## test/java/math/BigInteger/BigIntegerTest.java

```rev 10672 : 8058505: BigIntegerTest does not exercise Burnikel-Ziegler division
Modify divideLarge() method such that the w/z division exercises the B-Z branch.
Reviewed-by: TBD
Contributed-by: robbiexgibson@yahoo.com```
```*** 69,78 ****
--- 69,79 ----
static final int BITS_TOOM_COOK = 7680;
static final int BITS_KARATSUBA_SQUARE = 4096;
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;
// #bits for testing Karatsuba
static final int ORDER_KARATSUBA = 2760;
*** 286,308 ****
* a specified number of ints in its representation.  This test is based on
* the observation that if {@code w = u*pow(2,a)} and {@code z = v*pow(2,b)}
* 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.
*/
public static void divideLarge() {
int failCount = 0;

!         BigInteger base = BigInteger.ONE.shiftLeft(BITS_BURNIKEL_ZIEGLER - 33);
for (int i=0; i<SIZE; i++) {
!             BigInteger addend = new BigInteger(BITS_BURNIKEL_ZIEGLER - 34, rnd);

BigInteger u = v.multiply(BigInteger.valueOf(2 + rnd.nextInt(Short.MAX_VALUE - 1)));

if(rnd.nextBoolean()) {
--- 287,309 ----
* a specified number of ints in its representation.  This test is based on
* the observation that if {@code w = u*pow(2,a)} and {@code z = v*pow(2,b)}
* 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, 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 + BITS_BURNIKEL_ZIEGLER_OFFSET - 33);
for (int i=0; i<SIZE; i++) {
!             BigInteger addend = new BigInteger(BITS_BURNIKEL_ZIEGLER + BITS_BURNIKEL_ZIEGLER_OFFSET - 34, rnd);

BigInteger u = v.multiply(BigInteger.valueOf(2 + rnd.nextInt(Short.MAX_VALUE - 1)));

if(rnd.nextBoolean()) {
*** 310,327 ****
}
if(rnd.nextBoolean()) {
v = v.negate();
}

!             int a = 17 + 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[] divideResult = u.divideAndRemainder(v);
!             divideResult[0] = divideResult[0].multiply(BigInteger.valueOf(1L << (a - b)));
!             divideResult[1] = divideResult[1].multiply(BigInteger.valueOf(1L << b));
BigInteger[] bzResult = w.divideAndRemainder(z);

if (divideResult[0].compareTo(bzResult[0]) != 0 ||
divideResult[1].compareTo(bzResult[1]) != 0) {
failCount++;
--- 311,328 ----
}
if(rnd.nextBoolean()) {
v = v.negate();
}

!             int a = BITS_BURNIKEL_ZIEGLER_OFFSET + rnd.nextInt(16);
int b = 1 + rnd.nextInt(16);
!             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.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 ||
divideResult[1].compareTo(bzResult[1]) != 0) {
failCount++;
```