Find area of circle on a grid using euclidean distance?

Question

I would like to have a function where I can input a radius value and have said function spit out the area for that size circle. The catch is I want it to do so for integer b

Answer 1

This is an old question but I was recently working on the same thing. What you are trying to do is as you said, Gauss's circle problem, which is sort of described here

While I too have difficulty understaning the serious maths behind it all, what it more or less pans out to when not using wierd alien symbols is this:

1 + 4 * sum(i=0, r^2/4, r^2/(4*i+1) - r^2/(4*i+3))

which in java at least is:

int sum = 0;
for(int i = 0; i <= (radius*radius)/4; i++)
  sum += (radius*radius)/(4*i+1) - (radius*radius)/(4*i+3);
sum = sum * 4 + 1;

I have no idea why or how this works and to be honest Im a bit bummed I have to use a loop to get this out rather than a single line, as it means the performance is O(r^2/4) rather than O(1).

Since the math wizards can't seem to do better than a loop, I decided to see whether I could get it down to O(r + 1) performance, which I did. So don't use the above, use the below. O(r^2/4) is terrible and will be slower even despite mine using square roots.

int sum = 0;
for(int x = 0; x <= radius; x++)
  sum += Math.sqrt(radius * radius - x * x);
sum = sum * 4 + 1;

What this code does is loop from centre out to the edge along an orthogonal line, and at each point adding the distance from line to edge in a perpendicualr direction. At the end it will have the number of points in a quater, so it quadruples the result and adds one because there is also central point. I feel like the wolfram equation does something similar, since it also multiplies by 4 and adds one, but IDK why it loops r^2/4.

Honestly these aren't great solution, but it seems to be the best there is. If you are calling a function which does this regularly then as new radii come up save the results in a look-up table rather than doing a full calc each time.

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Its not a part of your question, but it may be relevant to someone maybe so I'll add it in anyway. I was personally working on finding all the points within a circle with cells defined by:

(centreX - cellX)^2 + (centreY - cellY)^2 <= radius^2 + radius

Which puts the whole thing out of whack because the extra +radius makes this not exactly the pythagorean theorem. That extra bit makes the circles look a whole lot more visually appealing on a grid though, as they don't have those little pimples on the orthogonal edges. It turns out that, yes my shape is still a circle, but its using sqrt(r^2+r) as radius instead of r, which apparently works but dont ask me how. Anyway that means that for me, my code is slightly different and looks more like this:

int sum = 0;
int compactR = ((radius * radius) + radius) //Small performance boost I suppose
for(int j = 0; j <= compactR  / 4; j++)
  sum += compactR / (4 * j + 1) - compactR / (4 * j + 3);
sum = sum * 4 + 1;