O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

and logn < n, but what about logn^2?

Answer 1

n is only less than (log n)² for values of n less than 0.49...

So in general (log n)² is better for large n...

But since these O(something)-notations always leave out constant factors, in your case it might not be possible to say for sure which algorithm is better...

Here's a graph:

(The blue line is n and the green line is (log n)²)

Notice, how the difference for small values of n isn't so big and might easily be dwarfed by the constant factors not included in the Big-O notation.

But for large n, (log n)² wins hands down:

Answer 2

For each constant k asymptotically log(n)^k < n.

Proof is simple, do log on both sides of the equation, and you get:

log(log(n))*k < log(n)

It is easy to see that asymptotically, this is correct.

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Semantic note: Assuming here log(n)^k == log(n) * log(n) * ... * log(n) (k times) and NOT log(log(log(...log(n)))..) (k times) as it is sometimes also used.