1 /* @(#)e_sqrt.c 1.3 95/01/18 */
2 /*
3  * ====================================================
4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5  *
6  * Developed at SunSoft, 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 /* __ieee754_sqrt(x)
14  * Return correctly rounded sqrt.
15  *           ------------------------------------------
16  *	     |  Use the hardware sqrt if you have one |
17  *           ------------------------------------------
18  * Method:
19  *   Bit by bit method using integer arithmetic. (Slow, but portable)
20  *   1. Normalization
21  *	Scale x to y in [1,4) with even powers of 2:
22  *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
23  *		sqrt(x) = 2^k * sqrt(y)
24  *   2. Bit by bit computation
25  *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
26  *	     i							 0
27  *                                     i+1         2
28  *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
29  *	     i      i            i                 i
30  *
31  *	To compute q    from q , one checks whether
32  *		    i+1       i
33  *
34  *			      -(i+1) 2
35  *			(q + 2      ) <= y.			(2)
36  *     			  i
37  *							      -(i+1)
38  *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
39  *		 	       i+1   i             i+1   i
40  *
41  *	With some algebric manipulation, it is not difficult to see
42  *	that (2) is equivalent to
43  *                             -(i+1)
44  *			s  +  2       <= y			(3)
45  *			 i                i
46  *
47  *	The advantage of (3) is that s  and y  can be computed by
48  *				      i      i
49  *	the following recurrence formula:
50  *	    if (3) is false
51  *
52  *	    s     =  s  ,	y    = y   ;			(4)
53  *	     i+1      i		 i+1    i
54  *
55  *	    otherwise,
56  *                         -i                     -(i+1)
57  *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
58  *           i+1      i          i+1    i     i
59  *
60  *	One may easily use induction to prove (4) and (5).
61  *	Note. Since the left hand side of (3) contain only i+2 bits,
62  *	      it does not necessary to do a full (53-bit) comparison
63  *	      in (3).
64  *   3. Final rounding
65  *	After generating the 53 bits result, we compute one more bit.
66  *	Together with the remainder, we can decide whether the
67  *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
68  *	(it will never equal to 1/2ulp).
69  *	The rounding mode can be detected by checking whether
70  *	huge + tiny is equal to huge, and whether huge - tiny is
71  *	equal to huge for some floating point number "huge" and "tiny".
72  *
73  * Special cases:
74  *	sqrt(+-0) = +-0 	... exact
75  *	sqrt(inf) = inf
76  *	sqrt(-ve) = NaN		... with invalid signal
77  *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
78  *
79  * Other methods : see the appended file at the end of the program below.
80  *---------------
81  */
82 
83 #include "fdlibm.h"
84 
85 #ifndef _DOUBLE_IS_32BITS
86 
87 #ifdef __STDC__
88 static	const double	one	= 1.0, tiny=1.0e-300;
89 #else
90 static	double	one	= 1.0, tiny=1.0e-300;
91 #endif
92 
93 #ifdef __STDC__
__ieee754_sqrt(double x)94 	double __ieee754_sqrt(double x)
95 #else
96 	double __ieee754_sqrt(x)
97 	double x;
98 #endif
99 {
100 	double z;
101 	int32_t 	sign = (int)0x80000000;
102 	uint32_t r,t1,s1,ix1,q1;
103 	int32_t ix0,s0,q,m,t,i;
104 
105 	EXTRACT_WORDS(ix0,ix1,x);
106 
107     /* take care of Inf and NaN */
108 	if((ix0&0x7ff00000)==0x7ff00000) {
109 	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
110 					   sqrt(-inf)=sNaN */
111 	}
112     /* take care of zero */
113 	if(ix0<=0) {
114 	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
115 	    else if(ix0<0)
116 		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
117 	}
118     /* normalize x */
119 	m = (ix0>>20);
120 	if(m==0) {				/* subnormal x */
121 	    while(ix0==0) {
122 		m -= 21;
123 		ix0 |= (ix1>>11); ix1 <<= 21;
124 	    }
125 	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
126 	    m -= i-1;
127 	    ix0 |= (ix1>>(32-i));
128 	    ix1 <<= i;
129 	}
130 	m -= 1023;	/* unbias exponent */
131 	ix0 = (ix0&0x000fffff)|0x00100000;
132 	if(m&1){	/* odd m, double x to make it even */
133 	    ix0 += ix0 + ((ix1&sign)>>31);
134 	    ix1 += ix1;
135 	}
136 	m >>= 1;	/* m = [m/2] */
137 
138     /* generate sqrt(x) bit by bit */
139 	ix0 += ix0 + ((ix1&sign)>>31);
140 	ix1 += ix1;
141 	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
142 	r = 0x00200000;		/* r = moving bit from right to left */
143 
144 	while(r!=0) {
145 	    t = s0+r;
146 	    if(t<=ix0) {
147 		s0   = t+r;
148 		ix0 -= t;
149 		q   += r;
150 	    }
151 	    ix0 += ix0 + ((ix1&sign)>>31);
152 	    ix1 += ix1;
153 	    r>>=1;
154 	}
155 
156 	r = sign;
157 	while(r!=0) {
158 	    t1 = s1+r;
159 	    t  = s0;
160 	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
161 		s1  = t1+r;
162 		if(((t1&sign)==(uint32_t)sign)&&(s1&sign)==0) s0 += 1;
163 		ix0 -= t;
164 		if (ix1 < t1) ix0 -= 1;
165 		ix1 -= t1;
166 		q1  += r;
167 	    }
168 	    ix0 += ix0 + ((ix1&sign)>>31);
169 	    ix1 += ix1;
170 	    r>>=1;
171 	}
172 
173     /* use floating add to find out rounding direction */
174 	if((ix0|ix1)!=0) {
175 	    z = one-tiny; /* trigger inexact flag */
176 	    if (z>=one) {
177 	        z = one+tiny;
178 	        if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;}
179 		else if (z>one) {
180 		    if (q1==(uint32_t)0xfffffffe) q+=1;
181 		    q1+=2;
182 		} else
183 	            q1 += (q1&1);
184 	    }
185 	}
186 	ix0 = (q>>1)+0x3fe00000;
187 	ix1 =  q1>>1;
188 	if ((q&1)==1) ix1 |= sign;
189 	ix0 += (m <<20);
190 	INSERT_WORDS(z,ix0,ix1);
191 	return z;
192 }
193 #endif /* defined(_DOUBLE_IS_32BITS) */
194 
195 /*
196 Other methods  (use floating-point arithmetic)
197 -------------
198 (This is a copy of a drafted paper by Prof W. Kahan
199 and K.C. Ng, written in May, 1986)
200 
201 	Two algorithms are given here to implement sqrt(x)
202 	(IEEE double precision arithmetic) in software.
203 	Both supply sqrt(x) correctly rounded. The first algorithm (in
204 	Section A) uses newton iterations and involves four divisions.
205 	The second one uses reciproot iterations to avoid division, but
206 	requires more multiplications. Both algorithms need the ability
207 	to chop results of arithmetic operations instead of round them,
208 	and the INEXACT flag to indicate when an arithmetic operation
209 	is executed exactly with no roundoff error, all part of the
210 	standard (IEEE 754-1985). The ability to perform shift, add,
211 	subtract and logical AND operations upon 32-bit words is needed
212 	too, though not part of the standard.
213 
214 A.  sqrt(x) by Newton Iteration
215 
216    (1)	Initial approximation
217 
218 	Let x0 and x1 be the leading and the trailing 32-bit words of
219 	a floating point number x (in IEEE double format) respectively
220 
221 	    1    11		     52				  ...widths
222 	   ------------------------------------------------------
223 	x: |s|	  e     |	      f				|
224 	   ------------------------------------------------------
225 	      msb    lsb  msb				      lsb ...order
226 
227 
228 	     ------------------------  	     ------------------------
229 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
230 	     ------------------------  	     ------------------------
231 
232 	By performing shifts and subtracts on x0 and x1 (both regarded
233 	as integers), we obtain an 8-bit approximation of sqrt(x) as
234 	follows.
235 
236 		k  := (x0>>1) + 0x1ff80000;
237 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
238 	Here k is a 32-bit integer and T1[] is an integer array containing
239 	correction terms. Now magically the floating value of y (y's
240 	leading 32-bit word is y0, the value of its trailing word is 0)
241 	approximates sqrt(x) to almost 8-bit.
242 
243 	Value of T1:
244 	static int T1[32]= {
245 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
246 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
247 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
248 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
249 
250     (2)	Iterative refinement
251 
252 	Apply Heron's rule three times to y, we have y approximates
253 	sqrt(x) to within 1 ulp (Unit in the Last Place):
254 
255 		y := (y+x/y)/2		... almost 17 sig. bits
256 		y := (y+x/y)/2		... almost 35 sig. bits
257 		y := y-(y-x/y)/2	... within 1 ulp
258 
259 
260 	Remark 1.
261 	    Another way to improve y to within 1 ulp is:
262 
263 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
264 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
265 
266 				2
267 			    (x-y )*y
268 		y := y + 2* ----------	...within 1 ulp
269 			       2
270 			     3y  + x
271 
272 
273 	This formula has one division fewer than the one above; however,
274 	it requires more multiplications and additions. Also x must be
275 	scaled in advance to avoid spurious overflow in evaluating the
276 	expression 3y*y+x. Hence it is not recommended uless division
277 	is slow. If division is very slow, then one should use the
278 	reciproot algorithm given in section B.
279 
280     (3) Final adjustment
281 
282 	By twiddling y's last bit it is possible to force y to be
283 	correctly rounded according to the prevailing rounding mode
284 	as follows. Let r and i be copies of the rounding mode and
285 	inexact flag before entering the square root program. Also we
286 	use the expression y+-ulp for the next representable floating
287 	numbers (up and down) of y. Note that y+-ulp = either fixed
288 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
289 	mode.
290 
291 		I := FALSE;	... reset INEXACT flag I
292 		R := RZ;	... set rounding mode to round-toward-zero
293 		z := x/y;	... chopped quotient, possibly inexact
294 		If(not I) then {	... if the quotient is exact
295 		    if(z=y) {
296 		        I := i;	 ... restore inexact flag
297 		        R := r;  ... restore rounded mode
298 		        return sqrt(x):=y.
299 		    } else {
300 			z := z - ulp;	... special rounding
301 		    }
302 		}
303 		i := TRUE;		... sqrt(x) is inexact
304 		If (r=RN) then z=z+ulp	... rounded-to-nearest
305 		If (r=RP) then {	... round-toward-+inf
306 		    y = y+ulp; z=z+ulp;
307 		}
308 		y := y+z;		... chopped sum
309 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
310 	        I := i;	 		... restore inexact flag
311 	        R := r;  		... restore rounded mode
312 	        return sqrt(x):=y.
313 
314     (4)	Special cases
315 
316 	Square root of +inf, +-0, or NaN is itself;
317 	Square root of a negative number is NaN with invalid signal.
318 
319 
320 B.  sqrt(x) by Reciproot Iteration
321 
322    (1)	Initial approximation
323 
324 	Let x0 and x1 be the leading and the trailing 32-bit words of
325 	a floating point number x (in IEEE double format) respectively
326 	(see section A). By performing shifs and subtracts on x0 and y0,
327 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
328 
329 	    k := 0x5fe80000 - (x0>>1);
330 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
331 
332 	Here k is a 32-bit integer and T2[] is an integer array
333 	containing correction terms. Now magically the floating
334 	value of y (y's leading 32-bit word is y0, the value of
335 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
336 	to almost 7.8-bit.
337 
338 	Value of T2:
339 	static int T2[64]= {
340 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
341 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
342 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
343 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
344 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
345 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
346 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
347 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
348 
349     (2)	Iterative refinement
350 
351 	Apply Reciproot iteration three times to y and multiply the
352 	result by x to get an approximation z that matches sqrt(x)
353 	to about 1 ulp. To be exact, we will have
354 		-1ulp < sqrt(x)-z<1.0625ulp.
355 
356 	... set rounding mode to Round-to-nearest
357 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
358 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
359 	... special arrangement for better accuracy
360 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
361 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
362 
363 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
364 	(a) the term z*y in the final iteration is always less than 1;
365 	(b) the error in the final result is biased upward so that
366 		-1 ulp < sqrt(x) - z < 1.0625 ulp
367 	    instead of |sqrt(x)-z|<1.03125ulp.
368 
369     (3)	Final adjustment
370 
371 	By twiddling y's last bit it is possible to force y to be
372 	correctly rounded according to the prevailing rounding mode
373 	as follows. Let r and i be copies of the rounding mode and
374 	inexact flag before entering the square root program. Also we
375 	use the expression y+-ulp for the next representable floating
376 	numbers (up and down) of y. Note that y+-ulp = either fixed
377 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
378 	mode.
379 
380 	R := RZ;		... set rounding mode to round-toward-zero
381 	switch(r) {
382 	    case RN:		... round-to-nearest
383 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
384 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
385 	       break;
386 	    case RZ:case RM:	... round-to-zero or round-to--inf
387 	       R:=RP;		... reset rounding mod to round-to-+inf
388 	       if(x<z*z ... rounded up) z = z - ulp; else
389 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
390 	       break;
391 	    case RP:		... round-to-+inf
392 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
393 	       if(x>z*z ...chopped) z = z+ulp;
394 	       break;
395 	}
396 
397 	Remark 3. The above comparisons can be done in fixed point. For
398 	example, to compare x and w=z*z chopped, it suffices to compare
399 	x1 and w1 (the trailing parts of x and w), regarding them as
400 	two's complement integers.
401 
402 	...Is z an exact square root?
403 	To determine whether z is an exact square root of x, let z1 be the
404 	trailing part of z, and also let x0 and x1 be the leading and
405 	trailing parts of x.
406 
407 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
408 	    I := 1;		... Raise Inexact flag: z is not exact
409 	else {
410 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
411 	    k := z1 >> 26;		... get z's 25-th and 26-th
412 					    fraction bits
413 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
414 	}
415 	R:= r		... restore rounded mode
416 	return sqrt(x):=z.
417 
418 	If multiplication is cheaper then the foregoing red tape, the
419 	Inexact flag can be evaluated by
420 
421 	    I := i;
422 	    I := (z*z!=x) or I.
423 
424 	Note that z*z can overwrite I; this value must be sensed if it is
425 	True.
426 
427 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
428 	zero.
429 
430 		    --------------------
431 		z1: |        f2        |
432 		    --------------------
433 		bit 31		   bit 0
434 
435 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
436 	or even of logb(x) have the following relations:
437 
438 	-------------------------------------------------
439 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
440 	-------------------------------------------------
441 	00			00		odd and even
442 	01			01		even
443 	10			10		odd
444 	10			00		even
445 	11			01		even
446 	-------------------------------------------------
447 
448     (4)	Special cases (see (4) of Section A).
449 
450  */
451 
452