Generate a random point within a circle (uniformly)

Question

I need to generate a uniformly random point within a circle of radius R.

I realize that by just picking a uniformly random angle in the interval [0 ... 2π),

Answer 1

A programmer solution:

• Create a bit map (a matrix of boolean values). It can be as large as you want.
• Draw a circle in that bit map.
• Create a lookup table of the circle's points.
• Choose a random index in this lookup table.

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

bool matrix[MATRIX_SIZE][MATRIX_SIZE] = {0};

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        matrix[x][y] = true;

        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;

      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

The bitmap is only necessary for the explanation of the logic. This is the code without the bitmap:

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;
      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

Answer 2

I think that in this case using polar coordinates is a way of complicate the problem, it would be much easier if you pick random points into a square with sides of length 2R and then select the points (x,y) such that x^2+y^2<=R^2.

Answer 3

1) Choose a random X between -1 and 1.

var X:Number = Math.random() * 2 - 1;

2) Using the circle formula, calculate the maximum and minimum values of Y given that X and a radius of 1:

var YMin:Number = -Math.sqrt(1 - X * X);
var YMax:Number = Math.sqrt(1 - X * X);

3) Choose a random Y between those extremes:

var Y:Number = Math.random() * (YMax - YMin) + YMin;

4) Incorporate your location and radius values in the final value:

var finalX:Number = X * radius + pos.x;
var finalY:Number = Y * radois + pos.y;

Answer 4

Solution in Java and the distribution example (2000 points)

public void getRandomPointInCircle() {
    double t = 2 * Math.PI * Math.random();
    double r = Math.sqrt(Math.random());
    double x = r * Math.cos(t);
    double y = r * Math.sin(t);
    System.out.println(x);
    System.out.println(y);
}

based on previus solution https://stackoverflow.com/a/5838055/5224246 from @sigfpe

Answer 5

First we generate a cdf[x] which is

The probability that a point is less than distance x from the centre of the circle. Assume the circle has a radius of R.

obviously if x is zero then cdf[0] = 0

obviously if x is R then the cdf[R] = 1

obviously if x = r then the cdf[r] = (Pi r^2)/(Pi R^2)

This is because each "small area" on the circle has the same probability of being picked, So the probability is proportionally to the area in question. And the area given a distance x from the centre of the circle is Pi r^2

so cdf[x] = x^2/R^2 because the Pi cancel each other out

we have cdf[x]=x^2/R^2 where x goes from 0 to R

So we solve for x

R^2 cdf[x] = x^2

x = R Sqrt[ cdf[x] ]

We can now replace cdf with a random number from 0 to 1

x = R Sqrt[ RandomReal[{0,1}] ]

Finally

r = R Sqrt[  RandomReal[{0,1}] ];
theta = 360 deg * RandomReal[{0,1}];
{r,theta}

we get the polar coordinates {0.601168 R, 311.915 deg}

Answer 6

Here is a fast and simple solution.

Pick two random numbers in the range (0, 1), namely a and b. If b < a, swap them. Your point is (b*R*cos(2*pi*a/b), b*R*sin(2*pi*a/b)).

You can think about this solution as follows. If you took the circle, cut it, then straightened it out, you'd get a right-angled triangle. Scale that triangle down, and you'd have a triangle from (0, 0) to (1, 0) to (1, 1) and back again to (0, 0). All of these transformations change the density uniformly. What you've done is uniformly picked a random point in the triangle and reversed the process to get a point in the circle.