---
title: "On the computation of pi"
author: "Jayashri Govindan"
date: "23 Octobre 2023"
output: html_document
---

{r setup, include=FALSE}
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, we get the approximation

{r}
set.seed(42)
N = 100000
x = runif(N)
theta = pi/2*runif(N)
2/(mean(x+sin(theta)>1))

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:

{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)
ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()

It is therefore 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)