Suppose the following scenario:
Check out the image here - Red areas show expected evenly distributed coordinates
How can we randomly generate coordinates outside the circle within the rectangular area that are also evenly distributed? (within the blue region shown in the image below)
Produce two random values within specified ranges (following a uniform distribution in a rectangle)
a = Math.random() * width
b = Math.random() * height
Determine if the point is situated outside the circle:
(a-center_x)*(a-center_x)+(b-center_y)*(b-center_y) > r*r
If not, repeat the random generation process until the condition is met (utilizing the rejection method)
https://i.sstatic.net/VUEb7.png
This Delphi code was used to generate the above:
var
Hgt, Wdt, i, N, CX, CY, R, x, y: Integer;
begin
Hgt := 300;
Wdt := 400;
CX := 220;
CY := 120;
R := 100;
N := 15000;
for i := 0 to N-1 do begin
x := Round(Random() * Wdt);
y := Round(Random() * Hgt);
while (x-CX)*(x-CX)+(y-CY)*(y-CY) < R*R do begin
x := Round(Random() * Wdt);
y := Round(Random() * Hgt);
end;
Canvas.Pixels[x, y] := clRed;
end;
Begin by creating a random distribution of x and y values within a square, discarding any pairs falling outside the confines of a centered rectangle.
Procedure
max_dimension = max(l, b)
min_x = (max_dimension-l)/2;
max_x = min_x + l;
min_y = (max_dimension-b)/2;
max_y = min_y + b;
while true
x = Math.random()*max_dimension
if (x < min_x || x > max_x) continue;
y = Math.random()*max_dimension
if (y < min_y || x > max_y) continue;
Next, verify if x and y fall within the circular boundary.
xc = x - l/2;
yc = y - b/2;
if (xc*xc + yc*yc < r*r) continue
break
}
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