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Unformatted text preview: Introduction to Algorithms Solution Set 5 CS 482, Spring 2008 (1) (a) If e is contained in a min-cut ( A,B ), let C be the capacity of this cut. Every max-flow f has value v ( f ) = C , and such a flow must saturate every edge of the min-cut, including edge e . In more detail, the argument goes as follows. Let E ( A,B ) denote the set of edges e = ( u,v ) with u ∈ A, v ∈ B and let E ( B,A ) denote the set of edges e = ( u,v ) with u ∈ B, v ∈ A . Then C = v ( f ) = X e ∈ E ( A,B ) f ( e )- X e ∈ E ( B,A ) f ( e ) ≤ X e ∈ E ( A,B ) c e- X e ∈ E ( B,A ) 0 = C- . The left side is equal to the right side, and the only way this can happen is if f ( e ) = 0 for every e ∈ E ( B,A ) and f ( e ) = c e for every e ∈ E ( A,B ). Since e ∈ E ( A,B ), this implies f ( e ) = c e , i.e. edge e is saturated by f . If e is not contained in a min-cut, then decreasing the capacity of edge e by 1 doesn’t change the min-cut capacity of the flow network, hence it also doesn’t change the max-flow value. Let G denote the flow network obtained from G by decreasing the capacity of edge e by 1, and let f be a max-flow in G . Then f is also a max-flow in G (since its value equals the min-cut capacity in G , which equals the min-cut capacity in G , which equals the max-flow value in G ) and f doesn’t saturate edge e in the original network G . (b) We will prove that any flow which contains a directed cycle of flow-carrying edges can be transformed into another flow with no such cycles, without changing the flow value. The idea is to use “augmenting cycles” in the same sort of way that the Ford-Fulkerson algorithm uses augmenting paths....
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This note was uploaded on 10/02/2008 for the course CS 482 taught by Professor Kleinberg during the Spring '08 term at Cornell University (Engineering School).
- Spring '08