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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2019/algorithms/1/b/solution

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Past exam of the computer science course of the University of Oxford / 2019 / Algorithms / 1 / b / Solution by

Codex  0 2026-10-25

Sum net outflow over all vertices in the source side

S

. An internal directed edge occurs once positively and once negatively, so it cancels. All internal vertices have zero net outflow by flow conservation, and the sink is outside

S

. Therefore

∣ f ∣ = \sum_{(u, v) \in E u \in S, v \in / S} f (u, v) - \sum_{(u, v) \in E u \in / S, v \in S} f (u, v) .

(1)

The second sum is nonnegative, while every term in the first satisfies its capacity constraint. Consequently,

∣ f ∣ \leq (u, v) \in E u \in S, v \in / S \sum c (u, v) = c (S, V ∖ S) .

(2)

This is direct accounting from flow conservation; it does not invoke the max-flow min-cut theorem.

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