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h4_sol

# h4_sol - Problem 1 Chapter 7-3(a The value of the flow 10...

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Problem 1 Chapter 7-3 (a) The value of the flow: 10 Maximum flow: 11 (b) Min cut from s, a, b, c to t, d. Capacity of the min cut is 11. Problem 2 Chapter 7 -12 Idea: Run max-flow algorithm on G, and find a min cut of G, which separates the vertices set V into two sets S and T. Remove any k edges from S to T. if Max-flow(G) > k, the maximum s-t flow on G’ will be Max-flow(G) – k; otherwise the new max flow will be 0. Problem 3 Chapter 7-13 We are given a graph G with node capacity, and we can build a new graph G’ = (V’, E’), which has only edge capacities as follows: For each node v in G, we separate it into two nodes v and v’, then define the capacity of the new added edge e(v, v’) = capacity(v). And assign all the edges in G with capacity infinity. Clearly, G and G’ could hold the same amount of maximum flow, where G with node capacity but G’ with edge capacity. Now we reduced the problem of finding the maximum flow in Graph with node capacity to the maximum flow problem with edge capacity we talked in class. We can simply run the Max-flow algorithm on the new graph G’, to find an s-t max flow. How to write an algorithm for such problem? Algorithm Max-Flow-Node-Capacity Input: A graph G= (V, E), and each node v with capacity c(v) Output: The maximum flow value Step 1: Construct the new graph G’ following the instructions above; Step 2: Run Max-flow algorithm on G’, get the max-flow F; Step 3: Output the max-flow F. End of the algorithm. We can see the problems of finding the max-flow with node capacity and finding the max-flow with edge capacity are equivalent. So, the Max-flow and Min-cut Theorem also hold for the node capacity case. Problem 4 Chapter 7-16 To solve this problem, we show it can be reduced to the max-flow problem. The technique used is very similar to the reduction from Bipartite Matching to Max-flow.

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h4_sol - Problem 1 Chapter 7-3(a The value of the flow 10...

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