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Until yesterday, I was under the naive impression that constructing a weighted graph where the nearest-neighbour algorithm gives the worst possible route, would have the property that any other strategy would be better; in particular the furthest-neighbour algorithm. This was of course a mistake: indeed, for $n=4$ or $5$ cities, forcing the nearest-neighbor to produce a bad route, one also forces, in parallel, an equally bad route for the furthest-neighbour strategy. Things get complicated and fiddly for $n>5$, so I am thinking this is not a good approach, which brings me back to square one:

Question

Is it possible to construct a weighted graph such that, given a starting city, choosing the furthest away city at each stage yields a shorter route than picking the closest?

TSP
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  • Do you want to still have the property that the nearest-neighbor algorithm gives the worst possible result? If not, it's easy. – Robert Israel Sep 17 '14 at 23:14
  • @RobertIsrael: I think the way the question is formulated implies that nearest-neighbor need not be best. The alternative would make an interesting follow-up question. I would like to see your solution to the former. – Samuel Sep 17 '14 at 23:56
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    Oops, sorry, not so easy. – Robert Israel Sep 18 '14 at 00:38
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    Now posted to MO, http://mathoverflow.net/questions/222637/travelling-salesman-can-the-furthest-neighbour-algorithm-beat-the-nearest-neigh – Gerry Myerson Nov 05 '15 at 02:10

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