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-\documentclass[a4paper]{article}
-
-\usepackage{a4wide,amsmath,amssymb,url,graphicx,comment,enumerate,color,array,inconsolata,minted,listings}
-
-\title{Homework $1$}
-\author{Camil Staps (s4498062)}
-
-\newcommand{\R}{\mathbb R}
-\newcommand{\Q}{\mathbb Q}
-\newcommand{\Z}{\mathbb Z}
-\newcommand{\N}{\mathbb N}
-\newcommand{\F}{\mathbb F}
-\newcommand{\Zstar}{\Z^{^*}}
-
-\begin{document}
-\maketitle
-
-\section*{Solution}
-
-\begin{enumerate}
-\item  \begin{enumerate}
-	\item 	This would work: $p_1$ and $p_2$ share $k_1$ and $k_1'$; $p_1$ and $p_3$ share $k_2$ and $k_2'$; and $p_2$ and $p_3$ share also $k_2$ and $k_2'$. Furthermore, none of the executives has two parts of the key ($k_i$ and $k_i'$ for some $i$).
-	\item 	$p_2$ and $p_3$ can't compute the key together: together they have only $k_1'$ and $k_2'$, they would need either $k_1$ or $k_2$ in addition to that to be able to compute $k$. Therefore, this is not a suitable solution.
-	\item 	$p_1$ can compute $k$ by himself with $k = k_2 \oplus k_2'$, so this is not suitable.
-	\end{enumerate}
-
-\item 	We first compute the one time pad key used, by XOR-ing the ciphertext with the plaintext ``attack by sea''. After that, we XOR the key we found with the plaintext ``attack by air''. Since the first eight bytes of both the plaintext and the key stay the same, we only have to consider the last three bytes.
-
-	\begin{table}[h]
-	\centering
-	{\renewcommand{\arraystretch}{1.2}
-	\begin{tabular}{r | >{\ttfamily}c >{\ttfamily}c >{\ttfamily}c}
-	Plaintext (8-bit ASCII) 			& s 			& e				& a				\\
-	Plaintext (hex)						& 73			& 65			& 61			\\\cline{2-4}
-	Ciphertext (hex)					& 2C			& F4		 	& 71			\\\cline{2-4}
-	Key (XOR of plaintext with ciphertext) & \textbf{5F}& \textbf{91}	& \textbf{10}	\\\cline{2-4}
-	New plaintext (8-bit ASCII)			& a				& i				& r				\\
-	New plaintext (hex)					& 61			& 69			& 72			\\\cline{2-4}
-	New ciphertext (XOR of new plaintext with key) & \textbf{3E} & \textbf{F8} & \textbf{62}
-	\end{tabular}
-	}
-	\end{table}
-
-	The encryption of ``attack by air'' using the same key is thus \texttt{4F 40 20 0D 18 E9 13 33 3E F8 62}.
-
-\item 	\begin{enumerate}
-	\item 	No, node 25 is a child of node 5, so encrypting with this key will give player 25 access.
-	\item 	Yes, this will give players 27 till 30 access. However, 6's parent 2 would give access to player 25 as well.
-	\item 	It would be better to use node 1 which gives access to more players whilst not exposing the movie to player 25.
-	\item 	Yes, per the reason stated before.
-	\item 	Yes, we can't use 26's parent 12 as this would expose the movie to player 25.
-	\item 	Yes, we can't use 11's parent 5 per the reason stated at (a), but using this key won't give player 25 access.
-	\item 	With node 6 we already cover all players with this key, so this is unnecessary.
-	\item 	This key has already been covered with the key in node 1.
-	\end{enumerate}
-
-	Summarising, we need keys $k_1, k_{11}, k_{26}$ and $k_6$.
-
-\item 	We notice that the first eleven letters could correspond to ``Barack Obama'' in plaintext. This gives us:
-
-		\begin{verbatim}BARAC KOBAM A____ ___R_ _A_R_ CA_AM _R_CA __R__ _____\end{verbatim}
-
-		Then we notice we could fit the word ``America(n)'' in, giving:
-
-		\begin{verbatim}BARAC KOBAM AI___ E_IR_ _A_RI CA_AM ERICA __RE_ I_E__\end{verbatim}
-
-		We then realise the plaintext could be ``Barack Obama is the first African American president''. 
-
-\item 	\begin{enumerate}
-	\item 	I wrote a Haskell program which can be found in appendix \ref{App:5a}. The used libraries are used only for basic operations (sorting and reversing lists, replacing substrings in a string) that aren't really part of the main functionality and could be easily written without the libraries. However, this is easier.
-
-		Lines 25 through 37 provide the \texttt{Freq} type which holds a string and an integer, and is used to store the number of occurrences of a substring in a string.
-
-		Lines 13 through 17 provide a method to count $n$-grams occurrences in a string, it returns a list of \texttt{Freq}s. 
-
-		Lines 19 through 23 provide a helper method for \texttt{countfreqs} which takes a list of String occurrences and their frequencies, and a String, and increments the frequency of that String in the list by one.
-
-		The \texttt{main} method reads the first command argument, removes spaces and counts and sorts letter, bigram and trigram occurrences in the string.
-
-	\item 	From the output of the program we find that the trigram \texttt{BPM} occurs the most (7 times) and that the letters \texttt{M}, \texttt{B} and \texttt{P} occur the most as well (respectively 34, 27 and 19 times). From this we guess that \texttt{BPM} could correspond to \texttt{THE} in the plaintext. This would mean a shift of 8.
-
-		We then use the program from appendix \ref{App:5b}, which lets you shift letters in a string. We let it shift $26-8=18$ times, giving:
-
-		\sloppy\texttt{AGOODGLASSINTHEBISHOPSHOSTELINTHEDEVILSSEATTWENTYONEDEGREESANDTHIRTEENMINUTE
-		SNORTHEASTANDBYNORTHMAINBRANCHSEVENTHLIMBEASTSIDESHOOTFROMTHELEFTEYEOFTHEDEA
-		THSHEADABEELINEFROMTHETREETHROUGHTHESHOTFIFTYFEETOUT}
-
-		Or:
-
-		\begin{quote}\itshape
-		A good glass in the bishop's hostel in the devil's seat\\
-		twenty-one degrees and thirteen minutes northeast and by north\\
-		main branch seventh limb east side shoot from the left eye of the death's-head\\
-		a bee line from the tree through the shot fifty feet out.\\
-		\par\raggedleft--- \textup{Edgar Allan Poe}, The Gold-Bug
-		\end{quote}
-	\end{enumerate}
-\end{enumerate}
-
-\appendix
-
-\newpage
-\section{Sourcecode for assignment 5a} \label{App:5a}
-\inputminted[breaklines=true,linenos=true,tabsize=4]{haskell}{CamilStaps-assignment1-freqs.hs}
-
-\newpage
-\section{Sourcecode for assignment 5b} \label{App:5b}
-\inputminted[breaklines=true,linenos=true,tabsize=4]{haskell}{CamilStaps-assignment1-shift.hs}
-
-\end{document}