Question

I had an assignment in school last week to implement a function for calculating the n:th number in the fibonacci sequence. A ‘sub-assignment’ was to implement it using accumulation(Might not be a correct translation) in order to give the function O(n) time complexity. This all worked fine until I tried making the function (Int -> Integer). By experimenting a bit I realised that the time complexity was close to O(n^2) for really large numbers.
It occurs to me that this must be because of the Integer implementation, which makes me a bit curious about how it works. I did a few Google searches and didn’t find anything that seemed useful so I am turning to you guys in hope of either getting an explanation or a link that explains it thoroughly.

My code:

ackfib 0 = 0
ackfib 1 = 1        
ackfib n = loop n 1 0 1
    where
        loop n n1 n2 i 
            | i < n     = loop n (n1+n2) n1 (i+1)
            | i == n    = n1
            | i > n     = error "n must be greater than or equal to 0"

I am thankful for all answers

Viktor

Answer 1

This has nothing to do with Haskell really, it’s just a result of the fact that the Fibonacci numbers grow exponentially quickly. Specifically, the nth Fibonacci number has about (log₂ φ) n or roughly 0.48 n bits where φ is the golden ratio (1 + sqrt 5) / 2. Since addition of k-bit integers takes O(k) time, your O(n) additions actually take a total of O(n^2) time, because on average the numbers you’re adding have O(n) bits.

(Note for sticklers: big O should really be big Theta in the above.)