premier exercice, première proposition

module2/exo1/toy_document_orgmode_python_fr.org

-#+TITLE:  Votre titre
-#+AUTHOR: Votre nom
-#+DATE:   La date du jour
+#+TITLE:  À propos du calcul de π
+#+AUTHOR: Paul Beziau
+#+DATE:   9/12/2021
 #+LANGUAGE: fr
 # #+PROPERTY: header-args :eval never-export
 
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 * Quelques explications
 
@@ -91,3 +91,54 @@ 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.
+* En demandant à la lib maths
+Mon ordinateur m'indique que π vaut approximativement: 
+
+#+begin_src python :exports both :session
+from math import *
+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 :exports both :session
+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
+
+#+begin_src python :results file :session :var matplot_lib_filename="figure.png" :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
+
+#+RESULTS:
+[[file:<matplotlib.collections.PathCollection object at 0x7f8ea60c6be0>]]