Simplest way to plot points randomly inside a circle
Asked Answered
S

4

8

I have a basic JSFiddle whereby I want to have random points plotted inside a circle.

But I do not know how to limit the points to be inside the circle.

This is what I currently have:

var ctx = canvas.getContext('2d'),
    count = 1000, // number of random  points
    cx = 150,
    cy = 150,
    radius = 148;

    ctx.beginPath();
    ctx.moveTo(cx, cy);
    ctx.arc(canvas.width/2, canvas.height/2, radius, 0, 2*Math.PI);
    ctx.closePath();
    ctx.fillStyle = '#00000';
    ctx.fill();

// create random points
    ctx.fillStyle = '#ffffff';

while(count) {
    // randomise x:y

    ctx.fillRect(x + canvas.width/2, y + canvas.height/2, 2, 2);
    count--;
}

How would i go about generating random (x,y) coordinates to plot random points inside the circle?

My current fiddle: http://jsfiddle.net/e8jqdxp3/

Sherronsherry answered 18/9, 2015 at 1:18 Comment(0)
C
17

To plot points randomly in a circle, you can pick a random value from the radius squared, then square root it, pick a random angle, and convert the polar coordinate to rectangular. The square / square root step ensures that we get a uniform distribution (otherwise most points would be near the center of the circle).

So the formula to plot a random point in the circle is the following, where r' is a random value between 0 and r2, and θ is a random value between 0 and :

Equation

Screenshot of result:

Screenshot of result

Live Demo:

var canvas = document.getElementById("thecanvas");

var ctx = canvas.getContext('2d'),
    count = 1000, // number of random  points
    cx = 150,
    cy = 150,
    radius = 148;

ctx.fillStyle = '#CCCCCC';
ctx.fillRect(0, 0, canvas.width, canvas.height);

ctx.fillStyle = '#000000';

ctx.beginPath();
ctx.moveTo(cx, cy);
ctx.arc(canvas.width / 2, canvas.height / 2, radius, 0, 2 * Math.PI);
ctx.closePath();

ctx.fill();

// create random points
ctx.fillStyle = '#ffffff';

while (count) {
    var pt_angle = Math.random() * 2 * Math.PI;
    var pt_radius_sq = Math.random() * radius * radius;
    var pt_x = Math.sqrt(pt_radius_sq) * Math.cos(pt_angle);
    var pt_y = Math.sqrt(pt_radius_sq) * Math.sin(pt_angle);
    ctx.fillRect(pt_x + canvas.width / 2, pt_y + canvas.width / 2, 2, 2);
    count--;
}
<canvas id="thecanvas" width="400" height="400"></canvas>

JSFiddle Version: https://jsfiddle.net/qc735bqw/

Capriole answered 18/9, 2015 at 1:32 Comment(7)
Nice, You can avoid square / squareRoot with: r=Math.random()*radius then x = cx + r * Math.cos(pt_angle); and y = cy + r * Math.sin(pt_angle); and then fillRect(x,y,2,2). Also note that if the full radius is randomly selected, a 2px rect can escape the boundary of the circle because rects are defined with their x,y at top-left. A simple fix is to use (radius-2) or a lengthier fix is to trap that edge case when the angle is between -PI/2 and PI/2.Pomposity
Thanks for this i was wondering if there is a simple way to do the same for an ellipse?Sherronsherry
@Dave A simple way is to just scale the output when you draw it. So if you want a horizontal elipse, just instead of drawing x and y, draw x and y/2 (or maybe Math.floor(y/2)).Capriole
So i just divide the final y calculation by the given scale.Sherronsherry
@Dave Yup. (minimum comment length filler text here)Capriole
@Pomposity using this some plots are outside of the circle. Even if I change radius to radius-2.Powers
@Pomposity removing the squareRoot will lead to a non uniform distribution.Bateau
E
3

Randomly pick dSquared (0..radius^2) and theta (0..2pi), then

x = sqrt(dSquared) cos(theta)
y = sqrt(dSquared) sin(theta)
Elastin answered 18/9, 2015 at 1:29 Comment(0)
C
1

This worked for me:

const getRandomCoordinateInCircle = radius => {
  var angle = Math.random() * Math.PI * 2;
  const x = Math.cos(angle) * radius * Math.random();
  const y = Math.sin(angle) * radius * Math.random();
  return { x, y };
};

console.log(getRandomCoordinateInCircle(1000);
// { x: 118.35662725763385, y: -52.60516556856313 }

Returns a random point with { x: 0, y: 0} as the centre of the circle.

Carpospore answered 25/1, 2022 at 1:45 Comment(0)
P
0

JSFiddle

var ctx = canvas.getContext('2d'),
    count = 1000, // number of random  points
    cx = canvas.width/2,
    cy = canvas.height/2,
    radius = 148;

    ctx.beginPath();
    ctx.moveTo(cx, cy);
    ctx.arc(0+canvas.width/2, 0+canvas.height/2, radius, 0, 2*Math.PI);
    ctx.closePath();
    ctx.fillStyle = '#00000';
    ctx.fill();

    ctx.fillStyle = '#ffffff';

while(count) {
    var x = Math.random() * canvas.width;
    var y = Math.random() * canvas.height;
    var xDiff = cx - x;
    var yDiff = cy - y;
    if(Math.sqrt(xDiff*xDiff+yDiff*yDiff)<radius)
    {
        ctx.fillRect(x, y, 2, 2);
        count--;
    }    
}
Pelite answered 18/9, 2015 at 1:51 Comment(0)

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