#  On the computation of $\pi$
##  Asking the maths library
My computer tells me that π is approximatively

from math import *
print(pi)

## Buffon’s needle
Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**

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)

## 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 $P[X2 + Y2 ≤ 1] = π/4$ (see ["Monte Carlo method"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:

%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)
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,  𝑋2+𝑌2
  is smaller than 1:
  
  4*np.mean(accept)