Edited by BluePhase, 24 October 2012 - 01:48 PM.
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Posted 24 October 2012 - 01:47 PM
Edited by BluePhase, 24 October 2012 - 01:48 PM.
Posted 24 October 2012 - 02:52 PM
Posted 12 November 2012 - 04:39 PM
Posted 12 November 2012 - 04:45 PM
Posted 12 November 2012 - 05:14 PM
Posted 12 November 2012 - 06:42 PM
Posted 12 November 2012 - 06:53 PM
You can probably do better than this by starting with a sphere large enough to hold all the points and shrinking it "appropriately", depending on how many points of contact you currently have. I think this method should take time O(N), but it could be tricky to implement.
Edited by max343, 12 November 2012 - 06:54 PM.
Posted 12 November 2012 - 08:26 PM
You can probably do better than this by starting with a sphere large enough to hold all the points and shrinking it "appropriately", depending on how many points of contact you currently have. I think this method should take time O(N), but it could be tricky to implement.
I don't think that it's possible to do it in O(n) time for dimensions larger than 1 (I think that it's possible to construct a reduction from the diameter problem to the CH problem, but I'm not 100% sure about this). On the other hand an O(n*log(k)) implementation is possible (n is the number of vertices and k is the number of vertices on the convex hull), though it's not that trivial.
Posted 13 November 2012 - 07:58 PM
FindMinSphere(S, origSphereVerts) 1. sphereVerts = origSphereVerts, remainingVerts = empty 2. for each i[k] in random pi(1,2,...,length(S)) 2.1 if S[i[k]] is not contained in the sphere spanned by sphereVerts 2.1.1 sphereVerts = FindMinSphere(remainingVerts, EliminateExtraVertex(origSphereVerts + S[i[k]])) 2.2 remainingVerts += S[i[k]] 3. return sphereVerts EliminateExtraVertex(S) 1. X=S 2. if length(S)==5 2.1 X = from the 5 possible spheres in X choose the one that contains all five points 3. return X
Edited by max343, 13 November 2012 - 07:59 PM.
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