{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "March 28, 2019"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Premier test jupyter"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# À propos du calcul de π $\\pi$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "hello world\n"
     ]
    }
   ],
   "source": [
    "import os\n",
    "print('hello world')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## En demandant à la lib maths\n",
    "Mon ordinateur m’indique que π vaut approximativement"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "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 [aiguilles de Buffon](https://fr.wikipedia.org/wiki/Aiguille_de_Buffon), on obtiendrait comme **approximation** :\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3.128911138923655"
      ]
     },
     "execution_count": 5,
     "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": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ut[2]: 3.1289111389236548\n",
    "1.3 Avec un argument \"fréquentiel\" de surface\n",
    "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 ∼ U(0, 1) et Y ∼ U(0, 1) alors P[X\n",
    "2 + Y\n",
    "2 ≤ 1] = π/4 (voir\n",
    "méthode de Monte Carlo sur Wikipedia). Le code suivant illustre ce fait :\n",
    "In [3]: %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)\n",
    "1\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')\n",
    "Il est alors aisé d’obtenir une approximation (pas terrible) de π en comptant combien de fois,\n",
    "en moyenne, X\n",
    "2 + Y\n",
    "2\n",
    "est inférieur à 1 :\n",
    "In [4]: 4*np.mean(accept)\n",
    "Out[4]: 3.1120000000000001\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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