1 /* @(#)e_sqrt.c 5.1 93/09/24 */
2 /*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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 */
12
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
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_libm.h"
89 #include "math_private.h"
90
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
96
97 #ifdef __STDC__
98 double attribute_hidden
__ieee754_sqrt(double x)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;
110
111 EXTRACT_WORDS(ix0, ix1, x);
112
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] */
146
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 */
152
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 }
164
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 }
183
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 }
208
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)
214
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.
227
228 A. sqrt(x) by Newton Iteration
229
230 (1) Initial approximation
231
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
234
235 1 11 52 ...widths
236 ------------------------------------------------------
237 x: |s| e | f |
238 ------------------------------------------------------
239 msb lsb msb lsb ...order
240
241
242 ------------------------ ------------------------
243 x0: |s| e | f1 | x1: | f2 |
244 ------------------------ ------------------------
245
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.
249
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.
256
257 Value of T1:
258 static int T1[32]= {
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,};
263
264 (2) Iterative refinement
265
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):
268
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
272
273
274 Remark 1.
275 Another way to improve y to within 1 ulp is:
276
277 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
278 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
279
280 2
281 (x-y )*y
282 y := y + 2* ---------- ...within 1 ulp
283 2
284 3y + x
285
286
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.
293
294 (3) Final adjustment
295
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.
304
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.
327
328 (4) Special cases
329
330 Square root of +inf, +-0, or NaN is itself;
331 Square root of a negative number is NaN with invalid signal.
332
333
334 B. sqrt(x) by Reciproot Iteration
335
336 (1) Initial approximation
337
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.
342
343 k := 0x5fe80000 - (x0>>1);
344 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
345
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.
351
352 Value of T2:
353 static int T2[64]= {
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,
359 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
360 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
361 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
362
363 (2) Iterative refinement
364
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.
369
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)
376
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.
382
383 (3) Final adjustment
384
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.
393
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 }
410
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.
415
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.
420
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.
431
432 If multiplication is cheaper then the foregoing red tape, the
433 Inexact flag can be evaluated by
434
435 I := i;
436 I := (z*z!=x) or I.
437
438 Note that z*z can overwrite I; this value must be sensed if it is
439 True.
440
441 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
442 zero.
443
444 --------------------
445 z1: | f2 |
446 --------------------
447 bit 31 bit 0
448
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:
451
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 -------------------------------------------------
461
462 (4) Special cases (see (4) of Section A).
463
464 */
465