xref: /openbsd-src/lib/libm/src/e_sqrt.c (revision 2b0358df1d88d06ef4139321dd05bd5e05d91eaf)
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 char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
15 #endif
16 
17 /* 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 <sys/cdefs.h>
88 #include <float.h>
89 #include <math.h>
90 
91 #include "math_private.h"
92 
93 static	const double	one	= 1.0, tiny=1.0e-300;
94 
95 double
96 sqrt(double x)
97 {
98 	double z;
99 	int32_t sign = (int)0x80000000;
100 	int32_t ix0,s0,q,m,t,i;
101 	u_int32_t r,t1,s1,ix1,q1;
102 
103 	EXTRACT_WORDS(ix0,ix1,x);
104 
105     /* take care of Inf and NaN */
106 	if((ix0&0x7ff00000)==0x7ff00000) {
107 	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
108 					   sqrt(-inf)=sNaN */
109 	}
110     /* take care of zero */
111 	if(ix0<=0) {
112 	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
113 	    else if(ix0<0)
114 		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
115 	}
116     /* normalize x */
117 	m = (ix0>>20);
118 	if(m==0) {				/* subnormal x */
119 	    while(ix0==0) {
120 		m -= 21;
121 		ix0 |= (ix1>>11); ix1 <<= 21;
122 	    }
123 	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
124 	    m -= i-1;
125 	    ix0 |= (ix1>>(32-i));
126 	    ix1 <<= i;
127 	}
128 	m -= 1023;	/* unbias exponent */
129 	ix0 = (ix0&0x000fffff)|0x00100000;
130 	if(m&1){	/* odd m, double x to make it even */
131 	    ix0 += ix0 + ((ix1&sign)>>31);
132 	    ix1 += ix1;
133 	}
134 	m >>= 1;	/* m = [m/2] */
135 
136     /* generate sqrt(x) bit by bit */
137 	ix0 += ix0 + ((ix1&sign)>>31);
138 	ix1 += ix1;
139 	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
140 	r = 0x00200000;		/* r = moving bit from right to left */
141 
142 	while(r!=0) {
143 	    t = s0+r;
144 	    if(t<=ix0) {
145 		s0   = t+r;
146 		ix0 -= t;
147 		q   += r;
148 	    }
149 	    ix0 += ix0 + ((ix1&sign)>>31);
150 	    ix1 += ix1;
151 	    r>>=1;
152 	}
153 
154 	r = sign;
155 	while(r!=0) {
156 	    t1 = s1+r;
157 	    t  = s0;
158 	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
159 		s1  = t1+r;
160 		if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
161 		ix0 -= t;
162 		if (ix1 < t1) ix0 -= 1;
163 		ix1 -= t1;
164 		q1  += r;
165 	    }
166 	    ix0 += ix0 + ((ix1&sign)>>31);
167 	    ix1 += ix1;
168 	    r>>=1;
169 	}
170 
171     /* use floating add to find out rounding direction */
172 	if((ix0|ix1)!=0) {
173 	    z = one-tiny; /* trigger inexact flag */
174 	    if (z>=one) {
175 	        z = one+tiny;
176 	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
177 		else if (z>one) {
178 		    if (q1==(u_int32_t)0xfffffffe) q+=1;
179 		    q1+=2;
180 		} else
181 	            q1 += (q1&1);
182 	    }
183 	}
184 	ix0 = (q>>1)+0x3fe00000;
185 	ix1 =  q1>>1;
186 	if ((q&1)==1) ix1 |= sign;
187 	ix0 += (m <<20);
188 	INSERT_WORDS(z,ix0,ix1);
189 	return z;
190 }
191 
192 /*
193 Other methods  (use floating-point arithmetic)
194 -------------
195 (This is a copy of a drafted paper by Prof W. Kahan
196 and K.C. Ng, written in May, 1986)
197 
198 	Two algorithms are given here to implement sqrt(x)
199 	(IEEE double precision arithmetic) in software.
200 	Both supply sqrt(x) correctly rounded. The first algorithm (in
201 	Section A) uses newton iterations and involves four divisions.
202 	The second one uses reciproot iterations to avoid division, but
203 	requires more multiplications. Both algorithms need the ability
204 	to chop results of arithmetic operations instead of round them,
205 	and the INEXACT flag to indicate when an arithmetic operation
206 	is executed exactly with no roundoff error, all part of the
207 	standard (IEEE 754-1985). The ability to perform shift, add,
208 	subtract and logical AND operations upon 32-bit words is needed
209 	too, though not part of the standard.
210 
211 A.  sqrt(x) by Newton Iteration
212 
213    (1)	Initial approximation
214 
215 	Let x0 and x1 be the leading and the trailing 32-bit words of
216 	a floating point number x (in IEEE double format) respectively
217 
218 	    1    11		     52				  ...widths
219 	   ------------------------------------------------------
220 	x: |s|	  e     |	      f				|
221 	   ------------------------------------------------------
222 	      msb    lsb  msb				      lsb ...order
223 
224 
225 	     ------------------------  	     ------------------------
226 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
227 	     ------------------------  	     ------------------------
228 
229 	By performing shifts and subtracts on x0 and x1 (both regarded
230 	as integers), we obtain an 8-bit approximation of sqrt(x) as
231 	follows.
232 
233 		k  := (x0>>1) + 0x1ff80000;
234 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
235 	Here k is a 32-bit integer and T1[] is an integer array containing
236 	correction terms. Now magically the floating value of y (y's
237 	leading 32-bit word is y0, the value of its trailing word is 0)
238 	approximates sqrt(x) to almost 8-bit.
239 
240 	Value of T1:
241 	static int T1[32]= {
242 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
243 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
244 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
245 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
246 
247     (2)	Iterative refinement
248 
249 	Apply Heron's rule three times to y, we have y approximates
250 	sqrt(x) to within 1 ulp (Unit in the Last Place):
251 
252 		y := (y+x/y)/2		... almost 17 sig. bits
253 		y := (y+x/y)/2		... almost 35 sig. bits
254 		y := y-(y-x/y)/2	... within 1 ulp
255 
256 
257 	Remark 1.
258 	    Another way to improve y to within 1 ulp is:
259 
260 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
261 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
262 
263 				2
264 			    (x-y )*y
265 		y := y + 2* ----------	...within 1 ulp
266 			       2
267 			     3y  + x
268 
269 
270 	This formula has one division fewer than the one above; however,
271 	it requires more multiplications and additions. Also x must be
272 	scaled in advance to avoid spurious overflow in evaluating the
273 	expression 3y*y+x. Hence it is not recommended uless division
274 	is slow. If division is very slow, then one should use the
275 	reciproot algorithm given in section B.
276 
277     (3) Final adjustment
278 
279 	By twiddling y's last bit it is possible to force y to be
280 	correctly rounded according to the prevailing rounding mode
281 	as follows. Let r and i be copies of the rounding mode and
282 	inexact flag before entering the square root program. Also we
283 	use the expression y+-ulp for the next representable floating
284 	numbers (up and down) of y. Note that y+-ulp = either fixed
285 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
286 	mode.
287 
288 		I := FALSE;	... reset INEXACT flag I
289 		R := RZ;	... set rounding mode to round-toward-zero
290 		z := x/y;	... chopped quotient, possibly inexact
291 		If(not I) then {	... if the quotient is exact
292 		    if(z=y) {
293 		        I := i;	 ... restore inexact flag
294 		        R := r;  ... restore rounded mode
295 		        return sqrt(x):=y.
296 		    } else {
297 			z := z - ulp;	... special rounding
298 		    }
299 		}
300 		i := TRUE;		... sqrt(x) is inexact
301 		If (r=RN) then z=z+ulp	... rounded-to-nearest
302 		If (r=RP) then {	... round-toward-+inf
303 		    y = y+ulp; z=z+ulp;
304 		}
305 		y := y+z;		... chopped sum
306 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
307 	        I := i;	 		... restore inexact flag
308 	        R := r;  		... restore rounded mode
309 	        return sqrt(x):=y.
310 
311     (4)	Special cases
312 
313 	Square root of +inf, +-0, or NaN is itself;
314 	Square root of a negative number is NaN with invalid signal.
315 
316 
317 B.  sqrt(x) by Reciproot Iteration
318 
319    (1)	Initial approximation
320 
321 	Let x0 and x1 be the leading and the trailing 32-bit words of
322 	a floating point number x (in IEEE double format) respectively
323 	(see section A). By performing shifs and subtracts on x0 and y0,
324 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
325 
326 	    k := 0x5fe80000 - (x0>>1);
327 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
328 
329 	Here k is a 32-bit integer and T2[] is an integer array
330 	containing correction terms. Now magically the floating
331 	value of y (y's leading 32-bit word is y0, the value of
332 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
333 	to almost 7.8-bit.
334 
335 	Value of T2:
336 	static int T2[64]= {
337 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
338 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
339 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
340 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
341 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
342 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
343 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
344 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
345 
346     (2)	Iterative refinement
347 
348 	Apply Reciproot iteration three times to y and multiply the
349 	result by x to get an approximation z that matches sqrt(x)
350 	to about 1 ulp. To be exact, we will have
351 		-1ulp < sqrt(x)-z<1.0625ulp.
352 
353 	... set rounding mode to Round-to-nearest
354 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
355 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
356 	... special arrangement for better accuracy
357 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
358 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
359 
360 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
361 	(a) the term z*y in the final iteration is always less than 1;
362 	(b) the error in the final result is biased upward so that
363 		-1 ulp < sqrt(x) - z < 1.0625 ulp
364 	    instead of |sqrt(x)-z|<1.03125ulp.
365 
366     (3)	Final adjustment
367 
368 	By twiddling y's last bit it is possible to force y to be
369 	correctly rounded according to the prevailing rounding mode
370 	as follows. Let r and i be copies of the rounding mode and
371 	inexact flag before entering the square root program. Also we
372 	use the expression y+-ulp for the next representable floating
373 	numbers (up and down) of y. Note that y+-ulp = either fixed
374 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
375 	mode.
376 
377 	R := RZ;		... set rounding mode to round-toward-zero
378 	switch(r) {
379 	    case RN:		... round-to-nearest
380 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
381 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
382 	       break;
383 	    case RZ:case RM:	... round-to-zero or round-to--inf
384 	       R:=RP;		... reset rounding mod to round-to-+inf
385 	       if(x<z*z ... rounded up) z = z - ulp; else
386 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
387 	       break;
388 	    case RP:		... round-to-+inf
389 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
390 	       if(x>z*z ...chopped) z = z+ulp;
391 	       break;
392 	}
393 
394 	Remark 3. The above comparisons can be done in fixed point. For
395 	example, to compare x and w=z*z chopped, it suffices to compare
396 	x1 and w1 (the trailing parts of x and w), regarding them as
397 	two's complement integers.
398 
399 	...Is z an exact square root?
400 	To determine whether z is an exact square root of x, let z1 be the
401 	trailing part of z, and also let x0 and x1 be the leading and
402 	trailing parts of x.
403 
404 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
405 	    I := 1;		... Raise Inexact flag: z is not exact
406 	else {
407 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
408 	    k := z1 >> 26;		... get z's 25-th and 26-th
409 					    fraction bits
410 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
411 	}
412 	R:= r		... restore rounded mode
413 	return sqrt(x):=z.
414 
415 	If multiplication is cheaper then the foregoing red tape, the
416 	Inexact flag can be evaluated by
417 
418 	    I := i;
419 	    I := (z*z!=x) or I.
420 
421 	Note that z*z can overwrite I; this value must be sensed if it is
422 	True.
423 
424 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
425 	zero.
426 
427 		    --------------------
428 		z1: |        f2        |
429 		    --------------------
430 		bit 31		   bit 0
431 
432 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
433 	or even of logb(x) have the following relations:
434 
435 	-------------------------------------------------
436 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
437 	-------------------------------------------------
438 	00			00		odd and even
439 	01			01		even
440 	10			10		odd
441 	10			00		even
442 	11			01		even
443 	-------------------------------------------------
444 
445     (4)	Special cases (see (4) of Section A).
446 
447  */
448 
449 #if LDBL_MANT_DIG == 53
450 #ifdef __weak_alias
451 __weak_alias(sqrtl, sqrt);
452 #endif /* __weak_alias */
453 #endif /* LDBL_MANT_DIG == 53 */
454