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d7c7db59c0b2f422c65ebb1d22a0db91
mooc-rr
Commits
310b68dd
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310b68dd
authored
May 10, 2020
by
Elias Bouacida
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Code de l'exercice 1 du module 2. Premier test.
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module2/exo1/toy_document_orgmode_python_fr.org
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310b68dd
#+TITLE:
Votre titre
#+AUTHOR:
Votre nom
#+DATE:
La date du jour
#+TITLE:
À propos de \pi
#+AUTHOR:
Konrad Hansen
#+DATE:
2019-03-28
#+LANGUAGE: fr
# #+PROPERTY: header-args :eval never-export
...
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@@ -11,6 +11,67 @@
#+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>
* En demandant à la lib maths
Mon ordinateur indique que \pi vaut /approximativement/:
#+begin_src python :results output :exports both
from math import *
pi
#+end_src
#+RESULTS:
: 3.141592653589793
* En utilisant la méthode des aiguilles de Buffon
Mais calculé avec la *méthode* des [[https://fr.wikipedia.org/wiki/Aiguille_de_Buffon][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=pi/2)
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∼U(0,1) et Y∼U(0,1) alors P[X^2+Y^2≤1]=π/4 (voir
[[https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80][méthode de Monte Carlo sur Wikipedia]]).
Le code suivant illustre ce fait :
#+begin_src python :results output :session :exports both
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')
plt.savefig(matplot_lib_filename)
print(matplot_lib_filename)
#+end_src
#+begin_src python :results output :session :exports both
4*np.mean(accept)
#+end_src
* Quelques explications
Ceci est un document org-mode avec quelques exemples de code
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