---
title: "On the computation of pi"
author: "Arnaud Legrand"
date: "25 juin 2018"
output: html_document
---
## Asking the maths library
My computer tells me that π 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
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∼U(0,1) and Y∼U(0,1), then P[X2+Y2≤1]=π/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
df = data.frame(X = runif(N), Y = runif(N))
df$Accept = (df$X**2 + df$Y**2 <=1)
library(ggplot2)
ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()
```
It is therefore straightforward to obtain a (not really good) approximation to π by counting how many times, on average, X2+Y2 is smaller than 1 :
```{r}
4*mean(df$Accept)
```