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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
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    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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\newcommand{\lecturenumber}{14}
\newcommand{\lecturetitle}{Matchings Continued}
\newcommand{\scribename}{Hunter Pearson}
\newcommand{\lecturedate}{March 9, 2026}

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\section*{Brief Detour Into Carlos' View of the world}

Acronym: A.C.O: Algorithms, Combinatorial, and Optimization. (The Barriers between these are blurred.
    Algorithms: Combinatorial algorithms like the augmenting path algorithm we saw last time. Common from a computer science lens.
    Combinatorics: graph theoretic properties, like Halls theorem, common from math lens.
    Optimization: "polyhedral approach" uses linear algebra and "math programming" for modeling and optimization. Common from operations research lens.
        Polyhedral approach: Idea is to associate to a set $\mathcal{C}$ of combinatorial objects you care about to points in space and use geometry of these points to find the "best".

        For example, consider matchings of a graph $G=(V,E)$. For each edge $e \in E$, associate to it a binary decision variable $x_e \in \{0,1\}$. Then finding a maximum in $G$ is the same as the following linear program: \begin{align*}
            \text{Maximize: }& \sum_{e \in E}x_e\\
            \text{Such That: }& \sum_{e \in \delta(\mu)} x_e \leq 1 \forall \mu \in V\\
            & x_e \in \{0,1\} \forall e \in E
        \end{align*}

        So a matching $M$ is a point $X_M \in \{0,1\}^E$. $X_{M_e} = 1 \iff e \in M$.

        \begin{itemize}
            \item Note: There are specialized computer software to solve problems like IP
            \item In this sense, it give you lots of modeling flexibility
            \item In general, problems with 0,1 variables are "Hard"
            \item If $G$ happens to be bipartite, you can replace $x_e \in \{0,1\}$ for $0 \leq x_e \leq 1$ to obtain a linear programming relaxation that happens to produce integer solutions. It is really special when this happens because LP's are "easy".
            \item If $G$ is not bipartite, the story is more complicate but still an exact LP formulation is possible. 
            \item Combinatorial optimization/mathematical programming study how to leverage LP's/IP's to get "good" solutions
        \end{itemize}

        \section*{What if Halls Condition Doesn't Hold?}

        Recall Halls Condition: \begin{equation*}
            |N(S)| \geq |S|, \forall S \subseteq X \tag{$\star$}
        \end{equation*}

        Halls Theorem tells us that if $(\star)$ holds for all $S \subseteq X$, then there exists a matching of size $|X|$.

        Question: What if the condition doesn't hold for some $S \subseteq X$? How large of a matching exists?

        For $\emptyset \not = S \subseteq X$, let the defect of $S$ be $$\df(S) \coloneqq |S| - |N(S)|$$.

        Halls condition becomes $\df(S) \leq 0 \forall S \subseteq X$.

        To answer our question, it is natural to think about the worst case, i.e. the largest defect.

        Define $$d^* = \max_{\emptyset \not = S \subseteq X} \df(S)$$

        Also, let $\alpha (G) $ be the size of the largest matching in $G = (X \sqcup Y,E)$.

        \begin{corollary}
            (Ore's Defect Formula) Let $G=(X \sqcup Y,E)$ be bipartite. Then $\alpha (G) = |X|-d^*$
        \end{corollary}

        \begin{proof}
            Every matching misses to match $\df(S)$ nodes from $S \subseteq X$. This implies $$|\alpha(G)| \leq |X|-\df(S) \forall S \subseteq X$$ Then 
            \begin{align*}
                \alpha(G) &\leq \min_{\emptyset \not = S \subseteq X}\{|X|-\df(S)\}\\
                &= |X|-\max_S \df(S)\\
                &= |x|-d^*
            \end{align*}

            Thus we have $$\alpha(G) \leq |x|-d^*$$

            Now we want to show that $\alpha(G) \geq |x|-d^*$. It suffices to produce a matching of $G$ of size $|x|-d^*$.

            Idea: Produce a new graph $G'$ by adding $d^*$ dummy nodes to $Y$, each of which is  connected to all of $x$.

            Claim: $G'$ meets Halls Condition $\forall S \subseteq X$.

            This holds since for each $S \subseteq X$, $N(S)$ increased by $d^*$ and $\df(s) \leq d^*$.

            The new defect of $S$ is less than or equal to 0.

            Then $G'$ has a matching $M$ of size $|X|$ (by Hall's Theorem)

            Now $M= M_1 \sqcup M_2$, where $M_2$ is the subset of edges of $M$ that connect to dummy nodes, and $M_1$ is the rest. Then $|M_2| \leq d^*$ since there are only $d^*$ dummy nodes.

            Finally, $M_1$ is a matching in $G$ (the original graph)

            Since it does not use dummy nodes, we know $$|M_1| = |M| - |M_2| = |X|-|M_2| \geq |X|-d^*$$
            
        \end{proof}

        \section*{The Minimum Vertex Cover Problem}

        A new problem: the minimum vertex cover problem.

        A vertex cover (i.e. a cover made up of vertices) of a graph $G=(V,E)$ is a subset of the nodes such that every edge  has at least one endpoint in $S$.

        Thus $e \cap S \not = \emptyset \forall e \in E$.

        \begin{figure}[h]
    \centering
\end{figure}
\begin{figure}[h]
    \centering
    \begin{minipage}{.333\textwidth}
    \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle, fill = skyblue] (m) at (0,.75) {};
            \node [draw, shape = circle] (x) at (0,0) {};
            \node [draw, shape = circle, fill = skyblue] (v) at (1,.75) {};
            \node [draw, shape = circle] (w) at (1,0) {};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{A Vertex Cover}
        \label{fig:Orig}
    \end{minipage}%
    \begin{minipage}{.333\textwidth}
        \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle, fill = skyblue] (m) at (0,.75) {};
            \node [draw, shape = circle, fill = skyblue] (x) at (0,0) {};
            \node [draw, shape = circle, fill = skyblue] (v) at (1,.75) {};
            \node [draw, shape = circle, fill = skyblue] (w) at (1,0) {};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{Another Vertex Cover}
        \label{fig:Proper}
    \end{minipage}%
    \begin{minipage}{.333\textwidth}
      \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle] (m) at (0,.75) {};
            \node [draw, shape = circle] (x) at (0,0) {};
            \node [draw, shape = circle, fill = skyblue] (v) at (1,.75) {};
            \node [draw, shape = circle] (w) at (1,0) {};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{Not A Vertex Cover}
        \label{fig:non Proper}
    \end{minipage}
\end{figure}

The goal is to find a vertex cover that is as small as possible. Let $\beta(G)$ be the minimum size of a vertex cover in $G$.

\begin{theorem}
    (K\"{o}nig-Egerv\'{a}ry) If $G$ is a bipartite graph, then $\alpha(G) = \beta(G)$
\end{theorem}

\begin{proof}
    $\alpha(G) \leq \beta(G)$: Let $M^*$  be a maximum matching. No node in a vertex cover can cover more than 1 edge in $M^*$ because the edges in $M^*$ are non adjacent. So you need at least $m^* = \alpha(G)$ nodes to cover $G$

    $\alpha(G) \geq \beta(G)$: It suffices to produce a vertex cover $S \subseteq X$ such that $|S| = \alpha(G)$. Since by definition of $\beta(G)$, we will have $\alpha(G) = |S| \geq \beta(G)$. By Ore's defect formula, we have that $$\alpha(G) = |x|-\df(T) = |X|-|T|+|N(T)|$$ for some $T \subseteq X$ with $\df(T) = d^*$. Note that $N(T) \sqcup (X-T)$ is a vertex cover.

    Finally, the size of $N(T) \sqcup (X-T)$ is $|N(T)| + |X|-|T| = \alpha(G)$.
\end{proof}

This theorem gives you a min-max relation. Thus it allows us a certificate of optimality.

For bipartite graphs, say $M^*$ is a matching and I claim it is a maximum matching. I can convince you of this by producing a vertex cover $S$ of size $|S|=|M^*|$, and vice versa. This kind of relation is in some ways akin to LP duality.




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