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     1.4 +\documentclass[a4paper, 10pt, pagesize, smallheadings]{article}
     1.5 +\usepackage{graphicx}
     1.6 +%\usepackage[latin1]{inputenc}
     1.7 +\usepackage{amsmath, amsthm, amssymb}
     1.8 +\usepackage{typearea}
     1.9 +\usepackage{algorithm}
    1.10 +\usepackage{algorithmic}
    1.11 +\usepackage{fullpage}
    1.12 +\usepackage{mathtools}
    1.13 +\usepackage[all]{xy}
    1.14 +\addtolength{\voffset}{-40pt}
    1.15 +\title{CSP Exercise 02 Solution}
    1.16 +\author{Eugen Sawin}
    1.17 +\renewcommand{\familydefault}{\sfdefault}
    1.18 +\newcommand{\E}{\mathcal{E}}
    1.19 +\newcommand{\R}{\mathcal{R}}
    1.20 +
    1.21 +%\include{pythonlisting}
    1.22 +
    1.23 +\pagestyle{empty}
    1.24 +\begin{document}
    1.25 +\maketitle
    1.26 +
    1.27 +\section*{Exercise 2.1}
    1.28 +(a) The strongly connected components are the subgraphs induced by following sets of vertices.
    1.29 +\begin{itemize}
    1.30 +  \item $V_1 = \{v_1,v_2,v_5,v_6\}$
    1.31 +  \item $V_2 = \{v_3,v_4\}$
    1.32 +\end{itemize}
    1.33 +(b) The cliques are the the complete subgraphs induced by the following sets of vertices.
    1.34 +\begin{itemize}
    1.35 +  \item $V_1 = \{v_1,v_2,v_5\}$
    1.36 +  \item $V_{2-8} \in E$
    1.37 +  \item $V_{9-14} \in \{\{v_i\} \mid v_i \in V\}$
    1.38 +\end{itemize}
    1.39 +
    1.40 +\section*{Exercise 2.2}
    1.41 +We define the following undirected simple graph $G = \langle W, E \rangle$ with $\{w_i,w_j\} \in E$ iff $w_i$ and $w_j$ dislike each other. Additionally we define two sets $B_1 = B_2 = \{\}$ for the buildings and an auxiliary set $F = \{\}$ for remembering \emph{expanded} nodes.
    1.42 +\begin{enumerate}
    1.43 +  \item For all $w_i \in W$ with $w_i \notin \bigcup E$ add $w_i$ to $B_1$ and $F$, i.e. $B_1 = B_1 \cup \{w_i\}$ and $F = F \cup \{w_i\}$. This way we assign all workers, who are liked by everyone to building $B_1$.
    1.44 +  \item If $F = W$ terminate, we have found a solution.\\
    1.45 +  Otherwise select any $w_i \notin F$ and add it to $B_1$ and $F$, i.e. $B_1 = B_1 \cup \{w_i\}$ and $F = F \cup \{w_i\}$.\\
    1.46 +  Add all its disliked neighbours to $B_2$, i.e. $B_2 = B_2 \cup \{w_j \mid \{w_i,w_j\} \in E\}$.\\
    1.47 +  If $B_1 \cap B_2 \neq \emptyset$ terminate, there was a conflict and therefore no possible solution.
    1.48 +  \item If there a $w_i \notin F$ with $w_i \in B_1 \cup B_2$ select it, otherwise continue with $2.$\\
    1.49 +  Add all its disliked neighbours to the other building, i.e. $w_i \in B_1 \implies B_2 = B_2 \cup \{w_j \mid (w_i,w_j) \in E\}$, $w_i \in B_2 \implies B_1 = B_1 \cup \{w_j \mid (w_i,w_j) \in E\}$ and $F = F \cup \{w_i\}$.\\
    1.50 +  If $B_1 \cap B_2 \neq \emptyset$ terminate, there was a conflict and therefore no possible solution.\\
    1.51 +  Otherwise continue with $3.$
    1.52 +\end{enumerate}
    1.53 +This is essentially a breadth-first search with random-walk used to jump to disconnected nodes. Assumming that set containment can be tested in $\mathcal{O}(1)$ and set intersection in $\mathcal{O}(n)$ we have an asymptotic complexity of $\mathcal{O}(|W| \cdot |E|)$.
    1.54 +\end{document}