How to generate uniformly distributed random coordinates inside a c... (2024)

Abhiram B R on 8 Dec 2021

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Edited: Walter Roberson on 21 Mar 2022

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I want to generate uniform randomly distributed points inside a cube. The points should also satisfy a condition that they should be atleast a constant distance apart. I have written a matlab code, but the end result is not uniformly distributed points. I think the issue is that i am creating one random point at a time. Please have a look at the attached script. Any suggestions?

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% matlab code to generate random coordinates for inclusions

% translation of inclusions into polymer box

% here uniform distribution is considered

% polymer box size 80*80*100

% inclusions should not overlap each other

% inclusions should not cross the boundary too

% in order to avoid inclusions from crossing the boundary, consider a

% shrinked box dimension

% all units in Angstroms

% size of inclusion 8*8*10

% RVE dimension

xlo = 0;

xhi = 80;

ylo = 0;

yhi = 80;

zlo = 0;

zhi = 100;

% inclusion dimension

xhi_incl = 8;

ylo_incl = 0;

yhi_incl = 8;

zlo_incl = 0;

zhi_incl = 10;

% half the diagonal distance of inclusion should be avoided from RVE size

% to prevent inclusions from overlapping the boundary

% tolerance of 1 Angstroms is provided

point1 = [xlo_incl ylo_incl zlo_incl];

point2 = [xhi_incl yhi_incl zhi_incl];

dia_dist = norm(point2-point1);

trim_tolerance = 1;

trim_dist = dia_dist*0.5+trim_tolerance;

% actual dimension to work with

xlo_trim = xlo+trim_dist;

xhi_trim = xhi-trim_dist;

ylo_trim = ylo+trim_dist;

yhi_trim = yhi-trim_dist;

zlo_trim = zlo+trim_dist;

zhi_trim = zhi-trim_dist;

% generate random numbers

num_incl = 8;

dia_tolerance = 3; % tolerance value for seperation of RVE(angstroms)

dia_tot = dia_dist+dia_tolerance;

x_rand = xlo_trim+(xhi_trim-xlo_trim)*rand(1,1);

y_rand = ylo_trim+(yhi_trim-ylo_trim)*rand(1,1);

z_rand = zlo_trim+(zhi_trim-zlo_trim)*rand(1,1);

rand_coord = zeros(num_incl,3);

rand_coord(1,:) = [x_rand y_rand z_rand];

i = 1;

while(i<8)

check_matrix =(dia_tot+1);

x_rand1 = xlo_trim+(xhi_trim-xlo_trim)*rand(1,1);

y_rand1 = ylo_trim+(yhi_trim-ylo_trim)*rand(1,1);

z_rand1 = zlo_trim+(zhi_trim-zlo_trim)*rand(1,1);

coord_new = [x_rand1 y_rand1 z_rand1];

for j=1:i

coord_new;

rand_coord(j,:);

check_dist(j) = norm(coord_new-rand_coord(j,:));

end

check_dist;

A = all(check_matrix(:) > dia_tot);

if A > 0

rand_coord(i+1,:) = coord_new;

i = i+1;

end

end

4 Comments

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Walter Roberson on 8 Dec 2021

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If the requirement is that each point be at least a particular distance from every other point, then the problem becomes equivalent to sphere packing in a cube that is increased in each dimension by the minimum distance. Or, more simply, just sphere packing where you ensure that the test coordinates are all inside the cube and you do not check for hitting the outside boundary of the cube.

There have been several posts about sphere packing, either with uniform radius, or with radius chosen from a set of possibilities, or with radius confined to a range of values. Some of the posts have been about packing in cuboids, and some of them have been about packing in cylinders.

Abhiram B R on 8 Dec 2021

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Thanks Walter for your comment. Yes it is similar to packing inside a cuboid. One major difference here is that, the points should also follow a distribution. (uniform or gaussian)

Thanks

Walter Roberson on 8 Dec 2021

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Check out the previous posts on the topic of sphere packing, and you will find that I (among other people) have posted with uniform distribution in space algorithms -- at least uniform distribution in the places that it attempts the place the points, with there being no certainty that the resulting packing will be uniformly distributed !

In particular if you look at the one that was modified not long ago for placement into a cylinder, you will see a specific demonstration there that the algorithm that I used for placing inside the cylinder results in uniform spatial distribution of attempt points (placing uniformly inside a circular cross-section requires a mathematical trick.)

John D'Errico on 16 Dec 2021

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Edited: John D'Errico on 16 Dec 2021

If the points are truly uniformly distributed, then there is no theortetical minimum distance between pairs of points. Does the fact that you are generating them one at a time matter? No. The fallacy lies in your expectations, that a uniform distribution has no points that lie closely together.

As Walter has pointed out, this becomes a sphere packing problem, but sphere packing is not so trivial.

Simple schemes like rejection sampling can work, but don't expect to get any kind of dense sampling that might be even close to optimal, certainly not in a reasonable amount of time. You can also try things like starting with a random sample, then perturbing points to move away from each other when they are too close or too far from their nearest neighbors. You could think of that in terms of a force on each point, that depends upon the distance to their nearest neighbor.

Of course that result will no longer be truly uniformly distributed. However, it may be the best thing you can do in a reasonable time if you want some sort of fairly dense packing subject to a minimum distance.

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