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moocrr-reproducibility-study
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moocrr-session3
moocrr-reproducibility-study
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721b7c56
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721b7c56
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Jul 10, 2024
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bef6cb6b56771551e9882774baaa3f04
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Add org file to generate PDF
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721b7c56
title: "On the computation of pi"
auteur : Meiling WU
output: html_notebook
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/htmlize.css"/>
#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="http://www.pirilampo.org/styles/readtheorg/css/readtheorg.css"/>
#+HTML_HEAD: <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
#+HTML_HEAD: <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.4/js/bootstrap.min.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
#+PROPERTY: header-args :session :exports both
* En demandant à la lib maths
Mon ordinateur m'indique que π vaut /approximativement:
#+begin_src python :results output :exports both
from math import *
print(pi)
#+end_src
#+RESULTS:
: 3.141592653589793
* En utilisant la méthode des aiguilles de Buffon
Mais calculé avec la *méthode des aiguilles de Buffon, on obtiendrait comme *approximation :
#+begin_src python :results output :exports both
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=np.pi/2)
print(2/(sum((x+np.sin(theta))>1)/N))
#+end_src
#+RESULTS:
: 3.128911138923655
* Avec un argument "fréquentiel" de surface
Sinon, une méthode plus simple à comprendre et ne faisant pas
intervenir d'appel à la fonction sinus se base sur le fait que si $X\sim
U(0,1)$, alors $P[X^2+Y^2\leq 1] = \pi/4$ (voir [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on
Wikipedia]]). Le code suivant illustre ce fait :
#+begin_src python :results output :exports both
import matplotlib.pyplot as plt
import numpy as np
matplot_lib_filename = "fig.png"
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')
plt.savefig(matplot_lib_filename)
print("./fig.png")
#+end_src
#+RESULTS:
: ./fig.png
\ No newline at end of file
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