While trying to analyse the runtime of an algorithm, I arrive to a recurrence of the following type :

$$\begin{cases} T(n) = \Theta(1), & \text{for small enough $n$;}\\ T(n) \leq T(a_n n + h(n)) + T((1-a_n)n+h(n)) + f(n), & \text{for larger $n$.} \end{cases}$$ Where:

  • $a_n$ is unknown and can depend on $n$ but is bounded by two constants $0<c_1\leq a_n \leq c_2 < 1$.
  • $h(n)$ is some "fudge term" which is negligeable compared to $n$ (say $O(n^\epsilon)$ for $0\leq \epsilon < 1$).

If $a_n$ was a constant, then I could use the Akra-Bazzi method to get a result. On the other hand, if the fudge term didn't exist, then some type of recursion-tree analysis would be straight-forward.

To make things a bit more concrete, here is the recurrence I want to get the asymptotic growth of:

$$\begin{cases} T(n) = 1, & \text{for n = 1;}\\ T(n) \leq T(a_n n + \sqrt n) + T((1-a_n)n+\sqrt n) + n, & \text{for $n \geq 2$} \end{cases}$$ Where $\frac{1}{4} \leq a_n \leq \frac{3}{4}$ for all $n\geq 1$.

I tried various guesses on upper bounds or transformations. Everything tells me this should work out to $O(n\log(n))$ and I can argue informaly that it does (although I might be wrong). But I can only prove $O(n^{1+\epsilon})$ (for any $\epsilon>0$), and this feels like something that some generalization of the Master theorem à la Akra-Bazzi should be able to take care of.

Any suggestions on how to tackle this type of recurrence?

  • $\begingroup$ Have you tried induction? $\endgroup$ May 31, 2020 at 18:12
  • $\begingroup$ I had and I just have again. In this attempt I got stuck at trying to show that $(a_n n + \sqrt n)^{(a_n+1/\sqrt n)}((1-a_n) n + \sqrt n)^{((1-a_n)+1/\sqrt n)} \leq n$ for large enough $n$, and I'm not certain it holds. $\endgroup$
    – Tassle
    May 31, 2020 at 19:10

1 Answer 1


According to the OP, to complete the proof we need to prove that for large enough $n$, $$ (a_n n + \sqrt{n})^{a_n + 1/\sqrt{n}}((1-a_n)n + \sqrt{n})^{1-a_n + 1/\sqrt{n}} \leq n. $$ Taking out a factor of $n^{1+2/\sqrt{n}}$, we get $$ (a_n + 1/\sqrt{n})^{a_n + 1/\sqrt{n}} (1-a_n + 1/\sqrt{n})^{1-a_n + 1/\sqrt{n}} \leq n^{-2/\sqrt{n}}. $$ Dividing by $(1+2/\sqrt{n})^{1+2/\sqrt{n}}$, we get $$ \left( \frac{a_n + 1/\sqrt{n}}{1 + 2/\sqrt{n}} \right)^{a_n + 1/\sqrt{n}} \left( \frac{1-a_n + 1/\sqrt{n}}{1 + 2/\sqrt{n}} \right)^{1-a_n + 1/\sqrt{n}} \leq \frac{1}{n^{2/\sqrt{n}} (1+2/\sqrt{n})^{1+2/\sqrt{n}}}. $$ Raising to the power $1/(1+2/\sqrt{n})$, and defining $p = (a_n + 1/\sqrt{n})/(1 + 2/\sqrt{n})$, we get $$ p^p (1-p)^{1-p} \leq \frac{1}{n^{(2/\sqrt{n})/(1+2/\sqrt{n})}(1+2/\sqrt{n})}. $$

The left-hand side is $1/\exp h(p)$, where $h(p)$ is the entropy function. Hence it is maximized when $a_n = 1/4$, at which point $p \approx 1/4$. Since $(1/4)^{1/4} (3/4)^{3/4} < 1$, for large enough $n$ we can bound the left-hand side by some $\theta < 1$. As for the right-hand side, as $n\to\infty$ it tends to $1$. Indeed, taking logarithm, we get $$ \frac{2\log n}{\sqrt{n} + 2} + O\left(\frac1{\sqrt{n}}\right) \longrightarrow 0. $$ In particular, for large enough $n$ it will be larger than $\theta$.

  • $\begingroup$ Thank you! I made a small mistake in my comment, as the inequality actually has to be strict, but of course your proof provides this as well. $\endgroup$
    – Tassle
    May 31, 2020 at 20:34

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