From 873f4387e3e730d7665a67e169084a853653059b Mon Sep 17 00:00:00 2001
From: 1ed6a8939c7714d38b1fe3c75581e0cf
 <1ed6a8939c7714d38b1fe3c75581e0cf@app-learninglab.inria.fr>
Date: Wed, 5 Oct 2022 19:48:15 +0000
Subject: [PATCH] Update toy_notebook_en.ipynb

---
 module2/exo1/toy_notebook_en.ipynb | 51 +++++++++++++++---------------
 1 file changed, 25 insertions(+), 26 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 5bca4a1..669b417 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -2,20 +2,20 @@
  "cells": [
   {
    "cell_type": "markdown",
-   "id": "93d7cca7",
+   "id": "fa916478",
    "metadata": {},
    "source": [
-    "# 1 On the computation of π\n",
+    "# On the computation of $\\pi$\n",
     "\n",
-    "## 1.1 Asking the maths library\n",
+    "## Asking the maths library\n",
     "\n",
-    "My computer tells me that π is approximativel"
+    "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
-   "id": "d0bc647a",
+   "execution_count": 1,
+   "id": "08c5c563",
    "metadata": {},
    "outputs": [
     {
@@ -33,18 +33,17 @@
   },
   {
    "cell_type": "markdown",
-   "id": "99049b47",
+   "id": "8c9199c1",
    "metadata": {},
    "source": [
-    "## 1.2 Buffon’s needle\n",
-    "\n",
-    "Applying the method of <span style=\"color:blue\"> Buffon’s needle </span>, we get the <b>approximation</b>"
+    "## Buffon's needle\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": 3,
-   "id": "d9073c83",
+   "execution_count": 2,
+   "id": "abff8880",
    "metadata": {},
    "outputs": [
     {
@@ -53,7 +52,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 3,
+     "execution_count": 2,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -69,18 +68,17 @@
   },
   {
    "cell_type": "markdown",
-   "id": "2b0f149b",
+   "id": "4bd4c67b",
    "metadata": {},
    "source": [
-    "## 1.3 Using a surface fraction argument\n",
-    "\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 ∼ U(0,1) and Y ∼ U(0,1), then P[X2 +Y2 ≤ 1] = π/4 (see <a style=\"text-decoration: none;\" href=\"https://en.wikipedia.org/wiki/Monte_Carlo_method\"> \"Monte Carlo method\" on Wikipedia </a>). The following code uses this approach:\n"
+    "## 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:"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
-   "id": "6b5b21e8",
+   "execution_count": 3,
+   "id": "5413190d",
    "metadata": {},
    "outputs": [
     {
@@ -97,9 +95,9 @@
     }
    ],
    "source": [
-    "%matplotlib inline\n",
-    "\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",
@@ -107,6 +105,7 @@
     "\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",
@@ -115,16 +114,16 @@
   },
   {
    "cell_type": "markdown",
-   "id": "ca88818e",
+   "id": "c1e500ca",
    "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": 5,
-   "id": "fe6e57c1",
+   "execution_count": 4,
+   "id": "fc8a2a3e",
    "metadata": {},
    "outputs": [
     {
@@ -133,7 +132,7 @@
        "3.112"
       ]
      },
-     "execution_count": 5,
+     "execution_count": 4,
      "metadata": {},
      "output_type": "execute_result"
     }
-- 
2.18.1