2 /* @(#)e_sqrt.c 1.3 95/01/18 */
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
7 * Developed at SunSoft, a Sun Microsystems, Inc. business.
8 * Permission to use, copy, modify, and distribute this
9 * software is freely granted, provided that this notice
11 * ====================================================
14 #include <sys/cdefs.h>
18 #include "math_private.h"
20 #ifdef USE_BUILTIN_SQRT
24 return (__builtin_sqrt(x));
28 * Return correctly rounded sqrt.
29 * ------------------------------------------
30 * | Use the hardware sqrt if you have one |
31 * ------------------------------------------
33 * Bit by bit method using integer arithmetic. (Slow, but portable)
35 * Scale x to y in [1,4) with even powers of 2:
36 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
37 * sqrt(x) = 2^k * sqrt(y)
38 * 2. Bit by bit computation
39 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
42 * s = 2*q , and y = 2 * ( y - q ). (1)
45 * To compute q from q , one checks whether
52 * If (2) is false, then q = q ; otherwise q = q + 2 .
55 * With some algebric manipulation, it is not difficult to see
56 * that (2) is equivalent to
61 * The advantage of (3) is that s and y can be computed by
63 * the following recurrence formula:
71 * s = s + 2 , y = y - s - 2 (5)
74 * One may easily use induction to prove (4) and (5).
75 * Note. Since the left hand side of (3) contain only i+2 bits,
76 * it does not necessary to do a full (53-bit) comparison
79 * After generating the 53 bits result, we compute one more bit.
80 * Together with the remainder, we can decide whether the
81 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
82 * (it will never equal to 1/2ulp).
83 * The rounding mode can be detected by checking whether
84 * huge + tiny is equal to huge, and whether huge - tiny is
85 * equal to huge for some floating point number "huge" and "tiny".
88 * sqrt(+-0) = +-0 ... exact
90 * sqrt(-ve) = NaN ... with invalid signal
91 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
93 * Other methods : see the appended file at the end of the program below.
97 static const double one = 1.0, tiny=1.0e-300;
103 int32_t sign = (int)0x80000000;
104 int32_t ix0,s0,q,m,t,i;
105 u_int32_t r,t1,s1,ix1,q1;
107 EXTRACT_WORDS(ix0,ix1,x);
109 /* take care of Inf and NaN */
110 if((ix0&0x7ff00000)==0x7ff00000) {
111 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
114 /* take care of zero */
116 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
118 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
122 if(m==0) { /* subnormal x */
125 ix0 |= (ix1>>11); ix1 <<= 21;
127 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
129 ix0 |= (ix1>>(32-i));
132 m -= 1023; /* unbias exponent */
133 ix0 = (ix0&0x000fffff)|0x00100000;
134 if(m&1){ /* odd m, double x to make it even */
135 ix0 += ix0 + ((ix1&sign)>>31);
138 m >>= 1; /* m = [m/2] */
140 /* generate sqrt(x) bit by bit */
141 ix0 += ix0 + ((ix1&sign)>>31);
143 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
144 r = 0x00200000; /* r = moving bit from right to left */
153 ix0 += ix0 + ((ix1&sign)>>31);
162 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
164 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
166 if (ix1 < t1) ix0 -= 1;
170 ix0 += ix0 + ((ix1&sign)>>31);
175 /* use floating add to find out rounding direction */
177 z = one-tiny; /* trigger inexact flag */
180 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
182 if (q1==(u_int32_t)0xfffffffe) q+=1;
188 ix0 = (q>>1)+0x3fe00000;
190 if ((q&1)==1) ix1 |= sign;
192 INSERT_WORDS(z,ix0,ix1);
197 #if (LDBL_MANT_DIG == 53)
198 __weak_reference(sqrt, sqrtl);
202 Other methods (use floating-point arithmetic)
204 (This is a copy of a drafted paper by Prof W. Kahan
205 and K.C. Ng, written in May, 1986)
207 Two algorithms are given here to implement sqrt(x)
208 (IEEE double precision arithmetic) in software.
209 Both supply sqrt(x) correctly rounded. The first algorithm (in
210 Section A) uses newton iterations and involves four divisions.
211 The second one uses reciproot iterations to avoid division, but
212 requires more multiplications. Both algorithms need the ability
213 to chop results of arithmetic operations instead of round them,
214 and the INEXACT flag to indicate when an arithmetic operation
215 is executed exactly with no roundoff error, all part of the
216 standard (IEEE 754-1985). The ability to perform shift, add,
217 subtract and logical AND operations upon 32-bit words is needed
218 too, though not part of the standard.
220 A. sqrt(x) by Newton Iteration
222 (1) Initial approximation
224 Let x0 and x1 be the leading and the trailing 32-bit words of
225 a floating point number x (in IEEE double format) respectively
228 ------------------------------------------------------
230 ------------------------------------------------------
231 msb lsb msb lsb ...order
234 ------------------------ ------------------------
235 x0: |s| e | f1 | x1: | f2 |
236 ------------------------ ------------------------
238 By performing shifts and subtracts on x0 and x1 (both regarded
239 as integers), we obtain an 8-bit approximation of sqrt(x) as
242 k := (x0>>1) + 0x1ff80000;
243 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
244 Here k is a 32-bit integer and T1[] is an integer array containing
245 correction terms. Now magically the floating value of y (y's
246 leading 32-bit word is y0, the value of its trailing word is 0)
247 approximates sqrt(x) to almost 8-bit.
251 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
252 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
253 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
254 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
256 (2) Iterative refinement
258 Apply Heron's rule three times to y, we have y approximates
259 sqrt(x) to within 1 ulp (Unit in the Last Place):
261 y := (y+x/y)/2 ... almost 17 sig. bits
262 y := (y+x/y)/2 ... almost 35 sig. bits
263 y := y-(y-x/y)/2 ... within 1 ulp
267 Another way to improve y to within 1 ulp is:
269 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
270 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
274 y := y + 2* ---------- ...within 1 ulp
279 This formula has one division fewer than the one above; however,
280 it requires more multiplications and additions. Also x must be
281 scaled in advance to avoid spurious overflow in evaluating the
282 expression 3y*y+x. Hence it is not recommended uless division
283 is slow. If division is very slow, then one should use the
284 reciproot algorithm given in section B.
288 By twiddling y's last bit it is possible to force y to be
289 correctly rounded according to the prevailing rounding mode
290 as follows. Let r and i be copies of the rounding mode and
291 inexact flag before entering the square root program. Also we
292 use the expression y+-ulp for the next representable floating
293 numbers (up and down) of y. Note that y+-ulp = either fixed
294 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
297 I := FALSE; ... reset INEXACT flag I
298 R := RZ; ... set rounding mode to round-toward-zero
299 z := x/y; ... chopped quotient, possibly inexact
300 If(not I) then { ... if the quotient is exact
302 I := i; ... restore inexact flag
303 R := r; ... restore rounded mode
306 z := z - ulp; ... special rounding
309 i := TRUE; ... sqrt(x) is inexact
310 If (r=RN) then z=z+ulp ... rounded-to-nearest
311 If (r=RP) then { ... round-toward-+inf
314 y := y+z; ... chopped sum
315 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
316 I := i; ... restore inexact flag
317 R := r; ... restore rounded mode
322 Square root of +inf, +-0, or NaN is itself;
323 Square root of a negative number is NaN with invalid signal.
326 B. sqrt(x) by Reciproot Iteration
328 (1) Initial approximation
330 Let x0 and x1 be the leading and the trailing 32-bit words of
331 a floating point number x (in IEEE double format) respectively
332 (see section A). By performing shifs and subtracts on x0 and y0,
333 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
335 k := 0x5fe80000 - (x0>>1);
336 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
338 Here k is a 32-bit integer and T2[] is an integer array
339 containing correction terms. Now magically the floating
340 value of y (y's leading 32-bit word is y0, the value of
341 its trailing word y1 is set to zero) approximates 1/sqrt(x)
346 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
347 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
348 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
349 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
350 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
351 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
352 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
353 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
355 (2) Iterative refinement
357 Apply Reciproot iteration three times to y and multiply the
358 result by x to get an approximation z that matches sqrt(x)
359 to about 1 ulp. To be exact, we will have
360 -1ulp < sqrt(x)-z<1.0625ulp.
362 ... set rounding mode to Round-to-nearest
363 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
364 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
365 ... special arrangement for better accuracy
366 z := x*y ... 29 bits to sqrt(x), with z*y<1
367 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
369 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
370 (a) the term z*y in the final iteration is always less than 1;
371 (b) the error in the final result is biased upward so that
372 -1 ulp < sqrt(x) - z < 1.0625 ulp
373 instead of |sqrt(x)-z|<1.03125ulp.
377 By twiddling y's last bit it is possible to force y to be
378 correctly rounded according to the prevailing rounding mode
379 as follows. Let r and i be copies of the rounding mode and
380 inexact flag before entering the square root program. Also we
381 use the expression y+-ulp for the next representable floating
382 numbers (up and down) of y. Note that y+-ulp = either fixed
383 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
386 R := RZ; ... set rounding mode to round-toward-zero
388 case RN: ... round-to-nearest
389 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
390 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
392 case RZ:case RM: ... round-to-zero or round-to--inf
393 R:=RP; ... reset rounding mod to round-to-+inf
394 if(x<z*z ... rounded up) z = z - ulp; else
395 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
397 case RP: ... round-to-+inf
398 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
399 if(x>z*z ...chopped) z = z+ulp;
403 Remark 3. The above comparisons can be done in fixed point. For
404 example, to compare x and w=z*z chopped, it suffices to compare
405 x1 and w1 (the trailing parts of x and w), regarding them as
406 two's complement integers.
408 ...Is z an exact square root?
409 To determine whether z is an exact square root of x, let z1 be the
410 trailing part of z, and also let x0 and x1 be the leading and
413 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
414 I := 1; ... Raise Inexact flag: z is not exact
416 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
417 k := z1 >> 26; ... get z's 25-th and 26-th
419 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
421 R:= r ... restore rounded mode
424 If multiplication is cheaper then the foregoing red tape, the
425 Inexact flag can be evaluated by
430 Note that z*z can overwrite I; this value must be sensed if it is
433 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
441 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
442 or even of logb(x) have the following relations:
444 -------------------------------------------------
445 bit 27,26 of z1 bit 1,0 of x1 logb(x)
446 -------------------------------------------------
452 -------------------------------------------------
454 (4) Special cases (see (4) of Section A).