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Subject: Re: nearest neighbour of a point



In comp.graphics.algorithms running.polygon@gmail.com wrote:

> I have 2 sets of points X & Y. The distribution of points in both the
> sets is such that points at a certain distance are concentrated and on
> either side of this distance, they start becoming sparce, something
> like a normal distribution.

This description is a bit hard to parse. You seem to be saying that
within set X, distances to the nearest neighbor cluster around a
certain value, rather separated from zero. And the same for set Y.
Or is this clustering true for *all* distances between points in one
set? I.e. are all points of set X expected to fall roughly on a
spherical shell of radius R around any point in set X one picks as a
reference?

The former case would indicate that your pointset is some kind of
regular grid embedded in that high-D space, and it may be a grid of
considerably fewer actual dimensions than the embedding space. You
could find out by counting the number of almost-equally-far neighbors
per point. E.g. a regular "grid" of points along a curve in 3D space
would yield 2 close neighbors per point, a gridded surface would have
4, and a volumetric grid would have 6.

E.g. if you find an average of 2*d close neighbors per vertex, your
pointset might be a grid on a d-dimensional manifold, rather than a
truly N-dimensional data set. If d is a good deal smaller than N,
that could ease the search algorithms a lot.

> The points are in higher dimentions > 3. They may be up to 50
> dimentions.

In that case, and assuming the argument above doesn't apply, odds are
that there's no algorithm that could improve upon the naive "compute
all distances and remember the shortes" approach. 2^50 is guaranteed
to be much larger than the number of points you have, so no kind of
subdivision method can really help you, if the dataset's actual
dimensionality is 50. The fact to remember is that in
high-dimensional spaces, everything is pretty much next to everything
else, so the only way to find the closest thing is to check
everything.

You can do the maths yourself: compare the volume of a 50-dimensional
hyper-spherical shell whose radius and thickness are the center and
width of your distance distribution, to that of a 50-dimensional
hypersphere of said radius. You'll find that the shell's volume is
not really much different from that of the full sphere. I.e. the
restriction to a thin spherical shell isn't going to help you much in
searching or organizing points.

--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

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