On the computation of pi
Table of Contents
1 Asking the maths library
My computer tells me that \(\pi\) is approximatively
pi
[1] 3.141593
2 Buffon's needle
Applying the method of Buffon's needle, we get the approximation
set.seed(42) N = 100000 x = runif(N) theta = pi/2*runif(N) 2/(mean(x+sin(theta)>1))
[1] 3.14327
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 \(X\sim U(0,1)\) and \(Y\sim U(0,1)\), then \(P[X^2+Y^2\leq 1] = \pi/4\) (see "Monte Carlo method" on Wikipedia). The following code uses this approach:
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:
4*mean(df$Accept)
[1] 3.156