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Nov 17, 2020 at 7:08 vote accept oarfish
Nov 23, 2015 at 15:59 vote accept oarfish
Nov 17, 2020 at 7:08
Nov 21, 2015 at 12:00 history edited oarfish CC BY-SA 3.0
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Nov 20, 2015 at 18:23 history edited oarfish CC BY-SA 3.0
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Nov 20, 2015 at 17:15 history tweeted twitter.com/StackCompSci/status/667753054855368704
Nov 20, 2015 at 17:08 history edited oarfish CC BY-SA 3.0
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Nov 20, 2015 at 11:18 answer added oarfish timeline score: 4
Nov 14, 2015 at 16:23 comment added oarfish @YuvalFilmus See e.g. here courses.engr.illinois.edu/cs598csc/sp2009/lectures/… First, compute an MST. Then double the edges of the MST, obtaining an eulerian tour. Now walk the tour and skip over all nodes already visited by using the edge to the next but one node (or skip as long as the next node is one that has already been visited).
Nov 14, 2015 at 16:16 comment added Yuval Filmus What is the MST heuristic?
Nov 14, 2015 at 15:18 comment added Raphael @oarfish The title is the biggest issue; it makes people expect a certain type of question. Then they read things that does not fit at all, they skip the rest and comment.
Nov 14, 2015 at 12:51 answer added Yuval Filmus timeline score: 4
Nov 14, 2015 at 8:42 history edited oarfish CC BY-SA 3.0
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Nov 14, 2015 at 8:40 comment added oarfish I do not see how this type of graph shows that the 2-approximation bound of MST heuristic is tight seems to make my intent clear. How would you suggest to edit the question?
Nov 14, 2015 at 8:39 comment added David Richerby I suggest you rephrase your question to make that clear.
Nov 14, 2015 at 8:37 comment added oarfish @DavidRicherby I know why it is no worse than a 2-approximation, this is supposed to be a worst-case instance showing the bound is tight. Or maybe a series of increasing n would have the ratio converge to two. And as far as I understand, the only black edges should be the spokes. Although having nodes in the circle connected with cost 1 does not change the tour I think.
Nov 14, 2015 at 8:35 comment added David Richerby No single example can "prove that the MST approximation to TSP is a 2-approximation", since being a 2-approximation requires being within a factor of two of optimal on all graphs. What this graph is supposed to do is prove that MST is not better than a 2-approximation, by showing a graph on which it's out by exactly a factor of two. By the way, there should presumably be a black edge from V1 to V6.
Nov 13, 2015 at 23:45 review First posts
Nov 14, 2015 at 8:39
Nov 13, 2015 at 23:44 history asked oarfish CC BY-SA 3.0