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May 8, 2015 at 14:59 answer added Willem Hagemann timeline score: 1
S Sep 9, 2013 at 20:18 history bounty ended CommunityBot
S Sep 9, 2013 at 20:18 history notice removed CommunityBot
S Sep 1, 2013 at 19:13 history bounty started Simd
S Sep 1, 2013 at 19:13 history notice added Simd Draw attention
Aug 31, 2013 at 15:15 history tweeted twitter.com/#!/StackCSTheory/status/373826328090648577
S Aug 31, 2013 at 4:02 history suggested D.W. CC BY-SA 3.0
Choose title and body that makes it more likely this will be found via search (to incorporate several different ways that people might search for this problem). Also add some more tags.
Aug 31, 2013 at 3:03 comment added D.W. (cont.) A warning, though: the half-space representation might be exponentially larger than the vertex representation, in the worst case. So, you might want to try this idea out on your real-world data to see whether it works in your situation or whether it goes exponential.
Aug 31, 2013 at 3:00 review Suggested edits
S Aug 31, 2013 at 4:02
Aug 31, 2013 at 2:58 comment added D.W. I see you're representing the convex polyhedron as the convex hull of a set of vertices (the vertex representation). If you convert this to half-space representation, the problem might be easier. I.e., convert to a a representation of the convex polyhedron as the intersection of a bunch of half-spaces, where each is specified by a linear inequality. Once it is in half-space representation, you can evaluate whether each of the linear inequalities is satisfied by the given point: if all inequalities are satisfied by the point, then the point is in the convex polyhedron; otherwise it isn't.
Aug 30, 2013 at 18:42 history asked Simd CC BY-SA 3.0