Question I was asked: Does it make a difference if I say "The worst case run time is $O(n^2)$ vs the worst case run time is $\Theta(n^2)$?"

To me, the only difference is that when we say $O(n^2)$, the function may also be $O(n)$, we do not know. But when we say $\Theta(n^2)$, we know for a fact the function is $O(n^2)$ and $\Omega(n^2)$, because it is bounded by $c_1n^2\leq f(n)\leq c_2n^2$ (correct me if I am wrong).

Therefore, can we not say that $\Theta(n^2)$ gives us a more accurate (or at least equal) sense of worst-case run time than $O(n^2)$?

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    $\begingroup$ In your last sentence, did you really mean O(n), not O(n^2)? O(n) is incompatible with Theta(n^2) (unlike vice versa), so no you can't say that. But that sentence construction in English is saying you think we can say that. $\endgroup$ Sep 13, 2022 at 15:59
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    $\begingroup$ Keep in mind that not all algorithms can have "Theta" complexity. There are algorithms that vary in complexity. Those can only be described with Big-O complexity. In that case to have a better understanding on the performance you want to define a statistical distribution of the inputs and then compute the average complexity. That complexity can be described in "theta" terms... unfortunately computing the average complexity is often really difficult/time consuming or even intractable. $\endgroup$
    – Bakuriu
    Sep 13, 2022 at 19:02
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    $\begingroup$ One minor thing that may help. When we say that something $O(n^2)$ or $\Theta(n^2)$ we typically have to prove it. Sometimes the proof for $\Theta$ is natural, sometimes its harder. And often the O is all we really need for our proofs, so we are often lazy in that regard. $\endgroup$
    – Cort Ammon
    Sep 14, 2022 at 0:46
  • $\begingroup$ I agree with @Peter Cordes. Perhaps that point should be responded to or even an edit made? $\endgroup$ Sep 14, 2022 at 13:22
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    $\begingroup$ @PeterCordes yes i meant n^2 $\endgroup$ Sep 15, 2022 at 13:10

3 Answers 3


Yes. Your understanding is correct on all points!


O(f(n)) is also used when there is no simple function that your runtime is close to. For example: Find the smallest prime factor of n by trial division, finishing when a factor is found: There are O(n^1/2) tests if you divide by all integers up to the square root of n. But for even n there is only one test, and similar if n has any small factor.

So you can’t give any reasonable Theta unless you say f(n) = Theta(f(n)) which is true but pointless.

There is also the possibility that I can prove that a function is for example $O(n^2)$. I may suspect it is actually smaller, it may actually be $O(n)$ but I cannot prove it. Obviously in that situation I can also not prove that it is $\Theta(n^2)$ - because it isn't true.

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    $\begingroup$ This example is a bit contrived because you used n as the value of the input, as opposed to the length of the input. If you used n as the length of the input, then there would be many different values of input for the same n, and typically one of them will achieve the worst-case. $\endgroup$
    – Stef
    Sep 13, 2022 at 8:17
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    $\begingroup$ Using n as the input isn't contrived. O is not a computer science concept and so length doesn't pertain to it in general. $\endgroup$ Sep 13, 2022 at 15:46
  • $\begingroup$ For moderately sized numbers, giving the runtime or number of divisions relative to the number n instead of the size of the number n is more useful. And of course who says we always want to know the worst case? The average case is also very useful. $\endgroup$
    – gnasher729
    Sep 15, 2022 at 13:53

You answer yourself by saying that $\Theta$ is the conjunction of $O$ and $\Omega$. So yes of course, $\Theta$ is more informative than $O$ alone.

Make sure anyway that you don't confuse the asymptotic behavior of the algorithm complexity, and that of its worst-case complexity.

E.g. the running-time of QuickSort is $O(n^2)$ and $\Omega(n\log n)$, while its worst-case running-time is $\Theta(n^2)$.


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