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   "source": [
    "# ** À propos du calcul de $\\pi$**\n",
    "\n",
    "## ** En demandant à la lib maths**\n",
    "\n",
    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import *\n",
    "print(pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "## ** En utilisant la méthode des aiguilles de Buffon**\n",
    "\n",
    "Mais calculé avec la **méthode** des https://fr.wikipedia.org/wiki/Aiguille_de_Buffon - automatic!, on obtiendrait comme **approximation** :\n"
   ]
  },
  {
   "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=np.pi/2)\n",
    "2/(sum((x+np.sin(theta))>1)/N)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "## ** 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 sinus se base sur le fait que si *$X \\sim U*(0,1)$ et *$Y \\sim U*(0,1)$ alors $P[X^2+Y^2<=1]=\\pi/4$ (voir 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 :\n"
   ]
  },
  {
   "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')\n"
   ]
  },
  {
   "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 :\n",
    "    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "4*np.mean(accept)"
   ]
  }
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