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MATH 239: added April 3, 2013 lecture.

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61 math239.tex
@@ -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}

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