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module2/exo1/module2/exo1/toy_document_en.Rmd

+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
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