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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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    \textbf{Lecture #1:} #2 \\
    \textbf{Date:} #3 \hfill     \textbf{Scribe:} #4
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%%% -+-+-+-+-+-+- BEGIN HERE -+-+-+-+-+-+- %%%
\newcommand{\lecturenumber}{13}
\newcommand{\lecturetitle}{The Bipartite Maximum Matching problem.}
\newcommand{\scribename}{Noura Allugmani}
\newcommand{\lecturedate}{March 4, 2026}

\begin{document}

\thispagestyle{firstpage}
\scribebox{\lecturenumber}{\lecturetitle}{\lecturedate}{\scribename}

\noindent Given: a bipartite graph $G=(L, R, E)$. \\
Find: a matching $M \subseteq E$ of maximum size.
\section{Some Terminology}
Given a matching $M$ and some $u \in V \Rightarrow u$ is "matched" if it belongs to some $e \in M$. Otherwise, $u$ is "free".
An augmenting path (with respect to $M$ ) is a path $P$ such that:
\begin{enumerate}
    \item the endpoints of $P$ are free w.r.t $M$
    \item each other $u \in P$ is matched w.r.t. $M$.
\end{enumerate}
%%

\begin{example}
\end{example}
\begin{center}
\begin{tikzpicture}[
  nodeStyle/.style={circle, draw=black, line width=1pt, minimum size=9mm, inner sep=0pt}, 
  solidEdge/.style={black, line width=1pt},
  spring/.style={red, line width=1.2pt, decorate, decoration={snake, amplitude=1.2mm, segment length=3mm}}
]
% helper: one row with 6 nodes, and choose which edges are red-spring
% edges are between (1-2), (2-3), (3-4), (4-5), (5-6)

% ---------------- Row 1 ----------------
\def\y{0}
\foreach \i in {1,...,6}{
  \node[nodeStyle] (r1-\i) at (2.0*\i, \y) {};
}
% edges: black except (2-3) and (4-5) are springs
\draw[solidEdge] (r1-1) -- (r1-2);
\draw[spring]    (r1-2) -- (r1-3);
\draw[solidEdge] (r1-3) -- (r1-4);
\draw[spring]    (r1-4) -- (r1-5);
\draw[solidEdge] (r1-5) -- (r1-6); 
\node[right] at (13,0) {An augmenting path};
% ---------------- Row 2 ----------------
\def\y{-2.0}
\foreach \i in {1,...,6}{
  \node[nodeStyle] (r2-\i) at (2.0*\i, \y) {};
}
% edges: springs at (1-2), (3-4), (5-6)
\draw[spring]    (r2-1) -- (r2-2);
\draw[solidEdge] (r2-2) -- (r2-3);
\draw[spring]    (r2-3) -- (r2-4);
\draw[solidEdge] (r2-4) -- (r2-5);
\draw[spring]    (r2-5) -- (r2-6);
\node[right] at (13,-2) {Not an aug path};
% ---------------- Row 3 ----------------
\def\y{-4.0}
\foreach \i in {1,...,6}{
  \node[nodeStyle] (r3-\i) at (2.0*\i, \y) {};
}
% edges: only (3-4) is spring
\draw[solidEdge] (r3-1) -- (r3-2);
\draw[solidEdge] (r3-2) -- (r3-3);
\draw[spring]    (r3-3) -- (r3-4);
\draw[solidEdge] (r3-4) -- (r3-5);
\draw[solidEdge] (r3-5) -- (r3-6);
\node[right] at (13,-4) {Not an aug path};
\end{tikzpicture} 
\end{center}


%%
Given tow sets $A, B$, let their symmetric difference be
$$
A \Delta B=(A \mid B) \cup(B \backslash A) .
$$
%%



\begin{center}
\begin{tikzpicture}[line width=1pt]

% --- Universe box ---
\draw[blue!70!black, line width=1.2pt] (-4,-2.4) rectangle (4,2.4);

% --- Circle centers and radius ---
\def\r{1.6}
\coordinate (A) at (-1.0,0);
\coordinate (B) at ( 1.0,0);

% --- Shade A and B (then remove intersection to get A \Delta B) ---
\begin{scope}
  \fill[pattern=north east lines] (A) circle (\r);
  \fill[pattern=north east lines] (B) circle (\r);
\end{scope}

% Remove the overlap (make it white)
\begin{scope}
  \clip (A) circle (\r);
  \clip (B) circle (\r);
  \fill[white] (-5,-5) rectangle (5,5);
\end{scope}

% --- Draw circle boundaries on top ---
\draw (A) circle (\r);
\draw (B) circle (\r);

% --- Labels inside circles ---
\node at ($(A)+(0,1.9)$) {\Large $A$};
\node at ($(B)+(0,1.9)$) {\Large $B$};

% --- A \Delta B label + arrows to both shaded parts ---
\node[anchor=west] at (2.2,2.0) {\Large $A\Delta B$};
\draw[-stealth, line width=1pt] (3.0,1.85) -- (-1.9,0.9);
\draw[-stealth, line width=1pt] (3.0,1.85) -- ( 1.9,0.9);

\end{tikzpicture}
\end{center}




%%
\begin{lemma} If $M$ is a matchiny and $P$ is an augmenting path w.r.t. $M$, the $M^{\prime}=P \Delta M$ is a matching with $\left|M^{\prime}\right|=|M|+1$. 
\end{lemma}
\begin{proof}
  Note that $P$ has an odd \# of edges, and its edges alternate between $e$ s.t. $e\notin M$ and $e$ s.t. $e \in M$. Its first and last edges, $e_0, e_k$ are s.t. $e_0, e_k \notin M$. \\
$\Rightarrow M^{\prime}=M \Delta P$ is s.t. $|M^{\prime}|=|M|+1.$ \\
$\Rightarrow M^{\prime}$ is a matching since:
\begin{itemize}
\item  any $a \in V\setminus P$ has the same match in M as in $M^{\prime}$.
\item  each $u \in P$ has exactly one neighbor in $M^\prime$.  
\end{itemize}
\end{proof}
%%
\begin{center}
\begin{tikzpicture}[
  vertex/.style={circle,draw,minimum size=7mm,inner sep=0pt},
  edge/.style={line width=1pt},
  match/.style={
    blue,
    line width=1.3pt,
    decorate,
    decoration={snake,amplitude=1mm,segment length=3mm}
  },
  augment/.style={red,line width=1.8pt}
]

% ================= LEFT GRAPH M =================
\begin{scope}

\node[vertex] (v1) at (0,2) {};
\node[vertex] (v2) at (1.2,3) {};
\node[vertex] (v3) at (2.4,2.3) {};
\node[vertex] (v4) at (3.8,2.5) {};
\node[vertex] (v5) at (3.8,0.8) {};
\node[vertex] (v6) at (2.4,1.2) {};
\node[vertex] (v7) at (1.0,1.1) {};
\node[vertex] (v8) at (2.0,-0.3) {};

\draw[match] (v1)--(v2);
\draw[edge]  (v1)--(v7);
\draw[edge]  (v2)--(v3);
\draw[match] (v3)--(v6);
\draw[edge]  (v3)--(v4);
\draw[edge]  (v4)--(v5);
\draw[edge]  (v5)--(v8);
\draw[match] (v7)--(v8);
\draw[edge]  (v7)--(v6);

\draw[augment]
(1.78,2.62)
.. controls (2.00,2.92) and (2.95,2.95) .. (4.18,2.98)
.. controls (4.55,2.88) and (4.55,2.28) .. (4.18,2.08)
.. controls (3.75,1.98) and (3.15,1.95) .. (2.78,1.92)
.. controls (2.40,1.88) and (2.18,1.72) .. (2.16,1.42)
.. controls (2.14,1.12) and (2.28,0.95) .. (2.58,0.93)
.. controls (2.95,0.92) and (3.25,1.00) .. (3.62,1.08)
.. controls (4.02,1.22) and (4.38,1.08) .. (4.40,0.76)
.. controls (4.30,0.36) and (3.86,-0.02) .. (3.35,-0.42)
.. controls (2.95,-0.72) and (2.38,-0.84) .. (2.05,-0.66)
.. controls (1.62,-0.46) and (1.30,-0.04) .. (0.98,0.22)
.. controls (0.60,0.52) and (0.22,0.95) .. (0.36,1.26)
.. controls (0.52,1.58) and (0.98,1.62) .. (1.38,1.58)
.. controls (1.74,1.55) and (2.00,1.58) .. (2.12,1.70)
.. controls (2.18,1.86) and (2.02,2.16) .. (1.78,2.62)
-- cycle;

\node at (2,-1.15) {$M$};

\end{scope}

% ================= RIGHT GRAPH M' =================
\begin{scope}[xshift=7cm]

\node[vertex] (u1) at (0,2) {};
\node[vertex] (u2) at (1.2,3) {};
\node[vertex] (u3) at (2.4,2.3) {};
\node[vertex] (u4) at (3.8,2.5) {};
\node[vertex] (u5) at (3.8,0.8) {};
\node[vertex] (u6) at (2.4,1.2) {};
\node[vertex] (u7) at (1.0,1.1) {};
\node[vertex] (u8) at (2.0,-0.3) {};

\draw[edge]  (u1)--(u2);
\draw[edge]  (u1)--(u7);
\draw[match] (u2)--(u3);
\draw[edge]  (u3)--(u6);
\draw[match] (u3)--(u4);
\draw[edge]  (u4)--(u5);
\draw[match] (u5)--(u8);
\draw[edge]  (u7)--(u8);
\draw[match] (u7)--(u6);

\draw[augment]
(1.78,2.62)
.. controls (2.00,2.92) and (2.95,2.95) .. (4.18,2.98)
.. controls (4.55,2.88) and (4.55,2.28) .. (4.18,2.08)
.. controls (3.75,1.98) and (3.15,1.95) .. (2.78,1.92)
.. controls (2.40,1.88) and (2.18,1.72) .. (2.16,1.42)
.. controls (2.14,1.12) and (2.28,0.95) .. (2.58,0.93)
.. controls (2.95,0.92) and (3.25,1.00) .. (3.62,1.08)
.. controls (4.02,1.22) and (4.38,1.08) .. (4.40,0.76)
.. controls (4.30,0.36) and (3.86,-0.02) .. (3.35,-0.42)
.. controls (2.95,-0.72) and (2.38,-0.84) .. (2.05,-0.66)
.. controls (1.62,-0.46) and (1.30,-0.04) .. (0.98,0.22)
.. controls (0.60,0.52) and (0.22,0.95) .. (0.36,1.26)
.. controls (0.52,1.58) and (0.98,1.62) .. (1.38,1.58)
.. controls (1.74,1.55) and (2.00,1.58) .. (2.12,1.70)
.. controls (2.18,1.86) and (2.02,2.16) .. (1.78,2.62)
-- cycle;

\node at (2,-1.15) {$M'$};

\end{scope}
\end{tikzpicture}
\end{center}
\begin{lemma}
 A matching $M$ is a maximum matching if and only if it contains no augmenting path.   
\end{lemma}
\begin{proof}
 $ \Leftarrow$ we proved the contrapositive. If $M$ has an augmenting path, then by Lemma1.2 $M$ can not be a maximum matching.\\
$\Rightarrow$ Suppose $M^{*}$ is a maximum matching but $M$ is not, $\Rightarrow \quad|M|<\left|M^*\right|.$
WTS that $M$ has an augmenting path (we will use $M$ and $M^{*}$to build an augments path w.r.t. $M$).
Consider $M \Delta M^{*}$, and the induced subgraph
$G^{\prime}=\left(V, M\Delta M^{*}\right).$ \\
\textbf{Claim:} Each $u \in V$ had $\operatorname{deg}_{G^{\prime}}(u) \leqslant 2$.
\begin{itemize}
\item  if $u$ is free in both $M$ and $M^*$, deg $(u)=0$.
\item  if $u$ is matched in $M$ but free in $M^*$, $\operatorname{deg}(u)=1$.
\begin{itemize}
\item $
 M \Delta M^\prime=\left(M \setminus M^\prime\right) \cup\left(M^* \backslash M\right)
$
\end{itemize}
\item  if $u$ is free in $M$ but matched in $M^{*}$, similarly $\operatorname{deg}(u)=1$.
\item  if $u$ is matched in both $M$ and $M^*$.
\begin{itemize}
\item  $u$ is matched to the same node in both $M$ and $M^*$ $\Rightarrow \operatorname{deg}(u)=0$
\end{itemize}
\item  $u$ is matched to different nodes in $M$ and $
M^* \Rightarrow \operatorname{deg}(u)=2$.
\end{itemize}
%%
Given the claim we just showed, $G^\prime$ can be decomposed into isolated nodes, cycles, and paths (similar to a proof we have seen a few lectures ago). Since $|M|<\left|M^*\right|$, there must exist at least on component of $G^{\prime}$ with more edges from $M^*$ than from $M$.
%%
\begin{itemize}
\item  Can not be the isolated nodes.
\item  Can not be the cycles (these have edges that alternative $M$ and $M^*$)\\ %(counts from each of $M$ and $M^*$ must be same).
\end{itemize}
$\Rightarrow$ the component must be a path.
\begin{itemize}
\item  Can not have an even \# edges (because
same counting argument as for cycle).
\item  must have an odd \# of edges.
\begin{itemize}
\item  can not start / end with edges from $M$ (opposite of what we are looking for, these have more edges from $M$ than from $M^*$).
\item  path must start with edges from $M^*$
$\Rightarrow P$ must be an augmenting path w.r.t. $M$.
\end{itemize}
\end{itemize}
\end{proof}
%%
\section{Algorithm}  

\begin{enumerate}
\item $M \leftarrow \varnothing$
\item while $M$ has an augmenting path do
\begin{enumerate}
\renewcommand{\labelenumii}{2\alph{enumii}}
\item Find an augmenting path $P$ w.r.t.\ $M$
\item $M \leftarrow M \Delta P$
\end{enumerate}
\end{enumerate}

return $M$


%%

\section{Feasibility}
By lemma1.2, ALG maintains the invariant that $M$ is a matching.

%%
\section{Termination}
On each it iteration, $|M|$ increase by $1$ (by lemma1.2).
The size of a matching can not exceed $n / 2$ where $n=|V|$, so ALG must terminate after $\leq \frac{n}{2}$ iterations.

\section{Optimality}
At some point, it exists the while loop
$\Rightarrow$ find $M$ has augmenting path. By lemma1.3, $M$ must be a maximum matching. \\

\noindent \textbf{Note:} ALG is under specified since we have not described how to do step $(*)$ 2a.\\

\noindent If $G$ is bipartite, $(*)$ is fairly easy. Given $G=(L, R, E)$ and an incumbent matching $M$, turn $G$ into a directed graph $D$ by orienting each edge $e=\{l, r\} \in E$ as:\\
$\rightarrow (l,r)$ if $\{l, r\} \notin M$.\\
$\rightarrow(r, l)$ if $\{l, r\} \in M$.
%%






\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[
v/.style={circle,draw,minimum size=8mm,inner sep=0pt},
e/.style={black,line width=1pt},
match/.style={blue,line width=1.4pt,decorate,
  decoration={snake,amplitude=1mm,segment length=3mm}},
dedge/.style={->,black,line width=1pt,>=Stealth},
dmatch/.style={->,red,line width=1.4pt,decorate,
  decoration={snake,amplitude=1mm,segment length=3mm},>=Stealth},
lbl/.style={font=\large}
]

%==================== LEFT: G ====================%
\begin{scope}

% vertices
\node[v] (L1) at (0,3) {};
\node[v] (R1) at (2,3) {};

\node[v] (L2) at (0,2) {};
\node[v] (R2) at (2,2) {};

\node[v] (L3) at (0,1) {};
\node[v] (R3) at (2,1) {};

\node[v] (L4) at (0,0) {};
\node[v] (R4) at (2,0) {};

% corrected edges
\draw[e]     (L1) -- (R1);
\draw[match] (L2) -- (R1);

\draw[e]     (L1) -- (R2);
\draw[e]     (L2) -- (R2);

\draw[e]     (L2) -- (R3);
\draw[e]     (L3) -- (R3);

\draw[e]     (L2) -- (R4);
\draw[match] (L4) -- (R3);

\node[lbl,blue!70!black] at (1,-0.75) {$G$};

\end{scope}

%==================== RIGHT: D ====================%
\begin{scope}[xshift=8cm]

% vertices
\node[v] (A1) at (0,3.2) {};
\node[v] (B1) at (2,3.2) {};

\node[v] (A2) at (0,2.1) {};
\node[v] (B2) at (2,2.0) {};

\node[v] (A3) at (0,1.0) {};
\node[v] (B3) at (2,1.0) {};

\node[v] (A4) at (0,0.0) {};
\node[v] (B4) at (2,0.0) {};

% directed edges as in sketch
\draw[dmatch] (A1) -- (B1);
\draw[dmatch] (A2) -- (B2);
\draw[dmatch] (B1) -- (A2);

\draw[dedge]  (A1) -- (B2);
\draw[dedge]  (A2) -- (B3);
\draw[dedge]  (A3) -- (B3);
\draw[dedge]  (B3) -- (A4);
\draw[dedge]  (A2) -- (B4);

% red circles + labels
\draw[red,line width=1.2pt] (A1) circle (0.48);
\node[red,font=\large] at (0.55,3.9) {$\in F\cap R$};

\draw[red,line width=1.2pt] (B2) circle (0.48);
\node[red,font=\large] at (3.0,2.2) {$\in F\cap R$};

\node[lbl,blue!70!black] at (1,-0.8) {$D$};

\end{scope}
\end{tikzpicture}
\end{center}
\end{figure}
\noindent If $F$ is the set of free nodes w.r.t. $M$, follow any directed path starting in $F \cap L$ and ending in $F \cap R$.\\
\noindent In step $(*)$, if $G$ is bipartite, one can use standard graph traversal algorithm in $D$ to find augmenting paths.

\section{Running time (w.r.t. size of $G$, i.e. $n=|V|$ and $m=|E|$)}
Given $f, g: \mathbb{N} \longrightarrow \mathbb{N}$, we say $f=O(g)$ if $\exists ~k \in \mathbb{N}$ and $C \in \mathbb{R}_{>0}$ s.t. $f(n) \leq C g(n), \quad \forall n \geqslant k$.\\

\noindent Building D takes $O(n+m)$. Traversing  $D$ to find augmenting path takes $O(n+m)$. We may assume $m \geq\frac{n}{2}$. Otherwise $\exists$ isolated nodes that we can remove
$
\Rightarrow O(n+m)=O(m) .
$\\

\noindent We saw ALG terminates after $\leq \frac{n}{2}=O(n)$ iterations. Each iteration takes $O(m)$ time. Overall, ALG terminates in $O(m n)$ time. \\

\noindent Can ALG be generalized to run faster or work in non-bipartite graphs?
\begin{itemize}
\item If you augment various augmenting paths at the same time, there is an algorithm that runs in time $O(m\sqrt{n})$ (Hopcroft–Karp algorithm ).
\item For non-bipartite graphs, $(*)$ is still possible but much more subtle. Edmonds' (1965) " Paths,
Trees, and Flowers" that works in $O(nm^2)$ time overall. (Final project idea.)

\end{itemize}


\end{document}