diff --git a/school/intro-crypto/uebung/03/03.tex b/school/intro-crypto/uebung/03/03.tex new file mode 100644 index 0000000..3e5aa40 --- /dev/null +++ b/school/intro-crypto/uebung/03/03.tex @@ -0,0 +1,221 @@ +\documentclass[12pt,a4paper,german]{article} +\usepackage{url} +%\usepackage{graphics} +\usepackage{times} +\usepackage[T1]{fontenc} +\usepackage{pifont} +\usepackage{ngerman} +\usepackage{float} +\usepackage{diagbox} +\usepackage[latin1]{inputenc} +\usepackage{geometry} +\usepackage{amsfonts} +\usepackage{amsmath} +\usepackage{csquotes} +\usepackage{graphicx} +\usepackage{epsfig} +\usepackage{paralist} +\geometry{left=2.0cm,textwidth=17cm,top=3.5cm,textheight=23cm} + +%%%%%%%%%% Fill out the the definitions %%%%%%%%% +\def \name {Valentin Brandl} % +\def \matrikel {108018274494} % +% \def \pname {Vorname2 Nachname2} % +% \def \pmatrikel {Matrikelnummer2} % +\def \gruppe {Gruppe 193} % +\def \uebung {1} % +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + + % DO NOT MODIFY THIS HEADER +\newcommand{\hwsol}{ +\vspace*{-2cm} +\noindent \matrikel \quad \name \hfill Gruppe:\gruppe \\ +% \noindent \pmatrikel \quad \pname \\ +\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center} +} + +\newcommand{\cmark}{\ding{51}}% +\newcommand{\xmark}{\ding{55}}% + +\begin{document} +%Import header +\hwsol + + +\section*{Aufgabe 1} + +Ciphertext: + +\begin{align*} + \text{Hex} && \text{Bin} \\ + (DC)_{16} &=& (11011100)_2 \\ + (48)_{16} &=& (01001000)_2 \\ + (13)_{16} &=& (00010011)_2 +\end{align*} + +Schlüssel: + +\begin{align*} + \text{Hex} && \text{Bin} \\ + (98)_{16} &=& (10011000)_2 \\ + (29)_{16} &=& (00101001)_2 \\ + (60)_{16} &=& (01100000)_2 +\end{align*} + +Binäre berechnung der ersten 3 Bytes: + +\begin{itemize} + + \item + \begin{tabular}{ccccccccc} + & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 \\ + xor & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ + \hline + & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 + \end{tabular} + + $(01000100)_2 = (44)_{16} = `D`_{ASCII}$ + + + \item + \begin{tabular}{ccccccccc} + & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ + xor & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\ + \hline + & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 + \end{tabular} + + $(01100001)_2 = (61)_{16} = `a`_{ASCII}$ + + \item + \begin{tabular}{ccccccccc} + & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\ + xor & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ + \hline + & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 1 + \end{tabular} + + $(01110011)_2 = (73)_{16} = `s`_{ASCII}$ + +\end{itemize} + +\begin{itemize} + + \item + \begin{tabular}{lcccccccccccc} + Ciphertext & DC & 48 & 13 & 3B & 9C & 4C & 49 & 80 & AC & A7 & B9 & 54 \\ + Schlüssel & 98 & 29 & 60 & 72 & F2 & 38 & 2C & F2 & C2 & C2 & CD & 1D \\ + \hline + Plaintext & 44 & 61 & 73 & 49 & 6E & 74 & 65 & 72 & 6E & 65 & 74 & 49 \\ + ASCII & D & a & s & I & n & t & e & r & n & e & t & I + \end{tabular} + + \item + \begin{tabular}{lcccccccccccc} + Ciphertext & F2 & 7C & 2B & 9E & D5 & DF & 0D & 05 & B3 & 1D & 4E & F8 \\ + Schlüssel & 81 & 08 & 65 & FB & A0 & B3 & 6C & 6B & D7 & 3C & 6F & D9 \\ + \hline + Plaintext & 73 & 74 & 4E & 65 & 75 & 6C & 61 & 6E & 64 & 21 & 21 & 21 \\ + ASCII & s & t & N & e & u & l & a & n & d & ! & ! & ! +\end{tabular} + +Plaintext: \enquote{DasInternetIstNeuland!!!}. + +\end{itemize} + +\section*{Aufgabe 2} + +Da jedes Bit im Ciphertext ein eigenes Bit im Schlüssel hat, lässt sich jeder beliebige Klartext (der selben Länge) aus +dem Ciphertext ableiten, zumindest wenn der Schlüssel richtig gewählt wird. Daher ist es nicht möglich, zu sagen, ob ein +Schlüssel korrekt ist, ohne den ursprünglichen Plaintext zu kennen. + +\section*{Aufgabe 3} + +\begin{enumerate}[(a)] + + \item Die Magic Number des PNG Formats ist $89$ $50$ $4e$ $47$ $0d$ $0a$ $1a$ $0a$. Diese 8 Bytes bilden den Anfang + jeder validen PNG Datei. Das eröffnet die Möglichkeit für einen Known-Plaintext Angriff. + + \item Für $a$, $b$ und $c$ mit $a \text{ xor } b = c$ gilt $b \text{ xor } c = a$ und $a \text{ xor } c = b$, also + lässt sich der Schlüssel berechnen, in dem man den Ciphertext mit dem Known-Plaintext xor'ed. + + \begin{tabular}{lcccccccc} + Ciphertext (hex) & 33 & fd & eb & 12 & cd & 0a & 0a & f5 \\ + Known Plaintext (hex) & 89 & 50 & 4e & 47 & 0d & 0a & 1a & 0a \\ + \hline + Schlüssel (hex) & ba & ad & a5 & 55 & c0 & 00 & 10 & ff \\ + \end{tabular} + + \item Die Süßigkeiten sind im Lageraum im ID (gegenüber vom Fachschaftsraum), in dem Regal direkt gegenüber der + Tür. Ca auf der Höhe zwischen Türe und Grill. + +\end{enumerate} + +\section*{Aufgabe 4} + +\begin{enumerate}[(a)] + + \item + \begin{enumerate}[1)] + + \item Berechnung der ersten 8 Bytes des Schlüsselstroms: + + Known Plaintext: \enquote{https://} $\Rightarrow$ $68$ $74$ $74$ $70$ $73$ $3a$ $2f$ $2f$ + + \begin{tabular}{lcccccccc} + Ciphertext & 4E & 7E & 3D & 88 & 8E & 01 & 0D & 84 \\ + Known-Plaintext & 68 & 74 & 74 & 70 & 73 & 3A & 2F & 2F \\ + \hline + Schlüssel & 26 & 0A & 49 & F8 & FD & 3B & 22 & AB \\ + \end{tabular} + + \item Aufstellen eines Gleichungssystems zur Bestimmung von $A$, $B$ und $C$ des LCG: + + $m = 257$ + + \begin{eqnarray*} + S_0 &= (26)_{16} = (38)_{10} \\ + S_1 &= (0A)_{16} = (10)_{10} \\ + S_2 &= (49)_{16} = (73)_{10} \\ + S_3 &= (F8)_{16} = (248)_{10} \\ + S_4 &= (FD)_{16} = (253)_{10} \\\\ + S_2 &\equiv A * S_1 + B * S_0 + C &\mod 257 \\ + S_3 &\equiv A * S_2 + B * S_1 + C &\mod 257 \\ + S_4 &\equiv A * S_3 + B * S_2 + C &\mod 257 \\ + \end{eqnarray*} + + \begin{align} + 73 &\equiv 10 * A + 38 * B + C &\mod 257 \\ + C &\equiv 73 - 10 * A - 38 * B &\mod 257 \\ + \\ + 248 &\equiv 73 * A + 10 * B + 73 - 10 * A - 38 * B &\mod 257 \\ + 248 &\equiv 63 * A - 28 * B + 73 &\mod 257 \\ + 63 * A &\equiv 175 + 28 * B &\mod 257 \\ + A &\equiv 29 * B + 117 &\mod 257 \\ + \\ + 253 &\equiv 248(29*B + 117) + 73 * B + 73 - 10(29*B + 117) - 38*B &\mod 257 \\ + 253 &\equiv 253*B + 232 + 73*B + 73 - 33*B - 142 - 38 *B &\mod 257 \\ + 253 &\equiv 255*B + 163 &\mod 257 \\ + 255*B &\equiv 90 &\mod 257 \\ + \underline{B} &\equiv 212 &\mod 257 \\ + \\ + A &\equiv 29 * 212 + 117 &\mod 257 \\ + \underline{A} &\equiv 97 &\mod 257 \\ + \\ + C &\equiv 73 - 10 * 97 - 38 * 212 &\mod 257 \\ + \underline{C} &\equiv 42 &\mod 257 + \end{align} + + $A = 97 \mod 257$, $B = 212 \mod 257$, $C = 42 \mod 257$ + + Plaintext: \url{https://youtu.be/VmUGe8KDdGI} + + \end{enumerate} + + \item Es werden $n+1$ Parameter und $n$ Seed-Werte benötigt. Da man also $n+1$ Unbekannte im aufzustellenden + Gleichungssystem hat, werden auch $n+1$ Klartext-Chiffretext-Paare für einen erfolgreichen Angriff benötigt. + +\end{enumerate} + +\end{document} + diff --git a/school/intro-crypto/uebung/03/ASCII Conversion Chart.pdf b/school/intro-crypto/uebung/03/ASCII Conversion Chart.pdf new file mode 100644 index 0000000..2d7b51b Binary files /dev/null and b/school/intro-crypto/uebung/03/ASCII Conversion Chart.pdf differ diff --git a/school/intro-crypto/uebung/03/lage.hex b/school/intro-crypto/uebung/03/lage.hex new file mode 100644 index 0000000..071c388 Binary files /dev/null and b/school/intro-crypto/uebung/03/lage.hex differ diff --git a/school/intro-crypto/uebung/03/plain.png b/school/intro-crypto/uebung/03/plain.png new file mode 100644 index 0000000..6dcd3ed Binary files /dev/null and b/school/intro-crypto/uebung/03/plain.png differ diff --git a/school/intro-crypto/uebung/03/u03/Cargo.lock b/school/intro-crypto/uebung/03/u03/Cargo.lock new file mode 100644 index 0000000..e7e39ae --- /dev/null +++ b/school/intro-crypto/uebung/03/u03/Cargo.lock @@ -0,0 +1,4 @@ +[[package]] +name = "u03" +version = "0.1.0" + diff --git a/school/intro-crypto/uebung/03/u03/Cargo.toml b/school/intro-crypto/uebung/03/u03/Cargo.toml new file mode 100644 index 0000000..32d1196 --- /dev/null +++ b/school/intro-crypto/uebung/03/u03/Cargo.toml @@ -0,0 +1,6 @@ +[package] +name = "u03" +version = "0.1.0" +authors = ["Valentin Brandl "] + +[dependencies] diff --git a/school/intro-crypto/uebung/03/u03/src/main.rs b/school/intro-crypto/uebung/03/u03/src/main.rs new file mode 100644 index 0000000..221046b --- /dev/null +++ b/school/intro-crypto/uebung/03/u03/src/main.rs @@ -0,0 +1,101 @@ +// use std::fs; +struct Prng { + a: u16, + b: u16, + c: u16, + s0: u16, + s1: u16, +} + +impl Iterator for Prng { + type Item = u8; + + fn next(&mut self) -> Option { + let r = self.s0; + self.s0 = self.s1; + self.s1 = ((self.a * self.s1) % 257 + (self.b * r) % 257 + self.c) % 257; + Some((r % 256) as u8) + } +} + +impl Prng { + fn new(s0: u16, s1: u16) -> Self { + Self { + s0: s0 % 257, + s1: s1 % 257, + a: 97, + b: 212, + c: 42, + } + } +} + +fn main() { + // let c: &[u8] = &[0x4E, 0x7E, 0x3D, 0x88, 0x8E, 0x01, 0x0D, 0x84]; + // let p: &[u8] = &[0x68, 0x74, 0x74, 0x70, 0x73, 0x3A, 0x2F, 0x2F]; + // c.iter() + // .zip(p.iter()) + // .map(|(c, p)| c ^ p) + // .for_each(|k| print!("{:02X} ", k)); + + let c: &[u8] = &[ + 0x4E, 0x7E, 0x3D, 0x88, 0x8E, 0x01, 0x0D, 0x84, 0xB8, 0x7E, 0xBF, 0x1A, 0x25, 0x37, 0xFA, + 0x4D, 0x89, 0x87, 0x91, 0xFA, 0x50, 0x51, 0xFC, 0x42, 0x7A, 0x9A, 0x6A, 0xE4, + ]; + let p = Prng::new(0x26, 0x0a); + + c.into_iter() + .zip(p) + .map(|(c, k)| c ^ k) + .map(|c| c as char) + .for_each(|c| print!("{}", c)); + println!(); + + // let key: &[u8] = &[0xba, 0xad, 0xa5, 0x55, 0xc0, 0x00, 0x10, 0xff]; + // // let cipher: &[u8] = &[0x33, 0xfd, 0xeb, 0x12, 0xcd, 0x0a, 0x0a, 0xf5]; + // // let key: &[u8] = &[0x89, 0x50, 0x4e, 0x47, 0x0d, 0x0a, 0x1a, 0x0a]; + // // cipher + // // .iter() + // // .zip(key.iter()) + // // .map(|(c, k)| c ^ k) + // // .for_each(|p| print!("0x{:x} ", p)); + + // let args: Vec<_> = std::env::args().skip(1).collect(); + // let data = fs::read(&args[0]).expect("Cannot read"); + // let plain: Vec<_> = data + // .into_iter() + // .zip(key.into_iter().cycle()) + // .map(|(c, k)| c ^ k) + // .collect(); + + // fs::write(&args[1], plain).expect("Cannot write"); +} + +// fn main() { +// let cipher: &[u8] = &[ +// 0xDC, 0x48, 0x13, 0x3B, 0x9C, 0x4C, 0x49, 0x80, 0xAC, 0xA7, 0xB9, 0x54, 0xF2, 0x7C, 0x2B, +// 0x9E, 0xD5, 0xDF, 0x0D, 0x05, 0xB3, 0x1D, 0x4E, 0xF8, +// ]; +// let key: &[u8] = &[ +// 0x98, 0x29, 0x60, 0x72, 0xF2, 0x38, 0x2C, 0xF2, 0xC2, 0xC2, 0xCD, 0x1D, 0x81, 0x08, 0x65, +// 0xFB, 0xA0, 0xB3, 0x6C, 0x6B, 0xD7, 0x3C, 0x6F, 0xD9, +// ]; +// cipher.iter().for_each(|c| print!("0x{:X} ", c)); +// println!(); +// key.iter().for_each(|c| print!("0x{:X} ", c)); +// println!(); +// cipher +// .iter() +// .zip(key.iter()) +// .map(|(c, k)| c ^ k) +// .for_each(|p| print!("{:X} ", p)); +// println!(); + +// cipher +// .iter() +// .zip(key.iter()) +// .map(|(c, k)| c ^ k) +// .map(|c| c as char) +// .for_each(|p| print!("{} ", p)); +// println!(); +// } diff --git a/school/intro-crypto/uebung/03/u03/target/.rustc_info.json b/school/intro-crypto/uebung/03/u03/target/.rustc_info.json new file mode 100644 index 0000000..f5035cb --- /dev/null +++ b/school/intro-crypto/uebung/03/u03/target/.rustc_info.json @@ -0,0 +1 @@ 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