This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CO250/CM340 INTRODUCTION TO OPTIMIZATION  HW 6 SOLUTIONS Exercise 1 (10 marks) . (a) If the answer is NO then the max flow has value at most M . The Max Flow  Min Cut theorem then implies that the min cut has capacity at most M . So there is a cut C of capacity at most M . This cut can be essentially the certificate. Essentially because we have not finished yet: we need to be able to check that C is a cut in polynomial time. This is not so obvious if C is given as a set of edges: how does one check quickly that this set intersects all stpaths? To make this easy we can take the certificate not to contain just the cut C but also the set of vertices U which are reachable from paths starting at s without using edges of C . We can check in polynomial time (say O ( n 2 ) that no edge of G goes from in U to outside U except for edges of C . We also check that t / C . This verifies C is a cut. Then we can compute the capacity of C also in time O ( n 2 ) . Then check the capacity is at most M . (b) Lets denote H by the graph obtained from G by deleting just the edge e . If all of G s maximum weight matchings contain e , then the weight of the max matching in H will be less than the one in G . This also holds vice versa. So the empty certificate is enough: to check a YES answer, we just have to use Xenias algorithm to find the max weight matching in G and that in H , and check that they have different weights. Exercise 2 (15 marks) . (a) (P) is min n X i =1 x i subject to x i + x j 1 for each i,j such that v i v j is one of the edges e k , 1 k m x i ( i { 1 , 2 ,...,n } ) . (b) Let ( i ) denote the set of k such that edge e k is incident with vertex i . The ith column of the constraint matrix contains 1 in row k whenever edge e k is incident with vertex i , that is, k ( i ) . So when we take the transpose, there are 1s in row i at each column k in ( i ) . Thus, (D) is 1 2 max m X k =1 y i subject to X k ( i ) y k 1 for each 1 i...
View
Full
Document
 Spring '10
 GUENIN

Click to edit the document details