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Computational complexity of Fibonacci Sequence

Asked 2008-12-11T20:20:25.107
402

I understand Big-O notation, but I don't know how to calculate it for many functions. In particular, I've been trying to figure out the computational complexity of the naive version of the Fibonacci sequence:

int Fibonacci(int n)
{
    if (n <= 1)
        return n;
    else
        return Fibonacci(n - 1) + Fibonacci(n - 2);
}

What is the computational complexity of the Fibonacci sequence and how is it calculated?

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2 Answers

171

Just ask yourself how many statements need to execute for F(n) to complete.

For F(1), the answer is 1 (the first part of the conditional).

For F(n), the answer is F(n-1) + F(n-2).

So what function satisfies these rules? Try an (a > 1):

an == a(n-1) + a(n-2)

Divide through by a(n-2):

a2 == a + 1

Solve for a and you get (1+sqrt(5))/2 = 1.6180339887, otherwise known as the golden ratio.

So it takes exponential time.

answered 2008-12-11T20:26:21.553
36

There's a very nice discussion of this specific problem over at MIT. On page 5, they make the point that, if you assume that an addition takes one computational unit, the time required to compute Fib(N) is very closely related to the result of Fib(N).

As a result, you can skip directly to the very close approximation of the Fibonacci series:

Fib(N) = (1/sqrt(5)) * 1.618^(N+1) (approximately)

and say, therefore, that the worst case performance of the naive algorithm is

O((1/sqrt(5)) * 1.618^(N+1)) = O(1.618^(N+1))

PS: There is a discussion of the closed form expression of the Nth Fibonacci number over at Wikipedia if you'd like more information.

answered 2008-12-11T21:10:03.057

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