/* @(#)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
#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
{
int n0;
double z;
int sign = (int)0x80000000;
unsigned r,t1,s1,ix1,q1;
int ix0,s0,q,m,t,i;
n0 = ((*(int*)&one)>>29)^1; /* index of high word */
ix0 = *(n0+(int*)&x); /* high word of x */
ix1 = *((1-n0)+(int*)&x); /* low word of 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>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==(unsigned)0xffffffff) { q1=0; q += 1;}
else if (z>one) {
if (q1==(unsigned)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);
*(n0+(int*)&z) = ix0;
*((1-n0)+(int*)&z) = ix1;
return z;
}
/*
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+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).
*/