initial check-in, version 0.5.0 v0.5.0

1 files changed, 442 insertions, 0 deletions

diff --git a/src/math/e_sqrt.c b/src/math/e_sqrt.c
new file mode 100644
index 00000000..2bc68747
--- /dev/null
+++ b/src/math/e_sqrt.c
@@ -0,0 +1,442 @@
+
+/* @(#)e_sqrt.c 1.3 95/01/18 */
+/*
+ * ====================================================
+ * 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.
+ * ====================================================
+ */
+
+/* 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"
+
+static  const double    one     = 1.0, tiny=1.0e-300;
+
+double
+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 */
+            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==(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;
+}
+
+/*
+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).   
+ 
+ */
+