249 lines
6.3 KiB
TeX
249 lines
6.3 KiB
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{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{csquotes}
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\usepackage{graphicx}
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\usepackage{epsfig}
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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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%%%%%%%%%% 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 {Gruppenkuerzel} %
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\def \uebung {1} %
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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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\begin{document}
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%Import header
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\hwsol
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\section*{Aufgabe 2}
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\begin{enumerate}[(a)]
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\item
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\begin{eqnarray*}
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A &\text{:}& 23 \to{} 0.02 \\
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B &\text{:}& 5 \to{} 0.00 \\
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C &\text{:}& 22 \to{} 0.02 \\
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D &\text{:}& 8 \to{} 0.01 \\
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E &\text{:}& 17 \to{} 0.02 \\
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F &\text{:}& 2 \to{} 0.00 \\
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G &\text{:}& 80 \to{} 0.07 \\
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J &\text{:}& 66 \to{} 0.06 \\
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K &\text{:}& 39 \to{} 0.04 \\
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L &\text{:}& 47 \to{} 0.04 \\
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M &\text{:}& 2 \to{} 0.00 \\
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N &\text{:}& 17 \to{} 0.02 \\
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O &\text{:}& 63 \to{} 0.06 \\
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P &\text{:}& 80 \to{} 0.07 \\
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Q &\text{:}& 62 \to{} 0.06 \\
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R &\text{:}& 57 \to{} 0.05 \\
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S &\text{:}& 136 \to{} 0.12 \\
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T &\text{:}& 6 \to{} 0.01 \\
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U &\text{:}& 57 \to{} 0.05 \\
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V &\text{:}& 6 \to{} 0.01 \\
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W &\text{:}& 76 \to{} 0.07 \\
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X &\text{:}& 25 \to{} 0.02 \\
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Y &\text{:}& 8 \to{} 0.01 \\
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Z &\text{:}& 83 \to{} 0.08 \\
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\text{\"A} &\text{:}& 43 \to{} 0.04 \\
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\text{\"O} &\text{:}& 14 \to{} 0.01 \\
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\text{\"U} &\text{:}& 28 \to{} 0.03 \\
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\text{<EFBFBD>} &\text{:}& 18 \to{} 0.02 \\
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sum &\text{:}& 1090
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\end{eqnarray*}
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\item
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\begin{displayquote}
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gr<67>ndlich durchgecheckt steht sie da \\
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und wartet auf den start - alles klar! \\
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experten streiten sich um ein paar daten \\
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die crew hat da noch ein paar fragen \\
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doch der countdown l<>uft \\
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\end{displayquote}
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\item
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\begin{tabular}{|c|c|}
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\hline
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von & zu \\\hline
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A & f \\
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B & <20> \\
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C & g \\
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D & j \\
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E & <20> \\
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F & x \\
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G & n \\
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J & i \\
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K & m \\
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L & c \\
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M & <20> \\
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N & v \\
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O & d \\
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P & r \\
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Q & a \\
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R & l \\
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S & e \\
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T & <20> \\
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U & h \\
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V & z \\
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W & t \\
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X & w \\
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Y & p \\
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Z & s \\
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<09> & o \\
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<09> & b \\
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<09> & u \\
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<09> & k \\
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H/I & q/y \\
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\hline
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\end{tabular}
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Weder H noch I kommen im Ciphertext vor, daher ist es nicht m<>glich, die Substitution von H/I eindeutig zu
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bestimmen. Da jedoch auch weder q noch y im Plaintext vorkommen, muss entweder $H \to q \text{ und } I \to y$
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oder $H \to y \text{ und } I \to q$ gelten.
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\item $30! = \vert \{A,B,...Z,\text{<EFBFBD>},\text{<EFBFBD>},\text{<EFBFBD>},\text{<EFBFBD>}\} \vert!$
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\item Name des Textes: Major Tom (v<>llig losgel<65>st) \\
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Erscheinungsjahr: 1982
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\end{enumerate}
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\section*{Aufgabe 3}
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\begin{enumerate}[(a)]
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\item $2^{63} = 9,223,372,036,854,775,808$
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\item $2^{63} * 0.03g = 276,701,161,105,643,260g \approx 276,701,161,106t$ \\
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$277 \cdot 10^9 / 460 \cdot 10^6 \approx 602 \Rightarrow$ mehr als das 600-fache der j<>hrlichen Reisernte
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\item $2^{10} * 0.1mm = 102.4mm = 1.204m$
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\item
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\begin{eqnarray*}
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2^n * 0.1mm &\geq{}& 1,000,000mm \\
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2^n &\geq{}& 10000000 \\
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n &\geq{}& 24
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\end{eqnarray*}
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Man muss das Blatt 24 mal falten.
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\item
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\begin{eqnarray*}
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2^n * 0.1mm &\geq{}& 384,400,000,000mm \\
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2^n &\geq{}& 3,844,000,000,000 \\
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n &\geq{}& 42
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\end{eqnarray*}
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Man muss das Blatt 42 mal falten.
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\item
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\begin{eqnarray*}
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2^n * 0.1mm &\geq{}& 9,460,730,472,580,800,000mm \\
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2^n &\geq{}& 94,607,304,725,808,000,000mm \\
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n &\geq{}& 67
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\end{eqnarray*}
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Man muss das Blatt 67 mal falten.
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\end{enumerate}
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\section*{Aufgabe 4}
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\begin{enumerate}[(a)]
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\item $2^n$
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\item $\frac{2^n}{2}$
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\item Gerechnet mit $1y = 365d$
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\begin{figure}[h]
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\begin{tabular}{|l|l|l|l|}
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\hline
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\diagbox{Case}{$n$} & 80 & 112 & 192 \\\hline
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Worst Case & $2.50 * 10^8y$ & $1.08 * 10^{18}y$ & $1.30 * 10^{42}y$ \\\hline
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Average Case & $1.25 * 10^8y$ & $5.38 * 10^{17}y$ & $6.50 * 10^{41}y$ \\\hline
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\end{tabular}
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\caption{GPU}
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\end{figure}
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\begin{figure}[h]
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\begin{tabular}{|l|l|l|l|}
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\hline
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\diagbox{Case}{$n$} & 80 & 112 & 192 \\\hline
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Worst Case & $5.81 * 10^7y$ & $2.49 * 10^{17}y$ & $3.02 * 10^{41}y$ \\\hline
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Average Case & $2.90 * 10^7y$ & $1.25 * 10^{17}y$ & $1.51 * 10^{41}y$ \\\hline
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\end{tabular}
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\caption{Amazon Cloud}
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\end{figure}
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\begin{figure}[h]
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\begin{tabular}{|l|l|l|l|}
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\hline
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\diagbox{Case}{$n$} & 80 & 112 & 192 \\\hline
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Worst Case & $3.41 * 10^7y$ & $1.46 * 10^{17}y$ & $1.77 * 10^{41}y$ \\\hline
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Average Case & $1.70 * 10^7y$ & $7.32 * 10^{16}y$ & $8.85 * 10^{40}y$ \\\hline
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\end{tabular}
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\caption{FPGA}
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\end{figure}
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\item Unter Anwendung der Erkenntnisse aus e):
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\begin{eqnarray*}
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n &=& 80 \\
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n + w &=& 112 \\
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w &=& 32 \\
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x &=& \frac{2^{112} - 2^{80}}{r} \\
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\end{eqnarray*}
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\begin{itemize}
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\item $r_{GPU} = 15.3 * 10^7 \frac{k}{s}$
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\item $r_{Amazon} = 66 * 10^7 \frac{k}{s}$
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\item $r_{FPGA} = 11.25 * 10^8 \frac{k}{s}$
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\end{itemize}
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\begin{eqnarray*}
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x_{GPU} &=& 1.06 * 10^{18}y \\
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x_{Amazon} &=& 2.49 * 10^{17}y \\
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x_{FPGA} &=& 1.46 * 10^{17}y \\
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\end{eqnarray*}
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\item
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\begin{itemize}
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\item $r$: Anzahl der Versuche pro Sekunde
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\item $n$: Aktuelle Bit L<>nge der Schl<68>ssel
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\item $w$: Verl<72>ngerung der Schl<68>ssell<6C>nge um $w$ Bit
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\item $x$: Gesucht: Wie viel langsamer der Angriff wird
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\end{itemize}
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\begin{eqnarray*}
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\frac{2^n}{r} + x &=& \frac{2^{n+w}}{r} \\
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x &=& \frac{2^{n+w} - 2^n}{r} \\
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x &=& \frac{2^n(2^w-1)}{r}
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\end{eqnarray*}
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\item
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\begin{eqnarray*}
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183d &=& 4392h \\\\
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\frac{4392}{2^n} &\leq{}& 1 \\
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n &\geq{}& 13 \Rightarrow \text{ 13 Verdoppelungen der Rechenleistung} \\\\
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13 * 18m &=& 234 m = 19.5y \approx 20y \\
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1992 + 20 &=& 2012
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\end{eqnarray*}
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Seit dem Jahr 2012 sollte es m<>glich sein, 50-bit Keys in weniger als 1 Stunde zu Bruteforcen.
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\end{enumerate}
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\end{document}
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