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Unformatted text preview: Problem 1 Chapter 73 (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 maxflow 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 Maxflow(G) > k, the maximum st flow on G will be Maxflow(G) k; otherwise the new max flow will be 0. Problem 3 Chapter 713 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 Maxflow algorithm on the new graph G, to find an st max flow. How to write an algorithm for such problem? Algorithm MaxFlowNodeCapacity 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 Maxflow algorithm on G, get the maxflow F; Step 3: Output the maxflow F. End of the algorithm. We can see the problems of finding the maxflow with node capacity and finding the maxflow with edge capacity are equivalent. So, the Maxflow and Mincut Theorem also hold for the node capacity case. Problem 4 Chapter 716 To solve this problem, we show it can be reduced to the maxflow problem. The technique used is very similar to the reduction from Bipartite Matching to Maxflow....
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This note was uploaded on 05/21/2008 for the course CS CO453 taught by Professor Graham during the Winter '08 term at UCSD.
 Winter '08
 Graham

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