From 41f39e0a63f9b60f0701c78e0ce416cee52ed8cb Mon Sep 17 00:00:00 2001
From: d673626e3823c315c4fb8df298c44ddd
 <d673626e3823c315c4fb8df298c44ddd@app-learninglab.inria.fr>
Date: Fri, 24 Apr 2020 21:20:17 +0000
Subject: [PATCH] save progress

---
 module2/exo1/toy_notebook_fr.ipynb | 55 ++++++++++++++++++++++++++++--
 1 file changed, 53 insertions(+), 2 deletions(-)

diff --git a/module2/exo1/toy_notebook_fr.ipynb b/module2/exo1/toy_notebook_fr.ipynb
index 0556842..5fe2f70 100644
--- a/module2/exo1/toy_notebook_fr.ipynb
+++ b/module2/exo1/toy_notebook_fr.ipynb
@@ -13,13 +13,64 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 1,
    "metadata": {},
-   "outputs": [],
+   "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",
+    "\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": 4,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "3.128911138923655"
+      ]
+     },
+     "execution_count": 4,
+     "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": [
+    "## Avec un argument \"fréquentiel\"  de surface\n",
+    "\n",
+    "Sinon une méthode plus simple à comprendre et ne faisant pas appel à la fonction sinus se base sur le fait que si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\le 1] =\\pi / 4$"
+   ]
   }
  ],
  "metadata": {
-- 
2.18.1