Update module2/exo1/toy_notebook_en.ipynb

parent 1d329275
{
"cells": [],
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# On the computation of $\\pi$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Asking the maths library"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"My computer tells me that $\\pi$ is _approximatively_"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"from math import *\n",
"print(pi)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Buffon’s needle"
]
},
{
"cell_type": "markdown",
"metadata": {},
"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": null,
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"np.random.seed(seed=42)\n",
"N = 10000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum((x+np.sin(theta))>1)/N)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Using a surface fraction argument"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"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 $P[X^2 + Y^2 ≤ 1] = \\pi/4$ (see [\"Monte Carlo method\"\n",
"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"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",
"y = np.random.uniform(size=N, low=0, high=1)\n",
"accept = (x*x+y*y) <= 1\n",
"reject = np.logical_not(accept)\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\n",
"many times, on average, $X^2 + Y^2$ is smaller than 1:"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"4*np.mean(accept)"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
......@@ -16,10 +130,9 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.3"
"version": "3.6.4"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
{
"cells": [],
"cells": [
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"average = 10"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": [
"sum = average + 10"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
......@@ -16,10 +49,9 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.3"
"version": "3.6.4"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
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