CO-351-1079-Midterm_solutions

CO-351-1079-Midterm_solutions - CO 351 Network Flow Theory...

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University of Waterloo Fall 2007 Instructor: Ashwin Nayak Solutions to Midterm Examination October 22, 2007 Note: these are one of many correct solutions. Question 1. Let D = ( N,A ) be a digraph with s,t N , s n = t , with arc-lengths w R A . Part 1.1 [4 marks] Prove that there is an st -dipath in D if and only if for every ( s,t )-separator S N , we have δ out ( S ) n = . Answer: ( ) Consider an st -dipath P in D , and any ( s,t )-separator S . Since s S , and t n∈ S , there is a ±rst node v on P that is not in S . The arc uv in P incident on v belongs to δ out ( S ), so it is non-empty. ( ) Consider the set S of nodes v reachable from s , i.e., is such that there is an sv -dipath. Since s S , if the set S does not contain t , we have an ( s,t )-separator such that δ out ( S ) = , a contradiction. Part 1.2 [3 marks] Suppose there is an sv -dipath for every v N . De±ne y v as the length of a shortest sv - dipath in D . Prove that if D has no negative dicycle, then y v is a feasible potential. Answer: We have to show that y v y u + w uv for every arc uv A . Since D has no negative dicycle, the length of any sv -diwalk is at least the length of the shortest sv -dipath. The quantity y u + w uv is the length of the sv -diwalk consisting of a shortest su -dipath followed by the arc uv . Therefore, y v y u + w uv . Part 1.3 [3 marks] Suppose the in-degree of every node in D is at least 1. Prove that D contains a dicycle. Answer: Consider a longest dipath P = ( v 1 ,...,v k ) in D . Since the in-degree of v 1 is at least one, there is a node z such that zv 1 A . If z does not lie on the dipath P , we get a longer dipath, a contradiction. Therefore z = v i for some i ∈ { v 2 ,... ,v k } , and we have a dicycle. Question 2. [6 marks] Let D = ( N,A ) be a digraph with arc-lengths w R A , and s,t N,s n = t . Describe the steps of the Dual Algorithm which, given a feasible potential y R N , ±nds either a shortest st -dipath in D , or an ( s,t )-separator Z such that δ out D ( Z ) = . Answer: Let D = ( y ) denote the equality subgraph. Step 1. If there is an st -dipath P in D = ( y ), then stop, and output P . Step 2. Find an st -separator Z in D = ( y ), with δ out D = ( y ) ( Z ) = . Step 3. If δ out D ( Z ) = , then stop, and output Z . Step 4. Let α = min b w uv ( y v y u ) : uv δ out D ( Z ) B . For each
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This note was uploaded on 11/18/2010 for the course CO 351 taught by Professor Various during the Fall '05 term at Waterloo.

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CO-351-1079-Midterm_solutions - CO 351 Network Flow Theory...

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