From 295f0564f96fd80969839240347cdb02f1db3979 Mon Sep 17 00:00:00 2001
From: 00bd05cb42ba6c2a491841c680c81c95
 <00bd05cb42ba6c2a491841c680c81c95@app-learninglab.inria.fr>
Date: Tue, 3 Jun 2025 15:18:14 +0000
Subject: [PATCH] v2

---
 module2/exo1/toy_notebook_en.ipynb | 46 ++++++++++++++----------------
 1 file changed, 21 insertions(+), 25 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 1c79f01..8c88c1b 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,19 +4,16 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "toy_notebook_en \n",
-    "March 28, 2019\n",
-    "\n",
     "## 1 On the computation of π\n",
     "\n",
     "### 1.1 Asking the maths library\n",
     "\n",
-    "My computer tells me that π is approximately"
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 7,
+   "execution_count": 14,
    "metadata": {},
    "outputs": [
     {
@@ -38,32 +35,32 @@
    "source": [
     "### 1.2 Buffon’s needle\n",
     "\n",
-    "Applying the method of Buffon’s needle, we get the approximation\n"
+    "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": 8,
+   "execution_count": 15,
    "metadata": {},
    "outputs": [
     {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3.128911138923655\n"
-     ]
+     "data": {
+      "text/plain": [
+       "3.128911138923655"
+      ]
+     },
+     "execution_count": 15,
+     "metadata": {},
+     "output_type": "execute_result"
     }
    ],
    "source": [
     "import numpy as np\n",
-    "\n",
     "np.random.seed(seed=42)\n",
     "N = 10000\n",
-    "x = np.random.uniform(low=0, high=1, size=N)\n",
-    "theta = np.random.uniform(low=0, high=pi/2, size=N)\n",
-    "\n",
-    "approx_pi_buffon = 2 / (np.sum((x + np.sin(theta)) > 1) / N)\n",
-    "print(approx_pi_buffon)\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)"
    ]
   },
   {
@@ -74,12 +71,12 @@
     "\n",
     "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 ∼ U(0, 1) and Y ∼ U(0, 1), then P[X² + Y² ≤ 1] = π/4 (see \"Monte Carlo method\" on Wikipedia).  \n",
-    "The following code uses this approach:\n"
+    "The following code uses this approach:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 9,
+   "execution_count": 18,
    "metadata": {},
    "outputs": [
     {
@@ -102,25 +99,24 @@
     "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"
+    "ax.set_aspect('equal')"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "It is then straightforward to obtain a (not really good) approximation to π by counting how many times, on average, X² + Y² is smaller than 1:"
+    "It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 10,
+   "execution_count": 19,
    "metadata": {},
    "outputs": [
     {
@@ -129,7 +125,7 @@
        "3.112"
       ]
      },
-     "execution_count": 10,
+     "execution_count": 19,
      "metadata": {},
      "output_type": "execute_result"
     }
-- 
2.18.1