src/libm/e_sqrt.c
author Ryan C. Gordon <icculus@icculus.org>
Tue, 26 May 2015 21:19:23 -0400
changeset 9649 d7762e30ba24
parent 6044 35448a5ea044
child 11683 48bcba563d9c
permissions -rw-r--r--
Stack hint should look for 0, not -1, and not care about environment variables.
     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 #if defined(LIBM_SCCS) && !defined(lint)
    14 static const char rcsid[] =
    15     "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc 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_libm.h"
    89 #include "math_private.h"
    90 
    91 #ifdef __STDC__
    92 static const double one = 1.0, tiny = 1.0e-300;
    93 #else
    94 static double one = 1.0, tiny = 1.0e-300;
    95 #endif
    96 
    97 #ifdef __STDC__
    98 double attribute_hidden
    99 __ieee754_sqrt(double x)
   100 #else
   101 double attribute_hidden
   102 __ieee754_sqrt(x)
   103      double x;
   104 #endif
   105 {
   106     double z;
   107     int32_t sign = (int) 0x80000000;
   108     int32_t ix0, s0, q, m, t, i;
   109     u_int32_t r, t1, s1, ix1, q1;
   110 
   111     EXTRACT_WORDS(ix0, ix1, x);
   112 
   113     /* take care of Inf and NaN */
   114     if ((ix0 & 0x7ff00000) == 0x7ff00000) {
   115         return x * x + x;       /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
   116                                    sqrt(-inf)=sNaN */
   117     }
   118     /* take care of zero */
   119     if (ix0 <= 0) {
   120         if (((ix0 & (~sign)) | ix1) == 0)
   121             return x;           /* sqrt(+-0) = +-0 */
   122         else if (ix0 < 0)
   123             return (x - x) / (x - x);   /* sqrt(-ve) = sNaN */
   124     }
   125     /* normalize x */
   126     m = (ix0 >> 20);
   127     if (m == 0) {               /* subnormal x */
   128         while (ix0 == 0) {
   129             m -= 21;
   130             ix0 |= (ix1 >> 11);
   131             ix1 <<= 21;
   132         }
   133         for (i = 0; (ix0 & 0x00100000) == 0; i++)
   134             ix0 <<= 1;
   135         m -= i - 1;
   136         ix0 |= (ix1 >> (32 - i));
   137         ix1 <<= i;
   138     }
   139     m -= 1023;                  /* unbias exponent */
   140     ix0 = (ix0 & 0x000fffff) | 0x00100000;
   141     if (m & 1) {                /* odd m, double x to make it even */
   142         ix0 += ix0 + ((ix1 & sign) >> 31);
   143         ix1 += ix1;
   144     }
   145     m >>= 1;                    /* m = [m/2] */
   146 
   147     /* generate sqrt(x) bit by bit */
   148     ix0 += ix0 + ((ix1 & sign) >> 31);
   149     ix1 += ix1;
   150     q = q1 = s0 = s1 = 0;       /* [q,q1] = sqrt(x) */
   151     r = 0x00200000;             /* r = moving bit from right to left */
   152 
   153     while (r != 0) {
   154         t = s0 + r;
   155         if (t <= ix0) {
   156             s0 = t + r;
   157             ix0 -= t;
   158             q += r;
   159         }
   160         ix0 += ix0 + ((ix1 & sign) >> 31);
   161         ix1 += ix1;
   162         r >>= 1;
   163     }
   164 
   165     r = sign;
   166     while (r != 0) {
   167         t1 = s1 + r;
   168         t = s0;
   169         if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) {
   170             s1 = t1 + r;
   171             if (((t1 & sign) == sign) && (s1 & sign) == 0)
   172                 s0 += 1;
   173             ix0 -= t;
   174             if (ix1 < t1)
   175                 ix0 -= 1;
   176             ix1 -= t1;
   177             q1 += r;
   178         }
   179         ix0 += ix0 + ((ix1 & sign) >> 31);
   180         ix1 += ix1;
   181         r >>= 1;
   182     }
   183 
   184     /* use floating add to find out rounding direction */
   185     if ((ix0 | ix1) != 0) {
   186         z = one - tiny;         /* trigger inexact flag */
   187         if (z >= one) {
   188             z = one + tiny;
   189             if (q1 == (u_int32_t) 0xffffffff) {
   190                 q1 = 0;
   191                 q += 1;
   192             } else if (z > one) {
   193                 if (q1 == (u_int32_t) 0xfffffffe)
   194                     q += 1;
   195                 q1 += 2;
   196             } else
   197                 q1 += (q1 & 1);
   198         }
   199     }
   200     ix0 = (q >> 1) + 0x3fe00000;
   201     ix1 = q1 >> 1;
   202     if ((q & 1) == 1)
   203         ix1 |= sign;
   204     ix0 += (m << 20);
   205     INSERT_WORDS(z, ix0, ix1);
   206     return z;
   207 }
   208 
   209 /*
   210 Other methods  (use floating-point arithmetic)
   211 -------------
   212 (This is a copy of a drafted paper by Prof W. Kahan
   213 and K.C. Ng, written in May, 1986)
   214 
   215 	Two algorithms are given here to implement sqrt(x)
   216 	(IEEE double precision arithmetic) in software.
   217 	Both supply sqrt(x) correctly rounded. The first algorithm (in
   218 	Section A) uses newton iterations and involves four divisions.
   219 	The second one uses reciproot iterations to avoid division, but
   220 	requires more multiplications. Both algorithms need the ability
   221 	to chop results of arithmetic operations instead of round them,
   222 	and the INEXACT flag to indicate when an arithmetic operation
   223 	is executed exactly with no roundoff error, all part of the
   224 	standard (IEEE 754-1985). The ability to perform shift, add,
   225 	subtract and logical AND operations upon 32-bit words is needed
   226 	too, though not part of the standard.
   227 
   228 A.  sqrt(x) by Newton Iteration
   229 
   230    (1)	Initial approximation
   231 
   232 	Let x0 and x1 be the leading and the trailing 32-bit words of
   233 	a floating point number x (in IEEE double format) respectively
   234 
   235 	    1    11		     52				  ...widths
   236 	   ------------------------------------------------------
   237 	x: |s|	  e     |	      f				|
   238 	   ------------------------------------------------------
   239 	      msb    lsb  msb				      lsb ...order
   240 
   241 
   242 	     ------------------------  	     ------------------------
   243 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
   244 	     ------------------------  	     ------------------------
   245 
   246 	By performing shifts and subtracts on x0 and x1 (both regarded
   247 	as integers), we obtain an 8-bit approximation of sqrt(x) as
   248 	follows.
   249 
   250 		k  := (x0>>1) + 0x1ff80000;
   251 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
   252 	Here k is a 32-bit integer and T1[] is an integer array containing
   253 	correction terms. Now magically the floating value of y (y's
   254 	leading 32-bit word is y0, the value of its trailing word is 0)
   255 	approximates sqrt(x) to almost 8-bit.
   256 
   257 	Value of T1:
   258 	static int T1[32]= {
   259 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
   260 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
   261 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
   262 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
   263 
   264     (2)	Iterative refinement
   265 
   266 	Apply Heron's rule three times to y, we have y approximates
   267 	sqrt(x) to within 1 ulp (Unit in the Last Place):
   268 
   269 		y := (y+x/y)/2		... almost 17 sig. bits
   270 		y := (y+x/y)/2		... almost 35 sig. bits
   271 		y := y-(y-x/y)/2	... within 1 ulp
   272 
   273 
   274 	Remark 1.
   275 	    Another way to improve y to within 1 ulp is:
   276 
   277 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
   278 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
   279 
   280 				2
   281 			    (x-y )*y
   282 		y := y + 2* ----------	...within 1 ulp
   283 			       2
   284 			     3y  + x
   285 
   286 
   287 	This formula has one division fewer than the one above; however,
   288 	it requires more multiplications and additions. Also x must be
   289 	scaled in advance to avoid spurious overflow in evaluating the
   290 	expression 3y*y+x. Hence it is not recommended uless division
   291 	is slow. If division is very slow, then one should use the
   292 	reciproot algorithm given in section B.
   293 
   294     (3) Final adjustment
   295 
   296 	By twiddling y's last bit it is possible to force y to be
   297 	correctly rounded according to the prevailing rounding mode
   298 	as follows. Let r and i be copies of the rounding mode and
   299 	inexact flag before entering the square root program. Also we
   300 	use the expression y+-ulp for the next representable floating
   301 	numbers (up and down) of y. Note that y+-ulp = either fixed
   302 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
   303 	mode.
   304 
   305 		I := FALSE;	... reset INEXACT flag I
   306 		R := RZ;	... set rounding mode to round-toward-zero
   307 		z := x/y;	... chopped quotient, possibly inexact
   308 		If(not I) then {	... if the quotient is exact
   309 		    if(z=y) {
   310 		        I := i;	 ... restore inexact flag
   311 		        R := r;  ... restore rounded mode
   312 		        return sqrt(x):=y.
   313 		    } else {
   314 			z := z - ulp;	... special rounding
   315 		    }
   316 		}
   317 		i := TRUE;		... sqrt(x) is inexact
   318 		If (r=RN) then z=z+ulp	... rounded-to-nearest
   319 		If (r=RP) then {	... round-toward-+inf
   320 		    y = y+ulp; z=z+ulp;
   321 		}
   322 		y := y+z;		... chopped sum
   323 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
   324 	        I := i;	 		... restore inexact flag
   325 	        R := r;  		... restore rounded mode
   326 	        return sqrt(x):=y.
   327 
   328     (4)	Special cases
   329 
   330 	Square root of +inf, +-0, or NaN is itself;
   331 	Square root of a negative number is NaN with invalid signal.
   332 
   333 
   334 B.  sqrt(x) by Reciproot Iteration
   335 
   336    (1)	Initial approximation
   337 
   338 	Let x0 and x1 be the leading and the trailing 32-bit words of
   339 	a floating point number x (in IEEE double format) respectively
   340 	(see section A). By performing shifs and subtracts on x0 and y0,
   341 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
   342 
   343 	    k := 0x5fe80000 - (x0>>1);
   344 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
   345 
   346 	Here k is a 32-bit integer and T2[] is an integer array
   347 	containing correction terms. Now magically the floating
   348 	value of y (y's leading 32-bit word is y0, the value of
   349 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
   350 	to almost 7.8-bit.
   351 
   352 	Value of T2:
   353 	static int T2[64]= {
   354 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
   355 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
   356 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
   357 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
   358 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
   359 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
   360 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
   361 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
   362 
   363     (2)	Iterative refinement
   364 
   365 	Apply Reciproot iteration three times to y and multiply the
   366 	result by x to get an approximation z that matches sqrt(x)
   367 	to about 1 ulp. To be exact, we will have
   368 		-1ulp < sqrt(x)-z<1.0625ulp.
   369 
   370 	... set rounding mode to Round-to-nearest
   371 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
   372 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
   373 	... special arrangement for better accuracy
   374 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
   375 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
   376 
   377 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
   378 	(a) the term z*y in the final iteration is always less than 1;
   379 	(b) the error in the final result is biased upward so that
   380 		-1 ulp < sqrt(x) - z < 1.0625 ulp
   381 	    instead of |sqrt(x)-z|<1.03125ulp.
   382 
   383     (3)	Final adjustment
   384 
   385 	By twiddling y's last bit it is possible to force y to be
   386 	correctly rounded according to the prevailing rounding mode
   387 	as follows. Let r and i be copies of the rounding mode and
   388 	inexact flag before entering the square root program. Also we
   389 	use the expression y+-ulp for the next representable floating
   390 	numbers (up and down) of y. Note that y+-ulp = either fixed
   391 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
   392 	mode.
   393 
   394 	R := RZ;		... set rounding mode to round-toward-zero
   395 	switch(r) {
   396 	    case RN:		... round-to-nearest
   397 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
   398 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
   399 	       break;
   400 	    case RZ:case RM:	... round-to-zero or round-to--inf
   401 	       R:=RP;		... reset rounding mod to round-to-+inf
   402 	       if(x<z*z ... rounded up) z = z - ulp; else
   403 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
   404 	       break;
   405 	    case RP:		... round-to-+inf
   406 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
   407 	       if(x>z*z ...chopped) z = z+ulp;
   408 	       break;
   409 	}
   410 
   411 	Remark 3. The above comparisons can be done in fixed point. For
   412 	example, to compare x and w=z*z chopped, it suffices to compare
   413 	x1 and w1 (the trailing parts of x and w), regarding them as
   414 	two's complement integers.
   415 
   416 	...Is z an exact square root?
   417 	To determine whether z is an exact square root of x, let z1 be the
   418 	trailing part of z, and also let x0 and x1 be the leading and
   419 	trailing parts of x.
   420 
   421 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
   422 	    I := 1;		... Raise Inexact flag: z is not exact
   423 	else {
   424 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
   425 	    k := z1 >> 26;		... get z's 25-th and 26-th
   426 					    fraction bits
   427 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
   428 	}
   429 	R:= r		... restore rounded mode
   430 	return sqrt(x):=z.
   431 
   432 	If multiplication is cheaper then the foregoing red tape, the
   433 	Inexact flag can be evaluated by
   434 
   435 	    I := i;
   436 	    I := (z*z!=x) or I.
   437 
   438 	Note that z*z can overwrite I; this value must be sensed if it is
   439 	True.
   440 
   441 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
   442 	zero.
   443 
   444 		    --------------------
   445 		z1: |        f2        |
   446 		    --------------------
   447 		bit 31		   bit 0
   448 
   449 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
   450 	or even of logb(x) have the following relations:
   451 
   452 	-------------------------------------------------
   453 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
   454 	-------------------------------------------------
   455 	00			00		odd and even
   456 	01			01		even
   457 	10			10		odd
   458 	10			00		even
   459 	11			01		even
   460 	-------------------------------------------------
   461 
   462     (4)	Special cases (see (4) of Section A).
   463 
   464  */