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I have a problem that I solved, but I'm unsure if my answers are correct or not. I've never seen the implementation of time complexities as elements of others.

f(n) element of O(log(n))
g(n) element of theta(n)

Prove if the below are true are false, provide a counter example if it's false.

1) g(n) element of omega(f(n))
2) f(n)g(n) element of O(nlog(n))
3) f(n)g(n) element of theta(nlog(n))

Below are my attempts, which was just because I thought it was the logical solution. Any ideas on how I can go about solving these types of questions?

1) False, g(n) has a tight bound at (n). If it had a lower bound of f(n), then g(n) should allow log(n), but that's too high of a complexity.

2) True

3) False, f(n) having an upper bound of log(n) implies it could be even faster. Setting a tight bound contradicts that wherein it cant go faster.

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    $\begingroup$ You posted the same thing almost literally, got feedback about why your question is bad, and got a link to a comprehensive answer to your question. Deleting and reposting is not appropriate -- look, you already wasted poor Denis' time! $\endgroup$ – Raphael May 28 '16 at 1:33
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    $\begingroup$ Your titular question does not seem to have anything to do with your post. Note that your question has nothing to do with algorithms or "time complexities" -- it's all about asymptotic bounds. $\endgroup$ – Raphael May 28 '16 at 1:34
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Actually the 1) is true: omega means "at least", and $O$ means "at most". You now that $g(n)$ is in $n$, which is at least $\log(n)$.

2) is indeed true

For 3) you don't have enough information: it is only true if $f$ is in $\theta(log (n))$ which you don't know, so 3) could be true or false. You can give a counter example where it is false: $f(n)=\log(\log (n)))$ and $g(n)=n$.

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  • $\begingroup$ Thank you for the reply! I was just wondering how your counter example works, why is it log(log(n))? $\endgroup$ – Andrew Raleigh May 28 '16 at 1:07
  • $\begingroup$ Anythin in $O(\log n)$ but not $\theta (log n)$ is ok, you can also take $\sqrt{\log n}$ for instance, or even a constant. $\endgroup$ – Denis May 28 '16 at 9:26

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