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Rev 207 Rev 345
 
 
/* @(#)e_sqrt.c 5.1 93/09/24 */
/* @(#)e_sqrt.c 5.1 93/09/24 */
/*
/*
 * ====================================================
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 *
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * software is freely granted, provided that this notice
 * is preserved.
 * is preserved.
 * ====================================================
 * ====================================================
 */
 */
 
 
/* __ieee754_sqrt(x)
/* __ieee754_sqrt(x)
 * Return correctly rounded sqrt.
 * Return correctly rounded sqrt.
 *           ------------------------------------------
 *           ------------------------------------------
 *           |  Use the hardware sqrt if you have one |
 *           |  Use the hardware sqrt if you have one |
 *           ------------------------------------------
 *           ------------------------------------------
 * Method:
 * Method:
 *   Bit by bit method using integer arithmetic. (Slow, but portable)
 *   Bit by bit method using integer arithmetic. (Slow, but portable)
 *   1. Normalization
 *   1. Normalization
 *      Scale x to y in [1,4) with even powers of 2:
 *      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
 *      find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
 *              sqrt(x) = 2^k * sqrt(y)
 *              sqrt(x) = 2^k * sqrt(y)
 *   2. Bit by bit computation
 *   2. Bit by bit computation
 *      Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
 *      Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
 *           i                                                   0
 *           i                                                   0
 *                                     i+1         2
 *                                     i+1         2
 *          s  = 2*q , and      y  =  2   * ( y - q  ).         (1)
 *          s  = 2*q , and      y  =  2   * ( y - q  ).         (1)
 *           i      i            i                 i
 *           i      i            i                 i
 *
 *
 *      To compute q    from q , one checks whether
 *      To compute q    from q , one checks whether
 *                  i+1       i
 *                  i+1       i
 *
 *
 *                            -(i+1) 2
 *                            -(i+1) 2
 *                      (q + 2      ) <= y.                     (2)
 *                      (q + 2      ) <= y.                     (2)
 *                        i
 *                        i
 *                                                            -(i+1)
 *                                                            -(i+1)
 *      If (2) is false, then q   = q ; otherwise q   = q  + 2      .
 *      If (2) is false, then q   = q ; otherwise q   = q  + 2      .
 *                             i+1   i             i+1   i
 *                             i+1   i             i+1   i
 *
 *
 *      With some algebric manipulation, it is not difficult to see
 *      With some algebric manipulation, it is not difficult to see
 *      that (2) is equivalent to
 *      that (2) is equivalent to
 *                             -(i+1)
 *                             -(i+1)
 *                      s  +  2       <= y                      (3)
 *                      s  +  2       <= y                      (3)
 *                       i                i
 *                       i                i
 *
 *
 *      The advantage of (3) is that s  and y  can be computed by
 *      The advantage of (3) is that s  and y  can be computed by
 *                                    i      i
 *                                    i      i
 *      the following recurrence formula:
 *      the following recurrence formula:
 *          if (3) is false
 *          if (3) is false
 *
 *
 *          s     =  s  ,       y    = y   ;                    (4)
 *          s     =  s  ,       y    = y   ;                    (4)
 *           i+1      i          i+1    i
 *           i+1      i          i+1    i
 *
 *
 *          otherwise,
 *          otherwise,
 *                         -i                     -(i+1)
 *                         -i                     -(i+1)
 *          s     =  s  + 2  ,  y    = y  -  s  - 2             (5)
 *          s     =  s  + 2  ,  y    = y  -  s  - 2             (5)
 *           i+1      i          i+1    i     i
 *           i+1      i          i+1    i     i
 *
 *
 *      One may easily use induction to prove (4) and (5).
 *      One may easily use induction to prove (4) and (5).
 *      Note. Since the left hand side of (3) contain only i+2 bits,
 *      Note. Since the left hand side of (3) contain only i+2 bits,
 *            it does not necessary to do a full (53-bit) comparison
 *            it does not necessary to do a full (53-bit) comparison
 *            in (3).
 *            in (3).
 *   3. Final rounding
 *   3. Final rounding
 *      After generating the 53 bits result, we compute one more bit.
 *      After generating the 53 bits result, we compute one more bit.
 *      Together with the remainder, we can decide whether the
 *      Together with the remainder, we can decide whether the
 *      result is exact, bigger than 1/2ulp, or less than 1/2ulp
 *      result is exact, bigger than 1/2ulp, or less than 1/2ulp
 *      (it will never equal to 1/2ulp).
 *      (it will never equal to 1/2ulp).
 *      The rounding mode can be detected by checking whether
 *      The rounding mode can be detected by checking whether
 *      huge + tiny is equal to huge, and whether huge - tiny is
 *      huge + tiny is equal to huge, and whether huge - tiny is
 *      equal to huge for some floating point number "huge" and "tiny".
 *      equal to huge for some floating point number "huge" and "tiny".
 *
 *
 * Special cases:
 * Special cases:
 *      sqrt(+-0) = +-0         ... exact
 *      sqrt(+-0) = +-0         ... exact
 *      sqrt(inf) = inf
 *      sqrt(inf) = inf
 *      sqrt(-ve) = NaN         ... with invalid signal
 *      sqrt(-ve) = NaN         ... with invalid signal
 *      sqrt(NaN) = NaN         ... with invalid signal for signaling NaN
 *      sqrt(NaN) = NaN         ... with invalid signal for signaling NaN
 *
 *
 * Other methods : see the appended file at the end of the program below.
 * Other methods : see the appended file at the end of the program below.
 *---------------
 *---------------
 */
 */
 
 
#include "fdlibm.h"
#include "fdlibm.h"
 
 
#ifndef _DOUBLE_IS_32BITS
#ifndef _DOUBLE_IS_32BITS
 
 
#ifdef __STDC__
#ifdef __STDC__
static  const double    one     = 1.0, tiny=1.0e-300;
static  const double    one     = 1.0, tiny=1.0e-300;
#else
#else
static  double  one     = 1.0, tiny=1.0e-300;
static  double  one     = 1.0, tiny=1.0e-300;
#endif
#endif
 
 
#ifdef __STDC__
#ifdef __STDC__
        double __ieee754_sqrt(double x)
        double __ieee754_sqrt(double x)
#else
#else
        double __ieee754_sqrt(x)
        double __ieee754_sqrt(x)
        double x;
        double x;
#endif
#endif
{
{
        double z;
        double z;
        __int32_t sign = (int)0x80000000;
        __int32_t sign = (int)0x80000000;
        __uint32_t r,t1,s1,ix1,q1;
        __uint32_t r,t1,s1,ix1,q1;
        __int32_t ix0,s0,q,m,t,i;
        __int32_t ix0,s0,q,m,t,i;
 
 
        EXTRACT_WORDS(ix0,ix1,x);
        EXTRACT_WORDS(ix0,ix1,x);
 
 
    /* take care of Inf and NaN */
    /* take care of Inf and NaN */
        if((ix0&0x7ff00000)==0x7ff00000) {
        if((ix0&0x7ff00000)==0x7ff00000) {
            return x*x+x;               /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
            return x*x+x;               /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
                                           sqrt(-inf)=sNaN */
                                           sqrt(-inf)=sNaN */
        }
        }
    /* take care of zero */
    /* take care of zero */
        if(ix0<=0) {
        if(ix0<=0) {
            if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
            if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
            else if(ix0<0)
            else if(ix0<0)
                return (x-x)/(x-x);             /* sqrt(-ve) = sNaN */
                return (x-x)/(x-x);             /* sqrt(-ve) = sNaN */
        }
        }
    /* normalize x */
    /* normalize x */
        m = (ix0>>20);
        m = (ix0>>20);
        if(m==0) {                               /* subnormal x */
        if(m==0) {                               /* subnormal x */
            while(ix0==0) {
            while(ix0==0) {
                m -= 21;
                m -= 21;
                ix0 |= (ix1>>11); ix1 <<= 21;
                ix0 |= (ix1>>11); ix1 <<= 21;
            }
            }
            for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
            for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
            m -= i-1;
            m -= i-1;
            ix0 |= (ix1>>(32-i));
            ix0 |= (ix1>>(32-i));
            ix1 <<= i;
            ix1 <<= i;
        }
        }
        m -= 1023;      /* unbias exponent */
        m -= 1023;      /* unbias exponent */
        ix0 = (ix0&0x000fffff)|0x00100000;
        ix0 = (ix0&0x000fffff)|0x00100000;
        if(m&1){        /* odd m, double x to make it even */
        if(m&1){        /* odd m, double x to make it even */
            ix0 += ix0 + ((ix1&sign)>>31);
            ix0 += ix0 + ((ix1&sign)>>31);
            ix1 += ix1;
            ix1 += ix1;
        }
        }
        m >>= 1;        /* m = [m/2] */
        m >>= 1;        /* m = [m/2] */
 
 
    /* generate sqrt(x) bit by bit */
    /* generate sqrt(x) bit by bit */
        ix0 += ix0 + ((ix1&sign)>>31);
        ix0 += ix0 + ((ix1&sign)>>31);
        ix1 += ix1;
        ix1 += ix1;
        q = q1 = s0 = s1 = 0;    /* [q,q1] = sqrt(x) */
        q = q1 = s0 = s1 = 0;    /* [q,q1] = sqrt(x) */
        r = 0x00200000;         /* r = moving bit from right to left */
        r = 0x00200000;         /* r = moving bit from right to left */
 
 
        while(r!=0) {
        while(r!=0) {
            t = s0+r;
            t = s0+r;
            if(t<=ix0) {
            if(t<=ix0) {
                s0   = t+r;
                s0   = t+r;
                ix0 -= t;
                ix0 -= t;
                q   += r;
                q   += r;
            }
            }
            ix0 += ix0 + ((ix1&sign)>>31);
            ix0 += ix0 + ((ix1&sign)>>31);
            ix1 += ix1;
            ix1 += ix1;
            r>>=1;
            r>>=1;
        }
        }
 
 
        r = sign;
        r = sign;
        while(r!=0) {
        while(r!=0) {
            t1 = s1+r;
            t1 = s1+r;
            t  = s0;
            t  = s0;
            if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
            if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
                s1  = t1+r;
                s1  = t1+r;
                if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
                if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
                ix0 -= t;
                ix0 -= t;
                if (ix1 < t1) ix0 -= 1;
                if (ix1 < t1) ix0 -= 1;
                ix1 -= t1;
                ix1 -= t1;
                q1  += r;
                q1  += r;
            }
            }
            ix0 += ix0 + ((ix1&sign)>>31);
            ix0 += ix0 + ((ix1&sign)>>31);
            ix1 += ix1;
            ix1 += ix1;
            r>>=1;
            r>>=1;
        }
        }
 
 
    /* use floating add to find out rounding direction */
    /* use floating add to find out rounding direction */
        if((ix0|ix1)!=0) {
        if((ix0|ix1)!=0) {
            z = one-tiny; /* trigger inexact flag */
            z = one-tiny; /* trigger inexact flag */
            if (z>=one) {
            if (z>=one) {
                z = one+tiny;
                z = one+tiny;
                if (q1==(__uint32_t)0xffffffff) { q1=0; q += 1;}
                if (q1==(__uint32_t)0xffffffff) { q1=0; q += 1;}
                else if (z>one) {
                else if (z>one) {
                    if (q1==(__uint32_t)0xfffffffe) q+=1;
                    if (q1==(__uint32_t)0xfffffffe) q+=1;
                    q1+=2;
                    q1+=2;
                } else
                } else
                    q1 += (q1&1);
                    q1 += (q1&1);
            }
            }
        }
        }
        ix0 = (q>>1)+0x3fe00000;
        ix0 = (q>>1)+0x3fe00000;
        ix1 =  q1>>1;
        ix1 =  q1>>1;
        if ((q&1)==1) ix1 |= sign;
        if ((q&1)==1) ix1 |= sign;
        ix0 += (m <<20);
        ix0 += (m <<20);
        INSERT_WORDS(z,ix0,ix1);
        INSERT_WORDS(z,ix0,ix1);
        return z;
        return z;
}
}
 
 
#endif /* defined(_DOUBLE_IS_32BITS) */
#endif /* defined(_DOUBLE_IS_32BITS) */
 
 
/*
/*
Other methods  (use floating-point arithmetic)
Other methods  (use floating-point arithmetic)
-------------
-------------
(This is a copy of a drafted paper by Prof W. Kahan
(This is a copy of a drafted paper by Prof W. Kahan
and K.C. Ng, written in May, 1986)
and K.C. Ng, written in May, 1986)
 
 
        Two algorithms are given here to implement sqrt(x)
        Two algorithms are given here to implement sqrt(x)
        (IEEE double precision arithmetic) in software.
        (IEEE double precision arithmetic) in software.
        Both supply sqrt(x) correctly rounded. The first algorithm (in
        Both supply sqrt(x) correctly rounded. The first algorithm (in
        Section A) uses newton iterations and involves four divisions.
        Section A) uses newton iterations and involves four divisions.
        The second one uses reciproot iterations to avoid division, but
        The second one uses reciproot iterations to avoid division, but
        requires more multiplications. Both algorithms need the ability
        requires more multiplications. Both algorithms need the ability
        to chop results of arithmetic operations instead of round them,
        to chop results of arithmetic operations instead of round them,
        and the INEXACT flag to indicate when an arithmetic operation
        and the INEXACT flag to indicate when an arithmetic operation
        is executed exactly with no roundoff error, all part of the
        is executed exactly with no roundoff error, all part of the
        standard (IEEE 754-1985). The ability to perform shift, add,
        standard (IEEE 754-1985). The ability to perform shift, add,
        subtract and logical AND operations upon 32-bit words is needed
        subtract and logical AND operations upon 32-bit words is needed
        too, though not part of the standard.
        too, though not part of the standard.
 
 
A.  sqrt(x) by Newton Iteration
A.  sqrt(x) by Newton Iteration
 
 
   (1)  Initial approximation
   (1)  Initial approximation
 
 
        Let x0 and x1 be the leading and the trailing 32-bit words of
        Let x0 and x1 be the leading and the trailing 32-bit words of
        a floating point number x (in IEEE double format) respectively
        a floating point number x (in IEEE double format) respectively
 
 
            1    11                  52                           ...widths
            1    11                  52                           ...widths
           ------------------------------------------------------
           ------------------------------------------------------
        x: |s|    e     |             f                         |
        x: |s|    e     |             f                         |
           ------------------------------------------------------
           ------------------------------------------------------
              msb    lsb  msb                                 lsb ...order
              msb    lsb  msb                                 lsb ...order
 
 
 
 
             ------------------------        ------------------------
             ------------------------        ------------------------
        x0:  |s|   e    |    f1     |    x1: |          f2           |
        x0:  |s|   e    |    f1     |    x1: |          f2           |
             ------------------------        ------------------------
             ------------------------        ------------------------
 
 
        By performing shifts and subtracts on x0 and x1 (both regarded
        By performing shifts and subtracts on x0 and x1 (both regarded
        as integers), we obtain an 8-bit approximation of sqrt(x) as
        as integers), we obtain an 8-bit approximation of sqrt(x) as
        follows.
        follows.
 
 
                k  := (x0>>1) + 0x1ff80000;
                k  := (x0>>1) + 0x1ff80000;
                y0 := k - T1[31&(k>>15)].       ... y ~ sqrt(x) to 8 bits
                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
        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
        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)
        leading 32-bit word is y0, the value of its trailing word is 0)
        approximates sqrt(x) to almost 8-bit.
        approximates sqrt(x) to almost 8-bit.
 
 
        Value of T1:
        Value of T1:
        static int T1[32]= {
        static int T1[32]= {
        0,      1024,   3062,   5746,   9193,   13348,  18162,  23592,
        0,      1024,   3062,   5746,   9193,   13348,  18162,  23592,
        29598,  36145,  43202,  50740,  58733,  67158,  75992,  85215,
        29598,  36145,  43202,  50740,  58733,  67158,  75992,  85215,
        83599,  71378,  60428,  50647,  41945,  34246,  27478,  21581,
        83599,  71378,  60428,  50647,  41945,  34246,  27478,  21581,
        16499,  12183,  8588,   5674,   3403,   1742,   661,    130,};
        16499,  12183,  8588,   5674,   3403,   1742,   661,    130,};
 
 
    (2) Iterative refinement
    (2) Iterative refinement
 
 
        Apply Heron's rule three times to y, we have y approximates
        Apply Heron's rule three times to y, we have y approximates
        sqrt(x) to within 1 ulp (Unit in the Last Place):
        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 17 sig. bits
                y := (y+x/y)/2          ... almost 35 sig. bits
                y := (y+x/y)/2          ... almost 35 sig. bits
                y := y-(y-x/y)/2        ... within 1 ulp
                y := y-(y-x/y)/2        ... within 1 ulp
 
 
 
 
        Remark 1.
        Remark 1.
            Another way to improve y to within 1 ulp is:
            Another way to improve y to within 1 ulp is:
 
 
                y := (y+x/y)            ... almost 17 sig. bits to 2*sqrt(x)
                y := (y+x/y)            ... almost 17 sig. bits to 2*sqrt(x)
                y := y - 0x00100006     ... almost 18 sig. bits to sqrt(x)
                y := y - 0x00100006     ... almost 18 sig. bits to sqrt(x)
 
 
                                2
                                2
                            (x-y )*y
                            (x-y )*y
                y := y + 2* ----------  ...within 1 ulp
                y := y + 2* ----------  ...within 1 ulp
                               2
                               2
                             3y  + x
                             3y  + x
 
 
 
 
        This formula has one division fewer than the one above; however,
        This formula has one division fewer than the one above; however,
        it requires more multiplications and additions. Also x must be
        it requires more multiplications and additions. Also x must be
        scaled in advance to avoid spurious overflow in evaluating the
        scaled in advance to avoid spurious overflow in evaluating the
        expression 3y*y+x. Hence it is not recommended uless division
        expression 3y*y+x. Hence it is not recommended uless division
        is slow. If division is very slow, then one should use the
        is slow. If division is very slow, then one should use the
        reciproot algorithm given in section B.
        reciproot algorithm given in section B.
 
 
    (3) Final adjustment
    (3) Final adjustment
 
 
        By twiddling y's last bit it is possible to force y to be
        By twiddling y's last bit it is possible to force y to be
        correctly rounded according to the prevailing rounding mode
        correctly rounded according to the prevailing rounding mode
        as follows. Let r and i be copies of the rounding mode and
        as follows. Let r and i be copies of the rounding mode and
        inexact flag before entering the square root program. Also we
        inexact flag before entering the square root program. Also we
        use the expression y+-ulp for the next representable floating
        use the expression y+-ulp for the next representable floating
        numbers (up and down) of y. Note that y+-ulp = either fixed
        numbers (up and down) of y. Note that y+-ulp = either fixed
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
        mode.
        mode.
 
 
                I := FALSE;     ... reset INEXACT flag I
                I := FALSE;     ... reset INEXACT flag I
                R := RZ;        ... set rounding mode to round-toward-zero
                R := RZ;        ... set rounding mode to round-toward-zero
                z := x/y;       ... chopped quotient, possibly inexact
                z := x/y;       ... chopped quotient, possibly inexact
                If(not I) then {        ... if the quotient is exact
                If(not I) then {        ... if the quotient is exact
                    if(z=y) {
                    if(z=y) {
                        I := i;  ... restore inexact flag
                        I := i;  ... restore inexact flag
                        R := r;  ... restore rounded mode
                        R := r;  ... restore rounded mode
                        return sqrt(x):=y.
                        return sqrt(x):=y.
                    } else {
                    } else {
                        z := z - ulp;   ... special rounding
                        z := z - ulp;   ... special rounding
                    }
                    }
                }
                }
                i := TRUE;              ... sqrt(x) is inexact
                i := TRUE;              ... sqrt(x) is inexact
                If (r=RN) then z=z+ulp  ... rounded-to-nearest
                If (r=RN) then z=z+ulp  ... rounded-to-nearest
                If (r=RP) then {        ... round-toward-+inf
                If (r=RP) then {        ... round-toward-+inf
                    y = y+ulp; z=z+ulp;
                    y = y+ulp; z=z+ulp;
                }
                }
                y := y+z;               ... chopped sum
                y := y+z;               ... chopped sum
                y0:=y0-0x00100000;      ... y := y/2 is correctly rounded.
                y0:=y0-0x00100000;      ... y := y/2 is correctly rounded.
                I := i;                 ... restore inexact flag
                I := i;                 ... restore inexact flag
                R := r;                 ... restore rounded mode
                R := r;                 ... restore rounded mode
                return sqrt(x):=y.
                return sqrt(x):=y.
 
 
    (4) Special cases
    (4) Special cases
 
 
        Square root of +inf, +-0, or NaN is itself;
        Square root of +inf, +-0, or NaN is itself;
        Square root of a negative number is NaN with invalid signal.
        Square root of a negative number is NaN with invalid signal.
 
 
 
 
B.  sqrt(x) by Reciproot Iteration
B.  sqrt(x) by Reciproot Iteration
 
 
   (1)  Initial approximation
   (1)  Initial approximation
 
 
        Let x0 and x1 be the leading and the trailing 32-bit words of
        Let x0 and x1 be the leading and the trailing 32-bit words of
        a floating point number x (in IEEE double format) respectively
        a floating point number x (in IEEE double format) respectively
        (see section A). By performing shifs and subtracts on x0 and y0,
        (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.
        we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
 
 
            k := 0x5fe80000 - (x0>>1);
            k := 0x5fe80000 - (x0>>1);
            y0:= k - T2[63&(k>>14)].    ... y ~ 1/sqrt(x) to 7.8 bits
            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
        Here k is a 32-bit integer and T2[] is an integer array
        containing correction terms. Now magically the floating
        containing correction terms. Now magically the floating
        value of y (y's leading 32-bit word is y0, the value of
        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)
        its trailing word y1 is set to zero) approximates 1/sqrt(x)
        to almost 7.8-bit.
        to almost 7.8-bit.
 
 
        Value of T2:
        Value of T2:
        static int T2[64]= {
        static int T2[64]= {
        0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
        0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
        0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
        0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
        0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
        0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
        0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
        0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
        0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
        0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
        0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
        0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
        0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
        0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
        0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
        0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
 
 
    (2) Iterative refinement
    (2) Iterative refinement
 
 
        Apply Reciproot iteration three times to y and multiply the
        Apply Reciproot iteration three times to y and multiply the
        result by x to get an approximation z that matches sqrt(x)
        result by x to get an approximation z that matches sqrt(x)
        to about 1 ulp. To be exact, we will have
        to about 1 ulp. To be exact, we will have
                -1ulp < sqrt(x)-z<1.0625ulp.
                -1ulp < sqrt(x)-z<1.0625ulp.
 
 
        ... set rounding mode to Round-to-nearest
        ... 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*(1.5-0.5*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)
           y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
        ... special arrangement for better accuracy
        ... special arrangement for better accuracy
           z := x*y                     ... 29 bits to sqrt(x), with z*y<1
           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)
           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
        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;
        (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
        (b) the error in the final result is biased upward so that
                -1 ulp < sqrt(x) - z < 1.0625 ulp
                -1 ulp < sqrt(x) - z < 1.0625 ulp
            instead of |sqrt(x)-z|<1.03125ulp.
            instead of |sqrt(x)-z|<1.03125ulp.
 
 
    (3) Final adjustment
    (3) Final adjustment
 
 
        By twiddling y's last bit it is possible to force y to be
        By twiddling y's last bit it is possible to force y to be
        correctly rounded according to the prevailing rounding mode
        correctly rounded according to the prevailing rounding mode
        as follows. Let r and i be copies of the rounding mode and
        as follows. Let r and i be copies of the rounding mode and
        inexact flag before entering the square root program. Also we
        inexact flag before entering the square root program. Also we
        use the expression y+-ulp for the next representable floating
        use the expression y+-ulp for the next representable floating
        numbers (up and down) of y. Note that y+-ulp = either fixed
        numbers (up and down) of y. Note that y+-ulp = either fixed
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
        mode.
        mode.
 
 
        R := RZ;                ... set rounding mode to round-toward-zero
        R := RZ;                ... set rounding mode to round-toward-zero
        switch(r) {
        switch(r) {
            case RN:            ... round-to-nearest
            case RN:            ... round-to-nearest
               if(x<= z*(z-ulp)...chopped) z = z - ulp; else
               if(x<= z*(z-ulp)...chopped) z = z - ulp; else
               if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
               if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
               break;
               break;
            case RZ:case RM:    ... round-to-zero or round-to--inf
            case RZ:case RM:    ... round-to-zero or round-to--inf
               R:=RP;           ... reset rounding mod to 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*z ... rounded up) z = z - ulp; else
               if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
               if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
               break;
               break;
            case RP:            ... round-to-+inf
            case RP:            ... round-to-+inf
               if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
               if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
               if(x>z*z ...chopped) z = z+ulp;
               if(x>z*z ...chopped) z = z+ulp;
               break;
               break;
        }
        }
 
 
        Remark 3. The above comparisons can be done in fixed point. For
        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
        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
        x1 and w1 (the trailing parts of x and w), regarding them as
        two's complement integers.
        two's complement integers.
 
 
        ...Is z an exact square root?
        ...Is z an exact square root?
        To determine whether z is an exact square root of x, let z1 be the
        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 part of z, and also let x0 and x1 be the leading and
        trailing parts of x.
        trailing parts of x.
 
 
        If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
        If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
            I := 1;             ... Raise Inexact flag: z is not exact
            I := 1;             ... Raise Inexact flag: z is not exact
        else {
        else {
            j := 1 - [(x0>>20)&1]       ... j = logb(x) mod 2
            j := 1 - [(x0>>20)&1]       ... j = logb(x) mod 2
            k := z1 >> 26;              ... get z's 25-th and 26-th
            k := z1 >> 26;              ... get z's 25-th and 26-th
                                            fraction bits
                                            fraction bits
            I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
            I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
        }
        }
        R:= r           ... restore rounded mode
        R:= r           ... restore rounded mode
        return sqrt(x):=z.
        return sqrt(x):=z.
 
 
        If multiplication is cheaper then the foregoing red tape, the
        If multiplication is cheaper then the foregoing red tape, the
        Inexact flag can be evaluated by
        Inexact flag can be evaluated by
 
 
            I := i;
            I := i;
            I := (z*z!=x) or I.
            I := (z*z!=x) or I.
 
 
        Note that z*z can overwrite I; this value must be sensed if it is
        Note that z*z can overwrite I; this value must be sensed if it is
        True.
        True.
 
 
        Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
        Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
        zero.
        zero.
 
 
                    --------------------
                    --------------------
                z1: |        f2        |
                z1: |        f2        |
                    --------------------
                    --------------------
                bit 31             bit 0
                bit 31             bit 0
 
 
        Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
        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:
        or even of logb(x) have the following relations:
 
 
        -------------------------------------------------
        -------------------------------------------------
        bit 27,26 of z1         bit 1,0 of x1   logb(x)
        bit 27,26 of z1         bit 1,0 of x1   logb(x)
        -------------------------------------------------
        -------------------------------------------------
        00                      00              odd and even
        00                      00              odd and even
        01                      01              even
        01                      01              even
        10                      10              odd
        10                      10              odd
        10                      00              even
        10                      00              even
        11                      01              even
        11                      01              even
        -------------------------------------------------
        -------------------------------------------------
 
 
    (4) Special cases (see (4) of Section A).
    (4) Special cases (see (4) of Section A).
 
 
 */
 */
 
 

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