src/libm/e_sqrt.c
 author Ryan C. Gordon Tue, 26 May 2015 21:19:23 -0400 changeset 9649 d7762e30ba24 parent 6044 35448a5ea044 child 11683 48bcba563d9c permissions -rw-r--r--
Stack hint should look for 0, not -1, and not care about environment variables.
1 /* @(#)e_sqrt.c 5.1 93/09/24 */
2 /*
3  * ====================================================
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  */
13 #if defined(LIBM_SCCS) && !defined(lint)
14 static const char rcsid[] =
15     "\$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp \$";
16 #endif
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  */
88 #include "math_libm.h"
89 #include "math_private.h"
91 #ifdef __STDC__
92 static const double one = 1.0, tiny = 1.0e-300;
93 #else
94 static double one = 1.0, tiny = 1.0e-300;
95 #endif
97 #ifdef __STDC__
98 double attribute_hidden
99 __ieee754_sqrt(double x)
100 #else
101 double attribute_hidden
102 __ieee754_sqrt(x)
103      double x;
104 #endif
105 {
106     double z;
107     int32_t sign = (int) 0x80000000;
108     int32_t ix0, s0, q, m, t, i;
109     u_int32_t r, t1, s1, ix1, q1;
111     EXTRACT_WORDS(ix0, ix1, x);
113     /* take care of Inf and NaN */
114     if ((ix0 & 0x7ff00000) == 0x7ff00000) {
115         return x * x + x;       /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
116                                    sqrt(-inf)=sNaN */
117     }
118     /* take care of zero */
119     if (ix0 <= 0) {
120         if (((ix0 & (~sign)) | ix1) == 0)
121             return x;           /* sqrt(+-0) = +-0 */
122         else if (ix0 < 0)
123             return (x - x) / (x - x);   /* sqrt(-ve) = sNaN */
124     }
125     /* normalize x */
126     m = (ix0 >> 20);
127     if (m == 0) {               /* subnormal x */
128         while (ix0 == 0) {
129             m -= 21;
130             ix0 |= (ix1 >> 11);
131             ix1 <<= 21;
132         }
133         for (i = 0; (ix0 & 0x00100000) == 0; i++)
134             ix0 <<= 1;
135         m -= i - 1;
136         ix0 |= (ix1 >> (32 - i));
137         ix1 <<= i;
138     }
139     m -= 1023;                  /* unbias exponent */
140     ix0 = (ix0 & 0x000fffff) | 0x00100000;
141     if (m & 1) {                /* odd m, double x to make it even */
142         ix0 += ix0 + ((ix1 & sign) >> 31);
143         ix1 += ix1;
144     }
145     m >>= 1;                    /* m = [m/2] */
147     /* generate sqrt(x) bit by bit */
148     ix0 += ix0 + ((ix1 & sign) >> 31);
149     ix1 += ix1;
150     q = q1 = s0 = s1 = 0;       /* [q,q1] = sqrt(x) */
151     r = 0x00200000;             /* r = moving bit from right to left */
153     while (r != 0) {
154         t = s0 + r;
155         if (t <= ix0) {
156             s0 = t + r;
157             ix0 -= t;
158             q += r;
159         }
160         ix0 += ix0 + ((ix1 & sign) >> 31);
161         ix1 += ix1;
162         r >>= 1;
163     }
165     r = sign;
166     while (r != 0) {
167         t1 = s1 + r;
168         t = s0;
169         if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) {
170             s1 = t1 + r;
171             if (((t1 & sign) == sign) && (s1 & sign) == 0)
172                 s0 += 1;
173             ix0 -= t;
174             if (ix1 < t1)
175                 ix0 -= 1;
176             ix1 -= t1;
177             q1 += r;
178         }
179         ix0 += ix0 + ((ix1 & sign) >> 31);
180         ix1 += ix1;
181         r >>= 1;
182     }
184     /* use floating add to find out rounding direction */
185     if ((ix0 | ix1) != 0) {
186         z = one - tiny;         /* trigger inexact flag */
187         if (z >= one) {
188             z = one + tiny;
189             if (q1 == (u_int32_t) 0xffffffff) {
190                 q1 = 0;
191                 q += 1;
192             } else if (z > one) {
193                 if (q1 == (u_int32_t) 0xfffffffe)
194                     q += 1;
195                 q1 += 2;
196             } else
197                 q1 += (q1 & 1);
198         }
199     }
200     ix0 = (q >> 1) + 0x3fe00000;
201     ix1 = q1 >> 1;
202     if ((q & 1) == 1)
203         ix1 |= sign;
204     ix0 += (m << 20);
205     INSERT_WORDS(z, ix0, ix1);
206     return z;
207 }
209 /*
210 Other methods  (use floating-point arithmetic)
211 -------------
212 (This is a copy of a drafted paper by Prof W. Kahan
213 and K.C. Ng, written in May, 1986)
215 	Two algorithms are given here to implement sqrt(x)
216 	(IEEE double precision arithmetic) in software.
217 	Both supply sqrt(x) correctly rounded. The first algorithm (in
218 	Section A) uses newton iterations and involves four divisions.
219 	The second one uses reciproot iterations to avoid division, but
220 	requires more multiplications. Both algorithms need the ability
221 	to chop results of arithmetic operations instead of round them,
222 	and the INEXACT flag to indicate when an arithmetic operation
223 	is executed exactly with no roundoff error, all part of the
224 	standard (IEEE 754-1985). The ability to perform shift, add,
225 	subtract and logical AND operations upon 32-bit words is needed
226 	too, though not part of the standard.
228 A.  sqrt(x) by Newton Iteration
230    (1)	Initial approximation
232 	Let x0 and x1 be the leading and the trailing 32-bit words of
233 	a floating point number x (in IEEE double format) respectively
235 	    1    11		     52				  ...widths
236 	   ------------------------------------------------------
237 	x: |s|	  e     |	      f				|
238 	   ------------------------------------------------------
239 	      msb    lsb  msb				      lsb ...order
242 	     ------------------------  	     ------------------------
243 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
244 	     ------------------------  	     ------------------------
246 	By performing shifts and subtracts on x0 and x1 (both regarded
247 	as integers), we obtain an 8-bit approximation of sqrt(x) as
248 	follows.
250 		k  := (x0>>1) + 0x1ff80000;
251 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
252 	Here k is a 32-bit integer and T1[] is an integer array containing
253 	correction terms. Now magically the floating value of y (y's
254 	leading 32-bit word is y0, the value of its trailing word is 0)
255 	approximates sqrt(x) to almost 8-bit.
257 	Value of T1:
258 	static int T1= {
259 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
260 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
261 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
262 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
264     (2)	Iterative refinement
266 	Apply Heron's rule three times to y, we have y approximates
267 	sqrt(x) to within 1 ulp (Unit in the Last Place):
269 		y := (y+x/y)/2		... almost 17 sig. bits
270 		y := (y+x/y)/2		... almost 35 sig. bits
271 		y := y-(y-x/y)/2	... within 1 ulp
274 	Remark 1.
275 	    Another way to improve y to within 1 ulp is:
277 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
278 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
280 				2
281 			    (x-y )*y
282 		y := y + 2* ----------	...within 1 ulp
283 			       2
284 			     3y  + x
287 	This formula has one division fewer than the one above; however,
288 	it requires more multiplications and additions. Also x must be
289 	scaled in advance to avoid spurious overflow in evaluating the
290 	expression 3y*y+x. Hence it is not recommended uless division
291 	is slow. If division is very slow, then one should use the
292 	reciproot algorithm given in section B.
294     (3) Final adjustment
296 	By twiddling y's last bit it is possible to force y to be
297 	correctly rounded according to the prevailing rounding mode
298 	as follows. Let r and i be copies of the rounding mode and
299 	inexact flag before entering the square root program. Also we
300 	use the expression y+-ulp for the next representable floating
301 	numbers (up and down) of y. Note that y+-ulp = either fixed
302 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
303 	mode.
305 		I := FALSE;	... reset INEXACT flag I
306 		R := RZ;	... set rounding mode to round-toward-zero
307 		z := x/y;	... chopped quotient, possibly inexact
308 		If(not I) then {	... if the quotient is exact
309 		    if(z=y) {
310 		        I := i;	 ... restore inexact flag
311 		        R := r;  ... restore rounded mode
312 		        return sqrt(x):=y.
313 		    } else {
314 			z := z - ulp;	... special rounding
315 		    }
316 		}
317 		i := TRUE;		... sqrt(x) is inexact
318 		If (r=RN) then z=z+ulp	... rounded-to-nearest
319 		If (r=RP) then {	... round-toward-+inf
320 		    y = y+ulp; z=z+ulp;
321 		}
322 		y := y+z;		... chopped sum
323 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
324 	        I := i;	 		... restore inexact flag
325 	        R := r;  		... restore rounded mode
326 	        return sqrt(x):=y.
328     (4)	Special cases
330 	Square root of +inf, +-0, or NaN is itself;
331 	Square root of a negative number is NaN with invalid signal.
334 B.  sqrt(x) by Reciproot Iteration
336    (1)	Initial approximation
338 	Let x0 and x1 be the leading and the trailing 32-bit words of
339 	a floating point number x (in IEEE double format) respectively
340 	(see section A). By performing shifs and subtracts on x0 and y0,
341 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
343 	    k := 0x5fe80000 - (x0>>1);
344 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
346 	Here k is a 32-bit integer and T2[] is an integer array
347 	containing correction terms. Now magically the floating
348 	value of y (y's leading 32-bit word is y0, the value of
349 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
350 	to almost 7.8-bit.
352 	Value of T2:
353 	static int T2= {
354 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
355 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
356 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
357 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
358 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
360 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
361 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
363     (2)	Iterative refinement
365 	Apply Reciproot iteration three times to y and multiply the
366 	result by x to get an approximation z that matches sqrt(x)
367 	to about 1 ulp. To be exact, we will have
368 		-1ulp < sqrt(x)-z<1.0625ulp.
370 	... set rounding mode to Round-to-nearest
371 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
372 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
373 	... special arrangement for better accuracy
374 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
375 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
377 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
378 	(a) the term z*y in the final iteration is always less than 1;
379 	(b) the error in the final result is biased upward so that
380 		-1 ulp < sqrt(x) - z < 1.0625 ulp
381 	    instead of |sqrt(x)-z|<1.03125ulp.
383     (3)	Final adjustment
385 	By twiddling y's last bit it is possible to force y to be
386 	correctly rounded according to the prevailing rounding mode
387 	as follows. Let r and i be copies of the rounding mode and
388 	inexact flag before entering the square root program. Also we
389 	use the expression y+-ulp for the next representable floating
390 	numbers (up and down) of y. Note that y+-ulp = either fixed
391 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
392 	mode.
394 	R := RZ;		... set rounding mode to round-toward-zero
395 	switch(r) {
396 	    case RN:		... round-to-nearest
397 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
398 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
399 	       break;
400 	    case RZ:case RM:	... round-to-zero or round-to--inf
401 	       R:=RP;		... reset rounding mod to round-to-+inf
402 	       if(x<z*z ... rounded up) z = z - ulp; else
403 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
404 	       break;
405 	    case RP:		... round-to-+inf
406 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
407 	       if(x>z*z ...chopped) z = z+ulp;
408 	       break;
409 	}
411 	Remark 3. The above comparisons can be done in fixed point. For
412 	example, to compare x and w=z*z chopped, it suffices to compare
413 	x1 and w1 (the trailing parts of x and w), regarding them as
414 	two's complement integers.
416 	...Is z an exact square root?
417 	To determine whether z is an exact square root of x, let z1 be the
418 	trailing part of z, and also let x0 and x1 be the leading and
419 	trailing parts of x.
421 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
422 	    I := 1;		... Raise Inexact flag: z is not exact
423 	else {
424 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
425 	    k := z1 >> 26;		... get z's 25-th and 26-th
426 					    fraction bits
427 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
428 	}
429 	R:= r		... restore rounded mode
430 	return sqrt(x):=z.
432 	If multiplication is cheaper then the foregoing red tape, the
433 	Inexact flag can be evaluated by
435 	    I := i;
436 	    I := (z*z!=x) or I.
438 	Note that z*z can overwrite I; this value must be sensed if it is
439 	True.
441 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
442 	zero.
444 		    --------------------
445 		z1: |        f2        |
446 		    --------------------
447 		bit 31		   bit 0
449 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
450 	or even of logb(x) have the following relations:
452 	-------------------------------------------------
453 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
454 	-------------------------------------------------
455 	00			00		odd and even
456 	01			01		even
457 	10			10		odd
458 	10			00		even
459 	11			01		even
460 	-------------------------------------------------
462     (4)	Special cases (see (4) of Section A).
464  */