We often don’t root phylogenetic trees. This reduces the number of possible trees and is described by the equation:

Text: (2n-5)!/[2n-3*(n-3)!]

Rendered:

\(\frac{(2*n-5)!}{2^{n-3} * (n-3)!}\)

Modify the function used in the “number of phylogenetic trees” tutorial to work for unrooted trees.

Compare your results to http://carrot.mcb.uconn.edu/mcb396_41/tree_number.html

You can use the simplest form of the function which doesn’t have any additional argument, eg

The function is here; make the necessary changeas

```
#code below work for an un-rooted tree
tree_count <- function(n = 3){
numerator <- factorial(2*n-5)
denominator <- 2^(n-3)*factorial(n-3)
trees <- numerator / denominator
print(trees)
}
```

`tree_count(n=20)`

`## [1] 2.216431e+20`

Create a function that will work for rooted OR unrooted trees. Do this by adding an additional argument like

type = “rooted”

and conditional additional statements like

if(type == "rooted){ #do this }

if(type == "unrooted){ #do something else }

Again, you can use the simplest form of the argument.

```
tree_count_choice <- function(n, type){
if (type=="unrooted"){
numerator <- factorial(2*n-5)
denominator <- 2^(n-3)*factorial(n-3)
trees <- numerator / denominator
}
if (type=="rooted"){
numerator <- factorial(2*n-3)
denominator <- 2^(n-2)*factorial(n-2)
trees <- numerator / denominator
}
print(trees)
}
```

`tree_count_choice(n=11,"rooted")`

`## [1] 654729075`

`tree_count_choice(n=11,"unrooted")`

`## [1] 34459425`

For this assignment, render to RRubps, submit the link, and submit a screen grab that contains

- the code for the function
- the function returning the value for a rooted tree with 11 taxa
- the function returning the value for an unrooted tree with 11 taxa.