You have to discard option d because , according to definition of small o
notation if f(n) =o(g(n)) then for ALL constants c> 0 you have f(n) <
cg(n). or Lim(n->infinite) f(n)/g(n) = 0.

On Sat, Aug 25, 2012 at 10:24 PM, vishal yadav <vishalyada...@gmail.com>wrote:

> Because you can always find a positive constant c for which following
> inequality hold true.
>   A(n) <= cW(n) i.e. the avg. case time complexity always upper bounded by
> worst case time complexity. Which is the definition of Big O.
>
> On Sat, Aug 25, 2012 at 7:11 PM, rahul sharma <rahul23111...@gmail.com>wrote:
>
>> *Let w(n) and A(n) denote respectively, the worst case and average case
>> running time of an algorithm executed on an input of size n. which of the
>> following is ALWAYS TRUE?*
>> (A) [image: A(n) = \Omega(W(n))]
>> (B) [image: A(n) = \Theta(W(n))]
>> (C) [image: A(n) = O(W(n))]
>> (D) [image: A(n) = o(W(n))]
>>
>> answer is c
>>
>> plz explain y???
>>
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>

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