2 * ====================================================
3 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 * Developed at SunPro, a Sun Microsystems, Inc. business.
6 * Permission to use, copy, modify, and distribute this
7 * software is freely granted, provided that this notice
9 * ====================================================
13 * Return correctly rounded sqrt.
14 * ------------------------------------------
15 * | Use the hardware sqrt if you have one |
16 * ------------------------------------------
18 * Bit by bit method using integer arithmetic. (Slow, but portable)
20 * Scale x to y in [1,4) with even powers of 2:
21 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
22 * sqrt(x) = 2^k * sqrt(y)
23 * 2. Bit by bit computation
24 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
27 * s = 2*q , and y = 2 * ( y - q ). (1)
30 * To compute q from q , one checks whether
37 * If (2) is false, then q = q ; otherwise q = q + 2 .
40 * With some algebric manipulation, it is not difficult to see
41 * that (2) is equivalent to
46 * The advantage of (3) is that s and y can be computed by
48 * the following recurrence formula:
56 * s = s + 2 , y = y - s - 2 (5)
59 * One may easily use induction to prove (4) and (5).
60 * Note. Since the left hand side of (3) contain only i+2 bits,
61 * it does not necessary to do a full (53-bit) comparison
64 * After generating the 53 bits result, we compute one more bit.
65 * Together with the remainder, we can decide whether the
66 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
67 * (it will never equal to 1/2ulp).
68 * The rounding mode can be detected by checking whether
69 * huge + tiny is equal to huge, and whether huge - tiny is
70 * equal to huge for some floating point number "huge" and "tiny".
73 * sqrt(+-0) = +-0 ... exact
75 * sqrt(-ve) = NaN ... with invalid signal
76 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
78 * Other methods : see the appended file at the end of the program below.
83 #include "math_private.h"
85 static const double one
= 1.0, tiny
= 1.0e-300;
87 double __ieee754_sqrt(double x
)
90 int32_t sign
= (int)0x80000000;
91 int32_t ix0
,s0
,q
,m
,t
,i
;
92 u_int32_t r
,t1
,s1
,ix1
,q1
;
94 EXTRACT_WORDS(ix0
,ix1
,x
);
96 /* take care of Inf and NaN */
97 if((ix0
&0x7ff00000)==0x7ff00000) {
98 return x
*x
+x
; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
101 /* take care of zero */
103 if(((ix0
&(~sign
))|ix1
)==0) return x
;/* sqrt(+-0) = +-0 */
105 return (x
-x
)/(x
-x
); /* sqrt(-ve) = sNaN */
109 if(m
==0) { /* subnormal x */
112 ix0
|= (ix1
>>11); ix1
<<= 21;
114 for(i
=0;(ix0
&0x00100000)==0;i
++) ix0
<<=1;
116 ix0
|= (ix1
>>(32-i
));
119 m
-= 1023; /* unbias exponent */
120 ix0
= (ix0
&0x000fffff)|0x00100000;
121 if(m
&1){ /* odd m, double x to make it even */
122 ix0
+= ix0
+ ((ix1
&sign
)>>31);
125 m
>>= 1; /* m = [m/2] */
127 /* generate sqrt(x) bit by bit */
128 ix0
+= ix0
+ ((ix1
&sign
)>>31);
130 q
= q1
= s0
= s1
= 0; /* [q,q1] = sqrt(x) */
131 r
= 0x00200000; /* r = moving bit from right to left */
140 ix0
+= ix0
+ ((ix1
&sign
)>>31);
149 if((t
<ix0
)||((t
==ix0
)&&(t1
<=ix1
))) {
151 if(((t1
&sign
)==sign
)&&(s1
&sign
)==0) s0
+= 1;
153 if (ix1
< t1
) ix0
-= 1;
157 ix0
+= ix0
+ ((ix1
&sign
)>>31);
162 /* use floating add to find out rounding direction */
164 z
= one
-tiny
; /* trigger inexact flag */
167 if (q1
==(u_int32_t
)0xffffffff) { q1
=0; q
+= 1;}
169 if (q1
==(u_int32_t
)0xfffffffe) q
+=1;
175 ix0
= (q
>>1)+0x3fe00000;
177 if ((q
&1)==1) ix1
|= sign
;
179 INSERT_WORDS(z
,ix0
,ix1
);
185 Other methods (use floating-point arithmetic)
187 (This is a copy of a drafted paper by Prof W. Kahan
188 and K.C. Ng, written in May, 1986)
190 Two algorithms are given here to implement sqrt(x)
191 (IEEE double precision arithmetic) in software.
192 Both supply sqrt(x) correctly rounded. The first algorithm (in
193 Section A) uses newton iterations and involves four divisions.
194 The second one uses reciproot iterations to avoid division, but
195 requires more multiplications. Both algorithms need the ability
196 to chop results of arithmetic operations instead of round them,
197 and the INEXACT flag to indicate when an arithmetic operation
198 is executed exactly with no roundoff error, all part of the
199 standard (IEEE 754-1985). The ability to perform shift, add,
200 subtract and logical AND operations upon 32-bit words is needed
201 too, though not part of the standard.
203 A. sqrt(x) by Newton Iteration
205 (1) Initial approximation
207 Let x0 and x1 be the leading and the trailing 32-bit words of
208 a floating point number x (in IEEE double format) respectively
211 ------------------------------------------------------
213 ------------------------------------------------------
214 msb lsb msb lsb ...order
217 ------------------------ ------------------------
218 x0: |s| e | f1 | x1: | f2 |
219 ------------------------ ------------------------
221 By performing shifts and subtracts on x0 and x1 (both regarded
222 as integers), we obtain an 8-bit approximation of sqrt(x) as
225 k := (x0>>1) + 0x1ff80000;
226 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
227 Here k is a 32-bit integer and T1[] is an integer array containing
228 correction terms. Now magically the floating value of y (y's
229 leading 32-bit word is y0, the value of its trailing word is 0)
230 approximates sqrt(x) to almost 8-bit.
234 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
235 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
236 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
237 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
239 (2) Iterative refinement
241 Apply Heron's rule three times to y, we have y approximates
242 sqrt(x) to within 1 ulp (Unit in the Last Place):
244 y := (y+x/y)/2 ... almost 17 sig. bits
245 y := (y+x/y)/2 ... almost 35 sig. bits
246 y := y-(y-x/y)/2 ... within 1 ulp
250 Another way to improve y to within 1 ulp is:
252 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
253 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
257 y := y + 2* ---------- ...within 1 ulp
262 This formula has one division fewer than the one above; however,
263 it requires more multiplications and additions. Also x must be
264 scaled in advance to avoid spurious overflow in evaluating the
265 expression 3y*y+x. Hence it is not recommended uless division
266 is slow. If division is very slow, then one should use the
267 reciproot algorithm given in section B.
271 By twiddling y's last bit it is possible to force y to be
272 correctly rounded according to the prevailing rounding mode
273 as follows. Let r and i be copies of the rounding mode and
274 inexact flag before entering the square root program. Also we
275 use the expression y+-ulp for the next representable floating
276 numbers (up and down) of y. Note that y+-ulp = either fixed
277 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
280 I := FALSE; ... reset INEXACT flag I
281 R := RZ; ... set rounding mode to round-toward-zero
282 z := x/y; ... chopped quotient, possibly inexact
283 If(not I) then { ... if the quotient is exact
285 I := i; ... restore inexact flag
286 R := r; ... restore rounded mode
289 z := z - ulp; ... special rounding
292 i := TRUE; ... sqrt(x) is inexact
293 If (r=RN) then z=z+ulp ... rounded-to-nearest
294 If (r=RP) then { ... round-toward-+inf
297 y := y+z; ... chopped sum
298 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
299 I := i; ... restore inexact flag
300 R := r; ... restore rounded mode
305 Square root of +inf, +-0, or NaN is itself;
306 Square root of a negative number is NaN with invalid signal.
309 B. sqrt(x) by Reciproot Iteration
311 (1) Initial approximation
313 Let x0 and x1 be the leading and the trailing 32-bit words of
314 a floating point number x (in IEEE double format) respectively
315 (see section A). By performing shifs and subtracts on x0 and y0,
316 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
318 k := 0x5fe80000 - (x0>>1);
319 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
321 Here k is a 32-bit integer and T2[] is an integer array
322 containing correction terms. Now magically the floating
323 value of y (y's leading 32-bit word is y0, the value of
324 its trailing word y1 is set to zero) approximates 1/sqrt(x)
329 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
330 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
331 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
332 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
333 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
334 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
335 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
336 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
338 (2) Iterative refinement
340 Apply Reciproot iteration three times to y and multiply the
341 result by x to get an approximation z that matches sqrt(x)
342 to about 1 ulp. To be exact, we will have
343 -1ulp < sqrt(x)-z<1.0625ulp.
345 ... set rounding mode to Round-to-nearest
346 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
347 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
348 ... special arrangement for better accuracy
349 z := x*y ... 29 bits to sqrt(x), with z*y<1
350 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
352 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
353 (a) the term z*y in the final iteration is always less than 1;
354 (b) the error in the final result is biased upward so that
355 -1 ulp < sqrt(x) - z < 1.0625 ulp
356 instead of |sqrt(x)-z|<1.03125ulp.
360 By twiddling y's last bit it is possible to force y to be
361 correctly rounded according to the prevailing rounding mode
362 as follows. Let r and i be copies of the rounding mode and
363 inexact flag before entering the square root program. Also we
364 use the expression y+-ulp for the next representable floating
365 numbers (up and down) of y. Note that y+-ulp = either fixed
366 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
369 R := RZ; ... set rounding mode to round-toward-zero
371 case RN: ... round-to-nearest
372 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
373 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
375 case RZ:case RM: ... round-to-zero or round-to--inf
376 R:=RP; ... reset rounding mod to round-to-+inf
377 if(x<z*z ... rounded up) z = z - ulp; else
378 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
380 case RP: ... round-to-+inf
381 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
382 if(x>z*z ...chopped) z = z+ulp;
386 Remark 3. The above comparisons can be done in fixed point. For
387 example, to compare x and w=z*z chopped, it suffices to compare
388 x1 and w1 (the trailing parts of x and w), regarding them as
389 two's complement integers.
391 ...Is z an exact square root?
392 To determine whether z is an exact square root of x, let z1 be the
393 trailing part of z, and also let x0 and x1 be the leading and
396 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
397 I := 1; ... Raise Inexact flag: z is not exact
399 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
400 k := z1 >> 26; ... get z's 25-th and 26-th
402 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
404 R:= r ... restore rounded mode
407 If multiplication is cheaper then the foregoing red tape, the
408 Inexact flag can be evaluated by
413 Note that z*z can overwrite I; this value must be sensed if it is
416 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
424 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
425 or even of logb(x) have the following relations:
427 -------------------------------------------------
428 bit 27,26 of z1 bit 1,0 of x1 logb(x)
429 -------------------------------------------------
435 -------------------------------------------------
437 (4) Special cases (see (4) of Section A).