1
2 /* @(#)e_sqrt.c 5.1 93/09/24 */
3 /*
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 *
7 * Developed at SunPro, 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 /* sqrt(x)
15 * Return correctly rounded sqrt.
16 * ------------------------------------------
17 * | Use the hardware sqrt if you have one |
18 * ------------------------------------------
19 * Method:
20 * Bit by bit method using integer arithmetic. (Slow, but portable)
21 * 1. Normalization
22 * Scale x to y in [1,4) with even powers of 2:
23 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
24 * sqrt(x) = 2^k * sqrt(y)
25 * 2. Bit by bit computation
26 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
27 * i 0
28 * i+1 2
29 * s = 2*q , and y = 2 * ( y - q ). (1)
30 * i i i i
31 *
32 * To compute q from q , one checks whether
33 * i+1 i
34 *
35 * -(i+1) 2
36 * (q + 2 ) <= y. (2)
37 * i
38 * -(i+1)
39 * If (2) is false, then q = q ; otherwise q = q + 2 .
40 * i+1 i i+1 i
41 *
42 * With some algebric manipulation, it is not difficult to see
43 * that (2) is equivalent to
44 * -(i+1)
45 * s + 2 <= y (3)
46 * i i
47 *
48 * The advantage of (3) is that s and y can be computed by
49 * i i
50 * the following recurrence formula:
51 * if (3) is false
52 *
53 * s = s , y = y ; (4)
54 * i+1 i i+1 i
55 *
56 * otherwise,
57 * -i -(i+1)
58 * s = s + 2 , y = y - s - 2 (5)
59 * i+1 i i+1 i i
60 *
61 * One may easily use induction to prove (4) and (5).
62 * Note. Since the left hand side of (3) contain only i+2 bits,
63 * it does not necessary to do a full (53-bit) comparison
64 * in (3).
65 * 3. Final rounding
66 * After generating the 53 bits result, we compute one more bit.
67 * Together with the remainder, we can decide whether the
68 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
69 * (it will never equal to 1/2ulp).
70 * The rounding mode can be detected by checking whether
71 * huge + tiny is equal to huge, and whether huge - tiny is
72 * equal to huge for some floating point number "huge" and "tiny".
73 *
74 * Special cases:
75 * sqrt(+-0) = +-0 ... exact
76 * sqrt(inf) = inf
77 * sqrt(-ve) = NaN ... with invalid signal
78 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
79 *
80 * Other methods : see the appended file at the end of the program below.
81 *---------------
82 */
83
84 #include "fdlibm.h"
85
86 #ifdef _NEED_FLOAT64
87
88 __float64
sqrt64(__float64 x)89 sqrt64(__float64 x)
90 {
91 __float64 z;
92 __uint32_t sign = 0x80000000;
93 __uint32_t r, t1, s1, ix1, q1;
94 __int32_t ix0, s0, q, m, t, i;
95
96 EXTRACT_WORDS(ix0, ix1, x);
97
98 /* take care of Inf and NaN */
99 if ((ix0 & 0x7ff00000) == 0x7ff00000) {
100 if (ix0 < 0 && !isnan(x))
101 return __math_invalid(x); /* sqrt(-inf)=sNaN */
102 return x + x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf */
103 }
104 /* take care of zero */
105 if (ix0 <= 0) {
106 if (((ix0 & (~sign)) | ix1) == 0)
107 return x; /* sqrt(+-0) = +-0 */
108 else if (ix0 < 0)
109 return __math_invalid(x); /* sqrt(-ve) = sNaN */
110 }
111 /* normalize x */
112 m = (ix0 >> 20);
113 if (m == 0) { /* subnormal x */
114 while (ix0 == 0) {
115 m -= 21;
116 ix0 |= (ix1 >> 11);
117 ix1 <<= 21;
118 }
119 for (i = 0; (ix0 & 0x00100000) == 0; i++)
120 ix0 <<= 1;
121 m -= i - 1;
122 ix0 |= (ix1 >> (32 - i));
123 ix1 <<= i;
124 }
125 m -= 1023; /* unbias exponent */
126 ix0 = (ix0 & 0x000fffff) | 0x00100000;
127 if (m & 1) { /* odd m, double x to make it even */
128 ix0 += ix0 + ((ix1 & sign) >> 31);
129 ix1 += ix1;
130 }
131 m >>= 1; /* m = [m/2] */
132
133 /* generate sqrt(x) bit by bit */
134 ix0 += ix0 + ((ix1 & sign) >> 31);
135 ix1 += ix1;
136 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
137 r = 0x00200000; /* r = moving bit from right to left */
138
139 while (r != 0) {
140 t = s0 + r;
141 if (t <= ix0) {
142 s0 = t + r;
143 ix0 -= t;
144 q += r;
145 }
146 ix0 += ix0 + ((ix1 & sign) >> 31);
147 ix1 += ix1;
148 r >>= 1;
149 }
150
151 r = sign;
152 while (r != 0) {
153 t1 = s1 + r;
154 t = s0;
155 if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) {
156 s1 = t1 + r;
157 if (((t1 & sign) == sign) && (s1 & sign) == 0)
158 s0 += 1;
159 ix0 -= t;
160 if (ix1 < t1)
161 ix0 -= 1;
162 ix1 -= t1;
163 q1 += r;
164 }
165 ix0 += ix0 + ((ix1 & sign) >> 31);
166 ix1 += ix1;
167 r >>= 1;
168 }
169
170 if ((ix0 | ix1) != 0) {
171 FE_DECL_ROUND(rnd);
172 if (__is_nearest(rnd) || __is_upward(rnd)) {
173 if (q1 == (__uint32_t)0xffffffff) {
174 q1 = 0;
175 q += 1;
176 } else if (__is_upward(rnd)) {
177 if (q1 == (__uint32_t)0xfffffffe)
178 q += 1;
179 q1 += 2;
180 } else
181 q1 += (q1 & 1);
182 }
183 }
184 ix0 = (q >> 1) + 0x3fe00000;
185 ix1 = q1 >> 1;
186 if ((q & 1) == 1)
187 ix1 |= sign;
188 ix0 += (m << 20);
189 INSERT_WORDS(z, ix0, ix1);
190 return z;
191 }
192
193 _MATH_ALIAS_d_d(sqrt)
194
195 #endif /* _NEED_FLOAT64 */
196
197 /*
198 Other methods (use floating-point arithmetic)
199 -------------
200 (This is a copy of a drafted paper by Prof W. Kahan
201 and K.C. Ng, written in May, 1986)
202
203 Two algorithms are given here to implement sqrt(x)
204 (IEEE double precision arithmetic) in software.
205 Both supply sqrt(x) correctly rounded. The first algorithm (in
206 Section A) uses newton iterations and involves four divisions.
207 The second one uses reciproot iterations to avoid division, but
208 requires more multiplications. Both algorithms need the ability
209 to chop results of arithmetic operations instead of round them,
210 and the INEXACT flag to indicate when an arithmetic operation
211 is executed exactly with no roundoff error, all part of the
212 standard (IEEE 754-1985). The ability to perform shift, add,
213 subtract and logical AND operations upon 32-bit words is needed
214 too, though not part of the standard.
215
216 A. sqrt(x) by Newton Iteration
217
218 (1) Initial approximation
219
220 Let x0 and x1 be the leading and the trailing 32-bit words of
221 a floating point number x (in IEEE double format) respectively
222
223 1 11 52 ...widths
224 ------------------------------------------------------
225 x: |s| e | f |
226 ------------------------------------------------------
227 msb lsb msb lsb ...order
228
229
230 ------------------------ ------------------------
231 x0: |s| e | f1 | x1: | f2 |
232 ------------------------ ------------------------
233
234 By performing shifts and subtracts on x0 and x1 (both regarded
235 as integers), we obtain an 8-bit approximation of sqrt(x) as
236 follows.
237
238 k := (x0>>1) + 0x1ff80000;
239 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
240 Here k is a 32-bit integer and T1[] is an integer array containing
241 correction terms. Now magically the floating value of y (y's
242 leading 32-bit word is y0, the value of its trailing word is 0)
243 approximates sqrt(x) to almost 8-bit.
244
245 Value of T1:
246 static int T1[32]= {
247 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
248 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
249 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
250 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
251
252 (2) Iterative refinement
253
254 Apply Heron's rule three times to y, we have y approximates
255 sqrt(x) to within 1 ulp (Unit in the Last Place):
256
257 y := (y+x/y)/2 ... almost 17 sig. bits
258 y := (y+x/y)/2 ... almost 35 sig. bits
259 y := y-(y-x/y)/2 ... within 1 ulp
260
261
262 Remark 1.
263 Another way to improve y to within 1 ulp is:
264
265 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
266 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
267
268 2
269 (x-y )*y
270 y := y + 2* ---------- ...within 1 ulp
271 2
272 3y + x
273
274
275 This formula has one division fewer than the one above; however,
276 it requires more multiplications and additions. Also x must be
277 scaled in advance to avoid spurious overflow in evaluating the
278 expression 3y*y+x. Hence it is not recommended uless division
279 is slow. If division is very slow, then one should use the
280 reciproot algorithm given in section B.
281
282 (3) Final adjustment
283
284 By twiddling y's last bit it is possible to force y to be
285 correctly rounded according to the prevailing rounding mode
286 as follows. Let r and i be copies of the rounding mode and
287 inexact flag before entering the square root program. Also we
288 use the expression y+-ulp for the next representable floating
289 numbers (up and down) of y. Note that y+-ulp = either fixed
290 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
291 mode.
292
293 I := FALSE; ... reset INEXACT flag I
294 R := RZ; ... set rounding mode to round-toward-zero
295 z := x/y; ... chopped quotient, possibly inexact
296 If(not I) then { ... if the quotient is exact
297 if(z=y) {
298 I := i; ... restore inexact flag
299 R := r; ... restore rounded mode
300 return sqrt(x):=y.
301 } else {
302 z := z - ulp; ... special rounding
303 }
304 }
305 i := TRUE; ... sqrt(x) is inexact
306 If (r=RN) then z=z+ulp ... rounded-to-nearest
307 If (r=RP) then { ... round-toward-+inf
308 y = y+ulp; z=z+ulp;
309 }
310 y := y+z; ... chopped sum
311 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
312 I := i; ... restore inexact flag
313 R := r; ... restore rounded mode
314 return sqrt(x):=y.
315
316 (4) Special cases
317
318 Square root of +inf, +-0, or NaN is itself;
319 Square root of a negative number is NaN with invalid signal.
320
321
322 B. sqrt(x) by Reciproot Iteration
323
324 (1) Initial approximation
325
326 Let x0 and x1 be the leading and the trailing 32-bit words of
327 a floating point number x (in IEEE double format) respectively
328 (see section A). By performing shifs and subtracts on x0 and y0,
329 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
330
331 k := 0x5fe80000 - (x0>>1);
332 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
333
334 Here k is a 32-bit integer and T2[] is an integer array
335 containing correction terms. Now magically the floating
336 value of y (y's leading 32-bit word is y0, the value of
337 its trailing word y1 is set to zero) approximates 1/sqrt(x)
338 to almost 7.8-bit.
339
340 Value of T2:
341 static int T2[64]= {
342 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
343 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
344 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
345 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
346 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
347 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
348 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
349 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
350
351 (2) Iterative refinement
352
353 Apply Reciproot iteration three times to y and multiply the
354 result by x to get an approximation z that matches sqrt(x)
355 to about 1 ulp. To be exact, we will have
356 -1ulp < sqrt(x)-z<1.0625ulp.
357
358 ... set rounding mode to Round-to-nearest
359 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
360 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
361 ... special arrangement for better accuracy
362 z := x*y ... 29 bits to sqrt(x), with z*y<1
363 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
364
365 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
366 (a) the term z*y in the final iteration is always less than 1;
367 (b) the error in the final result is biased upward so that
368 -1 ulp < sqrt(x) - z < 1.0625 ulp
369 instead of |sqrt(x)-z|<1.03125ulp.
370
371 (3) Final adjustment
372
373 By twiddling y's last bit it is possible to force y to be
374 correctly rounded according to the prevailing rounding mode
375 as follows. Let r and i be copies of the rounding mode and
376 inexact flag before entering the square root program. Also we
377 use the expression y+-ulp for the next representable floating
378 numbers (up and down) of y. Note that y+-ulp = either fixed
379 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
380 mode.
381
382 R := RZ; ... set rounding mode to round-toward-zero
383 switch(r) {
384 case RN: ... round-to-nearest
385 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
386 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
387 break;
388 case RZ:case RM: ... round-to-zero or round-to--inf
389 R:=RP; ... reset rounding mod to round-to-+inf
390 if(x<z*z ... rounded up) z = z - ulp; else
391 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
392 break;
393 case RP: ... round-to-+inf
394 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
395 if(x>z*z ...chopped) z = z+ulp;
396 break;
397 }
398
399 Remark 3. The above comparisons can be done in fixed point. For
400 example, to compare x and w=z*z chopped, it suffices to compare
401 x1 and w1 (the trailing parts of x and w), regarding them as
402 two's complement integers.
403
404 ...Is z an exact square root?
405 To determine whether z is an exact square root of x, let z1 be the
406 trailing part of z, and also let x0 and x1 be the leading and
407 trailing parts of x.
408
409 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
410 I := 1; ... Raise Inexact flag: z is not exact
411 else {
412 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
413 k := z1 >> 26; ... get z's 25-th and 26-th
414 fraction bits
415 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
416 }
417 R:= r ... restore rounded mode
418 return sqrt(x):=z.
419
420 If multiplication is cheaper then the foregoing red tape, the
421 Inexact flag can be evaluated by
422
423 I := i;
424 I := (z*z!=x) or I.
425
426 Note that z*z can overwrite I; this value must be sensed if it is
427 True.
428
429 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
430 zero.
431
432 --------------------
433 z1: | f2 |
434 --------------------
435 bit 31 bit 0
436
437 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
438 or even of logb(x) have the following relations:
439
440 -------------------------------------------------
441 bit 27,26 of z1 bit 1,0 of x1 logb(x)
442 -------------------------------------------------
443 00 00 odd and even
444 01 01 even
445 10 10 odd
446 10 00 even
447 11 01 even
448 -------------------------------------------------
449
450 (4) Special cases (see (4) of Section A).
451
452 */
453
