5.4k views

Let $P$ be quicksort program to sort numbers in ascending order using the first element as the pivot. Let $t_1$ and $t_2$ be the number of comparisons made by P for the inputs  $[1 \ 2 \ 3 \ 4 \ 5]$ and $[4 \ 1 \ 5 \ 3 \ 2]$ respectively. Which one of the following holds?

1. $t_1 = 5$
2. $t_1 < t_2$
3. $t_1>t_2$
4. $t_1 = t_2$

edited | 5.4k views

it would be $t_1>t_2$, because the first case is the worst case of quicksort i.e. minimum number is chosen as pivot. Hence in the worst case the comparisons are high.

The splitting occurs as
$ $
$ $
$ $
$ $

and

$ $
$  $
$ $

Number of recursive calls remain the same, but in second case the number of elements passed for the recursive call is less and hence the number of comparisons also less.

Correct Answer: $C$
by Junior (795 points)
edited
+3
+1
why will the no of recursive calls be same??
+28

Total number of comparisons made by partition algorithm for input array of size $n = n+1$
(See 2nd page here http://homepages.math.uic.edu/~leon/cs-mcs401-s08/handouts/quicksort.pdf)
           6
        5
          4
             3
 
Total = 6+5+4+3 = 18

            6
         4+3
    3
 

total = 6+4+3+3 = 16
Hence $t_1 \gt t_2$

+2
How does the splitting take place this way? For the ist case u hv taken elements less than or equal to pivot in left list and elements greater than pivot in right list and in 2nd case less than pivot in left list and elements greater or equal to  pivot in right list ?
0

#Edit #Suggestion

Correct me if i am wrong .?

@Sachin Mittal 1 Sir in your explanatory comment ,

 in the 2nd case do you mean to say $[4 1 5 3 2] instead of [1 2 3 4 5]$  ..????

 Minor mistake in calculation in case 2 , the calculation comes out to be 16 instead of 14.

@Arjun Sir plz have a look at this and also

other @Admins can edit that  comment and delete this comment. (after proper action)

0

No. of comparisons in partition algorithm are  $\Theta(n)$

It is not always n+1. In the same pdf at 6th page--

partition() performs n comparisons (possibly −1 or  n+1, depending on the implementation).  In fact,  −1 is the lower bound on the number of comparisons that any partitioning algorithm can perform.

First case [1 2 3 4 5]

1 [2 3 4 5]    ->  4 comparisons
2 [3 4 5]  -> 3 comparisons
3 [4 5] -> 2 comparisons
 -> 1 comparison

Second case [4 1 5 3 2]
[1 3 2]  -> 4 comparisons
[3 2]  -> 2 comparisons
3   -> 1 comparison

Hence, in second case number of comparisons is less. => t1 > t2.

by (401 points)
+3

Second case [4 1 5 3 2]
[1 3 2]  -> 4 comparisons
2315 -> 2 comparisons
135 -> 0 comparisons

Total 6

0
it should be 4 1 2 3 5 in second  case.
+3
this seems intuitively correct to me though selected answer has different logic. I wonder which is correct. :(
We dont even need to compare the number of comparisions we know that Quick sort gives worst result

list is already sorted ascending or decending and when all element are equal it has time complexity of O($n^{2}$)

in rest cases it is O($nlogn$)  hence t1>t2
by (321 points)

When first element or last element is chosen as pivot, Quick Sort's worst case occurs for the sorted arrays. In every step of quick sort, numbers are divided as per the following recurrence. T(n) = T(n-1) + O(n)

So, t1>t2

by Loyal (9.6k points)
both arrays are worst case example of Quick sort .∴ t1=t2.
by Active (1k points)
0
Array 2 is not worst case example.

i think equal number of comparisons are required as the pivot element is always the highest or lwest element in the current array being worked on

by Boss (14.4k points)