# Solving the big-Oh notation for $T(n) = 2 T(n/2) + O(n)$ [duplicate]

Possible Duplicate:
Solving or approximating recurrence relations for sequences of numbers

I know that the solution for $T(n) = 2 T(n/2) + O(n)$ is $T(n) = O(n \log(n))$

But how do you get to that point? I don't understand when it says put t into the equation repeatedly until it drops out...

Any help?

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## marked as duplicate by Raphael♦Nov 23 '12 at 22:41

It means "open" the recursion.

For simplicity - denote $O(n)$ as $c\cdot n$:

\begin{align} T(n) &= 2T(n/2) + cn \\ &= 2(2T(n/4) + cn/2) + cn\\ &= 2 (2 (2T(n/8) + cn/4) + cn/2) + cn\\ & \vdots \end{align}

It might give you intuition, but it is NOT a proof. To prove it, you will need mathematical induction or the master theorem.

Proving with induction (assuming O(n) component is n for simplicity):

Claim: T(n) <= n*logn + n

Base:
T(1) = 1 (assumption)
Assumption: the claim is correct for all k < n.
Proof:

T(n) = 2T(n/2) + n = (assumption) <= 2* (n/2 * log(n/2) + n/2) + n
= n*log(n/2) + 2n = n*(log(n)-log(2)) + 2n = (assuming base 2 for log)
= n*(log(n) -1 ) + 2n = nlogn -n + 2n = n*logn +n


QED

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How would you prove with master theorem? – user1846486 Nov 23 '12 at 9:25
@user1846486 just apply the formula. – UmNyobe Nov 23 '12 at 9:26
what is the master theorem formula? – user1846486 Nov 23 '12 at 9:28
just google it :D – UmNyobe Nov 23 '12 at 9:28
Let me take it back—I think I just explained why instructors tell you this isn't a proof. It's because it's too hard to teach people how to do this properly in introductory algorithms courses. – Peter Shor Nov 23 '12 at 14:04