If I have some algorithm that runs at best n time and at worst n^2 time, is it fair to say that the algorithm is Big Omega (n)?
Does this mean that the algorithm will run at least n (time)? I am just not sure if I have the right idea here.
Thanks.
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Algorithms are usually quoted as Big Oh in the worst case, so yours is O(n^2) – Mitch Wheat Sep 23 '11 at 04:43
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I mean if I wanted to use big Omega is this correct? Thats what I'm not sure about – Spencer Sep 23 '11 at 04:49
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Also, make sure you cast your vote on the answers you like and that you accept any answer that correctly answers your question. Make sure you do this for your older questions as well to encourage more answers. – Pål Brattberg Sep 23 '11 at 04:49
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No, it's fair to say it's Big Omega n^2, you always assume the worst with Big Oh. – Pål Brattberg Sep 23 '11 at 04:51
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Okay will do. And thanks for being helpful. – Spencer Sep 23 '11 at 04:53
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No worries, have fun! :) – Pål Brattberg Sep 23 '11 at 05:16
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And, as Jim Lewis points out: Big *Omega* is `f(n)` while Big *O* is `O(n^2)`. My comment mixes Big O and Big Omega, sorry about that. – Pål Brattberg Sep 23 '11 at 06:07
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For Big Oh, you will state the worst time as the time it takes, in your case O(n^2). For Big Omega, you state the smallest time, in this case f(n).
Also see this guide to Big O and this discussion of Big O and Big Omega.
Pål Brattberg
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Yes. If the runtime f(n) is asymptotically bounded below by g(n) = n, then f(n) is BigOmega(n).
Edit: Most of the time, algorithms are analyzed in terms of their worst-case behavior -- which corresponds to "Big O" notation. In your case, the runtime is O(n^2).
But for those rare occasions when you need to talk about a lower bound, or best case behavior, "Big Omega" notation is used. And in your case, the runtime is at least n, so it is correct to describe it as BigOmega(n).
Jim Lewis
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