Uniform random points in a circle using polar coordinates

I need this seldom enough to forget how it's done - but then it's annoying to have to think/google for the solution again when I do need it... So I'll document here.

The task is to generate uniformly distributed numbers within a circle of radius R in the (x,y) plane. At first polar coordinates seems like a great idea, and the naive solution is to pick a radius r uniformly distributed in [0, R], and then an angle theta uniformly distributed in [0, 2pi]. BUT, you end up with an exess of points near the origin (0, 0)!  This is wrong because if we look at a certain angle interval, say [theta, theta+dtheta], there needs to be more points generated further out (at large r), than close to zero. The radius must not be picked from a uniform distribution, but one that goes as

pdf_r = (2/R^2)*r

That's easy enough to do by calculating the inverse of the cumulative distribution, and we get for r:

r = R*sqrt( rand() )

where rand() is a uniform random number in [0, 1]. Here is a picture:

fig2

some matlab code here.

The thinking for generating random points on the surface of a sphere in 3D is very similar. If I get inspired I will do a post on that later, meanwhile you can go read these lecture notes.

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9 Responses to “Uniform random points in a circle using polar coordinates”

  1. admin Says:

    Why is pdf_r= (2/R^2)*r ?

    In polar coordinates an infinitesimal area-element is: dr * r * dtheta

    if we want points uniformly distributed in this area the pdf for r should "clearly" be proportional to r.

    we also want the pdf to be normalized so that the integral from 0 to R of the pdf is 1. That leads to the pre-factor (2/R^2)

  2. Evan Says:

    Thanks. I too made the mistake but realized later that it would be more concentrated closer to the origin.

    Your code could be a lot more efficient by vectorizing instead of looping:

    % prepare
    Nmax = 1e3;
    R = 5;
    % Wrong
    r1 = R*rand(Nmax,1);
    theta1 = 2*pi*rand(Nmax,1);
    x1 = r1.*cos(theta1);
    y1 = r1.*sin(theta1);
    % Right
    r2 = R*sqrt(rand(Nmax,1));
    theta2 = 2*pi*rand(Nmax,1);
    x2 = r2.*cos(theta2);
    y2 = r2.*sin(theta2);
    % plot it
    subplot(1,2,1);
    plot(x1, y1, 'r.');
    axis square
    title('Left')
    subplot(1,2,2);
    plot(x2, y2, 'g.');
    axis square
    title('Right')

  3. Mariam Says:

    the vectoring instead of looping as shown by Evan is actually very effective to decrease the FLOPs of your code ...how ever i am getting the warning "Warning: Size vector should be a row vector with integer elements. "Warning: Size vector should be a row vector with integer elements. " for
    r1 = R*rand(Nmax,1);
    theta1 = 2*pi*rand(Nmax,1);
    and
    r2 = R*sqrt(rand(Nmax,1));
    theta2 = 2*pi*rand(Nmax,1);
    i know it can be ignored but its annoying me as to why am i getting it. any suggestions to overcome it?

  4. anders Says:

    Mariam: It's hard to say but my guess is you have something non-integer in the "Nmax" variable.

  5. Harold Says:

    As the rest of you here sound to be either math professors/majors or at least much more mathmatically versed than myself can someone explain this in layman's terms and/or write a step by step guide as to where to put the radius, amount of points, etc an how to calculate it.

  6. Branch Says:

    Thanks for the help! I had need of a uniform distribution inside an annulus (see the 'half-moon' classification in Neural Networks and Learning Machines by Haykin). Generalizing your result to the case of an annulus of inner radius A and outer radius B, I get

    r = sqrt( rand() × (B^2 − A^2) + A^2 )

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