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 * ====================================================
16 * Return correctly rounded sqrt.
17 * ------------------------------------------
18 * | Use the hardware sqrt if you have one |
19 * ------------------------------------------
21 * Bit by bit method using integer arithmetic. (Slow, but portable)
23 * Scale x to y in [1,4) with even powers of 2:
24 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
25 * sqrt(x) = 2^k * sqrt(y)
26 * 2. Bit by bit computation
27 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
30 * s = 2*q , and y = 2 * ( y - q ). (1)
33 * To compute q from q , one checks whether
40 * If (2) is false, then q = q ; otherwise q = q + 2 .
43 * With some algebric manipulation, it is not difficult to see
44 * that (2) is equivalent to
49 * The advantage of (3) is that s and y can be computed by
51 * the following recurrence formula:
59 * s = s + 2 , y = y - s - 2 (5)
62 * One may easily use induction to prove (4) and (5).
63 * Note. Since the left hand side of (3) contain only i+2 bits,
64 * it does not necessary to do a full (53-bit) comparison
67 * After generating the 53 bits result, we compute one more bit.
68 * Together with the remainder, we can decide whether the
69 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
70 * (it will never equal to 1/2ulp).
71 * The rounding mode can be detected by checking whether
72 * huge + tiny is equal to huge, and whether huge - tiny is
73 * equal to huge for some floating point number "huge" and "tiny".
76 * sqrt(+-0) = +-0 ... exact
78 * sqrt(-ve) = NaN ... with invalid signal
79 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
81 * Other methods : see the appended file at the end of the program below.
88 #include "math_private.h"
90 static const double one = 1.0, tiny=1.0e-300;
93 __ieee754_sqrt(double x)
96 int32_t sign = (int)0x80000000;
97 int32_t ix0,s0,q,m,t,i;
98 u_int32_t r,t1,s1,ix1,q1;
100 EXTRACT_WORDS(ix0,ix1,x);
102 /* take care of Inf and NaN */
103 if((ix0&0x7ff00000)==0x7ff00000) {
104 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
107 /* take care of zero */
109 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
111 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
115 if(m==0) { /* subnormal x */
118 ix0 |= (ix1>>11); ix1 <<= 21;
120 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
122 ix0 |= (ix1>>(32-i));
125 m -= 1023; /* unbias exponent */
126 ix0 = (ix0&0x000fffff)|0x00100000;
127 if(m&1){ /* odd m, double x to make it even */
128 ix0 += ix0 + ((ix1&sign)>>31);
131 m >>= 1; /* m = [m/2] */
133 /* generate sqrt(x) bit by bit */
134 ix0 += ix0 + ((ix1&sign)>>31);
136 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
137 r = 0x00200000; /* r = moving bit from right to left */
146 ix0 += ix0 + ((ix1&sign)>>31);
155 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
157 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
159 if (ix1 < t1) ix0 -= 1;
163 ix0 += ix0 + ((ix1&sign)>>31);
168 /* use floating add to find out rounding direction */
170 z = one-tiny; /* trigger inexact flag */
173 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
175 if (q1==(u_int32_t)0xfffffffe) q+=1;
181 ix0 = (q>>1)+0x3fe00000;
183 if ((q&1)==1) ix1 |= sign;
185 INSERT_WORDS(z,ix0,ix1);
189 #if (LDBL_MANT_DIG == 53)
190 __weak_reference(sqrt, sqrtl);
194 Other methods (use floating-point arithmetic)
196 (This is a copy of a drafted paper by Prof W. Kahan
197 and K.C. Ng, written in May, 1986)
199 Two algorithms are given here to implement sqrt(x)
200 (IEEE double precision arithmetic) in software.
201 Both supply sqrt(x) correctly rounded. The first algorithm (in
202 Section A) uses newton iterations and involves four divisions.
203 The second one uses reciproot iterations to avoid division, but
204 requires more multiplications. Both algorithms need the ability
205 to chop results of arithmetic operations instead of round them,
206 and the INEXACT flag to indicate when an arithmetic operation
207 is executed exactly with no roundoff error, all part of the
208 standard (IEEE 754-1985). The ability to perform shift, add,
209 subtract and logical AND operations upon 32-bit words is needed
210 too, though not part of the standard.
212 A. sqrt(x) by Newton Iteration
214 (1) Initial approximation
216 Let x0 and x1 be the leading and the trailing 32-bit words of
217 a floating point number x (in IEEE double format) respectively
220 ------------------------------------------------------
222 ------------------------------------------------------
223 msb lsb msb lsb ...order
226 ------------------------ ------------------------
227 x0: |s| e | f1 | x1: | f2 |
228 ------------------------ ------------------------
230 By performing shifts and subtracts on x0 and x1 (both regarded
231 as integers), we obtain an 8-bit approximation of sqrt(x) as
234 k := (x0>>1) + 0x1ff80000;
235 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
236 Here k is a 32-bit integer and T1[] is an integer array containing
237 correction terms. Now magically the floating value of y (y's
238 leading 32-bit word is y0, the value of its trailing word is 0)
239 approximates sqrt(x) to almost 8-bit.
243 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
244 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
245 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
246 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
248 (2) Iterative refinement
250 Apply Heron's rule three times to y, we have y approximates
251 sqrt(x) to within 1 ulp (Unit in the Last Place):
253 y := (y+x/y)/2 ... almost 17 sig. bits
254 y := (y+x/y)/2 ... almost 35 sig. bits
255 y := y-(y-x/y)/2 ... within 1 ulp
259 Another way to improve y to within 1 ulp is:
261 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
262 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
266 y := y + 2* ---------- ...within 1 ulp
271 This formula has one division fewer than the one above; however,
272 it requires more multiplications and additions. Also x must be
273 scaled in advance to avoid spurious overflow in evaluating the
274 expression 3y*y+x. Hence it is not recommended uless division
275 is slow. If division is very slow, then one should use the
276 reciproot algorithm given in section B.
280 By twiddling y's last bit it is possible to force y to be
281 correctly rounded according to the prevailing rounding mode
282 as follows. Let r and i be copies of the rounding mode and
283 inexact flag before entering the square root program. Also we
284 use the expression y+-ulp for the next representable floating
285 numbers (up and down) of y. Note that y+-ulp = either fixed
286 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
289 I := FALSE; ... reset INEXACT flag I
290 R := RZ; ... set rounding mode to round-toward-zero
291 z := x/y; ... chopped quotient, possibly inexact
292 If(not I) then { ... if the quotient is exact
294 I := i; ... restore inexact flag
295 R := r; ... restore rounded mode
298 z := z - ulp; ... special rounding
301 i := TRUE; ... sqrt(x) is inexact
302 If (r=RN) then z=z+ulp ... rounded-to-nearest
303 If (r=RP) then { ... round-toward-+inf
306 y := y+z; ... chopped sum
307 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
308 I := i; ... restore inexact flag
309 R := r; ... restore rounded mode
314 Square root of +inf, +-0, or NaN is itself;
315 Square root of a negative number is NaN with invalid signal.
318 B. sqrt(x) by Reciproot Iteration
320 (1) Initial approximation
322 Let x0 and x1 be the leading and the trailing 32-bit words of
323 a floating point number x (in IEEE double format) respectively
324 (see section A). By performing shifs and subtracts on x0 and y0,
325 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
327 k := 0x5fe80000 - (x0>>1);
328 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
330 Here k is a 32-bit integer and T2[] is an integer array
331 containing correction terms. Now magically the floating
332 value of y (y's leading 32-bit word is y0, the value of
333 its trailing word y1 is set to zero) approximates 1/sqrt(x)
338 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
339 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
340 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
341 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
342 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
343 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
344 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
345 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
347 (2) Iterative refinement
349 Apply Reciproot iteration three times to y and multiply the
350 result by x to get an approximation z that matches sqrt(x)
351 to about 1 ulp. To be exact, we will have
352 -1ulp < sqrt(x)-z<1.0625ulp.
354 ... set rounding mode to Round-to-nearest
355 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
356 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
357 ... special arrangement for better accuracy
358 z := x*y ... 29 bits to sqrt(x), with z*y<1
359 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
361 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
362 (a) the term z*y in the final iteration is always less than 1;
363 (b) the error in the final result is biased upward so that
364 -1 ulp < sqrt(x) - z < 1.0625 ulp
365 instead of |sqrt(x)-z|<1.03125ulp.
369 By twiddling y's last bit it is possible to force y to be
370 correctly rounded according to the prevailing rounding mode
371 as follows. Let r and i be copies of the rounding mode and
372 inexact flag before entering the square root program. Also we
373 use the expression y+-ulp for the next representable floating
374 numbers (up and down) of y. Note that y+-ulp = either fixed
375 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
378 R := RZ; ... set rounding mode to round-toward-zero
380 case RN: ... round-to-nearest
381 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
382 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
384 case RZ:case RM: ... round-to-zero or round-to--inf
385 R:=RP; ... reset rounding mod to round-to-+inf
386 if(x<z*z ... rounded up) z = z - ulp; else
387 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
389 case RP: ... round-to-+inf
390 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
391 if(x>z*z ...chopped) z = z+ulp;
395 Remark 3. The above comparisons can be done in fixed point. For
396 example, to compare x and w=z*z chopped, it suffices to compare
397 x1 and w1 (the trailing parts of x and w), regarding them as
398 two's complement integers.
400 ...Is z an exact square root?
401 To determine whether z is an exact square root of x, let z1 be the
402 trailing part of z, and also let x0 and x1 be the leading and
405 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
406 I := 1; ... Raise Inexact flag: z is not exact
408 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
409 k := z1 >> 26; ... get z's 25-th and 26-th
411 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
413 R:= r ... restore rounded mode
416 If multiplication is cheaper then the foregoing red tape, the
417 Inexact flag can be evaluated by
422 Note that z*z can overwrite I; this value must be sensed if it is
425 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
433 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
434 or even of logb(x) have the following relations:
436 -------------------------------------------------
437 bit 27,26 of z1 bit 1,0 of x1 logb(x)
438 -------------------------------------------------
444 -------------------------------------------------
446 (4) Special cases (see (4) of Section A).