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 * ====================================================
15 * Return correctly rounded sqrt.
16 * ------------------------------------------
17 * | Use the hardware sqrt if you have one |
18 * ------------------------------------------
20 * Bit by bit method using integer arithmetic. (Slow, but portable)
22 * Scale x to y in [1,4) with even powers of 2:
23 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
24 * sqrt(x) = 2^k * sqrt(y)
25 * 2. Bit by bit computation
26 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
29 * s = 2*q , and y = 2 * ( y - q ). (1)
32 * To compute q from q , one checks whether
39 * If (2) is false, then q = q ; otherwise q = q + 2 .
42 * With some algebric manipulation, it is not difficult to see
43 * that (2) is equivalent to
48 * The advantage of (3) is that s and y can be computed by
50 * the following recurrence formula:
58 * s = s + 2 , y = y - s - 2 (5)
61 * One may easily use induction to prove (4) and (5).
62 * Note. Since the left hand side of (3) contain only i+2 bits,
63 * it does not necessary to do a full (53-bit) comparison
66 * After generating the 53 bits result, we compute one more bit.
67 * Together with the remainder, we can decide whether the
68 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
69 * (it will never equal to 1/2ulp).
70 * The rounding mode can be detected by checking whether
71 * huge + tiny is equal to huge, and whether huge - tiny is
72 * equal to huge for some floating point number "huge" and "tiny".
75 * sqrt(+-0) = +-0 ... exact
77 * sqrt(-ve) = NaN ... with invalid signal
78 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
80 * Other methods : see the appended file at the end of the program below.
85 #include "math_private.h"
87 static const double one = 1.0, tiny=1.0e-300;
93 int32_t sign = (int)0x80000000;
94 int32_t ix0,s0,q,m,t,i;
95 uint32_t r,t1,s1,ix1,q1;
97 EXTRACT_WORDS(ix0,ix1,x);
99 /* take care of Inf and NaN */
100 if((ix0&0x7ff00000)==0x7ff00000) {
101 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
104 /* take care of zero */
106 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
108 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
112 if(m==0) { /* subnormal x */
115 ix0 |= (ix1>>11); ix1 <<= 21;
117 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
119 ix0 |= (ix1>>(32-i));
122 m -= 1023; /* unbias exponent */
123 ix0 = (ix0&0x000fffff)|0x00100000;
124 if(m&1){ /* odd m, double x to make it even */
125 ix0 += ix0 + ((ix1&sign)>>31);
128 m >>= 1; /* m = [m/2] */
130 /* generate sqrt(x) bit by bit */
131 ix0 += ix0 + ((ix1&sign)>>31);
133 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
134 r = 0x00200000; /* r = moving bit from right to left */
143 ix0 += ix0 + ((ix1&sign)>>31);
152 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
154 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
156 if (ix1 < t1) ix0 -= 1;
160 ix0 += ix0 + ((ix1&sign)>>31);
165 /* use floating add to find out rounding direction */
167 z = one-tiny; /* trigger inexact flag */
170 if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;}
172 if (q1==(uint32_t)0xfffffffe) q+=1;
178 ix0 = (q>>1)+0x3fe00000;
180 if ((q&1)==1) ix1 |= sign;
182 INSERT_WORDS(z,ix0,ix1);
187 Other methods (use floating-point arithmetic)
189 (This is a copy of a drafted paper by Prof W. Kahan
190 and K.C. Ng, written in May, 1986)
192 Two algorithms are given here to implement sqrt(x)
193 (IEEE double precision arithmetic) in software.
194 Both supply sqrt(x) correctly rounded. The first algorithm (in
195 Section A) uses newton iterations and involves four divisions.
196 The second one uses reciproot iterations to avoid division, but
197 requires more multiplications. Both algorithms need the ability
198 to chop results of arithmetic operations instead of round them,
199 and the INEXACT flag to indicate when an arithmetic operation
200 is executed exactly with no roundoff error, all part of the
201 standard (IEEE 754-1985). The ability to perform shift, add,
202 subtract and logical AND operations upon 32-bit words is needed
203 too, though not part of the standard.
205 A. sqrt(x) by Newton Iteration
207 (1) Initial approximation
209 Let x0 and x1 be the leading and the trailing 32-bit words of
210 a floating point number x (in IEEE double format) respectively
213 ------------------------------------------------------
215 ------------------------------------------------------
216 msb lsb msb lsb ...order
219 ------------------------ ------------------------
220 x0: |s| e | f1 | x1: | f2 |
221 ------------------------ ------------------------
223 By performing shifts and subtracts on x0 and x1 (both regarded
224 as integers), we obtain an 8-bit approximation of sqrt(x) as
227 k := (x0>>1) + 0x1ff80000;
228 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
229 Here k is a 32-bit integer and T1[] is an integer array containing
230 correction terms. Now magically the floating value of y (y's
231 leading 32-bit word is y0, the value of its trailing word is 0)
232 approximates sqrt(x) to almost 8-bit.
236 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
237 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
238 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
239 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
241 (2) Iterative refinement
243 Apply Heron's rule three times to y, we have y approximates
244 sqrt(x) to within 1 ulp (Unit in the Last Place):
246 y := (y+x/y)/2 ... almost 17 sig. bits
247 y := (y+x/y)/2 ... almost 35 sig. bits
248 y := y-(y-x/y)/2 ... within 1 ulp
252 Another way to improve y to within 1 ulp is:
254 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
255 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
259 y := y + 2* ---------- ...within 1 ulp
264 This formula has one division fewer than the one above; however,
265 it requires more multiplications and additions. Also x must be
266 scaled in advance to avoid spurious overflow in evaluating the
267 expression 3y*y+x. Hence it is not recommended uless division
268 is slow. If division is very slow, then one should use the
269 reciproot algorithm given in section B.
273 By twiddling y's last bit it is possible to force y to be
274 correctly rounded according to the prevailing rounding mode
275 as follows. Let r and i be copies of the rounding mode and
276 inexact flag before entering the square root program. Also we
277 use the expression y+-ulp for the next representable floating
278 numbers (up and down) of y. Note that y+-ulp = either fixed
279 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
282 I := FALSE; ... reset INEXACT flag I
283 R := RZ; ... set rounding mode to round-toward-zero
284 z := x/y; ... chopped quotient, possibly inexact
285 If(not I) then { ... if the quotient is exact
287 I := i; ... restore inexact flag
288 R := r; ... restore rounded mode
291 z := z - ulp; ... special rounding
294 i := TRUE; ... sqrt(x) is inexact
295 If (r=RN) then z=z+ulp ... rounded-to-nearest
296 If (r=RP) then { ... round-toward-+inf
299 y := y+z; ... chopped sum
300 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
301 I := i; ... restore inexact flag
302 R := r; ... restore rounded mode
307 Square root of +inf, +-0, or NaN is itself;
308 Square root of a negative number is NaN with invalid signal.
311 B. sqrt(x) by Reciproot Iteration
313 (1) Initial approximation
315 Let x0 and x1 be the leading and the trailing 32-bit words of
316 a floating point number x (in IEEE double format) respectively
317 (see section A). By performing shifs and subtracts on x0 and y0,
318 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
320 k := 0x5fe80000 - (x0>>1);
321 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
323 Here k is a 32-bit integer and T2[] is an integer array
324 containing correction terms. Now magically the floating
325 value of y (y's leading 32-bit word is y0, the value of
326 its trailing word y1 is set to zero) approximates 1/sqrt(x)
331 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
332 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
333 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
334 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
335 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
336 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
337 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
338 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
340 (2) Iterative refinement
342 Apply Reciproot iteration three times to y and multiply the
343 result by x to get an approximation z that matches sqrt(x)
344 to about 1 ulp. To be exact, we will have
345 -1ulp < sqrt(x)-z<1.0625ulp.
347 ... set rounding mode to Round-to-nearest
348 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
349 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
350 ... special arrangement for better accuracy
351 z := x*y ... 29 bits to sqrt(x), with z*y<1
352 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
354 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
355 (a) the term z*y in the final iteration is always less than 1;
356 (b) the error in the final result is biased upward so that
357 -1 ulp < sqrt(x) - z < 1.0625 ulp
358 instead of |sqrt(x)-z|<1.03125ulp.
362 By twiddling y's last bit it is possible to force y to be
363 correctly rounded according to the prevailing rounding mode
364 as follows. Let r and i be copies of the rounding mode and
365 inexact flag before entering the square root program. Also we
366 use the expression y+-ulp for the next representable floating
367 numbers (up and down) of y. Note that y+-ulp = either fixed
368 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
371 R := RZ; ... set rounding mode to round-toward-zero
373 case RN: ... round-to-nearest
374 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
375 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
377 case RZ:case RM: ... round-to-zero or round-to--inf
378 R:=RP; ... reset rounding mod to round-to-+inf
379 if(x<z*z ... rounded up) z = z - ulp; else
380 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
382 case RP: ... round-to-+inf
383 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
384 if(x>z*z ...chopped) z = z+ulp;
388 Remark 3. The above comparisons can be done in fixed point. For
389 example, to compare x and w=z*z chopped, it suffices to compare
390 x1 and w1 (the trailing parts of x and w), regarding them as
391 two's complement integers.
393 ...Is z an exact square root?
394 To determine whether z is an exact square root of x, let z1 be the
395 trailing part of z, and also let x0 and x1 be the leading and
398 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
399 I := 1; ... Raise Inexact flag: z is not exact
401 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
402 k := z1 >> 26; ... get z's 25-th and 26-th
404 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
406 R:= r ... restore rounded mode
409 If multiplication is cheaper then the foregoing red tape, the
410 Inexact flag can be evaluated by
415 Note that z*z can overwrite I; this value must be sensed if it is
418 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
426 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
427 or even of logb(x) have the following relations:
429 -------------------------------------------------
430 bit 27,26 of z1 bit 1,0 of x1 logb(x)
431 -------------------------------------------------
437 -------------------------------------------------
439 (4) Special cases (see (4) of Section A).