The problem was from an exam, I spent much time wrapping my head up around this kind of problems, so I decided to ask for help ;(


We implement a merge sort algorithm to sort $n$ items. The algorithm will divide the set into 2 roughly equal-size halves, and merge the 2 halves after each half set is recursively sorted. Because the item comparison is complicated, the merge process takes $\theta(m\sqrt{m})$ steps for input size $m$. What is the time complexity for this algorithm?

(a) $\theta{(n\log{n})}$
(b) $\theta{(n)}$
(c) $\theta{(n^2)}$
(d) $\theta{(n\sqrt{n})}$
(e) $\theta{(n\sqrt{n}\log{n})}$

What I tried:

  1. I know the merge sort normally can be written $S(n) = 2S(\frac{n}{2}) + n, S(1) = 1$, where the recursive function $S$ is the step cost of the merge sort for the size $n$.

  2. But the problem specifies the step costs to merge is $\theta(n\sqrt{n})$, I have to rewrite it as $S(n) = 2S(\frac{n}{2}) + \theta(n\sqrt{n}) = 2S(\frac{n}{2}) + n\sqrt{n}$. I have no idea how to transform it into the general solution...

Please help me, I will learn a lot from this problem! Thanks :)

  • $\begingroup$ Is the answer D? $\endgroup$
    – Gokul
    Commented Dec 10, 2018 at 1:09
  • $\begingroup$ Sorry, I don't know the answer, welcome to share any idea! $\endgroup$ Commented Dec 10, 2018 at 3:48

1 Answer 1


You are on the right track.

Now that we have the recurrence relation $$S(n) = 2S(\frac{n}{2}) + \Theta(n\sqrt{n}),$$ Applying the case three of the master theorem, where $a=2$, $b=2$, $\epsilon=\frac12$, we will have $S(n)= \Theta( n\sqrt n)$.

I was stretching a bit when we were applying the master theorem since the regularity condition, the existence of such $c>0$ required for case three is not necessarily satisfied. What I really meant is that we can find two constant $0<c_1<c_2$ such that $c_1n\sqrt n\le S(n)-2S(\frac n2) \le c_2n\sqrt n$ for all $n$. Define $S_1$ by $S_1(n) = 2S_1(\frac{n}{2}) + c_1n\sqrt{n}$ and $S_2$ by $S_2(n) = 2S_2(\frac{n}{2}) + c_2 n\sqrt{n}$ with the same initial condition as $S$. Now we can apply case three of the master theorem to $S_1$ and $S_2$ to get the same $\Theta$ bound, since the regularity condition is satisfied with $1>c=0.8>\sqrt2/2$. Since $S_1(n)\le S(n)\le S_2(n)$, S(n) has the same $\Theta$ bound.

In case you have not learned/are not allowed to use the master theorem, you may have to more or less prove case three of the master theorem by yourself. You could find a proof, for example, in CLRS.

  • $\begingroup$ Welcome! $\quad$ $\endgroup$
    – John L.
    Commented Dec 10, 2018 at 12:24

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