222 lines
6.3 KiB
TeX
222 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{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{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 {Gruppe 193} %
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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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\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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Ciphertext:
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\begin{align*}
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\text{Hex} && \text{Bin} \\
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(DC)_{16} &=& (11011100)_2 \\
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(48)_{16} &=& (01001000)_2 \\
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(13)_{16} &=& (00010011)_2
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\end{align*}
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Schl<EFBFBD>ssel:
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\begin{align*}
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\text{Hex} && \text{Bin} \\
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(98)_{16} &=& (10011000)_2 \\
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(29)_{16} &=& (00101001)_2 \\
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(60)_{16} &=& (01100000)_2
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\end{align*}
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Bin<EFBFBD>re berechnung der ersten 3 Bytes:
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\begin{itemize}
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\item
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\begin{tabular}{ccccccccc}
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& 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 \\
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xor & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
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\hline
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& 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0
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\end{tabular}
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$(01000100)_2 = (44)_{16} = `D`_{ASCII}$
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\item
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\begin{tabular}{ccccccccc}
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& 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\
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xor & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
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\hline
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& 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1
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\end{tabular}
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$(01100001)_2 = (61)_{16} = `a`_{ASCII}$
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\item
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\begin{tabular}{ccccccccc}
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& 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
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xor & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
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\hline
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& 0 & 1 & 1 & 1 & 0 & 0 & 1 & 1
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\end{tabular}
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$(01110011)_2 = (73)_{16} = `s`_{ASCII}$
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\end{itemize}
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\begin{itemize}
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\item
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\begin{tabular}{lcccccccccccc}
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Ciphertext & DC & 48 & 13 & 3B & 9C & 4C & 49 & 80 & AC & A7 & B9 & 54 \\
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Schl<68>ssel & 98 & 29 & 60 & 72 & F2 & 38 & 2C & F2 & C2 & C2 & CD & 1D \\
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\hline
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Plaintext & 44 & 61 & 73 & 49 & 6E & 74 & 65 & 72 & 6E & 65 & 74 & 49 \\
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ASCII & D & a & s & I & n & t & e & r & n & e & t & I
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\end{tabular}
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\item
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\begin{tabular}{lcccccccccccc}
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Ciphertext & F2 & 7C & 2B & 9E & D5 & DF & 0D & 05 & B3 & 1D & 4E & F8 \\
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Schl<68>ssel & 81 & 08 & 65 & FB & A0 & B3 & 6C & 6B & D7 & 3C & 6F & D9 \\
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\hline
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Plaintext & 73 & 74 & 4E & 65 & 75 & 6C & 61 & 6E & 64 & 21 & 21 & 21 \\
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ASCII & s & t & N & e & u & l & a & n & d & ! & ! & !
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\end{tabular}
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Plaintext: \enquote{DasInternetIstNeuland!!!}.
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\end{itemize}
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\section*{Aufgabe 2}
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Da jedes Bit im Ciphertext ein eigenes Bit im Schl<68>ssel hat, l<>sst sich jeder beliebige Klartext (der selben L<>nge) aus
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dem Ciphertext ableiten, zumindest wenn der Schl<68>ssel richtig gew<65>hlt wird. Daher ist es nicht m<>glich, zu sagen, ob ein
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Schl<EFBFBD>ssel korrekt ist, ohne den urspr<70>nglichen Plaintext zu kennen.
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\section*{Aufgabe 3}
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\begin{enumerate}[(a)]
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\item Die Magic Number des PNG Formats ist $89$ $50$ $4e$ $47$ $0d$ $0a$ $1a$ $0a$. Diese 8 Bytes bilden den Anfang
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jeder validen PNG Datei. Das er<65>ffnet die M<>glichkeit f<>r einen Known-Plaintext Angriff.
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\item F<>r $a$, $b$ und $c$ mit $a \text{ xor } b = c$ gilt $b \text{ xor } c = a$ und $a \text{ xor } c = b$, also
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l<>sst sich der Schl<68>ssel berechnen, in dem man den Ciphertext mit dem Known-Plaintext xor'ed.
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\begin{tabular}{lcccccccc}
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Ciphertext (hex) & 33 & fd & eb & 12 & cd & 0a & 0a & f5 \\
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Known Plaintext (hex) & 89 & 50 & 4e & 47 & 0d & 0a & 1a & 0a \\
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\hline
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Schl<68>ssel (hex) & ba & ad & a5 & 55 & c0 & 00 & 10 & ff \\
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\end{tabular}
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\item Die S<><53>igkeiten sind im Lageraum im ID (gegen<65>ber vom Fachschaftsraum), in dem Regal direkt gegen<65>ber der
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T<>r. Ca auf der H<>he zwischen T<>re und Grill.
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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
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\begin{enumerate}[1)]
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\item Berechnung der ersten 8 Bytes des Schl<68>sselstroms:
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Known Plaintext: \enquote{https://} $\Rightarrow$ $68$ $74$ $74$ $70$ $73$ $3a$ $2f$ $2f$
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\begin{tabular}{lcccccccc}
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Ciphertext & 4E & 7E & 3D & 88 & 8E & 01 & 0D & 84 \\
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Known-Plaintext & 68 & 74 & 74 & 70 & 73 & 3A & 2F & 2F \\
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\hline
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Schl<68>ssel & 26 & 0A & 49 & F8 & FD & 3B & 22 & AB \\
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\end{tabular}
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\item Aufstellen eines Gleichungssystems zur Bestimmung von $A$, $B$ und $C$ des LCG:
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$m = 257$
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\begin{eqnarray*}
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S_0 &= (26)_{16} = (38)_{10} \\
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S_1 &= (0A)_{16} = (10)_{10} \\
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S_2 &= (49)_{16} = (73)_{10} \\
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S_3 &= (F8)_{16} = (248)_{10} \\
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S_4 &= (FD)_{16} = (253)_{10} \\\\
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S_2 &\equiv A * S_1 + B * S_0 + C &\mod 257 \\
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S_3 &\equiv A * S_2 + B * S_1 + C &\mod 257 \\
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S_4 &\equiv A * S_3 + B * S_2 + C &\mod 257 \\
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\end{eqnarray*}
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\begin{align}
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73 &\equiv 10 * A + 38 * B + C &\mod 257 \\
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C &\equiv 73 - 10 * A - 38 * B &\mod 257 \\
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\\
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248 &\equiv 73 * A + 10 * B + 73 - 10 * A - 38 * B &\mod 257 \\
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248 &\equiv 63 * A - 28 * B + 73 &\mod 257 \\
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63 * A &\equiv 175 + 28 * B &\mod 257 \\
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A &\equiv 29 * B + 117 &\mod 257 \\
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\\
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253 &\equiv 248(29*B + 117) + 73 * B + 73 - 10(29*B + 117) - 38*B &\mod 257 \\
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253 &\equiv 253*B + 232 + 73*B + 73 - 33*B - 142 - 38 *B &\mod 257 \\
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253 &\equiv 255*B + 163 &\mod 257 \\
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255*B &\equiv 90 &\mod 257 \\
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\underline{B} &\equiv 212 &\mod 257 \\
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\\
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A &\equiv 29 * 212 + 117 &\mod 257 \\
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\underline{A} &\equiv 97 &\mod 257 \\
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\\
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C &\equiv 73 - 10 * 97 - 38 * 212 &\mod 257 \\
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\underline{C} &\equiv 42 &\mod 257
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\end{align}
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$A = 97 \mod 257$, $B = 212 \mod 257$, $C = 42 \mod 257$
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Plaintext: \url{https://youtu.be/VmUGe8KDdGI}
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\end{enumerate}
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\item Es werden $n+1$ Parameter und $n$ Seed-Werte ben<65>tigt. Da man also $n+1$ Unbekannte im aufzustellenden
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Gleichungssystem hat, werden auch $n+1$ Klartext-Chiffretext-Paare f<>r einen erfolgreichen Angriff ben<65>tigt.
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\end{enumerate}
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\end{document}
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