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Jan 30, 2011 at 23:21 answer added sclv timeline score: 0
Jan 30, 2011 at 21:15 comment added user3548 Hi Oosterwal, SqrRoot((Xc-Xa)^2 + (Yc-Ya)^2) < r Where: Xc = (set of x coordinates of all points) - specific (Xa) coordinate Yc = (set of y coordinates of all points) - specific (Ya) coordinate r = radius distance required. Note that the formula using '<' is obviously meaning 'Within'. Use '<=' if the points can be equal to the radius in question. Progammatically: for each s in S Select * from S Where sqrt((S(Xc)-s(Xa)^2 + (S(Yc)-s(Ya))^2)) < r This will work for two-Dimensional set of points with a radial distance.
Jan 20, 2011 at 4:13 answer added Sariel Har-Peled timeline score: 4
Jan 18, 2011 at 22:55 history tweeted twitter.com/#!/StackCSTheory/status/27499383931011072
Jan 18, 2011 at 16:06 answer added oosterwal timeline score: 2
Jan 18, 2011 at 15:40 answer added Vinicius dos Santos timeline score: 8
Jan 18, 2011 at 15:31 comment added Radu GRIGore en.wikipedia.org/wiki/Nearest_neighbor_search
Jan 18, 2011 at 14:52 comment added user3317 Another thing: Other assumptions are possible for us, as are approximations. Most of all, I just want to know what kind of solutions are already out there, and where to look for them.
Jan 18, 2011 at 14:34 comment added user3317 That's true. I should have mentioned that for us usually $\rho$ is much smaller than the diameter of $S$. If the maximum number (over $s\in S$ of points in $S$ within $\rho$ of $s$ is $M$, then we'd ideally like an algorithm that is $O(M|S|)$. Or something like that.
Jan 18, 2011 at 14:29 history edited user3317 CC BY-SA 2.5
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Jan 18, 2011 at 14:26 comment added Peter Taylor Isn't the output potentially of size $|S|^2$?
Jan 18, 2011 at 14:22 history asked user3317 CC BY-SA 2.5