diff --git a/module2/exo1/module2/exo1/toy_document_en.Rmd b/module2/exo1/module2/exo1/toy_document_en.Rmd
index d2d0fb1828d9be2ea212112ca237379b65b0c9c1..b7c1a551aa4245f193e9fc6ab830ae6702905496 100644
--- a/module2/exo1/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/module2/exo1/toy_document_en.Rmd
@@ -1,43 +1,28 @@
-1 On the computation of π
-1.1 Asking the maths library
-My computer tells me that π is approximatively
-In [1]: from math import *
-print(pi)
-3.141592653589793
-1.2 Buffon’s needle
-Applying the method of Buffon’s needle, we get the approximation
-In [2]: 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)
-Out[2]: 3.1289111389236548
+# On the computation of π
+# Nicholas
+# October 21, 2025
 
-1.3 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 ∼ U(0, 1) and Y ∼ U(0, 1), then P[X
-2 + Y
-2 ≤ 1] = π/4 (see "Monte Carlo method"
-on Wikipedia). The following code uses this approach:
-In [3]: %matplotlib inline
-import matplotlib.pyplot as plt
-np.random.seed(seed=42)
+# Asking the math library
+pi
+## [1] 3.141593
+
+# Buffon’s needle
+set.seed(42)
+N = 100000
+x = runif(N)
+theta = pi/2 * runif(N)
+2 / (mean(x + sin(theta) > 1))
+## [1] 3.14327
+
+# Using an area fraction argument
+set.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')
+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 π by counting how
-many times, on average, X
-2 + Y
-2
-is smaller than 1:
-In [4]: 4*np.mean(accept)
-Out[4]: 3.1120000000000001
-{r}## [1] 3.156
\ No newline at end of file
+4 * mean(df$Accept)
+## [1] 3.156