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+
+/* @(#)e_sqrt.c 1.3 95/01/18 */
+/*
+ * ====================================================
+ * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
+ *
+ * Developed at SunSoft, a Sun Microsystems, Inc. business.
+ * Permission to use, copy, modify, and distribute this
+ * software is freely granted, provided that this notice
+ * is preserved.
+ * ====================================================
+ */
+
+#ifndef lint
+static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_sqrt.c,v 1.10 2005/02/04 18:26:06 das Exp $";
+#endif
+
+/* __ieee754_sqrt(x)
+ * Return correctly rounded sqrt.
+ * ------------------------------------------
+ * | Use the hardware sqrt if you have one |
+ * ------------------------------------------
+ * Method:
+ * Bit by bit method using integer arithmetic. (Slow, but portable)
+ * 1. Normalization
+ * Scale x to y in [1,4) with even powers of 2:
+ * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
+ * sqrt(x) = 2^k * sqrt(y)
+ * 2. Bit by bit computation
+ * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
+ * i 0
+ * i+1 2
+ * s = 2*q , and y = 2 * ( y - q ). (1)
+ * i i i i
+ *
+ * To compute q from q , one checks whether
+ * i+1 i
+ *
+ * -(i+1) 2
+ * (q + 2 ) <= y. (2)
+ * i
+ * -(i+1)
+ * If (2) is false, then q = q ; otherwise q = q + 2 .
+ * i+1 i i+1 i
+ *
+ * With some algebric manipulation, it is not difficult to see
+ * that (2) is equivalent to
+ * -(i+1)
+ * s + 2 <= y (3)
+ * i i
+ *
+ * The advantage of (3) is that s and y can be computed by
+ * i i
+ * the following recurrence formula:
+ * if (3) is false
+ *
+ * s = s , y = y ; (4)
+ * i+1 i i+1 i
+ *
+ * otherwise,
+ * -i -(i+1)
+ * s = s + 2 , y = y - s - 2 (5)
+ * i+1 i i+1 i i
+ *
+ * One may easily use induction to prove (4) and (5).
+ * Note. Since the left hand side of (3) contain only i+2 bits,
+ * it does not necessary to do a full (53-bit) comparison
+ * in (3).
+ * 3. Final rounding
+ * After generating the 53 bits result, we compute one more bit.
+ * Together with the remainder, we can decide whether the
+ * result is exact, bigger than 1/2ulp, or less than 1/2ulp
+ * (it will never equal to 1/2ulp).
+ * The rounding mode can be detected by checking whether
+ * huge + tiny is equal to huge, and whether huge - tiny is
+ * equal to huge for some floating point number "huge" and "tiny".
+ *
+ * Special cases:
+ * sqrt(+-0) = +-0 ... exact
+ * sqrt(inf) = inf
+ * sqrt(-ve) = NaN ... with invalid signal
+ * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
+ *
+ * Other methods : see the appended file at the end of the program below.
+ *---------------
+ */
+
+#include "math.h"
+#include "math_private.h"
+
+static const double one = 1.0, tiny=1.0e-300;
+
+double
+__ieee754_sqrt(double x)
+{
+ double z;
+ int32_t sign = (int)0x80000000;
+ int32_t ix0,s0,q,m,t,i;
+ u_int32_t r,t1,s1,ix1,q1;
+
+ EXTRACT_WORDS(ix0,ix1,x);
+
+ /* take care of Inf and NaN */
+ if((ix0&0x7ff00000)==0x7ff00000) {
+ return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
+ sqrt(-inf)=sNaN */
+ }
+ /* take care of zero */
+ if(ix0<=0) {
+ if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
+ else if(ix0<0)
+ return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
+ }
+ /* normalize x */
+ m = (ix0>>20);
+ if(m==0) { /* subnormal x */
+ while(ix0==0) {
+ m -= 21;
+ ix0 |= (ix1>>11); ix1 <<= 21;
+ }
+ for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
+ m -= i-1;
+ ix0 |= (ix1>>(32-i));
+ ix1 <<= i;
+ }
+ m -= 1023; /* unbias exponent */
+ ix0 = (ix0&0x000fffff)|0x00100000;
+ if(m&1){ /* odd m, double x to make it even */
+ ix0 += ix0 + ((ix1&sign)>>31);
+ ix1 += ix1;
+ }
+ m >>= 1; /* m = [m/2] */
+
+ /* generate sqrt(x) bit by bit */
+ ix0 += ix0 + ((ix1&sign)>>31);
+ ix1 += ix1;
+ q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
+ r = 0x00200000; /* r = moving bit from right to left */
+
+ while(r!=0) {
+ t = s0+r;
+ if(t<=ix0) {
+ s0 = t+r;
+ ix0 -= t;
+ q += r;
+ }
+ ix0 += ix0 + ((ix1&sign)>>31);
+ ix1 += ix1;
+ r>>=1;
+ }
+
+ r = sign;
+ while(r!=0) {
+ t1 = s1+r;
+ t = s0;
+ if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
+ s1 = t1+r;
+ if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
+ ix0 -= t;
+ if (ix1 < t1) ix0 -= 1;
+ ix1 -= t1;
+ q1 += r;
+ }
+ ix0 += ix0 + ((ix1&sign)>>31);
+ ix1 += ix1;
+ r>>=1;
+ }
+
+ /* use floating add to find out rounding direction */
+ if((ix0|ix1)!=0) {
+ z = one-tiny; /* trigger inexact flag */
+ if (z>=one) {
+ z = one+tiny;
+ if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
+ else if (z>one) {
+ if (q1==(u_int32_t)0xfffffffe) q+=1;
+ q1+=2;
+ } else
+ q1 += (q1&1);
+ }
+ }
+ ix0 = (q>>1)+0x3fe00000;
+ ix1 = q1>>1;
+ if ((q&1)==1) ix1 |= sign;
+ ix0 += (m <<20);
+ INSERT_WORDS(z,ix0,ix1);
+ return z;
+}
+
+/*
+Other methods (use floating-point arithmetic)
+-------------
+(This is a copy of a drafted paper by Prof W. Kahan
+and K.C. Ng, written in May, 1986)
+
+ Two algorithms are given here to implement sqrt(x)
+ (IEEE double precision arithmetic) in software.
+ Both supply sqrt(x) correctly rounded. The first algorithm (in
+ Section A) uses newton iterations and involves four divisions.
+ The second one uses reciproot iterations to avoid division, but
+ requires more multiplications. Both algorithms need the ability
+ to chop results of arithmetic operations instead of round them,
+ and the INEXACT flag to indicate when an arithmetic operation
+ is executed exactly with no roundoff error, all part of the
+ standard (IEEE 754-1985). The ability to perform shift, add,
+ subtract and logical AND operations upon 32-bit words is needed
+ too, though not part of the standard.
+
+A. sqrt(x) by Newton Iteration
+
+ (1) Initial approximation
+
+ Let x0 and x1 be the leading and the trailing 32-bit words of
+ a floating point number x (in IEEE double format) respectively
+
+ 1 11 52 ...widths
+ ------------------------------------------------------
+ x: |s| e | f |
+ ------------------------------------------------------
+ msb lsb msb lsb ...order
+
+
+ ------------------------ ------------------------
+ x0: |s| e | f1 | x1: | f2 |
+ ------------------------ ------------------------
+
+ By performing shifts and subtracts on x0 and x1 (both regarded
+ as integers), we obtain an 8-bit approximation of sqrt(x) as
+ follows.
+
+ k := (x0>>1) + 0x1ff80000;
+ y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
+ Here k is a 32-bit integer and T1[] is an integer array containing
+ correction terms. Now magically the floating value of y (y's
+ leading 32-bit word is y0, the value of its trailing word is 0)
+ approximates sqrt(x) to almost 8-bit.
+
+ Value of T1:
+ static int T1[32]= {
+ 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
+ 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
+ 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
+ 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
+
+ (2) Iterative refinement
+
+ Apply Heron's rule three times to y, we have y approximates
+ sqrt(x) to within 1 ulp (Unit in the Last Place):
+
+ y := (y+x/y)/2 ... almost 17 sig. bits
+ y := (y+x/y)/2 ... almost 35 sig. bits
+ y := y-(y-x/y)/2 ... within 1 ulp
+
+
+ Remark 1.
+ Another way to improve y to within 1 ulp is:
+
+ y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
+ y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
+
+ 2
+ (x-y )*y
+ y := y + 2* ---------- ...within 1 ulp
+ 2
+ 3y + x
+
+
+ This formula has one division fewer than the one above; however,
+ it requires more multiplications and additions. Also x must be
+ scaled in advance to avoid spurious overflow in evaluating the
+ expression 3y*y+x. Hence it is not recommended uless division
+ is slow. If division is very slow, then one should use the
+ reciproot algorithm given in section B.
+
+ (3) Final adjustment
+
+ By twiddling y's last bit it is possible to force y to be
+ correctly rounded according to the prevailing rounding mode
+ as follows. Let r and i be copies of the rounding mode and
+ inexact flag before entering the square root program. Also we
+ use the expression y+-ulp for the next representable floating
+ numbers (up and down) of y. Note that y+-ulp = either fixed
+ point y+-1, or multiply y by nextafter(1,+-inf) in chopped
+ mode.
+
+ I := FALSE; ... reset INEXACT flag I
+ R := RZ; ... set rounding mode to round-toward-zero
+ z := x/y; ... chopped quotient, possibly inexact
+ If(not I) then { ... if the quotient is exact
+ if(z=y) {
+ I := i; ... restore inexact flag
+ R := r; ... restore rounded mode
+ return sqrt(x):=y.
+ } else {
+ z := z - ulp; ... special rounding
+ }
+ }
+ i := TRUE; ... sqrt(x) is inexact
+ If (r=RN) then z=z+ulp ... rounded-to-nearest
+ If (r=RP) then { ... round-toward-+inf
+ y = y+ulp; z=z+ulp;
+ }
+ y := y+z; ... chopped sum
+ y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
+ I := i; ... restore inexact flag
+ R := r; ... restore rounded mode
+ return sqrt(x):=y.
+
+ (4) Special cases
+
+ Square root of +inf, +-0, or NaN is itself;
+ Square root of a negative number is NaN with invalid signal.
+
+
+B. sqrt(x) by Reciproot Iteration
+
+ (1) Initial approximation
+
+ Let x0 and x1 be the leading and the trailing 32-bit words of
+ a floating point number x (in IEEE double format) respectively
+ (see section A). By performing shifs and subtracts on x0 and y0,
+ we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
+
+ k := 0x5fe80000 - (x0>>1);
+ y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
+
+ Here k is a 32-bit integer and T2[] is an integer array
+ containing correction terms. Now magically the floating
+ value of y (y's leading 32-bit word is y0, the value of
+ its trailing word y1 is set to zero) approximates 1/sqrt(x)
+ to almost 7.8-bit.
+
+ Value of T2:
+ static int T2[64]= {
+ 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
+ 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
+ 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
+ 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
+ 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
+ 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
+ 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
+ 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
+
+ (2) Iterative refinement
+
+ Apply Reciproot iteration three times to y and multiply the
+ result by x to get an approximation z that matches sqrt(x)
+ to about 1 ulp. To be exact, we will have
+ -1ulp < sqrt(x)-z<1.0625ulp.
+
+ ... set rounding mode to Round-to-nearest
+ y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
+ y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
+ ... special arrangement for better accuracy
+ z := x*y ... 29 bits to sqrt(x), with z*y<1
+ z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
+
+ Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
+ (a) the term z*y in the final iteration is always less than 1;
+ (b) the error in the final result is biased upward so that
+ -1 ulp < sqrt(x) - z < 1.0625 ulp
+ instead of |sqrt(x)-z|<1.03125ulp.
+
+ (3) Final adjustment
+
+ By twiddling y's last bit it is possible to force y to be
+ correctly rounded according to the prevailing rounding mode
+ as follows. Let r and i be copies of the rounding mode and
+ inexact flag before entering the square root program. Also we
+ use the expression y+-ulp for the next representable floating
+ numbers (up and down) of y. Note that y+-ulp = either fixed
+ point y+-1, or multiply y by nextafter(1,+-inf) in chopped
+ mode.
+
+ R := RZ; ... set rounding mode to round-toward-zero
+ switch(r) {
+ case RN: ... round-to-nearest
+ if(x<= z*(z-ulp)...chopped) z = z - ulp; else
+ if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
+ break;
+ case RZ:case RM: ... round-to-zero or round-to--inf
+ R:=RP; ... reset rounding mod to round-to-+inf
+ if(x<z*z ... rounded up) z = z - ulp; else
+ if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
+ break;
+ case RP: ... round-to-+inf
+ if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
+ if(x>z*z ...chopped) z = z+ulp;
+ break;
+ }
+
+ Remark 3. The above comparisons can be done in fixed point. For
+ example, to compare x and w=z*z chopped, it suffices to compare
+ x1 and w1 (the trailing parts of x and w), regarding them as
+ two's complement integers.
+
+ ...Is z an exact square root?
+ To determine whether z is an exact square root of x, let z1 be the
+ trailing part of z, and also let x0 and x1 be the leading and
+ trailing parts of x.
+
+ 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).
+
+ */
+
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