From dedcad21543dcf256f9c0afb1ca9e61c8a06682d Mon Sep 17 00:00:00 2001
From: 3f03c2c14c4cd9d0ee79b67f37ac25fe
 <3f03c2c14c4cd9d0ee79b67f37ac25fe@app-learninglab.inria.fr>
Date: Thu, 18 Mar 2021 09:06:09 +0000
Subject: [PATCH] Update toy_document_en.Rmd

---
 module2/exo1/toy_document_en.Rmd | 7 ++++++-
 1 file changed, 6 insertions(+), 1 deletion(-)

diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index f55186c..02a6e7c 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -11,12 +11,14 @@ knitr::opts_chunk$set(echo = TRUE)
 
 ## Asking the maths library
 My computer tells me that $\pi$ is *approximatively*
+
 ```{r}
 pi
 ```
 
 ## Buffon's needle
 Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__
+
 ```{r}
 set.seed(42)
 N = 100000
@@ -27,6 +29,7 @@ theta = pi/2*runif(N)
 
 ## 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 $Y\sim U(0,1)$, 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
@@ -34,10 +37,12 @@ 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)
 ```
 
-When you click on the button **Knit**, the document will be compiled in order to re-execute the R code and to include the results into the final document.
-- 
2.18.1