Add intro-crypto exercise 03

school/intro-crypto/uebung/03/03.tex

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+\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*}
+
+Schl<EFBFBD>ssel:
+
+\begin{align*}
+	\text{Hex} && \text{Bin} \\
+	(98)_{16} &=& (10011000)_2 \\
+	(29)_{16} &=& (00101001)_2 \\
+	(60)_{16} &=& (01100000)_2
+\end{align*}
+
+Bin<EFBFBD>re 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 \\
+			Schl<68>ssel  & 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 \\
+			Schl<68>ssel  & 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 Schl<68>ssel hat, l<>sst sich jeder beliebige Klartext (der selben L<>nge) aus
+dem Ciphertext ableiten, zumindest wenn der Schl<68>ssel richtig gew<65>hlt wird. Daher ist es nicht m<>glich, zu sagen, ob ein
+Schl<EFBFBD>ssel korrekt ist, ohne den urspr<70>nglichen 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 er<65>ffnet die M<>glichkeit f<>r einen Known-Plaintext Angriff.
+
+	\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
+		l<>sst sich der Schl<68>ssel 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
+			Schl<68>ssel (hex) & ba & ad & a5 & 55 & c0 & 00 & 10 & ff \\
+		\end{tabular}
+
+	\item Die S<><53>igkeiten sind im Lageraum im ID (gegen<65>ber vom Fachschaftsraum), in dem Regal direkt gegen<65>ber der
+		T<>r. Ca auf der H<>he zwischen T<>re und Grill.
+
+\end{enumerate}
+
+\section*{Aufgabe 4}
+
+\begin{enumerate}[(a)]
+
+	\item
+		\begin{enumerate}[1)]
+
+			\item Berechnung der ersten 8 Bytes des Schl<68>sselstroms:
+
+				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
+					Schl<68>ssel & 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 ben<65>tigt. Da man also $n+1$ Unbekannte im aufzustellenden
+		Gleichungssystem hat, werden auch $n+1$ Klartext-Chiffretext-Paare f<>r einen erfolgreichen Angriff ben<65>tigt.
+
+\end{enumerate}
+
+\end{document}
+

school/intro-crypto/uebung/03/u03/Cargo.lock

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+[[package]]
+name = "u03"
+version = "0.1.0"
+

school/intro-crypto/uebung/03/u03/Cargo.toml

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+[package]
+name = "u03"
+version = "0.1.0"
+authors = ["Valentin Brandl <vbrandl@riseup.net>"]
+
+[dependencies]

school/intro-crypto/uebung/03/u03/src/main.rs

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+// 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<u8> {
+        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!();
+// }

school/intro-crypto/uebung/03/u03/target/.rustc_info.json

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+{"rustc_fingerprint":12819272735836075344,"outputs":{"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",""],"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",""],"1164083562126845933":["rustc 1.29.2 (17a9dc751 2018-10-05)\nbinary: rustc\ncommit-hash: 17a9dc7513b9fea883dc9505f09f97c63d1d601b\ncommit-date: 2018-10-05\nhost: x86_64-unknown-linux-gnu\nrelease: 1.29.2\nLLVM version: 7.0\n",""]}}

school/intro-crypto/uebung/03/u03/target/debug/.fingerprint/u03-4d5f92818f15e465/bin-u03-4d5f92818f15e465

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+c9aea05b4f7bab6b

school/intro-crypto/uebung/03/u03/target/debug/.fingerprint/u03-4d5f92818f15e465/bin-u03-4d5f92818f15e465.json

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school/intro-crypto/uebung/03/u03/target/debug/.fingerprint/u03-560dcfd44619ed7e/bin-u03-560dcfd44619ed7e

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+d00b19fc5c9b0c78

school/intro-crypto/uebung/03/u03/target/debug/.fingerprint/u03-560dcfd44619ed7e/bin-u03-560dcfd44619ed7e.json

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+{"rustc":2049182171942789226,"features":"[]","target":8092603168892422263,"profile":8925656243208791261,"path":1036222786711178230,"deps":[],"local":[{"MtimeBased":[[1541015093,869731229],".fingerprint/u03-560dcfd44619ed7e/dep-bin-u03-560dcfd44619ed7e"]}],"rustflags":[],"edition":"Edition2015"}

school/intro-crypto/uebung/03/u03/target/debug/deps/u03-4d5f92818f15e465.d

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+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/deps/u03-4d5f92818f15e465: src/main.rs
+
+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/deps/u03-4d5f92818f15e465.d: src/main.rs
+
+src/main.rs:

school/intro-crypto/uebung/03/u03/target/debug/deps/u03-560dcfd44619ed7e.d

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+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/deps/u03-560dcfd44619ed7e.rmeta: src/main.rs
+
+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/deps/u03-560dcfd44619ed7e.d: src/main.rs
+
+src/main.rs:

school/intro-crypto/uebung/03/u03/target/debug/libu03.d

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+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/libu03.rmeta: /home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/src/main.rs

school/intro-crypto/uebung/03/u03/target/debug/u03.d

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+/home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/target/debug/u03: /home/me/Dokumente/notes/school/intro-crypto/uebung/03/u03/src/main.rs