From d9c2f96f929d83e5a63534e509e4a1bc8e6c6ffa Mon Sep 17 00:00:00 2001
From: 289230ab50cb5ba43830fe6a40fb0cae
 <289230ab50cb5ba43830fe6a40fb0cae@app-learninglab.inria.fr>
Date: Thu, 19 Oct 2023 12:23:35 +0000
Subject: [PATCH] (Useless) Edit metadata

---
 module2/exo1/toy_notebook_en.ipynb | 40 ++++++++++++++++--------------
 1 file changed, 22 insertions(+), 18 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index e58cc6f..3f97c46 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,22 +4,26 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "#  On the computation of $\\pi$"
+    "# On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "hideCode": false
+   },
    "source": [
     "## Asking the maths library\n",
-    "\n",
     "My computer tells me that π is *approximatively*"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 1,
-   "metadata": {},
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
    "outputs": [
     {
      "name": "stdout",
@@ -36,17 +40,21 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "hidePrompt": false
+   },
    "source": [
     "## Buffon's needle\n",
-    "\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": 2,
-   "metadata": {},
+   "metadata": {
+    "hideCode": false,
+    "hidePrompt": false
+   },
    "outputs": [
     {
      "data": {
@@ -70,10 +78,11 @@
   },
   {
    "cell_type": "markdown",
-   "metadata": {},
+   "metadata": {
+    "hideCode": false
+   },
    "source": [
     "## 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 $X \\sim U(0, 1)$ 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:"
    ]
   },
@@ -98,11 +107,12 @@
    "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",
-    "1\n",
+    "\n",
     "accept = (x*x+y*y) <= 1\n",
     "reject = np.logical_not(accept)\n",
     "fig, ax = plt.subplots(1)\n",
@@ -137,16 +147,10 @@
    "source": [
     "4*np.mean(accept)"
    ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {
+  "celltoolbar": "Hide code",
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -162,7 +166,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.6.4"
+   "version": "3.6.2"
   }
  },
  "nbformat": 4,
-- 
2.18.1