Assume we have: $$T(n)=7T(\frac{n}{2})+n^2\lg{n}$$ Can we solve it using master theorem?

As we know $n^{\lg_2{7}}\approx n^{2.81}$. On the other hand, we have $f(n)=n^2\lg n$. So we should compare $n^.081$ and $\lg{n}$. As I know, $\lg{n}=\mathcal{O}(n^{0.81})$. So the answer should be $T(n)=\Theta(n^{\lg7})$. Am I right? Can we use Master Theorem here? the logarithm make me unsure about it.


1 Answer 1


Yes, that recurrence can be solved using the master theorem, and you got the correct answer but your justification is missing on a detail.

In particular it is false (in general) that a recurrence of the form $T(n) = aT(n/b) +f(n)$ has solution $T(n) \in \Theta(n^{\log_b a})$ when $f(n) \in O(n^{\log_b a})$. As a concrete example, $T(n) = 4T(n/2) + n^2$ satisfies $f(n) = n^2 \in O(n^{\log_2 4})$ but it has solution $T(n) \in \Theta(n^2 \log n)$ (and not $T(n) \in \Theta(n^2)$). A trickier example is $T(n)=2T(n/2) + \frac{n}{\log \log n}$, where $\frac{n}{\log \log n} \in o(n) \subset O(n)$, but the master theorem does not apply.

What you actually need is $f(n) \in O(n^{\log_b a - \varepsilon})$ for some constant $\varepsilon>0$. This is clearly the case in your recurrence since $\log n \in O(n^c)$ for all constants $c>0$ (so you can choose any $\varepsilon$ in the non-empty open interval $(0, \log_2 7 - 2)$).

  • $\begingroup$ So the answer is $T(n)=\Theta(n^{\lg_2{7}})$? $\endgroup$ Nov 14, 2023 at 13:56
  • $\begingroup$ Yes.$\phantom{}$ $\endgroup$
    – Steven
    Nov 14, 2023 at 13:57

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