是n log n = O（n）且是n log n = Omega（n）

Question

Hey guys i have been asked the above 2 questions with an explanation as to why. I'm quite confused, i know that the time it takes for an O(nlogn) algorithm to run grows faster then a linear O(n) algorithm but not quite sure the answer to these. I'm fairly sure that n log n is not equivalent to O(n) but im not quite sure how to explain it. (i don't think we need to do an exact proof)

Answer 1

Claim: n*log(n) is not O(n) Proof: the proof is by contradiction. Suppose that n*log(n) were O(n) . Then by the definition of O there must exist constants n0 and c such that for all n > n0 , n*log(n) <= c * n . Dividing both sides by n gives log(n) <= c . However, there is no constant c such that log(n) <= c for all n > n0 ; consider the sequence n = 2^0, 2^1, 2^2, …, 2^k, … This is a contradiction; so, the supposition was incorrect. In other words, n*log(n) cannot be O(n) .

Claim: n*log(n) is Omega(n) Proof: the proof is direct. Choose n0 = 1 and c = 1 . Then n*log(n) >= n = 1*n = c*n , as required.

Answer 2

Simple answer:

With ignoring some details you can say that f(n) = O(g(n)) means f(n) <= g(n) (not an actual inequality but just in terms of growth). Also, you can say that f(n)=Omega(g(n)) means f(n) >= g(n). (We can extend this to Theta, little o and little omega as well).

So basically, if you have an intuition that nlogn is a bigger function than n, then you can say: n log n = Omega(n) or equivalently n = O(n log n). But not the other way around.

Therefore, nlogn=O(n) is a wrong statement.

Sophisticated answer:

If you want to prove that f(n)=O(g(n)), then you need to find certain constants that satisfy the Big-O definition. You can look it up but I'm not gonna get into the details of that.

However, if you want to prove that f(n) is not in O(g(n)), then simply said you want to prove that f(n) <= g(n) is a wrong statement. As a result, it should be the case that f(n) > g(n). Now, you can express the strictly larger (smaller) comparison as little-o or little-omega. So, to prove that f(n) is not in O(g(n)) you should actually prove that f(n) > g(n) which translates to f(n) = little-omega(g(n)). To prove little o and little omega relationships you have to use limits.

A very good reference to learn about these formal proofs in details is "Introduction to Algorithms, Third Edition" by Cormen.

Answer 3

n logn is not O(n) because it increases faster than n meaning that n is not an upper bound for it.

On the other hand n log n is Ω(n) because n is a lower bound for it. The definition of Ω is that for f(n)=Ω(g(n)) means that for any c there exists an n such that 0<=cg(n)<=f(n) .