---
title: "On the computation of pi"
author: "Arnaud Legrand"
date: "25 juin 2018"
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](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 $X\sim U(0,1)$ and $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1] = \pi/4$ (see ["Monte Carlo method" on Wikipedia](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 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:

```{r}
4*mean(df$Accept)
```

```{r}

data <- c(
  14.0, 7.6, 11.2, 12.8, 12.5, 9.9, 14.9, 9.4, 16.9, 10.2, 14.9, 18.1, 7.3, 9.8, 10.9,12.2, 9.9, 2.9, 2.8, 15.4, 15.7, 9.7, 13.1, 13.2, 12.3, 11.7, 16.0, 12.4, 17.9, 12.2, 16.2, 18.7, 8.9, 11.9, 12.1, 14.6, 12.1, 4.7, 3.9, 16.9, 16.8, 11.3, 14.4, 15.7, 14.0, 13.6, 18.0, 13.6, 19.9, 13.7, 17.0, 20.5, 9.9, 12.5, 13.2, 16.1, 13.5, 6.3, 6.4, 17.6, 19.1, 12.8, 15.5, 16.3, 15.2, 14.6, 19.1, 14.4, 21.4, 15.1, 19.6, 21.7, 11.3, 15.0, 14.3, 16.8, 14.0, 6.8, 8.2, 19.9, 20.4, 14.6, 16.4, 18.7, 16.8, 15.8, 20.4, 15.8, 22.4, 16.2, 20.3, 23.4, 12.1, 15.5, 15.4, 18.4, 15.7, 10.2, 8.9, 21.0  
)

```

```{r}
mean_value <- mean(data)
mean_value
```

```{r}

minimum <- min(data)
minimum

```

```{r}
maximum <- max(data)
maximum
```

\

```{r}

median <- median(data)
median
```

\

```{r}
sd <- sd(data)
sd

```

```{r}
# Line plot
plot(data, type = "o", col = "blue", 
     xlab = "Index", ylab = "Value", 
     main = "Line Plot of Data")
```

```{r}
# Histogram
hist(data, breaks = 10, col = "lightgreen", 
     xlab = "Value", main = "Histogram of Data")
```

\