A priori and asymptotic complexity level

How to determine开发者_如何学编程 the a priori and asymptotic complexity of following program code?

#include<stdio.h>


int br_nacina_zaba(int br_lopoca, int tren_poz, int korak) {  
    if (korak == 18) return 0;
    else if (tren_poz == br_lopoca) return 1;
    else if (tren_poz <= 0 && korak != 0) return 0;
    else if (tren_poz > br_lopoca) return 0;
    else return
               br_nacina_zaba(br_lopoca, tren_poz + 2, korak + 1)
             + br_nacina_zaba(br_lopoca, tren_poz + 3, korak + 1)
             + br_nacina_zaba(br_lopoca, tren_poz - 2, korak + 1)
             + br_nacina_zaba(br_lopoca, tren_poz - 3, korak + 1);
}

So I need to know the complexity of function br_nacina_zaba(n,0,0).

In my opinion, br_nacina_zaba(n,0,0) is in O(1). The maximum depth of the (quaternary) call tree is limited by 19 in the function's first LOC:

korak gets incremented in each recursive call. If you start with korak=0 and call the function at most 4 times in each recursive step, you'll have at most 4^18 recursive calls. 4^18 doesn't depend on n, hence the function is in O(1).

I don't know what you mean "complexity of function", but I run your function on codepad ( http://codepad.org/jFUW1ATj ) and got this result

br_nacina_zaba(1, 0, 0) was called 5 times.
br_nacina_zaba(2, 0, 0) was called 5 times.
br_nacina_zaba(3, 0, 0) was called 9 times.
br_nacina_zaba(4, 0, 0) was called 77 times.
br_nacina_zaba(5, 0, 0) was called 33445 times.
br_nacina_zaba(6, 0, 0) was called 1048573 times.
br_nacina_zaba(7, 0, 0) was called 15530681 times.