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 if (i)
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)
159 s0 += 1;
160 ix0 -= t;
161 if (ix1 < t1)
162 ix0 -= 1;
163 ix1 -= t1;
164 q1 += r;
165 }
166 ix0 += ix0 + ((ix1 & sign) >> 31);
167 ix1 += ix1;
168 r >>= 1;
169 }
170
171 if ((ix0 | ix1) != 0) {
172 FE_DECL_ROUND(rnd);
173 if (__is_nearest(rnd) || __is_upward(rnd)) {
174 if (q1 == (__uint32_t)0xffffffff) {
175 q1 = 0;
176 q += 1;
177 } else if (__is_upward(rnd)) {
178 if (q1 == (__uint32_t)0xfffffffe)
179 q += 1;
180 q1 += 2;
181 } else
182 q1 += (q1 & 1);
183 }
184 }
185 ix0 = (q >> 1) + 0x3fe00000;
186 ix1 = q1 >> 1;
187 if ((q & 1) == 1)
188 ix1 |= sign;
189 ix0 += (m << 20);
190 INSERT_WORDS(z, ix0, ix1);
191 return z;
192 }
193
194 _MATH_ALIAS_d_d(sqrt)
195
196 #endif /* _NEED_FLOAT64 */
197
198 /*
199 Other methods (use floating-point arithmetic)
200 -------------
201 (This is a copy of a drafted paper by Prof W. Kahan
202 and K.C. Ng, written in May, 1986)
203
204 Two algorithms are given here to implement sqrt(x)
205 (IEEE double precision arithmetic) in software.
206 Both supply sqrt(x) correctly rounded. The first algorithm (in
207 Section A) uses newton iterations and involves four divisions.
208 The second one uses reciproot iterations to avoid division, but
209 requires more multiplications. Both algorithms need the ability
210 to chop results of arithmetic operations instead of round them,
211 and the INEXACT flag to indicate when an arithmetic operation
212 is executed exactly with no roundoff error, all part of the
213 standard (IEEE 754-1985). The ability to perform shift, add,
214 subtract and logical AND operations upon 32-bit words is needed
215 too, though not part of the standard.
216
217 A. sqrt(x) by Newton Iteration
218
219 (1) Initial approximation
220
221 Let x0 and x1 be the leading and the trailing 32-bit words of
222 a floating point number x (in IEEE double format) respectively
223
224 1 11 52 ...widths
225 ------------------------------------------------------
226 x: |s| e | f |
227 ------------------------------------------------------
228 msb lsb msb lsb ...order
229
230
231 ------------------------ ------------------------
232 x0: |s| e | f1 | x1: | f2 |
233 ------------------------ ------------------------
234
235 By performing shifts and subtracts on x0 and x1 (both regarded
236 as integers), we obtain an 8-bit approximation of sqrt(x) as
237 follows.
238
239 k := (x0>>1) + 0x1ff80000;
240 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
241 Here k is a 32-bit integer and T1[] is an integer array containing
242 correction terms. Now magically the floating value of y (y's
243 leading 32-bit word is y0, the value of its trailing word is 0)
244 approximates sqrt(x) to almost 8-bit.
245
246 Value of T1:
247 static int T1[32]= {
248 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
249 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
250 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
251 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
252
253 (2) Iterative refinement
254
255 Apply Heron's rule three times to y, we have y approximates
256 sqrt(x) to within 1 ulp (Unit in the Last Place):
257
258 y := (y+x/y)/2 ... almost 17 sig. bits
259 y := (y+x/y)/2 ... almost 35 sig. bits
260 y := y-(y-x/y)/2 ... within 1 ulp
261
262
263 Remark 1.
264 Another way to improve y to within 1 ulp is:
265
266 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
267 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
268
269 2
270 (x-y )*y
271 y := y + 2* ---------- ...within 1 ulp
272 2
273 3y + x
274
275
276 This formula has one division fewer than the one above; however,
277 it requires more multiplications and additions. Also x must be
278 scaled in advance to avoid spurious overflow in evaluating the
279 expression 3y*y+x. Hence it is not recommended uless division
280 is slow. If division is very slow, then one should use the
281 reciproot algorithm given in section B.
282
283 (3) Final adjustment
284
285 By twiddling y's last bit it is possible to force y to be
286 correctly rounded according to the prevailing rounding mode
287 as follows. Let r and i be copies of the rounding mode and
288 inexact flag before entering the square root program. Also we
289 use the expression y+-ulp for the next representable floating
290 numbers (up and down) of y. Note that y+-ulp = either fixed
291 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
292 mode.
293
294 I := FALSE; ... reset INEXACT flag I
295 R := RZ; ... set rounding mode to round-toward-zero
296 z := x/y; ... chopped quotient, possibly inexact
297 If(not I) then { ... if the quotient is exact
298 if(z=y) {
299 I := i; ... restore inexact flag
300 R := r; ... restore rounded mode
301 return sqrt(x):=y.
302 } else {
303 z := z - ulp; ... special rounding
304 }
305 }
306 i := TRUE; ... sqrt(x) is inexact
307 If (r=RN) then z=z+ulp ... rounded-to-nearest
308 If (r=RP) then { ... round-toward-+inf
309 y = y+ulp; z=z+ulp;
310 }
311 y := y+z; ... chopped sum
312 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
313 I := i; ... restore inexact flag
314 R := r; ... restore rounded mode
315 return sqrt(x):=y.
316
317 (4) Special cases
318
319 Square root of +inf, +-0, or NaN is itself;
320 Square root of a negative number is NaN with invalid signal.
321
322
323 B. sqrt(x) by Reciproot Iteration
324
325 (1) Initial approximation
326
327 Let x0 and x1 be the leading and the trailing 32-bit words of
328 a floating point number x (in IEEE double format) respectively
329 (see section A). By performing shifs and subtracts on x0 and y0,
330 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
331
332 k := 0x5fe80000 - (x0>>1);
333 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
334
335 Here k is a 32-bit integer and T2[] is an integer array
336 containing correction terms. Now magically the floating
337 value of y (y's leading 32-bit word is y0, the value of
338 its trailing word y1 is set to zero) approximates 1/sqrt(x)
339 to almost 7.8-bit.
340
341 Value of T2:
342 static int T2[64]= {
343 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
344 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
345 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
346 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
347 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
348 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
349 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
350 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
351
352 (2) Iterative refinement
353
354 Apply Reciproot iteration three times to y and multiply the
355 result by x to get an approximation z that matches sqrt(x)
356 to about 1 ulp. To be exact, we will have
357 -1ulp < sqrt(x)-z<1.0625ulp.
358
359 ... set rounding mode to Round-to-nearest
360 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
361 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
362 ... special arrangement for better accuracy
363 z := x*y ... 29 bits to sqrt(x), with z*y<1
364 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
365
366 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
367 (a) the term z*y in the final iteration is always less than 1;
368 (b) the error in the final result is biased upward so that
369 -1 ulp < sqrt(x) - z < 1.0625 ulp
370 instead of |sqrt(x)-z|<1.03125ulp.
371
372 (3) Final adjustment
373
374 By twiddling y's last bit it is possible to force y to be
375 correctly rounded according to the prevailing rounding mode
376 as follows. Let r and i be copies of the rounding mode and
377 inexact flag before entering the square root program. Also we
378 use the expression y+-ulp for the next representable floating
379 numbers (up and down) of y. Note that y+-ulp = either fixed
380 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
381 mode.
382
383 R := RZ; ... set rounding mode to round-toward-zero
384 switch(r) {
385 case RN: ... round-to-nearest
386 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
387 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
388 break;
389 case RZ:case RM: ... round-to-zero or round-to--inf
390 R:=RP; ... reset rounding mod to round-to-+inf
391 if(x<z*z ... rounded up) z = z - ulp; else
392 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
393 break;
394 case RP: ... round-to-+inf
395 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
396 if(x>z*z ...chopped) z = z+ulp;
397 break;
398 }
399
400 Remark 3. The above comparisons can be done in fixed point. For
401 example, to compare x and w=z*z chopped, it suffices to compare
402 x1 and w1 (the trailing parts of x and w), regarding them as
403 two's complement integers.
404
405 ...Is z an exact square root?
406 To determine whether z is an exact square root of x, let z1 be the
407 trailing part of z, and also let x0 and x1 be the leading and
408 trailing parts of x.
409
410 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
411 I := 1; ... Raise Inexact flag: z is not exact
412 else {
413 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
414 k := z1 >> 26; ... get z's 25-th and 26-th
415 fraction bits
416 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
417 }
418 R:= r ... restore rounded mode
419 return sqrt(x):=z.
420
421 If multiplication is cheaper then the foregoing red tape, the
422 Inexact flag can be evaluated by
423
424 I := i;
425 I := (z*z!=x) or I.
426
427 Note that z*z can overwrite I; this value must be sensed if it is
428 True.
429
430 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
431 zero.
432
433 --------------------
434 z1: | f2 |
435 --------------------
436 bit 31 bit 0
437
438 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
439 or even of logb(x) have the following relations:
440
441 -------------------------------------------------
442 bit 27,26 of z1 bit 1,0 of x1 logb(x)
443 -------------------------------------------------
444 00 00 odd and even
445 01 01 even
446 10 10 odd
447 10 00 even
448 11 01 even
449 -------------------------------------------------
450
451 (4) Special cases (see (4) of Section A).
452
453 */
454
