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1 /* @(#)e_sqrt.c 5.1 93/09/24 */
2 /*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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 * ====================================================
13 #if defined(LIBM_SCCS) && !defined(lint)
14 static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
15 #endif
17 /* __ieee754_sqrt(x)
18 * Return correctly rounded sqrt.
19 * ------------------------------------------
20 * | Use the hardware sqrt if you have one |
21 * ------------------------------------------
22 * Method:
23 * Bit by bit method using integer arithmetic. (Slow, but portable)
24 * 1. Normalization
25 * Scale x to y in [1,4) with even powers of 2:
26 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
27 * sqrt(x) = 2^k * sqrt(y)
28 * 2. Bit by bit computation
29 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
30 * i 0
31 * i+1 2
32 * s = 2*q , and y = 2 * ( y - q ). (1)
33 * i i i i
35 * To compute q from q , one checks whether
36 * i+1 i
38 * -(i+1) 2
39 * (q + 2 ) <= y. (2)
40 * i
41 * -(i+1)
42 * If (2) is false, then q = q ; otherwise q = q + 2 .
43 * i+1 i i+1 i
45 * With some algebraic manipulation, it is not difficult to see
46 * that (2) is equivalent to
47 * -(i+1)
48 * s + 2 <= y (3)
49 * i i
51 * The advantage of (3) is that s and y can be computed by
52 * i i
53 * the following recurrence formula:
54 * if (3) is false
56 * s = s , y = y ; (4)
57 * i+1 i i+1 i
59 * otherwise,
60 * -i -(i+1)
61 * s = s + 2 , y = y - s - 2 (5)
62 * i+1 i i+1 i i
64 * One may easily use induction to prove (4) and (5).
65 * Note. Since the left hand side of (3) contain only i+2 bits,
66 * it does not necessary to do a full (53-bit) comparison
67 * in (3).
68 * 3. Final rounding
69 * After generating the 53 bits result, we compute one more bit.
70 * Together with the remainder, we can decide whether the
71 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
72 * (it will never equal to 1/2ulp).
73 * The rounding mode can be detected by checking whether
74 * huge + tiny is equal to huge, and whether huge - tiny is
75 * equal to huge for some floating point number "huge" and "tiny".
77 * Special cases:
78 * sqrt(+-0) = +-0 ... exact
79 * sqrt(inf) = inf
80 * sqrt(-ve) = NaN ... with invalid signal
81 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
83 * Other methods : see the appended file at the end of the program below.
84 *---------------
87 #include "math.h"
88 #include "math_private.h"
90 #ifdef __STDC__
91 static const double one = 1.0, tiny=1.0e-300;
92 #else
93 static double one = 1.0, tiny=1.0e-300;
94 #endif
96 #ifdef __STDC__
97 double __ieee754_sqrt(double x)
98 #else
99 double __ieee754_sqrt(x)
100 double x;
101 #endif
103 double z;
104 int32_t sign = (int)0x80000000;
105 int32_t ix0,s0,q,m,t,i;
106 u_int32_t r,t1,s1,ix1,q1;
108 EXTRACT_WORDS(ix0,ix1,x);
110 /* take care of Inf and NaN */
111 if((ix0&0x7ff00000)==0x7ff00000) {
112 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
113 sqrt(-inf)=sNaN */
115 /* take care of zero */
116 if(ix0<=0) {
117 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
118 else if(ix0<0)
119 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
121 /* normalize x */
122 m = (ix0>>20);
123 if(m==0) { /* subnormal x */
124 while(ix0==0) {
125 m -= 21;
126 ix0 |= (ix1>>11); ix1 <<= 21;
128 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
129 m -= i-1;
130 ix0 |= (ix1>>(32-i));
131 ix1 <<= i;
133 m -= 1023; /* unbias exponent */
134 ix0 = (ix0&0x000fffff)|0x00100000;
135 if(m&1){ /* odd m, double x to make it even */
136 ix0 += ix0 + ((ix1&sign)>>31);
137 ix1 += ix1;
139 m >>= 1; /* m = [m/2] */
141 /* generate sqrt(x) bit by bit */
142 ix0 += ix0 + ((ix1&sign)>>31);
143 ix1 += ix1;
144 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
145 r = 0x00200000; /* r = moving bit from right to left */
147 while(r!=0) {
148 t = s0+r;
149 if(t<=ix0) {
150 s0 = t+r;
151 ix0 -= t;
152 q += r;
154 ix0 += ix0 + ((ix1&sign)>>31);
155 ix1 += ix1;
156 r>>=1;
159 r = sign;
160 while(r!=0) {
161 t1 = s1+r;
162 t = s0;
163 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
164 s1 = t1+r;
165 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
166 ix0 -= t;
167 if (ix1 < t1) ix0 -= 1;
168 ix1 -= t1;
169 q1 += r;
171 ix0 += ix0 + ((ix1&sign)>>31);
172 ix1 += ix1;
173 r>>=1;
176 /* use floating add to find out rounding direction */
177 if((ix0|ix1)!=0) {
178 z = one-tiny; /* trigger inexact flag */
179 if (z>=one) {
180 z = one+tiny;
181 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
182 else if (z>one) {
183 if (q1==(u_int32_t)0xfffffffe) q+=1;
184 q1+=2;
185 } else
186 q1 += (q1&1);
189 ix0 = (q>>1)+0x3fe00000;
190 ix1 = q1>>1;
191 if ((q&1)==1) ix1 |= sign;
192 ix0 += (m <<20);
193 INSERT_WORDS(z,ix0,ix1);
194 return z;
198 Other methods (use floating-point arithmetic)
199 -------------
200 (This is a copy of a drafted paper by Prof W. Kahan
201 and K.C. Ng, written in May, 1986)
203 Two algorithms are given here to implement sqrt(x)
204 (IEEE double precision arithmetic) in software.
205 Both supply sqrt(x) correctly rounded. The first algorithm (in
206 Section A) uses newton iterations and involves four divisions.
207 The second one uses reciproot iterations to avoid division, but
208 requires more multiplications. Both algorithms need the ability
209 to chop results of arithmetic operations instead of round them,
210 and the INEXACT flag to indicate when an arithmetic operation
211 is executed exactly with no roundoff error, all part of the
212 standard (IEEE 754-1985). The ability to perform shift, add,
213 subtract and logical AND operations upon 32-bit words is needed
214 too, though not part of the standard.
216 A. sqrt(x) by Newton Iteration
218 (1) Initial approximation
220 Let x0 and x1 be the leading and the trailing 32-bit words of
221 a floating point number x (in IEEE double format) respectively
223 1 11 52 ...widths
224 ------------------------------------------------------
225 x: |s| e | f |
226 ------------------------------------------------------
227 msb lsb msb lsb ...order
230 ------------------------ ------------------------
231 x0: |s| e | f1 | x1: | f2 |
232 ------------------------ ------------------------
234 By performing shifts and subtracts on x0 and x1 (both regarded
235 as integers), we obtain an 8-bit approximation of sqrt(x) as
236 follows.
238 k := (x0>>1) + 0x1ff80000;
239 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
240 Here k is a 32-bit integer and T1[] is an integer array containing
241 correction terms. Now magically the floating value of y (y's
242 leading 32-bit word is y0, the value of its trailing word is 0)
243 approximates sqrt(x) to almost 8-bit.
245 Value of T1:
246 static int T1[32]= {
247 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
248 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
249 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
250 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
252 (2) Iterative refinement
254 Apply Heron's rule three times to y, we have y approximates
255 sqrt(x) to within 1 ulp (Unit in the Last Place):
257 y := (y+x/y)/2 ... almost 17 sig. bits
258 y := (y+x/y)/2 ... almost 35 sig. bits
259 y := y-(y-x/y)/2 ... within 1 ulp
262 Remark 1.
263 Another way to improve y to within 1 ulp is:
265 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
266 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
269 (x-y )*y
270 y := y + 2* ---------- ...within 1 ulp
272 3y + x
275 This formula has one division fewer than the one above; however,
276 it requires more multiplications and additions. Also x must be
277 scaled in advance to avoid spurious overflow in evaluating the
278 expression 3y*y+x. Hence it is not recommended uless division
279 is slow. If division is very slow, then one should use the
280 reciproot algorithm given in section B.
282 (3) Final adjustment
284 By twiddling y's last bit it is possible to force y to be
285 correctly rounded according to the prevailing rounding mode
286 as follows. Let r and i be copies of the rounding mode and
287 inexact flag before entering the square root program. Also we
288 use the expression y+-ulp for the next representable floating
289 numbers (up and down) of y. Note that y+-ulp = either fixed
290 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
291 mode.
293 I := FALSE; ... reset INEXACT flag I
294 R := RZ; ... set rounding mode to round-toward-zero
295 z := x/y; ... chopped quotient, possibly inexact
296 If(not I) then { ... if the quotient is exact
297 if(z=y) {
298 I := i; ... restore inexact flag
299 R := r; ... restore rounded mode
300 return sqrt(x):=y.
301 } else {
302 z := z - ulp; ... special rounding
305 i := TRUE; ... sqrt(x) is inexact
306 If (r=RN) then z=z+ulp ... rounded-to-nearest
307 If (r=RP) then { ... round-toward-+inf
308 y = y+ulp; z=z+ulp;
310 y := y+z; ... chopped sum
311 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
312 I := i; ... restore inexact flag
313 R := r; ... restore rounded mode
314 return sqrt(x):=y.
316 (4) Special cases
318 Square root of +inf, +-0, or NaN is itself;
319 Square root of a negative number is NaN with invalid signal.
322 B. sqrt(x) by Reciproot Iteration
324 (1) Initial approximation
326 Let x0 and x1 be the leading and the trailing 32-bit words of
327 a floating point number x (in IEEE double format) respectively
328 (see section A). By performing shifs and subtracts on x0 and y0,
329 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
331 k := 0x5fe80000 - (x0>>1);
332 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
334 Here k is a 32-bit integer and T2[] is an integer array
335 containing correction terms. Now magically the floating
336 value of y (y's leading 32-bit word is y0, the value of
337 its trailing word y1 is set to zero) approximates 1/sqrt(x)
338 to almost 7.8-bit.
340 Value of T2:
341 static int T2[64]= {
342 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
343 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
344 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
345 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
346 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
347 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
348 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
349 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
351 (2) Iterative refinement
353 Apply Reciproot iteration three times to y and multiply the
354 result by x to get an approximation z that matches sqrt(x)
355 to about 1 ulp. To be exact, we will have
356 -1ulp < sqrt(x)-z<1.0625ulp.
358 ... set rounding mode to Round-to-nearest
359 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
360 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
361 ... special arrangement for better accuracy
362 z := x*y ... 29 bits to sqrt(x), with z*y<1
363 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
365 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
366 (a) the term z*y in the final iteration is always less than 1;
367 (b) the error in the final result is biased upward so that
368 -1 ulp < sqrt(x) - z < 1.0625 ulp
369 instead of |sqrt(x)-z|<1.03125ulp.
371 (3) Final adjustment
373 By twiddling y's last bit it is possible to force y to be
374 correctly rounded according to the prevailing rounding mode
375 as follows. Let r and i be copies of the rounding mode and
376 inexact flag before entering the square root program. Also we
377 use the expression y+-ulp for the next representable floating
378 numbers (up and down) of y. Note that y+-ulp = either fixed
379 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
380 mode.
382 R := RZ; ... set rounding mode to round-toward-zero
383 switch(r) {
384 case RN: ... round-to-nearest
385 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
386 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
387 break;
388 case RZ:case RM: ... round-to-zero or round-to--inf
389 R:=RP; ... reset rounding mod to round-to-+inf
390 if(x<z*z ... rounded up) z = z - ulp; else
391 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
392 break;
393 case RP: ... round-to-+inf
394 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
395 if(x>z*z ...chopped) z = z+ulp;
396 break;
399 Remark 3. The above comparisons can be done in fixed point. For
400 example, to compare x and w=z*z chopped, it suffices to compare
401 x1 and w1 (the trailing parts of x and w), regarding them as
402 two's complement integers.
404 ...Is z an exact square root?
405 To determine whether z is an exact square root of x, let z1 be the
406 trailing part of z, and also let x0 and x1 be the leading and
407 trailing parts of x.
409 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
410 I := 1; ... Raise Inexact flag: z is not exact
411 else {
412 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
413 k := z1 >> 26; ... get z's 25-th and 26-th
414 fraction bits
415 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
417 R:= r ... restore rounded mode
418 return sqrt(x):=z.
420 If multiplication is cheaper then the foregoing red tape, the
421 Inexact flag can be evaluated by
423 I := i;
424 I := (z*z!=x) or I.
426 Note that z*z can overwrite I; this value must be sensed if it is
427 True.
429 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
430 zero.
432 --------------------
433 z1: | f2 |
434 --------------------
435 bit 31 bit 0
437 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
438 or even of logb(x) have the following relations:
440 -------------------------------------------------
441 bit 27,26 of z1 bit 1,0 of x1 logb(x)
442 -------------------------------------------------
443 00 00 odd and even
444 01 01 even
445 10 10 odd
446 10 00 even
447 11 01 even
448 -------------------------------------------------
450 (4) Special cases (see (4) of Section A).