If you can't do one-point-removal with four (because, for two opposite points, the origin cannot be on both sides of a line connecting them), then how can you do two-point-removal with a hexagon (two opposite points, origin can't be on both sides)? Don't you mean removing 1 point takes 5, removing two points takes 7, removing three points takes 9, and so on? On Tue, Nov 21, 2017 at 3:39 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Suppose you want to distribute a small given finite number of points around the origin of the Cartesian plane so that the origin is in the interior of the convex closure of the points. It's easy - you can't do it with two points, but you can do it with three.
Now suppose you want to ensure that the origin will still be inside the convex closure even if one of the points is removed by an evil entity. I have pretty much convinced myself that you can't do it with four points, though I would appreciate a clear proof. You can do it with five; a regular pentagon centered on the origin clearly works.
Now what if the evil entity is allowed to remove any two points? How many points do you need to be safe? I think the answer is six (a regular hexagon works).
Seven points defends against three removals, but I think you might need nine to defend against four. I suspect the answer in general is that you need more than twice as many points as the evil entity is allowed to remove, but I don't have a proof. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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