1 /* @(#)e_sqrt.c 5.1 93/09/24 */
2 /*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunPro, 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 /* sqrt(x)
14 * Return correctly rounded sqrt.
15 * ------------------------------------------
16 * | Use the hardware sqrt if you have one |
17 * ------------------------------------------
18 * Method:
19 * Bit by bit method using integer arithmetic. (Slow, but portable)
20 * 1. Normalization
21 * Scale x to y in [1,4) with even powers of 2:
22 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
23 * sqrt(x) = 2^k * sqrt(y)
24 * 2. Bit by bit computation
25 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
26 * i 0
27 * i+1 2
28 * s = 2*q , and y = 2 * ( y - q ). (1)
29 * i i i i
30 *
31 * To compute q from q , one checks whether
32 * i+1 i
33 *
34 * -(i+1) 2
35 * (q + 2 ) <= y. (2)
36 * i
37 * -(i+1)
38 * If (2) is false, then q = q ; otherwise q = q + 2 .
39 * i+1 i i+1 i
40 *
41 * With some algebric manipulation, it is not difficult to see
42 * that (2) is equivalent to
43 * -(i+1)
44 * s + 2 <= y (3)
45 * i i
46 *
47 * The advantage of (3) is that s and y can be computed by
48 * i i
49 * the following recurrence formula:
50 * if (3) is false
51 *
52 * s = s , y = y ; (4)
53 * i+1 i i+1 i
54 *
55 * otherwise,
56 * -i -(i+1)
57 * s = s + 2 , y = y - s - 2 (5)
58 * i+1 i i+1 i i
59 *
60 * One may easily use induction to prove (4) and (5).
61 * Note. Since the left hand side of (3) contain only i+2 bits,
62 * it does not necessary to do a full (53-bit) comparison
63 * in (3).
64 * 3. Final rounding
65 * After generating the 53 bits result, we compute one more bit.
66 * Together with the remainder, we can decide whether the
67 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
68 * (it will never equal to 1/2ulp).
69 * The rounding mode can be detected by checking whether
70 * huge + tiny is equal to huge, and whether huge - tiny is
71 * equal to huge for some floating point number "huge" and "tiny".
72 *
73 * Special cases:
74 * sqrt(+-0) = +-0 ... exact
75 * sqrt(inf) = inf
76 * sqrt(-ve) = NaN ... with invalid signal
77 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
78 *
79 * Other methods : see the appended file at the end of the program below.
80 *---------------
81 */
82
83 #include <float.h>
84 #include <math.h>
85
86 #include "math_private.h"
87
88 static const double one = 1.0, tiny=1.0e-300;
89
90 double
sqrt(double x)91 sqrt(double x)
92 {
93 double z;
94 int32_t sign = (int)0x80000000;
95 int32_t ix0,s0,q,m,t,i;
96 u_int32_t r,t1,s1,ix1,q1;
97
98 EXTRACT_WORDS(ix0,ix1,x);
99
100 /* take care of Inf and NaN */
101 if((ix0&0x7ff00000)==0x7ff00000) {
102 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
103 sqrt(-inf)=sNaN */
104 }
105 /* take care of zero */
106 if(ix0<=0) {
107 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
108 else if(ix0<0)
109 return (x-x)/(x-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); ix1 <<= 21;
117 }
118 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
119 m -= i-1;
120 ix0 |= (ix1>>(32-i));
121 ix1 <<= i;
122 }
123 m -= 1023; /* unbias exponent */
124 ix0 = (ix0&0x000fffff)|0x00100000;
125 if(m&1){ /* odd m, double x to make it even */
126 ix0 += ix0 + ((ix1&sign)>>31);
127 ix1 += ix1;
128 }
129 m >>= 1; /* m = [m/2] */
130
131 /* generate sqrt(x) bit by bit */
132 ix0 += ix0 + ((ix1&sign)>>31);
133 ix1 += ix1;
134 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
135 r = 0x00200000; /* r = moving bit from right to left */
136
137 while(r!=0) {
138 t = s0+r;
139 if(t<=ix0) {
140 s0 = t+r;
141 ix0 -= t;
142 q += r;
143 }
144 ix0 += ix0 + ((ix1&sign)>>31);
145 ix1 += ix1;
146 r>>=1;
147 }
148
149 r = sign;
150 while(r!=0) {
151 t1 = s1+r;
152 t = s0;
153 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
154 s1 = t1+r;
155 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
156 ix0 -= t;
157 if (ix1 < t1) ix0 -= 1;
158 ix1 -= t1;
159 q1 += r;
160 }
161 ix0 += ix0 + ((ix1&sign)>>31);
162 ix1 += ix1;
163 r>>=1;
164 }
165
166 /* use floating add to find out rounding direction */
167 if((ix0|ix1)!=0) {
168 z = one-tiny; /* trigger inexact flag */
169 if (z>=one) {
170 z = one+tiny;
171 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
172 else if (z>one) {
173 if (q1==(u_int32_t)0xfffffffe) q+=1;
174 q1+=2;
175 } else
176 q1 += (q1&1);
177 }
178 }
179 ix0 = (q>>1)+0x3fe00000;
180 ix1 = q1>>1;
181 if ((q&1)==1) ix1 |= sign;
182 ix0 += (m <<20);
183 INSERT_WORDS(z,ix0,ix1);
184 return z;
185 }
186
187 /*
188 Other methods (use floating-point arithmetic)
189 -------------
190 (This is a copy of a drafted paper by Prof W. Kahan
191 and K.C. Ng, written in May, 1986)
192
193 Two algorithms are given here to implement sqrt(x)
194 (IEEE double precision arithmetic) in software.
195 Both supply sqrt(x) correctly rounded. The first algorithm (in
196 Section A) uses newton iterations and involves four divisions.
197 The second one uses reciproot iterations to avoid division, but
198 requires more multiplications. Both algorithms need the ability
199 to chop results of arithmetic operations instead of round them,
200 and the INEXACT flag to indicate when an arithmetic operation
201 is executed exactly with no roundoff error, all part of the
202 standard (IEEE 754-1985). The ability to perform shift, add,
203 subtract and logical AND operations upon 32-bit words is needed
204 too, though not part of the standard.
205
206 A. sqrt(x) by Newton Iteration
207
208 (1) Initial approximation
209
210 Let x0 and x1 be the leading and the trailing 32-bit words of
211 a floating point number x (in IEEE double format) respectively
212
213 1 11 52 ...widths
214 ------------------------------------------------------
215 x: |s| e | f |
216 ------------------------------------------------------
217 msb lsb msb lsb ...order
218
219
220 ------------------------ ------------------------
221 x0: |s| e | f1 | x1: | f2 |
222 ------------------------ ------------------------
223
224 By performing shifts and subtracts on x0 and x1 (both regarded
225 as integers), we obtain an 8-bit approximation of sqrt(x) as
226 follows.
227
228 k := (x0>>1) + 0x1ff80000;
229 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
230 Here k is a 32-bit integer and T1[] is an integer array containing
231 correction terms. Now magically the floating value of y (y's
232 leading 32-bit word is y0, the value of its trailing word is 0)
233 approximates sqrt(x) to almost 8-bit.
234
235 Value of T1:
236 static int T1[32]= {
237 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
238 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
239 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
240 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
241
242 (2) Iterative refinement
243
244 Apply Heron's rule three times to y, we have y approximates
245 sqrt(x) to within 1 ulp (Unit in the Last Place):
246
247 y := (y+x/y)/2 ... almost 17 sig. bits
248 y := (y+x/y)/2 ... almost 35 sig. bits
249 y := y-(y-x/y)/2 ... within 1 ulp
250
251
252 Remark 1.
253 Another way to improve y to within 1 ulp is:
254
255 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
256 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
257
258 2
259 (x-y )*y
260 y := y + 2* ---------- ...within 1 ulp
261 2
262 3y + x
263
264
265 This formula has one division fewer than the one above; however,
266 it requires more multiplications and additions. Also x must be
267 scaled in advance to avoid spurious overflow in evaluating the
268 expression 3y*y+x. Hence it is not recommended uless division
269 is slow. If division is very slow, then one should use the
270 reciproot algorithm given in section B.
271
272 (3) Final adjustment
273
274 By twiddling y's last bit it is possible to force y to be
275 correctly rounded according to the prevailing rounding mode
276 as follows. Let r and i be copies of the rounding mode and
277 inexact flag before entering the square root program. Also we
278 use the expression y+-ulp for the next representable floating
279 numbers (up and down) of y. Note that y+-ulp = either fixed
280 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
281 mode.
282
283 I := FALSE; ... reset INEXACT flag I
284 R := RZ; ... set rounding mode to round-toward-zero
285 z := x/y; ... chopped quotient, possibly inexact
286 If(not I) then { ... if the quotient is exact
287 if(z=y) {
288 I := i; ... restore inexact flag
289 R := r; ... restore rounded mode
290 return sqrt(x):=y.
291 } else {
292 z := z - ulp; ... special rounding
293 }
294 }
295 i := TRUE; ... sqrt(x) is inexact
296 If (r=RN) then z=z+ulp ... rounded-to-nearest
297 If (r=RP) then { ... round-toward-+inf
298 y = y+ulp; z=z+ulp;
299 }
300 y := y+z; ... chopped sum
301 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
302 I := i; ... restore inexact flag
303 R := r; ... restore rounded mode
304 return sqrt(x):=y.
305
306 (4) Special cases
307
308 Square root of +inf, +-0, or NaN is itself;
309 Square root of a negative number is NaN with invalid signal.
310
311
312 B. sqrt(x) by Reciproot Iteration
313
314 (1) Initial approximation
315
316 Let x0 and x1 be the leading and the trailing 32-bit words of
317 a floating point number x (in IEEE double format) respectively
318 (see section A). By performing shifs and subtracts on x0 and y0,
319 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
320
321 k := 0x5fe80000 - (x0>>1);
322 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
323
324 Here k is a 32-bit integer and T2[] is an integer array
325 containing correction terms. Now magically the floating
326 value of y (y's leading 32-bit word is y0, the value of
327 its trailing word y1 is set to zero) approximates 1/sqrt(x)
328 to almost 7.8-bit.
329
330 Value of T2:
331 static int T2[64]= {
332 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
333 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
334 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
335 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
336 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
337 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
338 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
339 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
340
341 (2) Iterative refinement
342
343 Apply Reciproot iteration three times to y and multiply the
344 result by x to get an approximation z that matches sqrt(x)
345 to about 1 ulp. To be exact, we will have
346 -1ulp < sqrt(x)-z<1.0625ulp.
347
348 ... set rounding mode to Round-to-nearest
349 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
350 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
351 ... special arrangement for better accuracy
352 z := x*y ... 29 bits to sqrt(x), with z*y<1
353 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
354
355 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
356 (a) the term z*y in the final iteration is always less than 1;
357 (b) the error in the final result is biased upward so that
358 -1 ulp < sqrt(x) - z < 1.0625 ulp
359 instead of |sqrt(x)-z|<1.03125ulp.
360
361 (3) Final adjustment
362
363 By twiddling y's last bit it is possible to force y to be
364 correctly rounded according to the prevailing rounding mode
365 as follows. Let r and i be copies of the rounding mode and
366 inexact flag before entering the square root program. Also we
367 use the expression y+-ulp for the next representable floating
368 numbers (up and down) of y. Note that y+-ulp = either fixed
369 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
370 mode.
371
372 R := RZ; ... set rounding mode to round-toward-zero
373 switch(r) {
374 case RN: ... round-to-nearest
375 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
376 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
377 break;
378 case RZ:case RM: ... round-to-zero or round-to--inf
379 R:=RP; ... reset rounding mod to round-to-+inf
380 if(x<z*z ... rounded up) z = z - ulp; else
381 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
382 break;
383 case RP: ... round-to-+inf
384 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
385 if(x>z*z ...chopped) z = z+ulp;
386 break;
387 }
388
389 Remark 3. The above comparisons can be done in fixed point. For
390 example, to compare x and w=z*z chopped, it suffices to compare
391 x1 and w1 (the trailing parts of x and w), regarding them as
392 two's complement integers.
393
394 ...Is z an exact square root?
395 To determine whether z is an exact square root of x, let z1 be the
396 trailing part of z, and also let x0 and x1 be the leading and
397 trailing parts of x.
398
399 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
400 I := 1; ... Raise Inexact flag: z is not exact
401 else {
402 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
403 k := z1 >> 26; ... get z's 25-th and 26-th
404 fraction bits
405 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
406 }
407 R:= r ... restore rounded mode
408 return sqrt(x):=z.
409
410 If multiplication is cheaper then the foregoing red tape, the
411 Inexact flag can be evaluated by
412
413 I := i;
414 I := (z*z!=x) or I.
415
416 Note that z*z can overwrite I; this value must be sensed if it is
417 True.
418
419 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
420 zero.
421
422 --------------------
423 z1: | f2 |
424 --------------------
425 bit 31 bit 0
426
427 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
428 or even of logb(x) have the following relations:
429
430 -------------------------------------------------
431 bit 27,26 of z1 bit 1,0 of x1 logb(x)
432 -------------------------------------------------
433 00 00 odd and even
434 01 01 even
435 10 10 odd
436 10 00 even
437 11 01 even
438 -------------------------------------------------
439
440 (4) Special cases (see (4) of Section A).
441
442 */
443
444 #if LDBL_MANT_DIG == DBL_MANT_DIG
445 __strong_alias(sqrtl, sqrt);
446 #endif /* LDBL_MANT_DIG == DBL_MANT_DIG */
447