Creating random points in a circle

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Stephen Johnson
Stephen Johnson on 20 Apr 2013
Commented: Walter Roberson on 15 Aug 2023
I am new to MatLab but I am trying to write code for a problem and a couple of things are sticking me up. I am trying to create X number of random points within a circle. The purpose being to model a camp fire and map temperature from the center of the circle at the hottest to the outside. And I am lost. And I can't find an example anywhere I might be able to pull apart and work with. And help would be very much appreciated.
Thanks!

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Accepted Answer

Roger Stafford
Roger Stafford on 19 May 2013
Let the circle be of radius R and center at (x0,y0). Let n be the desired number of points within.
t = 2*pi*rand(n,1);
r = R*sqrt(rand(n,1));
x = x0 + r.*cos(t);
y = y0 + r.*sin(t);
plot(x,y)
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Wen Wei
Wen Wei on 15 Aug 2023
Edited: Wen Wei on 15 Aug 2023
@Jan Whether you're sampling from a half-closed [0,1) or open (0,1) or closed [0,1] interval is irrelevant, since measure-theoretically the measure of the end points are 0. That is to say, the event (exactly t = 0 or t = 1) occuring has probability 0.
So your statement that the difference is "tiny" is not quite right: the difference is ignorable, in a rigorous mathematical sense. (See Wiki page on measure theory/probability theory)
Walter Roberson
Walter Roberson on 15 Aug 2023
@Wen Wei
measure-theoretically the measure of the end points is not zero. None of the algorithms for rand() generates to infinite precision: they generate to at most 53 bits. The results of rand() are, at best, the same as the result of randi([1, 2^53-1]) / 2^53 . In terms of 53 bit precision, the 0 would be 1 value out of exactly 2^53 if you exclude 2^53 itself (because 0 to 2^53 is (2^53 + 1) distinct values); likewise if you exclude 0 but include 2^53 itself in the random integer to allow random value of exactly 1.0 to be generated, then that would be probability 1/2^53 . If you were to allow both exactly 0 and exactly 1 to be generated from rand, then the probability of 0 would be 1/(2^53+1) and likewise the probability of 1 would be the same.
The events (exactly t = 0 or t = 1) are only probability 0 if you are drawing from an infinite number of values... but you are not, since finite (fixed) precision results are being returned.

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More Answers (2)

Image Analyst
Image Analyst on 20 Apr 2013
Here, see my demo:
% M-file to place multiple points inside a big circle.
% Clean up
close all;
clc;
fontSize = 15;
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
% Initialize some parameters.
numberOfPoints = 25; % Number of small circles
bigImageWidth = 500;
bigImageHeight = 500; % square area 0f 500*500
bigCircleRadius = 250; % big circle radius
% Initialize an image to hold one single big circle.
bigCircleImage = zeros(bigImageHeight, bigImageWidth, 'uint8');
[x, y] = meshgrid(1:bigImageWidth, 1:bigImageHeight);
bigCircleImage((x - bigImageWidth/2).^2 + (y - bigImageHeight/2).^2 <= bigCircleRadius.^2) = 1;
clear('x', 'y'); % Release these variables, they're not needed anymore.
% Display it in the upper left plot.
subplot(2,2, 1);
imshow(bigCircleImage, []);
title('Big Circle Mask', 'FontSize', fontSize);
set(gcf, 'Position', get(0,'Screensize')); % Maximize figure.
% Initialize an output image to hold many small overlapping circles.
pointsImage = zeros(bigImageHeight, bigImageWidth, 'uint8');
% Get linear indexes of 500 randomly located in the rectangle.
numberOfPointsToPlace = 5000;
linearIndexes = randi(numel(pointsImage), numberOfPointsToPlace, 1);
% Set those points in the image
pointsImage(linearIndexes) = 255;
% Get locations in terms of row and columns:
[rows, columns] = ind2sub(size(pointsImage), linearIndexes);
% Display it in the lower left plot.
subplot(2,2, 2);
imshow(pointsImage);
title('Many Points', 'FontSize', fontSize);
% Multiply the big circle mask by the points image to clip
% those points that lie outside the big circle.
maskedByBigCircle = bigCircleImage .* pointsImage;
% Display it in the lower right plot.
subplot(2,2, 3);
imshow(maskedByBigCircle);
title('Many Points Masked by Big Circle', 'FontSize', fontSize);
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Oogboga
Oogboga on 5 Aug 2021
Hello, this is similar to a problem i am facing, maybe you can help?
The code below is a function that asks for a radius, and then generates a circle with random points within +/-10% of the radius. It also generates 2 circles +and-5% from the original circle.
function missrate=Final_radiationring(r)
clc
theta=0:0.1:2*pi;
%r=20;
q='Enter radius: ';
r=input(q);
xcenter=0;
ycenter=0;
xcord= xcenter+r*cos(theta);
ycord=ycenter+r*sin(theta);
plot(xcord,ycord,'b')
hold on
xchigh=xcenter+(r+(r/20))*cos(theta);
xclow=xcenter+(r-(r/20))*cos(theta);
ychigh=ycenter+(r+(r/20))*sin(theta);
yclow=ycenter+(r-(r/20))*sin(theta);
plot(xchigh,ychigh, 'm')
plot(xclow,yclow, 'g')
n = 300;
x0 = 0; % Center of the circle in the x direction.
y0 = 0; % Center of the circle in the y direction.
% Create set of points
rng default
t = 2*pi*rand(n,1);
g = 0.9*r+0.2*r*rand(n,1); %generate random points in boundary
x = x0 + g.*cos(t);
y = y0 + g.*sin(t);
% Now display our random set of points in a figure with circle
d=sqrt((x.^2)+(y.^2));
plot(x,y,'*r')
hold off
axis equal
grid on;
d
end
the "d" variable gives the distances of each point from the center of the circle. My goal is to somehow index and plot the ones outside of the circles seperately from the ones inside the circles.
any help is greatly appreciated.
Image Analyst
Image Analyst on 5 Aug 2021
Try this:
% Demo by Image Analyst.
clc; % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all; % Close all figures (except those of imtool.)
clearvars;
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;
r = 2;
missrate=Final_radiationring(r);
fprintf('Done running %s.m\n', mfilename);
function [missrate, radiusList] = Final_radiationring(r)
missrate = 0;
% Make 360 points around the circle.
numCirclePoints = 360;
theta = linspace(0, 2 * pi, numCirclePoints);
% Get radius, ignoring the one passed in.
%r=20;
q='Enter radius: ';
r=input(q);
xcenter=0;
ycenter=0;
radiusList = [r, r + r/20, r - r/20];
xcord = xcenter + r * cos(theta);
ycord = ycenter + r * sin(theta);
plot(xcord, ycord, 'b-', 'LineWidth', 2)
hold on
xchigh=xcenter + radiusList(2)*cos(theta);
xclow=xcenter + radiusList(3)*cos(theta);
ychigh=ycenter + radiusList(2)*sin(theta);
yclow=ycenter + radiusList(3)*sin(theta);
plot(xchigh,ychigh, 'm-', 'LineWidth', 2)
plot(xclow,yclow, 'g-', 'LineWidth', 2)
grid on;
numRandomPoints = 300;
x0 = 0; % Center of the circle in the x direction.
y0 = 0; % Center of the circle in the y direction.
% Create a noisy set of points
rng default
t = 2 * pi * rand(numRandomPoints,1);
g = 0.9 * r + 0.2 * r * rand(numRandomPoints,1); %generate random points in boundary
x = x0 + g.*cos(t);
y = y0 + g.*sin(t);
% Now display our random set of points in a figure with circle
radialDistances = sqrt(x.^2 + y.^2);
colors = ['r', 'g', 'b', 'm']
radiusList = [0, sort(radiusList, 'ascend'), inf];
for k = 1 : length(radiusList) - 1
r1 = radiusList(k);
r2 = radiusList(k+1);
logicalIndexes = radialDistances >= r1 & radialDistances < r2;
thisx = x(logicalIndexes);
thisy = y(logicalIndexes);
thisColor = colors(k);
plot(thisx, thisy, '.', 'Color', thisColor, 'MarkerSize', 20)
end
hold off
axis equal
grid on;
radialDistances
end

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Irfan
Irfan on 31 Jul 2013
Hi, Roger, Why did you use sqrt in r = R*sqrt(rand(n,1)); it should be without sqrt, is'nt ?
  1 Comment
Roger Stafford
Roger Stafford on 31 Jul 2013
That is untrue! Without the square root operation, the distribution of points within the circle will not be uniformly distributed throughout its interior. Just try it out with a large value of n, say, n = 1000 and see.
Think of it this way, Irfan. The area of a circle with the same center and with half the given radius, R, is one-fourth that of the larger circle. Therefore the probability that we get an r value less than or equal to R/2 should be one-fourth. That is, a 'rand' value of 1/4 should give rise to a value r = R/2. This is accomplished by taking the square root of the 'rand' value. A similar reasoning applies to other fractional circles.

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