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8302040: Port fdlibm sqrt to Java
Reviewed-by: bpb, thartmann, aturbanov
This commit is contained in:
Joe Darcy
parent
6423065b7d
commit
61e8867591
8 changed files with 875 additions and 6 deletions
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@ -104,6 +104,15 @@ class FdLibm {
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( ((long)high)) << 32 );
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}
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/**
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* Return a double with its high-order bits of the first argument
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* and the low-order bits of the second argument..
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*/
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private static double __HI_LO(int high, int low) {
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return Double.longBitsToDouble(((long)high << 32) |
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(low & 0xffff_ffffL));
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}
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/** Returns the arcsine of x.
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*
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* Method :
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@ -504,6 +513,449 @@ class FdLibm {
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}
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}
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/**
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* Return correctly rounded sqrt.
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* ------------------------------------------
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* | Use the hardware sqrt if you have one |
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* ------------------------------------------
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* Method:
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* Bit by bit method using integer arithmetic. (Slow, but portable)
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* 1. Normalization
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* Scale x to y in [1,4) with even powers of 2:
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* find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
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* sqrt(x) = 2^k * sqrt(y)
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* 2. Bit by bit computation
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* Let q = sqrt(y) truncated to i bit after binary point (q = 1),
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* i 0
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* i+1 2
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* s = 2*q , and y = 2 * ( y - q ). (1)
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* i i i i
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*
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* To compute q from q , one checks whether
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* i+1 i
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*
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* -(i+1) 2
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* (q + 2 ) <= y. (2)
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* i
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* -(i+1)
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* If (2) is false, then q = q ; otherwise q = q + 2 .
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* i+1 i i+1 i
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*
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* With some algebraic manipulation, it is not difficult to see
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* that (2) is equivalent to
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* -(i+1)
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* s + 2 <= y (3)
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* i i
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*
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* The advantage of (3) is that s and y can be computed by
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* i i
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* the following recurrence formula:
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* if (3) is false
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*
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* s = s , y = y ; (4)
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* i+1 i i+1 i
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*
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* otherwise,
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* -i -(i+1)
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* s = s + 2 , y = y - s - 2 (5)
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* i+1 i i+1 i i
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*
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* One may easily use induction to prove (4) and (5).
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* Note. Since the left hand side of (3) contain only i+2 bits,
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* it does not necessary to do a full (53-bit) comparison
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* in (3).
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* 3. Final rounding
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* After generating the 53 bits result, we compute one more bit.
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* Together with the remainder, we can decide whether the
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* result is exact, bigger than 1/2ulp, or less than 1/2ulp
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* (it will never equal to 1/2ulp).
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* The rounding mode can be detected by checking whether
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* huge + tiny is equal to huge, and whether huge - tiny is
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* equal to huge for some floating point number "huge" and "tiny".
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*
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* Special cases:
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* sqrt(+-0) = +-0 ... exact
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* sqrt(inf) = inf
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* sqrt(-ve) = NaN ... with invalid signal
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* sqrt(NaN) = NaN ... with invalid signal for signaling NaN
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*
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* Other methods : see the appended file at the end of the program below.
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*---------------
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*/
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static class Sqrt {
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private Sqrt() {throw new UnsupportedOperationException();}
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private static final double tiny = 1.0e-300;
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static double compute(double x) {
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double z = 0.0;
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int sign = 0x8000_0000;
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/*unsigned*/ int r, t1, s1, ix1, q1;
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int ix0, s0, q, m, t, i;
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ix0 = __HI(x); // high word of x
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ix1 = __LO(x); // low word of x
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// take care of Inf and NaN
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if ((ix0 & 0x7ff0_0000) == 0x7ff0_0000) {
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return x*x + x; // sqrt(NaN)=NaN, sqrt(+inf)=+inf, sqrt(-inf)=sNaN
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}
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// take care of zero
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if (ix0 <= 0) {
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if (((ix0 & (~sign)) | ix1) == 0)
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return x; // sqrt(+-0) = +-0
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else if (ix0 < 0)
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return (x-x)/(x-x); // sqrt(-ve) = sNaN
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}
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// normalize x
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m = (ix0 >> 20);
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if (m == 0) { // subnormal x
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while (ix0 == 0) {
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m -= 21;
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ix0 |= (ix1 >>> 11); // unsigned shift
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ix1 <<= 21;
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}
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for(i = 0; (ix0 & 0x0010_0000) == 0; i++) {
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ix0 <<= 1;
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}
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m -= i-1;
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ix0 |= (ix1 >>> (32 - i)); // unsigned shift
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ix1 <<= i;
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}
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m -= 1023; // unbias exponent */
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ix0 = (ix0 & 0x000f_ffff) | 0x0010_0000;
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if ((m & 1) != 0){ // odd m, double x to make it even
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ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
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ix1 += ix1;
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}
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m >>= 1; // m = [m/2]
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// generate sqrt(x) bit by bit
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ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
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ix1 += ix1;
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q = q1 = s0 = s1 = 0; // [q,q1] = sqrt(x)
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r = 0x0020_0000; // r = moving bit from right to left
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while (r != 0) {
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t = s0 + r;
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if (t <= ix0) {
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s0 = t + r;
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ix0 -= t;
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q += r;
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}
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ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
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ix1 += ix1;
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r >>>= 1; // unsigned shift
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}
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r = sign;
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while (r != 0) {
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t1 = s1 + r;
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t = s0;
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if ((t < ix0) ||
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((t == ix0) && (Integer.compareUnsigned(t1, ix1) <= 0 ))) { // t1 <= ix1
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s1 = t1 + r;
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if (((t1 & sign) == sign) && (s1 & sign) == 0) {
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s0 += 1;
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}
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ix0 -= t;
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if (Integer.compareUnsigned(ix1, t1) < 0) { // ix1 < t1
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ix0 -= 1;
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}
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ix1 -= t1;
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q1 += r;
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}
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ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
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ix1 += ix1;
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r >>>= 1; // unsigned shift
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}
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// use floating add to find out rounding direction
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if ((ix0 | ix1) != 0) {
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z = 1.0 - tiny; // trigger inexact flag
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if (z >= 1.0) {
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z = 1.0 + tiny;
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if (q1 == 0xffff_ffff) {
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q1 = 0;
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q += 1;
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} else if (z > 1.0) {
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if (q1 == 0xffff_fffe) {
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q += 1;
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}
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q1 += 2;
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} else {
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q1 += (q1 & 1);
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}
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}
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}
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ix0 = (q >> 1) + 0x3fe0_0000;
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ix1 = q1 >>> 1; // unsigned shift
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if ((q & 1) == 1) {
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ix1 |= sign;
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}
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ix0 += (m << 20);
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return __HI_LO(ix0, ix1);
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}
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}
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// The following comment is supplementary information from the FDLIBM sources.
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/*
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* Other methods (use floating-point arithmetic)
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* -------------
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* (This is a copy of a drafted paper by Prof W. Kahan
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* and K.C. Ng, written in May, 1986)
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*
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* Two algorithms are given here to implement sqrt(x)
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* (IEEE double precision arithmetic) in software.
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* Both supply sqrt(x) correctly rounded. The first algorithm (in
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* Section A) uses newton iterations and involves four divisions.
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* The second one uses reciproot iterations to avoid division, but
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* requires more multiplications. Both algorithms need the ability
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* to chop results of arithmetic operations instead of round them,
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* and the INEXACT flag to indicate when an arithmetic operation
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* is executed exactly with no roundoff error, all part of the
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* standard (IEEE 754-1985). The ability to perform shift, add,
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* subtract and logical AND operations upon 32-bit words is needed
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* too, though not part of the standard.
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*
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* A. sqrt(x) by Newton Iteration
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*
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* (1) Initial approximation
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*
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* Let x0 and x1 be the leading and the trailing 32-bit words of
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* a floating point number x (in IEEE double format) respectively
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*
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* 1 11 52 ...widths
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* ------------------------------------------------------
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* x: |s| e | f |
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* ------------------------------------------------------
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* msb lsb msb lsb ...order
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*
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*
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* ------------------------ ------------------------
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* x0: |s| e | f1 | x1: | f2 |
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* ------------------------ ------------------------
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*
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* By performing shifts and subtracts on x0 and x1 (both regarded
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* as integers), we obtain an 8-bit approximation of sqrt(x) as
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* follows.
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*
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* k := (x0>>1) + 0x1ff80000;
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* y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
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* Here k is a 32-bit integer and T1[] is an integer array containing
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* correction terms. Now magically the floating value of y (y's
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* leading 32-bit word is y0, the value of its trailing word is 0)
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* approximates sqrt(x) to almost 8-bit.
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*
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* Value of T1:
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* static int T1[32]= {
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* 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
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* 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
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* 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
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* 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
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*
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* (2) Iterative refinement
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*
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* Apply Heron's rule three times to y, we have y approximates
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* sqrt(x) to within 1 ulp (Unit in the Last Place):
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*
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* y := (y+x/y)/2 ... almost 17 sig. bits
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* y := (y+x/y)/2 ... almost 35 sig. bits
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* y := y-(y-x/y)/2 ... within 1 ulp
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*
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*
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* Remark 1.
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* Another way to improve y to within 1 ulp is:
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*
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* y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
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* y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
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*
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* 2
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* (x-y )*y
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* y := y + 2* ---------- ...within 1 ulp
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* 2
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* 3y + x
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*
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*
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* This formula has one division fewer than the one above; however,
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* it requires more multiplications and additions. Also x must be
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* scaled in advance to avoid spurious overflow in evaluating the
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* expression 3y*y+x. Hence it is not recommended uless division
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* is slow. If division is very slow, then one should use the
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* reciproot algorithm given in section B.
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*
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* (3) Final adjustment
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*
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* By twiddling y's last bit it is possible to force y to be
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* correctly rounded according to the prevailing rounding mode
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* as follows. Let r and i be copies of the rounding mode and
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* inexact flag before entering the square root program. Also we
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* use the expression y+-ulp for the next representable floating
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* numbers (up and down) of y. Note that y+-ulp = either fixed
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* point y+-1, or multiply y by nextafter(1,+-inf) in chopped
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* mode.
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*
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* I := FALSE; ... reset INEXACT flag I
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* R := RZ; ... set rounding mode to round-toward-zero
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* z := x/y; ... chopped quotient, possibly inexact
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* If(not I) then { ... if the quotient is exact
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* if(z=y) {
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* I := i; ... restore inexact flag
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* R := r; ... restore rounded mode
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* return sqrt(x):=y.
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* } else {
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* z := z - ulp; ... special rounding
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* }
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* }
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* i := TRUE; ... sqrt(x) is inexact
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* If (r=RN) then z=z+ulp ... rounded-to-nearest
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* If (r=RP) then { ... round-toward-+inf
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* y = y+ulp; z=z+ulp;
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* }
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* y := y+z; ... chopped sum
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* y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
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* I := i; ... restore inexact flag
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* R := r; ... restore rounded mode
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* return sqrt(x):=y.
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*
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* (4) Special cases
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*
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* Square root of +inf, +-0, or NaN is itself;
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* Square root of a negative number is NaN with invalid signal.
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*
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*
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* B. sqrt(x) by Reciproot Iteration
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*
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* (1) Initial approximation
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*
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* Let x0 and x1 be the leading and the trailing 32-bit words of
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* a floating point number x (in IEEE double format) respectively
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* (see section A). By performing shifs and subtracts on x0 and y0,
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* we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
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*
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* k := 0x5fe80000 - (x0>>1);
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* y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
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*
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* Here k is a 32-bit integer and T2[] is an integer array
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* containing correction terms. Now magically the floating
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* value of y (y's leading 32-bit word is y0, the value of
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* its trailing word y1 is set to zero) approximates 1/sqrt(x)
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* to almost 7.8-bit.
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*
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* Value of T2:
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* static int T2[64]= {
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* 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
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* 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
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* 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
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* 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
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* 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
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* 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
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* 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
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* 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
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*
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* (2) Iterative refinement
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*
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* Apply Reciproot iteration three times to y and multiply the
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* result by x to get an approximation z that matches sqrt(x)
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* to about 1 ulp. To be exact, we will have
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* -1ulp < sqrt(x)-z<1.0625ulp.
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*
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* ... set rounding mode to Round-to-nearest
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* y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
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* y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
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* ... special arrangement for better accuracy
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* z := x*y ... 29 bits to sqrt(x), with z*y<1
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* z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
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*
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* Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
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* (a) the term z*y in the final iteration is always less than 1;
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* (b) the error in the final result is biased upward so that
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* -1 ulp < sqrt(x) - z < 1.0625 ulp
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* instead of |sqrt(x)-z|<1.03125ulp.
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*
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* (3) Final adjustment
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*
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* By twiddling y's last bit it is possible to force y to be
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* correctly rounded according to the prevailing rounding mode
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* as follows. Let r and i be copies of the rounding mode and
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* inexact flag before entering the square root program. Also we
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* use the expression y+-ulp for the next representable floating
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* numbers (up and down) of y. Note that y+-ulp = either fixed
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* point y+-1, or multiply y by nextafter(1,+-inf) in chopped
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* mode.
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*
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* R := RZ; ... set rounding mode to round-toward-zero
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* switch(r) {
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* case RN: ... round-to-nearest
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* if(x<= z*(z-ulp)...chopped) z = z - ulp; else
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* if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
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* break;
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* case RZ:case RM: ... round-to-zero or round-to--inf
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* R:=RP; ... reset rounding mod to round-to-+inf
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* if(x<z*z ... rounded up) z = z - ulp; else
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* if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
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* break;
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* case RP: ... round-to-+inf
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* if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
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* if(x>z*z ...chopped) z = z+ulp;
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* break;
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* }
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*
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* Remark 3. The above comparisons can be done in fixed point. For
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* example, to compare x and w=z*z chopped, it suffices to compare
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* x1 and w1 (the trailing parts of x and w), regarding them as
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* two's complement integers.
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*
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* ...Is z an exact square root?
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* To determine whether z is an exact square root of x, let z1 be the
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* trailing part of z, and also let x0 and x1 be the leading and
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* trailing parts of x.
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*
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* If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
|
||||
* I := 1; ... Raise Inexact flag: z is not exact
|
||||
* else {
|
||||
* j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
|
||||
* k := z1 >> 26; ... get z's 25-th and 26-th
|
||||
* fraction bits
|
||||
* I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
|
||||
* }
|
||||
* R:= r ... restore rounded mode
|
||||
* return sqrt(x):=z.
|
||||
*
|
||||
* If multiplication is cheaper then the foregoing red tape, the
|
||||
* Inexact flag can be evaluated by
|
||||
*
|
||||
* I := i;
|
||||
* I := (z*z!=x) or I.
|
||||
*
|
||||
* Note that z*z can overwrite I; this value must be sensed if it is
|
||||
* True.
|
||||
*
|
||||
* Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
|
||||
* zero.
|
||||
*
|
||||
* --------------------
|
||||
* z1: | f2 |
|
||||
* --------------------
|
||||
* bit 31 bit 0
|
||||
*
|
||||
* Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
|
||||
* or even of logb(x) have the following relations:
|
||||
*
|
||||
* -------------------------------------------------
|
||||
* bit 27,26 of z1 bit 1,0 of x1 logb(x)
|
||||
* -------------------------------------------------
|
||||
* 00 00 odd and even
|
||||
* 01 01 even
|
||||
* 10 10 odd
|
||||
* 10 00 even
|
||||
* 11 01 even
|
||||
* -------------------------------------------------
|
||||
*
|
||||
* (4) Special cases (see (4) of Section A).
|
||||
*/
|
||||
|
||||
/**
|
||||
* cbrt(x)
|
||||
* Return cube root of x
|
||||
|
|
|
|||
|
|
@ -310,7 +310,9 @@ public final class StrictMath {
|
|||
* @return the positive square root of {@code a}.
|
||||
*/
|
||||
@IntrinsicCandidate
|
||||
public static native double sqrt(double a);
|
||||
public static double sqrt(double a) {
|
||||
return FdLibm.Sqrt.compute(a);
|
||||
}
|
||||
|
||||
/**
|
||||
* Returns the cube root of a {@code double} value. For
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue