453 lines
15 KiB
C
453 lines
15 KiB
C
|
|
/* @(#)e_sqrt.c 5.1 93/09/24 */
|
|
/*
|
|
* ====================================================
|
|
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
|
|
*
|
|
* Developed at SunPro, a Sun Microsystems, Inc. business.
|
|
* Permission to use, copy, modify, and distribute this
|
|
* software is freely granted, provided that this notice
|
|
* is preserved.
|
|
* ====================================================
|
|
*/
|
|
|
|
/* __ieee754_sqrt(x)
|
|
* Return correctly rounded sqrt.
|
|
* ------------------------------------------
|
|
* | Use the hardware sqrt if you have one |
|
|
* ------------------------------------------
|
|
* Method:
|
|
* Bit by bit method using integer arithmetic. (Slow, but portable)
|
|
* 1. Normalization
|
|
* Scale x to y in [1,4) with even powers of 2:
|
|
* find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
|
|
* sqrt(x) = 2^k * sqrt(y)
|
|
* 2. Bit by bit computation
|
|
* Let q = sqrt(y) truncated to i bit after binary point (q = 1),
|
|
* i 0
|
|
* i+1 2
|
|
* s = 2*q , and y = 2 * ( y - q ). (1)
|
|
* i i i i
|
|
*
|
|
* To compute q from q , one checks whether
|
|
* i+1 i
|
|
*
|
|
* -(i+1) 2
|
|
* (q + 2 ) <= y. (2)
|
|
* i
|
|
* -(i+1)
|
|
* If (2) is false, then q = q ; otherwise q = q + 2 .
|
|
* i+1 i i+1 i
|
|
*
|
|
* With some algebric manipulation, it is not difficult to see
|
|
* that (2) is equivalent to
|
|
* -(i+1)
|
|
* s + 2 <= y (3)
|
|
* i i
|
|
*
|
|
* The advantage of (3) is that s and y can be computed by
|
|
* i i
|
|
* the following recurrence formula:
|
|
* if (3) is false
|
|
*
|
|
* s = s , y = y ; (4)
|
|
* i+1 i i+1 i
|
|
*
|
|
* otherwise,
|
|
* -i -(i+1)
|
|
* s = s + 2 , y = y - s - 2 (5)
|
|
* i+1 i i+1 i i
|
|
*
|
|
* One may easily use induction to prove (4) and (5).
|
|
* Note. Since the left hand side of (3) contain only i+2 bits,
|
|
* it does not necessary to do a full (53-bit) comparison
|
|
* in (3).
|
|
* 3. Final rounding
|
|
* After generating the 53 bits result, we compute one more bit.
|
|
* Together with the remainder, we can decide whether the
|
|
* result is exact, bigger than 1/2ulp, or less than 1/2ulp
|
|
* (it will never equal to 1/2ulp).
|
|
* The rounding mode can be detected by checking whether
|
|
* huge + tiny is equal to huge, and whether huge - tiny is
|
|
* equal to huge for some floating point number "huge" and "tiny".
|
|
*
|
|
* Special cases:
|
|
* sqrt(+-0) = +-0 ... exact
|
|
* sqrt(inf) = inf
|
|
* sqrt(-ve) = NaN ... with invalid signal
|
|
* sqrt(NaN) = NaN ... with invalid signal for signaling NaN
|
|
*
|
|
* Other methods : see the appended file at the end of the program below.
|
|
*---------------
|
|
*/
|
|
|
|
#include "fdlibm.h"
|
|
|
|
#ifndef _DOUBLE_IS_32BITS
|
|
|
|
#ifdef __STDC__
|
|
static const double one = 1.0, tiny=1.0e-300;
|
|
#else
|
|
static double one = 1.0, tiny=1.0e-300;
|
|
#endif
|
|
|
|
#ifdef __STDC__
|
|
double __ieee754_sqrt(double x)
|
|
#else
|
|
double __ieee754_sqrt(x)
|
|
double x;
|
|
#endif
|
|
{
|
|
double z;
|
|
__int32_t sign = (int)0x80000000;
|
|
__uint32_t r,t1,s1,ix1,q1;
|
|
__int32_t ix0,s0,q,m,t,i;
|
|
|
|
EXTRACT_WORDS(ix0,ix1,x);
|
|
|
|
/* take care of Inf and NaN */
|
|
if((ix0&0x7ff00000)==0x7ff00000) {
|
|
return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
|
|
sqrt(-inf)=sNaN */
|
|
}
|
|
/* take care of zero */
|
|
if(ix0<=0) {
|
|
if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
|
|
else if(ix0<0)
|
|
return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
|
|
}
|
|
/* normalize x */
|
|
m = (ix0>>20);
|
|
if(m==0) { /* subnormal x */
|
|
while(ix0==0) {
|
|
m -= 21;
|
|
ix0 |= (ix1>>11); ix1 <<= 21;
|
|
}
|
|
for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
|
|
m -= i-1;
|
|
ix0 |= (ix1>>(32-i));
|
|
ix1 <<= i;
|
|
}
|
|
m -= 1023; /* unbias exponent */
|
|
ix0 = (ix0&0x000fffff)|0x00100000;
|
|
if(m&1){ /* odd m, double x to make it even */
|
|
ix0 += ix0 + ((ix1&sign)>>31);
|
|
ix1 += ix1;
|
|
}
|
|
m >>= 1; /* m = [m/2] */
|
|
|
|
/* generate sqrt(x) bit by bit */
|
|
ix0 += ix0 + ((ix1&sign)>>31);
|
|
ix1 += ix1;
|
|
q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
|
|
r = 0x00200000; /* r = moving bit from right to left */
|
|
|
|
while(r!=0) {
|
|
t = s0+r;
|
|
if(t<=ix0) {
|
|
s0 = t+r;
|
|
ix0 -= t;
|
|
q += r;
|
|
}
|
|
ix0 += ix0 + ((ix1&sign)>>31);
|
|
ix1 += ix1;
|
|
r>>=1;
|
|
}
|
|
|
|
r = sign;
|
|
while(r!=0) {
|
|
t1 = s1+r;
|
|
t = s0;
|
|
if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
|
|
s1 = t1+r;
|
|
if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
|
|
ix0 -= t;
|
|
if (ix1 < t1) ix0 -= 1;
|
|
ix1 -= t1;
|
|
q1 += r;
|
|
}
|
|
ix0 += ix0 + ((ix1&sign)>>31);
|
|
ix1 += ix1;
|
|
r>>=1;
|
|
}
|
|
|
|
/* use floating add to find out rounding direction */
|
|
if((ix0|ix1)!=0) {
|
|
z = one-tiny; /* trigger inexact flag */
|
|
if (z>=one) {
|
|
z = one+tiny;
|
|
if (q1==(__uint32_t)0xffffffff) { q1=0; q += 1;}
|
|
else if (z>one) {
|
|
if (q1==(__uint32_t)0xfffffffe) q+=1;
|
|
q1+=2;
|
|
} else
|
|
q1 += (q1&1);
|
|
}
|
|
}
|
|
ix0 = (q>>1)+0x3fe00000;
|
|
ix1 = q1>>1;
|
|
if ((q&1)==1) ix1 |= sign;
|
|
ix0 += (m <<20);
|
|
INSERT_WORDS(z,ix0,ix1);
|
|
return z;
|
|
}
|
|
|
|
#endif /* defined(_DOUBLE_IS_32BITS) */
|
|
|
|
/*
|
|
Other methods (use floating-point arithmetic)
|
|
-------------
|
|
(This is a copy of a drafted paper by Prof W. Kahan
|
|
and K.C. Ng, written in May, 1986)
|
|
|
|
Two algorithms are given here to implement sqrt(x)
|
|
(IEEE double precision arithmetic) in software.
|
|
Both supply sqrt(x) correctly rounded. The first algorithm (in
|
|
Section A) uses newton iterations and involves four divisions.
|
|
The second one uses reciproot iterations to avoid division, but
|
|
requires more multiplications. Both algorithms need the ability
|
|
to chop results of arithmetic operations instead of round them,
|
|
and the INEXACT flag to indicate when an arithmetic operation
|
|
is executed exactly with no roundoff error, all part of the
|
|
standard (IEEE 754-1985). The ability to perform shift, add,
|
|
subtract and logical AND operations upon 32-bit words is needed
|
|
too, though not part of the standard.
|
|
|
|
A. sqrt(x) by Newton Iteration
|
|
|
|
(1) Initial approximation
|
|
|
|
Let x0 and x1 be the leading and the trailing 32-bit words of
|
|
a floating point number x (in IEEE double format) respectively
|
|
|
|
1 11 52 ...widths
|
|
------------------------------------------------------
|
|
x: |s| e | f |
|
|
------------------------------------------------------
|
|
msb lsb msb lsb ...order
|
|
|
|
|
|
------------------------ ------------------------
|
|
x0: |s| e | f1 | x1: | f2 |
|
|
------------------------ ------------------------
|
|
|
|
By performing shifts and subtracts on x0 and x1 (both regarded
|
|
as integers), we obtain an 8-bit approximation of sqrt(x) as
|
|
follows.
|
|
|
|
k := (x0>>1) + 0x1ff80000;
|
|
y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
|
|
Here k is a 32-bit integer and T1[] is an integer array containing
|
|
correction terms. Now magically the floating value of y (y's
|
|
leading 32-bit word is y0, the value of its trailing word is 0)
|
|
approximates sqrt(x) to almost 8-bit.
|
|
|
|
Value of T1:
|
|
static int T1[32]= {
|
|
0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
|
|
29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
|
|
83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
|
|
16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
|
|
|
|
(2) Iterative refinement
|
|
|
|
Apply Heron's rule three times to y, we have y approximates
|
|
sqrt(x) to within 1 ulp (Unit in the Last Place):
|
|
|
|
y := (y+x/y)/2 ... almost 17 sig. bits
|
|
y := (y+x/y)/2 ... almost 35 sig. bits
|
|
y := y-(y-x/y)/2 ... within 1 ulp
|
|
|
|
|
|
Remark 1.
|
|
Another way to improve y to within 1 ulp is:
|
|
|
|
y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
|
|
y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
|
|
|
|
2
|
|
(x-y )*y
|
|
y := y + 2* ---------- ...within 1 ulp
|
|
2
|
|
3y + x
|
|
|
|
|
|
This formula has one division fewer than the one above; however,
|
|
it requires more multiplications and additions. Also x must be
|
|
scaled in advance to avoid spurious overflow in evaluating the
|
|
expression 3y*y+x. Hence it is not recommended uless division
|
|
is slow. If division is very slow, then one should use the
|
|
reciproot algorithm given in section B.
|
|
|
|
(3) Final adjustment
|
|
|
|
By twiddling y's last bit it is possible to force y to be
|
|
correctly rounded according to the prevailing rounding mode
|
|
as follows. Let r and i be copies of the rounding mode and
|
|
inexact flag before entering the square root program. Also we
|
|
use the expression y+-ulp for the next representable floating
|
|
numbers (up and down) of y. Note that y+-ulp = either fixed
|
|
point y+-1, or multiply y by nextafter(1,+-inf) in chopped
|
|
mode.
|
|
|
|
I := FALSE; ... reset INEXACT flag I
|
|
R := RZ; ... set rounding mode to round-toward-zero
|
|
z := x/y; ... chopped quotient, possibly inexact
|
|
If(not I) then { ... if the quotient is exact
|
|
if(z=y) {
|
|
I := i; ... restore inexact flag
|
|
R := r; ... restore rounded mode
|
|
return sqrt(x):=y.
|
|
} else {
|
|
z := z - ulp; ... special rounding
|
|
}
|
|
}
|
|
i := TRUE; ... sqrt(x) is inexact
|
|
If (r=RN) then z=z+ulp ... rounded-to-nearest
|
|
If (r=RP) then { ... round-toward-+inf
|
|
y = y+ulp; z=z+ulp;
|
|
}
|
|
y := y+z; ... chopped sum
|
|
y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
|
|
I := i; ... restore inexact flag
|
|
R := r; ... restore rounded mode
|
|
return sqrt(x):=y.
|
|
|
|
(4) Special cases
|
|
|
|
Square root of +inf, +-0, or NaN is itself;
|
|
Square root of a negative number is NaN with invalid signal.
|
|
|
|
|
|
B. sqrt(x) by Reciproot Iteration
|
|
|
|
(1) Initial approximation
|
|
|
|
Let x0 and x1 be the leading and the trailing 32-bit words of
|
|
a floating point number x (in IEEE double format) respectively
|
|
(see section A). By performing shifs and subtracts on x0 and y0,
|
|
we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
|
|
|
|
k := 0x5fe80000 - (x0>>1);
|
|
y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
|
|
|
|
Here k is a 32-bit integer and T2[] is an integer array
|
|
containing correction terms. Now magically the floating
|
|
value of y (y's leading 32-bit word is y0, the value of
|
|
its trailing word y1 is set to zero) approximates 1/sqrt(x)
|
|
to almost 7.8-bit.
|
|
|
|
Value of T2:
|
|
static int T2[64]= {
|
|
0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
|
|
0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
|
|
0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
|
|
0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
|
|
0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
|
|
0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
|
|
0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
|
|
0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
|
|
|
|
(2) Iterative refinement
|
|
|
|
Apply Reciproot iteration three times to y and multiply the
|
|
result by x to get an approximation z that matches sqrt(x)
|
|
to about 1 ulp. To be exact, we will have
|
|
-1ulp < sqrt(x)-z<1.0625ulp.
|
|
|
|
... set rounding mode to Round-to-nearest
|
|
y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
|
|
y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
|
|
... special arrangement for better accuracy
|
|
z := x*y ... 29 bits to sqrt(x), with z*y<1
|
|
z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
|
|
|
|
Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
|
|
(a) the term z*y in the final iteration is always less than 1;
|
|
(b) the error in the final result is biased upward so that
|
|
-1 ulp < sqrt(x) - z < 1.0625 ulp
|
|
instead of |sqrt(x)-z|<1.03125ulp.
|
|
|
|
(3) Final adjustment
|
|
|
|
By twiddling y's last bit it is possible to force y to be
|
|
correctly rounded according to the prevailing rounding mode
|
|
as follows. Let r and i be copies of the rounding mode and
|
|
inexact flag before entering the square root program. Also we
|
|
use the expression y+-ulp for the next representable floating
|
|
numbers (up and down) of y. Note that y+-ulp = either fixed
|
|
point y+-1, or multiply y by nextafter(1,+-inf) in chopped
|
|
mode.
|
|
|
|
R := RZ; ... set rounding mode to round-toward-zero
|
|
switch(r) {
|
|
case RN: ... round-to-nearest
|
|
if(x<= z*(z-ulp)...chopped) z = z - ulp; else
|
|
if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
|
|
break;
|
|
case RZ:case RM: ... round-to-zero or round-to--inf
|
|
R:=RP; ... reset rounding mod to round-to-+inf
|
|
if(x<z*z ... rounded up) z = z - ulp; else
|
|
if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
|
|
break;
|
|
case RP: ... round-to-+inf
|
|
if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
|
|
if(x>z*z ...chopped) z = z+ulp;
|
|
break;
|
|
}
|
|
|
|
Remark 3. The above comparisons can be done in fixed point. For
|
|
example, to compare x and w=z*z chopped, it suffices to compare
|
|
x1 and w1 (the trailing parts of x and w), regarding them as
|
|
two's complement integers.
|
|
|
|
...Is z an exact square root?
|
|
To determine whether z is an exact square root of x, let z1 be the
|
|
trailing part of z, and also let x0 and x1 be the leading and
|
|
trailing parts of x.
|
|
|
|
If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
|
|
I := 1; ... Raise Inexact flag: z is not exact
|
|
else {
|
|
j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
|
|
k := z1 >> 26; ... get z's 25-th and 26-th
|
|
fraction bits
|
|
I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
|
|
}
|
|
R:= r ... restore rounded mode
|
|
return sqrt(x):=z.
|
|
|
|
If multiplication is cheaper then the foregoing red tape, the
|
|
Inexact flag can be evaluated by
|
|
|
|
I := i;
|
|
I := (z*z!=x) or I.
|
|
|
|
Note that z*z can overwrite I; this value must be sensed if it is
|
|
True.
|
|
|
|
Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
|
|
zero.
|
|
|
|
--------------------
|
|
z1: | f2 |
|
|
--------------------
|
|
bit 31 bit 0
|
|
|
|
Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
|
|
or even of logb(x) have the following relations:
|
|
|
|
-------------------------------------------------
|
|
bit 27,26 of z1 bit 1,0 of x1 logb(x)
|
|
-------------------------------------------------
|
|
00 00 odd and even
|
|
01 01 even
|
|
10 10 odd
|
|
10 00 even
|
|
11 01 even
|
|
-------------------------------------------------
|
|
|
|
(4) Special cases (see (4) of Section A).
|
|
|
|
*/
|