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The cut capacity of $(S,S^c)$, with $S^c=V\setminus S$ the complement of a set, isOnly outward directed edges contribute; incoming and internal edges contribute nothing.^[ref]

Here is the accounting behind the bound, using nonnegative original edge amounts $f_{uv}$ in a finite flow network. Sum the divergence of a flow over the vertices in $S$:An edge with both endpoints in $S$ appears once with a plus sign at its tail and once with a minus sign at its head, so those contributions cancel. An outward edge appears only with a plus sign; an incoming edge appears only with a minus sign. Since $s\in S$, $t\notin S$ and every other vertex satisfies flow conservation, the left side is the flow value. ThusThe inequality uses two facts separately: each forward flow is at most its flow network edge capacity, and each backward flow is nonnegative. This is the weak duality bound underlying the max-flow min-cut theorem. It is an accounting proof, requiring no choice of augmenting paths.

For $S=\{s,b\}$ in the running example, explicitly add the two divergences of a flow:The internal $s\to b$ edge cancels. The $a\to b$ edge is a genuine original directed edge entering $S$, so its flow is subtracted; this is different from a backward edge in a residual network. Its flow network edge capacity is neither added to nor subtracted from cut capacity, which remains $c_{sa}+c_{bt}=6$.

Equality in the bound holds precisely when every outward edge is saturated and every incoming edge carries zero flow, becauseand every term is nonnegative. For $S=\{s\}$ the two outward edges are saturated and there are no incoming edges, so the gap is zero. For $S=\{s,b\}$ all outward edges are still saturated, but the incoming flow of $1$ gives a gap of $1$. This explains exactly why the second cut of a flow network is not a minimum cut.

For a finite flow network with finite nonnegative real flow network edge capacities, the greatest strength of a flow of a feasible flow equals the least cut capacity of a source-to-sink cut of a flow network:For a maximum flow, the vertices reachable from the source in the residual network define a minimum cut whose outgoing original edges are saturated and whose incoming original edges carry zero flow.^[ref]

Why is a cut of a flow network an upper bound? Sum flow conservation on its source side $S$. Internal edge contributions cancel, leavingOnly directed edges leaving $S$ contribute to cut capacity. Directed edges entering $S$ are subtracted in net flow, so adding their capacities to the upper bound would lose the intended bottleneck. This inequality is weak duality; the theorem says the bound is attained, an instance of strong duality. Its beauty is that a routing plan and a separating boundary independently witness each other's optimality.

In the running flow network, the two internal vertices give four possible source sides. For $S=\{s\}$, the cut capacity is $c_{sa}+c_{sb}=3+2=5$. For $S=\{s,a\}$ it is $c_{sb}+c_{ab}+c_{at}=2+1+2=5$. For $S=\{s,b\}$ it is $c_{sa}+c_{bt}=3+3=6$; the incoming edge $a\to b$ contributes nothing to that cut capacity. Finally, $S=\{s,a,b\}$ gives $c_{at}+c_{bt}=2+3=5$. Thus the minimum cut has cut capacity $5$, matching the explicitly computed maximum flow.

Ford and Fulkerson's 1956 paper opens with a railway-capacity question attributed there to T. Harris and proves a minimal-cut theorem as Theorem 1. Its original formulation uses graph paths carrying nonnegative amounts; the residual reachability certificate for maximum flow below gives the finite directed flow network proof used here.^[ref]

A minimum cut is a cut of a flow network of least cut capacity for the prescribed source $s$ and sink $t$. The max-flow min-cut theorem equates that least cut capacity with the greatest flow value.^[ref]

To certify a particular minimum cut, exhibit a feasible flow whose flow value equals its cut capacity. Every feasible flow is bounded above by every cut capacity, so this single equality proves both optimalities.^[ref]

In the running example, the four possible source sides $\{s\}$, $\{s,a\}$, $\{s,b\}$ and $\{s,a,b\}$ have cut capacities $5,5,6,5$, respectively. The feasible flow of flow value $5$ therefore certifies three different minimum cuts. A minimum cut need not be unique; the optimum cut capacity is the same for all of them. Merely saturating many edges does not give this certificate: flow conservation and the particular cut of a flow network must also be checked.

The vertex separator version of Menger's theorem says that, for distinct nonadjacent vertices $x,y$ of a finite graph, the maximum number of internally vertex-disjoint paths from $x$ to $y$ equals the minimum size of an $x$–$y$ vertex separator.

Use the following version of the max-flow min-cut theorem: in a finite directed flow network with nonnegative integer flow network edge capacities, the maximum flow value equals the minimum cut capacity, and a maximum flow can be chosen to have integer values on every directed edge. The last clause is the integral max-flow theorem.

Here is the reduction, including the correspondence with vertex separators. Put $N=|V(G)|$. Perform vertex splitting on each vertex $v\ne x,y$, replacing it by $v_{\mathrm{in}},v_{\mathrm{out}}$ with a directed edge $v_{\mathrm{in}}\to v_{\mathrm{out}}$ of flow network edge capacity one. Keep $x$ as the source and $y$ as the sink. For each original undirected edge $uv$, put directed edges from the output of $u$ to the input of $v$ and from the output of $v$ to the input of $u$, each of flow network edge capacity $N$; an unsplit endpoint uses $x$ or $y$ itself.

A family of $p$ internally vertex-disjoint paths gives a feasible integral flow of value $p$, by sending one unit down each graph path. Conversely, decompose an integral flow into unit source-to-sink graph paths and directed cycles, discarding the directed cycles. This decomposition can be obtained by following positive-flow directed edges and subtracting one unit along a source-to-sink graph path or a directed cycle, repeating until the flow value is exhausted. A capacity-one split edge can belong to at most one unit graph path. The projected graph paths in $G$ are therefore internally vertex-disjoint paths. Thus the largest number of such graph paths equals the maximum flow value.

Deleting a vertex separator $S$ corresponds to removing its split edges, of total flow network edge capacity $|S|$. This destroys every source-to-sink graph path; taking the source-reachable vertices after removal gives a cut of a flow network of cut capacity at most $|S|$. Thus the minimum cut capacity is no greater than the minimum vertex separator size.

For the reverse inequality, deleting all $N-2$ internal vertices separates $x,y$, because they are nonadjacent. Hence a minimum cut has cut capacity at most $N-2<N$ and cannot contain a directed edge of flow network edge capacity $N$. Its outgoing edges are therefore split edges. Let $S$ be the original vertices associated with them. Any $x$–$y$ graph path avoiding $S$ would lift to a source-to-sink graph path crossing this cut of a flow network without using an outgoing edge, an impossibility. Hence $S$ is a vertex separator, with size exactly the cut capacity. The two minima agree, and the max-flow min-cut theorem proves Menger's theorem.

Start with zero flow and apply the Ford-Fulkerson algorithm. The following augmenting paths and bottlenecks use the capacities in the original PDF diagram:For example, the second path is limited by the remaining capacity two on $6\to 8$ and by capacity two on $5\to 6$. After the third augmentation, $7\to 8$ has residual capacity one; the fourth path supplies it through $5\to 4\to 6\to 7$. Every displayed increment is the smallest residual capacity of its path.

The resulting arc flows areHere $uv$ denotes $u\to v$. These satisfy every capacity bound and flow conservation. The final residual network has source-reachable set $S=\{1,2,3,5\}$. Its outgoing arcs are $2\to 4$, $5\to 4$, $5\to 6$ and $5\to 7$, with capacities $2,1,2,1$, all saturated. ConsequentlyThis is a minimum cut, and the equal flow and cut values certify optimality by the max-flow min-cut theorem. The cut immediately before node eight also has capacity six, so the minimum cut is not unique.

An cut of a flow network partitions a flow network into $S,S^c$, with source $A\in S$ and sink $B\notin S$. Its cut capacity is the sum of capacities of arcs directed from $S$ to $S^c$; a minimum cut minimizes this sum.

Use the suggested nodes. Give $A\to a_e$ capacity $1$, each $a_e\to b_v$ capacity $M=m+kn+1$, and each $b_v\to B$ capacity $k$. By the max-flow min-cut theorem, a flow of value $kn$ exists once every cut capacity is at least $kn$.

Consider a cut of a flow network containing no crossing arc of capacity $M$, and let $U$ be the intersections on the sink side. Every street incident to $U$ must also be on that side, or its arc to an endpoint in $U$ would cross the cut of a flow network. Hence at least $d(U)$ source-to-street arcs cross, as do the $n-|U|$ intersection-to-sink arcs from the other intersections. Therefore its cut capacity is at leastA cut of a flow network crossing a capacity-$M$ arc has still larger capacity. The cut of a flow network just before the sink has capacity $kn$, proving the maximum flow value is exactly $kn$.

The integral max-flow theorem supplies an integral maximum flow. Every intersection sends $k$ units to the sink; each street can supply at most one unit, assigned to one endpoint. Orient each assigned street away from its assigned endpoint. Orient unassigned streets arbitrarily. Then each vertex has at least $k$ outgoing streets. An Ford-Fulkerson algorithm with integer capacities constructs this flow and therefore the street directions. Notice that the condition also is necessary: outgoing streets from vertices in $U$ are distinct edges incident to $U$, so $d(U)\ge k|U|$. This is an outdegree orientation criterion.