Algo doubt

Question

Assume f(n) and g(n)  are two functions such that f (n)=O(g(n))

which of the following always hold

F(n) = O (f(n)²)

F(n) = Ω (f(n)²)

G(n) = O (f(n)²)

G(n) = Ω (g(n))

Comments

Please recheck option D. since $g(n) = \Omega (g(n))$ is always true.

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option $A,D$ will alawys hold , option $B,C$ will not always hold

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Answer was d....Can u give some examples

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f(n) = O (f(n)2)   Take f(n)= 1/n  Here it is fail.

f(n) = Ω (f(n)2)    Take f(n) = n Here it is fail

g(n) = O (f(n)2)  Never hold true. it is saying complement of given question.

g(n) = Ω (g(n)) True

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@Prashant

g(n) = O(f(n)²) // though it is always not true but sometimes it may be true, take f(n)=n and g(n)=n² clearly, f(n)=O(g(n)) and also g(n)=O(f(n)²)..

secondaly, every function is both Big-Oh and Big-Omega of itslef, therefore g(n) = Ω (g(n)) is always true..

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nitish one case fail mean fail.
Ya  g(n) = Ω (g(n)) is ture i was thinking it is  g(n) = Ω (g(n)²)

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but you can not say that,

g(n) = O (f(n)2)  Never hold true.

sometimes it may be true(depending on f(n) and g(n))

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Question is following always hold .

Nowhere i said never hold true. Counterexample are given to show not hold not for hold. Hope you get it

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I think we have to consider all possible cases for something to always hold true. Therefore D should be correct.