# Solving recurrences of the form $T(n) = aT(n/a) + \Theta(n \log_2 a)$

The time complexity for the merge sort algorithm is $T(n) = 2T(n/2)+\Theta(n)$, and the solution to this recurrence is $\Theta(n\lg n)$.

However; assume you are not dividing the array in half for sorting, but instead in, say, $a$ pieces. Then the time would be something like $T(n) = aT(n/a) + \Theta(n \log_2 a)$. (The last term being the time to perform divide-and-merge for $a$ pieces with a total size of $n$. For positive, integer $a$, what is the solution to this recurrence? (assuming $T(n) = 1$ for $n = 1$ and that $n=a^x$ where x is a positive integer as well).

I have a hunch it would probably be $\Theta(n \log_2 n)$ for any $a$, but I'm not really sure how to prove it, with the logarithm being in a different base than $2$ .

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This is not a proof, but $n \log n$ seems to satisfy this equation quite nicely: $a (n/a) \log (n/a) + n \log a$ simplifies to the $n \log n$ we were looking for. Can we turn this into a proof by induction? –  Srivatsan Jul 28 '11 at 19:03

Ahh, yes. Obviously $\Theta(n \log_2 a) = \Theta(n)$ since $\log_2 a$ doesn't change when $n$ grows, didn't think about that... Thanks! –  Johan Sannemo Jul 28 '11 at 19:17
The behavior of the recurrence $T(n) = aT(n/a) + \Theta(n)$ could be a bit different than that of $T(n) = aT(n/a) + \Theta(n\log_a n)$ due to the larger overhead of the latter. In the former case (including $a=2$ in your first line), the time complexity works out to $$n^{\log_a(a)} \lg n = n \lg n,$$ as you have written. In the latter case, with overhead $\Theta(n \log_a n)$, we have to be more clever... [I'll write more later, as obligations are pulling me away from the computer at the moment].
Edited it a while ago: actually meant $n \log_2 a$, not $n \log_a n$, sorry! –  Johan Sannemo Jul 28 '11 at 19:12