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moocrr-session3
moocrr-reproducibility-study
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bbb97c83
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bbb97c83
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Oct 21, 2025
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0fb7116d59f3a379c99df4cd837c797e
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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
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