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Find n random points with a minimum distance r inside a 2D rectangular box

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jaydeep
jaydeep on 18 Jul 2014
Edited: Jan on 21 Feb 2017
I would like to generate the (x, y) coordinates of N randomly distributed points within a 2D Square box having 2000 m X 2000m. the points must have a minimum distance of 200m from each other, means a second point should not lies between the area of circle covered by the first point and so on. Any idea for a uniform random distribution?

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Answers (3)

Doug Hull
Doug Hull on 18 Jul 2014
Edited: Doug Hull on 18 Jul 2014
Here is something stupid that just might work.
While NOT_DONE
generate a point
If NOT too close to existing points
Place point
end
check to see if enough points are placed.
end

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Jan
Jan on 12 Feb 2017
Edited: Jan on 21 Feb 2017
function [X, Y, D] = GetPointsRandom(nWant, XWidth, YWidth, MinDist)
X = zeros(nWant, 1);
Y = zeros(nWant, 1);
dist_2 = MinDist ^ 2; % Squared once instead of SQRT each time
iLoop = 1; % Security break to yoid infinite loop
nValid = 0;
while nValid < nWant && iLoop < 1e6
newX = XWidth * rand;
newY = YWidth * rand;
if all(((X(1:nValid) - newX).^2 + (Y(1:nValid) - newY).^2) >= dist_2)
% Success: The new point does not touch existing points:
nValid = nValid + 1; % Append this point
X(nValid) = newX;
Y(nValid) = newY;
end
iLoop = iLoop + 1;
end
% Throw an error, if the area is filled too densely:
if nValid < nWant
error('Cannot find wanted number of points in %d iterations.', iLoop)
end
if nargout > 2
% D = pdist([X, Y]); % Faster with statistics toolbox
D = sqrt(bsxfun(@minus, X, X.') .^ 2 + bsxfun(@minus, Y, Y.') .^ 2);
end
end

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John BG
John BG on 12 Feb 2017
Edited: John BG on 20 Feb 2017
Hi Jaydeep
just completed a couple functions that may help with your question
1.
download the functions scatter_points7.m and scatter_points_saturate.m
from
2.
run this
[X,Y,Nmax,Dmatrix]=scatter_points7
3.
key in the menu the rectangle size, the amount of points you want and the minimum distance.
4.
scatter_points7 returns:
X Y coordinates of the random points
Nmax the maximum amount of points that would fit in the rectangle if placing them orderly.
Dmatrix the distances between all combinations of points.
5.
Check the minimum distance requirement is met with
% test 1
Ap=20;L=combinator(Ap,2,'c');
relD2=((X(L(:,2))-X(L(:,1))).^2+(Y(L(:,2))-Y(L(:,1))).^2).^.5
R0=200;find(relD2<R0)
=
Empty matrix: 1-by-0
or
% test 2
Ap=20;L2=combinator(Ap,2);
D2=Dmatrix+NaN*eye(Ap);
R0=200;D2(D2<R0)
=
Empty matrix: 0-by-1
.
'
6.
Now try this
[X,Y,Nmax,Dmatrix]=scatter_points_saturate(2000,2000,200)
.
.
to know the amount of points (below Nmax) actually generated
numel(X)
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG

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