Finding all the points within a circle in 2D space

Question

I am representing my 2D space (consider a window), where each pixel is shown as a cell in a 2D array. i.e. a 100x100 window is represented by the array of same dimensions.

Now given a point in the window, if I draw a circle of radius r, I want to find all the points lying in that circle.

I was thinking I'd check for the each point in the square region around the radius, with side = 2*r, if it lies in the circle or not. I'll use the normal distance formula maybe?

Hence, maybe the following:

for (x=center-radius ; x<center+radius ; x++){
    for (y=center-radius ; y<center+radius; y++) {
        if (inside) {
            // Do something
        }
    }
}

Will it serve my purpose? Can I make it faster?

Answer 1

Will it serve my purpose?

For your 100x100, yes.

Can I make it faster?

Yes. For example, you can:

• Check only 1 quadrant and get other points because of symmetry.
• Skip the square root when calculating the distance.

Code:

for (x = xCenter - radius ; x <= xCenter; x++)
{
    for (y = yCenter - radius ; y <= yCenter; y++)
    {
        // we don't have to take the square root, it's slow
        if ((x - xCenter)*(x - xCenter) + (y - yCenter)*(y - yCenter) <= r*r) 
        {
            xSym = xCenter - (x - xCenter);
            ySym = yCenter - (y - yCenter);
            // (x, y), (x, ySym), (xSym , y), (xSym, ySym) are in the circle
        }
    }
}

That's about 4x speed up.

JS tests for solutions presented here. Symmetry is the fastest on my computer. Trigonometry presented by Niet the Dark Absol is very clever, but it involves expensive mathematical functions like sin and acos, which have a negative impact on performance.

Answer 2

You can bypass the need for a conditional check:

for(x=center-radius; x<center+radius; x++) {
    yspan = radius*sin(acos((center-x)/radius));
    for(y=center-yspan; y<center+yspan; y++) {
        // (x,y) is inside the circle
    }
}

If needed, you can round(yspan).