If we have an algorithm with complexity $\Theta(m + n^2)$ and we know that $0 < m < n^2$ then its complexity becomes $\Theta(n^2)$. But if we had an algorithm with complexity $\Theta(m\log{n})$ and $0 < m < n^2$ could we conclude that its complexity is $\Theta(n^2\log{n})$?
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I assume that, by $\Theta$, you mean this notation. From $f(m,n) \in \Theta(m \log n)$ and $0 < m < n^2$, we can conclude that $f(m,n) \in O(n^2 \log n)$. We cannot conclude $f(m,n) \in \Theta(n^2 \log n)$, since it's possible that $m \in o(n^2)$. For example, if $m = n$, then $f(m,n) \in \Theta(n \log n)$. The reason we could conclude from $f(m,n) \in \Theta(m + n^2)$ and $0 < m < n^2$ that $f(m,n) \in \Theta(n^2)$ is that, even if $m \in o(n^2)$, the expression $m + n^2$ still contains an $n^2$ term which will dominate any smaller $m$. That is, we already know that $\Theta(m + n^2) \subset \Omega(n^2)$, and $m < n^2$ implies that $\Theta(m + n^2) \subset O(n^2)$, and thus $\Theta(m + n^2) \subset (\Omega(n^2) \cap O(n^2)) = \Theta(n^2)$. However, if we know that $m \in \Theta(n^2)$, then we can indeed conclude from $f(m,n) \in \Theta(m \log n)$ that $f(m,n) \in \Theta(n^2 \log n)$. |
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