{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# toy_notebook_fr\n",
    "## March 28, 2019"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. **A propos du calcul de** $\\pi$\n",
    " 1. **En demandant à la lib maths**\n",
    " Mon ordinateur m'ndique que $\\pi$ vaut _approximativement_"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import *\n",
    "print(pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " 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": null,
   "metadata": {},
   "outputs": [],
   "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": [
    " 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 sinus se base sur le fait que si \\X $\\sim U(0,1) et \\Y $\\sim U(0,1) alors \\P$[\\X^2 $\\oplus$ \\Y^2]$ = $\\pi$ $\\div$ 4 ( voir [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",
    "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, en moyenne,\\X^2 $\\oplus$ \\Y^2 est inférieur à 1 :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "4*np.mean(accept)"
   ]
  }
 ],
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