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   "source": [
    "# À propos du calcul de $\\pi$\n",
    "## En demandant à la lib maths\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",
    "Mais calculé avec la **méthode** des [https://fr.wikipedia.org/wiki/Aiguille_de_Buffon], on obtiendrait comme **approximation** :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.128911138923655"
      ]
     },
     "execution_count": 2,
     "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 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 $P[X^2+Y^2\\leq1]=\\pi/4$ (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 :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "ename": "SyntaxError",
     "evalue": "EOL while scanning string literal (<ipython-input-4-c8a26f7cb22f>, line 13)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-4-c8a26f7cb22f>\"\u001b[0;36m, line \u001b[0;32m13\u001b[0m\n\u001b[0;31m    ax.scatter(x[reject],y[reject], c='r’, alpha=0.2, edgecolor=None)\u001b[0m\n\u001b[0m                                                                      ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m EOL while scanning string literal\n"
     ]
    }
   ],
   "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)\n",
    "\n",
    "fig, ax=plt.subplots(1) \n",
    "ax.scatter(x[reject],y[reject], c='r’, 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 alor saisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inferoeir à 1 :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "ename": "SyntaxError",
     "evalue": "invalid syntax (<ipython-input-7-e627d624d667>, line 1)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-7-e627d624d667>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m    4.np.mean(accept)\u001b[0m\n\u001b[0m       ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
     ]
    }
   ],
   "source": [
    "4.np.mean(accept)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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