chap26-solutions

# chap26-solutions - Selected Solutions for Chapter 26...

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Selected Solutions for Chapter 26: Maximum Flow Solution to Exercise 26.2-11 For any two vertices u and ± in G , we can define a flow network G consisting of the directed version of G with s D u , t D ± , and all edge capacities set to 1 . (The flow network G has V vertices and 2 j E j edges, so that it has O.V / vertices and O.E/ edges, as required. We want all capacities to be 1 so that the number of edges of G crossing a cut equals the capacity of the cut in G .) Let f denote a maximum flow in G . We claim that for any u 2 V , the edge connectivity k equals min ± 2 V Nf u g fj f jg . We’ll show below that this claim holds. Assuming that it holds, we can find k as follows: EDGE-CONNECTIVITY .G/ k D 1 select any vertex u 2 G: V for each vertex ± 2 G: V N f u g set up the flow network G as described above find the maximum flow f on G k D min .k; j f j / return k The claim follows from the max-flow min-cut theorem and how we chose capac- ities so that the capacity of a cut is the number of edges crossing it. We prove that k D min ± 2 V Nf u g fj f jg , for any u 2 V by showing separately that k is at least this minimum and that k is at most this minimum. ± Proof that k ± min ± 2 V Nf u g fj f jg : Let m D min ± 2 V Nf u g fj f jg . Suppose we remove only m N 1 edges from G . For any vertex ± , by the max-flow min-cut theorem, u and ± are still connected.

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• Spring '10
• .
• Flow network, Maximum flow problem, Max-flow min-cut theorem, Menger's theorem, Ford–Fulkerson algorithm, min fjfu17 jg

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chap26-solutions - Selected Solutions for Chapter 26...

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