xref: /freebsd/lib/msun/src/e_sqrt.c (revision 3fb8f1272b50cb87cb624b321f7b81e76627c437)
1 
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 #include <float.h>
14 
15 #include "math.h"
16 #include "math_private.h"
17 
18 #ifdef USE_BUILTIN_SQRT
19 double
20 sqrt(double x)
21 {
22 	return (__builtin_sqrt(x));
23 }
24 #else
25 /* sqrt(x)
26  * Return correctly rounded sqrt.
27  *           ------------------------------------------
28  *	     |  Use the hardware sqrt if you have one |
29  *           ------------------------------------------
30  * Method:
31  *   Bit by bit method using integer arithmetic. (Slow, but portable)
32  *   1. Normalization
33  *	Scale x to y in [1,4) with even powers of 2:
34  *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
35  *		sqrt(x) = 2^k * sqrt(y)
36  *   2. Bit by bit computation
37  *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
38  *	     i							 0
39  *                                     i+1         2
40  *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
41  *	     i      i            i                 i
42  *
43  *	To compute q    from q , one checks whether
44  *		    i+1       i
45  *
46  *			      -(i+1) 2
47  *			(q + 2      ) <= y.			(2)
48  *     			  i
49  *							      -(i+1)
50  *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
51  *		 	       i+1   i             i+1   i
52  *
53  *	With some algebric manipulation, it is not difficult to see
54  *	that (2) is equivalent to
55  *                             -(i+1)
56  *			s  +  2       <= y			(3)
57  *			 i                i
58  *
59  *	The advantage of (3) is that s  and y  can be computed by
60  *				      i      i
61  *	the following recurrence formula:
62  *	    if (3) is false
63  *
64  *	    s     =  s  ,	y    = y   ;			(4)
65  *	     i+1      i		 i+1    i
66  *
67  *	    otherwise,
68  *                         -i                     -(i+1)
69  *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
70  *           i+1      i          i+1    i     i
71  *
72  *	One may easily use induction to prove (4) and (5).
73  *	Note. Since the left hand side of (3) contain only i+2 bits,
74  *	      it does not necessary to do a full (53-bit) comparison
75  *	      in (3).
76  *   3. Final rounding
77  *	After generating the 53 bits result, we compute one more bit.
78  *	Together with the remainder, we can decide whether the
79  *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
80  *	(it will never equal to 1/2ulp).
81  *	The rounding mode can be detected by checking whether
82  *	huge + tiny is equal to huge, and whether huge - tiny is
83  *	equal to huge for some floating point number "huge" and "tiny".
84  *
85  * Special cases:
86  *	sqrt(+-0) = +-0 	... exact
87  *	sqrt(inf) = inf
88  *	sqrt(-ve) = NaN		... with invalid signal
89  *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
90  *
91  * Other methods : see the appended file at the end of the program below.
92  *---------------
93  */
94 
95 static	const double	one	= 1.0, tiny=1.0e-300;
96 
97 double
98 sqrt(double x)
99 {
100 	double z;
101 	int32_t sign = (int)0x80000000;
102 	int32_t ix0,s0,q,m,t,i;
103 	u_int32_t r,t1,s1,ix1,q1;
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)==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==(u_int32_t)0xffffffff) { q1=0; q += 1;}
179 		else if (z>one) {
180 		    if (q1==(u_int32_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
194 
195 #if (LDBL_MANT_DIG == 53)
196 __weak_reference(sqrt, sqrtl);
197 #endif
198 
199 /*
200 Other methods  (use floating-point arithmetic)
201 -------------
202 (This is a copy of a drafted paper by Prof W. Kahan
203 and K.C. Ng, written in May, 1986)
204 
205 	Two algorithms are given here to implement sqrt(x)
206 	(IEEE double precision arithmetic) in software.
207 	Both supply sqrt(x) correctly rounded. The first algorithm (in
208 	Section A) uses newton iterations and involves four divisions.
209 	The second one uses reciproot iterations to avoid division, but
210 	requires more multiplications. Both algorithms need the ability
211 	to chop results of arithmetic operations instead of round them,
212 	and the INEXACT flag to indicate when an arithmetic operation
213 	is executed exactly with no roundoff error, all part of the
214 	standard (IEEE 754-1985). The ability to perform shift, add,
215 	subtract and logical AND operations upon 32-bit words is needed
216 	too, though not part of the standard.
217 
218 A.  sqrt(x) by Newton Iteration
219 
220    (1)	Initial approximation
221 
222 	Let x0 and x1 be the leading and the trailing 32-bit words of
223 	a floating point number x (in IEEE double format) respectively
224 
225 	    1    11		     52				  ...widths
226 	   ------------------------------------------------------
227 	x: |s|	  e     |	      f				|
228 	   ------------------------------------------------------
229 	      msb    lsb  msb				      lsb ...order
230 
231 
232 	     ------------------------  	     ------------------------
233 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
234 	     ------------------------  	     ------------------------
235 
236 	By performing shifts and subtracts on x0 and x1 (both regarded
237 	as integers), we obtain an 8-bit approximation of sqrt(x) as
238 	follows.
239 
240 		k  := (x0>>1) + 0x1ff80000;
241 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
242 	Here k is a 32-bit integer and T1[] is an integer array containing
243 	correction terms. Now magically the floating value of y (y's
244 	leading 32-bit word is y0, the value of its trailing word is 0)
245 	approximates sqrt(x) to almost 8-bit.
246 
247 	Value of T1:
248 	static int T1[32]= {
249 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
250 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
251 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
252 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
253 
254     (2)	Iterative refinement
255 
256 	Apply Heron's rule three times to y, we have y approximates
257 	sqrt(x) to within 1 ulp (Unit in the Last Place):
258 
259 		y := (y+x/y)/2		... almost 17 sig. bits
260 		y := (y+x/y)/2		... almost 35 sig. bits
261 		y := y-(y-x/y)/2	... within 1 ulp
262 
263 
264 	Remark 1.
265 	    Another way to improve y to within 1 ulp is:
266 
267 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
268 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
269 
270 				2
271 			    (x-y )*y
272 		y := y + 2* ----------	...within 1 ulp
273 			       2
274 			     3y  + x
275 
276 
277 	This formula has one division fewer than the one above; however,
278 	it requires more multiplications and additions. Also x must be
279 	scaled in advance to avoid spurious overflow in evaluating the
280 	expression 3y*y+x. Hence it is not recommended uless division
281 	is slow. If division is very slow, then one should use the
282 	reciproot algorithm given in section B.
283 
284     (3) Final adjustment
285 
286 	By twiddling y's last bit it is possible to force y to be
287 	correctly rounded according to the prevailing rounding mode
288 	as follows. Let r and i be copies of the rounding mode and
289 	inexact flag before entering the square root program. Also we
290 	use the expression y+-ulp for the next representable floating
291 	numbers (up and down) of y. Note that y+-ulp = either fixed
292 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
293 	mode.
294 
295 		I := FALSE;	... reset INEXACT flag I
296 		R := RZ;	... set rounding mode to round-toward-zero
297 		z := x/y;	... chopped quotient, possibly inexact
298 		If(not I) then {	... if the quotient is exact
299 		    if(z=y) {
300 		        I := i;	 ... restore inexact flag
301 		        R := r;  ... restore rounded mode
302 		        return sqrt(x):=y.
303 		    } else {
304 			z := z - ulp;	... special rounding
305 		    }
306 		}
307 		i := TRUE;		... sqrt(x) is inexact
308 		If (r=RN) then z=z+ulp	... rounded-to-nearest
309 		If (r=RP) then {	... round-toward-+inf
310 		    y = y+ulp; z=z+ulp;
311 		}
312 		y := y+z;		... chopped sum
313 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
314 	        I := i;	 		... restore inexact flag
315 	        R := r;  		... restore rounded mode
316 	        return sqrt(x):=y.
317 
318     (4)	Special cases
319 
320 	Square root of +inf, +-0, or NaN is itself;
321 	Square root of a negative number is NaN with invalid signal.
322 
323 
324 B.  sqrt(x) by Reciproot Iteration
325 
326    (1)	Initial approximation
327 
328 	Let x0 and x1 be the leading and the trailing 32-bit words of
329 	a floating point number x (in IEEE double format) respectively
330 	(see section A). By performing shifs and subtracts on x0 and y0,
331 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
332 
333 	    k := 0x5fe80000 - (x0>>1);
334 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
335 
336 	Here k is a 32-bit integer and T2[] is an integer array
337 	containing correction terms. Now magically the floating
338 	value of y (y's leading 32-bit word is y0, the value of
339 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
340 	to almost 7.8-bit.
341 
342 	Value of T2:
343 	static int T2[64]= {
344 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
345 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
346 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
347 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
348 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
349 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
350 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
351 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
352 
353     (2)	Iterative refinement
354 
355 	Apply Reciproot iteration three times to y and multiply the
356 	result by x to get an approximation z that matches sqrt(x)
357 	to about 1 ulp. To be exact, we will have
358 		-1ulp < sqrt(x)-z<1.0625ulp.
359 
360 	... set rounding mode to Round-to-nearest
361 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
362 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
363 	... special arrangement for better accuracy
364 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
365 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
366 
367 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
368 	(a) the term z*y in the final iteration is always less than 1;
369 	(b) the error in the final result is biased upward so that
370 		-1 ulp < sqrt(x) - z < 1.0625 ulp
371 	    instead of |sqrt(x)-z|<1.03125ulp.
372 
373     (3)	Final adjustment
374 
375 	By twiddling y's last bit it is possible to force y to be
376 	correctly rounded according to the prevailing rounding mode
377 	as follows. Let r and i be copies of the rounding mode and
378 	inexact flag before entering the square root program. Also we
379 	use the expression y+-ulp for the next representable floating
380 	numbers (up and down) of y. Note that y+-ulp = either fixed
381 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
382 	mode.
383 
384 	R := RZ;		... set rounding mode to round-toward-zero
385 	switch(r) {
386 	    case RN:		... round-to-nearest
387 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
388 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
389 	       break;
390 	    case RZ:case RM:	... round-to-zero or round-to--inf
391 	       R:=RP;		... reset rounding mod to round-to-+inf
392 	       if(x<z*z ... rounded up) z = z - ulp; else
393 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
394 	       break;
395 	    case RP:		... round-to-+inf
396 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
397 	       if(x>z*z ...chopped) z = z+ulp;
398 	       break;
399 	}
400 
401 	Remark 3. The above comparisons can be done in fixed point. For
402 	example, to compare x and w=z*z chopped, it suffices to compare
403 	x1 and w1 (the trailing parts of x and w), regarding them as
404 	two's complement integers.
405 
406 	...Is z an exact square root?
407 	To determine whether z is an exact square root of x, let z1 be the
408 	trailing part of z, and also let x0 and x1 be the leading and
409 	trailing parts of x.
410 
411 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
412 	    I := 1;		... Raise Inexact flag: z is not exact
413 	else {
414 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
415 	    k := z1 >> 26;		... get z's 25-th and 26-th
416 					    fraction bits
417 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
418 	}
419 	R:= r		... restore rounded mode
420 	return sqrt(x):=z.
421 
422 	If multiplication is cheaper then the foregoing red tape, the
423 	Inexact flag can be evaluated by
424 
425 	    I := i;
426 	    I := (z*z!=x) or I.
427 
428 	Note that z*z can overwrite I; this value must be sensed if it is
429 	True.
430 
431 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
432 	zero.
433 
434 		    --------------------
435 		z1: |        f2        |
436 		    --------------------
437 		bit 31		   bit 0
438 
439 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
440 	or even of logb(x) have the following relations:
441 
442 	-------------------------------------------------
443 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
444 	-------------------------------------------------
445 	00			00		odd and even
446 	01			01		even
447 	10			10		odd
448 	10			00		even
449 	11			01		even
450 	-------------------------------------------------
451 
452     (4)	Special cases (see (4) of Section A).
453 
454  */
455 
456