{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# toy_notebook_fr\n",
    "\n",
    "## March 28, 2019\n",
    "\n",
    "## 1 À propos du calcul de p\n",
    "\n",
    "## 1.1 En demandant à la lib maths\n",
    "\n",
    "Mon ordinateur m’indique que _p_ vautapproximativement\n",
    "\n",
    "In [ 1 ]: frommathimport *\n",
    "print(pi)\n",
    "\n",
    "3.\n",
    "\n",
    "## 1.2 En utilisant la méthode des aiguilles de Buffon\n",
    "\n",
    "Mais calculé avec la **méthode** desaiguilles de Buffon, on obtiendrait comme **approximation** :\n",
    "\n",
    "In [ 2 ]: import numpyas 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",
    "\n",
    "Out[ 2 ]: 3.\n",
    "\n",
    "## 1.3 Avec un argument \"fréquentiel\" de surface\n",
    "\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 siX \u0018U(0, 1)etY \u0018U(0, 1)alorsP[X^2 +Y^2 \u0014 1 ]= _p_ /4 (voir\n",
    "méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :\n",
    "\n",
    "In [ 3 ]: %matplotlib inline\n",
    "import matplotlib.pyplotas plt\n",
    "\n",
    "```\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",
    "```\n",
    "### 1\n",
    "\n",
    "\n",
    "```\n",
    "accept =(x*x+y*y) <= 1\n",
    "reject =np.logical_not(accept)\n",
    "```\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')\n",
    "```\n",
    "Il est alors aisé d’obtenir une approximation (pas terrible) de _p_ en comptant combien de fois,\n",
    "en moyenne,X^2 +Y^2 est inférieur à 1 :\n",
    "\n",
    "In [ 4 ]: 4 *np.mean(accept)\n",
    "\n",
    "Out[ 4 ]: 3.\n",
    "\n",
    "### 2\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.6.4"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}