From a0d55f617f87618046bae343d39f9f5b9fa6347b Mon Sep 17 00:00:00 2001
From: dc97a6904245e0d07c1302ee90fceec3
 <dc97a6904245e0d07c1302ee90fceec3@app-learninglab.inria.fr>
Date: Wed, 18 Jun 2025 15:34:59 +0000
Subject: [PATCH] no commit message

---
 module2/exo1/toy_notebook_en.ipynb | 64 ------------------------------
 1 file changed, 64 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 87c2a9f..e0ba403 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,24 +1,5 @@
 {
  "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# On the computation of $\\pi$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "## Asking the maths library\n",
-    "My computer tells me that $\\pi$ is *approximatively*"
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 1,
@@ -37,24 +18,6 @@
     " print(pi)"
    ]
   },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "## Buffon's needle"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "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,
@@ -80,24 +43,6 @@
     " 2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "## Using a surface fraction argument"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "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:"
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 3,
@@ -132,15 +77,6 @@
     " ax.set_aspect('equal')"
    ]
   },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "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": 4,
-- 
2.18.1