570 homework6 sol

# 570 homework6 sol - 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

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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 Ford-Fulkerson algorithm in the text book, there is a s-t 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 Ford-Fulkerson algorithm, we can get the max-flow and the max-flow 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 s-t-flow 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.

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570 homework6 sol - 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

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