From fd7b540d1c49608cad300900fdfbdbaa0066e451 Mon Sep 17 00:00:00 2001
From: 2967c9aae1a2f4cee2d1ad632cabf260
 <2967c9aae1a2f4cee2d1ad632cabf260@app-learninglab.inria.fr>
Date: Thu, 2 Jun 2022 22:40:28 +0000
Subject: [PATCH] Update toy_notebook_en.ipynb

---
 module2/exo1/toy_notebook_en.ipynb | 98 ++++++++++++++++++++++--------
 1 file changed, 74 insertions(+), 24 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 0bbbe37..a4d20f3 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,25 +1,75 @@
-{
- "cells": [],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.6.3"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
+# On the computation of $\pi$
+
+## Asking the maths library
+My computer tells me that $\pi$ is _approximatively_
+
+
+```python
+from math import *
+print(pi)
+```
+
+    3.141592653589793
+
+
+## Buffon’s needle
+Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
+
+
+```python
+import numpy as np
+np.random.seed(seed=42)
+N = 10000
+x = np.random.uniform(size=N, low=0, high=1)
+theta = np.random.uniform(size=N, low=0, high=pi/2)
+2/(sum((x+np.sin(theta))>1)/N)
+```
+
+
+
+
+    3.128911138923655
+
+
+
+## Using a surface fraction argument
+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 $P[X^2 + Y^2\le 1] = \pi/4$ (see ["Monte Carlo method"
+on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
+
+
+```python
+%matplotlib inline
+import matplotlib.pyplot as plt
+
+np.random.seed(seed=42)
+N = 1000
+x = np.random.uniform(size=N, low=0, high=1)
+y = np.random.uniform(size=N, low=0, high=1)
+accept = (x*x+y*y) <= 1
+reject = np.logical_not(accept)
+
+fig, ax = plt.subplots(1)
+ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)
+ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)
+ax.set_aspect('equal')
+```
+
+
+![png](output_6_0.png)
+
+
+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:
+
+
+```python
+4*np.mean(accept)
+```
+
+
+
+
+    3.112
+
 
-- 
2.18.1