vx32

Local 9vx git repository for patches.
git clone git://r-36.net/vx32
Log | Files | Refs

e_sqrt.c (14308B)


      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 #ifndef lint
     14 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_sqrt.c,v 1.9 2003/07/23 04:53:46 peter Exp $";
     15 #endif
     16 
     17 /* __ieee754_sqrt(x)
     18  * Return correctly rounded sqrt.
     19  *           ------------------------------------------
     20  *	     |  Use the hardware sqrt if you have one |
     21  *           ------------------------------------------
     22  * Method:
     23  *   Bit by bit method using integer arithmetic. (Slow, but portable)
     24  *   1. Normalization
     25  *	Scale x to y in [1,4) with even powers of 2:
     26  *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
     27  *		sqrt(x) = 2^k * sqrt(y)
     28  *   2. Bit by bit computation
     29  *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
     30  *	     i							 0
     31  *                                     i+1         2
     32  *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
     33  *	     i      i            i                 i
     34  *
     35  *	To compute q    from q , one checks whether
     36  *		    i+1       i
     37  *
     38  *			      -(i+1) 2
     39  *			(q + 2      ) <= y.			(2)
     40  *     			  i
     41  *							      -(i+1)
     42  *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
     43  *		 	       i+1   i             i+1   i
     44  *
     45  *	With some algebric manipulation, it is not difficult to see
     46  *	that (2) is equivalent to
     47  *                             -(i+1)
     48  *			s  +  2       <= y			(3)
     49  *			 i                i
     50  *
     51  *	The advantage of (3) is that s  and y  can be computed by
     52  *				      i      i
     53  *	the following recurrence formula:
     54  *	    if (3) is false
     55  *
     56  *	    s     =  s  ,	y    = y   ;			(4)
     57  *	     i+1      i		 i+1    i
     58  *
     59  *	    otherwise,
     60  *                         -i                     -(i+1)
     61  *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
     62  *           i+1      i          i+1    i     i
     63  *
     64  *	One may easily use induction to prove (4) and (5).
     65  *	Note. Since the left hand side of (3) contain only i+2 bits,
     66  *	      it does not necessary to do a full (53-bit) comparison
     67  *	      in (3).
     68  *   3. Final rounding
     69  *	After generating the 53 bits result, we compute one more bit.
     70  *	Together with the remainder, we can decide whether the
     71  *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
     72  *	(it will never equal to 1/2ulp).
     73  *	The rounding mode can be detected by checking whether
     74  *	huge + tiny is equal to huge, and whether huge - tiny is
     75  *	equal to huge for some floating point number "huge" and "tiny".
     76  *
     77  * Special cases:
     78  *	sqrt(+-0) = +-0 	... exact
     79  *	sqrt(inf) = inf
     80  *	sqrt(-ve) = NaN		... with invalid signal
     81  *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
     82  *
     83  * Other methods : see the appended file at the end of the program below.
     84  *---------------
     85  */
     86 
     87 #include "math.h"
     88 #include "math_private.h"
     89 
     90 static	const double	one	= 1.0, tiny=1.0e-300;
     91 
     92 double
     93 __ieee754_sqrt(double x)
     94 {
     95 	double z;
     96 	int32_t sign = (int)0x80000000;
     97 	int32_t ix0,s0,q,m,t,i;
     98 	u_int32_t r,t1,s1,ix1,q1;
     99 
    100 	EXTRACT_WORDS(ix0,ix1,x);
    101 
    102     /* take care of Inf and NaN */
    103 	if((ix0&0x7ff00000)==0x7ff00000) {
    104 	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
    105 					   sqrt(-inf)=sNaN */
    106 	}
    107     /* take care of zero */
    108 	if(ix0<=0) {
    109 	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
    110 	    else if(ix0<0)
    111 		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
    112 	}
    113     /* normalize x */
    114 	m = (ix0>>20);
    115 	if(m==0) {				/* subnormal x */
    116 	    while(ix0==0) {
    117 		m -= 21;
    118 		ix0 |= (ix1>>11); ix1 <<= 21;
    119 	    }
    120 	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
    121 	    m -= i-1;
    122 	    ix0 |= (ix1>>(32-i));
    123 	    ix1 <<= i;
    124 	}
    125 	m -= 1023;	/* unbias exponent */
    126 	ix0 = (ix0&0x000fffff)|0x00100000;
    127 	if(m&1){	/* odd m, double x to make it even */
    128 	    ix0 += ix0 + ((ix1&sign)>>31);
    129 	    ix1 += ix1;
    130 	}
    131 	m >>= 1;	/* m = [m/2] */
    132 
    133     /* generate sqrt(x) bit by bit */
    134 	ix0 += ix0 + ((ix1&sign)>>31);
    135 	ix1 += ix1;
    136 	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
    137 	r = 0x00200000;		/* r = moving bit from right to left */
    138 
    139 	while(r!=0) {
    140 	    t = s0+r;
    141 	    if(t<=ix0) {
    142 		s0   = t+r;
    143 		ix0 -= t;
    144 		q   += r;
    145 	    }
    146 	    ix0 += ix0 + ((ix1&sign)>>31);
    147 	    ix1 += ix1;
    148 	    r>>=1;
    149 	}
    150 
    151 	r = sign;
    152 	while(r!=0) {
    153 	    t1 = s1+r;
    154 	    t  = s0;
    155 	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
    156 		s1  = t1+r;
    157 		if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
    158 		ix0 -= t;
    159 		if (ix1 < t1) ix0 -= 1;
    160 		ix1 -= t1;
    161 		q1  += r;
    162 	    }
    163 	    ix0 += ix0 + ((ix1&sign)>>31);
    164 	    ix1 += ix1;
    165 	    r>>=1;
    166 	}
    167 
    168     /* use floating add to find out rounding direction */
    169 	if((ix0|ix1)!=0) {
    170 	    z = one-tiny; /* trigger inexact flag */
    171 	    if (z>=one) {
    172 	        z = one+tiny;
    173 	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
    174 		else if (z>one) {
    175 		    if (q1==(u_int32_t)0xfffffffe) q+=1;
    176 		    q1+=2;
    177 		} else
    178 	            q1 += (q1&1);
    179 	    }
    180 	}
    181 	ix0 = (q>>1)+0x3fe00000;
    182 	ix1 =  q1>>1;
    183 	if ((q&1)==1) ix1 |= sign;
    184 	ix0 += (m <<20);
    185 	INSERT_WORDS(z,ix0,ix1);
    186 	return z;
    187 }
    188 
    189 /*
    190 Other methods  (use floating-point arithmetic)
    191 -------------
    192 (This is a copy of a drafted paper by Prof W. Kahan
    193 and K.C. Ng, written in May, 1986)
    194 
    195 	Two algorithms are given here to implement sqrt(x)
    196 	(IEEE double precision arithmetic) in software.
    197 	Both supply sqrt(x) correctly rounded. The first algorithm (in
    198 	Section A) uses newton iterations and involves four divisions.
    199 	The second one uses reciproot iterations to avoid division, but
    200 	requires more multiplications. Both algorithms need the ability
    201 	to chop results of arithmetic operations instead of round them,
    202 	and the INEXACT flag to indicate when an arithmetic operation
    203 	is executed exactly with no roundoff error, all part of the
    204 	standard (IEEE 754-1985). The ability to perform shift, add,
    205 	subtract and logical AND operations upon 32-bit words is needed
    206 	too, though not part of the standard.
    207 
    208 A.  sqrt(x) by Newton Iteration
    209 
    210    (1)	Initial approximation
    211 
    212 	Let x0 and x1 be the leading and the trailing 32-bit words of
    213 	a floating point number x (in IEEE double format) respectively
    214 
    215 	    1    11		     52				  ...widths
    216 	   ------------------------------------------------------
    217 	x: |s|	  e     |	      f				|
    218 	   ------------------------------------------------------
    219 	      msb    lsb  msb				      lsb ...order
    220 
    221 
    222 	     ------------------------  	     ------------------------
    223 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
    224 	     ------------------------  	     ------------------------
    225 
    226 	By performing shifts and subtracts on x0 and x1 (both regarded
    227 	as integers), we obtain an 8-bit approximation of sqrt(x) as
    228 	follows.
    229 
    230 		k  := (x0>>1) + 0x1ff80000;
    231 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
    232 	Here k is a 32-bit integer and T1[] is an integer array containing
    233 	correction terms. Now magically the floating value of y (y's
    234 	leading 32-bit word is y0, the value of its trailing word is 0)
    235 	approximates sqrt(x) to almost 8-bit.
    236 
    237 	Value of T1:
    238 	static int T1[32]= {
    239 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
    240 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
    241 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
    242 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
    243 
    244     (2)	Iterative refinement
    245 
    246 	Apply Heron's rule three times to y, we have y approximates
    247 	sqrt(x) to within 1 ulp (Unit in the Last Place):
    248 
    249 		y := (y+x/y)/2		... almost 17 sig. bits
    250 		y := (y+x/y)/2		... almost 35 sig. bits
    251 		y := y-(y-x/y)/2	... within 1 ulp
    252 
    253 
    254 	Remark 1.
    255 	    Another way to improve y to within 1 ulp is:
    256 
    257 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
    258 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
    259 
    260 				2
    261 			    (x-y )*y
    262 		y := y + 2* ----------	...within 1 ulp
    263 			       2
    264 			     3y  + x
    265 
    266 
    267 	This formula has one division fewer than the one above; however,
    268 	it requires more multiplications and additions. Also x must be
    269 	scaled in advance to avoid spurious overflow in evaluating the
    270 	expression 3y*y+x. Hence it is not recommended uless division
    271 	is slow. If division is very slow, then one should use the
    272 	reciproot algorithm given in section B.
    273 
    274     (3) Final adjustment
    275 
    276 	By twiddling y's last bit it is possible to force y to be
    277 	correctly rounded according to the prevailing rounding mode
    278 	as follows. Let r and i be copies of the rounding mode and
    279 	inexact flag before entering the square root program. Also we
    280 	use the expression y+-ulp for the next representable floating
    281 	numbers (up and down) of y. Note that y+-ulp = either fixed
    282 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
    283 	mode.
    284 
    285 		I := FALSE;	... reset INEXACT flag I
    286 		R := RZ;	... set rounding mode to round-toward-zero
    287 		z := x/y;	... chopped quotient, possibly inexact
    288 		If(not I) then {	... if the quotient is exact
    289 		    if(z=y) {
    290 		        I := i;	 ... restore inexact flag
    291 		        R := r;  ... restore rounded mode
    292 		        return sqrt(x):=y.
    293 		    } else {
    294 			z := z - ulp;	... special rounding
    295 		    }
    296 		}
    297 		i := TRUE;		... sqrt(x) is inexact
    298 		If (r=RN) then z=z+ulp	... rounded-to-nearest
    299 		If (r=RP) then {	... round-toward-+inf
    300 		    y = y+ulp; z=z+ulp;
    301 		}
    302 		y := y+z;		... chopped sum
    303 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
    304 	        I := i;	 		... restore inexact flag
    305 	        R := r;  		... restore rounded mode
    306 	        return sqrt(x):=y.
    307 
    308     (4)	Special cases
    309 
    310 	Square root of +inf, +-0, or NaN is itself;
    311 	Square root of a negative number is NaN with invalid signal.
    312 
    313 
    314 B.  sqrt(x) by Reciproot Iteration
    315 
    316    (1)	Initial approximation
    317 
    318 	Let x0 and x1 be the leading and the trailing 32-bit words of
    319 	a floating point number x (in IEEE double format) respectively
    320 	(see section A). By performing shifs and subtracts on x0 and y0,
    321 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
    322 
    323 	    k := 0x5fe80000 - (x0>>1);
    324 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
    325 
    326 	Here k is a 32-bit integer and T2[] is an integer array
    327 	containing correction terms. Now magically the floating
    328 	value of y (y's leading 32-bit word is y0, the value of
    329 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
    330 	to almost 7.8-bit.
    331 
    332 	Value of T2:
    333 	static int T2[64]= {
    334 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
    335 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
    336 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
    337 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
    338 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
    339 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
    340 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
    341 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
    342 
    343     (2)	Iterative refinement
    344 
    345 	Apply Reciproot iteration three times to y and multiply the
    346 	result by x to get an approximation z that matches sqrt(x)
    347 	to about 1 ulp. To be exact, we will have
    348 		-1ulp < sqrt(x)-z<1.0625ulp.
    349 
    350 	... set rounding mode to Round-to-nearest
    351 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
    352 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
    353 	... special arrangement for better accuracy
    354 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
    355 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
    356 
    357 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
    358 	(a) the term z*y in the final iteration is always less than 1;
    359 	(b) the error in the final result is biased upward so that
    360 		-1 ulp < sqrt(x) - z < 1.0625 ulp
    361 	    instead of |sqrt(x)-z|<1.03125ulp.
    362 
    363     (3)	Final adjustment
    364 
    365 	By twiddling y's last bit it is possible to force y to be
    366 	correctly rounded according to the prevailing rounding mode
    367 	as follows. Let r and i be copies of the rounding mode and
    368 	inexact flag before entering the square root program. Also we
    369 	use the expression y+-ulp for the next representable floating
    370 	numbers (up and down) of y. Note that y+-ulp = either fixed
    371 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
    372 	mode.
    373 
    374 	R := RZ;		... set rounding mode to round-toward-zero
    375 	switch(r) {
    376 	    case RN:		... round-to-nearest
    377 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
    378 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
    379 	       break;
    380 	    case RZ:case RM:	... round-to-zero or round-to--inf
    381 	       R:=RP;		... reset rounding mod to round-to-+inf
    382 	       if(x<z*z ... rounded up) z = z - ulp; else
    383 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
    384 	       break;
    385 	    case RP:		... round-to-+inf
    386 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
    387 	       if(x>z*z ...chopped) z = z+ulp;
    388 	       break;
    389 	}
    390 
    391 	Remark 3. The above comparisons can be done in fixed point. For
    392 	example, to compare x and w=z*z chopped, it suffices to compare
    393 	x1 and w1 (the trailing parts of x and w), regarding them as
    394 	two's complement integers.
    395 
    396 	...Is z an exact square root?
    397 	To determine whether z is an exact square root of x, let z1 be the
    398 	trailing part of z, and also let x0 and x1 be the leading and
    399 	trailing parts of x.
    400 
    401 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
    402 	    I := 1;		... Raise Inexact flag: z is not exact
    403 	else {
    404 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
    405 	    k := z1 >> 26;		... get z's 25-th and 26-th
    406 					    fraction bits
    407 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
    408 	}
    409 	R:= r		... restore rounded mode
    410 	return sqrt(x):=z.
    411 
    412 	If multiplication is cheaper then the foregoing red tape, the
    413 	Inexact flag can be evaluated by
    414 
    415 	    I := i;
    416 	    I := (z*z!=x) or I.
    417 
    418 	Note that z*z can overwrite I; this value must be sensed if it is
    419 	True.
    420 
    421 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
    422 	zero.
    423 
    424 		    --------------------
    425 		z1: |        f2        |
    426 		    --------------------
    427 		bit 31		   bit 0
    428 
    429 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
    430 	or even of logb(x) have the following relations:
    431 
    432 	-------------------------------------------------
    433 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
    434 	-------------------------------------------------
    435 	00			00		odd and even
    436 	01			01		even
    437 	10			10		odd
    438 	10			00		even
    439 	11			01		even
    440 	-------------------------------------------------
    441 
    442     (4)	Special cases (see (4) of Section A).
    443 
    444  */
    445