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+29 votes

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?

  1. $A(n) = \Omega (W(n))$
  2. $A(n) = \Theta (W(n))$
  3. $A(n) = \text{O} (W(n))$
  4. $A(n) = \text{o} (W(n))$
in Algorithms by
edited by | 4.5k views
As arjun sir explained dats the reason c is the answer
Answer will be C

Average case of quick sort is $O(nlogn)$ and worst case is $O(n^2)$.
$$Best Case \leq AvgCase \leq WorstCase$$
Hence, (C) is the correct option.

is this always holds true?

BC<=Avg Case<=WC

5 Answers

+42 votes
Best answer
Worst case complexity can never be lower than the average case complexity, but it can be higher. So, $(C)$ is the answer.
$A(n) = O(W(n))$.
edited by
O(W(n)) is tightest upper bound in worst case. Worst case tightest upper bound is average case. right?
what does B option mean?
B means both have the same asymptotic growth rate.
where can I find more ques of this type for practice?
what does small 0 denote????
'o'(small Oh) denotes an upper bound which is not asymptotically tight.
Sir can u tell me wat is wrong with option D ??
sir we use average case only when both worst case and best case are equal .. so in this case even option 2 must be right
@Puja D option tells about small-o, which requires strictly lower time complexity. Say for merge sort, D is false as Average and Worst case complexities are the same.

@Venkat "=" in the case of Asymptotic notations are not the usual "equal to". You should see this portion from Cormen - just 2-3 pages.
yes sir i understood it with an example average case running time for randomized quick sort is O(nlogn) but the worst case is O(n^2) so the average case is made through approximation of all the possible input sequences it doesnt mean that thetha notation is used for average case theta gives a tight bound for running time when the function is same for all the input sequences theta notation is used for describing such functions ... as i understood correct me if i am wrong @Arjun sir i went through coremen

yes @Venkat you are correct.

Comman mistake   :thinking  O is only for "worst", omega for "best case" and theta for "average case".

Not to be confused with worst, best and average cases analysis: all three (Omega, O, Theta) notation are not related to the best, worst and average cases analysis of algorithms. Each one of these can be applied to each analysis.

for eg:

+12 votes

Let F(n)=A(n) , G(n)=W(n)

Rule :- For Omega :-   F(n)>=cG(n)

Option A -  A(n)=Ω(W(n))   it means A(n)>=cW(n) , which is false becoz Worst case time  can not be less than avg case time.

Rule :-  For Theta :-     c1G(n) <= F(n) <= c2G(n)

Option B -  A(n)=Θ(W(n))  it means c1W(n)<=A(n)<=c2W(n) , Which can not be always true. (i.e it is false due to Bold markd line.. )

Rule :- For Big Oh :- F(n)<=cG(n)

option C-  A(n)=O(W(n)) it means A(n)<=cW(n) , which is always true........

options D:- I dont know..

+1 vote
option C
Explain otherwise upvote correct answer ...
0 votes
Option C is correct.
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Explain otherwise upvote correct answer ...
0 votes


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