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When we have $F(n) = \Omega(H(n))$ and $G(n)=\mathcal{O}(H(n))$.

Can we prove that $G(n)/F(n) = \mathcal{O}(1)$?

I tired to use the definitions of $\mathcal{O}$ and $\Omega$ but all I ended up with were two inequalities that I couldn't use.

Thanks

Shankara Pailoor
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ammoun
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1 Answers1

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What you likely found in the inequalities you mention is that we have $F(n)\geq C_F H(n)$ and $G(n)\leq C_G H(n)$. So $$G(n)/F(n)\leq G(n)/(C_F H(n))\leq C_G H(n)/(C_F H_n)=C_G/C_F$$ Is this the application of the definitional inequalities you hadn't yet seen?

Kevin Carlson
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  • Kevin! I never thought about using two opposite inequalities this way, Thank you so much!!! – ammoun Sep 10 '12 at 02:44
  • Happy to help! Incidentally, welcome to StackExchange-the thing to do here if you're satisfied with an answer is to click the check mark to the left to "accept" it. That way others don't come to the question thinking it still needs a solution. – Kevin Carlson Sep 10 '12 at 03:19