Re:Big O Notation – Explained as easily as possible
by Al-Khwarizmi on Sunday January 17, 2021 @06:14AM
(#25809830)
Again, I think from a shallow perspective it makes sense.
If an algorithm always runs in O(n^3), then it's guaranteed that it runs in O(n^3) in the worst case. And if an algorithm runs in O(n^3) in the worst case, then it's guaranteed to always run in O(n^3) (but not in Θ(n^3), of course). So if you only care about worst-case performance, it's reasonable to only use the big O.
Of course, what your parent comment says is also true - you could say that mergesort is O(2^n), in the worst case or in any other case, and be correct because it's an upper bound. But people using Big Oh informally don't say that because you typically want to show how good your algorithm is, so you use the tightest upper bound possible (i.e. the big theta of the worst case).
If an algorithm always runs in O(n^3), then it's guaranteed that it runs in O(n^3) in the worst case. And if an algorithm runs in O(n^3) in the worst case, then it's guaranteed to always run in O(n^3) (but not in Θ(n^3), of course). So if you only care about worst-case performance, it's reasonable to only use the big O.
Of course, what your parent comment says is also true - you could say that mergesort is O(2^n), in the worst case or in any other case, and be correct because it's an upper bound. But people using Big Oh informally don't say that because you typically want to show how good your algorithm is, so you use the tightest upper bound possible (i.e. the big theta of the worst case).