From 101c3085e1c6b49d4863838b9b041db2b20caeb0 Mon Sep 17 00:00:00 2001 From: 3a8a08e9391701ee31513ee7080f15a0 <3a8a08e9391701ee31513ee7080f15a0@app-learninglab.inria.fr> Date: Wed, 8 Jun 2022 12:04:53 +0000 Subject: [PATCH] Replace toy_notebook_fr.ipynb --- module2/exo1/toy_notebook_fr.ipynb | 129 ++++++++++++++++++++++++++++- 1 file changed, 125 insertions(+), 4 deletions(-) diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb index 0bbbe37..5f40f74 100644 --- a/module2/exo1/toy_notebook_fr.ipynb +++ b/module2/exo1/toy_notebook_fr.ipynb @@ -1,8 +1,130 @@ { - "cells": [], + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "toy_notebook_fr\n", + "\n", + "March 28,2019" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 1 A propos du calcul de $\\pi$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.1 En demandant à la lib maths" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Mon ordinateur m'indique que $\\pi$ vaut _approximativement_" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "from math import *\n", + "print(pi)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 1.2 En utilisant la méthode des aiguilles de Buffon" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "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": null, + "metadata": {}, + "outputs": [], + "source": [ + "import numpy as np\n", + "np.random.seed(seed=42)\n", + "N = 1000\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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d'appel à la fonction sinus se base sur le fait que si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\leq 1] = \\pi/4$ ([voir méthode Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)\n", + "). Le code suivant illustre ce fait :" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "np.random.seed(seed=42)\n", + "N = 100\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", + "\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, en moyenne, $X^2$ + $Y^2$ est inférieur à 1:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "4*np.mean(accept)" + ] + } + ], "metadata": { "kernelspec": { - "display_name": "Python 3", + "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, @@ -16,10 +138,9 @@ "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", - "version": "3.6.3" + "version": "3.7.13" } }, "nbformat": 4, "nbformat_minor": 2 } - -- 2.18.1