Valentin Brandl
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school/intro-crypto/uebung/04/04.tex
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school/intro-crypto/uebung/04/04.tex
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\documentclass[12pt,a4paper,german]{article}
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\usepackage{url}
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%\usepackage{graphics}
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\usepackage{times}
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\usepackage[T1]{fontenc}
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\usepackage{pifont}
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\usepackage{ngerman}
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\usepackage{float}
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\usepackage{diagbox}
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\usepackage[latin1]{inputenc}
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\usepackage{geometry}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{delarray}
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% \usepackage{minted}
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\usepackage{csquotes}
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\usepackage{graphicx}
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\usepackage{epsfig}
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\usepackage{longtable}
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\usepackage{paralist}
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\geometry{left=2.0cm,textwidth=17cm,top=3.5cm,textheight=23cm}
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\graphicspath{.}
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%%%%%%%%%% Fill out the the definitions %%%%%%%%%
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\def \name {Valentin Brandl} %
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\def \matrikel {108018274494} %
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% \def \pname {Vorname2 Nachname2} %
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% \def \pmatrikel {Matrikelnummer2} %
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\def \gruppe {Gruppe 193} %
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\def \uebung {4} %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% DO NOT MODIFY THIS HEADER
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\newcommand{\hwsol}{
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\vspace*{-2cm}
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\noindent \matrikel \quad \name \hfill Gruppe:\gruppe \\
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% \noindent \pmatrikel \quad \pname \\
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\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center}
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}
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\newcommand{\cmark}{\ding{51}}%
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\newcommand{\xmark}{\ding{55}}%
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\begin{document}
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%Import header
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\hwsol
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\section*{Aufgabe 1}
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Grad $m = 6 \Rightarrow 2^m - 1 = 2^6 - 1 = 63$.
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Zustände und Tabellen wurden mit dem angehängten Code generiert.
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%{{{ a1
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\begin{enumerate}[a)]
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\item $x^5 + x^4 + x^2 + x + 1$
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\begin{figure}[h]
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\includegraphics[width=\textwidth]{1a}
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\caption{Schaltbild des Schieberegisters für 1a)}
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\end{figure}
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IV: $1 0 0 0 0 0$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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1 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 1 & 0 & 1 & 1 \\
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0 & 1 & 0 & 0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 & 1 & 1 \\
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1 & 0 & 0 & 1 & 0 & 0 & 0 \\
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1 & 1 & 0 & 0 & 1 & 0 & 0 \\
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0 & 1 & 1 & 0 & 0 & 1 & 1 \\
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0 & 0 & 1 & 1 & 0 & 0 & 0 \\
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1 & 0 & 0 & 1 & 1 & 0 & 0 \\
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0 & 1 & 0 & 0 & 1 & 1 & 1 \\
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1 & 0 & 1 & 0 & 0 & 1 & 1 \\
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1 & 1 & 0 & 1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 1 & 0 & 1 & 1 \\
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0 & 0 & 0 & 1 & 1 & 0 & 0 \\
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0 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 0 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{tabular}
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Wiederholung nach 21 Iterationen.
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Neuer IV: $1 1 1 1 1 1$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 0 & 1 & 1 & 1 & 1 & 1 \\
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1 & 0 & 0 & 1 & 1 & 1 & 1 \\
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1 & 1 & 0 & 0 & 1 & 1 & 1 \\
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1 & 1 & 1 & 0 & 0 & 1 & 1 \\
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0 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 1 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 & 1 & 1 & 1 \\
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1 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 1 & 0 & 0 & 0 & 1 & 1 \\
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0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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1 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 1 & 0 & 0 & 0 & 1 & 1 \\
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0 & 1 & 1 & 0 & 0 & 0 & 0 \\
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1 & 0 & 1 & 1 & 0 & 0 & 0 \\
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1 & 1 & 0 & 1 & 1 & 0 & 0 \\
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1 & 1 & 1 & 0 & 1 & 1 & 1 \\
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1 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 1 & 1 & 1 & 1 & 0 & 0 \\
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\underline{1} & \underline{1} & \underline{1} & \underline{1} & \underline{1} & \underline{1} & 1 \\
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\end{tabular}
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Wiederholung wieder nach 21 Iterationen.
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$63 - 21 - 21 = 21$ Fehlende Zustände.
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Neuer IV: $1 1 0 0 0 0$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 1 & 0 & 0 & 0 & 0 & 0 \\
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1 & 1 & 1 & 0 & 0 & 0 & 0 \\
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1 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 1 & 1 & 0 & 0 \\
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1 & 0 & 1 & 1 & 1 & 1 & 1 \\
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1 & 1 & 0 & 1 & 1 & 1 & 1 \\
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0 & 1 & 1 & 0 & 1 & 1 & 1 \\
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1 & 0 & 1 & 1 & 0 & 1 & 1 \\
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0 & 1 & 0 & 1 & 1 & 0 & 0 \\
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1 & 0 & 1 & 0 & 1 & 1 & 1 \\
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0 & 1 & 0 & 1 & 0 & 1 & 1 \\
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1 & 0 & 1 & 0 & 1 & 0 & 0 \\
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1 & 1 & 0 & 1 & 0 & 1 & 1 \\
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1 & 1 & 1 & 0 & 1 & 0 & 0 \\
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0 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 0 & 1 & 1 & 1 & 0 & 0 \\
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0 & 1 & 0 & 1 & 1 & 1 & 1 \\
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0 & 0 & 1 & 0 & 1 & 1 & 1 \\
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0 & 0 & 0 & 1 & 0 & 1 & 1 \\
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0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{1} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{tabular}
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Wiederholung wieder nach 21 Iterationen.
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$63 - 21 - 21 - 21 = 0$ Fehlende Zustände. Alle möglichen Zustände wurden erzeugt.
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Keine Sequenz maximaler Länge aber Länge unabhängig von IV$\Rightarrow$ es liegt ein irreduzibles Polynom
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zugrunde
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\item $x^5 + x + 1$
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\begin{figure}[h]
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\includegraphics[width=\textwidth]{1b}
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\caption{Schaltbild des Schieberegisters für 1b)}
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\end{figure}
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IV: $1 0 0 0 0 0$
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\begin{longtable}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 & 0 & 1 & 1 \\
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1 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 1 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 1 & 0 & 0 & 0 & 1 & 1 \\
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1 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 1 & 0 & 1 & 1 \\
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1 & 1 & 0 & 0 & 1 & 0 & 0 \\
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1 & 1 & 1 & 0 & 0 & 1 & 1 \\
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1 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 1 & 1 & 0 & 0 \\
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1 & 0 & 1 & 1 & 1 & 1 & 1 \\
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0 & 1 & 0 & 1 & 1 & 1 & 1 \\
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0 & 0 & 1 & 0 & 1 & 1 & 1 \\
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0 & 0 & 0 & 1 & 0 & 1 & 1 \\
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1 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 1 & 0 & 0 & 0 & 1 & 1 \\
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1 & 1 & 1 & 0 & 0 & 0 & 0 \\
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0 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 1 & 1 & 0 & 0 \\
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1 & 0 & 0 & 1 & 1 & 1 & 1 \\
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0 & 1 & 0 & 0 & 1 & 1 & 1 \\
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0 & 0 & 1 & 0 & 0 & 1 & 1 \\
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1 & 0 & 0 & 1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 1 & 0 & 0 \\
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1 & 0 & 1 & 0 & 0 & 1 & 1 \\
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1 & 1 & 0 & 1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 0 & 1 & 0 & 0 \\
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1 & 0 & 1 & 1 & 0 & 1 & 1 \\
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1 & 1 & 0 & 1 & 1 & 0 & 0 \\
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1 & 1 & 1 & 0 & 1 & 1 & 1 \\
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0 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 0 & 1 & 1 & 1 & 0 & 0 \\
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1 & 1 & 0 & 1 & 1 & 1 & 1 \\
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0 & 1 & 1 & 0 & 1 & 1 & 1 \\
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0 & 0 & 1 & 1 & 0 & 1 & 1 \\
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1 & 0 & 0 & 1 & 1 & 0 & 0 \\
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1 & 1 & 0 & 0 & 1 & 1 & 1 \\
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0 & 1 & 1 & 0 & 0 & 1 & 1 \\
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1 & 0 & 1 & 1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 1 & 1 & 0 & 0 \\
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1 & 0 & 1 & 0 & 1 & 1 & 1 \\
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0 & 1 & 0 & 1 & 0 & 1 & 1 \\
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1 & 0 & 1 & 0 & 1 & 0 & 0 \\
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1 & 1 & 0 & 1 & 0 & 1 & 1 \\
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1 & 1 & 1 & 0 & 1 & 0 & 0 \\
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1 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 1 & 1 & 1 & 1 & 0 & 0 \\
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1 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 0 & 1 & 1 & 1 & 1 & 1 \\
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0 & 0 & 0 & 1 & 1 & 1 & 1 \\
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0 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 0 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{longtable}
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Wiederholung nach 63 Iterationen, es wurden also eine Sequenz maximaler Länge erzeugt $\Rightarrow$ primitives
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Polynom liegt zugrunde.
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\item $x^5 + x^3 + x^2 + x + 1$
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\begin{figure}[h]
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\includegraphics[width=\textwidth]{1c}
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\caption{Schaltbild des Schieberegisters für 1c)}
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\end{figure}
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IV: $1 0 0 0 0 0$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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1 & 0 & 0 & 1 & 0 & 0 & 0 \\
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1 & 1 & 0 & 0 & 1 & 0 & 0 \\
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1 & 1 & 1 & 0 & 0 & 1 & 1 \\
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0 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 1 & 1 & 0 & 0 \\
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1 & 0 & 0 & 1 & 1 & 1 & 1 \\
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1 & 1 & 0 & 0 & 1 & 1 & 1 \\
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0 & 1 & 1 & 0 & 0 & 1 & 1 \\
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0 & 0 & 1 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 1 & 0 & 0 \\
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0 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 0 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{tabular}
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Wiederholung nach 15 Iterationen.
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Neuer IV: $1 1 1 1 1 1$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 1 & 1 & 1 & 1 & 1 & 1 \\
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0 & 0 & 1 & 1 & 1 & 1 & 1 \\
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0 & 0 & 0 & 1 & 1 & 1 & 1 \\
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1 & 0 & 0 & 0 & 1 & 1 & 1 \\
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0 & 1 & 0 & 0 & 0 & 1 & 1 \\
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1 & 0 & 1 & 0 & 0 & 0 & 0 \\
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1 & 1 & 0 & 1 & 0 & 0 & 0 \\
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1 & 1 & 1 & 0 & 1 & 0 & 0 \\
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0 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 0 & 1 & 1 & 1 & 0 & 0 \\
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1 & 1 & 0 & 1 & 1 & 1 & 1 \\
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1 & 1 & 1 & 0 & 1 & 1 & 1 \\
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1 & 1 & 1 & 1 & 0 & 1 & 1 \\
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1 & 1 & 1 & 1 & 1 & 0 & 0 \\
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\underline{1} & \underline{1} & \underline{1} & \underline{1} & \underline{1} & \underline{1} & 1 \\
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\end{tabular}
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Wiederholung nach 15 Iterationen. $63 - 15 - 15 = 33$ Fehlende Zustände.
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Neuer IV: $1 1 0 0 0 0$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 1 & 0 & 0 & 0 & 0 \\
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1 & 0 & 1 & 1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 1 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0 & 1 & 1 & 1 \\
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1 & 0 & 0 & 1 & 0 & 1 & 1 \\
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0 & 1 & 0 & 0 & 1 & 0 & 0 \\
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1 & 0 & 1 & 0 & 0 & 1 & 1 \\
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0 & 1 & 0 & 1 & 0 & 0 & 0 \\
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1 & 0 & 1 & 0 & 1 & 0 & 0 \\
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0 & 1 & 0 & 1 & 0 & 1 & 1 \\
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0 & 0 & 1 & 0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 & 0 & 1 & 1 \\
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0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{1} & \underline{0} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{tabular}
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Wiederholung nach 15 Iterationen. $63 - 15 - 15 - 15 = 18$ Fehlende Zustände.
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Neuer IV: $1 1 1 0 0 0$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 1 & 1 & 0 & 0 & 0 & 0 \\
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1 & 1 & 1 & 1 & 0 & 0 & 0 \\
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0 & 1 & 1 & 1 & 1 & 0 & 0 \\
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1 & 0 & 1 & 1 & 1 & 1 & 1 \\
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0 & 1 & 0 & 1 & 1 & 1 & 1 \\
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1 & 0 & 1 & 0 & 1 & 1 & 1 \\
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1 & 1 & 0 & 1 & 0 & 1 & 1 \\
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0 & 1 & 1 & 0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 1 & 0 & 1 & 1 \\
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1 & 0 & 0 & 1 & 1 & 0 & 0 \\
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0 & 1 & 0 & 0 & 1 & 1 & 1 \\
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0 & 0 & 1 & 0 & 0 & 1 & 1 \\
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0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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1 & 0 & 0 & 0 & 1 & 0 & 0 \\
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1 & 1 & 0 & 0 & 0 & 1 & 1 \\
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\underline{1} & \underline{1} & \underline{1} & \underline{0} & \underline{0} & \underline{0} & 0 \\
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\end{tabular}
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Wiederholung nach 15 Iterationen. $63 - 15 - 15 - 15 - 15 = 3$ Fehlende Zustände.
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Letzter IV: $1 0 1 1 0 1$
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\begin{tabular}{|cccccc|c|}
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$x_5$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ & Output \\\hline
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1 & 0 & 1 & 1 & 0 & 1 & 1 \\
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1 & 1 & 0 & 1 & 1 & 0 & 0 \\
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0 & 1 & 1 & 0 & 1 & 1 & 1 \\
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\underline{1} & \underline{0} & \underline{1} & \underline{1} & \underline{0} & \underline{1} & 1 \\
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\end{tabular}
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Wiederholung nach 3 Iterathonen. $63 - 15 - 15 - 15 - 15 -3 = 0$ fehlende Zustände.
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Keine Sequenz maximaler Länge und Länge ist abhängig von IV $\Rightarrow$ reduzibles Polynom liegt zugrunde.
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\end{enumerate} %}}}
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\section*{Aufgabe 2}
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\begin{eqnarray*}
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s = 155 \text{ Mbits/sec} = 155 * 2^{20} \text{ bit/sec} \\
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12h = 12 * 60 * 60 sec = 43200 sec \\
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155 * 2^{20} \frac{bit}{sec} * 43200 sec = 7021264896000 bit
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\end{eqnarray*}
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Gesucht $m \in \mathbb{N}$, so dass $2^m - 1 > 7021264896000$ (gelöst mit Wolframalpha)
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\begin{eqnarray*}
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2^m - 1 &> 7021264896000 \\
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m &> 42
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\end{eqnarray*}
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Vorausgesetzt, es handelt sich um ein primitives Polynom, ist der minimale Grad, der benötigt wird, dass eine
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Wiederholung in der Schlüsselfolge frühestens nach 12 Stunden passiert $m = 43$.
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|
||||
\section*{Aufgabe 3}
|
||||
|
||||
\begin{enumerate}[a)]
|
||||
|
||||
\item $m = 8 \Rightarrow 2^m - 1 = 2^8 - 1 = 255$
|
||||
|
||||
\item $2*m - 1 = 2 * 8 - 1 = 15$
|
||||
|
||||
\item
|
||||
\begin{align*}
|
||||
y_i &\equiv x_i + s_i &\mod 2 \\
|
||||
s_i &\equiv y_i + x_i &\mod 2
|
||||
\end{align*}
|
||||
|
||||
Rekonstruieren der ersten 15 Bit des Schlüsselstroms mit Hilfe des known-plaintext \enquote{Mo} $\Rightarrow$
|
||||
0x4d, 0x6f $\Rightarrow (01001101)_2, (01101111)_2$
|
||||
|
||||
Die ersten 2 Bytes des Ciphertext sind 0xEC, 0xD4 $\Rightarrow (11101100)_2, (11010100)_2$
|
||||
|
||||
Mit Hilfe des angehängten Programms wurden die folgenden 2 Schlüsselstrom Bytes berechnet: $(10100001)_2,
|
||||
(10111011)_2 \Rightarrow (A1)_{16}, (BB)_{16}$
|
||||
|
||||
\item Folgendes System $(A\mid b)$ gilt es zu lösen:
|
||||
|
||||
Die Matrix wurde mit Hilfe von \url{https://planetcalc.com/3324/} invertiert.
|
||||
|
||||
\begin{align*}
|
||||
\begin{array}({@{}cccccccc|c@{}})
|
||||
1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \\
|
||||
1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
|
||||
0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\
|
||||
1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\
|
||||
1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \\
|
||||
1 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\
|
||||
0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 1 \\
|
||||
1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1
|
||||
\end{array} \\
|
||||
Ax = b \Rightarrow A^{-1}b = x \\
|
||||
A^{-1} =
|
||||
\begin{matrix}
|
||||
0 & 1 & 0 & 0 & 1 & 1 & 1 & 0 \\
|
||||
0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
|
||||
1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\
|
||||
1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \\
|
||||
1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\
|
||||
0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\
|
||||
0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
|
||||
1 & 0 & 0 & 1 & 1 & 1 & 0 & 0
|
||||
\end{matrix} \\
|
||||
b =
|
||||
\begin{matrix}
|
||||
1\\
|
||||
0\\
|
||||
1\\
|
||||
1\\
|
||||
1\\
|
||||
0\\
|
||||
1\\
|
||||
1\\
|
||||
\end{matrix}\\
|
||||
x =
|
||||
\begin{array}({@{}cccccccc@{}})
|
||||
0 & 1 & 1 & 0 & 0 & 0 & 1 & 1
|
||||
\end{array} \\
|
||||
p_0 = 1 \\
|
||||
p_1 = 1 \\
|
||||
p_2 = 0 \\
|
||||
p_3 = 0 \\
|
||||
p_4 = 0 \\
|
||||
p_5 = 1 \\
|
||||
p_6 = 1 \\
|
||||
p_7 = 0 \\
|
||||
\end{align*}
|
||||
|
||||
\item Der Klartext ist \enquote{Mondl4ndunG}.
|
||||
|
||||
Berechnet mit dem Code im Anhang.
|
||||
|
||||
\item Die erste unsanfte und unbemannte Mondlandung war am 13.09.1959 (Lunik 2).
|
||||
|
||||
Die erste sanfte und unbemannte Mondlandung am 03.02.1966 (Luna 9)
|
||||
|
||||
Die erste bemannte Mondlandung war am 21.07.1969 (Apollo 11).
|
||||
|
||||
(Quelle: \url{https://de.wikipedia.org/wiki/Mondlandung})
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
% \section*{Code}
|
||||
|
||||
% \inputminted{rust}{./school/intro-crypto/uebung/04/lfsr/src/main.rs}
|
||||
|
||||
\end{document}
|
||||
|
||||
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|
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4
school/intro-crypto/uebung/04/lfsr/Cargo.lock
generated
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4
school/intro-crypto/uebung/04/lfsr/Cargo.lock
generated
Normal file
@ -0,0 +1,4 @@
|
||||
[[package]]
|
||||
name = "lfsr"
|
||||
version = "0.1.0"
|
||||
|
||||
6
school/intro-crypto/uebung/04/lfsr/Cargo.toml
Normal file
6
school/intro-crypto/uebung/04/lfsr/Cargo.toml
Normal file
@ -0,0 +1,6 @@
|
||||
[package]
|
||||
name = "lfsr"
|
||||
version = "0.1.0"
|
||||
authors = ["Valentin Brandl <vbrandl@riseup.net>"]
|
||||
|
||||
[dependencies]
|
||||
177
school/intro-crypto/uebung/04/lfsr/src/main.rs
Normal file
177
school/intro-crypto/uebung/04/lfsr/src/main.rs
Normal file
@ -0,0 +1,177 @@
|
||||
struct Lfsr {
|
||||
reg: Vec<bool>,
|
||||
p: Vec<bool>,
|
||||
}
|
||||
|
||||
impl Lfsr {
|
||||
fn new(iv: &[bool], p: &[bool]) -> Self {
|
||||
assert_eq!(iv.len(), p.len());
|
||||
Self {
|
||||
reg: iv.iter().cloned().collect(),
|
||||
p: p.iter().cloned().collect(),
|
||||
}
|
||||
}
|
||||
|
||||
fn next(&mut self) -> bool {
|
||||
// let res = self.reg[self.len - 1];
|
||||
let next = self
|
||||
.reg
|
||||
.iter()
|
||||
.zip(self.p.iter())
|
||||
.map(|(n, m)| n & m)
|
||||
.fold(false, |acc, v| acc ^ v);
|
||||
self.reg.insert(0, next);
|
||||
let ret = self.reg.pop().unwrap();
|
||||
assert_eq!(self.reg.len(), self.p.len());
|
||||
ret
|
||||
}
|
||||
|
||||
fn state(&self) -> &[bool] {
|
||||
&self.reg
|
||||
}
|
||||
}
|
||||
|
||||
fn convert_state(s: &[bool]) -> String {
|
||||
s.iter()
|
||||
.map(|x| if *x { "1".to_string() } else { "0".to_string() })
|
||||
.collect::<Vec<String>>()
|
||||
.join(" & ")
|
||||
}
|
||||
|
||||
/// helper for 3c)
|
||||
fn recover_keystream(plain: &[bool], cipher: &[bool]) -> Vec<bool> {
|
||||
assert_eq!(plain.len(), cipher.len());
|
||||
plain
|
||||
.iter()
|
||||
.zip(cipher.iter())
|
||||
.map(|(a, b)| a ^ b)
|
||||
.collect()
|
||||
}
|
||||
|
||||
fn decrypt(l: &mut Lfsr, c: &[bool]) -> u8 {
|
||||
let s: String = c
|
||||
.into_iter()
|
||||
.map(|c| c ^ l.next())
|
||||
.map(|c| if c { "1" } else { "0" })
|
||||
.collect();
|
||||
u8::from_str_radix(&s, 2).unwrap()
|
||||
}
|
||||
|
||||
/// main for 3d)
|
||||
fn main() {
|
||||
let p = &[false, true, true, false, false, false, true, true];
|
||||
// let s = &[true, false, true, true, true, false, true, true];
|
||||
// let s = &[true, false, true, false, false, false, false, true];
|
||||
// let s = &[true, false, false, false, false, true, false, true];
|
||||
let s = &[true, true, false, true, true, true, false, true];
|
||||
let mut l = Lfsr::new(s, p);
|
||||
// for _ in 0..17 {
|
||||
// println!("{}", if lfsr.next() { "1" } else { "0" });
|
||||
// }
|
||||
// 00010011
|
||||
// 11010100
|
||||
let c = &[true, true, false, true, false, true, false, false];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 00010011 = 0x13
|
||||
let c = &[false, false, false, true, false, false, true, true];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 00010000 = 0x10
|
||||
let c = &[false, false, false, true, false, false, false, false];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 01110101 = 0x75
|
||||
let c = &[false, true, true, true, false, true, false, true];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 01100000 = 0x60
|
||||
let c = &[false, true, true, false, false, false, false, false];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 00000101 = 0x05
|
||||
let c = &[false, false, false, false, false, true, false, true];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 01111100 = 0x7c
|
||||
let c = &[false, true, true, true, true, true, false, false];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 01010000 = 0x50
|
||||
let c = &[false, true, false, true, false, false, false, false];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 11011011 = 0xdb
|
||||
let c = &[true, true, false, true, true, false, true, true];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
|
||||
// 01110011 = 0x73
|
||||
let c = &[false, true, true, true, false, false, true, true];
|
||||
let d = decrypt(&mut l, c);
|
||||
println!("{}: {}", d, d as char);
|
||||
}
|
||||
|
||||
// fn main() {
|
||||
// let plain = &[
|
||||
// false, true, false, false, true, true, false, true, false, true, true, false, true, true,
|
||||
// true, true,
|
||||
// ];
|
||||
// let cipher = &[
|
||||
// true, true, true, false, true, true, false, false, true, true, false, true, false, true,
|
||||
// false, false,
|
||||
// ];
|
||||
// let key_stream = recover_keystream(plain, cipher);
|
||||
// println!(
|
||||
// "{}",
|
||||
// key_stream
|
||||
// .iter()
|
||||
// .map(|k| if *k { "1, " } else { "0, " })
|
||||
// .collect::<String>()
|
||||
// );
|
||||
// }
|
||||
|
||||
/// main for exercise 1
|
||||
fn main1() {
|
||||
// 1a)
|
||||
// let mut lfsr = Lfsr::new(
|
||||
// &[true, true, false, false, false, false],
|
||||
// // &[true, true, true, true, true, true],
|
||||
// // &[true, false, false, false, false, false],
|
||||
// &[false, true, false, true, true, true],
|
||||
// );
|
||||
|
||||
// 1b)
|
||||
// let mut lfsr = Lfsr::new(
|
||||
// // &[true, true, true, true, true, true],
|
||||
// &[true, false, false, false, false, false],
|
||||
// &[false, false, false, false, true, true],
|
||||
// );
|
||||
|
||||
// 1c)
|
||||
let mut lfsr = Lfsr::new(
|
||||
// &[true, true, true, false, false, false],
|
||||
// &[true, true, false, false, false, false],
|
||||
&[true, false, true, true, false, true],
|
||||
// &[true, true, true, true, true, true],
|
||||
// &[true, false, false, false, false, false],
|
||||
&[false, false, true, true, true, true],
|
||||
);
|
||||
let mut seen = Vec::new();
|
||||
let mut i = 0;
|
||||
while !seen.contains(&lfsr.state().iter().cloned().collect::<Vec<_>>()) {
|
||||
seen.push(lfsr.state().iter().cloned().collect::<Vec<_>>());
|
||||
print!("{}", convert_state(lfsr.state()));
|
||||
println!(" & {} \\\\", if lfsr.next() { 1 } else { 0 });
|
||||
i += 1;
|
||||
}
|
||||
eprintln!("Repetition after {} iterations", i);
|
||||
}
|
||||
@ -0,0 +1 @@
|
||||
{"rustc_fingerprint":6445661432976150582,"outputs":{"1164083562126845933":["rustc 1.30.0 (da5f414c2 2018-10-24)\nbinary: rustc\ncommit-hash: da5f414c2c0bfe5198934493f04c676e2b23ff2e\ncommit-date: 2018-10-24\nhost: x86_64-unknown-linux-gnu\nrelease: 1.30.0\nLLVM version: 8.0\n",""],"15337506775154344876":["___\nlib___.rlib\nlib___.so\nlib___.so\nlib___.a\nlib___.so\n/home/me/.rustup/toolchains/stable-x86_64-unknown-linux-gnu\ndebug_assertions\nproc_macro\ntarget_arch=\"x86_64\"\ntarget_endian=\"little\"\ntarget_env=\"gnu\"\ntarget_family=\"unix\"\ntarget_feature=\"fxsr\"\ntarget_feature=\"sse\"\ntarget_feature=\"sse2\"\ntarget_os=\"linux\"\ntarget_pointer_width=\"64\"\nunix\n",""],"1617349019360157463":["___\nlib___.rlib\nlib___.so\nlib___.so\nlib___.a\nlib___.so\n/home/me/.rustup/toolchains/stable-x86_64-unknown-linux-gnu\ndebug_assertions\nproc_macro\ntarget_arch=\"x86_64\"\ntarget_endian=\"little\"\ntarget_env=\"gnu\"\ntarget_family=\"unix\"\ntarget_feature=\"fxsr\"\ntarget_feature=\"sse\"\ntarget_feature=\"sse2\"\ntarget_os=\"linux\"\ntarget_pointer_width=\"64\"\nunix\n",""]},"successes":{}}
|
||||
@ -0,0 +1 @@
|
||||
87161daafed68c82
|
||||
@ -0,0 +1 @@
|
||||
{"rustc":7311099760834594431,"features":"[]","target":12005888738967920787,"profile":690535219432825423,"path":1036222786711178230,"deps":[],"local":[{"MtimeBased":[[1542232988,806345654],".fingerprint/lfsr-19130ef987a3303b/dep-bin-lfsr-19130ef987a3303b"]}],"rustflags":[],"edition":"Edition2015"}
|
||||
Binary file not shown.
@ -0,0 +1 @@
|
||||
58c9a3bc0629b4e9
|
||||
@ -0,0 +1 @@
|
||||
{"rustc":7311099760834594431,"features":"[]","target":12005888738967920787,"profile":8064701370884557241,"path":1036222786711178230,"deps":[],"local":[{"MtimeBased":[[1542232985,983443064],".fingerprint/lfsr-c5c66887bdea01a0/dep-bin-lfsr-c5c66887bdea01a0"]}],"rustflags":[],"edition":"Edition2015"}
|
||||
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BIN
school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-19130ef987a3303b
Executable file
BIN
school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-19130ef987a3303b
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|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-19130ef987a3303b: src/main.rs
|
||||
|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-19130ef987a3303b.d: src/main.rs
|
||||
|
||||
src/main.rs:
|
||||
@ -0,0 +1,5 @@
|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-c5c66887bdea01a0.rmeta: src/main.rs
|
||||
|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/deps/lfsr-c5c66887bdea01a0.d: src/main.rs
|
||||
|
||||
src/main.rs:
|
||||
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school/intro-crypto/uebung/04/lfsr/target/debug/lfsr
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school/intro-crypto/uebung/04/lfsr/target/debug/lfsr
Executable file
Binary file not shown.
1
school/intro-crypto/uebung/04/lfsr/target/debug/lfsr.d
Normal file
1
school/intro-crypto/uebung/04/lfsr/target/debug/lfsr.d
Normal file
@ -0,0 +1 @@
|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/lfsr: /home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/src/main.rs
|
||||
@ -0,0 +1 @@
|
||||
/home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/target/debug/liblfsr.rmeta: /home/me/Dokumente/notes/school/intro-crypto/uebung/04/lfsr/src/main.rs
|
||||
BIN
school/intro-crypto/uebung/04/lfsr_rust.zip
Normal file
BIN
school/intro-crypto/uebung/04/lfsr_rust.zip
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BIN
school/intro-crypto/uebung/04/Übung_4_EK1_WS1819.pdf
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school/intro-crypto/uebung/04/Übung_4_EK1_WS1819.pdf
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Loading…
Reference in New Issue
Block a user