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ote that any s-t cut C (a collection of edges whose removal separates |
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Note that any s-t cut C (a collection of edges whose removal separates |
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In fact, any minimum-capacity cut gives an optimal dual solution in this way. This is the gist of the max-flow min-cut theorem. |
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In fact, there is always an s-t cut that gives an optimal dual solution in this way. Here is a proof sketch: MAX-FLOW MIN-CUT THM: Given any feasible solution (x,y) to the dual LP, there exists an s-t cut whose capacity equals the value of the dual solution ∑u,v c(u,v) x(u,v). PROOF: Let (x,y) be a feasible solution to the dual. Without loss of generality assume y(s) = 0 (otherwise subtract y(s) from each y(v) to get an equivalent dual solution). Choose R uniformly at random in [0,1]. Let U={u ∈ V : y(u) ≤ R}. Then (U,V-U) forms an s-t cut. The probability that an edge (u,v) crosses the cut is : Pr[y(u) ≤ R < y(v)] ≤ max(0,y(v)-y(u)) ≤ x(u,v). Thus, the expected capacity of the cut is :∑(u,v)∈ E c(u,v)Pr[edge (u,v) cut] ≤ ∑(u,v)∈ E c(u,v) x(u,v). Thus, there exists an s-t cut with capacity at most ∑(u,v)∈ E c(u,v) x(u,v). QED It is also the case that if the capacities are integers, there is an integer-valued maximum flow f. |