4Soln - Computer Science C73 December 7, 2007 Scarborough...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 min-cut of the flow graph G , and f be a max-flow in G . By Proposition 7.6 in KT , V ( f ) = e out ( S ) f ( e ) e out ( T ) f ( e ). By the max-flow-min-cut 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 ....
View Full Document

Page1 / 5

4Soln - Computer Science C73 December 7, 2007 Scarborough...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online