Question

I have a simple function that calculates the n-th fibonnaci number below:

fibonacci :: Integer -> Integer
fibonacci 0 = 0
fibonacci 1 = 1
fibonacci n = (fibonacci (n-1) ) + (fibonacci (n-2))

But i am interested in a way to count the number of recursions of this function. Any ideas how to do it?

Answer 1

recursions :: Integer -> Integer
recursions 0 = 0
recursions 1 = 0
recursions n = recursions (n-1) + recursions (n-2) + 2

For the base cases, there are no recursions, for everything else, we have two direct recursive calls and those that are invoked from the two.

You can also re-use the fibonacci code,

recursions n = 2*fibonacci (n+1) - 2