Question

if x:
    for i in range(a):
       for z in range(a):
          for k in range(z):
             for p in range(i):
                c = (i * z) + (k * p)
else:
    for i in range(a):
       for z in range(a):
          for k in range(z):
              c = (i * z) + (k * p)

Would this be O(n^4)? Also, how many multiplications would occur?

EDIT: updated the code. Also, since the lower bound captures the max number of steps a valid input will force, wouldn’t big omega be n^4 as well?

Answer 1

Yes, the complexity is still O(n^4). To make things simple, here is the trick to rearrange your code

for i in range(a):
   for p in range(i):
       f(i, p)

where f(i, p) is

for z in range(a):
    for k in range(z):
        c = (i * z) + (k * p)

In the first part, f(i, p) has been executed for O(n^2/2) up to the largest order (because of the summation sum_i (i^2), do the math yourself). Similarly, the f(i, p) has the complexity of f(i, p) which is again equal to O(n^2/2).

So the combined resulting order is O(n^4/4). and there is two multiplications for each operation, so number of multiplication is O(n^4/2)