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    "# On the computation of $\\pi$\n",
    "\n",
    "## Asking the maths library\n",
    "My computer tells me that $\\pi$ is *approximatively*\n",
    "\n",
    "from math import *\n",
    "print(pi)\n",
    "\n",
    "## Buffon's needle\n",
    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__\n",
    "                        \n",
    "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"
   ]
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   "source": [
    "## Using a surface fraction argument\n",
    "A method that is easier to understand and does not make use of the $\\sin$ function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:\n",
    "    \n",
    "%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[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",
    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:\n",
    "\n",
    "4*np.mean(accept)"
   ]
  }
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