On the computation of pi
Table of Contents
1 Asking the math library
My computer tells me that π is approximatively
from math import * pi
3.141592653589793
2 * Buffon’s needle
Applying the method of Buffon’s needle, 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)
3.128911138923655
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 \(X \approx U(0,1)\) and \(Y \approx U(0,1)\), then \(P[X^2 + Y^2 \leq 1] = \pi/4\) (see “Monte Carlo method on Wikipedia”). The following code uses this approach:
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') plt.savefig(matplot_lib_filename) print(matplot_lib_filename)
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:
4*np.mean(accept)
3.112