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Unformatted text preview: HW6 SOLUTION 1. Chapter 7 Problem 3: Part (a) ν ( f ) = X e out of s f ( e ) = 6 + 3 + 1 = 10 Constructing the residual graph of the current flow network by the FordFulkerson algorithm in the text book, there is a st path( s → a → c → b → d → t ) in the residual graph of current flow network, thus the current net flow is not the maximum of the network Part (b) By the FordFulkerson algorithm, we can get the maxflow and the maxflow is 7 + 3 + 1 = 11. And the minimum s t cut is A = { s,a,b,c } ,B = { d,t } since C ( A,B ) = X e out of A c ( e ) = 5 + 5 + 1 = 11 Problem 5: The statement is false. We take Figure (7.3)(a) in Page 341 as the counter example by changing the capacity c u,v = 9 . 5. Clearly the minimum cut is A = { s,u } and B = { v,t } . After we add 1 to every capacity, ( A,B ) is not a minimum cut any more as the value of max flow is 32 now while the cut between A and B is 32 . 5. Problem 7: We first construct a bipartite graph G whose node are the customers c i (0 < i ≤ n ) and the base station b j (0 < j ≤ k ). We add edge e ij = ( c i ,b j ) in the graph G if and only if the distance of client c i and the base station b j is within r (the coordinates of both are given) and set c e ij = 1. then we add a source s and sink t in G . For each customer c i , we add e = ( s,c i ) in the graph G and set c e = 1.For each base station b j , we add e = ( b j ,t ) in the graph G and set c e = L . After building the graph G for the original problem, we find the maximum stflow value v in graph G by Fulkerson algorithm. If v = n , then every client can be connected simultaneously to a base station, otherwise it can’t. The correctness of the algorithm comes from the following lemma. Lemma 1: Every client can be connected simultaneous to a base station if and only the value of the maximum value of an s t flow in G is n . Proof: First if every client can be connected simultaneous to a base station, by the con struction of the graph G , it’s easy to see that this flow meets all capacity constraints and has value n . For the converse direction, assume that there is an s t flow of value n . The construction given in the algorithm makes sure that the base station is within the Date : Oct 29, 2006. 1 2 HW6 SOLUTION distance r of the customers since flow goes from c i to b j . So we only need to make sure that all customers are connected to some base station. Notice that { s } ,V{ s } is a minimum s t cut(it has value n ), so the flow must saturate all edges crossing the cut. Therefore, all customers must be connected to some base stations. This completes the correctness proof. Running time: The time complexity of constructing the graph is O ( nk ) since we need to see if the distance of each ( b i ,c j ) is below the range parameter r or not....
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This note was uploaded on 02/27/2008 for the course CSCI 570 taught by Professor Shamsian during the Fall '06 term at USC.
 Fall '06
 Shamsian
 Algorithms

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