From 13f3b9c578d5aae7d6c510525f17c9292792a2aa Mon Sep 17 00:00:00 2001
From: 318b38cbf16538baf6f4efa72de94b90
 <318b38cbf16538baf6f4efa72de94b90@app-learninglab.inria.fr>
Date: Mon, 17 Apr 2023 16:04:20 +0000
Subject: [PATCH] ok

---
 module2/exo1/toy_notebook_fr.ipynb | 149 ++++++++++++++++++++++++++---
 1 file changed, 134 insertions(+), 15 deletions(-)

diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0bbbe37..dbd72d3 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -1,25 +1,144 @@
 {
- "cells": [],
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# À propos du calcul de $\\pi$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## En demandant à la lib maths\n",
+    "Mon ordinateur m'indique que $\\pi$ vaut *approximativement*"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "scrolled": true
+   },
+   "outputs": [
+    {
+     "ename": "ERROR",
+     "evalue": "Error in parse(text = x, srcfile = src): <text>:1:6: unexpected symbol\n1: from math\n         ^\n",
+     "output_type": "error",
+     "traceback": [
+      "Error in parse(text = x, srcfile = src): <text>:1:6: unexpected symbol\n1: from math\n         ^\nTraceback:\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__ :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "ERROR",
+     "evalue": "Error in parse(text = x, srcfile = src): <text>:1:8: unexpected symbol\n1: import numpy\n           ^\n",
+     "output_type": "error",
+     "traceback": [
+      "Error in parse(text = x, srcfile = src): <text>:1:8: unexpected symbol\n1: import numpy\n           ^\nTraceback:\n"
+     ]
+    }
+   ],
+   "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": [
+    "## Avec un argument \"fréquentiel\" de surface\n",
+    "Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d'appel à la fonction sinus se base sur le fait que  si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\leq 1] = \\pi/4$ (voir [méthode de Monte Carlo sur Wikipedia](https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80)). Le code suivant illustre ce fait :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "ERROR",
+     "evalue": "Error in parse(text = x, srcfile = src): <text>:1:1: unexpected input\n1: %matplotlib inline  \n    ^\n",
+     "output_type": "error",
+     "traceback": [
+      "Error in parse(text = x, srcfile = src): <text>:1:1: unexpected input\n1: %matplotlib inline  \n    ^\nTraceback:\n"
+     ]
+    }
+   ],
+   "source": [
+    "%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')"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Il est alors aisé d'obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "4*np.mean(accept)"
+   ]
+  }
+ ],
  "metadata": {
   "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
+   "display_name": "R",
+   "language": "R",
+   "name": "ir"
   },
   "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.6.3"
+   "codemirror_mode": "r",
+   "file_extension": ".r",
+   "mimetype": "text/x-r-source",
+   "name": "R",
+   "pygments_lexer": "r",
+   "version": "3.4.1"
   }
  },
  "nbformat": 4,
  "nbformat_minor": 2
 }
-
-- 
2.18.1