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 #include <sys/cdefs.h>
15 __FBSDID("$FreeBSD$");
16
17 /* __ieee754_sqrt(x)
18 * Return correctly rounded sqrt.
19 * ------------------------------------------
20 * | Use the hardware sqrt if you have one |
21 * ------------------------------------------
22 * Method:
23 * Bit by bit method using integer arithmetic. (Slow, but portable)
24 * 1. Normalization
25 * Scale x to y in [1,4) with even powers of 2:
26 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
27 * sqrt(x) = 2^k * sqrt(y)
28 * 2. Bit by bit computation
29 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
30 * i 0
31 * i+1 2
32 * s = 2*q , and y = 2 * ( y - q ). (1)
33 * i i i i
34 *
35 * To compute q from q , one checks whether
36 * i+1 i
37 *
38 * -(i+1) 2
39 * (q + 2 ) <= y. (2)
40 * i
41 * -(i+1)
42 * If (2) is false, then q = q ; otherwise q = q + 2 .
43 * i+1 i i+1 i
44 *
45 * With some algebric manipulation, it is not difficult to see
46 * that (2) is equivalent to
47 * -(i+1)
48 * s + 2 <= y (3)
49 * i i
50 *
51 * The advantage of (3) is that s and y can be computed by
52 * i i
53 * the following recurrence formula:
54 * if (3) is false
55 *
56 * s = s , y = y ; (4)
57 * i+1 i i+1 i
58 *
59 * otherwise,
60 * -i -(i+1)
61 * s = s + 2 , y = y - s - 2 (5)
62 * i+1 i i+1 i i
63 *
64 * One may easily use induction to prove (4) and (5).
65 * Note. Since the left hand side of (3) contain only i+2 bits,
66 * it does not necessary to do a full (53-bit) comparison
67 * in (3).
68 * 3. Final rounding
69 * After generating the 53 bits result, we compute one more bit.
70 * Together with the remainder, we can decide whether the
71 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
72 * (it will never equal to 1/2ulp).
73 * The rounding mode can be detected by checking whether
74 * huge + tiny is equal to huge, and whether huge - tiny is
75 * equal to huge for some floating point number "huge" and "tiny".
76 *
77 * Special cases:
78 * sqrt(+-0) = +-0 ... exact
79 * sqrt(inf) = inf
80 * sqrt(-ve) = NaN ... with invalid signal
81 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
82 *
83 * Other methods : see the appended file at the end of the program below.
84 *---------------
85 */
86
87 #include <float.h>
88
89 #include "math.h"
90 #include "math_private.h"
91
92 static const double one = 1.0, tiny=1.0e-300;
93
94 double
__ieee754_sqrt(double x)95 __ieee754_sqrt(double x)
96 {
97 double z;
98 int32_t sign = (int)0x80000000;
99 int32_t ix0,s0,q,m,t,i;
100 u_int32_t r,t1,s1,ix1,q1;
101
102 EXTRACT_WORDS(ix0,ix1,x);
103
104 /* take care of Inf and NaN */
105 if((ix0&0x7ff00000)==0x7ff00000) {
106 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
107 sqrt(-inf)=sNaN */
108 }
109 /* take care of zero */
110 if(ix0<=0) {
111 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
112 else if(ix0<0)
113 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
114 }
115 /* normalize x */
116 m = (ix0>>20);
117 if(m==0) { /* subnormal x */
118 while(ix0==0) {
119 m -= 21;
120 ix0 |= (ix1>>11); ix1 <<= 21;
121 }
122 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
123 m -= i-1;
124 ix0 |= (ix1>>(32-i));
125 ix1 <<= i;
126 }
127 m -= 1023; /* unbias exponent */
128 ix0 = (ix0&0x000fffff)|0x00100000;
129 if(m&1){ /* odd m, double x to make it even */
130 ix0 += ix0 + ((ix1&sign)>>31);
131 ix1 += ix1;
132 }
133 m >>= 1; /* m = [m/2] */
134
135 /* generate sqrt(x) bit by bit */
136 ix0 += ix0 + ((ix1&sign)>>31);
137 ix1 += ix1;
138 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
139 r = 0x00200000; /* r = moving bit from right to left */
140
141 while(r!=0) {
142 t = s0+r;
143 if(t<=ix0) {
144 s0 = t+r;
145 ix0 -= t;
146 q += r;
147 }
148 ix0 += ix0 + ((ix1&sign)>>31);
149 ix1 += ix1;
150 r>>=1;
151 }
152
153 r = sign;
154 while(r!=0) {
155 t1 = s1+r;
156 t = s0;
157 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
158 s1 = t1+r;
159 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
160 ix0 -= t;
161 if (ix1 < t1) ix0 -= 1;
162 ix1 -= t1;
163 q1 += r;
164 }
165 ix0 += ix0 + ((ix1&sign)>>31);
166 ix1 += ix1;
167 r>>=1;
168 }
169
170 /* use floating add to find out rounding direction */
171 if((ix0|ix1)!=0) {
172 z = one-tiny; /* trigger inexact flag */
173 if (z>=one) {
174 z = one+tiny;
175 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
176 else if (z>one) {
177 if (q1==(u_int32_t)0xfffffffe) q+=1;
178 q1+=2;
179 } else
180 q1 += (q1&1);
181 }
182 }
183 ix0 = (q>>1)+0x3fe00000;
184 ix1 = q1>>1;
185 if ((q&1)==1) ix1 |= sign;
186 ix0 += (m <<20);
187 INSERT_WORDS(z,ix0,ix1);
188 return z;
189 }
190
191 #if (LDBL_MANT_DIG == 53)
192 __weak_reference(sqrt, sqrtl);
193 #endif
194
195 /*
196 Other methods (use floating-point arithmetic)
197 -------------
198 (This is a copy of a drafted paper by Prof W. Kahan
199 and K.C. Ng, written in May, 1986)
200
201 Two algorithms are given here to implement sqrt(x)
202 (IEEE double precision arithmetic) in software.
203 Both supply sqrt(x) correctly rounded. The first algorithm (in
204 Section A) uses newton iterations and involves four divisions.
205 The second one uses reciproot iterations to avoid division, but
206 requires more multiplications. Both algorithms need the ability
207 to chop results of arithmetic operations instead of round them,
208 and the INEXACT flag to indicate when an arithmetic operation
209 is executed exactly with no roundoff error, all part of the
210 standard (IEEE 754-1985). The ability to perform shift, add,
211 subtract and logical AND operations upon 32-bit words is needed
212 too, though not part of the standard.
213
214 A. sqrt(x) by Newton Iteration
215
216 (1) Initial approximation
217
218 Let x0 and x1 be the leading and the trailing 32-bit words of
219 a floating point number x (in IEEE double format) respectively
220
221 1 11 52 ...widths
222 ------------------------------------------------------
223 x: |s| e | f |
224 ------------------------------------------------------
225 msb lsb msb lsb ...order
226
227
228 ------------------------ ------------------------
229 x0: |s| e | f1 | x1: | f2 |
230 ------------------------ ------------------------
231
232 By performing shifts and subtracts on x0 and x1 (both regarded
233 as integers), we obtain an 8-bit approximation of sqrt(x) as
234 follows.
235
236 k := (x0>>1) + 0x1ff80000;
237 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
238 Here k is a 32-bit integer and T1[] is an integer array containing
239 correction terms. Now magically the floating value of y (y's
240 leading 32-bit word is y0, the value of its trailing word is 0)
241 approximates sqrt(x) to almost 8-bit.
242
243 Value of T1:
244 static int T1[32]= {
245 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
246 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
247 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
248 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
249
250 (2) Iterative refinement
251
252 Apply Heron's rule three times to y, we have y approximates
253 sqrt(x) to within 1 ulp (Unit in the Last Place):
254
255 y := (y+x/y)/2 ... almost 17 sig. bits
256 y := (y+x/y)/2 ... almost 35 sig. bits
257 y := y-(y-x/y)/2 ... within 1 ulp
258
259
260 Remark 1.
261 Another way to improve y to within 1 ulp is:
262
263 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
264 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
265
266 2
267 (x-y )*y
268 y := y + 2* ---------- ...within 1 ulp
269 2
270 3y + x
271
272
273 This formula has one division fewer than the one above; however,
274 it requires more multiplications and additions. Also x must be
275 scaled in advance to avoid spurious overflow in evaluating the
276 expression 3y*y+x. Hence it is not recommended uless division
277 is slow. If division is very slow, then one should use the
278 reciproot algorithm given in section B.
279
280 (3) Final adjustment
281
282 By twiddling y's last bit it is possible to force y to be
283 correctly rounded according to the prevailing rounding mode
284 as follows. Let r and i be copies of the rounding mode and
285 inexact flag before entering the square root program. Also we
286 use the expression y+-ulp for the next representable floating
287 numbers (up and down) of y. Note that y+-ulp = either fixed
288 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
289 mode.
290
291 I := FALSE; ... reset INEXACT flag I
292 R := RZ; ... set rounding mode to round-toward-zero
293 z := x/y; ... chopped quotient, possibly inexact
294 If(not I) then { ... if the quotient is exact
295 if(z=y) {
296 I := i; ... restore inexact flag
297 R := r; ... restore rounded mode
298 return sqrt(x):=y.
299 } else {
300 z := z - ulp; ... special rounding
301 }
302 }
303 i := TRUE; ... sqrt(x) is inexact
304 If (r=RN) then z=z+ulp ... rounded-to-nearest
305 If (r=RP) then { ... round-toward-+inf
306 y = y+ulp; z=z+ulp;
307 }
308 y := y+z; ... chopped sum
309 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
310 I := i; ... restore inexact flag
311 R := r; ... restore rounded mode
312 return sqrt(x):=y.
313
314 (4) Special cases
315
316 Square root of +inf, +-0, or NaN is itself;
317 Square root of a negative number is NaN with invalid signal.
318
319
320 B. sqrt(x) by Reciproot Iteration
321
322 (1) Initial approximation
323
324 Let x0 and x1 be the leading and the trailing 32-bit words of
325 a floating point number x (in IEEE double format) respectively
326 (see section A). By performing shifs and subtracts on x0 and y0,
327 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
328
329 k := 0x5fe80000 - (x0>>1);
330 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
331
332 Here k is a 32-bit integer and T2[] is an integer array
333 containing correction terms. Now magically the floating
334 value of y (y's leading 32-bit word is y0, the value of
335 its trailing word y1 is set to zero) approximates 1/sqrt(x)
336 to almost 7.8-bit.
337
338 Value of T2:
339 static int T2[64]= {
340 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
341 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
342 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
343 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
344 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
345 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
346 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
347 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
348
349 (2) Iterative refinement
350
351 Apply Reciproot iteration three times to y and multiply the
352 result by x to get an approximation z that matches sqrt(x)
353 to about 1 ulp. To be exact, we will have
354 -1ulp < sqrt(x)-z<1.0625ulp.
355
356 ... set rounding mode to Round-to-nearest
357 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
358 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
359 ... special arrangement for better accuracy
360 z := x*y ... 29 bits to sqrt(x), with z*y<1
361 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
362
363 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
364 (a) the term z*y in the final iteration is always less than 1;
365 (b) the error in the final result is biased upward so that
366 -1 ulp < sqrt(x) - z < 1.0625 ulp
367 instead of |sqrt(x)-z|<1.03125ulp.
368
369 (3) Final adjustment
370
371 By twiddling y's last bit it is possible to force y to be
372 correctly rounded according to the prevailing rounding mode
373 as follows. Let r and i be copies of the rounding mode and
374 inexact flag before entering the square root program. Also we
375 use the expression y+-ulp for the next representable floating
376 numbers (up and down) of y. Note that y+-ulp = either fixed
377 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
378 mode.
379
380 R := RZ; ... set rounding mode to round-toward-zero
381 switch(r) {
382 case RN: ... round-to-nearest
383 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
384 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
385 break;
386 case RZ:case RM: ... round-to-zero or round-to--inf
387 R:=RP; ... reset rounding mod to round-to-+inf
388 if(x<z*z ... rounded up) z = z - ulp; else
389 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
390 break;
391 case RP: ... round-to-+inf
392 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
393 if(x>z*z ...chopped) z = z+ulp;
394 break;
395 }
396
397 Remark 3. The above comparisons can be done in fixed point. For
398 example, to compare x and w=z*z chopped, it suffices to compare
399 x1 and w1 (the trailing parts of x and w), regarding them as
400 two's complement integers.
401
402 ...Is z an exact square root?
403 To determine whether z is an exact square root of x, let z1 be the
404 trailing part of z, and also let x0 and x1 be the leading and
405 trailing parts of x.
406
407 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
408 I := 1; ... Raise Inexact flag: z is not exact
409 else {
410 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
411 k := z1 >> 26; ... get z's 25-th and 26-th
412 fraction bits
413 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
414 }
415 R:= r ... restore rounded mode
416 return sqrt(x):=z.
417
418 If multiplication is cheaper then the foregoing red tape, the
419 Inexact flag can be evaluated by
420
421 I := i;
422 I := (z*z!=x) or I.
423
424 Note that z*z can overwrite I; this value must be sensed if it is
425 True.
426
427 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
428 zero.
429
430 --------------------
431 z1: | f2 |
432 --------------------
433 bit 31 bit 0
434
435 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
436 or even of logb(x) have the following relations:
437
438 -------------------------------------------------
439 bit 27,26 of z1 bit 1,0 of x1 logb(x)
440 -------------------------------------------------
441 00 00 odd and even
442 01 01 even
443 10 10 odd
444 10 00 even
445 11 01 even
446 -------------------------------------------------
447
448 (4) Special cases (see (4) of Section A).
449
450 */
451
452