Eleven seconds to compute the 1,000,000th Fibonacci number. Compared with the naive method which takes 143 seconds.

Code below:

    def add(a, b): return (a[0]+b[0], a[1]+b[1])
    def sub(a, b): return (a[0]-b[0], a[1]-b[1])
    def divsq5(a): return (a[1], a[0]/5)
    def mul(a, b): return (a[0]*b[0]+5*a[1]*b[1],a[0]*b[1]+a[1]*b[0])
    def subi(x, a): return (x-a[0],-a[1])
    def pow(b, e):
        r = (1, 0)
        while e > 0:
            if e & 1 == 1:
                result = mul(r, b)
            e = e >> 1
            b = mul(b, b)
        return r

    twophi = (1,1)

    def fib0(n):
        return divsq5(sub(pow(twophi, n), pow(subi(2,twophi), n)))[0]>>n

    def fib1(n):
        a = 0
        b = 1
        while n > 0:
            (a,b) = (b, a+b)
            n -= 1
        return a