Fast Fibonacci computation

Question

I saw a comment on Google+ a few weeks ago in which someone demonstrated a straight-forward computation of Fibonacci numbers which was not based on recursion and didn\'t use mem

Answer 1

Since Fibonacci sequence is a linear recurrence, its members can be evaluated in closed form. This involves computing a power, which can be done in O(logn) similarly to the matrix-multiplication solution, but the constant overhead should be lower. That's the fastest algorithm I know.

EDIT

Sorry, I missed the "exact" part. Another exact O(log(n)) alternative for the matrix-multiplication can be calculated as follows

from functools import lru_cache

@lru_cache(None)
def fib(n):
    if n in (0, 1):
        return 1
    if n & 1:  # if n is odd, it's faster than checking with modulo
        return fib((n+1)//2 - 1) * (2*fib((n+1)//2) - fib((n+1)//2 - 1))
    a, b = fib(n//2 - 1), fib(n//2)
    return a**2 + b**2

This is based on the derivation from a note by Prof. Edsger Dijkstra. The solution exploits the fact that to calculate both F(2N) and F(2N-1) you only need to know F(N) and F(N-1). Nevertheless, you are still dealing with long-number arithmetics, though the overhead should be smaller than that of the matrix-based solution. In Python you'd better rewrite this in imperative style due to the slow memoization and recursion, though I wrote it this way for the clarity of the functional formulation.