Partition in Quickselect

Question

I have to implement an algorithm that returns the median of an array. So I chose to implement Quickselect which seems to be efficient for this and I saw that for the tripartition I can use the same partition algorithm used in Quicksort.

The partition algorithm reported in my lecture notes (in C code) is the following:

for (i = lo, j = hi;
     (i <= j);)
{
    while (a[i] < pivot)
    {
        i++;
    }

    while (a[j] > pivot)
    {
        j--;
    }

    if (i <= j)
    {
        if (i < j)
        {
            /* exchange values */
            tmp = a[i];  
            a[i] = a[j];
            a[j] = tmp;
        }

        i++;
        j--;
    }

}

Now, if my array is: a = [40, 20, 100, 60, 80] and I choose as pivot the number 80, the result is: a = [40, 20, 80, 60, 100], but as you can see the values in the right partition are not all > 80 (we have 60). If I choose any other number the algorithm works properly.

Is this algorithm wrong?

Thank you for your help in advance!

Regarding the basic quicksort dual-partition algorithm you have designed, you're missing an "..or equals" case in one of the two sweep loops. [See here](https://www.cs.auckland.ac.nz/software/AlgAnim/qsort1a.html). A general select-algorithm can be seen on wiki, which is, I believe what you're really after. — WhozCraig, Sep 09 '14 at 10:29

score 1 · Answer 1 · answered Sep 09 '14 at 10:18

1

[40, 20, 100, 60, 80]
          i ... #while (a[i] < pivot)

[40, 20, 100, 60, 80]
          i        j ... #while (a[j] > pivot)

[40, 20, 80, 60, 100]
          i        j ... #/* exchange values */

[40, 20, 80, 60, 100]
             ij ... #i++;j--;

[40, 20, 80, 60, 100]
              j   i  ... #while (a[i] < pivot)

[40, 20, 80, 60, 100] ... exit for-loop .. finish

answered Sep 09 '14 at 10:18

BLUEPIXY

39,699
7
33
70

Ok, but Quickselect requests to have 3 partitions: A1 with the elements < pivot, A2 with the elements = pivot, and A3 with the elements > pivot. If I use this partition algorithm, A3 is not correct because 60 < 80! Maybe this algorithm cannot be used in Quickselect? – LordFenerSSJ Sep 09 '14 at 10:22

score 1 · Accepted Answer · edited May 31 '20 at 03:17

For quickselect u need to use a recursive algorithm that will find the correct position of the pivot element, thus dividing the whole array in 2halves [elements to the right of the pivot position have value less than pivot, and the ones to the left of the pivot position have value more than the pivot]

your algorithm doesn't consider what should be done after the first loop has ended. it only finds the position of the firstly selected pivot element (that too erroneously). What will happen if the pivot position found is not the middle position (selected pivot is not the median)?

It should then move to the left part (if the newly found position of the pivot element is more than the middle position) else it should move to the right part, and perform the above step once again.

do comment if you don't understand anything

Partition in Quickselect

2 Answers2