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.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 __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 /*
185 Other methods (use floating-point arithmetic)
186 -------------
187 (This is a copy of a drafted paper by Prof W. Kahan
188 and K.C. Ng, written in May, 1986)
189
190 Two algorithms are given here to implement sqrt(x)
191 (IEEE double precision arithmetic) in software.
192 Both supply sqrt(x) correctly rounded. The first algorithm (in
193 Section A) uses newton iterations and involves four divisions.
194 The second one uses reciproot iterations to avoid division, but
195 requires more multiplications. Both algorithms need the ability
196 to chop results of arithmetic operations instead of round them,
197 and the INEXACT flag to indicate when an arithmetic operation
198 is executed exactly with no roundoff error, all part of the
199 standard (IEEE 754-1985). The ability to perform shift, add,
200 subtract and logical AND operations upon 32-bit words is needed
201 too, though not part of the standard.
202
203 A. sqrt(x) by Newton Iteration
204
205 (1) Initial approximation
206
207 Let x0 and x1 be the leading and the trailing 32-bit words of
208 a floating point number x (in IEEE double format) respectively
209
210 1 11 52 ...widths
211 ------------------------------------------------------
212 x: |s| e | f |
213 ------------------------------------------------------
214 msb lsb msb lsb ...order
215
216
217 ------------------------ ------------------------
218 x0: |s| e | f1 | x1: | f2 |
219 ------------------------ ------------------------
220
221 By performing shifts and subtracts on x0 and x1 (both regarded
222 as integers), we obtain an 8-bit approximation of sqrt(x) as
223 follows.
224
225 k := (x0>>1) + 0x1ff80000;
226 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
227 Here k is a 32-bit integer and T1[] is an integer array containing
228 correction terms. Now magically the floating value of y (y's
229 leading 32-bit word is y0, the value of its trailing word is 0)
230 approximates sqrt(x) to almost 8-bit.
231
232 Value of T1:
233 static int T1[32]= {
234 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
235 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
236 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
237 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
238
239 (2) Iterative refinement
240
241 Apply Heron's rule three times to y, we have y approximates
242 sqrt(x) to within 1 ulp (Unit in the Last Place):
243
244 y := (y+x/y)/2 ... almost 17 sig. bits
245 y := (y+x/y)/2 ... almost 35 sig. bits
246 y := y-(y-x/y)/2 ... within 1 ulp
247
248
249 Remark 1.
250 Another way to improve y to within 1 ulp is:
251
252 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
253 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
254
255 2
256 (x-y )*y
257 y := y + 2* ---------- ...within 1 ulp
258 2
259 3y + x
260
261
262 This formula has one division fewer than the one above; however,
263 it requires more multiplications and additions. Also x must be
264 scaled in advance to avoid spurious overflow in evaluating the
265 expression 3y*y+x. Hence it is not recommended uless division
266 is slow. If division is very slow, then one should use the
267 reciproot algorithm given in section B.
268
269 (3) Final adjustment
270
271 By twiddling y's last bit it is possible to force y to be
272 correctly rounded according to the prevailing rounding mode
273 as follows. Let r and i be copies of the rounding mode and
274 inexact flag before entering the square root program. Also we
275 use the expression y+-ulp for the next representable floating
276 numbers (up and down) of y. Note that y+-ulp = either fixed
277 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
278 mode.
279
280 I := FALSE; ... reset INEXACT flag I
281 R := RZ; ... set rounding mode to round-toward-zero
282 z := x/y; ... chopped quotient, possibly inexact
283 If(not I) then { ... if the quotient is exact
284 if(z=y) {
285 I := i; ... restore inexact flag
286 R := r; ... restore rounded mode
287 return sqrt(x):=y.
288 } else {
289 z := z - ulp; ... special rounding
290 }
291 }
292 i := TRUE; ... sqrt(x) is inexact
293 If (r=RN) then z=z+ulp ... rounded-to-nearest
294 If (r=RP) then { ... round-toward-+inf
295 y = y+ulp; z=z+ulp;
296 }
297 y := y+z; ... chopped sum
298 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
299 I := i; ... restore inexact flag
300 R := r; ... restore rounded mode
301 return sqrt(x):=y.
302
303 (4) Special cases
304
305 Square root of +inf, +-0, or NaN is itself;
306 Square root of a negative number is NaN with invalid signal.
307
308
309 B. sqrt(x) by Reciproot Iteration
310
311 (1) Initial approximation
312
313 Let x0 and x1 be the leading and the trailing 32-bit words of
314 a floating point number x (in IEEE double format) respectively
315 (see section A). By performing shifs and subtracts on x0 and y0,
316 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
317
318 k := 0x5fe80000 - (x0>>1);
319 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
320
321 Here k is a 32-bit integer and T2[] is an integer array
322 containing correction terms. Now magically the floating
323 value of y (y's leading 32-bit word is y0, the value of
324 its trailing word y1 is set to zero) approximates 1/sqrt(x)
325 to almost 7.8-bit.
326
327 Value of T2:
328 static int T2[64]= {
329 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
330 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
331 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
332 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
333 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
334 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
335 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
336 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
337
338 (2) Iterative refinement
339
340 Apply Reciproot iteration three times to y and multiply the
341 result by x to get an approximation z that matches sqrt(x)
342 to about 1 ulp. To be exact, we will have
343 -1ulp < sqrt(x)-z<1.0625ulp.
344
345 ... set rounding mode to Round-to-nearest
346 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
347 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
348 ... special arrangement for better accuracy
349 z := x*y ... 29 bits to sqrt(x), with z*y<1
350 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
351
352 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
353 (a) the term z*y in the final iteration is always less than 1;
354 (b) the error in the final result is biased upward so that
355 -1 ulp < sqrt(x) - z < 1.0625 ulp
356 instead of |sqrt(x)-z|<1.03125ulp.
357
358 (3) Final adjustment
359
360 By twiddling y's last bit it is possible to force y to be
361 correctly rounded according to the prevailing rounding mode
362 as follows. Let r and i be copies of the rounding mode and
363 inexact flag before entering the square root program. Also we
364 use the expression y+-ulp for the next representable floating
365 numbers (up and down) of y. Note that y+-ulp = either fixed
366 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
367 mode.
368
369 R := RZ; ... set rounding mode to round-toward-zero
370 switch(r) {
371 case RN: ... round-to-nearest
372 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
373 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
374 break;
375 case RZ:case RM: ... round-to-zero or round-to--inf
376 R:=RP; ... reset rounding mod to round-to-+inf
377 if(x<z*z ... rounded up) z = z - ulp; else
378 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
379 break;
380 case RP: ... round-to-+inf
381 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
382 if(x>z*z ...chopped) z = z+ulp;
383 break;
384 }
385
386 Remark 3. The above comparisons can be done in fixed point. For
387 example, to compare x and w=z*z chopped, it suffices to compare
388 x1 and w1 (the trailing parts of x and w), regarding them as
389 two's complement integers.
390
391 ...Is z an exact square root?
392 To determine whether z is an exact square root of x, let z1 be the
393 trailing part of z, and also let x0 and x1 be the leading and
394 trailing parts of x.
395
396 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
397 I := 1; ... Raise Inexact flag: z is not exact
398 else {
399 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
400 k := z1 >> 26; ... get z's 25-th and 26-th
401 fraction bits
402 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
403 }
404 R:= r ... restore rounded mode
405 return sqrt(x):=z.
406
407 If multiplication is cheaper then the foregoing red tape, the
408 Inexact flag can be evaluated by
409
410 I := i;
411 I := (z*z!=x) or I.
412
413 Note that z*z can overwrite I; this value must be sensed if it is
414 True.
415
416 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
417 zero.
418
419 --------------------
420 z1: | f2 |
421 --------------------
422 bit 31 bit 0
423
424 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
425 or even of logb(x) have the following relations:
426
427 -------------------------------------------------
428 bit 27,26 of z1 bit 1,0 of x1 logb(x)
429 -------------------------------------------------
430 00 00 odd and even
431 01 01 even
432 10 10 odd
433 10 00 even
434 11 01 even
435 -------------------------------------------------
436
437 (4) Special cases (see (4) of Section A).
438
439 */
440