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+\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} + \setbeamertemplate{navigation symbols}{} + \begin{frame}[plain] + \maketitle + \end{frame} + + \begin{frame} + \frametitle{Test} + Ich mag Züge. + \end{frame} +\end{document} \ No newline at end of file diff --git a/buch/papers/reedsolomon/RS presentation/RS.toc b/buch/papers/reedsolomon/RS presentation/RS.toc new file mode 100644 index 0000000..4cd1c86 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS.toc @@ -0,0 +1 @@ +\babel@toc {ngerman}{} diff --git a/buch/papers/reedsolomon/RS presentation/Thumbs.db b/buch/papers/reedsolomon/RS presentation/Thumbs.db new file mode 100644 index 0000000..1626e26 Binary files /dev/null and b/buch/papers/reedsolomon/RS presentation/Thumbs.db differ -- cgit v1.2.1 From 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presentation/RS.pdf index 459d7e8..10719b7 100644 Binary files a/buch/papers/reedsolomon/RS presentation/RS.pdf and b/buch/papers/reedsolomon/RS presentation/RS.pdf differ diff --git a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz index fe8adf5..2fe95de 100644 Binary files a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz and b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz differ diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 3d2be8f..fb822da 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -3,7 +3,9 @@ \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage[ngerman]{babel} +\usepackage{tikz} \usetheme{Hannover} + \begin{document} \author{Joshua Bär und Michael Steiner} \title{Reed-Solomon-Code} @@ -17,9 +19,51 @@ \begin{frame}[plain] \maketitle \end{frame} - + \section{Introduction} + \begin{frame} + \frametitle{Idee} + + \end{frame} + \begin{frame} - \frametitle{Test} - Ich mag Züge. + \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} + Übertragen von den Zahlen + \textcolor{blue}{2}, \textcolor{blue}{1}, \textcolor{blue}{5} + als $ p(x) = \textcolor{blue}{2}x^2 + \textcolor{blue}{1}x + \textcolor{blue}{5} $.\newline + Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, + \textcolor{green}{15}, \textcolor{green}{26}, + \textcolor{green}{ 41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})$ + \only<1>{ + \includegraphics[]{images/polynom1.pdf}} + \only<2>{ + \includegraphics[]{images/polynom2.pdf}} + \end{frame} + + \end{document} \ No newline at end of file diff --git a/buch/papers/reedsolomon/RS presentation/RS.toc b/buch/papers/reedsolomon/RS presentation/RS.toc index 4cd1c86..32e7e8d 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.toc +++ b/buch/papers/reedsolomon/RS presentation/RS.toc @@ -1 +1,2 @@ \babel@toc {ngerman}{} +\beamer@sectionintoc {1}{Introduction}{2}{0}{1} diff --git a/buch/papers/reedsolomon/RS presentation/Thumbs.db b/buch/papers/reedsolomon/RS presentation/Thumbs.db deleted file mode 100644 index 1626e26..0000000 Binary files a/buch/papers/reedsolomon/RS presentation/Thumbs.db and /dev/null differ diff --git a/buch/papers/reedsolomon/RS presentation/images/fig1.pdf b/buch/papers/reedsolomon/RS presentation/images/fig1.pdf index 5cff7fe..abde60c 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\textcolor{green}{83}, \textcolor{green}{110})$ + + + \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]; + } + \punkt{(1,8/\teiler)} + %\punkt{(2,15/\teiler)} + %\punkt{(3,26/\teiler)} + \punkt{(4,41/\teiler)} + \punkt{(5,60/\teiler)} + \punkt{(6,83/\teiler)} + \punkt{(7,110/\teiler)} + \draw[color=gray,line width=1pt,dashed] + plot[domain=0.5:7, samples=100] + ({\x},{(0.1958*\x^2-1.2875*\x+3.0417)}); + \def\erpunkt#1{ + \fill[color=red] #1 circle[radius=0.08]; + \draw #1 circle[radius=0.07]; + } + \erpunkt{(2,50/\teiler)} + \erpunkt{(3,0.9414)} + + + \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} -- cgit v1.2.1 From 10f3cdb829c001c341ea31415efb44ff6a2878b8 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Wed, 21 Apr 2021 17:30:50 +0200 Subject: Persentation stand 17:30 --- buch/papers/reedsolomon/RS presentation/RS.aux | 37 +++++- buch/papers/reedsolomon/RS presentation/RS.log | 145 +++++++++++---------- buch/papers/reedsolomon/RS presentation/RS.nav | 26 +++- buch/papers/reedsolomon/RS presentation/RS.out | 4 +- buch/papers/reedsolomon/RS presentation/RS.pdf | Bin 117082 -> 132691 bytes buch/papers/reedsolomon/RS presentation/RS.snm | 1 + .../reedsolomon/RS presentation/RS.synctex.gz | Bin 6763 -> 19501 bytes buch/papers/reedsolomon/RS presentation/RS.tex | 109 ++++++++++++++-- buch/papers/reedsolomon/RS presentation/RS.toc | 4 +- 9 files changed, 235 insertions(+), 91 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.aux b/buch/papers/reedsolomon/RS 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--git a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz index 2fe95de..96af4cc 100644 Binary files a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz and b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz differ diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index fb822da..9bdf947 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -19,10 +19,15 @@ \begin{frame}[plain] \maketitle \end{frame} - \section{Introduction} + \section{Einführung} \begin{frame} \frametitle{Idee} - + \begin{itemize} + \item Reed-Solomon-Code beschäftigt sich mit der Übertragung von Daten + und deren Fehler Erkennung. + \item Idee Fourier Transformieren und dann senden. + \item Danach Empfangen und Rücktransformieren. + \end{itemize} \end{frame} \begin{frame} @@ -50,20 +55,100 @@ } \end{figure} \end{frame} + \begin{frame} - Übertragen von den Zahlen - \textcolor{blue}{2}, \textcolor{blue}{1}, \textcolor{blue}{5} - als $ p(x) = \textcolor{blue}{2}x^2 + \textcolor{blue}{1}x + \textcolor{blue}{5} $.\newline - Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, - \textcolor{green}{15}, \textcolor{green}{26}, - \textcolor{green}{ 41}, \textcolor{green}{60}, - \textcolor{green}{83}, \textcolor{green}{110})$ + \uncover<1->{ + Wie ist die Anzahl 0 definiert zum mitgeben? + Indem die Polymereigenschaft genutzt werden. + } + \uncover<2->{ + Wie wird der Fehler lokalisiert? + Indem in einem Endlichen Körper gerechnet wird. + } + + \end{frame} + +\section{Polynom Ansatz} + \begin{frame} + Die Diskrite Fouren Transformation ist so gegeben + \[ + \label{ft_discrete} + \hat{c}_{k} + = \frac{1}{N} \sum_{n=0}^{N-1} + {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} + \]. + + \[ + w = e^{-\frac{2\pi j}{N} k} + \] + 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{frame} + + \begin{frame} + Beispiel 2, 1, 5 Versenden und auf 2 Fehler absichern. + \end{frame} + \begin{frame} + 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} $. + \only<1>{ - \includegraphics[]{images/polynom1.pdf}} + Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, + \textcolor{green}{15}, \textcolor{green}{26}, + \textcolor{green}{ 41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})$ + \includegraphics[scale = 1.2]{images/polynom1.pdf}} \only<2>{ - \includegraphics[]{images/polynom2.pdf}} + Versende $ (p(1),p(2),...,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} + \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 + "Nutzlast" & Fehler & Versenden \\ + \hline + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + &&\\ + k & t & k+2t Werte eines Polynoms vom Grad k-1 \\ + \hline + \end{tabular} + \end{center} + \end{frame} +\section{Diskrete Fourien Transformation} + \begin{frame} + \[ + \begin{pmatrix} + \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n + \end{pmatrix} + = + \begin{pmatrix} + w^0 & w^0 & w^0 & \dots &w^0 \\ + w^0 & w^1 &w^2 & \dots &w^n \\ + w^0 & w^2 &w^4 & \dots &w^{2n} \\ + \vdots & \vdots &\vdots &\ddots &\vdots \\ + w^0 & w^{1n}&w^{2n}& \dots &w^{n} \\ + \end{pmatrix} + \begin{pmatrix} + \textcolor{blue}{5} \\ + \textcolor{blue}{1} \\ + \textcolor{blue}{2} \\ + \vdots \\ + 0 \\ + \end{pmatrix} + \] \end{frame} - \end{document} \ No newline at end of file diff --git a/buch/papers/reedsolomon/RS presentation/RS.toc b/buch/papers/reedsolomon/RS presentation/RS.toc index 32e7e8d..ff200c6 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.toc +++ b/buch/papers/reedsolomon/RS presentation/RS.toc @@ -1,2 +1,4 @@ \babel@toc {ngerman}{} -\beamer@sectionintoc {1}{Introduction}{2}{0}{1} +\beamer@sectionintoc {1}{Einführung}{2}{0}{1} +\beamer@sectionintoc {2}{Polynom Ansatz}{12}{0}{2} +\beamer@sectionintoc {3}{Diskrete Fourien Transformation}{17}{0}{3} -- cgit v1.2.1 From 7c0937851938305c2bb760f3cd4c2084c4493217 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Wed, 21 Apr 2021 18:18:22 +0200 Subject: Presentation neu arangiert --- buch/papers/reedsolomon/RS presentation/RS.tex | 186 +++++++++++++------------ 1 file changed, 96 insertions(+), 90 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 9bdf947..1a1cefd 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -21,12 +21,60 @@ \end{frame} \section{Einführung} \begin{frame} - \frametitle{Idee} + \frametitle{Einführung} \begin{itemize} \item Reed-Solomon-Code beschäftigt sich mit der Übertragung von Daten und deren Fehler Erkennung. - \item Idee Fourier Transformieren und dann senden. - \item Danach Empfangen und Rücktransformieren. + \end{itemize} + \end{frame} +\section{Polynom Ansatz} + \begin{frame} + Beispiel 2, 1, 5 Versenden und auf 2 Fehler absichern. + \end{frame} + \begin{frame} + 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} $. + + \only<1>{ + Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, + \textcolor{green}{15}, \textcolor{green}{26}, + \textcolor{green}{ 41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})$ + \includegraphics[scale = 1.2]{images/polynom1.pdf}} + \only<2>{ + Versende $ (p(1),p(2),...,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} + \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 + "Nutzlast" & Fehler & Versenden \\ + \hline + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ + 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + &&\\ + k & t & k+2t Werte eines Polynoms vom Grad k-1 \\ + \hline + \end{tabular} + \end{center} + + Ausserdem können bis zu 2t Fehler erkannt werden! + \end{frame} +\section{Fourier Transformation} + \begin{frame} + \frametitle{Idee} + \begin{itemize} + \item Idee mit Fourier Transformieren und dann senden. + \item Danach Empfangen und Rücktransformieren. \end{itemize} \end{frame} @@ -56,99 +104,57 @@ \end{figure} \end{frame} - +\section{Diskrete Fourier Transformation} \begin{frame} - \uncover<1->{ - Wie ist die Anzahl 0 definiert zum mitgeben? - Indem die Polymereigenschaft genutzt werden. - } - \uncover<2->{ - Wie wird der Fehler lokalisiert? - Indem in einem Endlichen Körper gerechnet wird. - } - + \frametitle{Diskrete Fourier Transformation} + Die Diskrete Fourier Transformation ist so gegeben: + \[ + \label{ft_discrete} + \hat{c}_{k} + = \frac{1}{N} \sum_{n=0}^{N-1} + {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} + \]. + + \[ + w = e^{-\frac{2\pi j}{N} k} + \] + 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{frame} -\section{Polynom Ansatz} - \begin{frame} - Die Diskrite Fouren Transformation ist so gegeben - \[ - \label{ft_discrete} - \hat{c}_{k} - = \frac{1}{N} \sum_{n=0}^{N-1} - {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} - \]. - - \[ - w = e^{-\frac{2\pi j}{N} k} - \] - 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{frame} - - \begin{frame} - Beispiel 2, 1, 5 Versenden und auf 2 Fehler absichern. - \end{frame} - \begin{frame} - 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} $. - \only<1>{ - Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, - \textcolor{green}{15}, \textcolor{green}{26}, - \textcolor{green}{ 41}, \textcolor{green}{60}, - \textcolor{green}{83}, \textcolor{green}{110})$ - \includegraphics[scale = 1.2]{images/polynom1.pdf}} - \only<2>{ - Versende $ (p(1),p(2),...,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} - \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 - "Nutzlast" & Fehler & Versenden \\ - \hline - 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ - 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ - 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ - &&\\ - k & t & k+2t Werte eines Polynoms vom Grad k-1 \\ - \hline - \end{tabular} - \end{center} - \end{frame} -\section{Diskrete Fourien Transformation} \begin{frame} + \frametitle{Diskrete Fourier Transformation} \[ - \begin{pmatrix} - \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n - \end{pmatrix} - = - \begin{pmatrix} - w^0 & w^0 & w^0 & \dots &w^0 \\ - w^0 & w^1 &w^2 & \dots &w^n \\ - w^0 & w^2 &w^4 & \dots &w^{2n} \\ - \vdots & \vdots &\vdots &\ddots &\vdots \\ - w^0 & w^{1n}&w^{2n}& \dots &w^{n} \\ - \end{pmatrix} - \begin{pmatrix} - \textcolor{blue}{5} \\ - \textcolor{blue}{1} \\ - \textcolor{blue}{2} \\ - \vdots \\ - 0 \\ - \end{pmatrix} + \begin{pmatrix} + \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n + \end{pmatrix} + = + \begin{pmatrix} + w^0 & w^0 & w^0 & \dots &w^0 \\ + w^0 & w^1 &w^2 & \dots &w^n \\ + w^0 & w^2 &w^4 & \dots &w^{2n} \\ + \vdots & \vdots &\vdots &\ddots &\vdots \\ + w^0 & w^{1n}&w^{2n}& \dots &w^{n} \\ + \end{pmatrix} + \begin{pmatrix} + \textcolor{blue}{f_0} \\ + \textcolor{blue}{f_1} \\ + \textcolor{blue}{f_2} \\ + \vdots \\ + 0 \\ + \end{pmatrix} \] \end{frame} - +\section{Probleme und Fragen} + \begin{frame} + \frametitle{Probleme und Fragen} + + Wie wird der Fehler lokalisiert? + \only<2>{ + Indem in einem Endlichen Körper gerechnet wird. + } + \end{frame} \end{document} \ No newline at end of file -- cgit v1.2.1 From 264bd585ba37fcf0a8fed6c83b38edfe2495daef Mon Sep 17 00:00:00 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@@ -17,9 +17,715 @@ \begin{frame}[plain] \maketitle \end{frame} - +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Test} Ich mag Züge. \end{frame} + + \begin{frame} + \frametitle{Reed-Solomon in Endlichen Körpern} + + \begin{itemize} + \item Warum Endliche Körper? + + \qquad bessere Laufzeit + + \vspace{10pt} + + \item Nachricht = Nutzdaten + Fehlerkorrekturteil + + \vspace{10pt} + + \item den Fehlerkorrekturteil brauchen wir im Optimalfall nicht + + \vspace{10pt} + + \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 $q = 11$ + + \only<1->{ist eine Primzahl} + + \only<1->{beinhaltet die Zahlen $\mathbb{Z}_{11} = [0,1,2,3,4,5,6,7,8,9,10]$} + + \vspace{10pt} + + \only<1->{\item Nachrichtenblock $n = q-1$} + + wird an den Empfänger gesendet + + \vspace{10pt} + + \only<1->{\item max. Fehler $z = 2$} + + maximale Anzahl von Fehler, die wir noch korrigieren können + + \vspace{10pt} + + \only<1->{\item Nutzlast $k = n -2t = 6$ Zahlen} + + Fehlerstellen $2t = 4$ Zahlen + + \only<1->{Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$} + + \only<1->{als Polynom $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} + + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Codierung} + + \begin{itemize} + \item Ansatz aus den Komplexen Zahlen mit der Fouriertransformation + + \vspace{10pt} + + \item $\mathrm{e}$ existiert nicht in $\mathbb{Z}_{11}$ + + \vspace{10pt} + + \item wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{Z}_{11}$ abdeckt + + $\mathbb{Z}_{11}\setminus\{0\} = [a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9]$ + + \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} +%------------------------------------------------------------------------------- + \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 + + \end{itemize} + + \begin{columns}[t] + \begin{column}{0.50\textwidth} + + Inverse der Fouriertransformation + \vspace{10pt} + \[ + F(\omega) = \int_{-\infty}^{\infty} f(t) \mathrm{e}^{-j\omega t} dt + \] + \vspace{10pt} + \[ + f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega + \] + + \end{column} + \begin{column}{0.50\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 | c c |} + \hline + $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ + \hline + & & & & $1$& $0$\\ + $0$& $8$& $11$& $0$& $0$& $1$\\ + $1$& $11$& $8$& $1$& $1$& $0$\\ + $2$& $8$& $3$& $2$& $-1$& $1$\\ + $3$& $3$& $2$& $1$& $3$& $-2$\\ + $4$& $2$& $1$& $2$& $-4$& $3$\\ + $5$& $1$& $0$& & $11$& $-8$\\ + \hline + \end{tabular} + + \vspace{10pt} + + \begin{tabular}{rcl} + $-4\cdot 8 + 3 \cdot 11$ &$=$& $1$\\ + $7 \cdot 8 + 3 \cdot 11$ &$=$& $1$\\ + $8^{-1}$ &$=$& $7$ + + \end{tabular} + + \end{center} + + \end{column} + \end{columns} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Decodirung 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} + + \[ + m = \begin{pmatrix} + 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ + 7^0& 7^1& 7^2& 7^3& 7^4& 7^5& 7^6& 7^7& 7^8& 7^9\\ + 7^0& 7^2& 7^4& 7^6& 7^8& 7^{10}& 7^{12}& 7^{14}& 7^{16}& 7^{18}\\ + 7^0& 7^3& 7^6& 7^9& 7^{12}& 7^{15}& 7^{18}& 7^{21}& 7^{24}& 7^{27}\\ + 7^0& 7^4& 7^8& 7^{12}& 7^{16}& 7^{20}& 7^{24}& 7^{28}& 7^{32}& 7^{36}\\ + 7^0& 7^5& 7^{10}& 7^{15}& 7^{20}& 7^{25}& 7^{30}& 7^{35}& 7^{40}& 7^{45}\\ + 7^0& 7^6& 7^{12}& 7^{18}& 7^{24}& 7^{30}& 7^{36}& 7^{42}& 7^{48}& 7^{54}\\ + 7^0& 7^7& 7^{14}& 7^{21}& 7^{28}& 7^{35}& 7^{42}& 7^{49}& 7^{56}& 7^{63}\\ + 7^0& 7^8& 7^{16}& 7^{24}& 7^{32}& 7^{40}& 7^{48}& 7^{56}& 7^{64}& 7^{72}\\ + 7^0& 7^9& 7^{18}& 7^{27}& 7^{36}& 7^{45}& 7^{54}& 7^{63}& 7^{72}& 7^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 5 \\ 2 \\ 10 \\ 2 \\ 7 \\ 10 \\ 4 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item $m = [0,0,0,0,4,7,2,5,8,1]$ + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Decodierung mit Fehler - Ansatz} + + \begin{itemize} + \item Gesendet: $V = [5,3,6,5,2,10,2,7,10,4]$ + + \item Empfangen: $W = [5,3,6,8,2,10,2,7,1,4]$ + + \item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerstellen}5,4,5,7,6,7]$ + \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 $X = [1,2,3,4,5,6,7,8,9,10]$ + + \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 $e$?} + + \begin{itemize} + + \item $e$ 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 + & & $0$& $1$\\ + $0$& $9X + 5$& $1$& $0$\\ + $1$& $10X + 3$& $9X+5$& $1$\\ + $2$& & $2X^2 + 0X + 5$& $10X + 3$\\ + \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} + + $D = [3,8]$ + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \begin{itemize} + + \item $W = [5,3,6,8,2,10,2,7,1,4]$ + + \item $D = [3,8]$ + + \end{itemize} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 8 \\ 2 \\ 10 \\ 2 \\ 7 \\ 1 \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^3& 8^6& 8^9& 8^{12}& 8^{15}& 8^{18}& 8^{21}& 8^{24}& 8^{27}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^8& 8^{16}& 8^{24}& 8^{32}& 8^{40}& 8^{48}& 8^{56}& 8^{64}& 8^{72}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item Fehlerstellen entfernen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ + \end{pmatrix} + \] + + \begin{itemize} + \item Nullstellen entfernen + \end{itemize} + + \end{frame} +%------------------------------------------------------------------------------- + \begin{frame} + \frametitle{Rekonstruktion der Nachricht} + + \[ + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ + \end{pmatrix} + = + \begin{pmatrix} + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}\\ + \end{pmatrix} + \cdot + \begin{pmatrix} + m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ + \end{pmatrix} + \] + + \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} \ No newline at end of file -- cgit v1.2.1 From 8473571bc77425cd198b4bba515a3f5fe10c8cd2 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Wed, 21 Apr 2021 22:53:49 +0200 Subject: Style verbessert --- buch/papers/reedsolomon/RS presentation/RS.tex | 17 +++++++++++------ 1 file changed, 11 insertions(+), 6 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 1a1cefd..65f8431 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -64,12 +64,16 @@ &&\\ k & t & k+2t Werte eines Polynoms vom Grad k-1 \\ \hline + &&\\ + &&\\ + &Ausserdem können bis zu 2t Fehler erkannt werden!\\ \end{tabular} \end{center} - Ausserdem können bis zu 2t Fehler erkannt werden! + + \end{frame} -\section{Fourier Transformation} +\section{Diskrete Fourier Transformation} \begin{frame} \frametitle{Idee} \begin{itemize} @@ -104,7 +108,7 @@ \end{figure} \end{frame} -\section{Diskrete Fourier Transformation} + \begin{frame} \frametitle{Diskrete Fourier Transformation} Die Diskrete Fourier Transformation ist so gegeben: @@ -134,10 +138,10 @@ = \begin{pmatrix} w^0 & w^0 & w^0 & \dots &w^0 \\ - w^0 & w^1 &w^2 & \dots &w^n \\ - w^0 & w^2 &w^4 & \dots &w^{2n} \\ + w^0 & w^1 &w^2 & \dots &w^N \\ + w^0 & w^2 &w^4 & \dots &w^{2N} \\ \vdots & \vdots &\vdots &\ddots &\vdots \\ - w^0 & w^{1n}&w^{2n}& \dots &w^{n} \\ + w^0 & w^{1(N-1)}&w^{2(N-1)}& \dots &w^{(N-1)(N-1)} \\ \end{pmatrix} \begin{pmatrix} \textcolor{blue}{f_0} \\ @@ -154,6 +158,7 @@ Wie wird der Fehler lokalisiert? \only<2>{ + \newline Indem in einem Endlichen Körper gerechnet wird. } \end{frame} -- cgit v1.2.1 From 38d0c69842308be5f096375ff070c5233b395c4c Mon Sep 17 00:00:00 2001 From: JODBaer Date: Thu, 22 Apr 2021 16:01:46 +0200 Subject: kleine korrekturen --- buch/papers/reedsolomon/RS presentation/RS.tex | 45 +++++++++++++++----------- 1 file changed, 26 insertions(+), 19 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index eecd66b..618121c 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -19,14 +19,18 @@ \begin{frame}[plain] \maketitle \end{frame} - \section{Einführung} +%------------------------------------------------------------------------------- +\section{Einführung} \begin{frame} \frametitle{Einführung} \begin{itemize} \item Reed-Solomon-Code beschäftigt sich mit der Übertragung von Daten und deren Fehler Erkennung. + \item Wird verwendet in: + \only<2>{CD, QR-Codes, Voyager-Sonde, etc.} \end{itemize} \end{frame} +%------------------------------------------------------------------------------- \section{Polynom Ansatz} \begin{frame} Beispiel 2, 1, 5 Versenden und auf 2 Fehler absichern. @@ -50,7 +54,7 @@ \includegraphics[scale = 1.2]{images/polynom2.pdf} \textcolor{green}{7} Zahlen versenden, um \textcolor{blue}{3} Zahlen gegen \textcolor{red}{2} Fehlern abzusichern.} \end{frame} - +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Parameter} \begin{center} @@ -59,20 +63,24 @@ "Nutzlast" & Fehler & Versenden \\ \hline 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ - 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ - 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ + 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ +\only<2->{3}& +\only<2->{2}& +\only<2->{7 Werte eines Polynoms vom Grad 2} \\ &&\\ - k & t & k+2t Werte eines Polynoms vom Grad k-1 \\ +\only<3->{k} & +\only<3->{t} & +\only<3->{k+2t Werte eines Polynoms vom Grad k-1} \\ \hline &&\\ &&\\ - &Ausserdem können bis zu 2t Fehler erkannt werden!\\ + \multicolumn{3}{l} { + \only<4>{Ausserdem können bis zu 2t Fehler erkannt werden!} + } \end{tabular} - \end{center} - - - + \end{center} \end{frame} +%------------------------------------------------------------------------------- \section{Diskrete Fourier Transformation} \begin{frame} \frametitle{Idee} @@ -81,7 +89,7 @@ \item Danach Empfangen und Rücktransformieren. \end{itemize} \end{frame} - +%------------------------------------------------------------------------------- \begin{frame} \begin{figure} \only<1>{ @@ -107,8 +115,7 @@ } \end{figure} \end{frame} - - +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Diskrete Fourier Transformation} Die Diskrete Fourier Transformation ist so gegeben: @@ -117,8 +124,8 @@ \hat{c}_{k} = \frac{1}{N} \sum_{n=0}^{N-1} {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} - \]. - + \] + Ersetzten als: \[ w = e^{-\frac{2\pi j}{N} k} \] @@ -128,14 +135,14 @@ \] \end{frame} - +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Diskrete Fourier Transformation} \[ \begin{pmatrix} \hat{c}_1 \\\hat{c}_2 \\\hat{c}_3 \\ \vdots \\\hat{c}_n \end{pmatrix} - = + = \frac{1}{N} \begin{pmatrix} w^0 & w^0 & w^0 & \dots &w^0 \\ w^0 & w^1 &w^2 & \dots &w^N \\ @@ -152,7 +159,7 @@ \end{pmatrix} \] \end{frame} - +%------------------------------------------------------------------------------- \section{Probleme und Fragen} \begin{frame} \frametitle{Probleme und Fragen} @@ -163,7 +170,7 @@ Indem in einem Endlichen Körper gerechnet wird. } \end{frame} - +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Reed-Solomon in Endlichen Körpern} -- cgit v1.2.1 From 9ce4fb55792c297989d1c001a621793303f31689 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Thu, 22 Apr 2021 22:13:29 +0200 Subject: Verbesserungen und anmerkungen umgesetzt --- buch/papers/reedsolomon/RS presentation/RS.tex | 56 ++++++++++++++------------ 1 file changed, 31 insertions(+), 25 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 618121c..9811cf6 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -22,36 +22,38 @@ %------------------------------------------------------------------------------- \section{Einführung} \begin{frame} - \frametitle{Einführung} + \frametitle{Reed-Solomon-Code:} \begin{itemize} - \item Reed-Solomon-Code beschäftigt sich mit der Übertragung von Daten - und deren Fehler Erkennung. - \item Wird verwendet in: - \only<2>{CD, QR-Codes, Voyager-Sonde, etc.} + \item \only<1>{Für Übertragung von Daten} + \item \only<2->{Ermöglicht Korrektur von Übertragungsfehler} + \item \only<3->{Wird verwendet in: CD, QR-Codes, Voyager-Sonde, etc.} \end{itemize} \end{frame} %------------------------------------------------------------------------------- \section{Polynom Ansatz} \begin{frame} - Beispiel 2, 1, 5 Versenden und auf 2 Fehler absichern. + \begin{itemize} + \item Beispiel $2, 1, 5$ versenden und auf 2 Fehler absichern + \end{itemize} \end{frame} \begin{frame} Übertragen von - ${f}_2=$\textcolor{blue}{2}, ${f}_1$\textcolor{blue}{1}, ${f}_0$\textcolor{blue}{5} + ${f}_2=\textcolor{blue}{2}$, ${f}_1=\textcolor{blue}{1}$, ${f}_0=\textcolor{blue}{5}$ als $ p(w) = \textcolor{blue}{2}w^2 + \textcolor{blue}{1}w + \textcolor{blue}{5} $. \only<1>{ - Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, + Versende $ (p(1),p(2),\dots,p(7)) = (\textcolor{green}{8}, \textcolor{green}{15}, \textcolor{green}{26}, - \textcolor{green}{ 41}, \textcolor{green}{60}, + \textcolor{green}{41}, \textcolor{green}{60}, \textcolor{green}{83}, \textcolor{green}{110})$ \includegraphics[scale = 1.2]{images/polynom1.pdf}} \only<2>{ - Versende $ (p(1),p(2),...,p(7)) = (\textcolor{green}{8}, + 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}{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} %------------------------------------------------------------------------------- @@ -60,22 +62,22 @@ \begin{center} \begin{tabular}{ c c c } \hline - "Nutzlast" & Fehler & Versenden \\ + ``Nutzlas´´ & Fehler & Versenden \\ \hline 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ \only<2->{3}& -\only<2->{2}& -\only<2->{7 Werte eines Polynoms vom Grad 2} \\ +\only<2->{3}& +\only<3->{9 Werte eines Polynoms vom Grad 2} \\ &&\\ -\only<3->{k} & -\only<3->{t} & -\only<3->{k+2t Werte eines Polynoms vom Grad k-1} \\ +\only<4->{$k$} & +\only<4->{$t$} & +\only<4->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ \hline &&\\ &&\\ \multicolumn{3}{l} { - \only<4>{Ausserdem können bis zu 2t Fehler erkannt werden!} + \only<4>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} } \end{tabular} \end{center} @@ -85,8 +87,9 @@ \begin{frame} \frametitle{Idee} \begin{itemize} - \item Idee mit Fourier Transformieren und dann senden. - \item Danach Empfangen und Rücktransformieren. + \item Fourier-transformieren + \item Übertragung + \item Rücktransformieren \end{itemize} \end{frame} %------------------------------------------------------------------------------- @@ -118,14 +121,16 @@ %------------------------------------------------------------------------------- \begin{frame} \frametitle{Diskrete Fourier Transformation} - Die Diskrete Fourier Transformation ist so gegeben: + \begin{itemize} + \item Diskrete Fourier-Transformation gegeben durch: + \[ \label{ft_discrete} \hat{c}_{k} = \frac{1}{N} \sum_{n=0}^{N-1} {f}_n \cdot e^{-\frac{2\pi j}{N} \cdot kn} \] - Ersetzten als: + \item Ersetzte \[ w = e^{-\frac{2\pi j}{N} k} \] @@ -133,6 +138,7 @@ \[ \hat{c}_{k}=\frac{1}{N}( {f}_0 w^0 + {f}_1 w^1 + {f}_2 w^2 + \dots + {f}_{N-1} w^N) \] + \end{itemize} \end{frame} %------------------------------------------------------------------------------- @@ -145,8 +151,8 @@ = \frac{1}{N} \begin{pmatrix} w^0 & w^0 & w^0 & \dots &w^0 \\ - w^0 & w^1 &w^2 & \dots &w^N \\ - w^0 & w^2 &w^4 & \dots &w^{2N} \\ + 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} @@ -167,7 +173,7 @@ Wie wird der Fehler lokalisiert? \only<2>{ \newline - Indem in einem Endlichen Körper gerechnet wird. + Indem in einem endlichen Körper gerechnet wird. } \end{frame} %------------------------------------------------------------------------------- -- cgit v1.2.1 From 5bca0960f8c9635375d2ca53c93d2bc5a2e37c10 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Thu, 22 Apr 2021 22:59:07 +0200 Subject: Animation verbessert --- buch/papers/reedsolomon/RS presentation/RS.tex | 37 ++++++++++++++------------ 1 file changed, 20 insertions(+), 17 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 9811cf6..732cee5 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -24,9 +24,9 @@ \begin{frame} \frametitle{Reed-Solomon-Code:} \begin{itemize} - \item \only<1>{Für Übertragung von Daten} - \item \only<2->{Ermöglicht Korrektur von Übertragungsfehler} - \item \only<3->{Wird verwendet in: CD, QR-Codes, Voyager-Sonde, etc.} + \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} %------------------------------------------------------------------------------- @@ -37,6 +37,7 @@ \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} $. @@ -66,18 +67,18 @@ \hline 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ -\only<2->{3}& -\only<2->{3}& -\only<3->{9 Werte eines Polynoms vom Grad 2} \\ +\visible<2->{3}& +\visible<2->{3}& +\visible<3->{9 Werte eines Polynoms vom Grad 2} \\ &&\\ -\only<4->{$k$} & -\only<4->{$t$} & -\only<4->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ +\visible<4->{$k$} & +\visible<4->{$t$} & +\visible<4->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ \hline &&\\ &&\\ \multicolumn{3}{l} { - \only<4>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} + \visible<4>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} } \end{tabular} \end{center} @@ -123,21 +124,23 @@ \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 \[ w = e^{-\frac{2\pi j}{N} k} - \] - Wenn $N$ konstant: + \]} + \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} @@ -166,12 +169,12 @@ \] \end{frame} %------------------------------------------------------------------------------- -\section{Probleme und Fragen} + \begin{frame} \frametitle{Probleme und Fragen} Wie wird der Fehler lokalisiert? - \only<2>{ + \visible<2>{ \newline Indem in einem endlichen Körper gerechnet wird. } -- cgit v1.2.1 From 967ff1f33d3faaa1e344ff687aff6c07cde29b77 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Thu, 22 Apr 2021 23:33:02 +0200 Subject: Update RS.tex --- buch/papers/reedsolomon/RS presentation/RS.tex | 288 ++++++++++++++----------- 1 file changed, 165 insertions(+), 123 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 732cee5..61324f7 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -15,6 +15,7 @@ \date{26.04.2021} \subject{Mathematisches Seminar} \setbeamercovered{transparent} + %\setbeamercovered{invisible} \setbeamertemplate{navigation symbols}{} \begin{frame}[plain] \maketitle @@ -83,7 +84,11 @@ \end{tabular} \end{center} \end{frame} +<<<<<<< Updated upstream %------------------------------------------------------------------------------- +======= + +>>>>>>> Stashed changes \section{Diskrete Fourier Transformation} \begin{frame} \frametitle{Idee} @@ -179,26 +184,38 @@ Indem in einem endlichen Körper gerechnet wird. } \end{frame} +<<<<<<< Updated upstream %------------------------------------------------------------------------------- +======= + +\section{Reed-Solomon in Endlichen Körpern} + +>>>>>>> Stashed changes \begin{frame} \frametitle{Reed-Solomon in Endlichen Körpern} \begin{itemize} - \item Warum Endliche Körper? + \onslide<1->{\item Warum endliche Körper?} - \qquad bessere Laufzeit + \onslide<1->{\qquad konkrete Zahlen $\rightarrow$ keine Rundungsfehler} - \vspace{10pt} + \onslide<1->{\qquad digitale Fehlerkorrektur} - \item Nachricht = Nutzdaten + Fehlerkorrekturteil + \onslide<1->{\qquad bessere Laufzeit} \vspace{10pt} - \item den Fehlerkorrekturteil brauchen wir im Optimalfall nicht + \onslide<1->{\item Nachricht = Nutzdaten + Fehlerkorrekturteil} \vspace{10pt} - \item Im Fehlerfall sollen wir aus der Nachricht ein Lokatorpolynom berechnen können, welches die Fehlerhaften Stellen beinhaltet + \onslide<1->{\item aus Fehlerkorrekturteil die Fehlerstellen finden} + + \onslide<1->{\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 @@ -212,35 +229,35 @@ % 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 $q = 11$ + \only<1->{\item endlicher Körper $q = 11$} \only<1->{ist eine Primzahl} - \only<1->{beinhaltet die Zahlen $\mathbb{Z}_{11} = [0,1,2,3,4,5,6,7,8,9,10]$} + \only<1->{beinhaltet die Zahlen $\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}$} \vspace{10pt} - \only<1->{\item Nachrichtenblock $n = q-1$} + \only<1->{\item Nachrichtenblock $=$ Nutzlast $+$ Fehlerkorrekturstellen - wird an den Empfänger gesendet + $n = q - 1 = 10$ Zahlen} \vspace{10pt} - \only<1->{\item max. Fehler $z = 2$} + \only<1->{\item Max.~Fehler $z = 2$ - maximale Anzahl von Fehler, die wir noch korrigieren können + maximale Anzahl von Fehler, die wir noch korrigieren können} \vspace{10pt} \only<1->{\item Nutzlast $k = n -2t = 6$ Zahlen} - Fehlerstellen $2t = 4$ Zahlen + \only<1->{Fehlerkorrkturstellen $2t = 4$ Zahlen} \only<1->{Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$} @@ -250,52 +267,54 @@ \end{frame} %------------------------------------------------------------------------------- +\section{Codierung eines Beispiels} \begin{frame} \frametitle{Codierung} \begin{itemize} - \item Ansatz aus den Komplexen Zahlen mit der Fouriertransformation + \only<1->{\item Ansatz aus den komplexen Zahlen mit der diskreten Fouriertransformation} \vspace{10pt} - \item $\mathrm{e}$ existiert nicht in $\mathbb{Z}_{11}$ + \only<1->{\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 $\mathbb{Z}_{11}$ abdeckt + \only<1->{\item Wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken - $\mathbb{Z}_{11}\setminus\{0\} = [a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9]$ + $\mathbb{Z}_{11}\setminus\{0\} = \{a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9\}$} \vspace{10pt} - \item wir wählen $a = 8$ + \only<1->{\item Wir wählen $a = 8$} - $\mathbb{Z}_{11}\setminus\{0\} = [1,8,9,6,4,10,3,2,5,7]$ + \only<1->{$\mathbb{Z}_{11}\setminus\{0\} = \{1,8,9,6,4,10,3,2,5,7\}$} - 8 ist eine Primitive Einheitswurzel + \only<1->{$8$ ist eine primitive Einheitswurzel} \vspace{10pt} - \item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$ + \only<1->{\item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$} - $\Rightarrow$ \qquad können wir auch als Matrix schreiben + \only<1->{$\Rightarrow$ \qquad können wir auch als Matrix schreiben} \end{itemize} \end{frame} -%------------------------------------------------------------------------------- +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Codierung} \begin{itemize} - \item Übertragungsvektor $V$ + \only<1->{\item Übertragungsvektor $v$} - \item $V = A \cdot m$ + \only<1->{\item $v = A \cdot m$} \end{itemize} \[ - V = \begin{pmatrix} + \only<1->{ + v = \begin{pmatrix} 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ @@ -311,29 +330,34 @@ \begin{pmatrix} 1 \\ 8 \\ 5 \\ 2 \\ 7 \\ 4 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} + } \] - + \only<1->{ \begin{itemize} - \item $V = [5,3,6,5,2,10,2,7,10,4]$ + \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} - \item Der Empfänger erhält den unveränderten Vektor $V = [5,3,6,5,2,10,2,7,10,4]$ + \only<1->{\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 + \only<1->{\item Wir suchen die Inverse der Matrix $A$} + + \vspace{10pt} \end{itemize} \begin{columns}[t] \begin{column}{0.50\textwidth} - + \only<1->{ Inverse der Fouriertransformation \vspace{10pt} \[ @@ -341,25 +365,26 @@ \] \vspace{10pt} \[ - f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega + \mathfrak{F}^{-1}(F(\omega)) = f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \mathrm{e}^{j \omega t} d\omega \] - + } \end{column} \begin{column}{0.50\textwidth} - - Inverse von a + \only<1->{ + Inverse von $a$} \vspace{10pt} + \only<1->{ \[ 8^{1} \Rightarrow 8^{-1} \] - - Inverse finden wir über den Eulkidischen Algorithmus + } + \only<1->{Inverse finden wir über den Eulkidischen Algorithmus} \vspace{10pt} \end{column} \end{columns} \end{frame} -%------------------------------------------------------------------------------- +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Der Euklidische Algorithmus} @@ -385,8 +410,8 @@ \begin{column}{0.50\textwidth} \begin{center} - - \begin{tabular}{| c | c c | c | c c |} + \only<1->{ + \begin{tabular}{| c | c c | c | r r |} \hline $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ \hline @@ -395,17 +420,17 @@ $1$& $11$& $8$& $1$& $1$& $0$\\ $2$& $8$& $3$& $2$& $-1$& $1$\\ $3$& $3$& $2$& $1$& $3$& $-2$\\ - $4$& $2$& $1$& $2$& $-4$& $3$\\ + $4$& $2$& $1$& $2$& \textcolor<3->{blue}{$-4$}& \textcolor<3->{red}{$3$}\\ $5$& $1$& $0$& & $11$& $-8$\\ \hline \end{tabular} - + } \vspace{10pt} \begin{tabular}{rcl} - $-4\cdot 8 + 3 \cdot 11$ &$=$& $1$\\ - $7 \cdot 8 + 3 \cdot 11$ &$=$& $1$\\ - $8^{-1}$ &$=$& $7$ + \only<1->{$\textcolor{blue}{-4} \cdot 8 + \textcolor{red}{3} \cdot 11$ &$=$& $1$}\\ + \only<1->{$7 \cdot 8 + 3 \cdot 11$ &$=$& $1$}\\ + \only<1->{$8^{-1}$ &$=$& $7$} \end{tabular} @@ -417,17 +442,17 @@ \end{frame} %------------------------------------------------------------------------------- \begin{frame} - \frametitle{Decodirung mit Inverser Matrix} + \frametitle{Decodierung mit Inverser Matrix} \begin{itemize} - \item $V = [5,3,6,5,2,10,2,7,10,4]$ + \only<1->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} - \item $m = 1/10 \cdot A^{-1} \cdot V$ + \only<1->{\item $m = 1/10 \cdot A^{-1} \cdot v$} - \item $m = 10 \cdot A^{-1} \cdot V$ + \only<1->{\item $m = 10 \cdot A^{-1} \cdot v$} \end{itemize} - + \only<1->{ \[ m = \begin{pmatrix} 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ @@ -446,85 +471,95 @@ 5 \\ 3 \\ 6 \\ 5 \\ 2 \\ 10 \\ 2 \\ 7 \\ 10 \\ 4 \\ \end{pmatrix} \] - + } + \only<1->{ \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: $V = [5,3,6,5,2,10,2,7,10,4]$ + \only<1->{\item Gesendet: $v = [5,3,6,5,2,10,2,7,10,4]$} - \item Empfangen: $W = [5,3,6,8,2,10,2,7,1,4]$ + \only<1->{\item Empfangen: $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} + + \only<1->{\item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]$} - \item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerstellen}5,4,5,7,6,7]$ \end{itemize} - Wie finden wir die Fehler? + \only<1->{Wie finden wir die Fehler?} + \only<1->{ \begin{itemize} \item $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$ \item $r(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + 5X^5 + 4X^4 + 5X^3 + 7X^2 + 6X + 7$ + %\only<7->{\item $e(X) = r(X) - m(X)$} + \item $e(X) = r(X) - m(X)$ + \end{itemize} - + } + \begin{center} - + \only<1->{ \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 - $r(a^{i})$& $5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$\\ - $m(a^{i})$& $5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$\\ - $e(a^{i})$& $0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$\\ + $r(a^{i})$& \only<1->{$5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$}\\ + $m(a^{i})$& \only<1->{$5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$}\\ + $e(a^{i})$& \only<1->{$0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$}\\ \hline \end{tabular} - + } \end{center} - + + \only<1->{ \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: $f(X) = X^{q-1}-1=0$ + \only<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} \vspace{10pt} - \item $f(X) = X^{10}-1 = 0$ \qquad für $X = [1,2,3,4,5,6,7,8,9,10]$ + \only<1->{\item $f(X) = X^{10}-1 = 0$ \qquad für $X \in \{1,2,3,4,5,6,7,8,9,10\}$} \vspace{10pt} - \item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + \only<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 $(X-a^7)(X-a^8)(X-a^9)$ + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$} \vspace{10pt} - \item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + \only<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 $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$ + \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 + \only<1->{\item $\operatorname{ggT}$ gibt uns eine Liste der Nullstellen, an denen es keine Fehler gegeben hat} \vspace{10pt} - $\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + \only<1->{$\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ - \qquad \qquad \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9)$ + \qquad \qquad \qquad \qquad $(X-a^7) \qquad \qquad (X-a^9)$} \end{itemize} @@ -574,33 +609,33 @@ \end{frame} %------------------------------------------------------------------------------- \begin{frame} - \frametitle{kennen wir $e$?} + \frametitle{Kennen wir $e(X)$?} \begin{itemize} - \item $e$ ist unbekannt auf der Empfängerseite + \only<1->{\item $e(X)$ ist unbekannt auf der Empfängerseite} \vspace{10pt} - \item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt? + \only<1->{\item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt?} \vspace{10pt} - \item $m$ ist nicht gänzlich unbekannt: $m = [0,0,0,0,?,?,?,?,?,?]$ + \only<1->{\item $m$ ist nicht gänzlich unbekannt: $m = [0,0,0,0,?,?,?,?,?,?]$ - In den bekannten Stellen liegt auch die Information, wo es Fehler gegeben hat + In den bekannten Stellen liegt auch die Information, wo es Fehler gegeben hat} \vspace{10pt} - \item daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$ + \only<1->{\item Daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$} \vspace{10pt} - \item $f(X) = X^{10} - 1 = X^{10} + 10$ + \only<1->{\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 + \only<1->{\item Jetzt können wir den $\operatorname{ggT}$ von $f(X)$ und $e(X)$ berechnen} \end{itemize} \end{frame} @@ -608,8 +643,8 @@ \begin{frame} \frametitle{Der Euklidische Algorithmus (nochmal)} - $\operatorname{ggT}(f(X),e(X))$ hat den Grad 8 - + \only<1->{$\operatorname{ggT}(f(X),e(X))$ hat den Grad $8$} + \only<1->{ \[ \arraycolsep=1.4pt \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} @@ -620,7 +655,8 @@ & & & &6X^8&+&0X^7&+&p(X)& & & & & & & & & & & & \\ \end{array} \] - + } + \only<1->{ \[ \arraycolsep=1.4pt \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} @@ -629,14 +665,14 @@ && 7X^8&+& p(X)& & & & & & & & & & & & & & & & \\ \end{array} \] - + } \vspace{10pt} - $\operatorname{ggT}(f(X),e(X)) = 6X^8$ + \only<1->{$\operatorname{ggT}(f(X),e(X)) = 6X^8$} \vspace{10pt} - $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen + \only<1->{ $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen } \end{frame} @@ -653,7 +689,7 @@ & & $0$& $1$\\ $0$& $9X + 5$& $1$& $0$\\ $1$& $10X + 3$& $9X+5$& $1$\\ - $2$& & $2X^2 + 0X + 5$& $10X + 3$\\ + $2$& & \textcolor<2->{blue}{$2X^2 + 0X + 5$}& $10X + 3$\\ \hline \end{tabular} @@ -662,49 +698,54 @@ \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& $d(X) = (X-a^i)(X-a^i)$ + \only<1->{Somit erhalten wir den Faktor& $d(X) = 2X^2 + 5$\\} + \only<1->{Faktorisiert erhalten wir& $d(X) = 2(X-5)(X-6)$\\} + \only<1->{Lokatorpolynom& $d(X) = (X-a^i)(X-a^i)$} \end{tabular} \vspace{10pt} - + \only<1->{ \begin{center} $a^i = 5 \qquad \Rightarrow \qquad i = 3$ $a^i = 6 \qquad \Rightarrow \qquad i = 8$ \end{center} - - $D = [3,8]$ + } + \only<1->{$d(X) = (X-a^3)(X-a^8)$} \end{frame} -%------------------------------------------------------------------------------- +%------------------------------------------------------------------------------- +\section{Nachricht Rekonstruieren} \begin{frame} \frametitle{Rekonstruktion der Nachricht} \begin{itemize} - \item $W = [5,3,6,8,2,10,2,7,1,4]$ + \only<1->{\item $w = [5,3,6,8,2,10,2,7,1,4]$} - \item $D = [3,8]$ + \only<1->{\item $d(X) = (X-\textcolor<4->{red}{a^3})(X-\textcolor<4->{red}{a^8})$} \end{itemize} - + \only<1->{ \[ + \textcolor{gray}{ \begin{pmatrix} - 5 \\ 3 \\ 6 \\ 8 \\ 2 \\ 10 \\ 2 \\ 7 \\ 1 \\ 4 \\ + a^0 \\ a^1 \\ a^2 \\ \textcolor<4->{red}{a^3} \\ a^4 \\ a^5 \\ a^6 \\ a^7 \\ \textcolor<4->{red}{a^8} \\ a^9 \\ + \end{pmatrix}} + \begin{pmatrix} + 5 \\ 3 \\ 6 \\ \textcolor<4->{red}{8} \\ 2 \\ 10 \\ 2 \\ 7 \\ \textcolor<4->{red}{1} \\ 4 \\ \end{pmatrix} = \begin{pmatrix} 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& 8^{12}& 8^{14}& 8^{16}& 8^{18}\\ - 8^0& 8^3& 8^6& 8^9& 8^{12}& 8^{15}& 8^{18}& 8^{21}& 8^{24}& 8^{27}\\ + \textcolor<4->{red}{8^0}& \textcolor<4->{red}{8^3}& \textcolor<4->{red}{8^6}& \textcolor<4->{red}{8^9}& \textcolor<4->{red}{8^{12}}& \textcolor<4->{red}{8^{15}}& \textcolor<4->{red}{8^{18}}& \textcolor<4->{red}{8^{21}}& \textcolor<4->{red}{8^{24}}& \textcolor<4->{red}{8^{27}}\\ 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& 8^{24}& 8^{28}& 8^{32}& 8^{36}\\ 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& 8^{30}& 8^{35}& 8^{40}& 8^{45}\\ 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& 8^{36}& 8^{42}& 8^{48}& 8^{54}\\ 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& 8^{42}& 8^{49}& 8^{56}& 8^{63}\\ - 8^0& 8^8& 8^{16}& 8^{24}& 8^{32}& 8^{40}& 8^{48}& 8^{56}& 8^{64}& 8^{72}\\ + \textcolor<4->{red}{8^0}& \textcolor<4->{red}{8^8}& \textcolor<4->{red}{8^{16}}& \textcolor<4->{red}{8^{24}}& \textcolor<4->{red}{8^{32}}& \textcolor<4->{red}{8^{40}}& \textcolor<4->{red}{8^{48}}& \textcolor<4->{red}{8^{56}}& \textcolor<4->{red}{8^{64}}& \textcolor<4->{red}{8^{72}}\\ 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& 8^{54}& 8^{63}& 8^{72}& 8^{81}\\ \end{pmatrix} \cdot @@ -712,13 +753,14 @@ m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ \end{pmatrix} \] - + } + \only<1->{ \begin{itemize} \item Fehlerstellen entfernen \end{itemize} - + } \end{frame} -%------------------------------------------------------------------------------- +%------------------------------------------------------------------------------- \begin{frame} \frametitle{Rekonstruktion der Nachricht} @@ -728,25 +770,25 @@ \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}\\ + 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}\\ + 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& \textcolor<3->{green}{8^6}& \textcolor<3->{green}{8^7}& \textcolor<3->{green}{8^8}& \textcolor<3->{green}{8^9}\\ + 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& \textcolor<3->{green}{8^{12}}& \textcolor<3->{green}{8^{14}}& \textcolor<3->{green}{8^{16}}& \textcolor<3->{green}{8^{18}}\\ + 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& \textcolor<3->{green}{8^{24}}& \textcolor<3->{green}{8^{28}}& \textcolor<3->{green}{8^{32}}& \textcolor<3->{green}{8^{36}}\\ + 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& \textcolor<3->{green}{8^{30}}& \textcolor<3->{green}{8^{35}}& \textcolor<3->{green}{8^{40}}& \textcolor<3->{green}{8^{45}}\\ + 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& \textcolor<3->{green}{8^{36}}& \textcolor<3->{green}{8^{42}}& \textcolor<3->{green}{8^{48}}& \textcolor<3->{green}{8^{54}}\\ + 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& \textcolor<3->{green}{8^{42}}& \textcolor<3->{green}{8^{49}}& \textcolor<3->{green}{8^{56}}& \textcolor<3->{green}{8^{63}}\\ + 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& \textcolor<3->{green}{8^{54}}& \textcolor<3->{green}{8^{63}}& \textcolor<3->{green}{8^{72}}& \textcolor<3->{green}{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 \\ + 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} \] - + \only<1->{ \begin{itemize} \item Nullstellen entfernen \end{itemize} - + } \end{frame} %------------------------------------------------------------------------------- \begin{frame} @@ -754,7 +796,7 @@ \[ \begin{pmatrix} - 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ 7 \\ 4 \\ + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor<2->{red}{7} \\ \textcolor<2->{red}{4} \\ \end{pmatrix} = \begin{pmatrix} @@ -764,8 +806,8 @@ 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ - 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}\\ - 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}\\ + \textcolor<2->{red}{8^0}& \textcolor<2->{red}{8^7}& \textcolor<2->{red}{8^{14}}& \textcolor<2->{red}{8^{21}}& \textcolor<2->{red}{8^{28}}& \textcolor<2->{red}{8^{35}}\\ + \textcolor<2->{red}{8^0}& \textcolor<2->{red}{8^9}& \textcolor<2->{red}{8^{18}}& \textcolor<2->{red}{8^{27}}& \textcolor<2->{red}{8^{36}}& \textcolor<2->{red}{8^{45}}\\ \end{pmatrix} \cdot \begin{pmatrix} @@ -774,11 +816,11 @@ \] \vspace{5pt} - + \only<1->{ \begin{itemize} \item Matrix in eine Quadratische Form bringen \end{itemize} - + } \end{frame} %------------------------------------------------------------------------------- \begin{frame} -- cgit v1.2.1 From 8c6a8e56c125c238dc64c21d1269fcdc7542c5cd Mon Sep 17 00:00:00 2001 From: JODBaer Date: Thu, 22 Apr 2021 23:45:32 +0200 Subject: =?UTF-8?q?merge=20lines=20gel=C3=B6scht?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/reedsolomon/RS presentation/RS.tex | 9 +++------ 1 file changed, 3 insertions(+), 6 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 61324f7..943f2da 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -84,11 +84,9 @@ \end{tabular} \end{center} \end{frame} -<<<<<<< Updated upstream + %------------------------------------------------------------------------------- -======= ->>>>>>> Stashed changes \section{Diskrete Fourier Transformation} \begin{frame} \frametitle{Idee} @@ -184,13 +182,12 @@ Indem in einem endlichen Körper gerechnet wird. } \end{frame} -<<<<<<< Updated upstream + %------------------------------------------------------------------------------- -======= + \section{Reed-Solomon in Endlichen Körpern} ->>>>>>> Stashed changes \begin{frame} \frametitle{Reed-Solomon in Endlichen Körpern} -- cgit v1.2.1 From 179ea16b001b6640e9b720d53ffc06f3e2389ff2 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Fri, 23 Apr 2021 00:30:36 +0200 Subject: appostroph verbessert --- buch/papers/reedsolomon/RS presentation/RS.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 943f2da..d09d77d 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -64,7 +64,7 @@ \begin{center} \begin{tabular}{ c c c } \hline - ``Nutzlas´´ & Fehler & Versenden \\ + ``Nutzlast'' & Fehler & Versenden \\ \hline 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ -- cgit v1.2.1 From ded210e33924d4c078e5a0d899c0585d7f987565 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Fri, 23 Apr 2021 12:58:40 +0200 Subject: Folien Verbesserungen animation --- buch/papers/reedsolomon/RS presentation/RS.aux | 167 +++++++++++----- buch/papers/reedsolomon/RS presentation/RS.log | 212 ++++++++++++++++----- buch/papers/reedsolomon/RS presentation/RS.nav | 117 ++++++++---- buch/papers/reedsolomon/RS presentation/RS.out | 9 +- buch/papers/reedsolomon/RS presentation/RS.pdf | Bin 135643 -> 207741 bytes buch/papers/reedsolomon/RS presentation/RS.snm | 2 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[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 index 913bc42..d9d6693 100644 Binary files a/buch/papers/reedsolomon/RS presentation/RS.pdf and b/buch/papers/reedsolomon/RS presentation/RS.pdf differ diff --git a/buch/papers/reedsolomon/RS presentation/RS.snm b/buch/papers/reedsolomon/RS presentation/RS.snm index 6607ea8..86859c9 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.snm +++ b/buch/papers/reedsolomon/RS presentation/RS.snm @@ -1 +1 @@ -\beamer@slide {ft_discrete}{15} +\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 index 001b5c8..04bd239 100644 Binary files a/buch/papers/reedsolomon/RS presentation/RS.synctex.gz and b/buch/papers/reedsolomon/RS presentation/RS.synctex.gz differ diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index d09d77d..7b2c4da 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -43,18 +43,18 @@ ${f}_2=\textcolor{blue}{2}$, ${f}_1=\textcolor{blue}{1}$, ${f}_0=\textcolor{blue}{5}$ als $ p(w) = \textcolor{blue}{2}w^2 + \textcolor{blue}{1}w + \textcolor{blue}{5} $. - \only<1>{ - Versende $ (p(1),p(2),\dots,p(7)) = (\textcolor{green}{8}, - \textcolor{green}{15}, \textcolor{green}{26}, - \textcolor{green}{41}, \textcolor{green}{60}, - \textcolor{green}{83}, \textcolor{green}{110})$ - \includegraphics[scale = 1.2]{images/polynom1.pdf}} - \only<2>{ - 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} + + 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} diff --git a/buch/papers/reedsolomon/RS presentation/RS.toc b/buch/papers/reedsolomon/RS presentation/RS.toc index 44c06ab..095b5e6 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.toc +++ b/buch/papers/reedsolomon/RS presentation/RS.toc @@ -1,6 +1,9 @@ \babel@toc {ngerman}{} \beamer@sectionintoc {1}{Einführung}{2}{0}{1} -\beamer@sectionintoc {2}{Polynom Ansatz}{3}{0}{2} -\beamer@sectionintoc {3}{Fourier Transformation}{7}{0}{3} -\beamer@sectionintoc {4}{Diskrete Fourier Transformation}{15}{0}{4} -\beamer@sectionintoc {5}{Probleme und Fragen}{17}{0}{5} +\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} -- cgit v1.2.1 From 0a80be4477602e2d909e5eda40dae485ec6acd56 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Fri, 23 Apr 2021 13:02:38 +0200 Subject: Read me erstellt --- buch/papers/reedsolomon/RS presentation/README.txt | 1 + 1 file changed, 1 insertion(+) create mode 100644 buch/papers/reedsolomon/RS presentation/README.txt (limited to 'buch/papers') 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 \ No newline at end of file -- cgit v1.2.1 From d1b6d92a02d9c44b3860b73d5660c5c6863de0df Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Fri, 23 Apr 2021 21:19:34 +0200 Subject: handout added --- buch/papers/reedsolomon/RS presentation/RS.tex | 290 +++---- .../reedsolomon/RS presentation/RS_handout.tex | 921 +++++++++++++++++++++ 2 files changed, 1069 insertions(+), 142 deletions(-) create mode 100644 buch/papers/reedsolomon/RS presentation/RS_handout.tex (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/RS presentation/RS.tex b/buch/papers/reedsolomon/RS presentation/RS.tex index 943f2da..c215e66 100644 --- a/buch/papers/reedsolomon/RS presentation/RS.tex +++ b/buch/papers/reedsolomon/RS presentation/RS.tex @@ -14,8 +14,8 @@ \institute{OST Ostschweizer Fachhochschule} \date{26.04.2021} \subject{Mathematisches Seminar} - \setbeamercovered{transparent} - %\setbeamercovered{invisible} + %\setbeamercovered{transparent} + \setbeamercovered{invisible} \setbeamertemplate{navigation symbols}{} \begin{frame}[plain] \maketitle @@ -64,22 +64,22 @@ \begin{center} \begin{tabular}{ c c c } \hline - ``Nutzlas´´ & Fehler & Versenden \\ + Nutzlas & Fehler & Versenden \\ \hline 3 & 2 & 7 Werte eines Polynoms vom Grad 2 \\ 4 & 2 & 8 Werte eines Polynoms vom Grad 3 \\ -\visible<2->{3}& -\visible<2->{3}& -\visible<3->{9 Werte eines Polynoms vom Grad 2} \\ +\visible<1->{3}& +\visible<1->{3}& +\visible<1->{9 Werte eines Polynoms vom Grad 2} \\ &&\\ -\visible<4->{$k$} & -\visible<4->{$t$} & -\visible<4->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ +\visible<1->{$k$} & +\visible<1->{$t$} & +\visible<1->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ \hline &&\\ &&\\ \multicolumn{3}{l} { - \visible<4>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} + \visible<1>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} } \end{tabular} \end{center} @@ -194,21 +194,21 @@ \begin{itemize} \onslide<1->{\item Warum endliche Körper?} - \onslide<1->{\qquad konkrete Zahlen $\rightarrow$ keine Rundungsfehler} + \onslide<2->{\qquad konkrete Zahlen $\rightarrow$ keine Rundungsfehler} - \onslide<1->{\qquad digitale Fehlerkorrektur} + \onslide<3->{\qquad digitale Fehlerkorrektur} - \onslide<1->{\qquad bessere Laufzeit} + %\onslide<4->{\qquad bessere Laufzeit} \vspace{10pt} - \onslide<1->{\item Nachricht = Nutzdaten + Fehlerkorrekturteil} + \onslide<4->{\item Nachricht = Nutzdaten + Fehlerkorrekturteil} \vspace{10pt} - \onslide<1->{\item aus Fehlerkorrekturteil die Fehlerstellen finden} + \onslide<5->{\item aus Fehlerkorrekturteil die Fehlerstellen finden} - \onslide<1->{\qquad $\Rightarrow$ gesucht ist ein Lokatorpolynom} + \onslide<6->{\qquad $\Rightarrow$ gesucht ist ein Lokatorpolynom} % \vspace{10pt} @@ -232,33 +232,33 @@ \begin{itemize} - \only<1->{\item endlicher Körper $q = 11$} + \onslide<1->{\item endlicher Körper $q = 11$} - \only<1->{ist eine Primzahl} + \onslide<2->{ist eine Primzahl} - \only<1->{beinhaltet die Zahlen $\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}$} + \onslide<3->{beinhaltet die Zahlen $\mathbb{F}_{11} = \{0,1,2,3,4,5,6,7,8,9,10\}$} \vspace{10pt} - \only<1->{\item Nachrichtenblock $=$ Nutzlast $+$ Fehlerkorrekturstellen + \onslide<4->{\item Nachrichtenblock $=$ Nutzlast $+$ Fehlerkorrekturstellen} - $n = q - 1 = 10$ Zahlen} + \onslide<5->{$n = q - 1 = 10$ Zahlen} \vspace{10pt} - \only<1->{\item Max.~Fehler $z = 2$ + \onslide<6->{\item Max.~Fehler $t = 2$} - maximale Anzahl von Fehler, die wir noch korrigieren können} + \onslide<7->{maximale Anzahl von Fehler, die wir noch korrigieren können} \vspace{10pt} - \only<1->{\item Nutzlast $k = n -2t = 6$ Zahlen} + \onslide<8->{\item Nutzlast $k = n -2t = 6$ Zahlen} - \only<1->{Fehlerkorrkturstellen $2t = 4$ Zahlen} + \onslide<9->{Fehlerkorrkturstellen $2t = 4$ Zahlen} - \only<1->{Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$} + \onslide<10->{Nachricht $m = [0,0,0,0,4,7,2,5,8,1]$} - \only<1->{als Polynom $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} + \onslide<11->{als Polynom $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} \end{itemize} @@ -269,31 +269,31 @@ \frametitle{Codierung} \begin{itemize} - \only<1->{\item Ansatz aus den komplexen Zahlen mit der diskreten Fouriertransformation} + \onslide<1->{\item Ansatz aus den komplexen Zahlen mit der diskreten Fouriertransformation} \vspace{10pt} - \only<1->{\item Eulersche Zahl $\mathrm{e}$ existiert nicht in $\mathbb{F}_{11}$} + \onslide<2->{\item Eulersche Zahl $\mathrm{e}$ existiert nicht in $\mathbb{F}_{11}$} \vspace{10pt} - \only<1->{\item Wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken + \onslide<3->{\item Wir suchen $a$ so, dass $a^i$ den gesamten Zahlenbereich von $\mathbb{F}_{11}$ abdecken} - $\mathbb{Z}_{11}\setminus\{0\} = \{a^0, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9\}$} + \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} - \only<1->{\item Wir wählen $a = 8$} + \onslide<5->{\item Wir wählen $a = 8$} - \only<1->{$\mathbb{Z}_{11}\setminus\{0\} = \{1,8,9,6,4,10,3,2,5,7\}$} + \onslide<6->{$\mathbb{Z}_{11}\setminus\{0\} = \{1,8,9,6,4,10,3,2,5,7\}$} - \only<1->{$8$ ist eine primitive Einheitswurzel} + \onslide<7->{$8$ ist eine primitive Einheitswurzel} \vspace{10pt} - \only<1->{\item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$} + \onslide<8->{\item $m(8^0) = 4\cdot1 + 7\cdot1 + 2\cdot1 + 5\cdot1 + 8\cdot1 + 1 = 5$} - \only<1->{$\Rightarrow$ \qquad können wir auch als Matrix schreiben} + \onslide<9->{$\Rightarrow$ \qquad können wir auch als Matrix schreiben} \end{itemize} @@ -303,14 +303,14 @@ \frametitle{Codierung} \begin{itemize} - \only<1->{\item Übertragungsvektor $v$} + \onslide<1->{\item Übertragungsvektor $v$} - \only<1->{\item $v = A \cdot m$} + \onslide<2->{\item $v = A \cdot m$} \end{itemize} \[ - \only<1->{ + \onslide<3->{ v = \begin{pmatrix} 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& 8^0\\ 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& 8^6& 8^7& 8^8& 8^9\\ @@ -329,11 +329,11 @@ \end{pmatrix} } \] - \only<1->{ + \begin{itemize} - \item $v = [5,3,6,5,2,10,2,7,10,4]$ + \onslide<4->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- \section{Decodierung ohne Fehler} @@ -341,41 +341,44 @@ \frametitle{Decodierung ohne Fehler} \begin{itemize} - \only<1->{\item Der Empfänger erhält den unveränderten Vektor - $v = [5,3,6,5,2,10,2,7,10,4]$} + \onslide<1->{\item Der Empfänger erhält den unveränderten Vektor $v = [5,3,6,5,2,10,2,7,10,4]$} \vspace{10pt} - \only<1->{\item Wir suchen die Inverse der Matrix $A$} + \onslide<2->{\item Wir suchen die Inverse der Matrix $A$} \vspace{10pt} \end{itemize} \begin{columns}[t] - \begin{column}{0.50\textwidth} - \only<1->{ - Inverse der Fouriertransformation + \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}{0.50\textwidth} - \only<1->{ - Inverse von $a$} + \begin{column}{0.45\textwidth} + \onslide<6->{Inverse von $a$} + \vspace{10pt} - \only<1->{ + + \onslide<7->{ \[ 8^{1} \Rightarrow 8^{-1} \] } - \only<1->{Inverse finden wir über den Eulkidischen Algorithmus} + + \onslide<8->{Inverse finden wir über den Eulkidischen Algorithmus} \vspace{10pt} \end{column} \end{columns} @@ -407,7 +410,7 @@ \begin{column}{0.50\textwidth} \begin{center} - \only<1->{ + \onslide<1->{ \begin{tabular}{| c | c c | c | r r |} \hline $k$ & $a_i$ & $b_i$ & $q_i$ & $c_i$ & $d_i$\\ @@ -417,17 +420,18 @@ $1$& $11$& $8$& $1$& $1$& $0$\\ $2$& $8$& $3$& $2$& $-1$& $1$\\ $3$& $3$& $2$& $1$& $3$& $-2$\\ - $4$& $2$& $1$& $2$& \textcolor<3->{blue}{$-4$}& \textcolor<3->{red}{$3$}\\ + $4$& $2$& $1$& $2$& \textcolor<2->{blue}{$-4$}& \textcolor<2->{red}{$3$}\\ $5$& $1$& $0$& & $11$& $-8$\\ \hline \end{tabular} } + \vspace{10pt} \begin{tabular}{rcl} - \only<1->{$\textcolor{blue}{-4} \cdot 8 + \textcolor{red}{3} \cdot 11$ &$=$& $1$}\\ - \only<1->{$7 \cdot 8 + 3 \cdot 11$ &$=$& $1$}\\ - \only<1->{$8^{-1}$ &$=$& $7$} + \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} @@ -442,16 +446,16 @@ \frametitle{Decodierung mit Inverser Matrix} \begin{itemize} - \only<1->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} + \onslide<1->{\item $v = [5,3,6,5,2,10,2,7,10,4]$} - \only<1->{\item $m = 1/10 \cdot A^{-1} \cdot v$} + \onslide<2->{\item $m = 1/10 \cdot A^{-1} \cdot v$} - \only<1->{\item $m = 10 \cdot A^{-1} \cdot v$} + \onslide<3->{\item $m = 10 \cdot A^{-1} \cdot v$} \end{itemize} - \only<1->{ + \onslide<4->{ \[ - m = \begin{pmatrix} + m = 10 \cdot \begin{pmatrix} 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0& 7^0\\ 7^0& 7^1& 7^2& 7^3& 7^4& 7^5& 7^6& 7^7& 7^8& 7^9\\ 7^0& 7^2& 7^4& 7^6& 7^8& 7^{10}& 7^{12}& 7^{14}& 7^{16}& 7^{18}\\ @@ -469,11 +473,11 @@ \end{pmatrix} \] } - \only<1->{ + \begin{itemize} - \item $m = [0,0,0,0,4,7,2,5,8,1]$ + \onslide<5->{\item $m = [0,0,0,0,4,7,2,5,8,1]$} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- \section{Decodierung mit Fehler} @@ -481,48 +485,46 @@ \frametitle{Decodierung mit Fehler - Ansatz} \begin{itemize} - \only<1->{\item Gesendet: $v = [5,3,6,5,2,10,2,7,10,4]$} + \onslide<1->{\item Gesendet: $v = [5,3,6,5,2,10,2,7,10,4]$} - \only<1->{\item Empfangen: $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} + \onslide<2->{\item Empfangen: $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} - \only<1->{\item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]$} + \onslide<3->{\item Rücktransformation: $r = [\underbrace{5,7,4,10,}_{Fehlerinfo}5,4,5,7,6,7]$} \end{itemize} - \only<1->{Wie finden wir die Fehler?} + \onslide<4->{Wie finden wir die Fehler?} - \only<1->{ \begin{itemize} - \item $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$ + \onslide<5->{\item $m(X) = 4X^5 + 7X^4 + 2X^3 + 5X^2 + 8X + 1$} - \item $r(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + 5X^5 + 4X^4 + 5X^3 + 7X^2 + 6X + 7$ + \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 $e(X) = r(X) - m(X)$} - \item $e(X) = r(X) - m(X)$ + \onslide<7->{\item $e(X) = r(X) - m(X)$} \end{itemize} - } \begin{center} - \only<1->{ + \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 - $r(a^{i})$& \only<1->{$5$& $3$& $6$& $8$& $2$& $10$& $2$& $7$& $1$& $4$}\\ - $m(a^{i})$& \only<1->{$5$& $3$& $6$& $5$& $2$& $10$& $2$& $7$& $10$& $4$}\\ - $e(a^{i})$& \only<1->{$0$& $0$& $0$& $3$& $0$& $0$& $0$& $0$& $2$& $0$}\\ + $r(a^{i})$& \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} - \only<1->{ + \begin{itemize} - \item Alle Stellen, die nicht Null sind, sind Fehler + \onslide<12->{\item Alle Stellen, die nicht Null sind, sind Fehler} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- @@ -530,31 +532,31 @@ \frametitle{Nullstellen des Fehlerpolynoms finden} \begin{itemize} - \only<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} + \onslide<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} \vspace{10pt} - \only<1->{\item $f(X) = X^{10}-1 = 0$ \qquad für $X \in \{1,2,3,4,5,6,7,8,9,10\}$} + \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} - \only<1->{\item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + \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} - \only<1->{\item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + \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} - \only<1->{\item $\operatorname{ggT}$ gibt uns eine Liste der Nullstellen, an denen es keine Fehler gegeben hat} + \onslide<5->{\item $\operatorname{ggT}$ gibt uns eine Liste der Nullstellen, an denen es keine Fehler gegeben hat} \vspace{10pt} - \only<1->{$\operatorname{ggT}(f(X),e(X)) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + \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 $(X-a^7) \qquad \qquad (X-a^9)$} @@ -567,39 +569,39 @@ \begin{itemize} - \item Satz von Fermat: $f(X) = X^{q-1}-1=0$ + \onslide<1->{\item Satz von Fermat: $f(X) = X^{q-1}-1=0$} \vspace{10pt} - \item $f(X) = X^{10}-1 = 0$ \qquad für $X = [1,2,3,4,5,6,7,8,9,10]$ + \onslide<1->{\item $f(X) = X^{10}-1 = 0$ \qquad für $X = [1,2,3,4,5,6,7,8,9,10]$} \vspace{10pt} - \item $f(X) = (X-a^0)(X-a^1)(X-a^2)(X-a^3)(X-a^4)(X-a^5)(X-a^6) \cdot$ + \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 $(X-a^7)(X-a^8)(X-a^9)$ + \qquad \qquad $(X-a^7)(X-a^8)(X-a^9)$} \vspace{10pt} - \item $e(X) = (X-a^0)(X-a^1)(X-a^2) \qquad \qquad (X-a^4)(X-a^5)(X-a^6) \cdot$ + \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 $(X-a^7) \qquad \qquad (X-a^9) \cdot p(x)$ + \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 + \onslide<1->{\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 $ + \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 $(X-a^7)(X-a^8)(X-a^9) \cdot q(X)$ + \qquad \qquad \qquad \qquad $(X-a^7)(X-a^8)(X-a^9) \cdot q(X)$} - $= d(X) \cdot e(X)$ + \onslide<3->{$= d(X) \cdot e(X)$} \vspace{10pt} - \item Lokatorpolynom $d(X) = (X-a^3)(X-a^8)$ + \onslide<4->{\item Lokatorpolynom $d(X) = (X-a^3)(X-a^8)$} \end{itemize} @@ -610,29 +612,29 @@ \begin{itemize} - \only<1->{\item $e(X)$ ist unbekannt auf der Empfängerseite} + \onslide<1->{\item $e(X)$ ist unbekannt auf der Empfängerseite} \vspace{10pt} - \only<1->{\item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt?} + \onslide<2->{\item $e(X) = r(X) - m(X)$ \qquad $\rightarrow$ \qquad $m(X)$ ist unbekannt?} \vspace{10pt} - \only<1->{\item $m$ ist nicht gänzlich unbekannt: $m = [0,0,0,0,?,?,?,?,?,?]$ + \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} - \only<1->{\item Daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$} + \onslide<4->{\item Daraus folgt $e(X) = 5X^9 + 7X^8 + 4X^7 + 10X^6 + p(X)$} \vspace{10pt} - \only<1->{\item $f(X) = X^{10} - 1 = X^{10} + 10$} + \onslide<5->{\item $f(X) = X^{10} - 1 = X^{10} + 10$} \vspace{10pt} - \only<1->{\item Jetzt können wir den $\operatorname{ggT}$ von $f(X)$ und $e(X)$ berechnen} + \onslide<6->{\item Jetzt können wir den $\operatorname{ggT}$ von $f(X)$ und $e(X)$ berechnen} \end{itemize} \end{frame} @@ -640,8 +642,8 @@ \begin{frame} \frametitle{Der Euklidische Algorithmus (nochmal)} - \only<1->{$\operatorname{ggT}(f(X),e(X))$ hat den Grad $8$} - \only<1->{ + \onslide<1->{$\operatorname{ggT}(f(X),e(X))$ hat den Grad $8$} + \onslide<2->{ \[ \arraycolsep=1.4pt \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} @@ -653,7 +655,7 @@ \end{array} \] } - \only<1->{ + \onslide<3->{ \[ \arraycolsep=1.4pt \begin{array}{rcrcrcrcccrcrcrcrcrcrcrcrcr} @@ -665,11 +667,11 @@ } \vspace{10pt} - \only<1->{$\operatorname{ggT}(f(X),e(X)) = 6X^8$} + \onslide<4->{$\operatorname{ggT}(f(X),e(X)) = 6X^8$} \vspace{10pt} - \only<1->{ $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen } + \onslide<5->{ $\operatorname{kgV}$ durch den erweiterten Euklidischen Algorithmus bestimmen } \end{frame} @@ -695,20 +697,22 @@ \vspace{10pt} \begin{tabular}{ll} - \only<1->{Somit erhalten wir den Faktor& $d(X) = 2X^2 + 5$\\} - \only<1->{Faktorisiert erhalten wir& $d(X) = 2(X-5)(X-6)$\\} - \only<1->{Lokatorpolynom& $d(X) = (X-a^i)(X-a^i)$} + \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} - \only<1->{ + + \onslide<6->{ \begin{center} $a^i = 5 \qquad \Rightarrow \qquad i = 3$ $a^i = 6 \qquad \Rightarrow \qquad i = 8$ \end{center} - } - \only<1->{$d(X) = (X-a^3)(X-a^8)$} + } + + \onslide<7->{$d(X) = (X-a^3)(X-a^8)$} \end{frame} %------------------------------------------------------------------------------- @@ -718,12 +722,12 @@ \begin{itemize} - \only<1->{\item $w = [5,3,6,8,2,10,2,7,1,4]$} + \onslide<1->{\item $w = [5,3,6,\textcolor{red}{8},2,10,2,7,\textcolor{red}{1},4]$} - \only<1->{\item $d(X) = (X-\textcolor<4->{red}{a^3})(X-\textcolor<4->{red}{a^8})$} + \onslide<2->{\item $d(X) = (X-\textcolor<4->{red}{a^3})(X-\textcolor<4->{red}{a^8})$} \end{itemize} - \only<1->{ + \onslide<3->{ \[ \textcolor{gray}{ \begin{pmatrix} @@ -751,11 +755,11 @@ \end{pmatrix} \] } - \only<1->{ + \begin{itemize} - \item Fehlerstellen entfernen + \onslide<5->{\item Fehlerstellen entfernen} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- \begin{frame} @@ -767,25 +771,25 @@ \end{pmatrix} = \begin{pmatrix} - 8^0& 8^0& 8^0& 8^0& 8^0& 8^0& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}& \textcolor<3->{green}{8^0}\\ - 8^0& 8^1& 8^2& 8^3& 8^4& 8^5& \textcolor<3->{green}{8^6}& \textcolor<3->{green}{8^7}& \textcolor<3->{green}{8^8}& \textcolor<3->{green}{8^9}\\ - 8^0& 8^2& 8^4& 8^6& 8^8& 8^{10}& \textcolor<3->{green}{8^{12}}& \textcolor<3->{green}{8^{14}}& \textcolor<3->{green}{8^{16}}& \textcolor<3->{green}{8^{18}}\\ - 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}& \textcolor<3->{green}{8^{24}}& \textcolor<3->{green}{8^{28}}& \textcolor<3->{green}{8^{32}}& \textcolor<3->{green}{8^{36}}\\ - 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}& \textcolor<3->{green}{8^{30}}& \textcolor<3->{green}{8^{35}}& \textcolor<3->{green}{8^{40}}& \textcolor<3->{green}{8^{45}}\\ - 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}& \textcolor<3->{green}{8^{36}}& \textcolor<3->{green}{8^{42}}& \textcolor<3->{green}{8^{48}}& \textcolor<3->{green}{8^{54}}\\ - 8^0& 8^7& 8^{14}& 8^{21}& 8^{28}& 8^{35}& \textcolor<3->{green}{8^{42}}& \textcolor<3->{green}{8^{49}}& \textcolor<3->{green}{8^{56}}& \textcolor<3->{green}{8^{63}}\\ - 8^0& 8^9& 8^{18}& 8^{27}& 8^{36}& 8^{45}& \textcolor<3->{green}{8^{54}}& \textcolor<3->{green}{8^{63}}& \textcolor<3->{green}{8^{72}}& \textcolor<3->{green}{8^{81}}\\ + 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} \] - \only<1->{ + \begin{itemize} - \item Nullstellen entfernen + \onslide<3->{\item Nullstellen entfernen} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- \begin{frame} @@ -793,7 +797,7 @@ \[ \begin{pmatrix} - 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor<2->{red}{7} \\ \textcolor<2->{red}{4} \\ + 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \textcolor<3->{red}{7} \\ \textcolor<3->{red}{4} \\ \end{pmatrix} = \begin{pmatrix} @@ -803,8 +807,8 @@ 8^0& 8^4& 8^8& 8^{12}& 8^{16}& 8^{20}\\ 8^0& 8^5& 8^{10}& 8^{15}& 8^{20}& 8^{25}\\ 8^0& 8^6& 8^{12}& 8^{18}& 8^{24}& 8^{30}\\ - \textcolor<2->{red}{8^0}& \textcolor<2->{red}{8^7}& \textcolor<2->{red}{8^{14}}& \textcolor<2->{red}{8^{21}}& \textcolor<2->{red}{8^{28}}& \textcolor<2->{red}{8^{35}}\\ - \textcolor<2->{red}{8^0}& \textcolor<2->{red}{8^9}& \textcolor<2->{red}{8^{18}}& \textcolor<2->{red}{8^{27}}& \textcolor<2->{red}{8^{36}}& \textcolor<2->{red}{8^{45}}\\ + \textcolor<3->{red}{8^0}& \textcolor<3->{red}{8^7}& \textcolor<3->{red}{8^{14}}& \textcolor<3->{red}{8^{21}}& \textcolor<3->{red}{8^{28}}& \textcolor<3->{red}{8^{35}}\\ + \textcolor<3->{red}{8^0}& \textcolor<3->{red}{8^9}& \textcolor<3->{red}{8^{18}}& \textcolor<3->{red}{8^{27}}& \textcolor<3->{red}{8^{36}}& \textcolor<3->{red}{8^{45}}\\ \end{pmatrix} \cdot \begin{pmatrix} @@ -813,11 +817,11 @@ \] \vspace{5pt} - \only<1->{ + \begin{itemize} - \item Matrix in eine Quadratische Form bringen + \onslide<2->{\item Matrix in eine Quadratische Form bringen} \end{itemize} - } + \end{frame} %------------------------------------------------------------------------------- \begin{frame} @@ -845,7 +849,7 @@ \vspace{5pt} \begin{itemize} - \item Matrix Invertieren + \onslide<2->{\item Matrix Invertieren} \end{itemize} \end{frame} @@ -873,9 +877,10 @@ \] \begin{center} - $\Downarrow$ + \onslide<2->{$\Downarrow$} \end{center} \[ + \onslide<3->{ \begin{pmatrix} m_0 \\ m_1 \\ m_2 \\ m_3 \\ m_4 \\ m_5 \\ \end{pmatrix} @@ -892,6 +897,7 @@ \begin{pmatrix} 5 \\ 3 \\ 6 \\ 2 \\ 10 \\ 2 \\ \end{pmatrix} + } \] \end{frame} @@ -919,7 +925,7 @@ \] \begin{itemize} - \item $m = [4,7,2,5,8,1]$ + \onslide<2->{\item $m = [4,7,2,5,8,1]$} \end{itemize} \end{frame} diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.tex b/buch/papers/reedsolomon/RS presentation/RS_handout.tex new file mode 100644 index 0000000..863b3a2 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS_handout.tex @@ -0,0 +1,921 @@ +\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} $. + + \only<1>{ + Versende $ (p(1),p(2),\dots,p(7)) = (\textcolor{green}{8}, + \textcolor{green}{15}, \textcolor{green}{26}, + \textcolor{green}{41}, \textcolor{green}{60}, + \textcolor{green}{83}, \textcolor{green}{110})$ + \includegraphics[scale = 1.2]{images/polynom1.pdf}} + \only<2>{ + 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 \\ +\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} + \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} -- cgit v1.2.1 From df810d1315cfb1c4b876d5145846d6ea70753141 Mon Sep 17 00:00:00 2001 From: JODBaer Date: Sat, 24 Apr 2021 15:27:05 +0200 Subject: Handout animation deleted --- .../reedsolomon/RS presentation/RS_handout.aux | 143 +++ 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a/buch/papers/reedsolomon/RS presentation/RS_handout.out b/buch/papers/reedsolomon/RS presentation/RS_handout.out new file mode 100644 index 0000000..364319e --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS_handout.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_handout.pdf b/buch/papers/reedsolomon/RS presentation/RS_handout.pdf new file mode 100644 index 0000000..382049d Binary files /dev/null and b/buch/papers/reedsolomon/RS presentation/RS_handout.pdf differ diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.snm b/buch/papers/reedsolomon/RS presentation/RS_handout.snm new file mode 100644 index 0000000..1796304 --- /dev/null +++ b/buch/papers/reedsolomon/RS presentation/RS_handout.snm @@ -0,0 +1 @@ +\beamer@slide {ft_discrete}{13} diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.synctex.gz b/buch/papers/reedsolomon/RS presentation/RS_handout.synctex.gz new file mode 100644 index 0000000..c28a28a Binary files /dev/null and b/buch/papers/reedsolomon/RS presentation/RS_handout.synctex.gz differ diff --git a/buch/papers/reedsolomon/RS presentation/RS_handout.tex b/buch/papers/reedsolomon/RS presentation/RS_handout.tex index 863b3a2..1cbb6ef 100644 --- a/buch/papers/reedsolomon/RS presentation/RS_handout.tex +++ b/buch/papers/reedsolomon/RS presentation/RS_handout.tex @@ -25,38 +25,29 @@ \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.} + \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 Beispiel $2, 1, 5$ versenden und auf 2 Fehler absichern + \item $2, 1, 5$ 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} $. - - \only<1>{ - Versende $ (p(1),p(2),\dots,p(7)) = (\textcolor{green}{8}, - \textcolor{green}{15}, \textcolor{green}{26}, - \textcolor{green}{41}, \textcolor{green}{60}, - \textcolor{green}{83}, \textcolor{green}{110})$ - \includegraphics[scale = 1.2]{images/polynom1.pdf}} - \only<2>{ - 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.} + \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} @@ -68,18 +59,14 @@ \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} \\ + 3& 3& 9 Werte eines Polynoms vom Grad 2 \\ &&\\ -\visible<1->{$k$} & -\visible<1->{$t$} & -\visible<1->{$k+2t$ Werte eines Polynoms vom Grad $k-1$} \\ + $k$ & $t$ & $k+2t$ Werte eines Polynoms vom Grad $k-1$ \\ \hline &&\\ &&\\ \multicolumn{3}{l} { - \visible<1>{Ausserdem können bis zu $2t$ Fehler erkannt werden!} + Ausserdem können bis zu $2t$ Fehler erkannt werden! } \end{tabular} \end{center} @@ -127,23 +114,23 @@ \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} @@ -177,10 +164,9 @@ \frametitle{Probleme und Fragen} Wie wird der Fehler lokalisiert? - \visible<2>{ \newline Indem in einem endlichen Körper gerechnet wird. - } + \end{frame} %------------------------------------------------------------------------------- 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} -- cgit v1.2.1 From dd7bd6ca3b6517435dfc6b740ab96f51aa15ac2e Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Sun, 16 May 2021 16:03:36 +0200 Subject: edit main.tex add chapters --- buch/papers/reedsolomon/main.tex | 10 +++++++++- 1 file changed, 9 insertions(+), 1 deletion(-) (limited to 'buch/papers') 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} -- cgit v1.2.1 From 898274b6cb5f825fe710eec58349799cdc5f6bc3 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Sun, 16 May 2021 16:04:13 +0200 Subject: create endlichekoerper.tex added chapter description --- buch/papers/reedsolomon/endlichekoerper.tex | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) create mode 100644 buch/papers/reedsolomon/endlichekoerper.tex (limited to 'buch/papers') 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. -- cgit v1.2.1 From 46fa4763d730b1312741eefb8a2981c73389ccae Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Mon, 17 May 2021 19:32:32 +0200 Subject: update of codebsp started, restetabelle 1&2 created --- buch/papers/reedsolomon/codebsp.tex | 71 +++++++++++++++++++++++++++++++ buch/papers/reedsolomon/restetabelle1.tex | 24 +++++++++++ buch/papers/reedsolomon/restetabelle2.tex | 24 +++++++++++ 3 files changed, 119 insertions(+) create mode 100644 buch/papers/reedsolomon/codebsp.tex create mode 100644 buch/papers/reedsolomon/restetabelle1.tex create mode 100644 buch/papers/reedsolomon/restetabelle2.tex (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/codebsp.tex b/buch/papers/reedsolomon/codebsp.tex new file mode 100644 index 0000000..e9359f9 --- /dev/null +++ b/buch/papers/reedsolomon/codebsp.tex @@ -0,0 +1,71 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Codierung eines Beispiels +\label{reedsolomon:section:codebsp}} +\rhead{Koerper Festlegen} + +Um die Funktionsweise eines Reed-Solomon-Codes besser zu verstehen werden wir die einzelnen Probleme und ihre Lösungen anhand eines Beispiels betrachten. +Da wir in Endlichen Körpern Rechnen werden wir zuerst solch ein Körper festlegen. 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. + +% die beiden Restetabellen von F_11 +%\input{papers/reedsolomon/restetabelle1} +%\input{papers/reedsolomon/restetabelle2} + + + + + +\textbf{DUMP} + +Da Körper laut der \textcolor{red}{Definition 4.6} eine Primzahl sein muss, + + +Dieser Körper sollte jedoch über eine nullteilerfreie Restetabelle verfügen. Somit kommen nur Primzahlen als Körper in frage. + + + Für das Beispiel wählen wir die Zahl $11$. + + uns zu aller erst auf ein sochen Körper festlegen. + +Um die Funktionsweise eines Reed-Solomon-Codes besser zu verstehen werden wir dies anhand eines Beispiels betrachten. + +Um die Nachfolgende Rechenwege besser zu verstehen, werden wir die einzelnen Rechenschritte anhand eines Beispiels betrachten. + + + + +Als erstes muss festgelegt werden, in welchem endlichen Körper gerechnet werden soll. +Da die Restetabelle eines Körpers nullteilerfrei sein soll, kommen so nur Primzahlen in Frage. +Für das Beispiel verwenden wir den Körper $\mathbb{F}_{11}$. So wählen wir + + +$q = 11$ + + +und beinhaltet die Zahlen + + +$Z_{11} = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ + +\subsection{De finibus bonorum et malorum +\label{reedsolomon: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/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 -- cgit v1.2.1 From 55fc006b2133da4f79eb6eb5179d584c130824a2 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Tue, 18 May 2021 18:29:59 +0200 Subject: updated codebsp.tex, created decohnefehler.tex (with blindtext) --- buch/papers/reedsolomon/codebsp.tex | 174 +++++++++++++++++++++--------- buch/papers/reedsolomon/decohnefehler.tex | 40 +++++++ 2 files changed, 161 insertions(+), 53 deletions(-) create mode 100644 buch/papers/reedsolomon/decohnefehler.tex (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/codebsp.tex b/buch/papers/reedsolomon/codebsp.tex index e9359f9..5b67c43 100644 --- a/buch/papers/reedsolomon/codebsp.tex +++ b/buch/papers/reedsolomon/codebsp.tex @@ -11,61 +11,129 @@ Um die Funktionsweise eines Reed-Solomon-Codes besser zu verstehen werden wir di Da wir in Endlichen Körpern Rechnen werden wir zuerst solch ein Körper festlegen. 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} - - - - -\textbf{DUMP} - -Da Körper laut der \textcolor{red}{Definition 4.6} eine Primzahl sein muss, - - -Dieser Körper sollte jedoch über eine nullteilerfreie Restetabelle verfügen. Somit kommen nur Primzahlen als Körper in frage. - - - Für das Beispiel wählen wir die Zahl $11$. - - uns zu aller erst auf ein sochen Körper festlegen. - -Um die Funktionsweise eines Reed-Solomon-Codes besser zu verstehen werden wir dies anhand eines Beispiels betrachten. - -Um die Nachfolgende Rechenwege besser zu verstehen, werden wir die einzelnen Rechenschritte anhand eines Beispiels betrachten. - - - - -Als erstes muss festgelegt werden, in welchem endlichen Körper gerechnet werden soll. -Da die Restetabelle eines Körpers nullteilerfrei sein soll, kommen so nur Primzahlen in Frage. -Für das Beispiel verwenden wir den Körper $\mathbb{F}_{11}$. So wählen wir - - -$q = 11$ - - -und beinhaltet die Zahlen - - -$Z_{11} = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ - -\subsection{De finibus bonorum et malorum -\label{reedsolomon: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. - - +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/decohnefehler.tex b/buch/papers/reedsolomon/decohnefehler.tex new file mode 100644 index 0000000..832d63f --- /dev/null +++ b/buch/papers/reedsolomon/decohnefehler.tex @@ -0,0 +1,40 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Decodierung ohne Fehler +\label{reedsolomon:section:decohnefehler}} +\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{reedsolomon: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. + + -- cgit v1.2.1 From 9c25485518e7f80050a8ee2a12b94abb009c9a58 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Tue, 18 May 2021 21:14:36 +0200 Subject: finished first final version of decohnefehler.tex --- buch/papers/reedsolomon/decohnefehler.tex | 128 ++++++++++++++++++++++-------- 1 file changed, 97 insertions(+), 31 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/decohnefehler.tex b/buch/papers/reedsolomon/decohnefehler.tex index 832d63f..90f8ba8 100644 --- a/buch/papers/reedsolomon/decohnefehler.tex +++ b/buch/papers/reedsolomon/decohnefehler.tex @@ -5,36 +5,102 @@ % \section{Decodierung ohne Fehler \label{reedsolomon:section:decohnefehler}} -\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{reedsolomon: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. +\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 den 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 -- cgit v1.2.1 From 5294c40d558e93a034d43846e98176291fb32692 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Mon, 24 May 2021 14:28:24 +0200 Subject: update decohnefehler.tex, create decmitfehler.tex --- buch/papers/reedsolomon/decmitfehler.tex | 16 ++++++++++++++++ buch/papers/reedsolomon/decohnefehler.tex | 2 +- 2 files changed, 17 insertions(+), 1 deletion(-) create mode 100644 buch/papers/reedsolomon/decmitfehler.tex (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/decmitfehler.tex b/buch/papers/reedsolomon/decmitfehler.tex new file mode 100644 index 0000000..fead10e --- /dev/null +++ b/buch/papers/reedsolomon/decmitfehler.tex @@ -0,0 +1,16 @@ +% +% 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} +moin + + +\subsection{Der Satz von Fermat +\label{reedsolomon:subsection:fermat}} +wer ist fermat? + + diff --git a/buch/papers/reedsolomon/decohnefehler.tex b/buch/papers/reedsolomon/decohnefehler.tex index 90f8ba8..6ca577a 100644 --- a/buch/papers/reedsolomon/decohnefehler.tex +++ b/buch/papers/reedsolomon/decohnefehler.tex @@ -80,7 +80,7 @@ Glücklicherweise lässt der sich analog wie bei der Inversen Fouriertransformat 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 den Nachrichtenvektor einfach bestimmen mit +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\\ -- cgit v1.2.1 From 60bfb41261f51cf20ce65a9242c2624b31d74e75 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Mon, 24 May 2021 17:17:56 +0200 Subject: decmitfehler.tex updated --- buch/papers/reedsolomon/decmitfehler.tex | 185 ++++++++++++++++++++++++++++++- 1 file changed, 183 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/decmitfehler.tex b/buch/papers/reedsolomon/decmitfehler.tex index fead10e..923c1c5 100644 --- a/buch/papers/reedsolomon/decmitfehler.tex +++ b/buch/papers/reedsolomon/decmitfehler.tex @@ -6,11 +6,192 @@ \section{Decodierung mit Fehler \label{reedsolomon:section:decmitfehler}} \rhead{fehlerhafte rekonstruktion} -moin +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}} -wer ist 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 -- cgit v1.2.1 From 81527bd39cb20969fa3a84c85a843bca511dcb51 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Mon, 24 May 2021 17:18:21 +0200 Subject: created rekonstruktion.tex --- buch/papers/reedsolomon/rekonstruktion.tex | 40 ++++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) create mode 100644 buch/papers/reedsolomon/rekonstruktion.tex (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/rekonstruktion.tex b/buch/papers/reedsolomon/rekonstruktion.tex new file mode 100644 index 0000000..a3edba4 --- /dev/null +++ b/buch/papers/reedsolomon/rekonstruktion.tex @@ -0,0 +1,40 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Nachricht Rekonstruieren +\label{reedsolomon:section:rekonstruktion}} +\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{reedsolomon: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. + + -- cgit v1.2.1 From e86e0ad0e4415450a9c8b28917024ee6d0d77da5 Mon Sep 17 00:00:00 2001 From: michael-OST <75078383+michael-OST@users.noreply.github.com> Date: Fri, 28 May 2021 15:23:51 +0200 Subject: text added --- buch/papers/reedsolomon/rekonstruktion.tex | 204 ++++++++++++++++++++++++----- 1 file changed, 174 insertions(+), 30 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/reedsolomon/rekonstruktion.tex b/buch/papers/reedsolomon/rekonstruktion.tex index a3edba4..8cb7744 100644 --- a/buch/papers/reedsolomon/rekonstruktion.tex +++ b/buch/papers/reedsolomon/rekonstruktion.tex @@ -5,36 +5,180 @@ % \section{Nachricht Rekonstruieren \label{reedsolomon:section:rekonstruktion}} -\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? +\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. -\subsection{De finibus bonorum et malorum -\label{reedsolomon: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. +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. -- cgit v1.2.1