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453 lines
15 KiB
C
453 lines
15 KiB
C
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/* @(#)e_sqrt.c 5.1 93/09/24 */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* __ieee754_sqrt(x)
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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 algebric 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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#include "fdlibm.h"
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#ifndef _DOUBLE_IS_32BITS
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#ifdef __STDC__
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static const double one = 1.0, tiny=1.0e-300;
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#else
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static double one = 1.0, tiny=1.0e-300;
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#endif
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#ifdef __STDC__
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double __ieee754_sqrt(double x)
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#else
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double __ieee754_sqrt(x)
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double x;
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#endif
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{
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double z;
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__int32_t sign = (int)0x80000000;
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__uint32_t r,t1,s1,ix1,q1;
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__int32_t ix0,s0,q,m,t,i;
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EXTRACT_WORDS(ix0,ix1,x);
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/* take care of Inf and NaN */
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if((ix0&0x7ff00000)==0x7ff00000) {
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return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
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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) 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); ix1 <<= 21;
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}
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for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
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m -= i-1;
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ix0 |= (ix1>>(32-i));
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ix1 <<= i;
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}
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m -= 1023; /* unbias exponent */
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ix0 = (ix0&0x000fffff)|0x00100000;
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if(m&1){ /* odd m, double x to make it even */
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ix0 += ix0 + ((ix1&sign)>>31);
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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);
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ix1 += ix1;
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q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
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r = 0x00200000; /* 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);
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ix1 += ix1;
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r>>=1;
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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)||((t==ix0)&&(t1<=ix1))) {
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s1 = t1+r;
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if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
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ix0 -= t;
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if (ix1 < t1) ix0 -= 1;
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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);
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ix1 += ix1;
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r>>=1;
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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 = one-tiny; /* trigger inexact flag */
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if (z>=one) {
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z = one+tiny;
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if (q1==(__uint32_t)0xffffffff) { q1=0; q += 1;}
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else if (z>one) {
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if (q1==(__uint32_t)0xfffffffe) q+=1;
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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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ix0 = (q>>1)+0x3fe00000;
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ix1 = q1>>1;
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if ((q&1)==1) ix1 |= sign;
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ix0 += (m <<20);
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INSERT_WORDS(z,ix0,ix1);
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return z;
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}
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#endif /* defined(_DOUBLE_IS_32BITS) */
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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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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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A. sqrt(x) by Newton Iteration
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(1) Initial approximation
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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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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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x0: |s| e | f1 | x1: | f2 |
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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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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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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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(2) Iterative refinement
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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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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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Remark 1.
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Another way to improve y to within 1 ulp is:
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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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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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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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(3) Final adjustment
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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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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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(4) Special cases
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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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B. sqrt(x) by Reciproot Iteration
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(1) Initial approximation
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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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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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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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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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(2) Iterative refinement
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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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... 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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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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(3) Final adjustment
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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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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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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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...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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If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
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I := 1; ... Raise Inexact flag: z is not exact
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else {
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j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
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k := z1 >> 26; ... get z's 25-th and 26-th
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fraction bits
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I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
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}
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R:= r ... restore rounded mode
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return sqrt(x):=z.
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If multiplication is cheaper then the foregoing red tape, the
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Inexact flag can be evaluated by
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I := i;
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I := (z*z!=x) or I.
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Note that z*z can overwrite I; this value must be sensed if it is
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True.
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Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
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zero.
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--------------------
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z1: | f2 |
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--------------------
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bit 31 bit 0
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Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
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or even of logb(x) have the following relations:
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-------------------------------------------------
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bit 27,26 of z1 bit 1,0 of x1 logb(x)
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-------------------------------------------------
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00 00 odd and even
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01 01 even
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10 10 odd
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10 00 even
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11 01 even
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-------------------------------------------------
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(4) Special cases (see (4) of Section A).
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*/
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