Why is the complexity of computing the Fibonacci series 2^n and not n^2?

Question

I am trying to find complexity of Fibonacci series using a recursion tree and concluded height of tree = O(n) worst case, cost of each level = cn,

Answer 1

You are correct that the depth of the tree is O(n), but you are not doing O(n) work at each level. At each level, you do O(1) work per recursive call, but each recursive call then contributes two new recursive calls, one at the level below it and one at the level two below it. This means that as you get further and further down the recursion tree, the number of calls per level grows exponentially.

Interestingly, you can actually establish the exact number of calls necessary to compute F(n) as 2F(n + 1) - 1, where F(n) is the nth Fibonacci number. We can prove this inductively. As a base case, to compute F(0) or F(1), we need to make exactly one call to the function, which terminates without making any new calls. Let's say that L(n) is the number of calls necessary to compute F(n). Then we have that

L(0) = 1 = 2*1 - 1 = 2F(1) - 1 = 2F(0 + 1) - 1

L(1) = 1 = 2*1 - 1 = 2F(2) - 1 = 2F(1 + 1) - 1

Now, for the inductive step, assume that for all n' < n, with n ≥ 2, that L(n') = 2F(n + 1) - 1. Then to compute F(n), we need to make 1 call to the initial function that computes F(n), which in turn fires off calls to F(n-2) and F(n-1). By the inductive hypothesis we know that F(n-1) and F(n-2) can be computed in L(n-1) and L(n-2) calls. Thus the total runtime is

1 + L(n - 1) + L(n - 2)

= 1 + 2F((n - 1) + 1) - 1 + 2F((n - 2) + 1) - 1

= 2F(n) + 2F(n - 1) - 1

= 2(F(n) + F(n - 1)) - 1

= 2(F(n + 1)) - 1

= 2F(n + 1) - 1

Which completes the induction.

At this point, you can use Binet's formula to show that

L(n) = 2(1/√5)(((1 + √5) / 2)ⁿ - ((1 - √5) / 2)ⁿ) - 1

And thus L(n) = O(((1 + √5) / 2)ⁿ). If we use the convention that

φ = (1 + √5) / 2 ≈ 1.6

We have that

L(n) = Θ(φⁿ)

And since φ < 2, this is o(2ⁿ) (using little-o notation).

Interestingly, I've chosen the name L(n) for this series because this series is called the Leonardo numbers. In addition to its use here, it arises in the analysis of the smoothsort algorithm.

Hope this helps!