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Apr 13, 2017 at 12:48 history edited CommunityBot
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Dec 16, 2016 at 0:28 answer added D.W. timeline score: 1
Dec 15, 2016 at 7:18 history tweeted twitter.com/StackCompSci/status/809296483572969472
Dec 14, 2016 at 16:13 comment added j_random_hacker Idea: If you have a pair of vertices $u, v$ such that $u$ is the nearest neighbour of $v$, then there is an optimal solution in which $u$ is not the endpoint of a path. (Suppose it was: Then delete any single edge adjacent to $v$ (there could be either 1 or 2 of these) and connect $v$ to $u$ instead for a solution with cost at least as good.)
Dec 14, 2016 at 15:14 comment added j_random_hacker This is certainly NP-hard: given any Euclidean Longest Path problem, add a new vertex $r$ sufficiently far away from every vertex, and make $b$ big enough to allow all vertices to be contained in a single path.
Dec 14, 2016 at 10:25 history edited D.W. CC BY-SA 3.0
deleted 23 characters in body; edited tags
Dec 14, 2016 at 10:18 history asked D.W. CC BY-SA 3.0