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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}{17}
\newcommand{\lecturetitle}{The max flow--min cut theorem.}
\newcommand{\scribename}{Hanbyul (Han) Lee}
\newcommand{\lecturedate}{March 30, 2026}

\begin{document}

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

\section{Review}

Last time:
\begin{itemize}
    \item definition of flows and valid flows;
    \item an algorithm for finding a maximum flow using augmenting paths in the residual graph $G_f$.
\end{itemize}

Recall:
\begin{lemma}
If $f$ is a valid flow in $G$, then $f$ is a maximum flow if and only if $G_f$ contains no augmenting $s$--$t$ path.
\end{lemma}

Today we show a min--max relation between $s$--$t$ flows and edge cuts separating $s$ from $t$.

We have $G=(V,E,c)$ directed and $s,t\in V$.

\begin{definition}
An $s$--$t$ cut is a partition
\[
V=S\sqcup T
\]
such that $s\in S$ and $t\in T$. If $e=(u,v)$ is a directed edge, then
\[
e\in \delta(S,T)
\quad\Longleftrightarrow\quad
u\in S,\ v\in T.
\]
Since $T=V\setminus S$, we also write $\delta(S)$ for $\delta(S,T)$.
\end{definition}

\begin{remark}
Since $G$ is directed, in general $\delta(S,T)\neq \delta(T,S)$.
\end{remark}

\begin{example}
The two sets $\delta(S,T)$ and $\delta(T,S)$ need not be the same.

\begin{center}
\begin{minipage}{0.45\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\Vertex[x=0.1,y=1.05,size=.55,color=white,NoLabel]{u1}
\Vertex[x=0.55,y=0.00,size=.55,color=white,NoLabel]{u2}
\Vertex[x=2.05,y=1.15,size=.55,color=white,NoLabel]{v1}
\Vertex[x=2.55,y=0.10,size=.55,color=white,NoLabel]{v2}
\Edge[Direct,lw=1.6pt](u1)(v1)
\Edge[Direct,lw=1.6pt](u2)(v1)
\Edge[Direct](v2)(u2)
\draw[blue,line width=1.2pt] plot[smooth cycle,tension=.9] coordinates {(-0.55,1.55) (-0.75,-0.70) (0.95,-0.95) (1.35,0.10) (1.10,1.55)};
\node at (-0.72,1.30) {$S$};
\node at (1.60,1.80) {$\delta(S,T)$};
\end{tikzpicture}
\end{minipage}
\hfill
\begin{minipage}{0.45\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\Vertex[x=0.10,y=1.05,size=.55,color=white,NoLabel]{a1}
\Vertex[x=0.55,y=0.00,size=.55,color=white,NoLabel]{a2}
\Vertex[x=2.05,y=1.15,size=.55,color=white,NoLabel]{b1}
\Vertex[x=2.55,y=0.10,size=.55,color=white,NoLabel]{b2}
\Edge[Direct](a1)(b1)
\Edge[Direct](a2)(b1)
\Edge[Direct,lw=1.6pt](b2)(a2)
\draw[blue,line width=1.2pt] plot[smooth cycle,tension=.9] coordinates {(1.30,1.60) (2.85,1.45) (2.95,-0.70) (1.65,-0.80) (1.35,0.15)};
\node at (2.90,1.35) {$T$};
\node at (1.90,-1.10) {$\delta(T,S)$};
\end{tikzpicture}
\end{minipage}
\end{center}
\end{example}

Let
\[
c(S,T)=\sum_{e\in\delta(S,T)} c_e,
\]
i.e., the total capacity of edges from $S$ to $T$. Also, for a flow $f\in\mathbb{R}^E_{\ge 0}$, let
\[
f(S,T)=\sum_{e\in\delta(S,T)} f_e,
\]
i.e., the total flow from $S$ to $T$.

\section{The max flow--min cut theorem}

\begin{theorem}[Max flow--min cut]
For any flow network, the value of any maximum flow is equal to the capacity of any minimum cut.
\end{theorem}

\begin{proof}
Let $(S,T)$ be any $s$--$t$ cut. Let $1_T\in\mathbb{R}^V$ be such that
\[
(1_T)_u=
\begin{cases}
1, & u\in T,\\
0, & \text{otherwise}.
\end{cases}
\]
Recall the incidence matrix $B\in\mathbb{R}^{V\times E}$. Put
\[
x=(1_T)^T B.
\]
Then
\[
x_e=
\begin{cases}
1, & e\in \delta(S,T),\\
-1, & e\in \delta(T,S),\\
0, & \text{otherwise}.
\end{cases}
\]

We first show that the value of any $s$--$t$ flow is upper bounded by the value of any $s$--$t$ cut. In particular,
\[
\text{max flow value} \le \text{min cut capacity}.
\]

Let $f\in\mathbb{R}^E_{\ge 0}$ be a valid $s$--$t$ flow. Then
\begin{align*}
(1_T)^T Bf
&=x^T f \\
&=f(S,T)-f(T,S) \\
&\le f(S,T) \qquad \text{since } f\in\mathbb{R}^E_{\ge 0},\\
&\le c(S,T) \qquad \text{since } f \text{ is a valid flow.}
\end{align*}
Also,
\[
(1_T)^T Bf=(1_T)^T(1_t-1_s)\,|f|=|f|.
\]
Therefore,
\[
|f|\le c(S,T).
\]

Now note where the two inequalities are tight:
\begin{itemize}
    \item $f(S,T)-f(T,S)\le f(S,T)$ is tight if and only if $f(T,S)=0$, i.e. $f_e=0$ for every $e\in\delta(T,S)$;
    \item $f(S,T)\le c(S,T)$ is tight if and only if $f(S,T)=c(S,T)$, i.e. $f_e=c_e$ for every $e\in\delta(S,T)$.
\end{itemize}

We now show the converse. Suppose $f$ is a maximum flow. Then there is no $s$--$t$ path with positive residual capacity in $G_f$. Partition
\[
V=S\sqcup T
\]
where $S$ is the set of vertices reachable from $s$ in $G_f$ and $T=V\setminus S$. Then $s\in S$ and $t\in T$, so $(S,T)$ is an $s$--$t$ cut. Also, $G_f$ contains no edge from $S$ to $T$ with positive residual capacity.

Hence, for every $e=(u,v)\in\delta(S,T)$,
\[
c_f(e)=c_e-f_e=0,
\]
so $f_e=c_e$. This makes the second inequality tight.

Now take $e=(u,v)\in\delta(T,S)$. Then $u\in T$ and $v\in S$, so the reverse residual edge $(v,u)$ goes from $S$ to $T$. Since no edge from $S$ to $T$ has positive residual capacity in $G_f$, the residual capacity of $(v,u)$ is $0$. But that residual capacity is exactly $f_e$. Hence $f_e=0$, and the first inequality is tight as well.

Therefore, for this choice of maximum flow $f$ and cut $(S,T)$,
\[
|f|=c(S,T).
\]
Since every flow value is at most every cut capacity, this cut is a minimum cut.
\end{proof}

\section{Applications with $\infty$-capacity edges}

We now cover applications of the theorem. For this it is useful to consider flow networks where
\[
c:E\to \mathbb{R}_{\ge 0}\cup\{\infty\}.
\]

\begin{fact}
\leavevmode
\begin{itemize}
    \item If $G$ contains an $s$--$t$ path made entirely of $\infty$-capacity edges, then the maximum flow is unbounded.
    \item Otherwise, the maximum flow value is finite, and the max flow--min cut theorem still applies.
\end{itemize}
\end{fact}

So in these applications we may design gadgets with $\infty$-capacity edges as needed, as long as we do not introduce an $\infty$-capacity $s$--$t$ path.

\section{K\"onig--Egerv\'ary}

Let us revisit something old.

\begin{theorem}[K\"onig--Egerv\'ary]
If $G$ is a bipartite undirected graph, then the cardinality of a maximum matching equals the cardinality of a minimum vertex cover.
\end{theorem}

\begin{proof}
Our gadget is a flow network $G'$ with a super-source $s$ and a super-target $t$.

Given $G=(X\sqcup Y,E)$, we let
\begin{itemize}
    \item $s$ be connected to all of $X$;
    \item all edges in $E$ be oriented from $X$ to $Y$;
    \item all of $Y$ be connected to $t$.
\end{itemize}
Set the capacities by
\[
c(s,x)=1 \quad (x\in X),
\qquad
c(y,t)=1 \quad (y\in Y),
\qquad
c(x,y)=\infty \quad ((x,y)\in E).
\]

\begin{center}
\begin{minipage}{0.30\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\node at (0.15,1.85) {$X$};
\node at (1.95,1.85) {$Y$};
\Vertex[x=0.20,y=1.15,size=.58,color=white,NoLabel]{lx1}
\Vertex[x=0.20,y=0.00,size=.58,color=white,NoLabel]{lx2}
\Vertex[x=1.95,y=1.10,size=.58,color=white,NoLabel]{ly1}
\Vertex[x=2.15,y=-0.05,size=.58,color=white,NoLabel]{ly2}
\Edge(lx1)(ly1)
\Edge(lx2)(ly1)
\Edge(lx2)(ly2)
\end{tikzpicture}
\end{minipage}
\hfill
\begin{minipage}{0.64\textwidth}
\centering
\begin{tikzpicture}[scale=0.92]
\Vertex[x=0.00,y=0.55,size=.58,color=white,NoLabel]{ss}
\Vertex[x=1.35,y=1.15,size=.58,color=white,NoLabel]{xx1}
\Vertex[x=1.35,y=-0.05,size=.58,color=white,NoLabel]{xx2}
\Vertex[x=3.45,y=1.15,size=.58,color=white,NoLabel]{yy1}
\Vertex[x=3.45,y=-0.05,size=.58,color=white,NoLabel]{yy2}
\Vertex[x=4.90,y=0.55,size=.58,color=white,NoLabel]{tt}
\Edge[Direct](ss)(xx1)
\Edge[Direct](ss)(xx2)
\Edge[Direct](xx1)(yy1)
\Edge[Direct](xx2)(yy1)
\Edge[Direct](xx2)(yy2)
\Edge[Direct](yy1)(tt)
\Edge[Direct](yy2)(tt)
\node at (-0.25,0.55) {$s$};
\node at (5.18,0.55) {$t$};
\node at (0.62,1.07) {$1$};
\node at (0.62,0.03) {$1$};
\node at (2.40,1.35) {$\infty$};
\node at (2.35,0.58) {$\infty$};
\node at (2.42,-0.28) {$\infty$};
\node at (4.18,1.07) {$1$};
\node at (4.18,0.03) {$1$};
\end{tikzpicture}
\end{minipage}
\end{center}

If we want only finite capacities, we may replace each $\infty$ by any integer $M>|X|+|Y|$; the same cuts remain optimal. In that version all capacities are integral.
However, having $\infty$ capacities will make proofs simpler.

Any integer flow $f$ in $G'$ gives a matching $M_f$ in $G$ by
\[
\{x,y\}\in M_f
\quad\Longleftrightarrow\quad
f(x,y)=1.
\]
Similarly, any matching in $G$ gives an integer flow in $G'$ of the same value. So maximum matching size is the same as maximum integer flow value.

At this point we have
\begin{center}
\begin{tikzpicture}[scale=1.0]
\node (flows) at (0,1.15) {integer flows};
\node at (2.65,1.15) {are to};
\node (match) at (5.30,1.15) {matchings};
\node (cuts) at (0,-1.10) {$s$--$t$ cuts};
\node (cover) at (5.30,-1.10) {vertex covers};
\draw[<->,line width=.8pt,shorten <=2pt,shorten >=2pt] (flows.south) -- (cuts.north);
\draw[<->,line width=.8pt,shorten <=2pt,shorten >=2pt] (match.south) -- (cover.north);
\draw[dotted,line width=.8pt,shorten <=6pt,shorten >=6pt] (cuts.east) -- (cover.west);
\node at (2.65,-1.42) {missing (yet)};
\end{tikzpicture}
\end{center}

Now consider any $s$--$t$ cut $(S,T)=(S,V\setminus S)$ of finite capacity, and let
\[
A=(X\cap T)\cup(Y\cap S)
  =(X\cap(V\setminus S))\cup(Y\cap S).
\]

\begin{center}
\begin{minipage}{0.30\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\node at (0.15,1.85) {$X$};
\node at (1.95,1.85) {$Y$};
\Vertex[x=0.20,y=1.15,size=.58,color=white,NoLabel]{clx1}
\Vertex[x=0.20,y=0.00,size=.58,color=white,NoLabel]{clx2}
\Vertex[x=1.95,y=1.10,size=.58,color=white,NoLabel]{cly1}
\Vertex[x=2.15,y=-0.05,size=.58,color=white,NoLabel]{cly2}
\Edge(clx1)(cly1)
\Edge(clx2)(cly1)
\Edge(clx2)(cly2)
\end{tikzpicture}
\end{minipage}
\hfill
\begin{minipage}{0.64\textwidth}
\centering
\begin{tikzpicture}[scale=0.92]
\Vertex[x=0.00,y=0.55,size=.58,color=white,NoLabel]{cs}
\Vertex[x=1.35,y=1.15,size=.58,color=reddishpurple,NoLabel]{cx1}
\Vertex[x=1.35,y=-0.05,size=.58,color=white,NoLabel]{cx2}
\Vertex[x=3.45,y=1.15,size=.58,color=bluishgreen,NoLabel]{cy1}
\Vertex[x=3.45,y=-0.05,size=.58,color=bluishgreen,NoLabel]{cy2}
\Vertex[x=4.90,y=0.55,size=.58,color=white,NoLabel]{ct}
\Edge[Direct](cs)(cx1)
\Edge[Direct](cs)(cx2)
\Edge[Direct](cx1)(cy1)
\Edge[Direct](cx2)(cy1)
\Edge[Direct](cx2)(cy2)
\Edge[Direct](cy1)(ct)
\Edge[Direct](cy2)(ct)
\node at (-0.25,0.55) {$s$};
\node at (5.18,0.55) {$t$};
\node at (0.62,1.07) {$1$};
\node at (0.62,0.03) {$1$};
\node at (2.40,1.35) {$\infty$};
\node at (2.35,0.58) {$\infty$};
\node at (2.42,-0.28) {$\infty$};
\node at (4.18,1.07) {$1$};
\node at (4.18,0.03) {$1$};
\draw[blue,line width=1.4pt] plot[smooth cycle,tension=.86] coordinates {(-0.55,0.90) (-0.65,-0.45) (1.15,-0.55) (2.75,-0.70) (4.10,-0.70) (4.00,1.70) (2.95,1.82) (1.70,0.65)};
\node[color=blue] at (3.90,1.95) {$S$};
\end{tikzpicture}
\end{minipage}
\end{center}

We claim that $A$ is a vertex cover. Indeed, if $(x,y)\in E$ has no endpoint in $A$, then
\[
x\notin X\cap T \quad\text{and}\quad y\notin Y\cap S.
\]
Since $x\in X$ and $y\in Y$, this means
\[
x\in X\cap S, \qquad y\in Y\cap T.
\]
So the edge $(x,y)$ crosses from $S$ to $T$, and in our gadget it has capacity $\infty$. This contradicts the assumption that $(S,T)$ has finite capacity.

Now the cut capacity $c(S,T)$ is exactly the number of edges
\begin{itemize}
    \item from $s$ to $X\cap(V\setminus S)$,
    \item from $Y\cap S$ to $t$.
\end{itemize}
Hence
\[
c(S,T)=|A|.
\]

Conversely, any vertex cover $A$ gives an $s$--$t$ cut of capacity $|A|$ by the reverse construction
\[
S=\{s\}\cup(X\setminus A)\cup(Y\cap A).
\]

\begin{center}
\begin{minipage}{0.30\textwidth}
\centering
\begin{tikzpicture}[scale=0.95]
\node at (0.15,1.85) {$X$};
\node at (1.95,1.85) {$Y$};
\Vertex[x=0.20,y=1.15,size=.58,color=white,NoLabel]{rlx1}
\Vertex[x=0.20,y=0.00,size=.58,color=white,NoLabel]{rlx2}
\Vertex[x=1.95,y=1.10,size=.58,color=white,NoLabel]{rly1}
\Vertex[x=2.15,y=-0.05,size=.58,color=white,NoLabel]{rly2}
\Edge(rlx1)(rly1)
\Edge(rlx2)(rly1)
\Edge(rlx2)(rly2)
\end{tikzpicture}
\end{minipage}
\hfill
\begin{minipage}{0.64\textwidth}
\centering
\begin{tikzpicture}[scale=0.92]
\Vertex[x=0.00,y=0.55,size=.58,color=white,NoLabel]{rs}
\Vertex[x=1.35,y=1.15,size=.58,color=reddishpurple,NoLabel]{rx1}
\Vertex[x=1.35,y=-0.05,size=.58,color=white,NoLabel]{rx2}
\Vertex[x=3.45,y=1.15,size=.58,color=bluishgreen,NoLabel]{ry1}
\Vertex[x=3.45,y=-0.05,size=.58,color=bluishgreen,NoLabel]{ry2}
\Vertex[x=4.90,y=0.55,size=.58,color=white,NoLabel]{rt}
\Edge[Direct](rs)(rx1)
\Edge[Direct](rs)(rx2)
\Edge[Direct](rx1)(ry1)
\Edge[Direct](rx2)(ry1)
\Edge[Direct](rx2)(ry2)
\Edge[Direct](ry1)(rt)
\Edge[Direct](ry2)(rt)
\node at (-0.25,0.55) {$s$};
\node at (5.18,0.55) {$t$};
\node at (0.62,1.07) {$1$};
\node at (0.62,0.03) {$1$};
\node at (2.40,1.35) {$\infty$};
\node at (2.35,0.58) {$\infty$};
\node at (2.42,-0.28) {$\infty$};
\node at (4.18,1.07) {$1$};
\node at (4.18,0.03) {$1$};
\draw[blue,line width=1.4pt] plot[smooth cycle,tension=.86] coordinates {(-0.55,0.90) (-0.65,-0.45) (1.15,-0.55) (2.75,-0.70) (4.10,-0.70) (4.00,1.70) (2.95,1.82) (1.70,0.65)};
\node[color=blue] at (3.90,1.95) {$S$};
\end{tikzpicture}
\end{minipage}
\end{center}

So
\[
\begin{array}{c}
\text{max flow value}=\text{max matching size}\\[0.5em]
\Updownarrow\ \text{max flow--min cut}\\[0.5em]
\text{min cut value}=\text{min vertex cover size}
\end{array}
\]
Therefore,
\[
\text{maximum matching size}=\text{minimum vertex cover size}.
\]
\end{proof}

\begin{remark}
Idea: min--max relation, duality, and the right gadget.
\end{remark}

\end{document}