From 46472757f97482842d307aed8a3f545b7ffce1bd Mon Sep 17 00:00:00 2001
From: 2967c9aae1a2f4cee2d1ad632cabf260
 <2967c9aae1a2f4cee2d1ad632cabf260@app-learninglab.inria.fr>
Date: Fri, 3 Jun 2022 11:18:07 +0000
Subject: [PATCH] Restored previous version

---
 module2/exo1/toy_notebook_en.ipynb | 99 ++++++++----------------------
 1 file changed, 24 insertions(+), 75 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index a4d20f3..a7e3998 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -1,75 +1,24 @@
-# 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
-
-
+{
+ "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
+}
\ No newline at end of file
-- 
2.18.1