Question

Let f(n) = Ω(n), and g(n) = O(f(n)). Then g(n) = _______ [Assume n > 0] 1. Ω(n) 2. O(n) 3. θ(n) 4. Ω(1)

Comments

@arjun sir, pls check my logic, f(n) = omega(n) means f(n) >= 10 (just taking a value) ,now g(n) = O(f(n)) means g(n) <= f(n) ,so the min value f(n) can take is 10 and g(n) is lower than that value,so as n given >0,the lowest value g(n) can have is 1. so omega(1).

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could it be done this way using constants?

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https://gateoverflow.in/11232/let-f-n-%CF%89-n-g-n-o-n-and-h-n-%D1%B3-n?show=11232#q11232 check this,and talk properly,you may do good or grt result,but pls learn how to talk with others,that is most important..i think your age is not < 20,so be mature.. and i asked that arjun sir..

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did you find it offensive?
i am so sorry.
its just that my sayantance forming is a little wierd.
I just meant to ask that is it possible to do like that?
:)

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yes. f(n) >= n, g(n) <= n, we cannot determine g accurately.

Answer 1

Actually none of the options are correct  check this

https://gateoverflow.in/36807/let-f-n-%CF%89-n-and-g-n-o-f-n-then-g-n-_______-assume-n-0-1-%CF%89-n-2-o-n-3-%CE%B8-n-4-%CF%89-1

Answer 2

NOTA
. g(n) can be 1/N

http://stackoverflow.com/questions/905551/are-there-any-o1-n-algorithms n

Answer 3

F(n)=omega(n)=n,$n^2$,$n^3$ etc  

So g(n)=O(n),O($n^2$),O($n^3$) etc , So option O(n) matches.

Comments

http://cs.stackexchange.com/questions/52069/nested-complexities

I asked this is CS.stackexchange and got the following reply!

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@Arjun sir:
to the gate people know about these 1/n algorithms?
i mean, is it considered in any of the books?

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GATE people are IIT profs - until they outsource it which I do not think they will do as they are intelligent. So, they will mostly stick to standard definitions. And for GATE they usually won't ask boundary questions.