CO-351-1079-Midterm_solutions

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

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/18/2010 for the course CO 351 taught by Professor Various during the Fall '05 term at Waterloo.

### Page1 / 6

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

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

View Full Document
Ask a homework question - tutors are online