From 065e47f208145d31c31d54d6ef695359191e07ce Mon Sep 17 00:00:00 2001
From: 71d5def4c7627f037492744421db8d5d
 <71d5def4c7627f037492744421db8d5d@app-learninglab.inria.fr>
Date: Sun, 29 Sep 2024 10:21:29 +0000
Subject: [PATCH] Minor corrections

---
 module2/exo1/toy_notebook_en.ipynb | 39 +++++++++---------------------
 1 file changed, 12 insertions(+), 27 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 60e78cb..5f135ce 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,21 +4,15 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# 1 On the computation of $\\pi$"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.1 Asking the maths library"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "My computer tells me that $\\pi$ is approximatively"
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
@@ -43,14 +37,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.2 Buffon's needle"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
+    "## Buffon's needle\n",
+    "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
    ]
   },
   {
@@ -82,14 +70,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## 1.3 Using a surface fraction argument"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "A method that is easier to understand and does not make use of the sin function is based on the fact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$, then P$[X^2 +Y^2 \\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach"
+    "## Using a surface fraction argument\n",
+    "A method that is easier to understand and does not make use of the sin function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then P$[X^2+Y^2\\leq 1] = \\pi/4$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
@@ -111,14 +93,17 @@
     }
    ],
    "source": [
-    "%matplotlib inline\n",
+    "%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",
@@ -129,7 +114,7 @@
    "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, $X^2 + Y^2$ 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:"
    ]
   },
   {
-- 
2.18.1