Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02
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The unit network for a bipartite graph identifies integral flows with matchings and finite cuts with vertex covers

Statement

For a finite bipartite graph with parts (X,Y), give the arcs s→x, each graph edge x→y, and y→t unit capacity. Integral flow paths correspond to matching edges xy; a vertex cover CX∪CY gives the cut {s}∪(X∖CX)∪CY of the same size. Consequently the maximum-flow/minimum-cut equality agrees with the matching/cover equality, and a flow of value ∣X∣ agrees with Hall's saturating matching criterion.

Facts & Assumptions

Given: The displayed unit network of a finite bipartite graph (X,Y).

[L1]

Integral Ford--Fulkerson gives equal maximum flow and minimum cut values (Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow).

[L2]
[L3]

In a finite bipartite graph, matching and minimum vertex-cover sizes agree (König's theorem: ν(G)=τ(G) for every finite bipartite graph).

[L4]

Hall's condition is equivalent to a matching saturating the left part (Hall's marriage theorem for a finite bipartite graph).

Proof

technique · direct
1.1

By [L2], each unit source-sink path has the form s,x,y,t, and arc-disjointness makes its middle edges a matching; conversely a matching gives these unit paths.

L2
1.2

A vertex cover CX∪CY yields the stated cut: no x→y arc crosses it, and its only cut arcs are the ∣CX∣ source arcs and ∣CY∣ sink arcs.

given
1.3

Thus [L1] makes maximum flow equal minimum cut, and [L3] identifies their common value with ν(G)=τ(G).

L1L3
2.1

A flow of value ∣X∣ uses every unit arc out of s, so step 1.1 makes a matching saturating X; [L4] gives the equivalent Hall condition.

L4
3.1

The path, cut, matching, cover, and Hall correspondences in steps 1.1--2.1 prove the proposition.

step 1.1step 1.2step 1.3step 2.1∎

Depends on

Used by

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