From fa22058dfde7fa9f77b4aada4489147f9b44a5ab Mon Sep 17 00:00:00 2001 From: 0a9dbb43cad0788758890fc5d2736551 <0a9dbb43cad0788758890fc5d2736551@app-learninglab.inria.fr> Date: Thu, 7 Aug 2025 14:38:16 +0000 Subject: [PATCH] Notebook pour calculer pi --- module2/exo1/toy_notebook_fr.ipynb | 131 ++++++++++++++++++++++++++++- 1 file changed, 128 insertions(+), 3 deletions(-) diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb index 0bbbe37..5e044e2 100644 --- a/module2/exo1/toy_notebook_fr.ipynb +++ b/module2/exo1/toy_notebook_fr.ipynb @@ -1,5 +1,131 @@ { - "cells": [], + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# A propos du calcul de $\\pi$ #\n", + "## 1.1 En demandant à la lib maths ##\n", + "Mon ordinateur m’indique que $\\pi$ vaut approximativement" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "3.141592653589793\n" + ] + } + ], + "source": [ + "from math import *\n", + "print(pi)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.2 En utilisant la méthode des aiguilles de Buffon ##\n", + "Mais calculé avec la méthode des [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon) , on obtiendrait comme approximation" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "3.128911138923655" + ] + }, + "execution_count": 2, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "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)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 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 $X \\sim U(0, 1)$ et $Y \\sim U(0, 1)$ alors $P[X^2 + Y^2 \\le 1] = \\pi/$ (voir\n", + "[méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "%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", + "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')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Il est alors aisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois,\n", + "en moyenne, $X^2 + Y^2$ est inférieur à 1 :" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "3.112" + ] + }, + "execution_count": 7, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "4*np.mean(accept)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [] + } + ], "metadata": { "kernelspec": { "display_name": "Python 3", @@ -16,10 +142,9 @@ "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", - "version": "3.6.3" + "version": "3.6.4" } }, "nbformat": 4, "nbformat_minor": 2 } - -- 2.18.1