1 On the computation of π
1.1 Asking the maths library
My computer tells me that π is approximatively
In [1]: from math import *
print(pi)
3.141592653589793
1.2 Buffon’s needle
Applying the method of Buffon’s needle, we get the approximation
In [2]: import numpy as np
np.random.seed(seed=42)
N = 10000
x = np.random.uniform(size=N, low=0, high=1)
theta = np.random.uniform(size=N, low=0, high=pi/2)
2/(sum((x+np.sin(theta))>1)/N)
Out[2]: 3.1289111389236548

1.3 Using a surface fraction argument
A method that is easier to understand and does not make use of the sin function is based on the
fact that if X ∼ U(0, 1) and Y ∼ U(0, 1), then P[X
2 + Y
2 ≤ 1] = π/4 (see "Monte Carlo method"
on Wikipedia). The following code uses this approach:
In [3]: %matplotlib inline
import matplotlib.pyplot as plt
np.random.seed(seed=42)
N = 1000
x = np.random.uniform(size=N, low=0, high=1)
y = np.random.uniform(size=N, low=0, high=1)
accept = (x*x+y*y) <= 1
reject = np.logical_not(accept)
fig, ax = plt.subplots(1)
ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)
ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)
ax.set_aspect('equal')

It is then straightforward to obtain a (not really good) approximation to π by counting how
many times, on average, X
2 + Y
2
is smaller than 1:
In [4]: 4*np.mean(accept)
Out[4]: 3.1120000000000001
{r}## [1] 3.156