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Oct 14, 2010 at 7:42 comment added Jose Capco I think there is a way if your polytopes were convex. You did tagged "convex-polytopes" but you did not mentioned in the post if they really were convex
Oct 13, 2010 at 20:59 vote accept Aaron
Oct 13, 2010 at 20:49 comment added Aaron Good point, Suresh. I should have mentioned that my constraints are defined by small constants, so I'm satisfied with an algorithm that is not strongly polynomial.
Oct 13, 2010 at 20:41 comment added Suresh Venkat I should point out that we don't even have LPs that run in the time bound you want. Getting a strongly polytime LP algorithm (i.e running in time poly in k,n and nothing else) is an open problem. so you're unlikely to get the answer you want.
Oct 13, 2010 at 20:35 answer added Noah Stein timeline score: 4
Oct 13, 2010 at 20:33 comment added David Eppstein Aaron: this same problem of NP-hardness to satisfy as many as possible but polynomial to satisfy all is shared by Max-2SAT. So that's where I would start looking for a reduction.
Oct 13, 2010 at 19:24 comment added Suvrit So your problem seems to be instance of a partial constraint satisfiability problem, but I am not at the moment, what to say beyond that.
Oct 13, 2010 at 19:14 comment added Aaron One difficulty with trying to reduce from max k-SAT is that it is NP-hard to determine whether every clause in a k-SAT formula is satisfiable or not, but it is easy to determine whether the intersection of every polytope is empty or not. Any reduction would have to somehow preserve this property.
Oct 13, 2010 at 19:00 comment added Andrew D. King I would guess it's NP-complete. Have you tried reducing the problem to minimum $n$-SAT?
Oct 13, 2010 at 17:38 history asked Aaron CC BY-SA 2.5