---
title: "On the computation of pi"
Author: Pedro Mota
Date: 14.01.2021
output:
  html_document:
    df_print: paged
---

## Asking the math library 

My computer 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 **aproximation**

```{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 use 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)
```