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"toy_notebook_fr\n",
"March 28, 2019\n",
"1 À propos du calcul de π\n",
"1.1 En demandant à la lib maths\n",
"Mon ordinateur m’indique que π vaut approximativement\n",
"In [1]: from math import *\n",
"print(pi)\n",
"3.141592653589793\n",
"1.2 En utilisant la méthode des aiguilles de Buffon\n",
"Mais calculé avec la méthode des aiguilles de Buffon, on obtiendrait comme approximation :\n",
"In [2]: import numpy as np\n",
"np.random.seed(seed=42)\n",
"N = 10000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum((x+np.sin(theta))>1)/N)\n",
"Out[2]: 3.1289111389236548\n",
"1.3 Avec un argument \"fréquentiel\" de surface\n",
"Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction\n",
"sinus se base sur le fait que si X ∼ U(0, 1) et Y ∼ U(0, 1) alors P[X2 + Y2 ≤ 1] = π/4 (voir\n",
"méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :\n",
"In [3]: %matplotlib inline\n",
"import matplotlib.pyplot as plt\n",
"np.random.seed(seed=42)\n",
"N = 1000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"y = np.random.uniform(size=N, low=0, high=1)\n",
"1\n",
"accept = (x*x+y*y) <= 1\n",
"reject = np.logical_not(accept)\n",
"fig, ax = plt.subplots(1)\n",
"ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
"ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
"ax.set_aspect('equal')\n",
"Il est alors aisé d’obtenir une approximation (pas terrible) de π en comptant combien de fois,\n",
"en moyenne, X2 + Y2 est inférieur à 1 :\n",
"In [4]: 4*np.mean(accept)\n",
"Out[4]: 3.1120000000000001\n",
"2"
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