Comparing big theta values

Question

I am trying to order these different big theta values from largest to smallest:

Θ(n2)
Θ(2n log n)
Θ(n log n2)
Θ(2n2)
Θ(log n)
Θ(n log 2n)
Θ(k2)
Θ(22n)
Θ(n3)
Θ(n)
Θ(2n)
Θ(n1.5)
Θ(√n)
Θ(2n2)

and some of the values are equivalent. Particularly, I want to know if a constant term makes one big-theta value larger than an identical big-theta term without the constant term (for example, are these two values equivalent: Θ(22n) & Θ(n)?).

I think your could answer the question yourself if you thought about what does the notation represent. — SJuan76, Aug 05 '13 at 08:04
Well, the main question I had was how to deal with values of the same order but different constant multipliers. I feel they are identical, but I wasn't sure how to order: Θ(n log n2) and Θ(n log 2n). The former is identical to log n3 and the latter is the same as log n2, but I wasn't sure if the former is considered larger than the latter? — adub3, Aug 05 '13 at 08:14

score 1 · Answer 1 · edited Jun 20 '20 at 09:12

1

Θ(log n)

Θ(√n) = Θ(n^1/2)

Θ(n) = Θ(2n) = Θ(22n)

Θ(n log n) = Θ(2n log n) = Θ(n log n²) = Θ(n log 2n)

Θ(n^1.5)

Θ(n²) = Θ(2n²)

Θ(n³)

Considering your comment:

n log 2n = n (log 2 + log n) = n log 2 + n log n

log 2 is a constant non-zero value, so:

Θ(n log 2n) = Θ(n log 2 + n log n) = Θ(n + n log n) = Θ(n log n)

See the sum and multiplication by a constant properties of the big-{O,Theta, Omega}-notations.

edited Jun 20 '20 at 09:12

Community

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answered Aug 05 '13 at 08:03

Alex Shesterov

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How is O(nlogn2) equal to O(nlogn)? – adub3 Aug 06 '13 at 05:41
@user2581779: A property of `log` function: `log(n*n) = 2*log(n)`, so `n*log(n*n) = 2*n*log(n)`. And multiplication by a constant does not matter for big-Theta. – Alex Shesterov Aug 06 '13 at 09:36
I was wondering how O(nlogn) appeared and O(k2) disappeared in your results? – adub3 Aug 07 '13 at 08:15

score 0 · Answer 2 · answered Aug 05 '13 at 08:06

0

If try replacing n with a huge value then you can figure it out yourself without even asking it to the forum:

o(1)
O(log log n)
O(log n)
O(n^c)
O(n)
O(n log n)
O(n^2)
O(c^n)
O(n!)

answered Aug 05 '13 at 08:06

sadaf2605

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Comparing big theta values

2 Answers2