3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
10 * ====================================================
13 #include <sys/cdefs.h>
17 #include "math_private.h"
19 #ifdef USE_BUILTIN_SQRT
23 return (__builtin_sqrt(x));
27 * Return correctly rounded sqrt.
28 * ------------------------------------------
29 * | Use the hardware sqrt if you have one |
30 * ------------------------------------------
32 * Bit by bit method using integer arithmetic. (Slow, but portable)
34 * Scale x to y in [1,4) with even powers of 2:
35 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
36 * sqrt(x) = 2^k * sqrt(y)
37 * 2. Bit by bit computation
38 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
41 * s = 2*q , and y = 2 * ( y - q ). (1)
44 * To compute q from q , one checks whether
51 * If (2) is false, then q = q ; otherwise q = q + 2 .
54 * With some algebric manipulation, it is not difficult to see
55 * that (2) is equivalent to
60 * The advantage of (3) is that s and y can be computed by
62 * the following recurrence formula:
70 * s = s + 2 , y = y - s - 2 (5)
73 * One may easily use induction to prove (4) and (5).
74 * Note. Since the left hand side of (3) contain only i+2 bits,
75 * it does not necessary to do a full (53-bit) comparison
78 * After generating the 53 bits result, we compute one more bit.
79 * Together with the remainder, we can decide whether the
80 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
81 * (it will never equal to 1/2ulp).
82 * The rounding mode can be detected by checking whether
83 * huge + tiny is equal to huge, and whether huge - tiny is
84 * equal to huge for some floating point number "huge" and "tiny".
87 * sqrt(+-0) = +-0 ... exact
89 * sqrt(-ve) = NaN ... with invalid signal
90 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
92 * Other methods : see the appended file at the end of the program below.
96 static const double one = 1.0, tiny=1.0e-300;
102 int32_t sign = (int)0x80000000;
103 int32_t ix0,s0,q,m,t,i;
104 u_int32_t r,t1,s1,ix1,q1;
106 EXTRACT_WORDS(ix0,ix1,x);
108 /* take care of Inf and NaN */
109 if((ix0&0x7ff00000)==0x7ff00000) {
110 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
113 /* take care of zero */
115 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
117 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
121 if(m==0) { /* subnormal x */
124 ix0 |= (ix1>>11); ix1 <<= 21;
126 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
128 ix0 |= (ix1>>(32-i));
131 m -= 1023; /* unbias exponent */
132 ix0 = (ix0&0x000fffff)|0x00100000;
133 if(m&1){ /* odd m, double x to make it even */
134 ix0 += ix0 + ((ix1&sign)>>31);
137 m >>= 1; /* m = [m/2] */
139 /* generate sqrt(x) bit by bit */
140 ix0 += ix0 + ((ix1&sign)>>31);
142 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
143 r = 0x00200000; /* r = moving bit from right to left */
152 ix0 += ix0 + ((ix1&sign)>>31);
161 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
163 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
165 if (ix1 < t1) ix0 -= 1;
169 ix0 += ix0 + ((ix1&sign)>>31);
174 /* use floating add to find out rounding direction */
176 z = one-tiny; /* trigger inexact flag */
179 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
181 if (q1==(u_int32_t)0xfffffffe) q+=1;
187 ix0 = (q>>1)+0x3fe00000;
189 if ((q&1)==1) ix1 |= sign;
191 INSERT_WORDS(z,ix0,ix1);
196 #if (LDBL_MANT_DIG == 53)
197 __weak_reference(sqrt, sqrtl);
201 Other methods (use floating-point arithmetic)
203 (This is a copy of a drafted paper by Prof W. Kahan
204 and K.C. Ng, written in May, 1986)
206 Two algorithms are given here to implement sqrt(x)
207 (IEEE double precision arithmetic) in software.
208 Both supply sqrt(x) correctly rounded. The first algorithm (in
209 Section A) uses newton iterations and involves four divisions.
210 The second one uses reciproot iterations to avoid division, but
211 requires more multiplications. Both algorithms need the ability
212 to chop results of arithmetic operations instead of round them,
213 and the INEXACT flag to indicate when an arithmetic operation
214 is executed exactly with no roundoff error, all part of the
215 standard (IEEE 754-1985). The ability to perform shift, add,
216 subtract and logical AND operations upon 32-bit words is needed
217 too, though not part of the standard.
219 A. sqrt(x) by Newton Iteration
221 (1) Initial approximation
223 Let x0 and x1 be the leading and the trailing 32-bit words of
224 a floating point number x (in IEEE double format) respectively
227 ------------------------------------------------------
229 ------------------------------------------------------
230 msb lsb msb lsb ...order
233 ------------------------ ------------------------
234 x0: |s| e | f1 | x1: | f2 |
235 ------------------------ ------------------------
237 By performing shifts and subtracts on x0 and x1 (both regarded
238 as integers), we obtain an 8-bit approximation of sqrt(x) as
241 k := (x0>>1) + 0x1ff80000;
242 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
243 Here k is a 32-bit integer and T1[] is an integer array containing
244 correction terms. Now magically the floating value of y (y's
245 leading 32-bit word is y0, the value of its trailing word is 0)
246 approximates sqrt(x) to almost 8-bit.
250 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
251 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
252 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
253 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
255 (2) Iterative refinement
257 Apply Heron's rule three times to y, we have y approximates
258 sqrt(x) to within 1 ulp (Unit in the Last Place):
260 y := (y+x/y)/2 ... almost 17 sig. bits
261 y := (y+x/y)/2 ... almost 35 sig. bits
262 y := y-(y-x/y)/2 ... within 1 ulp
266 Another way to improve y to within 1 ulp is:
268 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
269 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
273 y := y + 2* ---------- ...within 1 ulp
278 This formula has one division fewer than the one above; however,
279 it requires more multiplications and additions. Also x must be
280 scaled in advance to avoid spurious overflow in evaluating the
281 expression 3y*y+x. Hence it is not recommended uless division
282 is slow. If division is very slow, then one should use the
283 reciproot algorithm given in section B.
287 By twiddling y's last bit it is possible to force y to be
288 correctly rounded according to the prevailing rounding mode
289 as follows. Let r and i be copies of the rounding mode and
290 inexact flag before entering the square root program. Also we
291 use the expression y+-ulp for the next representable floating
292 numbers (up and down) of y. Note that y+-ulp = either fixed
293 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
296 I := FALSE; ... reset INEXACT flag I
297 R := RZ; ... set rounding mode to round-toward-zero
298 z := x/y; ... chopped quotient, possibly inexact
299 If(not I) then { ... if the quotient is exact
301 I := i; ... restore inexact flag
302 R := r; ... restore rounded mode
305 z := z - ulp; ... special rounding
308 i := TRUE; ... sqrt(x) is inexact
309 If (r=RN) then z=z+ulp ... rounded-to-nearest
310 If (r=RP) then { ... round-toward-+inf
313 y := y+z; ... chopped sum
314 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
315 I := i; ... restore inexact flag
316 R := r; ... restore rounded mode
321 Square root of +inf, +-0, or NaN is itself;
322 Square root of a negative number is NaN with invalid signal.
325 B. sqrt(x) by Reciproot Iteration
327 (1) Initial approximation
329 Let x0 and x1 be the leading and the trailing 32-bit words of
330 a floating point number x (in IEEE double format) respectively
331 (see section A). By performing shifs and subtracts on x0 and y0,
332 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
334 k := 0x5fe80000 - (x0>>1);
335 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
337 Here k is a 32-bit integer and T2[] is an integer array
338 containing correction terms. Now magically the floating
339 value of y (y's leading 32-bit word is y0, the value of
340 its trailing word y1 is set to zero) approximates 1/sqrt(x)
345 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
346 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
347 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
348 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
349 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
350 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
351 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
352 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
354 (2) Iterative refinement
356 Apply Reciproot iteration three times to y and multiply the
357 result by x to get an approximation z that matches sqrt(x)
358 to about 1 ulp. To be exact, we will have
359 -1ulp < sqrt(x)-z<1.0625ulp.
361 ... set rounding mode to Round-to-nearest
362 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
363 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
364 ... special arrangement for better accuracy
365 z := x*y ... 29 bits to sqrt(x), with z*y<1
366 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
368 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
369 (a) the term z*y in the final iteration is always less than 1;
370 (b) the error in the final result is biased upward so that
371 -1 ulp < sqrt(x) - z < 1.0625 ulp
372 instead of |sqrt(x)-z|<1.03125ulp.
376 By twiddling y's last bit it is possible to force y to be
377 correctly rounded according to the prevailing rounding mode
378 as follows. Let r and i be copies of the rounding mode and
379 inexact flag before entering the square root program. Also we
380 use the expression y+-ulp for the next representable floating
381 numbers (up and down) of y. Note that y+-ulp = either fixed
382 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
385 R := RZ; ... set rounding mode to round-toward-zero
387 case RN: ... round-to-nearest
388 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
389 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
391 case RZ:case RM: ... round-to-zero or round-to--inf
392 R:=RP; ... reset rounding mod to round-to-+inf
393 if(x<z*z ... rounded up) z = z - ulp; else
394 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
396 case RP: ... round-to-+inf
397 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
398 if(x>z*z ...chopped) z = z+ulp;
402 Remark 3. The above comparisons can be done in fixed point. For
403 example, to compare x and w=z*z chopped, it suffices to compare
404 x1 and w1 (the trailing parts of x and w), regarding them as
405 two's complement integers.
407 ...Is z an exact square root?
408 To determine whether z is an exact square root of x, let z1 be the
409 trailing part of z, and also let x0 and x1 be the leading and
412 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
413 I := 1; ... Raise Inexact flag: z is not exact
415 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
416 k := z1 >> 26; ... get z's 25-th and 26-th
418 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
420 R:= r ... restore rounded mode
423 If multiplication is cheaper then the foregoing red tape, the
424 Inexact flag can be evaluated by
429 Note that z*z can overwrite I; this value must be sensed if it is
432 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
440 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
441 or even of logb(x) have the following relations:
443 -------------------------------------------------
444 bit 27,26 of z1 bit 1,0 of x1 logb(x)
445 -------------------------------------------------
451 -------------------------------------------------
453 (4) Special cases (see (4) of Section A).