[issue8692] Use divide-and-conquer for faster factorials

[Daniel Stutzbach]

Daniel Stutzbach <daniel at stutzbachenterprises.com> added the comment:

On Wed, May 12, 2010 at 10:31 AM, Alexander Belopolsky
<report at bugs.python.org> wrote:
> if ((k & 1) != 1)
>          k = k - 1;
>
> looks odd to me. Maybe k += (k & 1) - 1?

If we change the logic slightly to put the odd entry on the right
instead of the left, we can do:

        k = (n + m) / 2;
        k |= 1; /* Ensure that k is odd */
        left = factorial_part_product(n, k-2);
        if (left == NULL) goto done;
        right = factorial_part_product(k, m);
        if (right == NULL) goto done;

That will split 1*3*5*7*9*11 into (1*3*5) * (7*9*11), just like the
old code.  It will split 1*3*5*7*9 into (1*3) * (5*7*9) while the old
code did (1*3*5) * (7*9), which is fine.

It's easier to read and fewer operations. :-)

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue8692>
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