%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lachaise Assignment % LaTeX Template % Version 1.0 (26/6/2018) % % This template originates from: % http://www.LaTeXTemplates.com % % Authors: % Marion Lachaise & François Févotte % Vel (vel@LaTeXTemplates.com) % % License: % CC BY-NC-SA 3.0 (http://creativecommons.org/licenses/by-nc-sa/3.0/) % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{article} \input{structure.tex} \title{CENG567: Homework \#2} \author{Yiğit Sever} \date{\today} %---------------------------------------------------------------------------------------- \begin{document} \maketitle \section{Checking Consistency of Judgements}% \label{sec:checking_consistency_of_judgements} Given the collection of $n$ butterflies and a potential judgement between every pair (or not if the judgement is ambiguous), we have a graph $G = (V, E)$ with $n = |V|$ vertices and $|E| = m \le \frac{n(n-1)}{2}$ edges, with every edge $(i, j) \in E$ labelled either \enquote{same} or \enquote{different}. At the end of our algorithm, the vertices should be \emph{consistently} labelled as either \texttt{A} and \texttt{B} or our algorithm should be able to prove that $G$ cannot be labelled as so. A modified graph traversal using either BFS or DFS (since a node can be discovered multiple times in both of them) will work. Here we will modify the graph traversal given on \emph{page 42} on our \nth{3} lecture slides that uses BFS. The input of the algorithm is a node $s \in E$. If, due to ambiguous (i.e.\ missing) nodes, the graph is not connected, the algorithm should be run until every connected component is discovered. % TODO: run until every cc is discovered is handwavey <15-11-20, yigit> % {\centering \begin{minipage}{.7\linewidth} \begin{algorithm}[H] \SetKwProg{Fn}{function}{}{end} \DontPrintSemicolon{} \SetAlgoLongEnd{} \Fn{consistency\_check(s: node)}{ \KwData{$K$ = data structure of discovered nodes} \KwResult{boolean = whether $G$ is consistent or not} \textbf{label} $s$ as \texttt{A} \tcc*[r]{the opposite label is B} put $s$ in $K$\; \While{$K$ is not empty}{ take a node $v$ from $K$\; \If{$v$ is not marked \enquote{explored}}{ mark $v$ \enquote{explored}\; \For{each edge $(v, w)$ incident to $v$}{ \uIf{$w$ is labelled}{ \If{the label of $w$ is not consistent with the label of $v$ with respect to the judgement $(v,w$)}{ terminate the algorithm; $G$ is \textbf{inconsistent}\; } } \Else{ \uIf{$(v,w)$ is labelled \enquote{same}}{ \textbf{label} $w$ with the label of $v$\; } \Else(\tcc*[f]{$(v,w)$ is labelled \enquote{different}}){ \textbf{label} $w$ with the opposite label of $v$\; } } put $w$ in $K$\; } } } the spanned connected component is \textbf{consistent} } \caption{Modified graph traversal algorithm so solve judgement consistency checking problem}% \label{alg:question_1} \end{algorithm} \end{minipage} \par } With the assumption that accessing the labels $(u, w)$ takes $\mathcal{O}(1)$ time this algorithm has the same running time as BFS; $\mathcal{O}(m+n)$. \section{Reachability}% \label{sec:reachability} \end{document}