JDK 9 RFR of 8058505: BigIntegerTest does not exercise Burnikel-Ziegler division

[Brian Burkhalter]

Hello,

This is a test-only change.

Issue:	https://bugs.openjdk.java.net/browse/JDK-8058505
Webrev:	http://cr.openjdk.java.net/~bpb/8058505/webrev.00/

I verified that the updated version passes (as mentioned below) and in fact exercises the B-Z division branch of the library code.

Thanks,

Brian

On Sep 15, 2014, at 7:17 AM, Robert Gibson <robbiexgibson at yahoo.com> wrote:

> Here is a patch to fix the test bug mentioned previously.  The Burnikel-Ziegler division code is now exercised, and you'll be glad to know that the tests still pass!
> Robert
> 
> --- BigIntegerTest.java 2014-09-15 15:55:47.632012000 +0200
> +++ BigIntegerTestPatched.java  2014-09-15 16:07:53.363563000 +0200
> @@ -71,6 +71,7 @@
>     static final int BITS_TOOM_COOK_SQUARE = 6912;
>     static final int BITS_SCHOENHAGE_BASE = 640;
>     static final int BITS_BURNIKEL_ZIEGLER = 2560;
> +    static final int BITS_BURNIKEL_ZIEGLER_OFFSET = 1280;
>     static final int ORDER_SMALL = 60;
>     static final int ORDER_MEDIUM = 100;
> @@ -288,19 +289,19 @@
>      * where {@code abs(u) > abs(v)} and {@code a > b && b > 0}, then if
>      * {@code w/z = q1*z + r1} and {@code u/v = q2*v + r2}, then
>      * {@code q1 = q2*pow(2,a-b)} and {@code r1 = r2*pow(2,b)}.  The test
> -     * ensures that {@code v} is just under the B-Z threshold and that {@code w}
> -     * and {@code z} are both over the threshold.  This implies that {@code u/v}
> -     * uses the standard division algorithm and {@code w/z} uses the B-Z
> -     * algorithm.  The results of the two algorithms are then compared using the
> -     * observation described in the foregoing and if they are not equal a
> -     * failure is logged.
> +     * ensures that {@code v} is just under the B-Z threshold, that {@code z} is
> +     * over the threshold and {@code w} is much larger than {@code z}. This
> +     * implies that {@code u/v} uses the standard division algorithm and
> +     * {@code w/z} uses the B-Z algorithm.  The results of the two algorithms
> +     * are then compared using the observation described in the foregoing and
> +     * if they are not equal a failure is logged.
>      */
>     public static void divideLarge() {
>         int failCount = 0;
> -        BigInteger base = BigInteger.ONE.shiftLeft(BITS_BURNIKEL_ZIEGLER - 33);
> +        BigInteger base = BigInteger.ONE.shiftLeft(BITS_BURNIKEL_ZIEGLER + BITS_BURNIKEL_ZIEGLER_OFFSET - 33);
>         for (int i=0; i<SIZE; i++) {
> -            BigInteger addend = new BigInteger(BITS_BURNIKEL_ZIEGLER - 34, rnd);
> +            BigInteger addend = new BigInteger(BITS_BURNIKEL_ZIEGLER + BITS_BURNIKEL_ZIEGLER_OFFSET - 34, rnd);
>             BigInteger v = base.add(addend);
>             BigInteger u = v.multiply(BigInteger.valueOf(2 + rnd.nextInt(Short.MAX_VALUE - 1)));
> @@ -312,14 +313,14 @@
>                 v = v.negate();
>             }
> -            int a = 17 + rnd.nextInt(16);
> +            int a = BITS_BURNIKEL_ZIEGLER_OFFSET + rnd.nextInt(16);
>             int b = 1 + rnd.nextInt(16);
> -            BigInteger w = u.multiply(BigInteger.valueOf(1L << a));
> -            BigInteger z = v.multiply(BigInteger.valueOf(1L << b));
> +            BigInteger w = u.multiply(BigInteger.ONE.shiftLeft(a));
> +            BigInteger z = v.multiply(BigInteger.ONE.shiftLeft(b));
>             BigInteger[] divideResult = u.divideAndRemainder(v);
> -            divideResult[0] = divideResult[0].multiply(BigInteger.valueOf(1L << (a - b)));
> -            divideResult[1] = divideResult[1].multiply(BigInteger.valueOf(1L << b));
> +            divideResult[0] = divideResult[0].multiply(BigInteger.ONE.shiftLeft(a - b));
> +            divideResult[1] = divideResult[1].multiply(BigInteger.ONE.shiftLeft(b));
>             BigInteger[] bzResult = w.divideAndRemainder(z);
>             if (divideResult[0].compareTo(bzResult[0]) != 0 ||