Create random points in a rectangular domain, but with minimum separation distance
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Hello,
I am not a regular user of this program. Hence this query.
I am interested to create random points in a 2D rectangular space, to start with, a uniform distribution. I know how to do this. However, I want to ensure that each of these points are separated by a certain minimum distance.
Can anyone help me with the script please.
Thank you.
Suresh
采纳的回答
Image Analyst
2014-10-12
You basically have to try and see. If it's too close, reject that point and get a new one.
x = rand(1, 10000);
y = rand(1, 10000);
minAllowableDistance = 0.05;
numberOfPoints = 300;
% Initialize first point.
keeperX = x(1);
keeperY = y(1);
% Try dropping down more points.
counter = 2;
for k = 2 : numberOfPoints
% Get a trial point.
thisX = x(k);
thisY = y(k);
% See how far is is away from existing keeper points.
distances = sqrt((thisX-keeperX).^2 + (thisY - keeperY).^2);
minDistance = min(distances);
if minDistance >= minAllowableDistance
keeperX(counter) = thisX;
keeperY(counter) = thisY;
counter = counter + 1;
end
end
plot(keeperX, keeperY, 'b*');
grid on;
7 个评论
SS
2014-10-12
Image Analyst
2014-10-12
See this code:
% Creates a random pattern of points with no point closer to another than some specified distance.
% Also puts a ray emanating from each point at random angle and random length.
% Initialize some things.
x = rand(1, 10000);
y = rand(1, 10000);
minAllowableDistance = 0.05;
numberOfPoints = 300;
rMax = 0.25;
% Initialize first point.
keeperX = x(1);
keeperY = y(1);
% Try dropping down more points.
counter = 2;
for k = 2 : numberOfPoints
% Get a trial point.
thisX = x(k);
thisY = y(k);
% See how far is is away from existing keeper points.
distances = sqrt((thisX-keeperX).^2 + (thisY - keeperY).^2);
minDistance = min(distances);
if minDistance >= minAllowableDistance
keeperX(counter) = thisX;
keeperY(counter) = thisY;
% Draw a line from it.
% First get a random angle
angle = 2*pi*rand(1);
% Next get a random length
r = rMax * rand(1);
% Now get endpoint of the ray.
x2 = thisX + r * cos(angle);
y2 = thisY + r * sin(angle);
% Plot the point
plot(keeperX(counter), keeperY(counter), 'bo', 'LineWidth', 3);
hold on;
% Draw the line.
line([thisX x2], [thisY, y2], 'Color', 'r', 'LineWidth', 3);
counter = counter + 1;
end
end
grid on;
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
SS
2014-10-13
Prashant Katiyar
2017-3-8
Is there anyway we can replace points to small circles of variable diameters? Also if we can variate the distance between each of these circles? Some thing like below?
William Harvie
2022-4-27
ngl this is probably the least effecient way to do this but it works
Image Analyst
2022-4-27
@William Harvie, please post a more efficient way to do it if you know of one.
@Prashant Katiyar, try this:
%============================================================================================================================================
% Initialization Steps.
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 25;
% Get a list of trial (x,y) circle centers.
x = 100 * rand(1, 1000000);
y = 100 * rand(1, 1000000);
% Specify how many circles are desired.
numberOfCircles = 300;
minRadius = 1;
maxRadius = 5;
% Specify how close the centers may be to each other.
% Should be at least twice the max radius if they are not to touch.
minAllowableDistance = max([11, 2 * maxRadius]);
% Initialize first point.
keeperX = x(1);
keeperY = y(1);
radii = minRadius;
% Try dropping down more points.
counter = 2;
for k = 2 : length(x)
% Get a trial point.
thisX = x(k);
thisY = y(k);
% See how far is is away from existing keeper points.
distances = sqrt((thisX-keeperX).^2 + (thisY - keeperY).^2);
minDistance = min(distances);
if minDistance >= minAllowableDistance
% This trial location works. Save it.
keeperX(counter) = thisX;
keeperY(counter) = thisY;
% Get a random radius.
radii(counter) = minRadius + (maxRadius - minRadius) * rand;
% Quit if we have enough or ran out of circles to try
if counter >= numberOfCircles
break;
end
counter = counter + 1;
end
end
% Plot a dot at the centers
plot(keeperX, keeperY, 'b+', 'MarkerSize', 9);
viscircles([keeperX(:), keeperY(:)], radii, 'Color', 'b');
grid on;
axis equal
numCirclesPlaced = length(keeperX);
caption = sprintf('Could place %d circles', numCirclesPlaced);
title(caption, 'fontSize', fontSize)
g = gcf;
g.WindowState = 'maximized'
Joan
2022-7-1
@Image Analyst, thank you very much for your code. It is showing me some direction. However, I am a bit stuck. Please add some code that would make the circles enclosed inside a polyshape of coordinates say,
xv=[0 0 20 35 66 85 105 105]; % X-Coordinates
yv=[0 20 20 40 40 20 16 0]; % Y-Coordinates
更多回答（2 个）
SS
2014-11-1
0 个投票
1 个评论
Image Analyst
2014-11-1
I suggest you post this in a brand new question. But try the ellipsoid function first using the "try and keep/reject" concept. If you can't get it working, post your code in a new question since it's a different topic than this one.
SS
2014-11-1
0 个投票
2 个评论
Image Analyst
2014-11-1
编辑：Image Analyst
2014-11-1
Yeah but I see you didn't take my advice, so expect a lot of questions like "Do the axes of the ellipsoids have to align with the xyz axes?", "What have you tried?", "Have you seen the ellipsoid function?", "Is this a homework problem?", and so on.
SS
2014-11-2
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