1
$\begingroup$

Hello I want to prove the recursion depth of merge sort, which is $O(\log(n))$. I think I can prove this by recurrence equation and the master theorem: $T(N)=2 T(n/2)+O(N) $ however i need to get $O(\log(n)) $. Basically I want only to calculate the complexity of the deviding and remove the conquer part from this equation. How can I do this? Or is there another way to prove the height of the recursion tree?

In the picture below you can see I want to prove the height from red to gray.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ refer to recrusion tree method of solving recurrences. $\endgroup$ – Navjot Waraich May 15 at 11:07
  • $\begingroup$ You can use the master theorem to prove the (worst-case) time complexity of mergesort. That does not give you a direct proof of the height of the comparison tree. $\endgroup$ – dkaeae May 15 at 11:23
  • $\begingroup$ solved with recrusion tree method thanks @NavjotWaraich $\endgroup$ – raviolican May 15 at 12:00
  • 1
    $\begingroup$ @raviolican Consider answering your own question. It might be helpful for future readers. $\endgroup$ – dkaeae May 15 at 12:12
2
$\begingroup$

First of all, let me correct the recursion for the running time of merge sort: $$ T(n) = \begin{cases} T(\lfloor n/2 \rfloor) + T(\lceil n/2 \rceil) + \Theta(n) & \text{if } n > 1, \\ \Theta(1) & \text{if } n = 1. \end{cases} $$ The corresponding recursion for depth is: $$ D(n) = \begin{cases} \max(D(\lfloor n/2 \rfloor), D(\lceil n/2 \rceil)) + 1 & \text{if } n > 1, \\ 0 & \text{if } n = 1. \end{cases} $$ (The value of $D(1)$ could also be $1$, depending on how you define depth.)

The solution of this recurrence is $D(n) =\lceil \log_2 n \rceil$.

When $n$ is a power of 2, you can calculate the depth of the recursion tree by noticing that the value of $n$ decreases by a factor of 2 at each level. For the general case, the main observation is that the depth is monotone in $n$, using which you can easily conclude $D(n) \leq \lceil \log_2 n \rceil$ by considering the smallest power of 2 which is at least $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.