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circuitpython/lib/libm_dbl/sqrt.c

185 lines
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/* origin: FreeBSD /usr/src/lib/msun/src/e_sqrt.c */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
/* 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
*/
#include "libm.h"
static const double tiny = 1.0e-300;
double sqrt(double x)
{
double z;
int32_t sign = (int)0x80000000;
int32_t ix0,s0,q,m,t,i;
uint32_t r,t1,s1,ix1,q1;
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 */
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++;
ix0 -= t;
if (ix1 < t1)
ix0--;
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 = 1.0 - tiny; /* raise inexact flag */
if (z >= 1.0) {
z = 1.0 + tiny;
if (q1 == (uint32_t)0xffffffff) {
q1 = 0;
q++;
} else if (z > 1.0) {
if (q1 == (uint32_t)0xfffffffe)
q++;
q1 += 2;
} else
q1 += q1 & 1;
}
}
ix0 = (q>>1) + 0x3fe00000;
ix1 = q1>>1;
if (q&1)
ix1 |= sign;
ix0 += m << 20;
INSERT_WORDS(z, ix0, ix1);
return z;
}