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<div id="content">
<h1 class="title">À propos du calcul de \(\pi\)</h1>
<div id="table-of-contents">
<h2>Table des matières</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org9e00bf0">1. En demandant à la lib maths</a></li>
<li><a href="#org7830e5e">2. En utilisant la méthodes des aiguilles de Buffon</a></li>
<li><a href="#orgec609c5">3. Avec un argument "fréquentiel" de surface</a></li>
</ul>
</div>
</div>
<div id="outline-container-org9e00bf0" class="outline-2">
<h2 id="org9e00bf0"><span class="section-number-2">1</span> En demandant à la lib maths</h2>
<div class="outline-text-2" id="text-1">
<p>
Mon ordinateur m'indique que \(\pi\) vaut approximativement :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #4d9391;">from</span> math <span style="color: #4d9391;">import</span> *
pi
</pre>
</div>
<pre class="example">
3.141592653589793
</pre>
</div>
</div>
<div id="outline-container-org7830e5e" class="outline-2">
<h2 id="org7830e5e"><span class="section-number-2">2</span> En utilisant la méthodes des aiguilles de Buffon</h2>
<div class="outline-text-2" id="text-2">
<p>
Mais calculé avec la <b>méthode</b> des <a href="https://fr.wikipedia.org/wiki/Aiguille_de_Buffon">aiguilles de Buffon</a>, on obtiendrait comme
<b>approximation</b> :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #4d9391;">import</span> numpy <span style="color: #4d9391;">as</span> np
np.random.seed<span style="color: #6bd9db;">(</span>seed=<span style="color: #cd5c60;">42</span><span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">N</span> = <span style="color: #cd5c60;">10000</span>
<span style="color: #807f96;">x</span> = np.random.uniform<span style="color: #6bd9db;">(</span>size=N, low=<span style="color: #cd5c60;">0</span>, high=<span style="color: #cd5c60;">1</span><span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">theta</span> = np.random.uniform<span style="color: #6bd9db;">(</span>size=N, low=<span style="color: #cd5c60;">0</span>, high=pi/<span style="color: #cd5c60;">2</span><span style="color: #6bd9db;">)</span>
<span style="color: #cd5c60;">2</span> / <span style="color: #6bd9db;">(</span><span style="color: #80bcb6;">sum</span><span style="color: #ab98b5;">(</span><span style="color: #5D8272;">(</span>x + np.sin<span style="color: #807f96;">(</span>theta<span style="color: #807f96;">)</span><span style="color: #5D8272;">)</span> &gt; <span style="color: #cd5c60;">1</span><span style="color: #ab98b5;">)</span> / N<span style="color: #6bd9db;">)</span>
</pre>
</div>
<pre class="example">
3.128911138923655
</pre>
</div>
</div>
<div id="outline-container-orgec609c5" class="outline-2">
<h2 id="orgec609c5"><span class="section-number-2">3</span> Avec un argument "fréquentiel" de surface</h2>
<div class="outline-text-2" id="text-3">
<p>
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 \(X \sim U(0, 1)\) et \(Y
\sim Y(0, 1)\) alors \(P[X^2 + Y^2 \le 1] = \pi / 4\) (voir méthode de <a href="https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80">Monte
Carlo sur Wikipedia</a>). Le code suivant illustre ce fait :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #4d9391;">import</span> matplotlib.pyplot <span style="color: #4d9391;">as</span> plt
np.random.seed<span style="color: #6bd9db;">(</span>seed=<span style="color: #cd5c60;">42</span><span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">N</span> = <span style="color: #cd5c60;">1000</span>
<span style="color: #807f96;">x</span> = np.random.uniform<span style="color: #6bd9db;">(</span>size=N, low=<span style="color: #cd5c60;">0</span>, high=<span style="color: #cd5c60;">1</span><span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">y</span> = np.random.uniform<span style="color: #6bd9db;">(</span>size=N, low=<span style="color: #cd5c60;">0</span>, high=<span style="color: #cd5c60;">1</span><span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">accept</span> = <span style="color: #6bd9db;">(</span>x*x+y*y<span style="color: #6bd9db;">)</span> &lt;= <span style="color: #cd5c60;">1</span>
<span style="color: #807f96;">reject</span> = np.logical_not<span style="color: #6bd9db;">(</span>accept<span style="color: #6bd9db;">)</span>
<span style="color: #807f96;">fig</span>, <span style="color: #807f96;">ax</span> = plt.subplots<span style="color: #6bd9db;">(</span><span style="color: #cd5c60;">1</span><span style="color: #6bd9db;">)</span>
ax.scatter<span style="color: #6bd9db;">(</span>x<span style="color: #ab98b5;">[</span>accept<span style="color: #ab98b5;">]</span>, y<span style="color: #ab98b5;">[</span>accept<span style="color: #ab98b5;">]</span>, c=<span style="color: #6fb593;">'b'</span>, alpha=<span style="color: #cd5c60;">0</span>.<span style="color: #cd5c60;">2</span>, edgecolor=<span style="color: #ab98b5;">None</span><span style="color: #6bd9db;">)</span>
ax.scatter<span style="color: #6bd9db;">(</span>x<span style="color: #ab98b5;">[</span>reject<span style="color: #ab98b5;">]</span>, y<span style="color: #ab98b5;">[</span>reject<span style="color: #ab98b5;">]</span>, c=<span style="color: #6fb593;">'r'</span>, alpha=<span style="color: #cd5c60;">0</span>.<span style="color: #cd5c60;">2</span>, edgecolor=<span style="color: #ab98b5;">None</span><span style="color: #6bd9db;">)</span>
ax.set_aspect<span style="color: #6bd9db;">(</span><span style="color: #6fb593;">'equal'</span><span style="color: #6bd9db;">)</span>
plt.savefig<span style="color: #6bd9db;">(</span>matplot_lib_filename<span style="color: #6bd9db;">)</span>
<span style="color: #4d9391;">print</span><span style="color: #6bd9db;">(</span>matplot_lib_filename<span style="color: #6bd9db;">)</span>
</pre>
</div>
<div class="figure">
<p><img src="pi_monte_carlo.png" alt="pi_monte_carlo.png" />
</p>
</div>
<p>
Il est alors aisé d'obtenir une approximation (pas terrible) de \(\pi\) en comptant combien de fois, en moyenne, \(X^2 + Y^2\) est inférieur à 1 :
</p>
<div class="org-src-container">
<pre class="src src-python"><span style="color: #cd5c60;">4</span> * np.mean<span style="color: #6bd9db;">(</span>accept<span style="color: #6bd9db;">)</span>
</pre>
</div>
<pre class="example">
3.112
</pre>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="date">Date: 2020-07-29 mer. 00:00</p>
<p class="author">Auteur: Pierre AYOUB</p>
<p class="date">Created: 2020-07-29 mer. 14:28</p>
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
</div>
</body>
</html>
module2/exo1/toy_document_orgmode_python_fr.org
View file @ 2c85c94a
#+TITLE: Votre titre #+TITLE: À propos du calcul de $\pi$
#+AUTHOR: Votre nom #+AUTHOR: Pierre AYOUB
#+DATE: La date du jour #+DATE: [2020-07-29 mer.]
#+LANGUAGE: fr #+LANGUAGE: fr
# #+PROPERTY: header-args :eval never-export # #+PROPERTY: header-args :eval never-export
...@@ -11,83 +11,72 @@ ...@@ -11,83 +11,72 @@
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#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script> #+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
* Quelques explications * En demandant à la lib maths
Ceci est un document org-mode avec quelques exemples de code Mon ordinateur m'indique que $\pi$ vaut approximativement :
python. Une fois ouvert dans emacs, ce document peut aisément être
exporté au format HTML, PDF, et Office. Pour plus de détails sur #+BEGIN_SRC python :results value :exports both :session *sess_python*
org-mode vous pouvez consulter https://orgmode.org/guide/. from math import *
pi
Lorsque vous utiliserez le raccourci =C-c C-e h o=, ce document sera #+END_SRC
compilé en html. Tout le code contenu sera ré-exécuté, les résultats
récupérés et inclus dans un document final. Si vous ne souhaitez pas #+RESULTS:
ré-exécuter tout le code à chaque fois, il vous suffit de supprimer : 3.141592653589793
le # et l'espace qui sont devant le ~#+PROPERTY:~ au début de ce
document. * En utilisant la méthodes des aiguilles de Buffon
Comme nous vous l'avons montré dans la vidéo, on inclue du code Mais calculé avec la *méthode* des [[https://fr.wikipedia.org/wiki/Aiguille_de_Buffon][aiguilles de Buffon]], on obtiendrait comme
python de la façon suivante (et on l'exécute en faisant ~C-c C-c~): *approximation* :
#+begin_src python :results output :exports both #+BEGIN_SRC python :results value :exports both :session *sess_python*
print("Hello world!") import numpy as np
#+end_src np.random.seed(seed=42)
N = 10000
#+RESULTS: x = np.random.uniform(size=N, low=0, high=1)
: Hello world! theta = np.random.uniform(size=N, low=0, high=pi/2)
2 / (sum((x + np.sin(theta)) > 1) / N)
Voici la même chose, mais avec une session python, donc une #+END_SRC
persistance d'un bloc à l'autre (et on l'exécute toujours en faisant
~C-c C-c~). #+RESULTS:
#+begin_src python :results output :session :exports both : 3.128911138923655
import numpy
x=numpy.linspace(-15,15) * Avec un argument "fréquentiel" de surface
print(x)
#+end_src 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 $X \sim U(0, 1)$ et $Y
#+RESULTS: \sim Y(0, 1)$ alors $P[X^2 + Y^2 \le 1] = \pi / 4$ (voir méthode de [[https://fr.wikipedia.org/wiki/M%25C3%25A9thode_de_Monte-Carlo#D%25C3%25A9termination_de_la_valeur_de_%25CF%2580][Monte
#+begin_example Carlo sur Wikipedia]]). Le code suivant illustre ce fait :
[-15. -14.3877551 -13.7755102 -13.16326531 -12.55102041
-11.93877551 -11.32653061 -10.71428571 -10.10204082 -9.48979592 #+HEADER: :var matplot_lib_filename="pi_monte_carlo.png"
-8.87755102 -8.26530612 -7.65306122 -7.04081633 -6.42857143 #+BEGIN_SRC python :results output file :exports both :session *sess_python*
-5.81632653 -5.20408163 -4.59183673 -3.97959184 -3.36734694 import matplotlib.pyplot as plt
-2.75510204 -2.14285714 -1.53061224 -0.91836735 -0.30612245
0.30612245 0.91836735 1.53061224 2.14285714 2.75510204 np.random.seed(seed=42)
3.36734694 3.97959184 4.59183673 5.20408163 5.81632653 N = 1000
6.42857143 7.04081633 7.65306122 8.26530612 8.87755102 x = np.random.uniform(size=N, low=0, high=1)
9.48979592 10.10204082 10.71428571 11.32653061 11.93877551 y = np.random.uniform(size=N, low=0, high=1)
12.55102041 13.16326531 13.7755102 14.3877551 15. ]
#+end_example accept = (x*x+y*y) <= 1
reject = np.logical_not(accept)
Et enfin, voici un exemple de sortie graphique:
#+begin_src python :results output file :session :var matplot_lib_filename="./cosxsx.png" :exports results fig, ax = plt.subplots(1)
import matplotlib.pyplot as plt 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)
plt.figure(figsize=(10,5)) ax.set_aspect('equal')
plt.plot(x,numpy.cos(x)/x)
plt.tight_layout() plt.savefig(matplot_lib_filename)
print(matplot_lib_filename)
plt.savefig(matplot_lib_filename) #+END_SRC
print(matplot_lib_filename)
#+end_src #+RESULTS:
[[file:pi_monte_carlo.png]]
#+RESULTS:
[[file:./cosxsx.png]] Il est alors aisé d'obtenir une approximation (pas terrible) de $\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inférieur à 1 :
Vous remarquerez le paramètre ~:exports results~ qui indique que le code #+BEGIN_SRC python :results value :exports both :session *sess_python*
ne doit pas apparaître dans la version finale du document. Nous vous 4 * np.mean(accept)
recommandons dans le cadre de ce MOOC de ne pas changer ce paramètre #+END_SRC
(indiquer ~both~) car l'objectif est que vos analyses de données soient
parfaitement transparentes pour être reproductibles. #+RESULTS:
: 3.112
Attention, la figure ainsi générée n'est pas stockée dans le document
org. C'est un fichier ordinaire, ici nommé ~cosxsx.png~. N'oubliez pas
de le committer si vous voulez que votre analyse soit lisible et
compréhensible sur GitLab.
Enfin, n'oubliez pas que nous vous fournissons dans les ressources de
ce MOOC une configuration avec un certain nombre de raccourcis
claviers permettant de créer rapidement les blocs de code python (en
faisant ~<p~, ~<P~ ou ~<PP~ suivi de ~Tab~).
Maintenant, à vous de jouer! Vous pouvez effacer toutes ces
informations et les remplacer par votre document computationnel.
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