Random point in circle method is not uniformly distributed

Question

I have the following method in Java:

public static Vector2d random(Circle circle) {
    // this returns a random number between 0 and Math.PI * 2
    double angle = MathUtils.random(0, Math.PI * 2);
    // give the point inside the unit circle
    // this returns a normalized vector from a given angle
    Vector2d point = new Vector2d(angle);
    // however, this is only along the edge
    // now add a random magnitude (because this is a normalized vector, we can just multiply it by the desired magnitude)
    double magnitude = Math.random();
    point = point.multiply(magnitude);
    // now expand this to fit the radius
    point = point.multiply(circle.getRadius());
    // now translate by circleCenter
    return point.add(circle.getCenter());
}

This does return a point in the defined circle, however, when you do this many times and plot the points, you can clearly see most points will be toward the center.

Why is this? I don't see how my math can do this.

Comment if you want me to add an image of the points on the plot, if you think that could be helpful.

Yeah, this happens for roughly the same reason that lines of longitude are closer together near the poles than near the equator. A really good way to get evenly distributed random points in a circle is to get random points in the circumscribed square, then reject the ones that fall outside the circle. — Dawood ibn Kareem, Aug 02 '17 at 03:24
Try this: Draw a circle (or imagine you are drawing one). Divide into 16 equal slices the way you would slice a pizza. Now, for each radius that goes from the center to the outer edge, divide the segment into 10 equal parts and put 10 evenly spaced dots along the radius. I think you'll notice that the points seem more bunched up near the center. That's effectively what your randomization method is doing. — ajb, Aug 02 '17 at 03:37
@ajb oh thanks, that helped me visualize it. now i understand why this happens — MCMastery, Aug 02 '17 at 03:39

score 2 · Answer 1 · answered Aug 02 '17 at 10:29

Of course, when r is small, the generated points are closer to each other.

As said by @DBrowne, you can adjust the density by the inverse CDF trick.

Alternatively, you can spare function evaluations by drawing uniform points in [-R,R]x[-R,R] and rejecting the ones such that X²+Y²>R² (about 21% of them). The method generalizes to any shape known by its implicit equation.

DBrowne · Answer 2 · 2017-08-02T05:30:27.907

Your math is flawed. Here's an explanation of why and the correct solution:

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]

http://www.anderswallin.net/2009/05/uniform-random-points-in-a-circle-using-polar-coordinates/

Random point in circle method is not uniformly distributed

2 Answers2