1
2/* @(#)e_sqrt.c 1.3 95/01/18 */
3/*
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 *
7 * Developed at SunSoft, a Sun Microsystems, Inc. business.
8 * Permission to use, copy, modify, and distribute this
9 * software is freely granted, provided that this notice
10 * is preserved.
11 * ====================================================
12 */
13
14#include <sys/cdefs.h>
15__FBSDID("$FreeBSD$");
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 <float.h>
88
89#include "math.h"
90#include "math_private.h"
91
92static	const double	one	= 1.0, tiny=1.0e-300;
93
94double
95__ieee754_sqrt(double x)
96{
97	double z;
98	int32_t sign = (int)0x80000000;
99	int32_t ix0,s0,q,m,t,i;
100	u_int32_t r,t1,s1,ix1,q1;
101
102	EXTRACT_WORDS(ix0,ix1,x);
103
104    /* take care of Inf and NaN */
105	if((ix0&0x7ff00000)==0x7ff00000) {
106	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
107					   sqrt(-inf)=sNaN */
108	}
109    /* take care of zero */
110	if(ix0<=0) {
111	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
112	    else if(ix0<0)
113		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
114	}
115    /* normalize x */
116	m = (ix0>>20);
117	if(m==0) {				/* subnormal x */
118	    while(ix0==0) {
119		m -= 21;
120		ix0 |= (ix1>>11); ix1 <<= 21;
121	    }
122	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
123	    m -= i-1;
124	    ix0 |= (ix1>>(32-i));
125	    ix1 <<= i;
126	}
127	m -= 1023;	/* unbias exponent */
128	ix0 = (ix0&0x000fffff)|0x00100000;
129	if(m&1){	/* odd m, double x to make it even */
130	    ix0 += ix0 + ((ix1&sign)>>31);
131	    ix1 += ix1;
132	}
133	m >>= 1;	/* m = [m/2] */
134
135    /* generate sqrt(x) bit by bit */
136	ix0 += ix0 + ((ix1&sign)>>31);
137	ix1 += ix1;
138	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
139	r = 0x00200000;		/* r = moving bit from right to left */
140
141	while(r!=0) {
142	    t = s0+r;
143	    if(t<=ix0) {
144		s0   = t+r;
145		ix0 -= t;
146		q   += r;
147	    }
148	    ix0 += ix0 + ((ix1&sign)>>31);
149	    ix1 += ix1;
150	    r>>=1;
151	}
152
153	r = sign;
154	while(r!=0) {
155	    t1 = s1+r;
156	    t  = s0;
157	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
158		s1  = t1+r;
159		if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
160		ix0 -= t;
161		if (ix1 < t1) ix0 -= 1;
162		ix1 -= t1;
163		q1  += r;
164	    }
165	    ix0 += ix0 + ((ix1&sign)>>31);
166	    ix1 += ix1;
167	    r>>=1;
168	}
169
170    /* use floating add to find out rounding direction */
171	if((ix0|ix1)!=0) {
172	    z = one-tiny; /* trigger inexact flag */
173	    if (z>=one) {
174	        z = one+tiny;
175	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
176		else if (z>one) {
177		    if (q1==(u_int32_t)0xfffffffe) q+=1;
178		    q1+=2;
179		} else
180	            q1 += (q1&1);
181	    }
182	}
183	ix0 = (q>>1)+0x3fe00000;
184	ix1 =  q1>>1;
185	if ((q&1)==1) ix1 |= sign;
186	ix0 += (m <<20);
187	INSERT_WORDS(z,ix0,ix1);
188	return z;
189}
190
191#if (LDBL_MANT_DIG == 53)
192__weak_reference(sqrt, sqrtl);
193#endif
194
195/*
196Other methods  (use floating-point arithmetic)
197-------------
198(This is a copy of a drafted paper by Prof W. Kahan
199and 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
214A.  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
320B.  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