Find the diameter of a set of n points in d-dimensional space

Question

I am interesting in finding the diameter of two points sets, in 128 dimensions. The first has 10000 points and the second 1000000. For that reason I would like to do something better than the naive approach which takes O(n²). The algorithm will be able to handle any number of points and dimensions, but I am currently very interested in these two particular data sets.

I am very interesting in gaining speed over accuracy, thus, based on this, I would find the (approximate) bounding box of the point set, by computing the min and max value per coordinate, thus O(n*d) time. Then, if I find the diameter of this box, the problem is solved.

In the 3d case, I could find the diameter of the one side, since I know the two edges and then, I could apply the Pythagorean theorem on the other, which is vertical to this side. I am not sure for this however and for sure, I can't see how to generalize it to d dimensions.

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An interesting answer can be found here, but it seems to be specific for 3 dimensions and I want a method for d dimensions.

Interesting paper: On computing the diameter of a point set in high dimensional Euclidean space. Link. However, implementing the algorithm seems too much for me in this phase.

Answer 1

The classic 2-approximation algorithm for this problem, with running time O(nd), is to choose an arbitrary point and then return the maximum distance to another point. The diameter is no smaller than this value and no larger than twice this value.

Answer 2

I would like to add a comment, but not enough reputation for that...

I just want to warn other readers that the "bounding box" solution is very inaccurate. Take for example the Euclidean ball of radius one. This set has diameter two, but its bounding box is [-1, 1]^d, which has diameter twice the square root of d. For d = 128, this is already a very bad approximation.

For a crude estimate, I would stay with David Eisenstat's answer.

Answer 3

There is a precision based algorithm which performs very well on any dimension, which is based on computing the dimension of an axial bounding box.

The idea is that it's possible to find the lower and upper boundaries of the axis bounding box length function since it's partial derivatives are limited, and depend on the angle between the axises.

The limit of the local maxima derivatives between two axises in 2d space can be computed as:

sin(a/2)*(1 + tan(a/2)) That means that, for example, for 90deg between axises the boundary is 1.42 (sqrt(2))

Which reduces to a/2 when a => 0, so the upper boundary is proportional to the angle.

For a multidimensional case the formula varies slightly, but still it's easy to compute.

So, the search of local minima convolves in logarithmic time.

The good news is that we can run the search of such local maxima in parallel. Also, we can filter out both the regions of the search based on the best achieved result so far, as well as the points themselves, which are belo the lower limit of the search in the worst region.

The worst case of the algorithm is where all of the points are placed on the surface of a sphere.

This can be firther improved: when we detect a local search which operates on just few points, we swap to bruteforce for this particular axis. It works fast, because we need only the points which are subject to that particular local search, which can be determined as points actually bound by two opposite spherical cones of a particular angle sharing the same axis.

It's hard to figure out the big O notation, because it depends on desired precision and the distribution of points (bad when most of the points are on a sphere's surface).

The algorithm i use is here:

1. Set the initial angle a = pi/2.
2. Take one axis for each dimension. The angle and the axises form the initial 'bucket'
3. For each axis, compute the span on that axis by projecting all the points onto the axis, and finding min and max of the coordinates on the axis.
4. Compute the upper and lower bounds of the diameter which is interesting. It's based on the formula: sin(a/2)*(1 + tan(a/2)) and multiplied by assimetry cooficient, computed from the length of the current axis projections.
5. For the next step, kill all of the points which fall under the lower bound in each dimension at the same time.
6. For each exis, If the amount of points above the upper bound is less then some reasonable amount (experimentally computed) then compute using a bruteforce (N^2) on the set of the points in question, and adjust the lower bound, and kill the axis for the next step.
7. For the next step, Kill all of the axises, which have all of their points under the lower bound.
8. If the precision is satisfactory (upper bound - lower bound) < epsilon, then return the upper bound as the result.
9. For all of the survived axises, there is a virtual cone on that axis (actually, the two opposite cones), which covers some area on a virtual sphere which encloses a face of the cube. If i'm not mistaken, it's angle would be a * sqrt(2). Set the new angle to a / sqrt(2). Create a whole bucket of new axises (2 * number of dimensions), so the new cone areas would cover the initial cone area. It's the hard part for me, as i have not enough imagination for n>3-dimensional case.
10. Continue from step (3).

You can paralellize the procedure, synchronizing the limits computed so far for the points from (5) through (7).