From 691378010fc877bed8038951f6b3810d5bd582c3 Mon Sep 17 00:00:00 2001
From: 3582521b868d7c148df1de482ec5a734
 <3582521b868d7c148df1de482ec5a734@app-learninglab.inria.fr>
Date: Wed, 25 Oct 2023 21:07:59 +0000
Subject: [PATCH] Update toy_document_en.Rmd

---
 module2/exo1/toy_document_en.Rmd | 19 ++++++++++---------
 1 file changed, 10 insertions(+), 9 deletions(-)

diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index 3015191..8ca3190 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -17,7 +17,7 @@ My computer tells me that $\pi$ is *approximatively*
 pi
 ```
 
-## Buffon's needle	
+## Buffon's needle
 
 Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
 
@@ -34,16 +34,17 @@ theta = pi/2*runif(N)
 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 $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1] = \pi/4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
 
 ```{r}
-set.seed(42)	
-N = 1000	
-df = data.frame(X = runif(N), Y = runif(N))	
-df$Accept = (df$X**2 + df$Y**2 <=1)	
-library(ggplot2)	
+set.seed(42)
+N = 1000
+df = data.frame(X = runif(N), Y = runif(N))
+df$Accept = (df$X**2 + df$Y**2 <=1)
+library(ggplot2)
 ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
+
 ```
 
 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:
 
-```{r}	
-4*mean(df$Accept)	
-```
+```{r}
+4*mean(df$Accept)
+```
\ No newline at end of file
-- 
2.18.1