COMP 360 Lecture Notes - Lecture 8: Minimum Cut, Max-Flow Min-Cut Theorem, Maximum Flow Problem

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Def: two paths are edge disjoint if they have no edge in common. Max flow formulation: assign unit capacity to every edge. Claim of the theorem: max number of edge disjoint path s -> t = value of max flow. Suppose there are "k" edge disjoint s -> t paths p1 . pk. => if you have a max flow then that is = number of edge disjoint path. Integrality theorem: there exists a 0-1 flow "f" of value "k". Consider edge (s, u) with f(s,u) = 1. By conservation, there exists an edge (u,v) with f(u,v) = 1. You reach some node "v" for the second time. Once you reach "t" you remove the simple path (set all flows to = 0). If you hit the same node twice that means you have a cycle. You remove the cycle by setting the flow of all cyclic edges to 0.

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