e_sqrt.c revision 141296
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#ifndef lint
15static char rcsid[] = "$FreeBSD: head/lib/msun/src/e_sqrt.c 141296 2005-02-04 18:26:06Z das $";
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.h"
89#include "math_private.h"
90
91static	const double	one	= 1.0, tiny=1.0e-300;
92
93double
94__ieee754_sqrt(double x)
95{
96	double z;
97	int32_t sign = (int)0x80000000;
98	int32_t ix0,s0,q,m,t,i;
99	u_int32_t r,t1,s1,ix1,q1;
100
101	EXTRACT_WORDS(ix0,ix1,x);
102
103    /* take care of Inf and NaN */
104	if((ix0&0x7ff00000)==0x7ff00000) {
105	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
106					   sqrt(-inf)=sNaN */
107	}
108    /* take care of zero */
109	if(ix0<=0) {
110	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
111	    else if(ix0<0)
112		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
113	}
114    /* normalize x */
115	m = (ix0>>20);
116	if(m==0) {				/* subnormal x */
117	    while(ix0==0) {
118		m -= 21;
119		ix0 |= (ix1>>11); ix1 <<= 21;
120	    }
121	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
122	    m -= i-1;
123	    ix0 |= (ix1>>(32-i));
124	    ix1 <<= i;
125	}
126	m -= 1023;	/* unbias exponent */
127	ix0 = (ix0&0x000fffff)|0x00100000;
128	if(m&1){	/* odd m, double x to make it even */
129	    ix0 += ix0 + ((ix1&sign)>>31);
130	    ix1 += ix1;
131	}
132	m >>= 1;	/* m = [m/2] */
133
134    /* generate sqrt(x) bit by bit */
135	ix0 += ix0 + ((ix1&sign)>>31);
136	ix1 += ix1;
137	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
138	r = 0x00200000;		/* r = moving bit from right to left */
139
140	while(r!=0) {
141	    t = s0+r;
142	    if(t<=ix0) {
143		s0   = t+r;
144		ix0 -= t;
145		q   += r;
146	    }
147	    ix0 += ix0 + ((ix1&sign)>>31);
148	    ix1 += ix1;
149	    r>>=1;
150	}
151
152	r = sign;
153	while(r!=0) {
154	    t1 = s1+r;
155	    t  = s0;
156	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
157		s1  = t1+r;
158		if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
159		ix0 -= t;
160		if (ix1 < t1) ix0 -= 1;
161		ix1 -= t1;
162		q1  += r;
163	    }
164	    ix0 += ix0 + ((ix1&sign)>>31);
165	    ix1 += ix1;
166	    r>>=1;
167	}
168
169    /* use floating add to find out rounding direction */
170	if((ix0|ix1)!=0) {
171	    z = one-tiny; /* trigger inexact flag */
172	    if (z>=one) {
173	        z = one+tiny;
174	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
175		else if (z>one) {
176		    if (q1==(u_int32_t)0xfffffffe) q+=1;
177		    q1+=2;
178		} else
179	            q1 += (q1&1);
180	    }
181	}
182	ix0 = (q>>1)+0x3fe00000;
183	ix1 =  q1>>1;
184	if ((q&1)==1) ix1 |= sign;
185	ix0 += (m <<20);
186	INSERT_WORDS(z,ix0,ix1);
187	return z;
188}
189
190/*
191Other methods  (use floating-point arithmetic)
192-------------
193(This is a copy of a drafted paper by Prof W. Kahan
194and K.C. Ng, written in May, 1986)
195
196	Two algorithms are given here to implement sqrt(x)
197	(IEEE double precision arithmetic) in software.
198	Both supply sqrt(x) correctly rounded. The first algorithm (in
199	Section A) uses newton iterations and involves four divisions.
200	The second one uses reciproot iterations to avoid division, but
201	requires more multiplications. Both algorithms need the ability
202	to chop results of arithmetic operations instead of round them,
203	and the INEXACT flag to indicate when an arithmetic operation
204	is executed exactly with no roundoff error, all part of the
205	standard (IEEE 754-1985). The ability to perform shift, add,
206	subtract and logical AND operations upon 32-bit words is needed
207	too, though not part of the standard.
208
209A.  sqrt(x) by Newton Iteration
210
211   (1)	Initial approximation
212
213	Let x0 and x1 be the leading and the trailing 32-bit words of
214	a floating point number x (in IEEE double format) respectively
215
216	    1    11		     52				  ...widths
217	   ------------------------------------------------------
218	x: |s|	  e     |	      f				|
219	   ------------------------------------------------------
220	      msb    lsb  msb				      lsb ...order
221
222
223	     ------------------------  	     ------------------------
224	x0:  |s|   e    |    f1     |	 x1: |          f2           |
225	     ------------------------  	     ------------------------
226
227	By performing shifts and subtracts on x0 and x1 (both regarded
228	as integers), we obtain an 8-bit approximation of sqrt(x) as
229	follows.
230
231		k  := (x0>>1) + 0x1ff80000;
232		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
233	Here k is a 32-bit integer and T1[] is an integer array containing
234	correction terms. Now magically the floating value of y (y's
235	leading 32-bit word is y0, the value of its trailing word is 0)
236	approximates sqrt(x) to almost 8-bit.
237
238	Value of T1:
239	static int T1[32]= {
240	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
241	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
242	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
243	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
244
245    (2)	Iterative refinement
246
247	Apply Heron's rule three times to y, we have y approximates
248	sqrt(x) to within 1 ulp (Unit in the Last Place):
249
250		y := (y+x/y)/2		... almost 17 sig. bits
251		y := (y+x/y)/2		... almost 35 sig. bits
252		y := y-(y-x/y)/2	... within 1 ulp
253
254
255	Remark 1.
256	    Another way to improve y to within 1 ulp is:
257
258		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
259		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
260
261				2
262			    (x-y )*y
263		y := y + 2* ----------	...within 1 ulp
264			       2
265			     3y  + x
266
267
268	This formula has one division fewer than the one above; however,
269	it requires more multiplications and additions. Also x must be
270	scaled in advance to avoid spurious overflow in evaluating the
271	expression 3y*y+x. Hence it is not recommended uless division
272	is slow. If division is very slow, then one should use the
273	reciproot algorithm given in section B.
274
275    (3) Final adjustment
276
277	By twiddling y's last bit it is possible to force y to be
278	correctly rounded according to the prevailing rounding mode
279	as follows. Let r and i be copies of the rounding mode and
280	inexact flag before entering the square root program. Also we
281	use the expression y+-ulp for the next representable floating
282	numbers (up and down) of y. Note that y+-ulp = either fixed
283	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
284	mode.
285
286		I := FALSE;	... reset INEXACT flag I
287		R := RZ;	... set rounding mode to round-toward-zero
288		z := x/y;	... chopped quotient, possibly inexact
289		If(not I) then {	... if the quotient is exact
290		    if(z=y) {
291		        I := i;	 ... restore inexact flag
292		        R := r;  ... restore rounded mode
293		        return sqrt(x):=y.
294		    } else {
295			z := z - ulp;	... special rounding
296		    }
297		}
298		i := TRUE;		... sqrt(x) is inexact
299		If (r=RN) then z=z+ulp	... rounded-to-nearest
300		If (r=RP) then {	... round-toward-+inf
301		    y = y+ulp; z=z+ulp;
302		}
303		y := y+z;		... chopped sum
304		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
305	        I := i;	 		... restore inexact flag
306	        R := r;  		... restore rounded mode
307	        return sqrt(x):=y.
308
309    (4)	Special cases
310
311	Square root of +inf, +-0, or NaN is itself;
312	Square root of a negative number is NaN with invalid signal.
313
314
315B.  sqrt(x) by Reciproot Iteration
316
317   (1)	Initial approximation
318
319	Let x0 and x1 be the leading and the trailing 32-bit words of
320	a floating point number x (in IEEE double format) respectively
321	(see section A). By performing shifs and subtracts on x0 and y0,
322	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
323
324	    k := 0x5fe80000 - (x0>>1);
325	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
326
327	Here k is a 32-bit integer and T2[] is an integer array
328	containing correction terms. Now magically the floating
329	value of y (y's leading 32-bit word is y0, the value of
330	its trailing word y1 is set to zero) approximates 1/sqrt(x)
331	to almost 7.8-bit.
332
333	Value of T2:
334	static int T2[64]= {
335	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
336	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
337	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
338	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
339	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
340	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
341	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
342	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
343
344    (2)	Iterative refinement
345
346	Apply Reciproot iteration three times to y and multiply the
347	result by x to get an approximation z that matches sqrt(x)
348	to about 1 ulp. To be exact, we will have
349		-1ulp < sqrt(x)-z<1.0625ulp.
350
351	... set rounding mode to Round-to-nearest
352	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
353	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
354	... special arrangement for better accuracy
355	   z := x*y			... 29 bits to sqrt(x), with z*y<1
356	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
357
358	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
359	(a) the term z*y in the final iteration is always less than 1;
360	(b) the error in the final result is biased upward so that
361		-1 ulp < sqrt(x) - z < 1.0625 ulp
362	    instead of |sqrt(x)-z|<1.03125ulp.
363
364    (3)	Final adjustment
365
366	By twiddling y's last bit it is possible to force y to be
367	correctly rounded according to the prevailing rounding mode
368	as follows. Let r and i be copies of the rounding mode and
369	inexact flag before entering the square root program. Also we
370	use the expression y+-ulp for the next representable floating
371	numbers (up and down) of y. Note that y+-ulp = either fixed
372	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
373	mode.
374
375	R := RZ;		... set rounding mode to round-toward-zero
376	switch(r) {
377	    case RN:		... round-to-nearest
378	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
379	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
380	       break;
381	    case RZ:case RM:	... round-to-zero or round-to--inf
382	       R:=RP;		... reset rounding mod to round-to-+inf
383	       if(x<z*z ... rounded up) z = z - ulp; else
384	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
385	       break;
386	    case RP:		... round-to-+inf
387	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
388	       if(x>z*z ...chopped) z = z+ulp;
389	       break;
390	}
391
392	Remark 3. The above comparisons can be done in fixed point. For
393	example, to compare x and w=z*z chopped, it suffices to compare
394	x1 and w1 (the trailing parts of x and w), regarding them as
395	two's complement integers.
396
397	...Is z an exact square root?
398	To determine whether z is an exact square root of x, let z1 be the
399	trailing part of z, and also let x0 and x1 be the leading and
400	trailing parts of x.
401
402	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
403	    I := 1;		... Raise Inexact flag: z is not exact
404	else {
405	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
406	    k := z1 >> 26;		... get z's 25-th and 26-th
407					    fraction bits
408	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
409	}
410	R:= r		... restore rounded mode
411	return sqrt(x):=z.
412
413	If multiplication is cheaper then the foregoing red tape, the
414	Inexact flag can be evaluated by
415
416	    I := i;
417	    I := (z*z!=x) or I.
418
419	Note that z*z can overwrite I; this value must be sensed if it is
420	True.
421
422	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
423	zero.
424
425		    --------------------
426		z1: |        f2        |
427		    --------------------
428		bit 31		   bit 0
429
430	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
431	or even of logb(x) have the following relations:
432
433	-------------------------------------------------
434	bit 27,26 of z1		bit 1,0 of x1	logb(x)
435	-------------------------------------------------
436	00			00		odd and even
437	01			01		even
438	10			10		odd
439	10			00		even
440	11			01		even
441	-------------------------------------------------
442
443    (4)	Special cases (see (4) of Section A).
444
445 */
446
447