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I was trying to solve the following question while reviewing master theorem.

Which of the following asymptotically grows faster.

(a) $ T(n) = 4T(n/2) + 10n $

(b) $ T(n) = 8T(n/3) + 24n^2 $

(c) $ T(n) = 16T(n/4) + 10n^2 $

(d) $ T(n) = 25T(n/5) + 20(n\log n)^{1.99} $

(e) They are all asymptotically the same

My calculation says, (a) is $\theta(n^2)$ (b) is $\theta(n^2)$ (c) is $\theta(n^2\log n)$. Now how can I evaluate (d)?

If $f(n)$ is smaller or larger than $n^{\log_b a}$by less than a polynomial factor, how can I solve $T(n)$?

Yuval Filmus
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Debasis Jana
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1 Answers1

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$ T(n) = 25T(n/5) + 20(n\log n)^{1.99} $

Note that $(\log n)^m=O(n^\epsilon)$ for all $m,\epsilon\in \Bbb R^+$. In plain words, all polynomials of $\log n$ grows slower asymptotically than all polynomials of $n$. Here we abuse "polynomials" to include functions like $x\to x^k$ for all $k\in\Bbb R^+$ such as $x\to x^{0.618}$ or $x\to x^{2019.46}$. This abuse of the term "polynomial" is common when we are talking about the asymptotic behavior of functions.

In particular, $(\log n)^{1.99}=O(n^{0.005})$.

Since $(n\log n)^{1.99}=n^{1.99}(\log n)^{1.99}= O(n^{1.995})=O(n^{log_525-0.005})$, you can apply the case 1 of Master theorem. $T(n)=\Theta(n^2)$.

If $f(n)$ is indeed smaller or larger than $n^{\log_b a}$by less than a polynomial factor, then we are running into the cases strictly between case 1 and case 2 or between case 2 and case 3. The master theorem you were reviewing cannot be applied any more.

You may look into this answer that handles some of those situations.

Exercise. Show that $(\log n)^m=O(n^\epsilon)$ for all $m,\epsilon\in \Bbb R^+$. (For example, $(\log n)^{10000000}=O(n^{0.000001}).$)

John L.
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