#include "libmin.h"
#include "libtarg.h"

/* 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
 */

static const double __tiny = 1.0e-300;

double
libmin_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;
}
