1
2 /*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunSoft, 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 #include <float.h>
14
15 #include "math.h"
16 #include "math_private.h"
17
18 #ifdef USE_BUILTIN_SQRT
19 double
sqrt(double x)20 sqrt(double x)
21 {
22 return (__builtin_sqrt(x));
23 }
24 #else
25 /* sqrt(x)
26 * Return correctly rounded sqrt.
27 * ------------------------------------------
28 * | Use the hardware sqrt if you have one |
29 * ------------------------------------------
30 * Method:
31 * Bit by bit method using integer arithmetic. (Slow, but portable)
32 * 1. Normalization
33 * Scale x to y in [1,4) with even powers of 2:
34 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
35 * sqrt(x) = 2^k * sqrt(y)
36 * 2. Bit by bit computation
37 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
38 * i 0
39 * i+1 2
40 * s = 2*q , and y = 2 * ( y - q ). (1)
41 * i i i i
42 *
43 * To compute q from q , one checks whether
44 * i+1 i
45 *
46 * -(i+1) 2
47 * (q + 2 ) <= y. (2)
48 * i
49 * -(i+1)
50 * If (2) is false, then q = q ; otherwise q = q + 2 .
51 * i+1 i i+1 i
52 *
53 * With some algebric manipulation, it is not difficult to see
54 * that (2) is equivalent to
55 * -(i+1)
56 * s + 2 <= y (3)
57 * i i
58 *
59 * The advantage of (3) is that s and y can be computed by
60 * i i
61 * the following recurrence formula:
62 * if (3) is false
63 *
64 * s = s , y = y ; (4)
65 * i+1 i i+1 i
66 *
67 * otherwise,
68 * -i -(i+1)
69 * s = s + 2 , y = y - s - 2 (5)
70 * i+1 i i+1 i i
71 *
72 * One may easily use induction to prove (4) and (5).
73 * Note. Since the left hand side of (3) contain only i+2 bits,
74 * it does not necessary to do a full (53-bit) comparison
75 * in (3).
76 * 3. Final rounding
77 * After generating the 53 bits result, we compute one more bit.
78 * Together with the remainder, we can decide whether the
79 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
80 * (it will never equal to 1/2ulp).
81 * The rounding mode can be detected by checking whether
82 * huge + tiny is equal to huge, and whether huge - tiny is
83 * equal to huge for some floating point number "huge" and "tiny".
84 *
85 * Special cases:
86 * sqrt(+-0) = +-0 ... exact
87 * sqrt(inf) = inf
88 * sqrt(-ve) = NaN ... with invalid signal
89 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
90 *
91 * Other methods : see the appended file at the end of the program below.
92 *---------------
93 */
94
95 static const double one = 1.0, tiny=1.0e-300;
96
97 double
sqrt(double x)98 sqrt(double x)
99 {
100 double z;
101 int32_t sign = (int)0x80000000;
102 int32_t ix0,s0,q,m,t,i;
103 u_int32_t r,t1,s1,ix1,q1;
104
105 EXTRACT_WORDS(ix0,ix1,x);
106
107 /* take care of Inf and NaN */
108 if((ix0&0x7ff00000)==0x7ff00000) {
109 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
110 sqrt(-inf)=sNaN */
111 }
112 /* take care of zero */
113 if(ix0<=0) {
114 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
115 else if(ix0<0)
116 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
117 }
118 /* normalize x */
119 m = (ix0>>20);
120 if(m==0) { /* subnormal x */
121 while(ix0==0) {
122 m -= 21;
123 ix0 |= (ix1>>11); ix1 <<= 21;
124 }
125 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
126 m -= i-1;
127 ix0 |= (ix1>>(32-i));
128 ix1 <<= i;
129 }
130 m -= 1023; /* unbias exponent */
131 ix0 = (ix0&0x000fffff)|0x00100000;
132 if(m&1){ /* odd m, double x to make it even */
133 ix0 += ix0 + ((ix1&sign)>>31);
134 ix1 += ix1;
135 }
136 m >>= 1; /* m = [m/2] */
137
138 /* generate sqrt(x) bit by bit */
139 ix0 += ix0 + ((ix1&sign)>>31);
140 ix1 += ix1;
141 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
142 r = 0x00200000; /* r = moving bit from right to left */
143
144 while(r!=0) {
145 t = s0+r;
146 if(t<=ix0) {
147 s0 = t+r;
148 ix0 -= t;
149 q += r;
150 }
151 ix0 += ix0 + ((ix1&sign)>>31);
152 ix1 += ix1;
153 r>>=1;
154 }
155
156 r = sign;
157 while(r!=0) {
158 t1 = s1+r;
159 t = s0;
160 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
161 s1 = t1+r;
162 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
163 ix0 -= t;
164 if (ix1 < t1) ix0 -= 1;
165 ix1 -= t1;
166 q1 += r;
167 }
168 ix0 += ix0 + ((ix1&sign)>>31);
169 ix1 += ix1;
170 r>>=1;
171 }
172
173 /* use floating add to find out rounding direction */
174 if((ix0|ix1)!=0) {
175 z = one-tiny; /* trigger inexact flag */
176 if (z>=one) {
177 z = one+tiny;
178 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
179 else if (z>one) {
180 if (q1==(u_int32_t)0xfffffffe) q+=1;
181 q1+=2;
182 } else
183 q1 += (q1&1);
184 }
185 }
186 ix0 = (q>>1)+0x3fe00000;
187 ix1 = q1>>1;
188 if ((q&1)==1) ix1 |= sign;
189 ix0 += (m <<20);
190 INSERT_WORDS(z,ix0,ix1);
191 return z;
192 }
193 #endif
194
195 #if (LDBL_MANT_DIG == 53)
196 __weak_reference(sqrt, sqrtl);
197 #endif
198
199 /*
200 Other methods (use floating-point arithmetic)
201 -------------
202 (This is a copy of a drafted paper by Prof W. Kahan
203 and K.C. Ng, written in May, 1986)
204
205 Two algorithms are given here to implement sqrt(x)
206 (IEEE double precision arithmetic) in software.
207 Both supply sqrt(x) correctly rounded. The first algorithm (in
208 Section A) uses newton iterations and involves four divisions.
209 The second one uses reciproot iterations to avoid division, but
210 requires more multiplications. Both algorithms need the ability
211 to chop results of arithmetic operations instead of round them,
212 and the INEXACT flag to indicate when an arithmetic operation
213 is executed exactly with no roundoff error, all part of the
214 standard (IEEE 754-1985). The ability to perform shift, add,
215 subtract and logical AND operations upon 32-bit words is needed
216 too, though not part of the standard.
217
218 A. sqrt(x) by Newton Iteration
219
220 (1) Initial approximation
221
222 Let x0 and x1 be the leading and the trailing 32-bit words of
223 a floating point number x (in IEEE double format) respectively
224
225 1 11 52 ...widths
226 ------------------------------------------------------
227 x: |s| e | f |
228 ------------------------------------------------------
229 msb lsb msb lsb ...order
230
231
232 ------------------------ ------------------------
233 x0: |s| e | f1 | x1: | f2 |
234 ------------------------ ------------------------
235
236 By performing shifts and subtracts on x0 and x1 (both regarded
237 as integers), we obtain an 8-bit approximation of sqrt(x) as
238 follows.
239
240 k := (x0>>1) + 0x1ff80000;
241 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
242 Here k is a 32-bit integer and T1[] is an integer array containing
243 correction terms. Now magically the floating value of y (y's
244 leading 32-bit word is y0, the value of its trailing word is 0)
245 approximates sqrt(x) to almost 8-bit.
246
247 Value of T1:
248 static int T1[32]= {
249 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
250 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
251 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
252 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
253
254 (2) Iterative refinement
255
256 Apply Heron's rule three times to y, we have y approximates
257 sqrt(x) to within 1 ulp (Unit in the Last Place):
258
259 y := (y+x/y)/2 ... almost 17 sig. bits
260 y := (y+x/y)/2 ... almost 35 sig. bits
261 y := y-(y-x/y)/2 ... within 1 ulp
262
263
264 Remark 1.
265 Another way to improve y to within 1 ulp is:
266
267 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
268 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
269
270 2
271 (x-y )*y
272 y := y + 2* ---------- ...within 1 ulp
273 2
274 3y + x
275
276
277 This formula has one division fewer than the one above; however,
278 it requires more multiplications and additions. Also x must be
279 scaled in advance to avoid spurious overflow in evaluating the
280 expression 3y*y+x. Hence it is not recommended uless division
281 is slow. If division is very slow, then one should use the
282 reciproot algorithm given in section B.
283
284 (3) Final adjustment
285
286 By twiddling y's last bit it is possible to force y to be
287 correctly rounded according to the prevailing rounding mode
288 as follows. Let r and i be copies of the rounding mode and
289 inexact flag before entering the square root program. Also we
290 use the expression y+-ulp for the next representable floating
291 numbers (up and down) of y. Note that y+-ulp = either fixed
292 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
293 mode.
294
295 I := FALSE; ... reset INEXACT flag I
296 R := RZ; ... set rounding mode to round-toward-zero
297 z := x/y; ... chopped quotient, possibly inexact
298 If(not I) then { ... if the quotient is exact
299 if(z=y) {
300 I := i; ... restore inexact flag
301 R := r; ... restore rounded mode
302 return sqrt(x):=y.
303 } else {
304 z := z - ulp; ... special rounding
305 }
306 }
307 i := TRUE; ... sqrt(x) is inexact
308 If (r=RN) then z=z+ulp ... rounded-to-nearest
309 If (r=RP) then { ... round-toward-+inf
310 y = y+ulp; z=z+ulp;
311 }
312 y := y+z; ... chopped sum
313 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
314 I := i; ... restore inexact flag
315 R := r; ... restore rounded mode
316 return sqrt(x):=y.
317
318 (4) Special cases
319
320 Square root of +inf, +-0, or NaN is itself;
321 Square root of a negative number is NaN with invalid signal.
322
323
324 B. sqrt(x) by Reciproot Iteration
325
326 (1) Initial approximation
327
328 Let x0 and x1 be the leading and the trailing 32-bit words of
329 a floating point number x (in IEEE double format) respectively
330 (see section A). By performing shifs and subtracts on x0 and y0,
331 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
332
333 k := 0x5fe80000 - (x0>>1);
334 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
335
336 Here k is a 32-bit integer and T2[] is an integer array
337 containing correction terms. Now magically the floating
338 value of y (y's leading 32-bit word is y0, the value of
339 its trailing word y1 is set to zero) approximates 1/sqrt(x)
340 to almost 7.8-bit.
341
342 Value of T2:
343 static int T2[64]= {
344 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
345 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
346 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
347 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
348 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
349 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
350 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
351 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
352
353 (2) Iterative refinement
354
355 Apply Reciproot iteration three times to y and multiply the
356 result by x to get an approximation z that matches sqrt(x)
357 to about 1 ulp. To be exact, we will have
358 -1ulp < sqrt(x)-z<1.0625ulp.
359
360 ... set rounding mode to Round-to-nearest
361 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
362 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
363 ... special arrangement for better accuracy
364 z := x*y ... 29 bits to sqrt(x), with z*y<1
365 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
366
367 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
368 (a) the term z*y in the final iteration is always less than 1;
369 (b) the error in the final result is biased upward so that
370 -1 ulp < sqrt(x) - z < 1.0625 ulp
371 instead of |sqrt(x)-z|<1.03125ulp.
372
373 (3) Final adjustment
374
375 By twiddling y's last bit it is possible to force y to be
376 correctly rounded according to the prevailing rounding mode
377 as follows. Let r and i be copies of the rounding mode and
378 inexact flag before entering the square root program. Also we
379 use the expression y+-ulp for the next representable floating
380 numbers (up and down) of y. Note that y+-ulp = either fixed
381 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
382 mode.
383
384 R := RZ; ... set rounding mode to round-toward-zero
385 switch(r) {
386 case RN: ... round-to-nearest
387 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
388 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
389 break;
390 case RZ:case RM: ... round-to-zero or round-to--inf
391 R:=RP; ... reset rounding mod to round-to-+inf
392 if(x<z*z ... rounded up) z = z - ulp; else
393 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
394 break;
395 case RP: ... round-to-+inf
396 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
397 if(x>z*z ...chopped) z = z+ulp;
398 break;
399 }
400
401 Remark 3. The above comparisons can be done in fixed point. For
402 example, to compare x and w=z*z chopped, it suffices to compare
403 x1 and w1 (the trailing parts of x and w), regarding them as
404 two's complement integers.
405
406 ...Is z an exact square root?
407 To determine whether z is an exact square root of x, let z1 be the
408 trailing part of z, and also let x0 and x1 be the leading and
409 trailing parts of x.
410
411 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
412 I := 1; ... Raise Inexact flag: z is not exact
413 else {
414 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
415 k := z1 >> 26; ... get z's 25-th and 26-th
416 fraction bits
417 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
418 }
419 R:= r ... restore rounded mode
420 return sqrt(x):=z.
421
422 If multiplication is cheaper then the foregoing red tape, the
423 Inexact flag can be evaluated by
424
425 I := i;
426 I := (z*z!=x) or I.
427
428 Note that z*z can overwrite I; this value must be sensed if it is
429 True.
430
431 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
432 zero.
433
434 --------------------
435 z1: | f2 |
436 --------------------
437 bit 31 bit 0
438
439 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
440 or even of logb(x) have the following relations:
441
442 -------------------------------------------------
443 bit 27,26 of z1 bit 1,0 of x1 logb(x)
444 -------------------------------------------------
445 00 00 odd and even
446 01 01 even
447 10 10 odd
448 10 00 even
449 11 01 even
450 -------------------------------------------------
451
452 (4) Special cases (see (4) of Section A).
453
454 */
455
456