{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": null,
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   "outputs": [],
   "source": [
    "toy_notebook_fr"
   ]
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  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "March 28, 2019"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#A propos du calcul de pi"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "hideCode": false
   },
   "outputs": [],
   "source": [
    "## En demandant à la lib maths"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "Mon ordi m'indique que pi vaut approximativement"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "In [1]: from math import *\n",
    "    print(pi)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "##En utilisant la méthode des aiguilles de Buffon"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Mais calculé avec la méthode des aiguilles de Buffon, on obtiendrait comme approximation:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "In [2]: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": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "## Avec un argument \"fréquentiel\" de surface"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d'apel à la fonction sinus se base sur le fait que si X ~ U(0,1) et Y ~ U(0,1) alors P[X² + Y² < 1] = pi/4 (voir méthode de Monte-Carlo sur Wikipedia). Le code suivant illustre ce fait :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "ename": "IndentationError",
     "evalue": "unexpected indent (<ipython-input-1-c47c06b12806>, line 2)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-1-c47c06b12806>\"\u001b[0;36m, line \u001b[0;32m2\u001b[0m\n\u001b[0;31m    import matplotlib.pyplot as plt\u001b[0m\n\u001b[0m    ^\u001b[0m\n\u001b[0;31mIndentationError\u001b[0m\u001b[0;31m:\u001b[0m unexpected indent\n"
     ]
    }
   ],
   "source": [
    "In [3]: %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",
    "    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')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Il est alors aisé d'obtenir une approximation (pas terrible) de pi en comptant combien de fois, en moyenne, X² + Y² est inférieur à 1:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "In [4] : 4*np.mean(accept)"
   ]
  }
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