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39 /* @(#)e_sqrt.c 1.3 95/01/18 */
41 * ====================================================
42 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
44 * Developed at SunSoft, a Sun Microsystems, Inc. business.
45 * Permission to use, copy, modify, and distribute this
46 * software is freely granted, provided that this notice
48 * ====================================================
52 * Return correctly rounded sqrt.
53 * ------------------------------------------
54 * | Use the hardware sqrt if you have one |
55 * ------------------------------------------
57 * Bit by bit method using integer arithmetic. (Slow, but portable)
59 * Scale x to y in [1,4) with even powers of 2:
60 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
61 * sqrt(y) = 2^k * sqrt(x)
62 * 2. Bit by bit computation
63 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
66 * s = 2*q , and y = 2 * ( y - q ). (1)
69 * To compute q from q , one checks whether
76 * If (2) is false, then q = q ; otherwise q = q + 2 .
79 * With some algebric manipulation, it is not difficult to see
80 * that (2) is equivalent to
85 * The advantage of (3) is that s and y can be computed by
87 * the following recurrence formula:
95 * s = s + 2 , y = y - s - 2 (5)
98 * One may easily use induction to prove (4) and (5).
99 * Note. Since the left hand side of (3) contain only i+2 bits,
100 * it does not necessary to do a full (53-bit) comparison
103 * After generating the 53 bits result, we compute one more bit.
104 * Together with the remainder, we can decide whether the
105 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
106 * (it will never equal to 1/2ulp).
107 * The rounding mode can be detected by checking whether
108 * huge + tiny is equal to huge, and whether huge - tiny is
109 * equal to huge for some floating point number "huge" and "tiny".
112 * sqrt(+-0) = +-0 ... exact
114 * sqrt(-ve) = NaN ... with invalid signal
115 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
117 * Other methods : see the appended file at the end of the program below.
123 #if defined(_MSC_VER)
124 /* Microsoft Compiler */
125 #pragma warning( disable : 4723 ) /* disables potential divide by 0 warning */
129 static const double one
= 1.0, tiny
=1.0e-300;
131 static double one
= 1.0, tiny
=1.0e-300;
135 double __ieee754_sqrt(double x
)
137 double __ieee754_sqrt(x
)
143 int sign
= (int)0x80000000;
144 unsigned r
,t1
,s1
,ix1
,q1
;
148 ix0
= __HI(u
); /* high word of x */
149 ix1
= __LO(u
); /* low word of x */
151 /* take care of Inf and NaN */
152 if((ix0
&0x7ff00000)==0x7ff00000) {
153 return x
*x
+x
; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
156 /* take care of zero */
158 if(((ix0
&(~sign
))|ix1
)==0) return x
;/* sqrt(+-0) = +-0 */
160 return (x
-x
)/(x
-x
); /* sqrt(-ve) = sNaN */
164 if(m
==0) { /* subnormal x */
167 ix0
|= (ix1
>>11); ix1
<<= 21;
169 for(i
=0;(ix0
&0x00100000)==0;i
++) ix0
<<=1;
171 ix0
|= (ix1
>>(32-i
));
174 m
-= 1023; /* unbias exponent */
175 ix0
= (ix0
&0x000fffff)|0x00100000;
176 if(m
&1){ /* odd m, double x to make it even */
177 ix0
+= ix0
+ ((ix1
&sign
)>>31);
180 m
>>= 1; /* m = [m/2] */
182 /* generate sqrt(x) bit by bit */
183 ix0
+= ix0
+ ((ix1
&sign
)>>31);
185 q
= q1
= s0
= s1
= 0; /* [q,q1] = sqrt(x) */
186 r
= 0x00200000; /* r = moving bit from right to left */
195 ix0
+= ix0
+ ((ix1
&sign
)>>31);
204 if((t
<ix0
)||((t
==ix0
)&&(t1
<=ix1
))) {
206 if(((int)(t1
&sign
)==sign
)&&(s1
&sign
)==0) s0
+= 1;
208 if (ix1
< t1
) ix0
-= 1;
212 ix0
+= ix0
+ ((ix1
&sign
)>>31);
217 /* use floating add to find out rounding direction */
219 z
= one
-tiny
; /* trigger inexact flag */
222 if (q1
==(unsigned)0xffffffff) { q1
=0; q
+= 1;}
224 if (q1
==(unsigned)0xfffffffe) q
+=1;
230 ix0
= (q
>>1)+0x3fe00000;
232 if ((q
&1)==1) ix1
|= sign
;
242 Other methods (use floating-point arithmetic)
244 (This is a copy of a drafted paper by Prof W. Kahan
245 and K.C. Ng, written in May, 1986)
247 Two algorithms are given here to implement sqrt(x)
248 (IEEE double precision arithmetic) in software.
249 Both supply sqrt(x) correctly rounded. The first algorithm (in
250 Section A) uses newton iterations and involves four divisions.
251 The second one uses reciproot iterations to avoid division, but
252 requires more multiplications. Both algorithms need the ability
253 to chop results of arithmetic operations instead of round them,
254 and the INEXACT flag to indicate when an arithmetic operation
255 is executed exactly with no roundoff error, all part of the
256 standard (IEEE 754-1985). The ability to perform shift, add,
257 subtract and logical AND operations upon 32-bit words is needed
258 too, though not part of the standard.
260 A. sqrt(x) by Newton Iteration
262 (1) Initial approximation
264 Let x0 and x1 be the leading and the trailing 32-bit words of
265 a floating point number x (in IEEE double format) respectively
268 ------------------------------------------------------
270 ------------------------------------------------------
271 msb lsb msb lsb ...order
274 ------------------------ ------------------------
275 x0: |s| e | f1 | x1: | f2 |
276 ------------------------ ------------------------
278 By performing shifts and subtracts on x0 and x1 (both regarded
279 as integers), we obtain an 8-bit approximation of sqrt(x) as
282 k := (x0>>1) + 0x1ff80000;
283 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
284 Here k is a 32-bit integer and T1[] is an integer array containing
285 correction terms. Now magically the floating value of y (y's
286 leading 32-bit word is y0, the value of its trailing word is 0)
287 approximates sqrt(x) to almost 8-bit.
291 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
292 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
293 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
294 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
296 (2) Iterative refinement
298 Apply Heron's rule three times to y, we have y approximates
299 sqrt(x) to within 1 ulp (Unit in the Last Place):
301 y := (y+x/y)/2 ... almost 17 sig. bits
302 y := (y+x/y)/2 ... almost 35 sig. bits
303 y := y-(y-x/y)/2 ... within 1 ulp
307 Another way to improve y to within 1 ulp is:
309 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
310 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
314 y := y + 2* ---------- ...within 1 ulp
319 This formula has one division fewer than the one above; however,
320 it requires more multiplications and additions. Also x must be
321 scaled in advance to avoid spurious overflow in evaluating the
322 expression 3y*y+x. Hence it is not recommended uless division
323 is slow. If division is very slow, then one should use the
324 reciproot algorithm given in section B.
328 By twiddling y's last bit it is possible to force y to be
329 correctly rounded according to the prevailing rounding mode
330 as follows. Let r and i be copies of the rounding mode and
331 inexact flag before entering the square root program. Also we
332 use the expression y+-ulp for the next representable floating
333 numbers (up and down) of y. Note that y+-ulp = either fixed
334 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
337 I := FALSE; ... reset INEXACT flag I
338 R := RZ; ... set rounding mode to round-toward-zero
339 z := x/y; ... chopped quotient, possibly inexact
340 If(not I) then { ... if the quotient is exact
342 I := i; ... restore inexact flag
343 R := r; ... restore rounded mode
346 z := z - ulp; ... special rounding
349 i := TRUE; ... sqrt(x) is inexact
350 If (r=RN) then z=z+ulp ... rounded-to-nearest
351 If (r=RP) then { ... round-toward-+inf
354 y := y+z; ... chopped sum
355 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
356 I := i; ... restore inexact flag
357 R := r; ... restore rounded mode
362 Square root of +inf, +-0, or NaN is itself;
363 Square root of a negative number is NaN with invalid signal.
366 B. sqrt(x) by Reciproot Iteration
368 (1) Initial approximation
370 Let x0 and x1 be the leading and the trailing 32-bit words of
371 a floating point number x (in IEEE double format) respectively
372 (see section A). By performing shifs and subtracts on x0 and y0,
373 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
375 k := 0x5fe80000 - (x0>>1);
376 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
378 Here k is a 32-bit integer and T2[] is an integer array
379 containing correction terms. Now magically the floating
380 value of y (y's leading 32-bit word is y0, the value of
381 its trailing word y1 is set to zero) approximates 1/sqrt(x)
386 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
387 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
388 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
389 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
390 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
391 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
392 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
393 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
395 (2) Iterative refinement
397 Apply Reciproot iteration three times to y and multiply the
398 result by x to get an approximation z that matches sqrt(x)
399 to about 1 ulp. To be exact, we will have
400 -1ulp < sqrt(x)-z<1.0625ulp.
402 ... set rounding mode to Round-to-nearest
403 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
404 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
405 ... special arrangement for better accuracy
406 z := x*y ... 29 bits to sqrt(x), with z*y<1
407 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
409 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
410 (a) the term z*y in the final iteration is always less than 1;
411 (b) the error in the final result is biased upward so that
412 -1 ulp < sqrt(x) - z < 1.0625 ulp
413 instead of |sqrt(x)-z|<1.03125ulp.
417 By twiddling y's last bit it is possible to force y to be
418 correctly rounded according to the prevailing rounding mode
419 as follows. Let r and i be copies of the rounding mode and
420 inexact flag before entering the square root program. Also we
421 use the expression y+-ulp for the next representable floating
422 numbers (up and down) of y. Note that y+-ulp = either fixed
423 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
426 R := RZ; ... set rounding mode to round-toward-zero
428 case RN: ... round-to-nearest
429 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
430 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
432 case RZ:case RM: ... round-to-zero or round-to--inf
433 R:=RP; ... reset rounding mod to round-to-+inf
434 if(x<z*z ... rounded up) z = z - ulp; else
435 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
437 case RP: ... round-to-+inf
438 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
439 if(x>z*z ...chopped) z = z+ulp;
443 Remark 3. The above comparisons can be done in fixed point. For
444 example, to compare x and w=z*z chopped, it suffices to compare
445 x1 and w1 (the trailing parts of x and w), regarding them as
446 two's complement integers.
448 ...Is z an exact square root?
449 To determine whether z is an exact square root of x, let z1 be the
450 trailing part of z, and also let x0 and x1 be the leading and
453 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
454 I := 1; ... Raise Inexact flag: z is not exact
456 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
457 k := z1 >> 26; ... get z's 25-th and 26-th
459 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
461 R:= r ... restore rounded mode
464 If multiplication is cheaper then the foregoing red tape, the
465 Inexact flag can be evaluated by
470 Note that z*z can overwrite I; this value must be sensed if it is
473 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
481 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
482 or even of logb(x) have the following relations:
484 -------------------------------------------------
485 bit 27,26 of z1 bit 1,0 of x1 logb(x)
486 -------------------------------------------------
492 -------------------------------------------------
494 (4) Special cases (see (4) of Section A).