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26_4 - Contents Maximum flow problem Minimum cut problem...

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1 Contents Maximum flow problem. Minimum cut problem. Max-flow min-cut theorem. Augmenting path algorithm. Shortest augmenting path. Capacity-scaling. Augmenting Path Augmenting path = path in residual graph. Max flow no augmenting paths ??? s 4 2 5 3 t 10 13 10 4 4 4 4 G s 4 2 5 3 t 10 10 10 4 4 4 4 3 G f 10 10 10 Flow value = 14 Max-Flow Min-Cut Theorem Augmenting path theorem (Ford-Fulkerson, 1956): A flow f is a max flow if and only if there are no augmenting paths. MAX-FLOW MIN-CUT THEOREM (Ford-Fulkerson, 1956): the value of the max flow is equal to the value of the min cut. We prove both theorems simultaneously by showing the TFAE: (i) f is a max flow. (ii) There is no augmenting path relative to f. (iii) There exists a cut (S, T) such that | f | = c(S, T). Proof We prove both simultaneously by showing the TFAE: (i) f is a max flow. (ii) There is no augmenting path relative to f. (iii) There exists a cut (S, T) such that |f| = c(S, T). (i) (ii) Let f be a flow. If there exists an augmenting path, then we can improve f by sending flow along path. (proof by contradiction) (iii) (i) This was the Corollary to Lemma 2. Proof (ii) (iii) Let f be a flow with no augmenting paths. Let S be set of vertices reachable from s in residual graph. Let T=V S. Is t T. ? (clearly s S, and t S by definition of f , o.w. augmenting path) For each u S, v T, f (u,v) = c(u,v). ? (o.w. c f (u,v)=c(u,v) f (u,v)>0, then v S). Therefore: ? | f | = f (S,T) = c(S,T). Residual Network s t S T Ford-Fulkerson algorithm Let G=(V,E) be a flow network. Ford-Fulkerson (G,s,t,c) 01 for each edge (u,v) in E do 02 f(u,v) f(v,u) 0 03 while there exists a path p from s to t in residual network G f do 04 c f = min{c f (u,v): (u,v) is in p} 05 for each edge (u,v) in p do 06 f(u,v) f(u,v) + c f 07 f(v,u) -f(u,v) 08 return f Cost?
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