verrät, auf welchen zeilen der Fehler aufgetreten ist + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Definition eines Beispiels} + + \begin{itemize} + + \item endlicher Körper $q = 11$ + + ist eine Primzahl + + beinhaltet die Zahlen $\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}$ + + \vspace{10pt} + + \item Nachrichtenblock $=$ Nutzlast $+$ Fehlerkorrekturstellen + + $n = q - 1 = 10$ Zahlen + + \vspace{10pt} + + \item Max.~Fehler $t = 2$ + + maximale Anzahl von Fehler, die wir noch korrigieren können + + \vspace{10pt} + + \item Nutzlast $k = n -2t = 6$ Zahlen + + Fehlerkorrkturstellen $2t = 4$ Zahlen + + Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$ + + als Polynom $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$ + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Codierung eines Beispiels} + \begin{frame} + \frametitle{Codierung} + + \begin{itemize} + \item Ansatz aus den komplexen Zahlen mit der diskreten Fouriertransformation + + \vspace{10pt} + + \item Eulersche Zahl $\mathrm{e}$ existiert nicht in $\mathbb{F}_{11}$ + + \vspace{10pt} + + \item Wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken + + $\mathbb{Z}_{11}\setminus\{0\} = \{a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9\}$ + + \vspace{10pt} + + \item Wir wählen $a = 8$ + + $\mathbb{Z}_{11}\setminus\{0\} = \{1,8,9,6,4,10,3,2,5,7\}$ + + $8$ ist eine primitive Einheitswurzel + + \vspace{10pt} + + \item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$ + + $\Rightarrow$ \qquad können wir auch als Matrix schreiben + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Codierung} + + \begin{itemize} + \item Übertragungsvektor $v$ + + \item $v = A \cdot m$ + + \end{itemize} + + \[ + v = \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^3& 8^6& 8^9& 8^{12}& 8^{15}& 8^{18}& 8^{21}& 8^{24}& 8^{27}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^8& 8^{16}& 8^{24}& 8^{32}& 8^{40}& 8^{48}& 8^{56}& 8^{64}& 8^{72}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 1 \\ 8 \\ 5 \\ 2 \\ 7 \\ 4 \\ 0 \\ 0 \\ 0 \\ 0 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item $v = [5,3,6,5,2,10,2,7,10,4]$ + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Decodierung ohne Fehler} + \begin{frame} + \frametitle{Decodierung ohne Fehler} + + \begin{itemize} + \item Der Empfänger erhält den unveränderten Vektor $v = [5,3,6,5,2,10,2,7,10,4]$ + + \vspace{10pt} + + \item Wir suchen die Inverse der Matrix $A$ + + \vspace{10pt} + + \end{itemize} + + \begin{columns}[t] + \begin{column}{0.55\textwidth} + Inverse der Fouriertransformation + \vspace{10pt} + + \[ + F(\omega) = \int_{-\infty}^{\infty} f(t) \mathrm{e}^{-j\omega t} dt + \] + + \vspace{10pt} + + \[ + \mathfrak{F}^{-1}(F(\omega)) = f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega + \] + + \end{column} + \begin{column}{0.45\textwidth} + Inverse von $a$ + + \vspace{10pt} + + \[ + 8^{1} \Rightarrow 8^{-1} + \] + + Inverse finden wir über den Eulkidischen Algorithmus + \vspace{10pt} + \end{column} + \end{columns} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Euklidische Algorithmus} + + \begin{columns}[t] + \begin{column}{0.50\textwidth} + + Recap aus der Vorlesung: + + Gegeben $a \in \mathbb{F}_p$, finde $b = a^{-1} \in \mathbb{F}_p$ + + \begin{tabular}{rcl} + $a b$ &$\equiv$& $1 \mod p$\\ + $a b$ &$=$& $1 + n p$\\ + $a b - n p$ &$=$& $1$\\ + &&\\ + $\operatorname{ggT}(a,p)$&$=$& $1$\\ + $sa + tp$&$=$& $1$\\ + $b$&$=$&$s$\\ + $n$&$=$&$-t$ + \end{tabular} + + \end{column} + \begin{column}{0.50\textwidth} + + \begin{center} + + \begin{tabular}{| c | c c | c | r r |} + \hline + $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ + \hline + & & & & $1$& $0$\\ + $0$& $8$& $11$& $0$& $0$& $1$\\ + $1$& $11$& $8$& $1$& $1$& $0$\\ + $2$& $8$& $3$& $2$& $-1$& $1$\\ + $3$& $3$& $2$& $1$& $3$& $-2$\\ + $4$& $2$& $1$& $2$& \textcolor{blue}{$-4$}& \textcolor{red}{$3$}\\ + $5$& $1$& $0$& & $11$& $-8$\\ + \hline + \end{tabular} + + + \vspace{10pt} + + \begin{tabular}{rcl} + $\textcolor{blue}{-4} \cdot 8 + \textcolor{red}{3} \cdot 11$ &$=$& $1$\\ + $7 \cdot 8 + 3 \cdot 11$ &$=$& $1$\\ + $8^{-1}$ &$=$& $7$ + + \end{tabular} + + \end{center} + + \end{column} + \end{columns} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Decodierung mit Inverser Matrix} + + \begin{itemize} + \item $v = [5,3,6,5,2,10,2,7,10,4]$ + + \item $m = 1/10 \cdot A^{-1} \cdot v$ + + \item $m = 10 \cdot A^{-1} \cdot v$ + + \end{itemize} + + \[ + m = 10 \cdot \begin{pmatrix} + 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ + 7^0& 7^1& 7^2& 7^3& 7^4& 7^5& 7^6& 7^7& 7^8& 7^9\\ + 7^0& 7^2& 7^4& 7^6& 7^8& 7^{10}& 7^{12}& 7^{14}& 7^{16}& 7^{18}\\ + 7^0& 7^3& 7^6& 7^9& 7^{12}& 7^{15}& 7^{18}& 7^{21}& 7^{24}& 7^{27}\\ + 7^0& 7^4& 7^8& 7^{12}& 7^{16}& 7^{20}& 7^{24}& 7^{28}& 7^{32}& 7^{36}\\ + 7^0& 7^5& 7^{10}& 7^{15}& 7^{20}& 7^{25}& 7^{30}& 7^{35}& 7^{40}& 7^{45}\\ + 7^0& 7^6& 7^{12}& 7^{18}& 7^{24}& 7^{30}& 7^{36}& 7^{42}& 7^{48}& 7^{54}\\ + 7^0& 7^7& 7^{14}& 7^{21}& 7^{28}& 7^{35}& 7^{42}& 7^{49}& 7^{56}& 7^{63}\\ + 7^0& 7^8& 7^{16}& 7^{24}& 7^{32}& 7^{40}& 7^{48}& 7^{56}& 7^{64}& 7^{72}\\ + 7^0& 7^9& 7^{18}& 7^{27}& 7^{36}& 7^{45}& 7^{54}& 7^{63}& 7^{72}& 7^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 5 \\ 2 \\ 10 \\ 2 \\ 7 \\ 10 \\ 4 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item $m = [0,0,0,0,4,7,2,5,8,1]$ + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- +\section{Decodierung mit Fehler} + \begin{frame} + \frametitle{Decodierung mit Fehler - Ansatz} + + \begin{itemize} + \item Gesendet: $v = [5,3,6,5,2,10,2,7,10,4]$ + + \item Empfangen: $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$ + + \item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]$ + + \end{itemize} + + Wie finden wir die Fehler? + + \begin{itemize} + \item $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$ + + \item $r(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + 5X^5 + 4X^4 + 5X^3 + 7X^2 + 6X + 7$ + + \item $e(X) = r(X) - m(X)$ + + \end{itemize} + + \begin{center} + + \begin{tabular}{c c c c c c c c c c c} + \hline + $i$& $0$& $1$& $2$& $3$& $4$& $5$& $6$& $7$& $8$& $9$\\ + \hline + $r(a^{i})$& $5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$\\ + $m(a^{i})$& $5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$\\ + $e(a^{i})$& $0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$\\ + \hline + \end{tabular} + + \end{center} + + \begin{itemize} + \item Alle Stellen, die nicht Null sind, sind Fehler + \end{itemize} + + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Nullstellen des Fehlerpolynoms finden} + + \begin{itemize} + \item Satz von Fermat: $f(X) = X^{q-1}-1=0$ + + \vspace{10pt} + + \item $f(X) = X^{10}-1 = 0$ \qquad für $X \in \{1,2,3,4,5,6,7,8,9,10\}$ + + \vspace{10pt} + + \item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$ + + \vspace{10pt} + + \item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$ + + \vspace{10pt} + + \item $\operatorname{ggT}$ gibt uns eine Liste der Nullstellen, an denen es keine Fehler gegeben hat + + \vspace{10pt} + + $\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9)$ + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Nullstellen des Fehlerpolynoms finden} + + \begin{itemize} + + \item Satz von Fermat: $f(X) = X^{q-1}-1=0$ + + \vspace{10pt} + + \item $f(X) = X^{10}-1 = 0$ \qquad für $X = [1,2,3,4,5,6,7,8,9,10]$ + + \vspace{10pt} + + \item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$ + + \vspace{10pt} + + \item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + + \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$ + + \vspace{10pt} + + \item $\operatorname{kgV}$ gibt uns eine Liste von aller Nullstellen, die wir in $e$ und $d$ zerlegen können + + \vspace{10pt} + + $\operatorname{kgV}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot $ + + \qquad \qquad \qquad \qquad $(X-a^7)(X-a^8)(X-a^9) \cdot q(X)$ + + $= d(X) \cdot e(X)$ + + \vspace{10pt} + + \item Lokatorpolynom $d(X) = (X-a^3)(X-a^8)$ + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Kennen wir $e(X)$?} + + \begin{itemize} + + \item $e(X)$ ist unbekannt auf der Empfängerseite + + \vspace{10pt} + + \item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt? + + \vspace{10pt} + + \item $m$ ist nicht gänzlich unbekannt: $m = [0,0,0,0,?,?,?,?,?,?]$ + + In den bekannten Stellen liegt auch die Information, wo es Fehler gegeben hat + + \vspace{10pt} + + \item Daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$ + + \vspace{10pt} + + \item $f(X) = X^{10} - 1 = X^{10} + 10$ + + \vspace{10pt} + + \item Jetzt können wir den $\operatorname{ggT}$ von $f(X)$ und $e(X)$ berechnen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Euklidische Algorithmus (nochmal)} + + $\operatorname{ggT}(f(X),e(X))$ hat den Grad $8$ + + \[ + \arraycolsep=1.4pt + \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + X^{10}& & & & & & &+& 10& & & & &:&5X^9&+&7X^8&+& 4X^7&+&10X^6&+&p(X)&=&9X&+&5\\ + X^{10}&+& 8X^9&+& 3X^8&+&2X^7&+& p(X)& & & & & & & & & & & & & & & & \\ \cline{1-9} + && 3X^9&+& 8X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ + && 3X^9&+& 2X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ \cline{3-9} + & & & &6X^8&+&0X^7&+&p(X)& & & & & & & & & & & & \\ + \end{array} + \] + + \[ + \arraycolsep=1.4pt + \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + 5X^9&+& 7X^8&+& 4X^7&+& 10X^6&+& p(X)& & & & &:&6X^8&+&0X^7& & & & & & &=&10X&+&3\\ + 5X^9&+& 0X^8&+& p(X)& & & & & & & & & & & & & & & & & & & & \\ \cline{1-5} + && 7X^8&+& p(X)& & & & & & & & & & & & & & & & \\ + \end{array} + \] + + \vspace{10pt} + + $\operatorname{ggT}(f(X),e(X)) = 6X^8$ + + \vspace{10pt} + + $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen + + \end{frame} + +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Der Erweiterte Euklidische Algorithmus} + + \begin{center} + + \begin{tabular}{| c | c | c c |} + \hline + $k$ & $q_i$ & $e_i$ & $f_i$\\ + \hline + & & $0$& $1$\\ + $0$& $9X + 5$& $1$& $0$\\ + $1$& $10X + 3$& $9X+5$& $1$\\ + $2$& & \textcolor{blue}{$2X^2 + 0X + 5$}& $10X + 3$\\ + \hline + \end{tabular} + + \end{center} + + \vspace{10pt} + + \begin{tabular}{ll} + Somit erhalten wir den Faktor& $d(X) = 2X^2 + 5$\\ + Faktorisiert erhalten wir& $d(X) = 2(X-5)(X-6)$\\ + Lokatorpolynom& $d(X) = (X-a^i)(X-a^i)$ + \end{tabular} + + \vspace{10pt} + + \begin{center} + $a^i = 5 \qquad \Rightarrow \qquad i = 3$ + + $a^i = 6 \qquad \Rightarrow \qquad i = 8$ + \end{center} + + + $d(X) = (X-a^3)(X-a^8)$ + + \end{frame} +%------------------------------------------------------------------------------- +\section{Nachricht Rekonstruieren} + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \begin{itemize} + + \item $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$ + + \item $d(X) = (X-\textcolor{red}{a^3})(X-\textcolor{red}{a^8})$ + + \end{itemize} + + \[ + \textcolor{gray}{ + \begin{pmatrix} + a^0 \\ a^1 \\ a^2 \\ \textcolor{red}{a^3} \\ a^4 \\ a^5 \\ a^6 \\ a^7 \\ \textcolor{red}{a^8} \\ a^9 \\ + \end{pmatrix}} + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ \textcolor{red}{8} \\ 2 \\ 10 \\ 2 \\ 7 \\ \textcolor{red}{1} \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^3}& \textcolor{red}{8^6}& \textcolor{red}{8^9}& \textcolor{red}{8^{12}}& \textcolor{red}{8^{15}}& \textcolor{red}{8^{18}}& \textcolor{red}{8^{21}}& \textcolor{red}{8^{24}}& \textcolor{red}{8^{27}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^8}& \textcolor{red}{8^{16}}& \textcolor{red}{8^{24}}& \textcolor{red}{8^{32}}& \textcolor{red}{8^{40}}& \textcolor{red}{8^{48}}& \textcolor{red}{8^{56}}& \textcolor{red}{8^{64}}& \textcolor{red}{8^{72}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item Fehlerstellen entfernen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& \textcolor{green}{8^0}& \textcolor{green}{8^0}& \textcolor{green}{8^0}& \textcolor{green}{8^0}\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& \textcolor{green}{8^6}& \textcolor{green}{8^7}& \textcolor{green}{8^8}& \textcolor{green}{8^9}\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& \textcolor{green}{8^{12}}& \textcolor{green}{8^{14}}& \textcolor{green}{8^{16}}& \textcolor{green}{8^{18}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& \textcolor{green}{8^{24}}& \textcolor{green}{8^{28}}& \textcolor{green}{8^{32}}& \textcolor{green}{8^{36}}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& \textcolor{green}{8^{30}}& \textcolor{green}{8^{35}}& \textcolor{green}{8^{40}}& \textcolor{green}{8^{45}}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& \textcolor{green}{8^{36}}& \textcolor{green}{8^{42}}& \textcolor{green}{8^{48}}& \textcolor{green}{8^{54}}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& \textcolor{green}{8^{42}}& \textcolor{green}{8^{49}}& \textcolor{green}{8^{56}}& \textcolor{green}{8^{63}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& \textcolor{green}{8^{54}}& \textcolor{green}{8^{63}}& \textcolor{green}{8^{72}}& \textcolor{green}{8^{81}}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ \textcolor{green}{m_6} \\ \textcolor{green}{m_7} \\ \textcolor{green}{m_8} \\ \textcolor{green}{m_9} \\ + \end{pmatrix} + \] + + \begin{itemize} + \item Nullstellen entfernen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor{red}{7} \\ \textcolor{red}{4} \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^7}& \textcolor{red}{8^{14}}& \textcolor{red}{8^{21}}& \textcolor{red}{8^{28}}& \textcolor{red}{8^{35}}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^9}& \textcolor{red}{8^{18}}& \textcolor{red}{8^{27}}& \textcolor{red}{8^{36}}& \textcolor{red}{8^{45}}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \vspace{5pt} + + \begin{itemize} + \item Matrix in eine Quadratische Form bringen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \vspace{5pt} + + \begin{itemize} + \item Matrix Invertieren + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + = + \begin{pmatrix} + 1& 1& 1& 1& 1& 1\\ + 1& 8& 9& 6& 4& 10\\ + 1& 9& 4& 3& 5& 1\\ + 1& 4& 5& 9& 3& 1\\ + 1& 10& 1& 10& 1& 10\\ + 1& 3& 9& 5& 4& 1\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \begin{center} + $\Downarrow$ + \end{center} + \[ + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + = + \begin{pmatrix} + 6& 4& 4& 6& 2& 1\\ + 2& 7& 10& 3& 4& 7\\ + 1& 8& 9& 8& 3& 4\\ + 3& 6& 6& 4& 5& 9\\ + 10& 10& 9& 8& 1& 6\\ + 1& 9& 6& 4& 7& 6\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + \] + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + = + \begin{pmatrix} + 6& 4& 4& 6& 2& 1\\ + 2& 7& 10& 3& 4& 7\\ + 1& 8& 9& 8& 3& 4\\ + 3& 6& 6& 4& 5& 9\\ + 10& 10& 9& 8& 1& 6\\ + 1& 9& 6& 4& 7& 6\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item $m = [4,7,2,5,8,1]$ + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + +\end{document} diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.toc b/buch/papers/reedsolomon/RS presentation/RS_handout.toc new file mode 100644 index 0000000..ce1bdc2 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS_handout.toc @@ -0,0 +1,9 @@ +\babel@toc {ngerman}{} +\beamer@sectionintoc {1}{Einführung}{2}{0}{1} +\beamer@sectionintoc {2}{Polynom Ansatz}{3}{0}{2} +\beamer@sectionintoc {3}{Diskrete Fourier Transformation}{5}{0}{3} +\beamer@sectionintoc {4}{Reed-Solomon in Endlichen Körpern}{16}{0}{4} +\beamer@sectionintoc {5}{Codierung eines Beispiels}{18}{0}{5} +\beamer@sectionintoc {6}{Decodierung ohne Fehler}{20}{0}{6} +\beamer@sectionintoc {7}{Decodierung mit Fehler}{23}{0}{7} +\beamer@sectionintoc {8}{Nachricht Rekonstruieren}{29}{0}{8} diff --git a/buch/papers/reedsolomon/RS presentation/images/fig1.pdf b/buch/papers/reedsolomon/RS presentation/images/fig1.pdf Binary files differnew file mode 100644 index 0000000..abde60c --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/images/fig1.pdf diff --git a/buch/papers/reedsolomon/RS presentation/images/fig1.svg b/buch/papers/reedsolomon/RS presentation/images/fig1.svg new file mode 100644 index 0000000..8682b56 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/images/fig1.svg @@ -0,0 +1,180 @@ +<?xml version="1.0"?> +<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.0//EN' + 'http://www.w3.org/TR/2001/REC-SVG-20010904/DTD/svg10.dtd'> +<svg xmlns:xlink="http://www.w3.org/1999/xlink" style="fill-opacity:1; color-rendering:auto; color-interpolation:auto; 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Dabei müssen wir die \textcolor{red}{Definition 4.6} berücksichtigen, die besagt, dass nur Primzahlen für endliche Körper in Frage kommen. +Wir legen für unser Beispiel den endlichen Körper $q = 11$ fest. +Alle folgenden Berechnungen wurden mit den beiden Restetabellen \textcolor{red}{xx} und \textcolor{red}{yy} durchgeführt. +Aus den Tabellen folgt auch, dass uns nur die Zahlen \[\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}\] zur Verfügung stehen. + +% die beiden Restetabellen von F_11 +%\input{papers/reedsolomon/restetabelle1} +%\input{papers/reedsolomon/restetabelle2} + +Die grösse des endlichen Körpers legt auch fest, wie gross unsere Nachricht $n$ bestehend aus Nutzdatenteil und Fehlerkorrekturteil sein kann und beträgt in unserem Beispiel +\[ +n = q - 1 = 10 \text{ Zahlen}. +\] + +Im nächsten Schritt bestimmen wir, wie viele Fehler $t$ maximal während der Übertragung auftreten dürfen, damit wir sie noch korrigieren können. +Unser Beispielcode sollte in der Lage sein +\[ +t = 2 +\] +Fehlerstellen korrigieren zu können. + +Die Grösse des Nutzdatenteils hängt von der Grösse der Nachricht sowie der Anzahl der Fehlerkorrekturstellen. Je robuster der Code sein muss, desto weniger Platz für Nutzdaten $k$ bleibt in der Nachricht übrig. +Bei maximal 2 Fehler können wir noch +\[ +k = n - 2t = 6\text{ Zahlen} +\] +übertragen. + +Zusammenfassend haben wir einen Codeblock mit der Länge von 10 Zahlen definiert, der 6 Zahlen als Nutzlast beinhaltet und in der Lage ist aus 2 fehlerhafte Stellen im Block die ursprünglichen Nutzdaten rekonstruieren kann. Zudem werden wir im weiteren feststellen, dass dieser Code maximal 4 Fehlerstellen erkennen, diese aber nicht rekonstruieren kann. + +Wir legen nun die Nachricht +\[ +m = [0,0,0,0,4,7,2,5,8,1] +\] +fest, die wir gerne an einen Empfänger übertragen möchten, wobei die vorderen vier Nullstellen für die Fehlerkorrektur zuständig sind. +Die Nachricht können wir auch als Polynom +\[ +m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1 +\] +darstellen. + +\subsection{Der Ansatz der diskreten Fouriertransformation + \label{reedsolomon:subsection:diskFT}} + +In einem vorherigen Kapitel (???) 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 $\mathrm{e}$ in $\mathbb{F}_{11}$ nicht existiert. +Wir suchen also eine Zahl $a^i$, die in endlichen Körpern existiert und den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken kann. +Dazu schreiben wir +\[ +\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\} +\] +um in +\[ +\mathbb{Z}_{11}\setminus\{0\} = \{a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9\}. +\] + +Wenn wir alle möglichen Werte für $a$ einsetzen, also + +%\begin{align} +%a = 0 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{0, 0, 0, 0, 0, 0, 0, 0, 0, 0\} \\ +%a = 1 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 1, 1, 1, 1, 1, 1, 1, 1, 1\} \\ +%a = 2 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 2, 4, 8, 5, 10, 9, 7, 3, 6\} \\ +%a = 3 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 3, 9, 5, 4, 1, 3, 9, 5, 4\} \\ +%a = 4 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 4, 5, 9, 3, 1, 4, 5, 9, 3\} \\ +%a = 5 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 5, 3, 4, 9, 1, 5, 3, 4, 9\} \\ +%a = 6 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 6, 3, 7, 9, 10, 5, 8, 4, 2\} \\ +%a = 7 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 7, 5, 2, 3, 10, 4, 6, 9, 8\} \\ +%a = 8 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 8, 9, 6, 4, 10, 3, 2, 5, 7\} \\ +%a = 9 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 9, 4, 3, 5, 1, 9, 4, 3, 5\} \\ +%a = 10 : \qquad \mathbb{Z}_{11}\setminus\{0\} = \{1, 10, 1, 10, 1, 10, 1, 10, 1, 10\} +%\end{align} + +\begin{center} +\begin{tabular}{c r c l} +%$a = 0 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{0, 0, 0, 0, 0, 0, 0, 0, 0, 0\}$ \\ +$a = 1 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 1, 1, 1, 1, 1, 1, 1, 1, 1\}$ \\ +$a = 2 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 2, 4, 8, 5, 10, 9, 7, 3, 6\}$ \\ +$a = 3 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 3, 9, 5, 4, 1, 3, 9, 5, 4\}$ \\ +$a = 4 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 4, 5, 9, 3, 1, 4, 5, 9, 3\}$ \\ +$a = 5 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 5, 3, 4, 9, 1, 5, 3, 4, 9\}$ \\ +$a = 6 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 6, 3, 7, 9, 10, 5, 8, 4, 2\}$ \\ +$a = 7 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 7, 5, 2, 3, 10, 4, 6, 9, 8\}$ \\ +$a = 8 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 8, 9, 6, 4, 10, 3, 2, 5, 7\}$ \\ +$a = 9 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 9, 4, 3, 5, 1, 9, 4, 3, 5\}$ \\ +$a = 10 :$& $\qquad \mathbb{Z}_{11}\setminus\{0\}$ &$=$& $\{1, 10, 1, 10, 1, 10, 1, 10, 1, 10\}$ +\end{tabular} +\end{center} + +so fällt uns auf, dass die Zahlen $2,6,7,8$ tatsächlich den gesamten Zahlenraum von $\mathbb{F}_{11}$ abbilden. Solche Zahlen werden \em Primitive Einheitswurzel \em genannt. +Für das Beispiel wählen wir die Zahl $a^i = 8$. +Damit wir unsere Nachricht codieren können, müssen wir $8^i$ in $m(X)$ einsetzen. + +\begin{center} + \begin{tabular}{c} + $m(8^0) = 4 \cdot 1 + 7 \cdot 1 + 2 \cdot 1 + 5 \cdot 1 + 8 \cdot 1 + 1 = 5$ \\ + $m(8^1) = 4 \cdot 8 + 7 \cdot 8 + 2 \cdot 8 + 5 \cdot 8 + 8 \cdot 8 + 1 = 3$ \\ + \vdots + \end{tabular} +\end{center} + +Für eine elegantere Formulierung stellen wir das ganze als Matrix dar, wobei $m$ unser Nachrichtenvektor, $A$ die Transformationsmatrix und $v$ unser Übertragungsvektor ist. + +\[ +v = A \cdot m \qquad \Rightarrow \qquad v = \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^3& 8^6& 8^9& 8^{12}& 8^{15}& 8^{18}& 8^{21}& 8^{24}& 8^{27}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^8& 8^{16}& 8^{24}& 8^{32}& 8^{40}& 8^{48}& 8^{56}& 8^{64}& 8^{72}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + 1 \\ 8 \\ 5 \\ 2 \\ 7 \\ 4 \\ 0 \\ 0 \\ 0 \\ 0 \\ +\end{pmatrix} +\] + +Somit bekommen wir für unseren Übertragungsvektor +\[ +v = [5,3,6,5,2,10,2,7,10,4], +\] +den wir jetzt über einen beliebigen Nachrichtenkanal versenden können. + +\textbf{NOTES} + +warum wird 0 weggelassen? diff --git a/buch/papers/reedsolomon/decmitfehler.tex b/buch/papers/reedsolomon/decmitfehler.tex new file mode 100644 index 0000000..923c1c5 --- /dev/null +++ b/buch/papers/reedsolomon/decmitfehler.tex @@ -0,0 +1,197 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Decodierung mit Fehler +\label{reedsolomon:section:decmitfehler}} +\rhead{fehlerhafte rekonstruktion} +Im zweiten Teil zur Decodierung betrachten wir den Fall, dass unser Übertragungskanal nicht fehlerfrei ist. +Wir legen daher den Fehlervektor +\[ +u = [0, 0, 0, 3, 0, 0, 0, 0, 2, 0] +\] +fest, den wir zu unserem Übertragungsvektor als Fehler dazu addieren und somit + +\begin{center} + +\begin{tabular}{c | c r } + $v$ & & $[5,3,6,5,2,10,2,7,10,4]$\\ + $u$ & $+$ & $[0,0,0,3,0,0,0,0,2,0]$\\ + \hline + $w$ & & $[5,3,6,8,2,10,2,7,1,4]$\\ +\end{tabular} + +% alternative design +%\begin{tabular}{c | c cccccccccccc } +% $v$ & & $[$&$5,$&$3,$&$6,$&$5,$&$2,$&$10,$&$2,$&$7,$&$10,$&$4$&$]$\\ +% $u$ & $+$ & $[$&$0,$&$0,$&$0,$&$3,$&$0,$&$0,$&$0,$&$0,$&$2,$&$0$&$]$\\ +% \hline +% $w$ & & $[$&$5,$&$3,$&$6,$&$8,$&$2,$&$10,$&$2,$&$7,$&$1,$&$4$&$]$\\ +%\end{tabular} + +\end{center} +als Übertragungsvektor auf der Empfängerseite erhalten. + +Wenn wir den Übertragungsvektor jetzt Rücktransformieren wie im vorherigen Kapitel erhalten wir +\[ +r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]. +\] +Im Vergleich zum vorherigen Kapitel sind die Fehlerkorrekturstellen jetzt $\neq 0$, was bedeutet, dass wir diesen Übertragungsvektor fehlerhaft empfangen haben und sich die Nachricht jetzt nicht mehr so einfach decodieren lässt. + +% warum wir die fehler suchen +Da Reed-Solomon-Codes in der Lage sind, eine Nachricht aus weniger Stellen zu rekonstruieren als wir ursprünglich haben, so müssen wir nur die Fehlerhaften Stellen finden und eliminieren, damit wir unsere Nutzdaten rekonstruieren können. +Damit stellt sich die Frage, wie wir die Fehlerstellen $e$ finden. +Dafür wählen wir einen Primitiven Ansatz mit +\begin{align} + m(X) & = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1 \\ + r(X) & = 5X^9 + 7X^8 + 4X^7 + 10X^6 + 5X^5 + 4X^4 + 5X^3 + 7X^2 + 6X + 7 \\ + e(X) & = r(X) - m(X). +\end{align} +Setzen wir jetzt unsere Einheitswurzel für $X$ ein, so erhalten wir +\begin{center} +\begin{tabular}{c c c c c c c c c c c} + \hline + $i$& $0$& $1$& $2$& $3$& $4$& $5$& $6$& $7$& $8$& $9$\\ + \hline + $r(a^{i})$& $5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$\\ + $m(a^{i})$& $5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$\\ + $e(a^{i})$& $0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$\\ + \hline +\end{tabular} +\end{center} +und damit die Information, dass an allen Stellen, die nicht Null sind, Fehler enthalten. +Um jetzt alle nicht Nullstellen zu finden, wenden wir den Satz von Fermat an. + +\subsection{Der Satz von Fermat +\label{reedsolomon:subsection:fermat}} +Der Satz von Fermat besagt, dass für +\[ +f(X) = X^{q-1} -1 = 0 +\] +gilt, egal was wir für $q$ einsetzen. + +Für unser Beispiel erhalten wir +\[ +f(X) = X^{10}-1 = 0 \qquad \text{für } X = \{1,2,3,4,5,6,7,8,9,10\} +\] +und können $f(X)$ auch umschreiben in +\[ +f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6)(X-a^7)(X-a^8)(X-a^9). +\] +Zur Überprüfung können wir unsere Einheitswurzel in $a$ einsetzen und werden sehen, dass wir für $f(X) = 0$ erhalten werden. +Nach der gleichen Überlegung können wir jetzt auch $e(X)$ darstellen als +\[ +e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6)(X-a^7) \qquad \qquad (X-a^9) \cdot p(x), +\] +wobei $p(X)$ das Restpolynom ist und die Fehlerstellen beinhaltet. +Wenn wir jetzt den grössten gemeinsamen Teiler von $f(X)$ und $e(X)$ berechnen, so erhalten wir mit +\[ +\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6)(X-a^7) \qquad \qquad (X-a^9) +\] +eine Liste von Nullstellen, an denen es keine Fehler gegeben hat. +Da wir uns jedoch für eine Liste mit Nullstellen interessieren, an denen es Fehler gegeben hat berechnen wir stattdessen das kgV von $f(X)$ und $e(X)$ als +\[ +\operatorname{kgV}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6)(X-a^7)(X-a^8)(X-a^9) \cdot q(X). +\] +Wir können das Resultat noch zerlegen in +\[ +\operatorname{kgV}(f(X),e(X)) = d(X) \cdot e(X). +\] +Somit muss $d(X)$ eine Liste von Nullstellen enthalten an denen es Fehler gegeben hat. +\[ +d(X) = (X-a^3)(X-a^8) +\] + + +und ist damit unser gesuchtes Lokatorpolynom. + +Das einzige Problem was jetzt noch bleibt ist, dass wir $e(X)$ berechnet haben aus +\[ +e(X) = r(X) - m(X), +\] +wobei $m(X)$ auf der Empfängerseite unbekannt ist. +Es sieht danach aus, das wir diesen Lösungsansatz nicht verwenden können, da uns ein entscheidender Teil fehlt. +Bei einer näheren Betrachtung von $m(X)$ fällt uns aber auf, dass wir doch etwas über $m(X)$ wissen. +Wir kennen nämlich die ersten vier Stellen, da diese für die Fehlerkorrektur zuständig sind und daher Null sein müssen. +\[ +m = [0,0,0,0,?,?,?,?,?,?] +\] +An genau diesen Stellen liegt auch die Information, wo unsere Fehlerstellen liegen, was uns ermöglicht, den Teil von $e(X)$ zu berechnen, der uns auch interessiert. + +Wir können $e(X)$ also bestimmen als +\[ +e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X) +\] +wobei $p(X)$ wiederum ein unbekanntes Restpolynom ist und +\[ +f(X) = X^{10} - 1 = X^{10} + 10 +\] +ist können wir so in einer ersten Instanz den grössten gemeinsamen Teiler von $f(X)$ und $e(X)$ berechnen. +Dafür nehmen wir uns wiederum den Euklidischen Algorithmus zur Hilfe und berechnen so + +\[ +\arraycolsep=1.4pt +\begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + X^{10}& & & & & & &+& 10& & & & &:&5X^9&+&7X^8&+& 4X^7&+&10X^6&+&p(X)&=&9X&+&5\\ + X^{10}&+& 8X^9&+& 3X^8&+&2X^7&+& p(X)& & & & & & & & & & & & & & & & \\ \cline{1-9} + && 3X^9&+& 8X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ + && 3X^9&+& 2X^8&+& 9X^7&+& p(X)& & & & & & & & & & & & \\ \cline{3-9} + & & & &6X^8&+&0X^7&+&p(X)& & & & & & & & & & & & \\ +\end{array} +\] + +\[ +\arraycolsep=1.4pt +\begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} + 5X^9&+& 7X^8&+& 4X^7&+& 10X^6&+& p(X)& & & & &:&6X^8&+&0X^7& & & & & & &=&10X&+&3\\ + 5X^9&+& 0X^8&+& p(X)& & & & & & & & & & & & & & & & & & & & \\ \cline{1-5} + && 7X^8&+& p(X)& & & & & & & & & & & & & & & & \\ +\end{array} +\] +und erhalten +\[ +\operatorname{ggT}(f(X),e(X)) = 6X^8 +\] +Mit den Resultaten, die wir vom Rechenweg des grössten gemeinsamen Teiler erhalten haben können wir jetzt auch das kleinste Gemeinsame Vielfache berechnen. Eine detailliertere Vorgehensweise findet man in Kapitel ???. + +Aus diesem erweiterten Euklidischen Algorithmus erhalten wir +\begin{center} + + \begin{tabular}{| c | c | c c |} + \hline + $k$ & $q_i$ & $e_i$ & $f_i$\\ + \hline + & & $0$& $1$\\ + $0$& $9X + 5$& $1$& $0$\\ + $1$& $10X + 3$& $9X+5$& $1$\\ + $2$& & \textcolor{blue}{$2X^2 + 0X + 5$}& $10X + 3$\\ + \hline + \end{tabular} + +\end{center} +und erhalten auf diesem Weg den Faktor +\[ +d(X) = 2X^2 + 5, +\] +den wir in +\[ +d(X) = 2(X-5)(X-6) +\] +zerlegen können. +Da die unbekannten Stellen im Lokatorpolynom +\[ +d(X) = (X-a^i)(X-a^i) +\] +sind, müssen wir nur noch $i$ berechnen als +\begin{center} + $a^i = 5 \qquad \Rightarrow \qquad i = 3$ + + $a^i = 6 \qquad \Rightarrow \qquad i = 8$. +\end{center} + +Somit erhalten wir schliesslich +\[ +d(X) = (X-a^3)(X-a^8) +\] +als unser Lokatorpolynom mit den Fehlerhaften Stellen.
\ No newline at end of file diff --git a/buch/papers/reedsolomon/decohnefehler.tex b/buch/papers/reedsolomon/decohnefehler.tex new file mode 100644 index 0000000..6ca577a --- /dev/null +++ b/buch/papers/reedsolomon/decohnefehler.tex @@ -0,0 +1,106 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Decodierung ohne Fehler +\label{reedsolomon:section:decohnefehler}} +\rhead{fehlerlose rekonstruktion} +Im ersten Teil zur Decodierung des Übertragungsvektor betrachten wir den Übertragungskanal als fehlerfrei. +Wir erhalten also unseren Übertragungsvektor +\[ +v = [5,3,6,5,2,10,2,7,10,4]. +\] + +Gesucht ist nun einen Weg, mit dem wir auf unseren Nachrichtenvektor zurückrechnen können. +Ein banaler Ansatz ist das Invertieren der Glechung +\[ +v = A \cdot m \qquad \Rightarrow \qquad m = A^{-1} \cdot v. +\] + +Nur stellt sich dann die Frage, wie wir auf die Inverse der Matix $A$ kommen. +Dazu können wir wiederum den Ansatz der Fouriertransformation uns zur Hilfe nehmen, +jedoch betrachten wir jetzt deren Inverse. +Definiert ist sie als +\[ +F(\omega) = \int_{-\infty}^{\infty} f(t) \mathrm{e}^{-j\omega t} dt \qquad \Rightarrow \qquad \mathfrak{F}^{-1}(F(\omega)) = f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega. +\] + +In unserem Fall suchen wir also eine inverse für die Primitive Einheitswurzel $a$, also +\[ +8^1 \qquad \Rightarrow \qquad 8^{-1}. +\] + +Im Abschnitt \textcolor{red}{4.1} haben wir den euklidischen Algorithmus kennengelernt, den wir auf unseren Fall anwenden können. + +\subsection{Der Euklidische Algorithmus +\label{reedsolomon:subsection:eukAlgo}} + +Die Funktionsweise des euklidischen Algorithmus ist im Kapitel \textcolor{red}{4.1} ausführlich beschrieben. +Für unsere Anwendung wählen wir die Parameter $a_i = 8$ und $b_i = 11$. +Daraus erhalten wir + +\begin{center} + +\begin{tabular}{| c | c c | c | r r |} + \hline + $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ + \hline + & & & & $1$& $0$\\ + $0$& $8$& $11$& $0$& $0$& $1$\\ + $1$& $11$& $8$& $1$& $1$& $0$\\ + $2$& $8$& $3$& $2$& $-1$& $1$\\ + $3$& $3$& $2$& $1$& $3$& $-2$\\ + $4$& $2$& $1$& $2$& \textcolor{blue}{$-4$}& \textcolor{red}{$3$}\\ + $5$& $1$& $0$& & $11$& $-8$\\ + \hline +\end{tabular} + +\end{center} +\begin{center} + +\begin{tabular}{rcl} + $\textcolor{blue}{-4} \cdot 8 + \textcolor{red}{3} \cdot 11$ &$=$& $1$\\ + $7 \cdot 8 + 3 \cdot 11$ &$=$& $1$\\ + $8^{-1}$ &$=$& $7$ + +\end{tabular} + +\end{center} + +als Inverse der Primitiven Einheitswurzel. + +Nun haben wir fast alles für die Rücktransformation beisammen. Wie auch bei der Inversen Fouriertransformation haben wir nun einen Vorfaktor +\[ +m = \textcolor{red}{s} \cdot A^{-1} \cdot v +\] +den wir noch bestimmen müssen. +Glücklicherweise lässt der sich analog wie bei der Inversen Fouriertransformation bestimmen und beträgt +\[ +s = \frac{1}{10}. +\] +Da $\frac{1}{10} = 10^{-1}$ entspricht können wir $s$ ebenfalls mit dem euklidischen Algorithmus bestimmen und stellen fest, dass $10^{-1} = 10$ ergibt. +Somit lässt sich der Nachrichtenvektor einfach bestimmen mit +\[ +m = 10 \cdot A^{-1} \cdot v \qquad \Rightarrow \qquad m = 10 \cdot \begin{pmatrix} + 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ + 7^0& 7^1& 7^2& 7^3& 7^4& 7^5& 7^6& 7^7& 7^8& 7^9\\ + 7^0& 7^2& 7^4& 7^6& 7^8& 7^{10}& 7^{12}& 7^{14}& 7^{16}& 7^{18}\\ + 7^0& 7^3& 7^6& 7^9& 7^{12}& 7^{15}& 7^{18}& 7^{21}& 7^{24}& 7^{27}\\ + 7^0& 7^4& 7^8& 7^{12}& 7^{16}& 7^{20}& 7^{24}& 7^{28}& 7^{32}& 7^{36}\\ + 7^0& 7^5& 7^{10}& 7^{15}& 7^{20}& 7^{25}& 7^{30}& 7^{35}& 7^{40}& 7^{45}\\ + 7^0& 7^6& 7^{12}& 7^{18}& 7^{24}& 7^{30}& 7^{36}& 7^{42}& 7^{48}& 7^{54}\\ + 7^0& 7^7& 7^{14}& 7^{21}& 7^{28}& 7^{35}& 7^{42}& 7^{49}& 7^{56}& 7^{63}\\ + 7^0& 7^8& 7^{16}& 7^{24}& 7^{32}& 7^{40}& 7^{48}& 7^{56}& 7^{64}& 7^{72}\\ + 7^0& 7^9& 7^{18}& 7^{27}& 7^{36}& 7^{45}& 7^{54}& 7^{63}& 7^{72}& 7^{81}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 5 \\ 2 \\ 10 \\ 2 \\ 7 \\ 10 \\ 4 \\ +\end{pmatrix} +\] +und wir erhalten +\[ +m = [0,0,0,0,4,7,2,5,8,1] +\] +als unsere Nachricht zurück.
\ No newline at end of file diff --git a/buch/papers/reedsolomon/endlichekoerper.tex b/buch/papers/reedsolomon/endlichekoerper.tex new file mode 100644 index 0000000..8ccd918 --- /dev/null +++ b/buch/papers/reedsolomon/endlichekoerper.tex @@ -0,0 +1,23 @@ +% +% teil1.tex -- Beispiel-File für das Paper +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Reed-Solomon in Endlichen Körpern +\label{reedsolomon:section:endlichekoerper}} +\rhead{Problemstellung} + +TODO: + +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/main.tex b/buch/papers/reedsolomon/main.tex index 8219b63..a7485cd 100644 --- a/buch/papers/reedsolomon/main.tex +++ b/buch/papers/reedsolomon/main.tex @@ -3,7 +3,7 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:reedsolomon}} +\chapter{Reed-Solomon-Code\label{chapter:reedsolomon}} \lhead{Thema} \begin{refsection} \chapterauthor{Joshua Bär und Michael Steiner} @@ -27,10 +27,18 @@ 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/teil0.tex} \input{papers/reedsolomon/teil1.tex} \input{papers/reedsolomon/teil2.tex} \input{papers/reedsolomon/teil3.tex} +% Michael +\input{papers/reedsolomon/endlichekoerper} +\input{papers/reedsolomon/codebsp} +\input{papers/reedsolomon/decohnefehler} +\input{papers/reedsolomon/decmitfehler} +\input{papers/reedsolomon/rekonstruktion} + \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/reedsolomon/rekonstruktion.tex b/buch/papers/reedsolomon/rekonstruktion.tex new file mode 100644 index 0000000..8cb7744 --- /dev/null +++ b/buch/papers/reedsolomon/rekonstruktion.tex @@ -0,0 +1,184 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Nachricht Rekonstruieren +\label{reedsolomon:section:rekonstruktion}} +\rhead{Rekonstruktion} +Im letzten Kapitel haben wir eine Möglichkeit gefunden, wie wir die Fehlerhaften Stellen lokalisieren können. +Mit diesen Stellen soll es uns nun möglich sein, aus dem fehlerhaften empfangenen Nachrichtenvektor wieder unsere Nachricht zu rekonstruieren. +Das Lokatorpolynom +\[ +d(X) = (X - a^3)(X-a^8) +\] +markiert dabei diese Fehlerhaften Stellen im Übertragungsvektor +\[ +w = [5,3,6,8,2,10,2,7,1,4]. +\] +Als Ausgangslage verwenden wir die Matrix, mit der wir den Nachrichtenvektor ursprünglich codiert haben. +Unser Ziel ist es wie auch schon im Kapitel X.X (Rekonstuktion ohne Fehler) eine Möglichkeit zu finden, wie wir den Übertragungsvektor decodieren können. +Aufgrund der Fehlerstellen müssen wir aber davon ausgehen, das wir nicht mehr den gleichen Weg verfolgen können wie wir im Kapitel X.X angewendet haben. + +Wir stellen also die Matrix auf und markieren gleichzeitig die Fehlerstellen. +\[ +\textcolor{gray}{ + \begin{pmatrix} + a^0 \\ a^1 \\ a^2 \\ \textcolor{red}{a^3} \\ a^4 \\ a^5 \\ a^6 \\ a^7 \\ \textcolor{red}{a^8} \\ a^9 \\ +\end{pmatrix}} +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ \textcolor{red}{8} \\ 2 \\ 10 \\ 2 \\ 7 \\ \textcolor{red}{1} \\ 4 \\ +\end{pmatrix} += +\begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^3}& \textcolor{red}{8^6}& \textcolor{red}{8^9}& \textcolor{red}{8^{12}}& \textcolor{red}{8^{15}}& \textcolor{red}{8^{18}}& \textcolor{red}{8^{21}}& \textcolor{red}{8^{24}}& \textcolor{red}{8^{27}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^8}& \textcolor{red}{8^{16}}& \textcolor{red}{8^{24}}& \textcolor{red}{8^{32}}& \textcolor{red}{8^{40}}& \textcolor{red}{8^{48}}& \textcolor{red}{8^{56}}& \textcolor{red}{8^{64}}& \textcolor{red}{8^{72}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ +\end{pmatrix} +\] +Die rot markierten Stellen im Übertragungsvektor enthalten Fehler und bringt uns daher kein weiterer Nutzen. +Aus diesem Grund werden diese Stellen aus dem Vektor entfernt, was wir hier ohne Probleme machen können, da dieser Code ja über Fehlerkorrekturstellen verfügt, deren Aufgabe es ist, eine bestimmte Anzahl an Fehler kompensieren zu können. +Die dazugehörigen Zeilen in der Matrix werden ebenfalls entfernt, da die Matrix gleich viele Zeilen wie im Übertragungsvektor aufweisen muss, damit man ihn decodieren kann. + +Daraus resultiert +\[ +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ +\end{pmatrix} += +\begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ +\end{pmatrix} +. +\] +Die Matrix ist jedoch nicht mehr quadratisch, was eine Rekonstruktion durch Inversion ausschliesst. +Um die quadratische Form wieder herzustellen müssen wir zwei Spalten aus der Matrix entfernen. +Wir kennen aber das Resultat aus den letzten vier Spalten, da wir wissen, das die Nachricht aus Nutzdatenteil und Fehlerkorrekturteil besteht, wobei der letzteres bekanntlich aus lauter Nullstellen besteht. +\[ +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ +\end{pmatrix} += +\begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& \textcolor{green}{8^0}& \textcolor{green}{8^0}& \textcolor{green}{8^0}& \textcolor{green}{8^0}\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& \textcolor{green}{8^6}& \textcolor{green}{8^7}& \textcolor{green}{8^8}& \textcolor{green}{8^9}\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& \textcolor{green}{8^{12}}& \textcolor{green}{8^{14}}& \textcolor{green}{8^{16}}& \textcolor{green}{8^{18}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& \textcolor{green}{8^{24}}& \textcolor{green}{8^{28}}& \textcolor{green}{8^{32}}& \textcolor{green}{8^{36}}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& \textcolor{green}{8^{30}}& \textcolor{green}{8^{35}}& \textcolor{green}{8^{40}}& \textcolor{green}{8^{45}}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& \textcolor{green}{8^{36}}& \textcolor{green}{8^{42}}& \textcolor{green}{8^{48}}& \textcolor{green}{8^{54}}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& \textcolor{green}{8^{42}}& \textcolor{green}{8^{49}}& \textcolor{green}{8^{56}}& \textcolor{green}{8^{63}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& \textcolor{green}{8^{54}}& \textcolor{green}{8^{63}}& \textcolor{green}{8^{72}}& \textcolor{green}{8^{81}}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ \textcolor{green}{m_6} \\ \textcolor{green}{m_7} \\ \textcolor{green}{m_8} \\ \textcolor{green}{m_9} \\ +\end{pmatrix} +\] +Wir nehmen die Entsprechenden Spalten aus der Matrix heraus und erhalten so das Überbestimmte Gleichungssystem +\[ +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor{red}{7} \\ \textcolor{red}{4} \\ +\end{pmatrix} += +\begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^7}& \textcolor{red}{8^{14}}& \textcolor{red}{8^{21}}& \textcolor{red}{8^{28}}& \textcolor{red}{8^{35}}\\ + \textcolor{red}{8^0}& \textcolor{red}{8^9}& \textcolor{red}{8^{18}}& \textcolor{red}{8^{27}}& \textcolor{red}{8^{36}}& \textcolor{red}{8^{45}}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ +\end{pmatrix} +. +\] +Die roten Zeilen können wir aufgrund der Überbestimmtheit ebenfalls entfernen und erhalten so die gesuchte quadratische Matrix +\[ +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ +\end{pmatrix} += +\begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ +\end{pmatrix} +. +\] +Nun können wir den Gauss-Algorithmus anwenden um die Matrix zu Invertieren. +\[ +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ +\end{pmatrix} += +\begin{pmatrix} + 1& 1& 1& 1& 1& 1\\ + 1& 8& 9& 6& 4& 10\\ + 1& 9& 4& 3& 5& 1\\ + 1& 4& 5& 9& 3& 1\\ + 1& 10& 1& 10& 1& 10\\ + 1& 3& 9& 5& 4& 1\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ +\end{pmatrix} +\qquad +\Rightarrow +\qquad +\begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ +\end{pmatrix} += +\begin{pmatrix} + 6& 4& 4& 6& 2& 1\\ + 2& 7& 10& 3& 4& 7\\ + 1& 8& 9& 8& 3& 4\\ + 3& 6& 6& 4& 5& 9\\ + 10& 10& 9& 8& 1& 6\\ + 1& 9& 6& 4& 7& 6\\ +\end{pmatrix} +\cdot +\begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ +\end{pmatrix} +\] +Multiplizieren wir nun aus, erhalten wir unseren Nutzdatenteil +\[ +m = [4,7,2,5,8,1] +\] +zurück, den wir ursprünglich versendet haben. + diff --git a/buch/papers/reedsolomon/restetabelle1.tex b/buch/papers/reedsolomon/restetabelle1.tex new file mode 100644 index 0000000..a5055c0 --- /dev/null +++ b/buch/papers/reedsolomon/restetabelle1.tex @@ -0,0 +1,24 @@ +% created by Michael Steiner +% +% Restetabelle von F_11: Addition +\begin{figure} +\begin{center} +\begin{tabular}{|>{$}c<{$}|>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}|} +\hline ++&0&1&2&3&4&5&6&7&8&9&10\\ +\hline +0&0&1&2&3&4&5&6&7&8&9&10\\ +1&1&2&3&4&5&6&7&8&9&10&0\\ +2&2&3&4&5&6&7&8&9&10&0&1\\ +3&3&4&5&6&7&8&9&10&0&1&2\\ +4&4&5&6&7&8&9&10&0&1&2&3\\ +5&5&6&7&8&9&10&0&1&2&3&4\\ +6&6&7&8&9&10&0&1&2&3&4&5\\ +7&7&8&9&10&0&1&2&3&4&5&6\\ +8&8&9&10&0&1&2&3&4&5&6&7\\ +9&9&10&0&1&2&3&4&5&6&7&8\\ +10&10&0&1&2&3&4&5&6&7&8&9\\ +\hline +\end{tabular} +\end{center} +\end{figure}
\ No newline at end of file diff --git a/buch/papers/reedsolomon/restetabelle2.tex b/buch/papers/reedsolomon/restetabelle2.tex new file mode 100644 index 0000000..887c981 --- /dev/null +++ b/buch/papers/reedsolomon/restetabelle2.tex @@ -0,0 +1,24 @@ +% created by Michael Steiner +% +% Restetabelle von F_11: Multiplikation +\begin{figure} +\begin{center} +\begin{tabular}{|>{$}c<{$}|>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}>{$}c<{$}|} +\hline +\cdot&0&1&2&3&4&5&6&7&8&9&10\\ +\hline +0&0&0&0&0&0&0&0&0&0&0&0\\ +1&0&1&2&3&4&5&6&7&8&9&10\\ +2&0&2&4&6&8&10&1&3&5&7&9\\ +3&0&3&6&9&1&4&7&10&2&5&8\\ +4&0&4&8&1&5&9&2&6&10&3&7\\ +5&0&5&10&4&9&3&8&2&7&1&6\\ +6&0&6&1&7&2&8&3&9&4&10&5\\ +7&0&7&3&10&6&2&9&5&1&8&4\\ +8&0&8&5&2&10&7&4&1&9&6&3\\ +9&0&9&7&5&3&1&10&8&6&4&2\\ +10&0&10&9&8&7&6&5&4&3&2&1\\ +\hline +\end{tabular} +\end{center} +\end{figure}
\ No newline at end of file |
