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Unformatted text preview: Computer Science C73 December 7, 2007 Scarborough Campus University of Toronto Solutions to Homework Assignment #4 Answer to Question 1. a. s t u v . 5 / 1 . 5 / 1 . 5 / 1 1 / 1 The label “ f/c ” on each edge indicates that the edge has capacity c , f of which is used by the flow. It is easy to check that the specified flow is proper (i.e., it satisfies the capacity and conservation constraints). It is a maximum flow because it has value 1, and the ( s,t )cut ( { s,u,v ] } , { t } ) has capacity 1. b. For any set of vertices S , let out ( S ) be the set of edges with tail in S and head not in S . Let ( S ∗ ,T ∗ ) be a mincut of the flow graph G , and f ∗ be a maxflow in G . By Proposition 7.6 in KT , V ( f ∗ ) = ∑ e ∈ out ( S ∗ ) f ∗ ( e ) − ∑ e ∈ out ( T ∗ ) f ∗ ( e ). By the maxflowmincut theorem, V ( f ∗ ) = ∑ e ∈ out ( S ∗ ) c ( e ). Thus, by the capacity constraint, every edge in out ( S ∗ ) is saturated in f ∗ . By assumption, V ( f ∗ ) > 0, and so there is some e ∈ out ( S ∗ ) such that f ∗ ( e ) > 0. As we just saw, such an edge is saturated in f ∗ , i.e. f ∗ ( e ) = c ( e ). Since all capacities are integers by assumption, f ∗ ( e ) is a positive integer, as wanted. c. We will prove, by induction on k , that for each integer k such that 0 ≤ k ≤ m , there is an integral flow f such that V ( f ) = k . Basis. k = 0. Take f ( e ) = 0 for each edge e . This trivially satisfies the capacity and conservation constraints. Thus it is an integral flow whose value is 0, as wanted. Induction Step. Step Let k be an arbitrary integer such that 0 < k ≤ m . Assume that there is an integral flow f ′ whose value is k − 1. We will prove that there is an integral flow f whose value is k . Since f ′ has value k − 1 < m , it is not a maximum flow. Therefore, the residual graph G f ′ has an ( s,t )path, say p . Because f ′ is an integral flow, the residual capacities of G f ′ are integers, and so the bottleneck of p has residual capacity at least 1. Let f ( e ) = f ′ ( e ) + 1 , if e is a forward edge on p f ′ ( e ) − 1 , if the reverse of e is a backward edge on p f ′ ( e ) , otherwise The function f is a flow in G : It satisfies the capacity constraint because the residual capacity of every edge on p in G f ′ is at least 1. It satisfies the conservation constraint because f ′ does: For each vertex on path p (other than s and t ), the incoming traffic changes by v , for some v ∈ {− 1 , , +1 } , if and only if the outgoing traffic also changes by v ; while for each vertex not on p , neither the incoming nor the outgoing traffic changes. Since f ′ is an integral flow, so is f . Furthermore, V ( f ) = V ( f ′ ) + 1 = k ....
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This note was uploaded on 02/10/2009 for the course RLG 100 taught by Professor Pearly during the Spring '08 term at University of Toronto.
 Spring '08
 Pearly

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