--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/libm/e_sqrt.c	Mon Sep 15 06:33:23 2008 +0000
@@ -0,0 +1,463 @@
+/* @(#)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.
+ * ====================================================
+ */
+
+#if defined(LIBM_SCCS) && !defined(lint)
+static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
+#endif
+
+/* __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 "math.h"
+#include "math_private.h"
+
+#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 attribute_hidden
+__ieee754_sqrt(double x)
+#else
+double attribute_hidden
+__ieee754_sqrt(x)
+     double x;
+#endif
+{
+    double z;
+    int32_t sign = (int) 0x80000000;
+    int32_t ix0, s0, q, m, t, i;
+    u_int32_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 */
+        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 == (u_int32_t) 0xffffffff) {
+                q1 = 0;
+                q += 1;
+            } else if (z > one) {
+                if (q1 == (u_int32_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;
+}
+
+/*
+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).
+
+ */