Find total area of all circles

How can we calculate the total area covered by all the circles together using Monte Carlo method?

[Askic V]

I gues the following code would do:

% Circle is given with (xc, yc, rc)
% (xc, yc) coordinates of center, rc is radius

% Monte Carlo algorithm works by creating a square having x and y  points in range [-rc ; rc]
% around center point (xc, yc)

xc = 5; yc = 4; rc = 2;
t = linspace(0, 360);

x = xc + rc*cosd(t);
y = yc + rc*sind(t);

plot(x,y)
axis equal

in_circle = 0;
total = 0;
nr_points = 1e6;
for i = 1:nr_points
    x = 2*rc*rand - rc + xc;
    y = 2*rc*rand - rc +yc;

    if (x-xc)^2+(y-yc)^2 <= rc^2
        in_circle = in_circle + 1;
    end
    total = total +1;

    area = (2*rc)^2 * in_circle/total;
end

Monte Carlo method estimates area by calculating ratio of total number of points inside a circle of radius rc and total number of generated points.

[Jan]

@Elysi Cochin: What are trhe given inputs?

[Torsten]

1. Enclose the area in which all circles are comprised by a rectangle with side lengths a and b.
2. Generate N random points uniformly distributed in the rectangle.
3. For each random point, loop over all circles and check the in_circle condition.
4. If the condition is true for at least one circle, set in_circle = in_circle + 1.
5. An approximation for the area of the circles is then area = in_circle/N * (a*b)

[Elysi Cochin]

Using the code in the comment by @Askic V I could compute the area of all the circles by inputting the x,y coordinate and radius using a loop. But my doubt is, is that code sufficient to compute the area of two intersecting circles?

[Image Analyst]

Did you see my Answer below?

Why would you want to compute the area of overlap? If you did, how do you think you'd handle it? Remove the break, right? What else?

Answer 1

[Image Analyst]

You need to generate an (x,y) point, like in an outer loop. Then in the inner loop you need to loop over all the known circles. If the point is inside a circle, increment the count, and skip checking the rest of the circles by calling "break" so that you don't count a point twice if it's in two circles. Here's a start, assuming xc and yc is a list of the circle center coordinates, and radii is a list of the corresponding radii.

numPoints = 100000; % Whatever
count = 0;
for k = 1 : numPoints
    x = whatever, get random number in the range.
    y = whatever
    for c = 1 : numCircles
        if sqrt(x - xc(c)).^ 2 + (y - yc(c))^2) < radii(c)
            % Then it's inside this circle
            count = count + 1
            break; % Skip checking the rest of the circles.
        end
    end
end
percentArea = count / numPoints