nlogn vs square root of n , Isn't square root slower?

Question

I was doing this quiz on an online course and one question stated;

The function nlogn + √n + 5 may set to belong to
A: nlogn
B: √n
C: n√n

The quiz said the correct answer is A, but isn't square root of n slower? I am new to finding the time complexity of an algorithm and could use an explanation. Or let me know if the answer is wrong.

in a nutshell, `n*sqrt(n)` > `n*log(n)` > `sqrt(n)` > `log(n)` — shripal mehta, Mar 20 '21 at 06:23
for better understandings, [plot on WolframAlpha](https://www.wolframalpha.com/input/?i=plot+%7Bsqrt%28x%29%2C+x+sqrt%28x%29%2C+x*log%28x%29%2C+log%28x%29%7D) — shripal mehta, Mar 20 '21 at 06:25

score 5 · Accepted Answer · edited Dec 21 '16 at 22:48

5

You should consider n to be an extremely large number. For any n>2, n>√n and logn>1. Thus, nlogn>√n.

edited Dec 21 '16 at 22:48

takendarkk

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answered Dec 21 '16 at 22:45

Mureinik

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score 1 · Answer 2 · answered Dec 26 '16 at 14:56

1

The correct Hierachy is this:

Superlinear[ n log n] > Linear [n] > sub polynomial [n^(1/a)]

Where a: a >= 1.

Thus n log n = O(n) = O(sqrt(n))

N does not need to be an 'Extremely large number" though Big-Oh deals with limits to infinity. In particular, you can set your n0 as `b+Where b is the base of the logarithm. In this case 10.

Test with n=11 yourself to find out.

answered Dec 26 '16 at 14:56

Tobi Alafin

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@Grez kev see my answer. – Tobi Alafin Dec 26 '16 at 16:36

score 0 · Answer 3 · answered Jun 27 '20 at 04:02

0

Yes, √n is surely more than logn. (cf)

But you're multiplying that logn with n.

n is obviously more than √n.

(If you would multiply √n with n too, then the answer is n√n.)

answered Jun 27 '20 at 04:02

Pacerier

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nlogn vs square root of n , Isn't square root slower?

3 Answers3