hw2_sol_long

# hw2_sol_long - Com S 511 Homework#2 Due on Friday Yuheng...

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Com S 511: Homework #2 Due on Friday, September 18, 2009 Yuheng Long 1

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Yuheng Long Com S 511 : Homework #2 Problem 1 Problem 1 (a) It is suffice to show that the flow satisfies the flow capacity constraint. Let c k ( e ) be the capacity of edge e in G k . Clearly, 2 c k - 1 ( e ) c k ( e ) 2 c k - 1 ( e ) + 1. Because, f * k - 1 is the maximum flow in G, for each edge e, f * k - 1 ( e ) c k - 1 ( e ). That is, 2 f * k - 1 ( e ) 2 c k - 1 ( e ) c k ( e ). (b) Let (A,B) be a minumum cut in G k - 1 , and c 0 k - 1 be its capacity. Then we have v ( f * k - 1 ) = c 0 k - 1 = e out of A c k - 1 ( e ). For any edge e, we have c k ( e ) 2 c k - 1 ( e ) + 1, thus e out of A c k ( e ) 2 e out of A ( c k - 1 ( e ) + 1) 2 v ( f * k - 1 ) + e out of A 1. Therefore, v ( f * k ) e out of A c k ( e ) 2 v ( f * k - 1 ) + m and v ( f * k ) - 2 v ( f * k - 1 ) m . (c) We have O ( lg C ) rounds to compute G 1 , G 2 , . . . , G B . In each round, we begin with 2 f * k - 1 and by (b), it could have at most m augmentations. In each augmentation, we use BFS/DFS to search a path, which is O(m+n) (3.13). Also, m n/ 2. Thus the overall running time is O ( m ( m + n ) lg C ) or O ( m 2 lg C ), which is polynomial in n, m and lg C .
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