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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 $O(n^2)$?

Carter Falkenberg
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Yes. Your understanding is correct on all points!

D.W.
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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.

gnasher729
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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 $\Theta(n^2)$.