Worst case running time of this algorithm (how to justify)?

问题

I have an algorithm i'm trying to implement. I've been asked to determine a function that describes its worst case running time. As input, it takes an array of some length (lets call it n). Then what it does is as follows:

if (n==0){ return 0;}
else if(n==1){return A[0];}
else{
     return f(n-1)+f(n-2)
}

Sorry if I'm a tad sparse on the implementation details, but in a sense, its rather similar to something like the fibbanoci sequence. I'm thinking the worst case running time of this algorithm is t(n)=2^n, because if n is large, it will decompose into 2 separate calculations, which in turn will split into 2 more and so on. I'm just not sure how to formally justify this

回答1:

Let's first get a recursion for the running time.

T(0) = T(1) = 1

since both just return a number (one is an array-lookup, but that's constant time too). And for n > 1 we have

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

since you evaluate f(n-1) and f(n-2) and add the two results. That's almost the same recurrence as the Fibonacci sequence itself, F(n) = F(n-1) + F(n-2), and the result is closely related.

 n | T(n) | F(n)
----------------
 0 |   1  |   0
 1 |   1  |   1
 2 |   3  |   1
 3 |   5  |   2
 4 |   9  |   3
 5 |  15  |   5
 6 |  25  |   8
 7 |  41  |  13
 8 |  67  |  21
 9 | 109  |  34
10 | 177  |  55
11 | 287  |  89

If you look at the values, you see that

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

and can prove that with induction, if you need to.

Since the terms of the Fibonacci sequence are given by F(n) = (φ^n - (1-φ)^n)/√5, where φ = (1 + √5)/2, you see that the complexity of your f is also Θ(φ^n), like that of the Fibonacci sequence. That's better than Θ(2^n), but still exponential, so calculation using this way is only feasible for small n.

来源：https://stackoverflow.com/questions/12655496/worst-case-running-time-of-this-algorithm-how-to-justify