From 8211c0c4e4017d24b0569022388c08d770c20d1f Mon Sep 17 00:00:00 2001
From: 616531ae6e3981887d590c5b8da5026e
<616531ae6e3981887d590c5b8da5026e@app-learninglab.inria.fr>
Date: Wed, 11 Jan 2023 08:36:17 +0000
Subject: [PATCH] update
---
module2/exo1/toy_notebook_en.ipynb | 16 ++++++++--------
1 file changed, 8 insertions(+), 8 deletions(-)
diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 3d4081e..78e09b3 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,7 +4,8 @@
"cell_type": "markdown",
"metadata": {},
"source": [
- "# On the computation of $\\pi$ \n",
+ "# On the computation of $\\pi$\n",
+ "\n",
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
@@ -31,7 +32,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
- "## Buffon's needle ##\n",
+ "## Buffon's needle\n",
"Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the _approximation_"
]
},
@@ -66,10 +67,9 @@
"cell_type": "markdown",
"metadata": {},
"source": [
- "## Using a surface fraction argument ##\n",
+ "## 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 \n",
- "fact that if X ~ *U*(0,1) and Y ~ *U*(0,1), then *P*[X2 + Y2 $\\le$ 1] = $\\pi$/4 (see [\"Monte Carlo method\" \n",
- "on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach."
+ "fact that if X ~ *U*(0,1) and Y ~ *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:"
]
},
{
@@ -100,9 +100,10 @@
"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",
+ "\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",
@@ -113,8 +114,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
- " It is then straightforward to obtain a (not really good) approximation to $\\pi$ by counting how \n",
- "many times, on average, X2 + Y2 is smaller than 1:"
+ "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:"
]
},
{
--
2.18.1