Timeline for closest equidistant point to N points in M dimensions
Current License: CC BY-SA 4.0
5 events
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Jan 20, 2022 at 22:19 | comment | added | Saúl RM | The algorithm you mention only uses that the distance from your point to the rest of the points is equal, so it will obtain infinitely many solutions if $N<M+1$. This is the case when you are given three non-colinear points in $\mathbb{R}^3$, then there will be a whole line of points at the same distance of the three points. But only one of them is closest to the three points: in this case it can be obtained by intersecting the line of points equidistant to the three given points with the plane containing those points (this is what my answer generalizes). | |
Jan 20, 2022 at 11:05 | comment | added | Menachem | More concretely, I think the algorithm referenced above, that involves solving a system of linear equations, will work just as well for the $N < M+1$ case, though the matrix is no longer square. Assuming the system has a solution, then the point $p$ obtained is the center of the (translated) sphere, and hence the equidistant point you seek. | |
Jan 20, 2022 at 11:02 | comment | added | Menachem | If three distinct points lie on a line, then no point is equidistant to them. If they don't lie on a line, then a unique circle passes through them; its center is the unique equidistant point. The same is true in general. We need to check that the $N$ points are in "general position", i.e., that they don't lie on a hyperplane of dimension lower than $N-1$. Three points shouldn't lie on a line, four points shouldn't lie in a plane etc. If the $N$ points are in general position, then a unique hypersphere passes through them; its center is the equidistant point. | |
Jan 18, 2022 at 18:56 | comment | added | Erik | I like this solution very much for cases where N >= (M+1)! But not sure if I could expand it to cases where N < (M+1)... where there are more than one equidistant points, but one of them is closest? | |
Jan 18, 2022 at 9:14 | history | answered | Menachem | CC BY-SA 4.0 |