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>
15 __FBSDID("$FreeBSD$");
20 #include "math_private.h"
22 #ifdef USE_BUILTIN_SQRT
26 return (__builtin_sqrt(x));
30 * Return correctly rounded sqrt.
31 * ------------------------------------------
32 * | Use the hardware sqrt if you have one |
33 * ------------------------------------------
35 * Bit by bit method using integer arithmetic. (Slow, but portable)
37 * Scale x to y in [1,4) with even powers of 2:
38 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
39 * sqrt(x) = 2^k * sqrt(y)
40 * 2. Bit by bit computation
41 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
44 * s = 2*q , and y = 2 * ( y - q ). (1)
47 * To compute q from q , one checks whether
54 * If (2) is false, then q = q ; otherwise q = q + 2 .
57 * With some algebric manipulation, it is not difficult to see
58 * that (2) is equivalent to
63 * The advantage of (3) is that s and y can be computed by
65 * the following recurrence formula:
73 * s = s + 2 , y = y - s - 2 (5)
76 * One may easily use induction to prove (4) and (5).
77 * Note. Since the left hand side of (3) contain only i+2 bits,
78 * it does not necessary to do a full (53-bit) comparison
81 * After generating the 53 bits result, we compute one more bit.
82 * Together with the remainder, we can decide whether the
83 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
84 * (it will never equal to 1/2ulp).
85 * The rounding mode can be detected by checking whether
86 * huge + tiny is equal to huge, and whether huge - tiny is
87 * equal to huge for some floating point number "huge" and "tiny".
90 * sqrt(+-0) = +-0 ... exact
92 * sqrt(-ve) = NaN ... with invalid signal
93 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
95 * Other methods : see the appended file at the end of the program below.
99 static const double one = 1.0, tiny=1.0e-300;
105 int32_t sign = (int)0x80000000;
106 int32_t ix0,s0,q,m,t,i;
107 u_int32_t r,t1,s1,ix1,q1;
109 EXTRACT_WORDS(ix0,ix1,x);
111 /* take care of Inf and NaN */
112 if((ix0&0x7ff00000)==0x7ff00000) {
113 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
116 /* take care of zero */
118 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
120 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
124 if(m==0) { /* subnormal x */
127 ix0 |= (ix1>>11); ix1 <<= 21;
129 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
131 ix0 |= (ix1>>(32-i));
134 m -= 1023; /* unbias exponent */
135 ix0 = (ix0&0x000fffff)|0x00100000;
136 if(m&1){ /* odd m, double x to make it even */
137 ix0 += ix0 + ((ix1&sign)>>31);
140 m >>= 1; /* m = [m/2] */
142 /* generate sqrt(x) bit by bit */
143 ix0 += ix0 + ((ix1&sign)>>31);
145 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
146 r = 0x00200000; /* r = moving bit from right to left */
155 ix0 += ix0 + ((ix1&sign)>>31);
164 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
166 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
168 if (ix1 < t1) ix0 -= 1;
172 ix0 += ix0 + ((ix1&sign)>>31);
177 /* use floating add to find out rounding direction */
179 z = one-tiny; /* trigger inexact flag */
182 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
184 if (q1==(u_int32_t)0xfffffffe) q+=1;
190 ix0 = (q>>1)+0x3fe00000;
192 if ((q&1)==1) ix1 |= sign;
194 INSERT_WORDS(z,ix0,ix1);
199 #if (LDBL_MANT_DIG == 53)
200 __weak_reference(sqrt, sqrtl);
204 Other methods (use floating-point arithmetic)
206 (This is a copy of a drafted paper by Prof W. Kahan
207 and K.C. Ng, written in May, 1986)
209 Two algorithms are given here to implement sqrt(x)
210 (IEEE double precision arithmetic) in software.
211 Both supply sqrt(x) correctly rounded. The first algorithm (in
212 Section A) uses newton iterations and involves four divisions.
213 The second one uses reciproot iterations to avoid division, but
214 requires more multiplications. Both algorithms need the ability
215 to chop results of arithmetic operations instead of round them,
216 and the INEXACT flag to indicate when an arithmetic operation
217 is executed exactly with no roundoff error, all part of the
218 standard (IEEE 754-1985). The ability to perform shift, add,
219 subtract and logical AND operations upon 32-bit words is needed
220 too, though not part of the standard.
222 A. sqrt(x) by Newton Iteration
224 (1) Initial approximation
226 Let x0 and x1 be the leading and the trailing 32-bit words of
227 a floating point number x (in IEEE double format) respectively
230 ------------------------------------------------------
232 ------------------------------------------------------
233 msb lsb msb lsb ...order
236 ------------------------ ------------------------
237 x0: |s| e | f1 | x1: | f2 |
238 ------------------------ ------------------------
240 By performing shifts and subtracts on x0 and x1 (both regarded
241 as integers), we obtain an 8-bit approximation of sqrt(x) as
244 k := (x0>>1) + 0x1ff80000;
245 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
246 Here k is a 32-bit integer and T1[] is an integer array containing
247 correction terms. Now magically the floating value of y (y's
248 leading 32-bit word is y0, the value of its trailing word is 0)
249 approximates sqrt(x) to almost 8-bit.
253 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
254 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
255 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
256 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
258 (2) Iterative refinement
260 Apply Heron's rule three times to y, we have y approximates
261 sqrt(x) to within 1 ulp (Unit in the Last Place):
263 y := (y+x/y)/2 ... almost 17 sig. bits
264 y := (y+x/y)/2 ... almost 35 sig. bits
265 y := y-(y-x/y)/2 ... within 1 ulp
269 Another way to improve y to within 1 ulp is:
271 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
272 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
276 y := y + 2* ---------- ...within 1 ulp
281 This formula has one division fewer than the one above; however,
282 it requires more multiplications and additions. Also x must be
283 scaled in advance to avoid spurious overflow in evaluating the
284 expression 3y*y+x. Hence it is not recommended uless division
285 is slow. If division is very slow, then one should use the
286 reciproot algorithm given in section B.
290 By twiddling y's last bit it is possible to force y to be
291 correctly rounded according to the prevailing rounding mode
292 as follows. Let r and i be copies of the rounding mode and
293 inexact flag before entering the square root program. Also we
294 use the expression y+-ulp for the next representable floating
295 numbers (up and down) of y. Note that y+-ulp = either fixed
296 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
299 I := FALSE; ... reset INEXACT flag I
300 R := RZ; ... set rounding mode to round-toward-zero
301 z := x/y; ... chopped quotient, possibly inexact
302 If(not I) then { ... if the quotient is exact
304 I := i; ... restore inexact flag
305 R := r; ... restore rounded mode
308 z := z - ulp; ... special rounding
311 i := TRUE; ... sqrt(x) is inexact
312 If (r=RN) then z=z+ulp ... rounded-to-nearest
313 If (r=RP) then { ... round-toward-+inf
316 y := y+z; ... chopped sum
317 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
318 I := i; ... restore inexact flag
319 R := r; ... restore rounded mode
324 Square root of +inf, +-0, or NaN is itself;
325 Square root of a negative number is NaN with invalid signal.
328 B. sqrt(x) by Reciproot Iteration
330 (1) Initial approximation
332 Let x0 and x1 be the leading and the trailing 32-bit words of
333 a floating point number x (in IEEE double format) respectively
334 (see section A). By performing shifs and subtracts on x0 and y0,
335 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
337 k := 0x5fe80000 - (x0>>1);
338 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
340 Here k is a 32-bit integer and T2[] is an integer array
341 containing correction terms. Now magically the floating
342 value of y (y's leading 32-bit word is y0, the value of
343 its trailing word y1 is set to zero) approximates 1/sqrt(x)
348 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
349 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
350 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
351 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
352 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
353 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
354 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
355 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
357 (2) Iterative refinement
359 Apply Reciproot iteration three times to y and multiply the
360 result by x to get an approximation z that matches sqrt(x)
361 to about 1 ulp. To be exact, we will have
362 -1ulp < sqrt(x)-z<1.0625ulp.
364 ... set rounding mode to Round-to-nearest
365 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
366 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
367 ... special arrangement for better accuracy
368 z := x*y ... 29 bits to sqrt(x), with z*y<1
369 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
371 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
372 (a) the term z*y in the final iteration is always less than 1;
373 (b) the error in the final result is biased upward so that
374 -1 ulp < sqrt(x) - z < 1.0625 ulp
375 instead of |sqrt(x)-z|<1.03125ulp.
379 By twiddling y's last bit it is possible to force y to be
380 correctly rounded according to the prevailing rounding mode
381 as follows. Let r and i be copies of the rounding mode and
382 inexact flag before entering the square root program. Also we
383 use the expression y+-ulp for the next representable floating
384 numbers (up and down) of y. Note that y+-ulp = either fixed
385 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
388 R := RZ; ... set rounding mode to round-toward-zero
390 case RN: ... round-to-nearest
391 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
392 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
394 case RZ:case RM: ... round-to-zero or round-to--inf
395 R:=RP; ... reset rounding mod to round-to-+inf
396 if(x<z*z ... rounded up) z = z - ulp; else
397 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
399 case RP: ... round-to-+inf
400 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
401 if(x>z*z ...chopped) z = z+ulp;
405 Remark 3. The above comparisons can be done in fixed point. For
406 example, to compare x and w=z*z chopped, it suffices to compare
407 x1 and w1 (the trailing parts of x and w), regarding them as
408 two's complement integers.
410 ...Is z an exact square root?
411 To determine whether z is an exact square root of x, let z1 be the
412 trailing part of z, and also let x0 and x1 be the leading and
415 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
416 I := 1; ... Raise Inexact flag: z is not exact
418 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
419 k := z1 >> 26; ... get z's 25-th and 26-th
421 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
423 R:= r ... restore rounded mode
426 If multiplication is cheaper then the foregoing red tape, the
427 Inexact flag can be evaluated by
432 Note that z*z can overwrite I; this value must be sensed if it is
435 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
443 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
444 or even of logb(x) have the following relations:
446 -------------------------------------------------
447 bit 27,26 of z1 bit 1,0 of x1 logb(x)
448 -------------------------------------------------
454 -------------------------------------------------
456 (4) Special cases (see (4) of Section A).