What would this resolve to?
$O(k \log k \log n)$
The logs are the same base here and for the life of me right now I can't remember how to multiply logarithms together....
Basically what is $\log k * \log n$?
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Why would you like to modify the expression ? – mounir ben salem Apr 30 '17 at 15:10
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I have an algorithm that gives a runtime of $O(k \log k \log n)$. And I just want to know if I can simplify $k \log k \log n$ any further... – gprime Apr 30 '17 at 15:13
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1Maybe you confused with $\log(kn)= \log k + \log n$? – Jay Zha Apr 30 '17 at 15:13
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Unless you know something about how $k$ relates to $n$ directly, there's probably not much more that can be done. You may be thinking of $\log(nk) = \log(n) + \log(k)$, but that's a different log rule.
In terms of big-O notation, it's clear as-is, unless there's some specific reason you wish to change it. For example you could also state it as $O(k \log(n^{\log(k)}))$ or $O(k \log(k^{\log(n)}))$, but these representations may be less clear.
Marcus Andrews
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Yes, that is what I was thinking of. So I was right and there is nothing more I can do. Thanks! – gprime Apr 30 '17 at 15:20