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Jul 17, 2014 at 1:20 answer added Douglas Zare timeline score: 8
Jul 16, 2014 at 23:05 vote accept Kellar
Jul 16, 2014 at 4:17 vote accept Kellar
Jul 16, 2014 at 4:17
Jul 16, 2014 at 0:23 answer added ofer zeitouni timeline score: 11
Jul 15, 2014 at 19:18 comment added Alex R. Using the known TSP bound that John Gunnar Carlsson mentioned, one can probably get a heuristic bound by first asking what the typical size is of the smallest square that countains $\sqrt{n}$ points in the unit square.
Jul 15, 2014 at 18:08 comment added John Gunnar Carlsson The quantity you're interested in is certainly bounded, because the length of the shortest path through all $n$ points (i.e. the TSP tour) has a length that is bounded by $\sqrt{2n} + 1.75$, as proven in [1], which we can approximate as simply $\sqrt{2n}$ since $n$ is large. If you take this tour and break it into segments consisting of $\sqrt{n}$ points each, you know that one of those segments has length of at most $\sqrt{2}$. [1] Few, L. "The shortest path and the shortest road through n points." Mathematika 2.02 (1955): 141-144.
Jul 15, 2014 at 15:30 history edited Ricardo Andrade
replaced deprecated tag 'geometry'
Jul 15, 2014 at 6:57 comment added The Masked Avenger Hmm. Maybe asking c greater than 1/2 is ambitious. I still think it should converge for c=1/2.
Jul 15, 2014 at 6:50 comment added The Masked Avenger I think it should decrease, even if it were n^c many points rounded up, for fixed positive c less than 1. Set d to satisfy 2d +c=1, and consider a kxk square partition for k = floor(n^d). One of those pieces will have n^c points in it, and the length of the path should be O(n^-d), or some appropriate power.
Jul 15, 2014 at 6:23 history edited Kellar CC BY-SA 3.0
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Jul 15, 2014 at 6:05 review First posts
Jul 15, 2014 at 6:51
Jul 15, 2014 at 6:05 history asked Kellar CC BY-SA 3.0