recursion tree and binary tree cost calculation

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)?

This can't make sense to anyone unless you can show us the Theta and small omega functions. — Puppy
– Puppy, Commented May 4, 2010 at 19:39

Yuliy · Accepted Answer · 2010-05-04 19:42:37Z

2

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)

answered May 4, 2010 at 19:42

Yuliy

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