amath/src/real/sqrt.c

272 lines
7.3 KiB
C

/*-
* Copyright (c) 2014-2017 Carsten Sonne Larsen <cs@innolan.net>
* All rights reserved.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
*
* THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR
* IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
* OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
* IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT,
* INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
* NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
* DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
* THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
* (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF
* THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
*
* Project homepage:
* https://amath.innolan.net
*
* The original source code can be obtained from:
* http://www.netlib.org/fdlibm/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.
* =================================================================
*/
/**
* @file sqrt.c
* @brief Square root function
*/
#include "prim.h"
static const double
one = 1.0,
tiny = 1.0e-300;
/**
* @brief Square root function
* @return Correctly rounded sqrt
* @details
* <pre>
* 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
* </pre>
*/
double sqrt(double x)
{
double z;
int32_t sign = (int32_t)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)
{
// sqrt(NaN)=NaN
// sqrt(+inf)=+inf
// sqrt(-inf)=sNaN
return x * x + x;
}
// take care of zero
if (ix0 <= 0)
{
if (((ix0 & (~sign)) | ix1) == 0)
{
// sqrt(+-0) = +-0
return x;
}
else if (ix0 < 0)
{
//sqrt(-ve) = NaN
return NAN;
}
}
// 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 = [m/2]
m >>= 1;
// 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) == (uint32_t)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)
{
// trigger inexact flag
z = one - tiny;
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;
}