From 98da18eae0f7cc7f2344a626b8a503697a870cad Mon Sep 17 00:00:00 2001
From: "Anton Y." <skycluster@gmail.com>
Date: Mon, 7 Jul 2025 02:09:54 +0300
Subject: [PATCH] task try 3

---
 module2/exo1/toy_document_orgmode_python_en.org | 11 -----------
 1 file changed, 11 deletions(-)

diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index 5f0f8a2..9dc0f60 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -18,9 +18,6 @@ from math import *
 pi
 #+end_src
 
-#+RESULTS:
-: 3.141592653589793
-
 * * Buffon's needle
 Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation*
 
@@ -33,9 +30,6 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
 2/(sum((x+np.sin(theta))>1)/N)
 #+end_src
 
-#+RESULTS:
-: 3.128911138923655
-
 * Using a surface fraction argument
 A method that is easier to understand and does not make use of the
 $\sin$ function is based on the fact that if $X\sim U(0,1)$ and $Y\sim
@@ -62,9 +56,6 @@ plt.savefig(matplot_lib_filename)
 print(matplot_lib_filename)
 #+end_src
 
-#+RESULTS:
-[[file:figure_pi_mc2.png]]
-
 It is then straightforward to obtain a (not really good) approximation
 to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller
 than 1: 
@@ -72,5 +63,3 @@ than 1:
 #+begin_src python :results output :session *python* :exports both
 4*np.mean(accept)
 #+end_src
-
-#+RESULTS:
-- 
2.18.1