{
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   "source": [
    "March 28, 2019"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 1 A propos du calcul de $\\pi$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1.1 En demandant à la lib maths"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "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)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## En utilisant la méthode des aiguilles de Buffon"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Mais calculé avec la **méthode** des ``aiguilles de Buffon``, on obtiendrait comme **approximation** :\n"
   ]
  },
  {
   "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)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Avec un argument \"fréquentiel\" de surface\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "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 si X $\\approx$ U(0, 1) et Y $\\approx$ U(0, 1) alors P[$X^2$ $\\sum$ $Y^2$ ≤ 1] = $^$\\pi$/_4$ (voir\n",
    "``méthode de Monte Carlo sur Wikipedia``). Le code suivant illustre ce fait :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "ename": "SyntaxError",
     "evalue": "invalid syntax (<ipython-input-4-23eb3c525cc9>, line 2)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-4-23eb3c525cc9>\"\u001b[0;36m, line \u001b[0;32m2\u001b[0m\n\u001b[0;31m    import matplotlib.pyplot as\u001b[0m\n\u001b[0m                                ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
     ]
    }
   ],
   "source": [
    "%matplotlib inline\n",
    "import matplotlib.pyplot as \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",
    "accept = (x*x+y*y) <= 1\n",
    "reject = np.logical_not(accept)\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"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Il est alors aisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois,\n",
    "en moyenne, $X^2$ $\\sum$ $Y^2$ est inférieur à 1 :\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "ename": "NameError",
     "evalue": "name 'accept' is not defined",
     "output_type": "error",
     "traceback": [
      "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[0;31mNameError\u001b[0m                                 Traceback (most recent call last)",
      "\u001b[0;32m<ipython-input-5-b03d6d9f8ffa>\u001b[0m in \u001b[0;36m<module>\u001b[0;34m\u001b[0m\n\u001b[0;32m----> 1\u001b[0;31m \u001b[0;36m4\u001b[0m\u001b[0;34m*\u001b[0m\u001b[0mnp\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mmean\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0maccept\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m",
      "\u001b[0;31mNameError\u001b[0m: name 'accept' is not defined"
     ]
    }
   ],
   "source": [
    "4*np.mean(accept)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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