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      1 /* @(#)e_sqrt.c 5.1 93/09/24 */
      2 /*
      3  * ====================================================
      4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
      5  *
      6  * Developed at SunPro, a Sun Microsystems, Inc. business.
      7  * Permission to use, copy, modify, and distribute this
      8  * software is freely granted, provided that this notice
      9  * is preserved.
     10  * ====================================================
     11  */
     12 
     13 #include <sys/cdefs.h>
     14 #if defined(LIBM_SCCS) && !defined(lint)
     15 __RCSID("$NetBSD: e_sqrt.c,v 1.13 2009/02/16 01:19:34 lukem Exp $");
     16 #endif
     17 
     18 /* __ieee754_sqrt(x)
     19  * Return correctly rounded sqrt.
     20  *           ------------------------------------------
     21  *	     |  Use the hardware sqrt if you have one |
     22  *           ------------------------------------------
     23  * Method:
     24  *   Bit by bit method using integer arithmetic. (Slow, but portable)
     25  *   1. Normalization
     26  *	Scale x to y in [1,4) with even powers of 2:
     27  *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
     28  *		sqrt(x) = 2^k * sqrt(y)
     29  *   2. Bit by bit computation
     30  *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
     31  *	     i							 0
     32  *                                     i+1         2
     33  *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
     34  *	     i      i            i                 i
     35  *
     36  *	To compute q    from q , one checks whether
     37  *		    i+1       i
     38  *
     39  *			      -(i+1) 2
     40  *			(q + 2      ) <= y.			(2)
     41  *     			  i
     42  *							      -(i+1)
     43  *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
     44  *		 	       i+1   i             i+1   i
     45  *
     46  *	With some algebric manipulation, it is not difficult to see
     47  *	that (2) is equivalent to
     48  *                             -(i+1)
     49  *			s  +  2       <= y			(3)
     50  *			 i                i
     51  *
     52  *	The advantage of (3) is that s  and y  can be computed by
     53  *				      i      i
     54  *	the following recurrence formula:
     55  *	    if (3) is false
     56  *
     57  *	    s     =  s  ,	y    = y   ;			(4)
     58  *	     i+1      i		 i+1    i
     59  *
     60  *	    otherwise,
     61  *                         -i                     -(i+1)
     62  *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
     63  *           i+1      i          i+1    i     i
     64  *
     65  *	One may easily use induction to prove (4) and (5).
     66  *	Note. Since the left hand side of (3) contain only i+2 bits,
     67  *	      it does not necessary to do a full (53-bit) comparison
     68  *	      in (3).
     69  *   3. Final rounding
     70  *	After generating the 53 bits result, we compute one more bit.
     71  *	Together with the remainder, we can decide whether the
     72  *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
     73  *	(it will never equal to 1/2ulp).
     74  *	The rounding mode can be detected by checking whether
     75  *	huge + tiny is equal to huge, and whether huge - tiny is
     76  *	equal to huge for some floating point number "huge" and "tiny".
     77  *
     78  * Special cases:
     79  *	sqrt(+-0) = +-0 	... exact
     80  *	sqrt(inf) = inf
     81  *	sqrt(-ve) = NaN		... with invalid signal
     82  *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
     83  *
     84  * Other methods : see the appended file at the end of the program below.
     85  *---------------
     86  */
     87 
     88 #include "math.h"
     89 #include "math_private.h"
     90 
     91 static	const double	one	= 1.0, tiny=1.0e-300;
     92 
     93 double
     94 __ieee754_sqrt(double x)
     95 {
     96 	double z;
     97 	int32_t sign = (int)0x80000000;
     98 	int32_t ix0,s0,q,m,t,i;
     99 	u_int32_t r,t1,s1,ix1,q1;
    100 
    101 	EXTRACT_WORDS(ix0,ix1,x);
    102 
    103     /* take care of Inf and NaN */
    104 	if((ix0&0x7ff00000)==0x7ff00000) {
    105 	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
    106 					   sqrt(-inf)=sNaN */
    107 	}
    108     /* take care of zero */
    109 	if(ix0<=0) {
    110 	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
    111 	    else if(ix0<0)
    112 		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
    113 	}
    114     /* normalize x */
    115 	m = (ix0>>20);
    116 	if(m==0) {				/* subnormal x */
    117 	    while(ix0==0) {
    118 		m -= 21;
    119 		ix0 |= (ix1>>11); ix1 <<= 21;
    120 	    }
    121 	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
    122 	    m -= i-1;
    123 	    ix0 |= (ix1>>(32-i));
    124 	    ix1 <<= i;
    125 	}
    126 	m -= 1023;	/* unbias exponent */
    127 	ix0 = (ix0&0x000fffff)|0x00100000;
    128 	if(m&1){	/* odd m, double x to make it even */
    129 	    ix0 += ix0 + ((ix1&sign)>>31);
    130 	    ix1 += ix1;
    131 	}
    132 	m >>= 1;	/* m = [m/2] */
    133 
    134     /* generate sqrt(x) bit by bit */
    135 	ix0 += ix0 + ((ix1&sign)>>31);
    136 	ix1 += ix1;
    137 	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
    138 	r = 0x00200000;		/* r = moving bit from right to left */
    139 
    140 	while(r!=0) {
    141 	    t = s0+r;
    142 	    if(t<=ix0) {
    143 		s0   = t+r;
    144 		ix0 -= t;
    145 		q   += r;
    146 	    }
    147 	    ix0 += ix0 + ((ix1&sign)>>31);
    148 	    ix1 += ix1;
    149 	    r>>=1;
    150 	}
    151 
    152 	r = sign;
    153 	while(r!=0) {
    154 	    t1 = s1+r;
    155 	    t  = s0;
    156 	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
    157 		s1  = t1+r;
    158 		if(((t1&sign)==(u_int32_t)sign)&&(s1&sign)==0) s0 += 1;
    159 		ix0 -= t;
    160 		if (ix1 < t1) ix0 -= 1;
    161 		ix1 -= t1;
    162 		q1  += r;
    163 	    }
    164 	    ix0 += ix0 + ((ix1&sign)>>31);
    165 	    ix1 += ix1;
    166 	    r>>=1;
    167 	}
    168 
    169     /* use floating add to find out rounding direction */
    170 	if((ix0|ix1)!=0) {
    171 	    z = one-tiny; /* trigger inexact flag */
    172 	    if (z>=one) {
    173 	        z = one+tiny;
    174 	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
    175 		else if (z>one) {
    176 		    if (q1==(u_int32_t)0xfffffffe) q+=1;
    177 		    q1+=2;
    178 		} else
    179 	            q1 += (q1&1);
    180 	    }
    181 	}
    182 	ix0 = (q>>1)+0x3fe00000;
    183 	ix1 =  q1>>1;
    184 	if ((q&1)==1) ix1 |= sign;
    185 	ix0 += (m <<20);
    186 	INSERT_WORDS(z,ix0,ix1);
    187 	return z;
    188 }
    189 
    190 /*
    191 Other methods  (use floating-point arithmetic)
    192 -------------
    193 (This is a copy of a drafted paper by Prof W. Kahan
    194 and K.C. Ng, written in May, 1986)
    195 
    196 	Two algorithms are given here to implement sqrt(x)
    197 	(IEEE double precision arithmetic) in software.
    198 	Both supply sqrt(x) correctly rounded. The first algorithm (in
    199 	Section A) uses newton iterations and involves four divisions.
    200 	The second one uses reciproot iterations to avoid division, but
    201 	requires more multiplications. Both algorithms need the ability
    202 	to chop results of arithmetic operations instead of round them,
    203 	and the INEXACT flag to indicate when an arithmetic operation
    204 	is executed exactly with no roundoff error, all part of the
    205 	standard (IEEE 754-1985). The ability to perform shift, add,
    206 	subtract and logical AND operations upon 32-bit words is needed
    207 	too, though not part of the standard.
    208 
    209 A.  sqrt(x) by Newton Iteration
    210 
    211    (1)	Initial approximation
    212 
    213 	Let x0 and x1 be the leading and the trailing 32-bit words of
    214 	a floating point number x (in IEEE double format) respectively
    215 
    216 	    1    11		     52				  ...widths
    217 	   ------------------------------------------------------
    218 	x: |s|	  e     |	      f				|
    219 	   ------------------------------------------------------
    220 	      msb    lsb  msb				      lsb ...order
    221 
    222 
    223 	     ------------------------  	     ------------------------
    224 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
    225 	     ------------------------  	     ------------------------
    226 
    227 	By performing shifts and subtracts on x0 and x1 (both regarded
    228 	as integers), we obtain an 8-bit approximation of sqrt(x) as
    229 	follows.
    230 
    231 		k  := (x0>>1) + 0x1ff80000;
    232 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
    233 	Here k is a 32-bit integer and T1[] is an integer array containing
    234 	correction terms. Now magically the floating value of y (y's
    235 	leading 32-bit word is y0, the value of its trailing word is 0)
    236 	approximates sqrt(x) to almost 8-bit.
    237 
    238 	Value of T1:
    239 	static int T1[32]= {
    240 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
    241 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
    242 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
    243 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
    244 
    245     (2)	Iterative refinement
    246 
    247 	Apply Heron's rule three times to y, we have y approximates
    248 	sqrt(x) to within 1 ulp (Unit in the Last Place):
    249 
    250 		y := (y+x/y)/2		... almost 17 sig. bits
    251 		y := (y+x/y)/2		... almost 35 sig. bits
    252 		y := y-(y-x/y)/2	... within 1 ulp
    253 
    254 
    255 	Remark 1.
    256 	    Another way to improve y to within 1 ulp is:
    257 
    258 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
    259 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
    260 
    261 				2
    262 			    (x-y )*y
    263 		y := y + 2* ----------	...within 1 ulp
    264 			       2
    265 			     3y  + x
    266 
    267 
    268 	This formula has one division fewer than the one above; however,
    269 	it requires more multiplications and additions. Also x must be
    270 	scaled in advance to avoid spurious overflow in evaluating the
    271 	expression 3y*y+x. Hence it is not recommended uless division
    272 	is slow. If division is very slow, then one should use the
    273 	reciproot algorithm given in section B.
    274 
    275     (3) Final adjustment
    276 
    277 	By twiddling y's last bit it is possible to force y to be
    278 	correctly rounded according to the prevailing rounding mode
    279 	as follows. Let r and i be copies of the rounding mode and
    280 	inexact flag before entering the square root program. Also we
    281 	use the expression y+-ulp for the next representable floating
    282 	numbers (up and down) of y. Note that y+-ulp = either fixed
    283 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
    284 	mode.
    285 
    286 		I := FALSE;	... reset INEXACT flag I
    287 		R := RZ;	... set rounding mode to round-toward-zero
    288 		z := x/y;	... chopped quotient, possibly inexact
    289 		If(not I) then {	... if the quotient is exact
    290 		    if(z=y) {
    291 		        I := i;	 ... restore inexact flag
    292 		        R := r;  ... restore rounded mode
    293 		        return sqrt(x):=y.
    294 		    } else {
    295 			z := z - ulp;	... special rounding
    296 		    }
    297 		}
    298 		i := TRUE;		... sqrt(x) is inexact
    299 		If (r=RN) then z=z+ulp	... rounded-to-nearest
    300 		If (r=RP) then {	... round-toward-+inf
    301 		    y = y+ulp; z=z+ulp;
    302 		}
    303 		y := y+z;		... chopped sum
    304 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
    305 	        I := i;	 		... restore inexact flag
    306 	        R := r;  		... restore rounded mode
    307 	        return sqrt(x):=y.
    308 
    309     (4)	Special cases
    310 
    311 	Square root of +inf, +-0, or NaN is itself;
    312 	Square root of a negative number is NaN with invalid signal.
    313 
    314 
    315 B.  sqrt(x) by Reciproot Iteration
    316 
    317    (1)	Initial approximation
    318 
    319 	Let x0 and x1 be the leading and the trailing 32-bit words of
    320 	a floating point number x (in IEEE double format) respectively
    321 	(see section A). By performing shifs and subtracts on x0 and y0,
    322 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
    323 
    324 	    k := 0x5fe80000 - (x0>>1);
    325 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
    326 
    327 	Here k is a 32-bit integer and T2[] is an integer array
    328 	containing correction terms. Now magically the floating
    329 	value of y (y's leading 32-bit word is y0, the value of
    330 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
    331 	to almost 7.8-bit.
    332 
    333 	Value of T2:
    334 	static int T2[64]= {
    335 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
    336 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
    337 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
    338 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
    339 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
    340 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
    341 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
    342 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
    343 
    344     (2)	Iterative refinement
    345 
    346 	Apply Reciproot iteration three times to y and multiply the
    347 	result by x to get an approximation z that matches sqrt(x)
    348 	to about 1 ulp. To be exact, we will have
    349 		-1ulp < sqrt(x)-z<1.0625ulp.
    350 
    351 	... set rounding mode to Round-to-nearest
    352 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
    353 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
    354 	... special arrangement for better accuracy
    355 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
    356 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
    357 
    358 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
    359 	(a) the term z*y in the final iteration is always less than 1;
    360 	(b) the error in the final result is biased upward so that
    361 		-1 ulp < sqrt(x) - z < 1.0625 ulp
    362 	    instead of |sqrt(x)-z|<1.03125ulp.
    363 
    364     (3)	Final adjustment
    365 
    366 	By twiddling y's last bit it is possible to force y to be
    367 	correctly rounded according to the prevailing rounding mode
    368 	as follows. Let r and i be copies of the rounding mode and
    369 	inexact flag before entering the square root program. Also we
    370 	use the expression y+-ulp for the next representable floating
    371 	numbers (up and down) of y. Note that y+-ulp = either fixed
    372 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
    373 	mode.
    374 
    375 	R := RZ;		... set rounding mode to round-toward-zero
    376 	switch(r) {
    377 	    case RN:		... round-to-nearest
    378 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
    379 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
    380 	       break;
    381 	    case RZ:case RM:	... round-to-zero or round-to--inf
    382 	       R:=RP;		... reset rounding mod to round-to-+inf
    383 	       if(x<z*z ... rounded up) z = z - ulp; else
    384 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
    385 	       break;
    386 	    case RP:		... round-to-+inf
    387 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
    388 	       if(x>z*z ...chopped) z = z+ulp;
    389 	       break;
    390 	}
    391 
    392 	Remark 3. The above comparisons can be done in fixed point. For
    393 	example, to compare x and w=z*z chopped, it suffices to compare
    394 	x1 and w1 (the trailing parts of x and w), regarding them as
    395 	two's complement integers.
    396 
    397 	...Is z an exact square root?
    398 	To determine whether z is an exact square root of x, let z1 be the
    399 	trailing part of z, and also let x0 and x1 be the leading and
    400 	trailing parts of x.
    401 
    402 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
    403 	    I := 1;		... Raise Inexact flag: z is not exact
    404 	else {
    405 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
    406 	    k := z1 >> 26;		... get z's 25-th and 26-th
    407 					    fraction bits
    408 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
    409 	}
    410 	R:= r		... restore rounded mode
    411 	return sqrt(x):=z.
    412 
    413 	If multiplication is cheaper than the foregoing red tape, the
    414 	Inexact flag can be evaluated by
    415 
    416 	    I := i;
    417 	    I := (z*z!=x) or I.
    418 
    419 	Note that z*z can overwrite I; this value must be sensed if it is
    420 	True.
    421 
    422 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
    423 	zero.
    424 
    425 		    --------------------
    426 		z1: |        f2        |
    427 		    --------------------
    428 		bit 31		   bit 0
    429 
    430 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
    431 	or even of logb(x) have the following relations:
    432 
    433 	-------------------------------------------------
    434 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
    435 	-------------------------------------------------
    436 	00			00		odd and even
    437 	01			01		even
    438 	10			10		odd
    439 	10			00		even
    440 	11			01		even
    441 	-------------------------------------------------
    442 
    443     (4)	Special cases (see (4) of Section A).
    444 
    445  */
    446 
    447