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I have the below recurrence: \begin{align} T(n, 1) &= 3n \\ T(1, m) &= 3m \\ T(n, m) &= 3n + T(\tfrac{n}{3}, \tfrac{m}{3}) \end{align}

How to get a tight asymptotic bound for $T(n, n^2)$ assuming that $n$ is a power of 3?

Using the substitution method for $T(m,n)$, I get a very weird relation:

$$T(n,m) = 3n + \frac{m}{3^{k-1}} + n\left(3 - \frac{1}{3^{k-1}}\right),$$ where $k = \log_3 n$.

Any suggestion would be helpful.

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  • $\begingroup$ Welcome to the site! What's $k$? $\endgroup$ Sep 21, 2021 at 19:46
  • $\begingroup$ Your solution cannot be right. Plugging $m=1$, you don't get $3n$. $\endgroup$ Jun 20, 2022 at 8:01

1 Answer 1

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Let $S(N,M) = T(3^N,3^M)/3$. If $N \geq M$ then \begin{align} S(N,M) &= 3^N + S(N-1,M-1) \\ &= 3^N + 3^{N-1} + S(N-2,M-2) + \\ &= \cdots \\ &= 3^N + \cdots + 3^{N-M+2} + S(N-M+1,1) \\ &= 3^N + \cdots + 3^{N-M+1}. \end{align} Similarly, if $N \leq M$ then \begin{align} S(N,M) &= 3^N + S(N-1,M-1) \\ &= 3^N + 3^{N-1} + S(N-2,M-2) + \\ &= \cdots \\ &= 3^N + \cdots + 3^{N-M+2} + S(1,M-N+1) \\ &= 3^N + \cdots + 3^{N-M+2} + 3^{M-N+1}. \end{align} It follows that $$ S(N,M)= \Theta(3^N + 3^{M-N}), $$ and so $$ T(n,m) = \Theta(n + m/n). $$ In particular, $$ T(n,n^2) = \Theta(n). $$

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  • $\begingroup$ The recurrence is really strange. It seems that T(n, n^4) = O(n^3), but T(n^2, n^4) is only O (n^2). Or T(x, y) = O (max (x, y/x)). $\endgroup$
    – gnasher729
    Sep 22, 2021 at 9:13
  • $\begingroup$ Indeed, $O(f + g)$ is the same as $O(\max(f,g))$. $\endgroup$ Sep 22, 2021 at 10:33

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