{
 "cells": [
  {
   "cell_type": "markdown",
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   "source": [
    "# On the computation of $\\pi$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "## Asking the maths library"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "hideCode": false,
    "hidePrompt": false
   },
   "source": [
    "My computer tells me that $\\pi$ *is approximatively*"
   ]
  },
  {
   "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": [
    "## Buffon’s needle"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get **the 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": [
    "##  Using a surface fraction argument"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A method that is easier to understand and does not make use of the sin function is based on the\n",
    "fact that if $$X ∼ \\mathcal{U}(0, 1)$$ and $$\\mathcal{Y} ∼ \\mathcal{U}(0, 1)$$, then $$\\mathcal{P}[X^2 + Y^2 ≤ 1] = π/4$$ (see \"[Monte Carlo method]\"(https://en.wikipedia.org/wiki/Monte_Carlo_method)\n",
    "on Wikipedia). The following code uses this approach:"
   ]
  },
  {
   "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)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
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