tag:blogger.com,1999:blog-6555947.post111031230481068175..comments2014-01-12T10:46:48.153-07:00Comments on The Geomblog: An old question, a new approachSuresh Venkatasubramaniannoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6555947.post-1110342236483899232005-03-08T21:23:00.000-07:002005-03-08T21:23:00.000-07:00Sorry about that Suresh. I was thinking of someth...Sorry about that Suresh. I was thinking of something like a dual or inverse but it isn't. <br /><br /><A></A><A></A>Posted by<A><B> </B></A>RafaelAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6555947.post-1110335648495147482005-03-08T19:34:00.000-07:002005-03-08T19:34:00.000-07:00Am not sure what you mean by that. If you replace ...Am not sure what you mean by that. If you replace vertices by edges, what are "vertices" in this new graph. Also, I am not sure what you mean by equivalent. These ar e both NP-hard, so there is a reduction from one to the other; apart from that they are quite different.<br /><br />A minimum vertex cover is a vertex cover of smallest cardinality.  <br /><br /><A></A><A></A>Posted by<A><B> </B></A>SureshAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6555947.post-1110332690753825632005-03-08T18:44:00.000-07:002005-03-08T18:44:00.000-07:00Is the vertex cover problem equivalent to the Trav...Is the vertex cover problem equivalent to the Traveling Salesman Problem with the vertices replaced by edges? <br /><br />What is the difference between a vertex cover V and a minimum vertex cover M? Can one find subsets M of V? <br /><br /><A></A><A></A>Posted by<A><B> </B></A>RafaelAnonymousnoreply@blogger.com