Re[4]: New portion of improvements for Dual-Pivot Quicksort

Thu May 13 17:34:54 UTC 2010

```Dmytro,

I've tested your suggested variants, and found that case "C"
(very interesting approach to find first position of zero
by counting negative elements) works slower than original
or two other cases.

Implementations "F" and "S" are very close to each other
and little bit faster than original. I prefer case "F":
it is shorter and more clear. Do you agree?

I'll prepare updated DualPivotQuicksort file and send it
tomorrow.

Thank you,

Wed, 12 May 2010 17:04:52 +0700 письмо от Dmytro Sheyko <dmytro_sheyko at hotmail.com>:

>
> Your changes are good for me.
>
>
> 1. Scanning for the first zero we can avoid range check (i >= left) if we have at least one negative value.
> --- DualPivotQuicksort.java    Tue May 11 09:04:19 2010
> +++ DualPivotQuicksortS.java    Wed May 12 12:10:46 2010
> @@ -1705,10 +1705,15 @@
>          }
>
>          // Find first zero element
> -        int zeroIndex = findAnyZero(a, left, n);
> +        int zeroIndex = 0;
>
> -        for (int i = zeroIndex - 1; i >= left && a[i] == 0.0f; i--) {
> -            zeroIndex = i;
> +        if (a[left] < 0.0f) {
> +            zeroIndex = findAnyZero(a, left, n);
> +
> +            // there is at least one negative value, so range check is not needed
> +            for (int i = zeroIndex - 1; /*i >= left &&*/ a[i] == 0.0f; i--) {
> +                zeroIndex = i;
> +            }
>          }
>
>          // Turn the right number of positive zeros back into negative zeros
>
> 2. We can find the position of the first zero by counting negative values during preprocessing phase.
> --- DualPivotQuicksort.java    Tue May 11 09:04:19 2010
> +++ DualPivotQuicksortC.java    Wed May 12 12:01:24 2010
> @@ -1678,7 +1678,7 @@
>           * Phase 1: Count negative zeros and move NaNs to end of array.
>           */
>          final int NEGATIVE_ZERO = Float.floatToIntBits(-0.0f);
> -        int numNegativeZeros = 0;
> +        int numNegativeZeros = 0, numNegativeValues = 0;
>          int n = right;
>
>          for (int k = left; k <= n; k++) {
> @@ -1689,6 +1689,8 @@
>              } else if (ak != ak) { // i.e., ak is NaN
>                  a[k--] = a[n];
>                  a[n--] = Float.NaN;
> +            } else if (ak < 0.0f) {
> +                numNegativeValues++;
>              }
>          }
>
> @@ -1705,7 +1707,7 @@
>          }
>
>          // Find first zero element
> -        int zeroIndex = findAnyZero(a, left, n);
> +        int zeroIndex = numNegativeValues;
>
>          for (int i = zeroIndex - 1; i >= left && a[i] == 0.0f; i--) {
>              zeroIndex = i;
>
> 3. We can use binary search to find the first zero and thus avoid linear scan.
> --- DualPivotQuicksort.java    Tue May 11 09:04:19 2010
> +++ DualPivotQuicksortF.java    Wed May 12 12:03:58 2010
> @@ -1705,11 +1705,7 @@
>          }
>
>          // Find first zero element
> -        int zeroIndex = findAnyZero(a, left, n);
> -
> -        for (int i = zeroIndex - 1; i >= left && a[i] == 0.0f; i--) {
> -            zeroIndex = i;
> -        }
> +        int zeroIndex = findFirstZero(a, left, n);
>
>          // Turn the right number of positive zeros back into negative zeros
>          for (int i = zeroIndex, m = zeroIndex + numNegativeZeros; i < m; i++) {
> @@ -1718,7 +1714,7 @@
>      }
>
>      /**
> -     * Returns the index of some zero element in the specified range via
> +     * Returns the index of the first zero element in the specified range via
>       * binary search. The range is assumed to be sorted, and must contain
>       * at least one zero.
>       *
> @@ -1726,18 +1722,17 @@
>       * @param low the index of the first element, inclusive, to be searched
>       * @param high the index of the last element, inclusive, to be searched
>       */
> -    private static int findAnyZero(float[] a, int low, int high) {
> -        while (true) {
> +    private static int findFirstZero(float[] a, int low, int high) {
> +        while (low < high) {
>              int middle = (low + high) >>> 1;
>              float middleValue = a[middle];
>
>              if (middleValue < 0.0f) {
>                  low = middle + 1;
> -            } else if (middleValue > 0.0f) {
> -                high = middle - 1;
> -            } else { // middleValue == 0.0f
> -                return middle;
> +            } else { // middleValue >= 0.0f
> +                high = middle;
>              }
> +            return low;
>          }
>      }
>
> Counting negative values appeared more expensive than any other variants.
> The last proposal seems to me as efficient as the current solution is in its worst case - when we have only one negative zero (in the half of array).
> And it shows the best result if we have many zeros.
>
> Regards,
> Dmytro Sheyko
>
> > From: iaroslavski at mail.ru
> > To: jjb at google.com; dmytro_sheyko at hotmail.com
> > CC: core-libs-dev at openjdk.java.net; iaroslavski at mail.ru
> > Subject: Re[2]: New portion of improvements for Dual-Pivot Quicksort
> > Date: Sun, 9 May 2010 23:51:27 +0400
> >
> > Josh,
> > Dmytro,
> >
> > I have done more thoroughly testing "great - less > 5 * seventh" vs. "less < e1 && great > e5",
> > and found that more symmetric code "less < e1 && great > e5" is little bit faster, ~0.5..0.7%
> > on both VMs. Other code has not been changed.
> >
> >
> >
> > Tue, 4 May 2010 21:57:42 -0700 письмо от Joshua Bloch <jjb at google.com>:
> >
> > >
> > > Old:
> > >
> > >298 if (less < e1 && great > e5) {
> > >
> > > New:
> > >
> > >256 if (great - less > 5 * seventh) {
> >
> > >Regards,
> > >Josh

```