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   "cell_type": "markdown",
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   "source": [
    "# A propos du calcul de $\\pi$\n",
    "\n",
    "## En demandant à la lib maths \n",
    "\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": [
    "## En utilisant la méthode des aiguilles de Buffon\n",
    "\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": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.128911138923655"
      ]
     },
     "execution_count": 3,
     "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": [
    "## Avec un argument \"fréquentiel\" de surfae \n",
    "Sinon, une méhode plus simple à comprendre et ne faisant pas intervenir d'appel à la fonction sinus se base sur le fait que si $X\\sim\\mathcal(U)(0,1)$ et $Y\\sim\\mathcal(U)(0,1)$ alors $P\\left[X^2+Y^2\\leq1\\right]=\\pi/4$ (voir la [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",
    "\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",
    "accept = (x*x+y*y) <= 1\n",
    "reject = np.logical_not(accept)"
   ]
  }
 ],
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