'# Monte Carlo Estimates of Pi

'Consider the unit circle centered at the origin.

'Consider the first quadrant: the unit circle quadrant and its $1 \times 1$ bounding unit square.

'$$\text{Area of unit circle quadrant: } \\\\ A_{quadrant} = \frac{\pi r^2}{4} = \frac{\pi}{4}$$

'$$\text{Area of unit square: } \\\\ A_{square} = 1$$

'$$\text{Compute } \pi \text{ via ratios: } \\\\ \frac{A_{quadrant}}{A_{square}} = \frac{\pi}{4}, \\; \pi = 4 \thinspace \frac{A_{quadrant}}{A_{square}} $$

'To compute $\pi$, randomly sample points in the first quadrant unit square to estimate the $\frac{A_{quadrant}}{A_{square}}$ ratio. Then, multiply by $4$.

def estimatePiArea(key:Key) -> Float =
  [k1, k2] = split_key(n=2, key)
  x = rand k1
  y = rand k2
  inBounds = (sq x + sq y) < 1.0
  4.0 * b_to_f inBounds

def estimatePiAvgVal(key:Key) -> Float =
  x = rand key
  4.0 * sqrt (1.0 - sq x)

def meanAndStdDev(n:Nat, f: (Key) -> Float, key:Key) -> (Float, Float) =
  samps = for i:(Fin n). many f key i
  (mean samps, std samps)

numSamps = 1000000

:p meanAndStdDev numSamps estimatePiArea (new_key 0)
> (3.141656, 1.642139)

:p meanAndStdDev numSamps estimatePiAvgVal (new_key 0)
> (3.145509, 0.8862508)