Calculate random points (pixel) within a circle (image)

Question

I have an image that contains a circles at a specific location, and of a specific diameter. What I need to do is to be able to calculate random points within the circle, and then manipulate the pixels said points correlate to. I have the following code already:

private Point CalculatePoint()
{
    var angle = _random.NextDouble() * ( Math.PI * 2 );
    var x = _originX + ( _radius * Math.Cos( angle ) );
    var y = _originY + ( _radius * Math.Sin( angle ) );
    return new Point( ( int )x, ( int )y );
}

And that works fine for finding all the points at the circumference of the circle, but I need all points from anywhere in the circle. If this doesn't make sense let me know and I will do my best to clarify.

Answer 1

If you want a simple solution just randomize the radius too:

private Point CalculatePoint()
{
    var angle = _random.NextDouble() * Math.PI * 2;
    var radius = _random.NextDouble() * _radius;
    var x = _originX + radius * Math.Cos(angle);
    var y = _originY + radius * Math.Sin(angle);
    return new Point((int)x,(int)y);
}

That however results in your points being more concentrated towards the center of the circle:

In order to get an uniform distribution make the following change to the algorithm:

var radius = Math.Sqrt(_random.NextDouble()) * _radius;

Which will give the following result:

For more information check the following link: MathWorld - Disk Point Picking.

And finally here's a simple JsFiddle demonstration comparing both version of the algorithm.

Answer 2

DO NOT just use random r and theta! This creates a weighted distribution with more points in the center. This page illustrates it well...

http://mathworld.wolfram.com/DiskPointPicking.html

Here's the method that creates a non-weighted distribution...

var r = rand(0,1)
var theta = rand(0,360)

var x = sqrt(r) * cos(theta)
var y = sqrt(r) * sin(theta)

Answer 3

You’re halfway there. In addition to generating a random angle, just generate a random distance, less than or equal to the radius, weighted so that you get a uniform distribution:

private Point CalculatePoint()
{
    var angle = _random.NextDouble() * Math.PI * 2;
    var distance = Math.Sqrt(_random.NextDouble()) * _radius;
    var x = _originX + (distance * Math.Cos(angle));
    var y = _originY + (distance * Math.Sin(angle));
    return new Point((int)x, (int)y);
}

Now you’re thinking with polar.

You can also weight the distance like so to avoid a square root:

var distance = _random.NextDouble() + _random.NextDouble();
distance = (distance <= 1 ? distance : 2 - distance) * _radius;

Answer 4

If performance is an issue then one alternative solution is to generate a random position in a box with the width/height of your circle and then throw away any points that are not in the area of the circle.

The advantage of this method is that you are doing no cos/sin/sqrt functions, which depending on your platform may be a big speed saving.

var x = _random.NextDouble();
var y = _random.NextDouble();
if (x*x + y*y < 1.0f)
{
    // we have a usable point inside a circle
    x = x * diameter - _radius + _OriginX;
    y = y * diameter - _radius + _OriginY;
    // use the point(x,y)
}

Answer 5

I have taken the approach of one of the listed comments, and extended the functionality to create a donut shaped point generation system.

            private Vector2 CalculatePosition()
            {
                double angle = _rnd.NextDouble() * Math.PI * 2;
                double radius = InnerCircleRadius + (Math.Sqrt(_rnd.NextDouble()) * (OuterCircleRadius - InnerCircleRadius));
                double x = (_context.ScreenLayout.Width * 0.5f) + (radius * Math.Cos(angle));
                double y = (_context.ScreenLayout.Height * 0.5f) + (radius * Math.Sin(angle));
                return new Vector2((int)x, (int)y);
            }

It is a similar approach as previously mentioned but provided different results. The inner part of the circle will be left blank with no points.