问题
I was reading this question Big-O notation's definition.
But I have less than 50 reputation to comment, so I hope someone help me.
My question is about this sentence:
There are many algorithms for which there is no single function g such that the complexity is both O(g) and Ω(g). For instance, insertion sort has a Big-O lower bound of O(n²) (meaning you can't find anything smaller than n²) and an Ω upper bound of Ω(n).
for large n the O(n²) is an upper bound and Ω(n) is a lower bound, or maybe I have misunderstood? could someone help me?
回答1:
has a Big-O lower bound of O(n²)
I don't really agree with the confusing way this was phrased (since big-O is itself an upper bound), but what I'm reading here is the following:
Big-O is an upper bound.
That is to say, f(n) ϵ O(g(n)) is true if |f(n)| <= k|g(n)| as n tends to infinity (by definition).
So let's say we have a function f(n) = n2 (which is, if we ignore constant factors, the worst-case for insertion sort). We can say n2 ϵ O(n2), but we can also say n2 ϵ O(n3) or n2 ϵ O(n4) or n2 ϵ O(n5) or ....
So the smallest g(n) we can find is n2.
But the answer you linked to is, as a whole, incorrect - insertion sort itself does not have upper or lower bounds, but rather it has best, average and worst cases, which have upper and lower bounds.
See the answer I posted there.
回答2:
maybe I have misunderstood?
No, you are right.
In general, the Big-O is for the upper bound and big-Ω for the lower bound.
For Insertion sort the worst case scenario, the upper bound is O(n2). Ω(n) is a lower bound.
It seems like you find a mistake in the other answer.
来源：https://stackoverflow.com/questions/46129461/big-o-and-omega-notations