Question

I’ve been asked to calculate the big theta for a homework assignment, but the lecture material has been a little sparse on this area.

Given the loops

for (x = 1; x <= n; x *= 2){
  for(y = 1; y <= n; y += 2)
    t++;

I’ve worked out an execution chart as

x       y
1       1, 3, 5, 7 ... n-2, n
2       1, 3, 5, 7 ... n-2, n
4       1, 3, 5, 7 ... n-2, n
8       1, 3, 5, 7 ... n-2, n
log n   (n+1)/2

Its the inner loop incrementor that’s throwing me off. It executes (n+1)/2 times, so the big theta must be (n log n + log n)/2.

Am I correct?

Answer 1

Your calculations appear correct, but you need to continue a few more steps.

Big theta ignores everything smaller than the biggest term and all constant factors (the equation might help with understanding this).

Theta((n log n + log n)/2)
= Theta(1/2 n log n + 1/2 log n)
= Theta(1/2 n log n)
= Theta(n log n)

Why it ignores constant factors is obvious from looking at the equation (you can just manipulate k appropriately).

Why it ignores everything smaller than the biggest term:

Assume g(x) <= f(x) (from any x onward, since the Theta equation only needs to hold from any n onward)
f(x) <= f(x) + g(x) <= 2.f(x)
Thus Theta(f(x) + g(x)) = Theta(2.f(x)) = Theta(f(x))