text: /* Copyright JS Foundation and other contributors, http://js.foundation
   *
   * Licensed under the Apache License, Version 2.0 (the "License");
   * you may not use this file except in compliance with the License.
   * You may obtain a copy of the License at
   *
   *     http://www.apache.org/licenses/LICENSE-2.0
   *
   * Unless required by applicable law or agreed to in writing, software
   * distributed under the License is distributed on an "AS IS" BASIS
   * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
   * See the License for the specific language governing permissions and
   * limitations under the License.
   *
   * This file is based on work under the following copyright and permission
   * notice:
   *
   *     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.
   *
   *     @(#)e_sqrt.c 1.3 95/01/18
   */
  
  #include "jerry-libm-internal.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
   *
   * Other methods: see the appended file at the end of the program below.
   */
  
  #define one  1.0
  #define tiny 1.0e-300
  
  double
  sqrt (double x)
  {
    int sign = (int) 0x80000000;
    unsigned r, t1, s1, ix1, q1;
    int ix0, s0, q, m, t, i;
  
    ix0 = __HI (x); /* high word of x */
    ix1 = __LO (x); /* low word of 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) /* sqrt(+-0) = +-0 */
      {
        return x;
      }
      else if (ix0 < 0) /* sqrt(-ve) = sNaN */
      {
        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 >>= 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;
    }
  
    double_accessor ret;
  
    /* use floating add to find out rounding direction */
    if ((ix0 | ix1) != 0)
    {
      ret.dbl = one - tiny; /* trigger inexact flag */
      if (ret.dbl >= one)
      {
        ret.dbl = one + tiny;
        if (q1 == (unsigned) 0xffffffff)
        {
          q1 = 0;
          q += 1;
        }
        else if (ret.dbl > one)
        {
          if (q1 == (unsigned) 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);
    ret.as_int.hi = ix0;
    ret.as_int.lo = ix1;
    return ret.dbl;
  } /* sqrt */
  
  #undef one
  #undef tiny
  
  /*
  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).
   */
  