1 /*
2 * ====================================================
3 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
4 *
5 * Developed at SunPro, a Sun Microsystems, Inc. business.
6 * Permission to use, copy, modify, and distribute this
7 * software is freely granted, provided that this notice
8 * is preserved.
9 * ====================================================
10 */
11
12 /* __ieee754_sqrt(x)
13 * Return correctly rounded sqrt.
14 * ------------------------------------------
15 * | Use the hardware sqrt if you have one |
16 * ------------------------------------------
17 * Method:
18 * Bit by bit method using integer arithmetic. (Slow, but portable)
19 * 1. Normalization
20 * Scale x to y in [1,4) with even powers of 2:
21 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
22 * sqrt(x) = 2^k * sqrt(y)
23 * 2. Bit by bit computation
24 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
25 * i 0
26 * i+1 2
27 * s = 2*q , and y = 2 * ( y - q ). (1)
28 * i i i i
29 *
30 * To compute q from q , one checks whether
31 * i+1 i
32 *
33 * -(i+1) 2
34 * (q + 2 ) <= y. (2)
35 * i
36 * -(i+1)
37 * If (2) is false, then q = q ; otherwise q = q + 2 .
38 * i+1 i i+1 i
39 *
40 * With some algebric manipulation, it is not difficult to see
41 * that (2) is equivalent to
42 * -(i+1)
43 * s + 2 <= y (3)
44 * i i
45 *
46 * The advantage of (3) is that s and y can be computed by
47 * i i
48 * the following recurrence formula:
49 * if (3) is false
50 *
51 * s = s , y = y ; (4)
52 * i+1 i i+1 i
53 *
54 * otherwise,
55 * -i -(i+1)
56 * s = s + 2 , y = y - s - 2 (5)
57 * i+1 i i+1 i i
58 *
59 * One may easily use induction to prove (4) and (5).
60 * Note. Since the left hand side of (3) contain only i+2 bits,
61 * it does not necessary to do a full (53-bit) comparison
62 * in (3).
63 * 3. Final rounding
64 * After generating the 53 bits result, we compute one more bit.
65 * Together with the remainder, we can decide whether the
66 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
67 * (it will never equal to 1/2ulp).
68 * The rounding mode can be detected by checking whether
69 * huge + tiny is equal to huge, and whether huge - tiny is
70 * equal to huge for some floating point number "huge" and "tiny".
71 *
72 * Special cases:
73 * sqrt(+-0) = +-0 ... exact
74 * sqrt(inf) = inf
75 * sqrt(-ve) = NaN ... with invalid signal
76 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
77 *
78 * Other methods : see the appended file at the end of the program below.
79 *---------------
80 */
81
82 #include "math_libm.h"
83 #include "math_private.h"
84
85 static const double one = 1.0, tiny = 1.0e-300;
86
__ieee754_sqrt(double x)87 double attribute_hidden __ieee754_sqrt(double x)
88 {
89 double z;
90 int32_t sign = (int)0x80000000;
91 int32_t ix0,s0,q,m,t,i;
92 u_int32_t r,t1,s1,ix1,q1;
93
94 EXTRACT_WORDS(ix0,ix1,x);
95
96 /* take care of Inf and NaN */
97 if((ix0&0x7ff00000)==0x7ff00000) {
98 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
99 sqrt(-inf)=sNaN */
100 }
101 /* take care of zero */
102 if(ix0<=0) {
103 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
104 else if(ix0<0)
105 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
106 }
107 /* normalize x */
108 m = (ix0>>20);
109 if(m==0) { /* subnormal x */
110 while(ix0==0) {
111 m -= 21;
112 ix0 |= (ix1>>11); ix1 <<= 21;
113 }
114 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
115 m -= i-1;
116 ix0 |= (ix1>>(32-i));
117 ix1 <<= i;
118 }
119 m -= 1023; /* unbias exponent */
120 ix0 = (ix0&0x000fffff)|0x00100000;
121 if(m&1){ /* odd m, double x to make it even */
122 ix0 += ix0 + ((ix1&sign)>>31);
123 ix1 += ix1;
124 }
125 m >>= 1; /* m = [m/2] */
126
127 /* generate sqrt(x) bit by bit */
128 ix0 += ix0 + ((ix1&sign)>>31);
129 ix1 += ix1;
130 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
131 r = 0x00200000; /* r = moving bit from right to left */
132
133 while(r!=0) {
134 t = s0+r;
135 if(t<=ix0) {
136 s0 = t+r;
137 ix0 -= t;
138 q += r;
139 }
140 ix0 += ix0 + ((ix1&sign)>>31);
141 ix1 += ix1;
142 r>>=1;
143 }
144
145 r = sign;
146 while(r!=0) {
147 t1 = s1+r;
148 t = s0;
149 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
150 s1 = t1+r;
151 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
152 ix0 -= t;
153 if (ix1 < t1) ix0 -= 1;
154 ix1 -= t1;
155 q1 += r;
156 }
157 ix0 += ix0 + ((ix1&sign)>>31);
158 ix1 += ix1;
159 r>>=1;
160 }
161
162 /* use floating add to find out rounding direction */
163 if((ix0|ix1)!=0) {
164 z = one-tiny; /* trigger inexact flag */
165 if (z>=one) {
166 z = one+tiny;
167 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
168 else if (z>one) {
169 if (q1==(u_int32_t)0xfffffffe) q+=1;
170 q1+=2;
171 } else
172 q1 += (q1&1);
173 }
174 }
175 ix0 = (q>>1)+0x3fe00000;
176 ix1 = q1>>1;
177 if ((q&1)==1) ix1 |= sign;
178 ix0 += (m <<20);
179 INSERT_WORDS(z,ix0,ix1);
180 return z;
181 }
182
183 /*
184 * wrapper sqrt(x)
185 */
186 #ifndef _IEEE_LIBM
sqrt(double x)187 double sqrt(double x)
188 {
189 double z = __ieee754_sqrt(x);
190 if (_LIB_VERSION == _IEEE_ || isnan(x))
191 return z;
192 if (x < 0.0)
193 return __kernel_standard(x, x, 26); /* sqrt(negative) */
194 return z;
195 }
196 #else
197 strong_alias(__ieee754_sqrt, sqrt)
198 #endif
199 libm_hidden_def(sqrt)
200
201
202 /*
203 Other methods (use floating-point arithmetic)
204 -------------
205 (This is a copy of a drafted paper by Prof W. Kahan
206 and K.C. Ng, written in May, 1986)
207
208 Two algorithms are given here to implement sqrt(x)
209 (IEEE double precision arithmetic) in software.
210 Both supply sqrt(x) correctly rounded. The first algorithm (in
211 Section A) uses newton iterations and involves four divisions.
212 The second one uses reciproot iterations to avoid division, but
213 requires more multiplications. Both algorithms need the ability
214 to chop results of arithmetic operations instead of round them,
215 and the INEXACT flag to indicate when an arithmetic operation
216 is executed exactly with no roundoff error, all part of the
217 standard (IEEE 754-1985). The ability to perform shift, add,
218 subtract and logical AND operations upon 32-bit words is needed
219 too, though not part of the standard.
220
221 A. sqrt(x) by Newton Iteration
222
223 (1) Initial approximation
224
225 Let x0 and x1 be the leading and the trailing 32-bit words of
226 a floating point number x (in IEEE double format) respectively
227
228 1 11 52 ...widths
229 ------------------------------------------------------
230 x: |s| e | f |
231 ------------------------------------------------------
232 msb lsb msb lsb ...order
233
234
235 ------------------------ ------------------------
236 x0: |s| e | f1 | x1: | f2 |
237 ------------------------ ------------------------
238
239 By performing shifts and subtracts on x0 and x1 (both regarded
240 as integers), we obtain an 8-bit approximation of sqrt(x) as
241 follows.
242
243 k := (x0>>1) + 0x1ff80000;
244 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
245 Here k is a 32-bit integer and T1[] is an integer array containing
246 correction terms. Now magically the floating value of y (y's
247 leading 32-bit word is y0, the value of its trailing word is 0)
248 approximates sqrt(x) to almost 8-bit.
249
250 Value of T1:
251 static int T1[32]= {
252 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
253 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
254 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
255 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
256
257 (2) Iterative refinement
258
259 Apply Heron's rule three times to y, we have y approximates
260 sqrt(x) to within 1 ulp (Unit in the Last Place):
261
262 y := (y+x/y)/2 ... almost 17 sig. bits
263 y := (y+x/y)/2 ... almost 35 sig. bits
264 y := y-(y-x/y)/2 ... within 1 ulp
265
266
267 Remark 1.
268 Another way to improve y to within 1 ulp is:
269
270 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
271 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
272
273 2
274 (x-y )*y
275 y := y + 2* ---------- ...within 1 ulp
276 2
277 3y + x
278
279
280 This formula has one division fewer than the one above; however,
281 it requires more multiplications and additions. Also x must be
282 scaled in advance to avoid spurious overflow in evaluating the
283 expression 3y*y+x. Hence it is not recommended uless division
284 is slow. If division is very slow, then one should use the
285 reciproot algorithm given in section B.
286
287 (3) Final adjustment
288
289 By twiddling y's last bit it is possible to force y to be
290 correctly rounded according to the prevailing rounding mode
291 as follows. Let r and i be copies of the rounding mode and
292 inexact flag before entering the square root program. Also we
293 use the expression y+-ulp for the next representable floating
294 numbers (up and down) of y. Note that y+-ulp = either fixed
295 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
296 mode.
297
298 I := FALSE; ... reset INEXACT flag I
299 R := RZ; ... set rounding mode to round-toward-zero
300 z := x/y; ... chopped quotient, possibly inexact
301 If(not I) then { ... if the quotient is exact
302 if(z=y) {
303 I := i; ... restore inexact flag
304 R := r; ... restore rounded mode
305 return sqrt(x):=y.
306 } else {
307 z := z - ulp; ... special rounding
308 }
309 }
310 i := TRUE; ... sqrt(x) is inexact
311 If (r=RN) then z=z+ulp ... rounded-to-nearest
312 If (r=RP) then { ... round-toward-+inf
313 y = y+ulp; z=z+ulp;
314 }
315 y := y+z; ... chopped sum
316 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
317 I := i; ... restore inexact flag
318 R := r; ... restore rounded mode
319 return sqrt(x):=y.
320
321 (4) Special cases
322
323 Square root of +inf, +-0, or NaN is itself;
324 Square root of a negative number is NaN with invalid signal.
325
326
327 B. sqrt(x) by Reciproot Iteration
328
329 (1) Initial approximation
330
331 Let x0 and x1 be the leading and the trailing 32-bit words of
332 a floating point number x (in IEEE double format) respectively
333 (see section A). By performing shifs and subtracts on x0 and y0,
334 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
335
336 k := 0x5fe80000 - (x0>>1);
337 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
338
339 Here k is a 32-bit integer and T2[] is an integer array
340 containing correction terms. Now magically the floating
341 value of y (y's leading 32-bit word is y0, the value of
342 its trailing word y1 is set to zero) approximates 1/sqrt(x)
343 to almost 7.8-bit.
344
345 Value of T2:
346 static int T2[64]= {
347 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
348 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
349 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
350 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
351 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
352 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
353 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
354 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
355
356 (2) Iterative refinement
357
358 Apply Reciproot iteration three times to y and multiply the
359 result by x to get an approximation z that matches sqrt(x)
360 to about 1 ulp. To be exact, we will have
361 -1ulp < sqrt(x)-z<1.0625ulp.
362
363 ... set rounding mode to Round-to-nearest
364 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
365 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
366 ... special arrangement for better accuracy
367 z := x*y ... 29 bits to sqrt(x), with z*y<1
368 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
369
370 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
371 (a) the term z*y in the final iteration is always less than 1;
372 (b) the error in the final result is biased upward so that
373 -1 ulp < sqrt(x) - z < 1.0625 ulp
374 instead of |sqrt(x)-z|<1.03125ulp.
375
376 (3) Final adjustment
377
378 By twiddling y's last bit it is possible to force y to be
379 correctly rounded according to the prevailing rounding mode
380 as follows. Let r and i be copies of the rounding mode and
381 inexact flag before entering the square root program. Also we
382 use the expression y+-ulp for the next representable floating
383 numbers (up and down) of y. Note that y+-ulp = either fixed
384 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
385 mode.
386
387 R := RZ; ... set rounding mode to round-toward-zero
388 switch(r) {
389 case RN: ... round-to-nearest
390 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
391 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
392 break;
393 case RZ:case RM: ... round-to-zero or round-to--inf
394 R:=RP; ... reset rounding mod to round-to-+inf
395 if(x<z*z ... rounded up) z = z - ulp; else
396 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
397 break;
398 case RP: ... round-to-+inf
399 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
400 if(x>z*z ...chopped) z = z+ulp;
401 break;
402 }
403
404 Remark 3. The above comparisons can be done in fixed point. For
405 example, to compare x and w=z*z chopped, it suffices to compare
406 x1 and w1 (the trailing parts of x and w), regarding them as
407 two's complement integers.
408
409 ...Is z an exact square root?
410 To determine whether z is an exact square root of x, let z1 be the
411 trailing part of z, and also let x0 and x1 be the leading and
412 trailing parts of x.
413
414 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
415 I := 1; ... Raise Inexact flag: z is not exact
416 else {
417 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
418 k := z1 >> 26; ... get z's 25-th and 26-th
419 fraction bits
420 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
421 }
422 R:= r ... restore rounded mode
423 return sqrt(x):=z.
424
425 If multiplication is cheaper then the foregoing red tape, the
426 Inexact flag can be evaluated by
427
428 I := i;
429 I := (z*z!=x) or I.
430
431 Note that z*z can overwrite I; this value must be sensed if it is
432 True.
433
434 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
435 zero.
436
437 --------------------
438 z1: | f2 |
439 --------------------
440 bit 31 bit 0
441
442 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
443 or even of logb(x) have the following relations:
444
445 -------------------------------------------------
446 bit 27,26 of z1 bit 1,0 of x1 logb(x)
447 -------------------------------------------------
448 00 00 odd and even
449 01 01 even
450 10 10 odd
451 10 00 even
452 11 01 even
453 -------------------------------------------------
454
455 (4) Special cases (see (4) of Section A).
456
457 */
458