Question

could you plese tell me the asymptotic complexity of this code?

f(n):
if (n<=2) then return 1;
else { 
     if (n>950) then { i=n/2; return f(i);}
     else return f(n-2);
}

I have thought of two solutions.

a)

O(1)         when n<=2
T(n/2) + 1   when n > 950
T(n-2) + 1   when 950>=n>2

and solving the recurrences:

O(1)         when n<=2
Θ(log n)     when n > 950
O(n^2)       when 950>=n>2

b) However I’m not so sure about the complexity of the last two statements, because if n is bigger than 950 the alghorithm will call f(i) until i is smaller than 950 and then proceed calling f(n-2).
So, the other solution is this one:

O(1)                  when n<=2
T(n/2) + T(n-2) + 1   otherwise

and solving the recurrences:

O(1)         when n<=2
O(n^2)       otherwise

I actually think the second one is right, but I’m not sure of it.
Thanks for your help.

Answer 1

okay, so start by thinking about what happens if n is really big. Eventually n will be big enough it will dominate everything else. And sure enough, until you get down to n=950 you’re going to get O(lg n) where “lg” denotes log base 2. (Why do I know this off hand?because the size of n decreases by a power of two every iteration.)

Once n is down to 950, then it’s going to decrease by 2 every time, so from 950 to 2 is O(n) — because you hit basically half the values and that 1/2 disappears into the constant.

But observe that there exists a value of n for which lg n > 950/2. Ergo the lg n term will dominate for some value of n. O(lg n).