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Questions about asymptotic notations such as Big-O, Omega, etc.

10 votes

n*log n and n/log n against polynomial running time

$\log n$ is the inverse of $2^n$. Just as $2^n$ grows faster than any polynomial $n^k$ regardless of how large a finite $k$ is, $\log n$ will grow slower than any polynomial functions $n^k$ regardles …
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