C randomized pivot quicksort (improving the partition function)

user3236524 picture user3236524 · Jan 26, 2014 · Viewed 18.7k times · Source

I'm a computer science student (just started), I was working on writing from pseudocode a randomized pivot version of Quicksort. I've written and tested it, and it all works perfectly however...

The partition part looks a bit too complicated, as it feels I have missed something or overthought it. I can't understand if it's ok or if I made some avoidable mistakes.

So long story short: it works, but how to do better?

Thanks in advance for all the help

void partition(int a[],int start,int end)
{
    srand (time(NULL));
    int pivotpos = 3;   //start + rand() % (end-start);
    int i = start;    // index 1
    int j = end;      // index 2
    int flag = 1;
    int pivot = a[pivotpos];   // sets the pivot's value
    while(i<j && flag)      // main loop
    {
        flag = 0;
        while (a[i]<pivot)
        {
            i++;
        }
        while (a[j]>pivot)
        {
            j--;
        }
        if(a[i]>a[j]) // swap && sets new pivot, and restores the flag
        {
            swap(&a[i],&a[j]);
            if(pivotpos == i)
                pivotpos = j;
            else if(pivotpos == j)
                pivotpos = i;
            flag++;
        }
        else if(a[i] == a[j])       // avoids getting suck on a mirror of values (fx pivot on pos 3 of : 1-0-0-1-1)
        {
            if(pivotpos == i) 
                j--;
            else if(pivotpos == j)
                i++;
            else
            {
                i++;
                j--;
            }
            flag++;
        }
    }
}

Answer

jfly picture jfly · Jan 26, 2014

This is the pseudo code of partition() from Introduction to Algorithms , which is called Lomuto's Partitioning Algorithm, and there's a good explanation below it in the book.

PARTITION(A, p, r)
1 x ← A[r]
2 i ← p - 1
3 for j ← p to r - 1
4   do if A[j] ≤ x
5       then i ←i + 1
6           exchange A[i] ↔ A[j]
7 exchange A[i + 1] ↔ A[r]
8 return i +1

You can implement a randomized partition implementation easily based on the pseudo code above. As the comment pointed out, move the srand() out of the partition.

// srand(time(NULL));
int partition(int* arr, int start, int end)
{
    int pivot_index = start + rand() % (end - start + 1);
    int pivot = arr[pivot_index ];

    swap(&arr[pivot_index ], &arr[end]); // swap random pivot to end.
    pivot_index = end;
    int i = start -1;

    for(int j = start; j <= end - 1; j++)
    {
        if(arr[j] <= pivot)
        {
            i++;
            swap(&arr[i], &arr[j]);
        }
    }
    swap(&arr[i + 1], &arr[pivot_index]); // place the pivot to right place

    return i + 1;
}

And there is another partition method mentioned in the book, which is called Hoare's Partitioning Algorithm, the pseudo code is as below:

Hoare-Partition(A, p, r)
x = A[p]
i = p - 1
j = r + 1
while true
    repeat
        j = j - 1
    until A[j] <= x
    repeat
        i = i + 1
    until A[i] >= x
    if i < j
        swap( A[i], A[j] )
    else
        return j

After the partition, every element in A[p...j] ≤ every element in A[j+1...r]. So the quicksort would be:

QUICKSORT (A, p, r)
if p < r then
 q = Hoare-Partition(A, p, r)
 QUICKSORT(A, p, q)
 QUICKSORT(A, q+1, r)