GATE CSE 2015 Set 1 | Question: 2

Question

Which one of the following is the recurrence equation for the worst case time complexity of the quick sort algorithm for sorting $n\;( \geq  2)$ numbers? In the recurrence equations given in the options below, $c$ is a constant.

• A. $T(n) = 2 T (n/2) + cn$
• B. $T(n) = T ( n - 1) + T(1) + cn$
• C. $T(n) = 2T ( n - 1) + cn$
• D. $T(n) = T (n/2) + cn$

Comments

What does cn indicates ?

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It means any constant which is multiplied to $\mathbf n$.

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In quick sort,

one thing to notice generally we don’t take pivot in recurrence relation but here we considered the pivot as well.

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• T(n-1): Time taken to recursively sort the sub array of size (n-1) when partitioning is performed.
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• T(1): Time taken to recursively sort the sub array of size 1 when partitioning is performed (base case).
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• cn: The time taken for the partitioning step, where "c" is a constant representing the time taken per element during partitioning, and "n" is the size of the array being partitioned.

Answer 1

Correct Option: B
Worst case for quick sort happens when $1$ element is on one list and $n-1$ elements on another list.

Comments

 But according to Cormen, worst case for Quicksort algo occurs when the partitioning produces one subproblem with n-1 elements and another with zero elements ( the nth element is the pivot which gets added to the end of the first subproblem). In fact, it states that even a partition with a 9:1 proportional split also runs in O(n lg n) time. 

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Yes. n-1 elements on one list- 0 on other list is because 'pivot' is ignored- which is anyway placed.

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So option B should really be

T(n)=T(n-1)+T(0) +cn

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Yes. But can't say the given is wrong either.

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Yes sir. Asymptotically they are the same.

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Isn't it option D for quick sort?

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Option D means we are dividing the elements in half which is not the case for worst condition.

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can't say the given is wrong because asymptotically they are the same.Is it?

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For Worst Case correct Recurrences are:

$\mathbf{T(n) = T(0) + T(n-1) + \theta (n)}$

which is the same as:

$\mathbf{T(n) = T(n-1) + \theta (n)}$

The solution of this recurrence is $\mathbf{\theta (n)}$

Maybe the examiner have something to do with $\mathbf{n \ge 2}$ case. If this isn't the case, then the options are definitely wrong $\color{blue} {\text{According to Cormen and Wikipedia.}}$

Answer 2

Quick sort worst case time complexity is n^2, when the array is sorted or almost sorted then Quicksort algorithm runs in O(n^2) time.

The recurrence relation for Quick sort worst case time complexity is

T(n) = T(n-1) + T(1) + cn.

Hence, B is Answer.

Answer 3

If the pivot is chosen as the last position then there is a left recursive of size (n-1) or right recursive call of size 0.

1.Partition function->O(n)

2.T(n-1) time in left recursive call

3.T(1) time in right recursive call

Recurrence relation: T(n-1)+T(1)+cn

Option B is correct