26_4 - 1 Contents ¡ Maximum flow problem. ¡ Minimum cut...

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Unformatted text preview: 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)...
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26_4 - 1 Contents ¡ Maximum flow problem. ¡ Minimum cut...

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