Question

I have a Haskell solution to Project Euler Problem 2 which works fine for the four million limit, as well as for limits up to 10^100000, taking only a few seconds on my machine.

But for anything bigger, e.g. 10^1000000, the computation does not return in good time, if at all (have tried leaving it for a couple of minutes). What is the limiting factor here?

evenFibonacciSum :: Integer -> Integer
evenFibonacciSum limit = 
  foldl' (\t (_,b) -> t + b) 0 . takeWhile ((<=limit) . snd) . iterate doIteration $ (1,2) where
    doIteration (a, b) = (twoAB - a, twoAB + b) where
      twoAB = 2*(a + b)

Answer 1

The problem is that you are summing the (even) Fibonacci numbers. That means you have to calculate them all. But

F(n) ≈ φ^n / √5, with  φ = (1 + √5)/2

So you are adding a lot of numbers of large size, Θ(n) bits for F(n). For a limit of 10^1000000, you need about 800000×2 additions of numbers larger than 10^500000. In general, you need Θ(n) additions of numbers with Θ(n) bits.

Adding numbers of d digits [in whatever base] is an O(d) operation. So your algorithm is quadratic in the exponent.

To avoid that, find a closed formula for the sum S(k) of the first k even Fibonacci numbers (hint: it’s a relatively easy formula involving one Fibonacci number), find the largest k so that F(3*k) <= limit, and compute the sum using the formula and the algorithm to compute F(n) in O(log n) steps e.g. here.