Question

I’ve got the following recursion:

T(n) = T(n/3) + T(2n/3) + O(n)

The height of the tree would be log3/2 of 2. Now the recursion tree for this recurrence is not a complete binary tree. It has missing nodes lower down. This makes sense to me, however I don’t understand how the following small omega notation relates to the cost of all leaves in the tree.

“… the total cost of all leaves would then be Theta (n^log3/2 of 2) which, since log3/2 of 2 is a constant strictly greater then 1, is small omega(n lg n).”

Can someone please help me understand how the Theta(n^log3/2 of 2) becomes small omega(n lg n)?

Answer 1

OK, to answer your explicit question about why n^(log_1.5(2)) is omega(n lg n):
For all k > 1, n^k grows faster than n lg n. (Powers grow faster than logs eventually). Therefore since 2 > 1.5, log_1.5(2) > 1, and thus n^(log_1.5(2)) grows faster than n lg n. And since our function is in Theta(n^(log_1.5(2))), it must also be in omega(n lg n)