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 <float.h>
15
16 #include "math.h"
17 #include "math_private.h"
18
19 #ifdef USE_BUILTIN_SQRT
20 double
sqrt(double x)21 sqrt(double x)
22 {
23 return (__builtin_sqrt(x));
24 }
25 #else
26 /* sqrt(x)
27 * Return correctly rounded sqrt.
28 * ------------------------------------------
29 * | Use the hardware sqrt if you have one |
30 * ------------------------------------------
31 * Method:
32 * Bit by bit method using integer arithmetic. (Slow, but portable)
33 * 1. Normalization
34 * Scale x to y in [1,4) with even powers of 2:
35 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
36 * sqrt(x) = 2^k * sqrt(y)
37 * 2. Bit by bit computation
38 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
39 * i 0
40 * i+1 2
41 * s = 2*q , and y = 2 * ( y - q ). (1)
42 * i i i i
43 *
44 * To compute q from q , one checks whether
45 * i+1 i
46 *
47 * -(i+1) 2
48 * (q + 2 ) <= y. (2)
49 * i
50 * -(i+1)
51 * If (2) is false, then q = q ; otherwise q = q + 2 .
52 * i+1 i i+1 i
53 *
54 * With some algebric manipulation, it is not difficult to see
55 * that (2) is equivalent to
56 * -(i+1)
57 * s + 2 <= y (3)
58 * i i
59 *
60 * The advantage of (3) is that s and y can be computed by
61 * i i
62 * the following recurrence formula:
63 * if (3) is false
64 *
65 * s = s , y = y ; (4)
66 * i+1 i i+1 i
67 *
68 * otherwise,
69 * -i -(i+1)
70 * s = s + 2 , y = y - s - 2 (5)
71 * i+1 i i+1 i i
72 *
73 * One may easily use induction to prove (4) and (5).
74 * Note. Since the left hand side of (3) contain only i+2 bits,
75 * it does not necessary to do a full (53-bit) comparison
76 * in (3).
77 * 3. Final rounding
78 * After generating the 53 bits result, we compute one more bit.
79 * Together with the remainder, we can decide whether the
80 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
81 * (it will never equal to 1/2ulp).
82 * The rounding mode can be detected by checking whether
83 * huge + tiny is equal to huge, and whether huge - tiny is
84 * equal to huge for some floating point number "huge" and "tiny".
85 *
86 * Special cases:
87 * sqrt(+-0) = +-0 ... exact
88 * sqrt(inf) = inf
89 * sqrt(-ve) = NaN ... with invalid signal
90 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
91 *
92 * Other methods : see the appended file at the end of the program below.
93 *---------------
94 */
95
96 static const double one = 1.0, tiny=1.0e-300;
97
98 double
sqrt(double x)99 sqrt(double x)
100 {
101 double z;
102 int32_t sign = (int)0x80000000;
103 int32_t ix0,s0,q,m,t,i;
104 u_int32_t r,t1,s1,ix1,q1;
105
106 EXTRACT_WORDS(ix0,ix1,x);
107
108 /* take care of Inf and NaN */
109 if((ix0&0x7ff00000)==0x7ff00000) {
110 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
111 sqrt(-inf)=sNaN */
112 }
113 /* take care of zero */
114 if(ix0<=0) {
115 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
116 else if(ix0<0)
117 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
118 }
119 /* normalize x */
120 m = (ix0>>20);
121 if(m==0) { /* subnormal x */
122 while(ix0==0) {
123 m -= 21;
124 ix0 |= (ix1>>11); ix1 <<= 21;
125 }
126 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
127 m -= i-1;
128 ix0 |= (ix1>>(32-i));
129 ix1 <<= i;
130 }
131 m -= 1023; /* unbias exponent */
132 ix0 = (ix0&0x000fffff)|0x00100000;
133 if(m&1){ /* odd m, double x to make it even */
134 ix0 += ix0 + ((ix1&sign)>>31);
135 ix1 += ix1;
136 }
137 m >>= 1; /* m = [m/2] */
138
139 /* generate sqrt(x) bit by bit */
140 ix0 += ix0 + ((ix1&sign)>>31);
141 ix1 += ix1;
142 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
143 r = 0x00200000; /* r = moving bit from right to left */
144
145 while(r!=0) {
146 t = s0+r;
147 if(t<=ix0) {
148 s0 = t+r;
149 ix0 -= t;
150 q += r;
151 }
152 ix0 += ix0 + ((ix1&sign)>>31);
153 ix1 += ix1;
154 r>>=1;
155 }
156
157 r = sign;
158 while(r!=0) {
159 t1 = s1+r;
160 t = s0;
161 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
162 s1 = t1+r;
163 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
164 ix0 -= t;
165 if (ix1 < t1) ix0 -= 1;
166 ix1 -= t1;
167 q1 += r;
168 }
169 ix0 += ix0 + ((ix1&sign)>>31);
170 ix1 += ix1;
171 r>>=1;
172 }
173
174 /* use floating add to find out rounding direction */
175 if((ix0|ix1)!=0) {
176 z = one-tiny; /* trigger inexact flag */
177 if (z>=one) {
178 z = one+tiny;
179 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
180 else if (z>one) {
181 if (q1==(u_int32_t)0xfffffffe) q+=1;
182 q1+=2;
183 } else
184 q1 += (q1&1);
185 }
186 }
187 ix0 = (q>>1)+0x3fe00000;
188 ix1 = q1>>1;
189 if ((q&1)==1) ix1 |= sign;
190 ix0 += (m <<20);
191 INSERT_WORDS(z,ix0,ix1);
192 return z;
193 }
194 #endif
195
196 #if (LDBL_MANT_DIG == 53)
197 __weak_reference(sqrt, sqrtl);
198 #endif
199
200 /*
201 Other methods (use floating-point arithmetic)
202 -------------
203 (This is a copy of a drafted paper by Prof W. Kahan
204 and K.C. Ng, written in May, 1986)
205
206 Two algorithms are given here to implement sqrt(x)
207 (IEEE double precision arithmetic) in software.
208 Both supply sqrt(x) correctly rounded. The first algorithm (in
209 Section A) uses newton iterations and involves four divisions.
210 The second one uses reciproot iterations to avoid division, but
211 requires more multiplications. Both algorithms need the ability
212 to chop results of arithmetic operations instead of round them,
213 and the INEXACT flag to indicate when an arithmetic operation
214 is executed exactly with no roundoff error, all part of the
215 standard (IEEE 754-1985). The ability to perform shift, add,
216 subtract and logical AND operations upon 32-bit words is needed
217 too, though not part of the standard.
218
219 A. sqrt(x) by Newton Iteration
220
221 (1) Initial approximation
222
223 Let x0 and x1 be the leading and the trailing 32-bit words of
224 a floating point number x (in IEEE double format) respectively
225
226 1 11 52 ...widths
227 ------------------------------------------------------
228 x: |s| e | f |
229 ------------------------------------------------------
230 msb lsb msb lsb ...order
231
232
233 ------------------------ ------------------------
234 x0: |s| e | f1 | x1: | f2 |
235 ------------------------ ------------------------
236
237 By performing shifts and subtracts on x0 and x1 (both regarded
238 as integers), we obtain an 8-bit approximation of sqrt(x) as
239 follows.
240
241 k := (x0>>1) + 0x1ff80000;
242 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
243 Here k is a 32-bit integer and T1[] is an integer array containing
244 correction terms. Now magically the floating value of y (y's
245 leading 32-bit word is y0, the value of its trailing word is 0)
246 approximates sqrt(x) to almost 8-bit.
247
248 Value of T1:
249 static int T1[32]= {
250 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
251 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
252 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
253 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
254
255 (2) Iterative refinement
256
257 Apply Heron's rule three times to y, we have y approximates
258 sqrt(x) to within 1 ulp (Unit in the Last Place):
259
260 y := (y+x/y)/2 ... almost 17 sig. bits
261 y := (y+x/y)/2 ... almost 35 sig. bits
262 y := y-(y-x/y)/2 ... within 1 ulp
263
264
265 Remark 1.
266 Another way to improve y to within 1 ulp is:
267
268 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
269 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
270
271 2
272 (x-y )*y
273 y := y + 2* ---------- ...within 1 ulp
274 2
275 3y + x
276
277
278 This formula has one division fewer than the one above; however,
279 it requires more multiplications and additions. Also x must be
280 scaled in advance to avoid spurious overflow in evaluating the
281 expression 3y*y+x. Hence it is not recommended uless division
282 is slow. If division is very slow, then one should use the
283 reciproot algorithm given in section B.
284
285 (3) Final adjustment
286
287 By twiddling y's last bit it is possible to force y to be
288 correctly rounded according to the prevailing rounding mode
289 as follows. Let r and i be copies of the rounding mode and
290 inexact flag before entering the square root program. Also we
291 use the expression y+-ulp for the next representable floating
292 numbers (up and down) of y. Note that y+-ulp = either fixed
293 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
294 mode.
295
296 I := FALSE; ... reset INEXACT flag I
297 R := RZ; ... set rounding mode to round-toward-zero
298 z := x/y; ... chopped quotient, possibly inexact
299 If(not I) then { ... if the quotient is exact
300 if(z=y) {
301 I := i; ... restore inexact flag
302 R := r; ... restore rounded mode
303 return sqrt(x):=y.
304 } else {
305 z := z - ulp; ... special rounding
306 }
307 }
308 i := TRUE; ... sqrt(x) is inexact
309 If (r=RN) then z=z+ulp ... rounded-to-nearest
310 If (r=RP) then { ... round-toward-+inf
311 y = y+ulp; z=z+ulp;
312 }
313 y := y+z; ... chopped sum
314 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
315 I := i; ... restore inexact flag
316 R := r; ... restore rounded mode
317 return sqrt(x):=y.
318
319 (4) Special cases
320
321 Square root of +inf, +-0, or NaN is itself;
322 Square root of a negative number is NaN with invalid signal.
323
324
325 B. sqrt(x) by Reciproot Iteration
326
327 (1) Initial approximation
328
329 Let x0 and x1 be the leading and the trailing 32-bit words of
330 a floating point number x (in IEEE double format) respectively
331 (see section A). By performing shifs and subtracts on x0 and y0,
332 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
333
334 k := 0x5fe80000 - (x0>>1);
335 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
336
337 Here k is a 32-bit integer and T2[] is an integer array
338 containing correction terms. Now magically the floating
339 value of y (y's leading 32-bit word is y0, the value of
340 its trailing word y1 is set to zero) approximates 1/sqrt(x)
341 to almost 7.8-bit.
342
343 Value of T2:
344 static int T2[64]= {
345 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
346 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
347 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
348 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
349 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
350 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
351 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
352 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
353
354 (2) Iterative refinement
355
356 Apply Reciproot iteration three times to y and multiply the
357 result by x to get an approximation z that matches sqrt(x)
358 to about 1 ulp. To be exact, we will have
359 -1ulp < sqrt(x)-z<1.0625ulp.
360
361 ... set rounding mode to Round-to-nearest
362 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
363 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
364 ... special arrangement for better accuracy
365 z := x*y ... 29 bits to sqrt(x), with z*y<1
366 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
367
368 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
369 (a) the term z*y in the final iteration is always less than 1;
370 (b) the error in the final result is biased upward so that
371 -1 ulp < sqrt(x) - z < 1.0625 ulp
372 instead of |sqrt(x)-z|<1.03125ulp.
373
374 (3) Final adjustment
375
376 By twiddling y's last bit it is possible to force y to be
377 correctly rounded according to the prevailing rounding mode
378 as follows. Let r and i be copies of the rounding mode and
379 inexact flag before entering the square root program. Also we
380 use the expression y+-ulp for the next representable floating
381 numbers (up and down) of y. Note that y+-ulp = either fixed
382 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
383 mode.
384
385 R := RZ; ... set rounding mode to round-toward-zero
386 switch(r) {
387 case RN: ... round-to-nearest
388 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
389 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
390 break;
391 case RZ:case RM: ... round-to-zero or round-to--inf
392 R:=RP; ... reset rounding mod to round-to-+inf
393 if(x<z*z ... rounded up) z = z - ulp; else
394 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
395 break;
396 case RP: ... round-to-+inf
397 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
398 if(x>z*z ...chopped) z = z+ulp;
399 break;
400 }
401
402 Remark 3. The above comparisons can be done in fixed point. For
403 example, to compare x and w=z*z chopped, it suffices to compare
404 x1 and w1 (the trailing parts of x and w), regarding them as
405 two's complement integers.
406
407 ...Is z an exact square root?
408 To determine whether z is an exact square root of x, let z1 be the
409 trailing part of z, and also let x0 and x1 be the leading and
410 trailing parts of x.
411
412 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
413 I := 1; ... Raise Inexact flag: z is not exact
414 else {
415 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
416 k := z1 >> 26; ... get z's 25-th and 26-th
417 fraction bits
418 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
419 }
420 R:= r ... restore rounded mode
421 return sqrt(x):=z.
422
423 If multiplication is cheaper then the foregoing red tape, the
424 Inexact flag can be evaluated by
425
426 I := i;
427 I := (z*z!=x) or I.
428
429 Note that z*z can overwrite I; this value must be sensed if it is
430 True.
431
432 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
433 zero.
434
435 --------------------
436 z1: | f2 |
437 --------------------
438 bit 31 bit 0
439
440 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
441 or even of logb(x) have the following relations:
442
443 -------------------------------------------------
444 bit 27,26 of z1 bit 1,0 of x1 logb(x)
445 -------------------------------------------------
446 00 00 odd and even
447 01 01 even
448 10 10 odd
449 10 00 even
450 11 01 even
451 -------------------------------------------------
452
453 (4) Special cases (see (4) of Section A).
454
455 */
456
457