# christhomson/lecture-notes

MATH 239: added April 3, 2013 lecture.

 @@ -14,6 +14,7 @@ \usepackage{ae,aecompl} \usepackage{tikz} \usepackage[lined]{algorithm2e} +\usepackage{enumerate} \newtheorem*{theorem}{Theorem} \newtheorem*{proposition}{Proposition} @@ -4739,6 +4740,66 @@ \end{enumerate} \end{defn} + \begin{defn} \lecture{April 3, 2013} + Let $X$ denote the special vertices in partition $A$, and let $Y$ denote the special vertices in partition $B$. + \end{defn} + + Let's now prove K\"onig's Theorem. + + \begin{proof} + Let $M$ be a maximum matching, and let $C = (A \backslash X) \cup Y$. + \\ \\ + We aim to show: + \begin{enumerate} + \item $C$ is a cover. + \item $|C| = |M|$. + \end{enumerate} + + Let's start with (1). Suppose $C$ is not a cover. Then there exists an edge $uv$ with $u \in X$ and $v \in B \backslash Y$. + \\ \\ + \underline{Case 1}: $u$ is unsaturated. The edge $uv$ implies that $v$ must be special, which implies that $v \in Y$, which is a contradiction. + \\ \\ + \underline{Case 2}: $u$ is the endpoint of $P(u)$. Then $P(u)$ together with $uv$ is an alternating path. This implies that $v$ is special, which means $v \in Y$, which is a contradiction. + \\ \\ + We've proven (1). Let's now prove (2). + \\ \\ + For (2), we want to show that $|C| = |M|$. Intuitively, we want to show that $|M| = |Y| + |A \backslash X|$, since $|Y| + |A \backslash X| = |C|$. In order to prove this, we really need to prove these three things: + \begin{enumerate}[(i)] + \item Edges of $M$ are between $X$ and $Y$ and between $A \backslash X$ and $B \backslash Y$. + \item Vertices in $A \backslash X$ are saturated. + \item Vertices in $Y$ are saturated. + \end{enumerate} + + Let's start with (i). Suppose (i) does not hold. Then there exists an edge $uv \in M$ where $u \in Y$ and $v \in A \backslash X$. Since $v$ is special, there exists an alternating path $P(u)$. The alternating path obtained by adding $uv$ to $P(u)$ implies that $v$ is special, which implies that $v \in X$, which is a contradiction. + \\ \\ + The proof for (ii) is very simple. (ii) holds because unsaturated vertices of $A$ are in $X$. + \\ \\ + Finally, let's prove (iii). Suppose there exists $u \in Y$ where $u$ is not saturated. Then $P(u)$ is an alternating path. This is a contradiction since $M$ is a \emph{maximum} matching. + \end{proof} + + We'll use a similar approach to that of this proof for the algorithm we need. + + \subsubsection{Algorithm For Finding a Maximum Matching and Minimal Vertex Cover} + \underline{Input}: a bipartite graph $G = (V, E)$ with partition $A, B$. + \\ \\ + \underline{Output}: a matching $M$ and a cover $C$ where $|M| = |C|$. \\ + + \begin{algorithm}[H] + $M = \emptyset$ (or any other matching)\; + \While{(loop)}{ + Construct $X, Y$ (and $P(v)$)\; + \uIf{there exists $u \in Y$ where $u$ is unsaturated}{ + $M := M \Delta P(u)$\; + } + \Else{ + $C = (A \backslash X) \cup Y$ is a cover\; + $|C| = |M|$, STOP\; + } + } + \end{algorithm} + + In this algorithm, $S_1 \Delta S_2$ (for any sets $S_1$ and $S_2$) is defined as the symmetric difference of $S_1$ and $S_2$ (the same way as earlier). That is, $S_1 \Delta S_2 := (S_1 \cup S_2) \backslash (S_1 \cap S_2)$. + \newpage \section*{Clicker Questions} \begin{itemize}