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TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02
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Ford-Fulkerson terminates for finite integer capacities and proves max-flow min-cut with an integral maximum flow

Statement

In a finite integral s-t network, begin with the zero flow and repeatedly augment along any residual s-t path by its integral bottleneck. This process terminates. Its final integral flow f∗ is maximum, and ∣f∗∣=min⁡{c(S):S is an s-t cut}.

Facts & Assumptions

Given: A finite integral s-t network and the augmenting-path process from its zero flow.

[L1]

Residual bottleneck augmentation preserves feasibility and raises the value by its positive integral bottleneck (Augmenting by the residual bottleneck preserves feasibility and increases the flow value).

[L2]

A flow with no residual s-t path has a cut of capacity equal to its value (If the residual network has no source-sink path, its reachable set gives a cut whose capacity equals the flow value).

[L3]

Every feasible flow has value at most every cut capacity (A flow's value is its net flow across every source-sink cut and never exceeds the cut capacity).

Proof

technique · constructive
1.1

The zero flow is feasible and integral, and every feasible flow has value at most the finite integer sum of capacities of arcs leaving s.

givenconstruct
1.2

Each augmentation increases the integral value by at least one by [L1], so no more than that finite source-capacity bound many augmentations occur.

L1
1.3

At termination the final flow has no residual path, so [L2] supplies a cut S with c(S)=∣f∗∣.

L2
1.4

By [L3], every feasible flow has value at most c(S)=∣f∗∣, and every cut has capacity at least ∣f∗∣.

L3
2.1

Steps 1.1--1.4 prove termination, integrality, maximality of f∗, and equality with the minimum cut capacity.

step 1.1step 1.2step 1.3step 1.4discharge-construct∎

Remarks

  • Integer bottlenecks are the termination mechanism. This theorem makes no assertion about arbitrary real capacities.

Depends on

Used by

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Sources