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I'm looking for efficient solution of following problem: For given set of points in n-dimensional euclidian space find such member of this set that minimizes total distance to other points in set.

The obvious naïve approach is quadratic, so I'm looking for something less than quadratic.

My first thought was that all I need is just to find the center of bounding sphere and then, find the closest point in set to this point. But this is actually not true, imagine right triangle - all its vertices are equidistant from such center, nevertheless, exactly one vertice meets our requirements.

It would be nice it one will shed some light on this issue.

shabunc
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    possible duplicate of [Algorithm to find point of minimum total distance from locations](http://stackoverflow.com/questions/8718086/algorithm-to-find-point-of-minimum-total-distance-from-locations) – BartoszKP Aug 31 '13 at 08:08
  • @BartoszKP, thank you for this link! Though I believe that it is a slightly different question. As far as I understand, in question asked there it is not supposed that the point should be the member of set. Though there is only one additional step - find closest point, it is asked about slightly different thing, and, therefore, there is a possibility that there are different solutions. – shabunc Aug 31 '13 at 08:19
  • Not sure either, however it's about the most convenient location from a set of given locations, so I guessed that's it's a point in the set. – BartoszKP Aug 31 '13 at 08:20
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    see http://cstheory.stackexchange.com/questions/9750/algorithm-to-find-a-point-among-n-points-in-plane-to-minimize-the-sum-of-distanc – Lior Kogan Sep 01 '13 at 19:13

2 Answers2

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What minimizes the distance to all of the points is their average. Only a guess, but after you'll find the average you could find a point closest to it. As correctly pointed out in comments, median instead of average will actually minimize the distance (average will minimize squared distance). Median can also be calculated in O(n). For high dimensional datasets this solution would be O(n*m) of course, where m is the number of dimensions.

Also some links:

See accepted answer here: Algorithm to find point of minimum total distance from locations

And link provided by mcdowella: http://en.wikipedia.org/wiki/Geometric_median

Community
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BartoszKP
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    The traditional average minimizes the sum of squared distances, as you can see by differentiating to solve the optimization problem. The geometric median minizies the sum of absolute differences - see http://en.wikipedia.org/wiki/Geometric_median – mcdowella Aug 31 '13 at 08:00
  • Getting the all-pair average is a quadratic problem, so you might go straight for the naïve approach. – Sergey Kalinichenko Aug 31 '13 at 08:01
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    Not all-pair avarage, just average over all coordinates. It's linear. – BartoszKP Aug 31 '13 at 08:02
  • I'm very thankful for you providing this links. But to be honest, it looks like I'm saying "I'm looking for algorithm for doing this kind of things" and you are telling me "Actually, you are looking for algorithm for finding N, here is an article about N on Wikipedia". But this does not answers how exactly I should solve the issue. The accepted answer in related question helps neither, since it is constructed in exactly same form. – shabunc Aug 31 '13 at 09:45
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I am making this up as I go along, but there appears to be a close connection between "best point of a set" and "best point" in convex optimization.

Your score function is a sum of distances. Each distance is convex U-shaped (OK V-shaped in this case) so their sum is convex U-shaped. In particular it has a perfectly good derivative everywhere except at points in the set, and this derivative is optimistic - if you take the value at a point and its derivative, neglecting any point at the point you are looking at, then predictions based on this will be optimistic - the line formed using the derivative lies almost entirely beneath the correct answer but grazes it at a single point.

This leads to the following algorithm:

Repeatedly

Pick a point at random and look to see if is the best point so far. If so, take note of it. Take the derivative of the sum of distances at this point. Use this, and the value at that point, to work out the predicted sum of distances at every other point and discard the points where this prediction is worse than the best answer so far as possible answers (although you still need to take them into account when working out distances and derivatives). These will be the points on the far side of a plane drawn through the chosen point normal to the derivative.

Now discard the chosen point as a contender as well and repeat if there are any points left to consider.

I would expect this to be something like n log n on randomly chosen points. However, if the set of points form the vertices of a regular polygon in n dimensions then it will cost N^2, discarding only the chosen point each time - any of the N points is in fact a correct answer and they all have the same sum of distances from each other.

I will of course up-vote anybody who can confirm or deny this general principle for finding the best of a set of given points under a convex objective function.

OK - I was interested enough in this to program this up - so I have 200+ lines of Java to dump in here if anybody cares. In 2 dimensions it's very fast, but at 20 dimensions you gain only a factor of two or so - this is reasonably understandable - each iteration cuts off points by projecting the problem down to a line and chopping off a fraction of the points outside the line. A randomly chosen point will be about half as far away from the centre as the other points - and very roughly you can expect the projection to cut off all but some multiple of the d-th root of 1/2 so as d increases the fraction of points you can discard in each iteration reduces.

mcdowella
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