Uses nothing but a uniform random number generator to approximate pi
Consider a circular dartboard of radius 1 centred at the origin on a cardboard square of area 4. If we throw darts at the board, then there should be a roughly equal density of darts on the dartboard, and of darts on the cardboard around the outside of the dartboard. Note that we are assuming a uniform distribution and considering only darts that are at least inside the square.
So the number of darts in the circle is approximately:
And the number of darts outside the circle (and thus in the square) is approximately:
If we divide the darts inside and outside the square:
In my program, we only consider numbers in the first quadrant to simplify calculations. Thus, we generate two floating point numbers in the interval . Since the ratio of areas remains the same, the formula does not change.
We consider the first number an x-coordinate and the second number a y-coordinate. This lets us check whether the point is on the circle by evaluating:
If it is in the circle, we count it, and otherwise it is in the square. After the desired number of iterations, the formula above calculates pi.
Upon running the program 10 times with 1,000,000 "darts", the value of pi was accurate to 4 decimals.