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Valentin Brandl
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school/intro-crypto/uebung/07/07.tex
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school/intro-crypto/uebung/07/07.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 {7} %
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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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\newcommand{\csquare}{\text{\rlap{$\checkmark$}}\square}%
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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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$P(x) = x^4 + x + 1$ und $grad(P(X)) = 4$
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\begin{enumerate}[a)]
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\item $GF(2) = GF(2^1)$. Maximaler Grad von $G(p^m)$ ist $(m-1)$, maximaler Grad von $GF(2^1)$ ist also $1-1=0$, es
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sind also nur die polynome $a_1(x) = 0$ und $a_2(x) = 1$ möglich. Beide haben einen Grad $< 4$, also gibt es $2$
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Polynome in $GF(2)$, deren Grad kleiner ist, als $grad(P(x))$.
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\item $GF(p^n)$: Alle $a_i$ des Polynoms sind Elemente der Menge $A = \{0,1,...,p-1\} \Rightarrow |A| = p$. Jedes
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Polynom besteht aus $n$ Koeffizienten, daher hat hat $GF(p^n)$ genau $p^n$ mögliche Polynome. Da der Grad der
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Polynome $< 4$ sein soll, muss $n$ also $\leq 4$ sein. Abhängig von $p$ sind also $4$ mögliche Werte für $n$
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denkbar, sodass der Grad $< 4$ bleibt
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\begin{tabular}{|c|c|}
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\hline
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$n$ & \# Polynome \\\hline
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1 & $p$ \\\hline
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2 & $p^2$ \\\hline
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3 & $p^3$ \\\hline
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4 & $p^4$ \\\hline
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\end{tabular}
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\end{enumerate}
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\section*{Aufgabe 2}
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\begin{tabular}{|c||c|c|c|c|c|c|c|c|}
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\hline
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$\cdot$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline\hline
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0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
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1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline
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2 & 0 & 2 & 4 & 6 & 5 & 7 & 1 & 3 \\\hline
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3 & 0 & 3 & 6 & 5 & 1 & 2 & 7 & 4 \\\hline
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4 & 0 & 4 & 5 & 1 & 7 & 3 & 2 & 5 \\\hline
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5 & 0 & 5 & 7 & 2 & 3 & 6 & 4 & 3 \\\hline
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6 & 0 & 6 & 1 & 7 & 2 & 4 & 6 & 5 \\\hline
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7 & 0 & 7 & 3 & 4 & 5 & 3 & 5 & 2 \\\hline
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\end{tabular}
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\section*{Aufgabe 3}
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$P(x) = x^8 + x^4 + x^3 + x + 1$
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\begin{enumerate}[a)]
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\item
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\begin{align*}
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A(x) &= x^7 + x^6 + x^4 + x^2 + x + 1 \\
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B(x) &= x^2 + x \\
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A(x) * B(x) &= x^9+x^7+x^5+x^4+x \equiv x^7+x^2+1 &\mod P(x) \\
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&= 0x85
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\end{align*}
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\item
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\begin{align*}
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A(x) &= x^4 + x^2 + x + 1 \\
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B(x) &= x^4 + 1 \\
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A(x) * B(x) &= x^8 + x^6 + x^5 + x^2 + x + 1 \equiv x^6 + x^5 + x^4 + x^3 + x^2 &\mod P(x) \\
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&= 0x7C
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\end{align*}
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\item
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\begin{align*}
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A(x) &= x^6 + x^5 + x^2 + x \\
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B(x) &= x^2 + x + 1 \\
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A(x) * B(x) &= x^5 + x^4 + x &\mod P(x) \\
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&= 0x32
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\end{align*}
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\end{enumerate}
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\section*{Aufgabe 4}
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Die multiplikativen Inversen wurden aus der Tabelle 4.2 im Buch abgelesen.
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$P(x) = x^8 + x^4 + x^3 + x + 1$
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\begin{enumerate}[a)]
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\item
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\begin{align*}
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A(x) &= x^5 + x^4 + x^3 + x^2 \\
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B(x) &= x^5 + x^2 + 1 \widehat{=} 0x25 \\
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B^{-1}(x) &= 0x4D \widehat{=} x^6 + x^3 + x^2 + 1 \\
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A(x) \div B(x) &= A(x) * B^{-1}(x) = \\
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&= x^{11} + x^{10} + x^9 + x^7 + x^5 + x^3 \equiv x^5 + x^3 + x &\mod P(x) \\
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&\widehat{=} 0x2A
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\end{align*}
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\item
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\begin{align*}
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A(x) &= x^7 + x^2 + x + 1 \\
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B(x) &= x^6 + x^4 + x^3 + x^2 \widehat{=} 0x5C \\
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B^{-1}(x) &= 0x51 \widehat{=} x^6 + x^4 + 1 \\
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A(x) \div B(x) &= A(x) * B^{-1}(x) = \\
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&= x^{13} + x^{11} + x^8 + x^5 + x^4 + x^2 + x + 1 \equiv x^7 + x^5 + x^4 + x^3 + 1 &\mod
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P(x) \\
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&\widehat{=} 0xB9
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\end{align*}
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\end{enumerate}
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\section*{Aufgabe 5}
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$P(x) = x^4 + x^2 + 3x + 5$
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\begin{enumerate}[a)]
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\item
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\begin{tabular}{|c||c|c|c|c|c|c|c|}
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\hline
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$+$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\hline\hline
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0 & 0 & 1 & 2 & 3 & 4 & 4 & 6 \\\hline
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1 & 1 & 2 & 3 & 4 & 5 & 6 & 0 \\\hline
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2 & 2 & 3 & 4 & 5 & 6 & 0 & 1 \\\hline
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3 & 3 & 4 & 5 & 6 & 0 & 1 & 2 \\\hline
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4 & 4 & 5 & 6 & 0 & 1 & 2 & 3 \\\hline
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5 & 5 & 6 & 0 & 1 & 2 & 3 & 4 \\\hline
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6 & 6 & 0 & 1 & 2 & 3 & 4 & 5 \\\hline
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\end{tabular}
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\item
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\begin{tabular}{|c||c|c|c|c|c|c|c|}
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\hline
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$\cdot$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\hline
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0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\hline
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1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\hline
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2 & 0 & 2 & 4 & 6 & 1 & 3 & 5 \\\hline
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3 & 0 & 3 & 6 & 2 & 5 & 1 & 4 \\\hline
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4 & 0 & 4 & 1 & 5 & 2 & 6 & 3 \\\hline
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5 & 0 & 5 & 3 & 1 & 6 & 4 & 2 \\\hline
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6 & 0 & 6 & 5 & 4 & 3 & 2 & 1 \\\hline
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\end{tabular}
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\item
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\begin{enumerate}[i)]
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\item
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\begin{align*}
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A(x) &= 3x^2 + x + 2 \\
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B(x) &= 6x^4 + 4x^2 + 3x + 5 \\
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A(x) + B(x) &= 6x^4 + 4x &\mod P(x)
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\end{align*}
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\item Nein, da $A(x) + B(x)$ auch ohne die Moduloreduktion keinen Koeffizienten mit einem Grad $> 6$ hat
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\end{enumerate}
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\item
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\begin{enumerate}[i)]
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\item
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\begin{align*}
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A(x) &= 3x^3 + x + 2 \\
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B(x) &= 6x^3 + 4x^2 + 3x + 5 \\
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A(x) - B(x) &= 4x^3 + 3x^2 + 5x + 4 &\mod P(x)
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\end{align*}
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\item Nein, da $A(x) - B(x)$ auch ohne die Moduloreduktion keinen Koeffizienten mit einem Grad $> 6$ hat
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\end{enumerate}
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\item
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\begin{enumerate}[i)]
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\item
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\begin{align*}
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A(x) &= 5x^4 + x + 2 \\
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B(x) &= 3x^3 + 5x^2 + 4 \\
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A(x) * B(x) &= x^7 + 4x^6 + 2x^4 + 4x^3 + 3x^2 + 4x + 1 &\mod P(x) \\
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&\equiv 2x^3 + 5x^2 + 3x + 5 &\mod P(x)
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\end{align*}
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\item
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\begin{align*}
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A(x) &= x^3 + 2x + 4 \\
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B(x) &= 4x^4 + 6x^3 + 3 \\
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A(x) * B(x) &= 4x^7 + 6x^6 + 6x^5 + 6x^3 + 6x + 5 &\mod P(x) \\
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&\equiv x^3 + 3x^2 + x + 4 &\mod P(x)
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\end{align*}
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\end{enumerate}
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\item
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\begin{align*}
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x^4 &\equiv 6x^2 + 4x + 2 &\mod P(x) \\
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x^5 &\equiv 6x^3 + 4x^2 + 2x &\mod P(x) \\
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x^6 &\equiv 4x^3 + x^2 + 5x + 5 &\mod P(x) \\
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x^7 &\equiv 3x^3 + 6x^2 + 1 &\mod P(x) \\
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\end{align*}
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\end{enumerate}
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\end{document}
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