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'# 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 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)
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