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#1 2006-10-10 03:50:27

claudio77
Member
Registered: 2006-10-10
Posts: 1

uniform point distribution in cube

Hi friends,

I've this kind of problems.
Supposing to have a cube volume in the (x,y,z) space with width equal to 1. Supposing also 10 points uniform division of each width along x,y,z respectively. By crossing the three points divisions this means that the cube contains 10x10x10=1000 points (I hope dizzy).

Given an arbitrary number N, among these 1000 points i must guess the N points which cover the volume in an "uniform" way which means, for instance:
- if N=1, i must guess the central point
- if N=2, ... 2 edges
- if N=3, ... 2 edges an the central point
- if N=4, ... 4 edges
- if N=5, ... 4 edges + central point
- and so on

If I correctly remember this should be related to some "maximum entropy" fundaments.
Does anyone know how can I develop a basic algorithm to implemet this procedure?
Considering also that I aim to deal with space with arbitrary dimension.
Thanks and regards.
Claudio

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#2 2006-10-10 16:51:34

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: uniform point distribution in cube

If you use 10 points, there are nine spaces between them, if that's okay with you, I like it.
If you use 11 points, then there is a central point, however.


igloo myrtilles fourmis

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#3 2006-10-10 16:51:49

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: uniform point distribution in cube

For small N, your points will be equivalent under translations/rotations to one or more of the basic crystal structures, see http://cst-www.nrl.navy.mil/lattice/

There are guaranteed to be equations you can find for the locations of the points of these structures if you poke around online.

Or, if you're trying to deal with large N and you want to use a computer, you could start out with your N points scattered at random around your cube. Then use some simple algorithm to move each point away from its nearest neighbors in small iterations until they're pretty uniformly distributed. I don't know what the best algorithm would be ... you'd probably have to play around with some ideas until you find one that converges.

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