From 11e725e583e107a9f261f28e0bd24c35ed0da8d9 Mon Sep 17 00:00:00 2001
From: d0fb1445df0f93dc28c67893dfa732c9
 <d0fb1445df0f93dc28c67893dfa732c9@app-learninglab.inria.fr>
Date: Mon, 26 Dec 2022 17:08:17 +0000
Subject: [PATCH] par1

---
 module2/exo1/toy_notebook_en.ipynb | 52 ++++++++++++++++++++++++++----
 1 file changed, 46 insertions(+), 6 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 84ddd87..56a0275 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,8 +1,23 @@
 {
  "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# 1 On the computation of π"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Asking the maths library\n",
+    "My computer tells me that $\\pi$ is *approximatively*"
+   ]
+  },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 7,
    "metadata": {},
    "outputs": [
     {
@@ -18,9 +33,17 @@
     "print(pi)"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 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,
+   "execution_count": 8,
    "metadata": {},
    "outputs": [
     {
@@ -29,7 +52,7 @@
        "3.128911138923655"
       ]
      },
-     "execution_count": 3,
+     "execution_count": 8,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -43,9 +66,17 @@
     "2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 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,
+   "execution_count": 9,
    "metadata": {},
    "outputs": [
     {
@@ -64,6 +95,7 @@
    "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",
@@ -71,15 +103,23 @@
     "\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",
     "ax.set_aspect('equal')"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "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": 5,
+   "execution_count": 11,
    "metadata": {},
    "outputs": [
     {
@@ -88,7 +128,7 @@
        "3.112"
       ]
      },
-     "execution_count": 5,
+     "execution_count": 11,
      "metadata": {},
      "output_type": "execute_result"
     }
-- 
2.18.1