src/video/e_sqrt.h
changeset 2756 a98604b691c8
parent 2755 2a3ec308d995
child 2757 0581f49c9294
equal deleted inserted replaced
2755:2a3ec308d995 2756:a98604b691c8
     1 /* @(#)e_sqrt.c 5.1 93/09/24 */
       
     2 /*
       
     3  * ====================================================
       
     4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
       
     5  *
       
     6  * Developed at SunPro, a Sun Microsystems, Inc. business.
       
     7  * Permission to use, copy, modify, and distribute this
       
     8  * software is freely granted, provided that this notice
       
     9  * is preserved.
       
    10  * ====================================================
       
    11  */
       
    12 
       
    13 #if defined(LIBM_SCCS) && !defined(lint)
       
    14 static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
       
    15 #endif
       
    16 
       
    17 /* __ieee754_sqrt(x)
       
    18  * Return correctly rounded sqrt.
       
    19  *           ------------------------------------------
       
    20  *	     |  Use the hardware sqrt if you have one |
       
    21  *           ------------------------------------------
       
    22  * Method:
       
    23  *   Bit by bit method using integer arithmetic. (Slow, but portable)
       
    24  *   1. Normalization
       
    25  *	Scale x to y in [1,4) with even powers of 2:
       
    26  *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
       
    27  *		sqrt(x) = 2^k * sqrt(y)
       
    28  *   2. Bit by bit computation
       
    29  *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
       
    30  *	     i							 0
       
    31  *                                     i+1         2
       
    32  *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
       
    33  *	     i      i            i                 i
       
    34  *
       
    35  *	To compute q    from q , one checks whether
       
    36  *		    i+1       i
       
    37  *
       
    38  *			      -(i+1) 2
       
    39  *			(q + 2      ) <= y.			(2)
       
    40  *     			  i
       
    41  *							      -(i+1)
       
    42  *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
       
    43  *		 	       i+1   i             i+1   i
       
    44  *
       
    45  *	With some algebric manipulation, it is not difficult to see
       
    46  *	that (2) is equivalent to
       
    47  *                             -(i+1)
       
    48  *			s  +  2       <= y			(3)
       
    49  *			 i                i
       
    50  *
       
    51  *	The advantage of (3) is that s  and y  can be computed by
       
    52  *				      i      i
       
    53  *	the following recurrence formula:
       
    54  *	    if (3) is false
       
    55  *
       
    56  *	    s     =  s  ,	y    = y   ;			(4)
       
    57  *	     i+1      i		 i+1    i
       
    58  *
       
    59  *	    otherwise,
       
    60  *                         -i                     -(i+1)
       
    61  *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
       
    62  *           i+1      i          i+1    i     i
       
    63  *
       
    64  *	One may easily use induction to prove (4) and (5).
       
    65  *	Note. Since the left hand side of (3) contain only i+2 bits,
       
    66  *	      it does not necessary to do a full (53-bit) comparison
       
    67  *	      in (3).
       
    68  *   3. Final rounding
       
    69  *	After generating the 53 bits result, we compute one more bit.
       
    70  *	Together with the remainder, we can decide whether the
       
    71  *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
       
    72  *	(it will never equal to 1/2ulp).
       
    73  *	The rounding mode can be detected by checking whether
       
    74  *	huge + tiny is equal to huge, and whether huge - tiny is
       
    75  *	equal to huge for some floating point number "huge" and "tiny".
       
    76  *
       
    77  * Special cases:
       
    78  *	sqrt(+-0) = +-0 	... exact
       
    79  *	sqrt(inf) = inf
       
    80  *	sqrt(-ve) = NaN		... with invalid signal
       
    81  *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
       
    82  *
       
    83  * Other methods : see the appended file at the end of the program below.
       
    84  *---------------
       
    85  */
       
    86 
       
    87 /*#include "math.h"*/
       
    88 #include "math_private.h"
       
    89 
       
    90 #ifdef __STDC__
       
    91 double SDL_NAME(copysign) (double x, double y)
       
    92 #else
       
    93 double SDL_NAME(copysign) (x, y)
       
    94      double
       
    95          x,
       
    96          y;
       
    97 #endif
       
    98 {
       
    99     u_int32_t hx, hy;
       
   100     GET_HIGH_WORD(hx, x);
       
   101     GET_HIGH_WORD(hy, y);
       
   102     SET_HIGH_WORD(x, (hx & 0x7fffffff) | (hy & 0x80000000));
       
   103     return x;
       
   104 }
       
   105 
       
   106 #ifdef __STDC__
       
   107 double SDL_NAME(scalbn) (double x, int n)
       
   108 #else
       
   109 double SDL_NAME(scalbn) (x, n)
       
   110      double
       
   111          x;
       
   112      int
       
   113          n;
       
   114 #endif
       
   115 {
       
   116     int32_t k, hx, lx;
       
   117     EXTRACT_WORDS(hx, lx, x);
       
   118     k = (hx & 0x7ff00000) >> 20;        /* extract exponent */
       
   119     if (k == 0) {               /* 0 or subnormal x */
       
   120         if ((lx | (hx & 0x7fffffff)) == 0)
       
   121             return x;           /* +-0 */
       
   122         x *= two54;
       
   123         GET_HIGH_WORD(hx, x);
       
   124         k = ((hx & 0x7ff00000) >> 20) - 54;
       
   125         if (n < -50000)
       
   126             return tiny * x;    /*underflow */
       
   127     }
       
   128     if (k == 0x7ff)
       
   129         return x + x;           /* NaN or Inf */
       
   130     k = k + n;
       
   131     if (k > 0x7fe)
       
   132         return huge * SDL_NAME(copysign) (huge, x);     /* overflow  */
       
   133     if (k > 0) {                /* normal result */
       
   134         SET_HIGH_WORD(x, (hx & 0x800fffff) | (k << 20));
       
   135         return x;
       
   136     }
       
   137     if (k <= -54) {
       
   138         if (n > 50000)          /* in case integer overflow in n+k */
       
   139             return huge * SDL_NAME(copysign) (huge, x); /*overflow */
       
   140         else
       
   141             return tiny * SDL_NAME(copysign) (tiny, x); /*underflow */
       
   142     }
       
   143     k += 54;                    /* subnormal result */
       
   144     SET_HIGH_WORD(x, (hx & 0x800fffff) | (k << 20));
       
   145     return x * twom54;
       
   146 }
       
   147 
       
   148 #ifdef __STDC__
       
   149 double
       
   150 __ieee754_sqrt(double x)
       
   151 #else
       
   152 double
       
   153 __ieee754_sqrt(x)
       
   154      double x;
       
   155 #endif
       
   156 {
       
   157     double z;
       
   158     int32_t sign = (int) 0x80000000;
       
   159     int32_t ix0, s0, q, m, t, i;
       
   160     u_int32_t r, t1, s1, ix1, q1;
       
   161 
       
   162     EXTRACT_WORDS(ix0, ix1, x);
       
   163 
       
   164     /* take care of Inf and NaN */
       
   165     if ((ix0 & 0x7ff00000) == 0x7ff00000) {
       
   166         return x * x + x;       /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
       
   167                                    sqrt(-inf)=sNaN */
       
   168     }
       
   169     /* take care of zero */
       
   170     if (ix0 <= 0) {
       
   171         if (((ix0 & (~sign)) | ix1) == 0)
       
   172             return x;           /* sqrt(+-0) = +-0 */
       
   173         else if (ix0 < 0)
       
   174             return (x - x) / (x - x);   /* sqrt(-ve) = sNaN */
       
   175     }
       
   176     /* normalize x */
       
   177     m = (ix0 >> 20);
       
   178     if (m == 0) {               /* subnormal x */
       
   179         while (ix0 == 0) {
       
   180             m -= 21;
       
   181             ix0 |= (ix1 >> 11);
       
   182             ix1 <<= 21;
       
   183         }
       
   184         for (i = 0; (ix0 & 0x00100000) == 0; i++)
       
   185             ix0 <<= 1;
       
   186         m -= i - 1;
       
   187         ix0 |= (ix1 >> (32 - i));
       
   188         ix1 <<= i;
       
   189     }
       
   190     m -= 1023;                  /* unbias exponent */
       
   191     ix0 = (ix0 & 0x000fffff) | 0x00100000;
       
   192     if (m & 1) {                /* odd m, double x to make it even */
       
   193         ix0 += ix0 + ((ix1 & sign) >> 31);
       
   194         ix1 += ix1;
       
   195     }
       
   196     m >>= 1;                    /* m = [m/2] */
       
   197 
       
   198     /* generate sqrt(x) bit by bit */
       
   199     ix0 += ix0 + ((ix1 & sign) >> 31);
       
   200     ix1 += ix1;
       
   201     q = q1 = s0 = s1 = 0;       /* [q,q1] = sqrt(x) */
       
   202     r = 0x00200000;             /* r = moving bit from right to left */
       
   203 
       
   204     while (r != 0) {
       
   205         t = s0 + r;
       
   206         if (t <= ix0) {
       
   207             s0 = t + r;
       
   208             ix0 -= t;
       
   209             q += r;
       
   210         }
       
   211         ix0 += ix0 + ((ix1 & sign) >> 31);
       
   212         ix1 += ix1;
       
   213         r >>= 1;
       
   214     }
       
   215 
       
   216     r = sign;
       
   217     while (r != 0) {
       
   218         t1 = s1 + r;
       
   219         t = s0;
       
   220         if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) {
       
   221             s1 = t1 + r;
       
   222             if (((int32_t) (t1 & sign) == sign) && (s1 & sign) == 0)
       
   223                 s0 += 1;
       
   224             ix0 -= t;
       
   225             if (ix1 < t1)
       
   226                 ix0 -= 1;
       
   227             ix1 -= t1;
       
   228             q1 += r;
       
   229         }
       
   230         ix0 += ix0 + ((ix1 & sign) >> 31);
       
   231         ix1 += ix1;
       
   232         r >>= 1;
       
   233     }
       
   234 
       
   235     /* use floating add to find out rounding direction */
       
   236     if ((ix0 | ix1) != 0) {
       
   237         z = one - tiny;         /* trigger inexact flag */
       
   238         if (z >= one) {
       
   239             z = one + tiny;
       
   240             if (q1 == (u_int32_t) 0xffffffff) {
       
   241                 q1 = 0;
       
   242                 q += 1;
       
   243             } else if (z > one) {
       
   244                 if (q1 == (u_int32_t) 0xfffffffe)
       
   245                     q += 1;
       
   246                 q1 += 2;
       
   247             } else
       
   248                 q1 += (q1 & 1);
       
   249         }
       
   250     }
       
   251     ix0 = (q >> 1) + 0x3fe00000;
       
   252     ix1 = q1 >> 1;
       
   253     if ((q & 1) == 1)
       
   254         ix1 |= sign;
       
   255     ix0 += (m << 20);
       
   256     INSERT_WORDS(z, ix0, ix1);
       
   257     return z;
       
   258 }
       
   259 
       
   260 /*
       
   261 Other methods  (use floating-point arithmetic)
       
   262 -------------
       
   263 (This is a copy of a drafted paper by Prof W. Kahan
       
   264 and K.C. Ng, written in May, 1986)
       
   265 
       
   266 	Two algorithms are given here to implement sqrt(x)
       
   267 	(IEEE double precision arithmetic) in software.
       
   268 	Both supply sqrt(x) correctly rounded. The first algorithm (in
       
   269 	Section A) uses newton iterations and involves four divisions.
       
   270 	The second one uses reciproot iterations to avoid division, but
       
   271 	requires more multiplications. Both algorithms need the ability
       
   272 	to chop results of arithmetic operations instead of round them,
       
   273 	and the INEXACT flag to indicate when an arithmetic operation
       
   274 	is executed exactly with no roundoff error, all part of the
       
   275 	standard (IEEE 754-1985). The ability to perform shift, add,
       
   276 	subtract and logical AND operations upon 32-bit words is needed
       
   277 	too, though not part of the standard.
       
   278 
       
   279 A.  sqrt(x) by Newton Iteration
       
   280 
       
   281    (1)	Initial approximation
       
   282 
       
   283 	Let x0 and x1 be the leading and the trailing 32-bit words of
       
   284 	a floating point number x (in IEEE double format) respectively
       
   285 
       
   286 	    1    11		     52				  ...widths
       
   287 	   ------------------------------------------------------
       
   288 	x: |s|	  e     |	      f				|
       
   289 	   ------------------------------------------------------
       
   290 	      msb    lsb  msb				      lsb ...order
       
   291 
       
   292 
       
   293 	     ------------------------  	     ------------------------
       
   294 	x0:  |s|   e    |    f1     |	 x1: |          f2           |
       
   295 	     ------------------------  	     ------------------------
       
   296 
       
   297 	By performing shifts and subtracts on x0 and x1 (both regarded
       
   298 	as integers), we obtain an 8-bit approximation of sqrt(x) as
       
   299 	follows.
       
   300 
       
   301 		k  := (x0>>1) + 0x1ff80000;
       
   302 		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
       
   303 	Here k is a 32-bit integer and T1[] is an integer array containing
       
   304 	correction terms. Now magically the floating value of y (y's
       
   305 	leading 32-bit word is y0, the value of its trailing word is 0)
       
   306 	approximates sqrt(x) to almost 8-bit.
       
   307 
       
   308 	Value of T1:
       
   309 	static int T1[32]= {
       
   310 	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
       
   311 	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
       
   312 	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
       
   313 	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
       
   314 
       
   315     (2)	Iterative refinement
       
   316 
       
   317 	Apply Heron's rule three times to y, we have y approximates
       
   318 	sqrt(x) to within 1 ulp (Unit in the Last Place):
       
   319 
       
   320 		y := (y+x/y)/2		... almost 17 sig. bits
       
   321 		y := (y+x/y)/2		... almost 35 sig. bits
       
   322 		y := y-(y-x/y)/2	... within 1 ulp
       
   323 
       
   324 
       
   325 	Remark 1.
       
   326 	    Another way to improve y to within 1 ulp is:
       
   327 
       
   328 		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
       
   329 		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
       
   330 
       
   331 				2
       
   332 			    (x-y )*y
       
   333 		y := y + 2* ----------	...within 1 ulp
       
   334 			       2
       
   335 			     3y  + x
       
   336 
       
   337 
       
   338 	This formula has one division fewer than the one above; however,
       
   339 	it requires more multiplications and additions. Also x must be
       
   340 	scaled in advance to avoid spurious overflow in evaluating the
       
   341 	expression 3y*y+x. Hence it is not recommended uless division
       
   342 	is slow. If division is very slow, then one should use the
       
   343 	reciproot algorithm given in section B.
       
   344 
       
   345     (3) Final adjustment
       
   346 
       
   347 	By twiddling y's last bit it is possible to force y to be
       
   348 	correctly rounded according to the prevailing rounding mode
       
   349 	as follows. Let r and i be copies of the rounding mode and
       
   350 	inexact flag before entering the square root program. Also we
       
   351 	use the expression y+-ulp for the next representable floating
       
   352 	numbers (up and down) of y. Note that y+-ulp = either fixed
       
   353 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
       
   354 	mode.
       
   355 
       
   356 		I := FALSE;	... reset INEXACT flag I
       
   357 		R := RZ;	... set rounding mode to round-toward-zero
       
   358 		z := x/y;	... chopped quotient, possibly inexact
       
   359 		If(not I) then {	... if the quotient is exact
       
   360 		    if(z=y) {
       
   361 		        I := i;	 ... restore inexact flag
       
   362 		        R := r;  ... restore rounded mode
       
   363 		        return sqrt(x):=y.
       
   364 		    } else {
       
   365 			z := z - ulp;	... special rounding
       
   366 		    }
       
   367 		}
       
   368 		i := TRUE;		... sqrt(x) is inexact
       
   369 		If (r=RN) then z=z+ulp	... rounded-to-nearest
       
   370 		If (r=RP) then {	... round-toward-+inf
       
   371 		    y = y+ulp; z=z+ulp;
       
   372 		}
       
   373 		y := y+z;		... chopped sum
       
   374 		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
       
   375 	        I := i;	 		... restore inexact flag
       
   376 	        R := r;  		... restore rounded mode
       
   377 	        return sqrt(x):=y.
       
   378 
       
   379     (4)	Special cases
       
   380 
       
   381 	Square root of +inf, +-0, or NaN is itself;
       
   382 	Square root of a negative number is NaN with invalid signal.
       
   383 
       
   384 
       
   385 B.  sqrt(x) by Reciproot Iteration
       
   386 
       
   387    (1)	Initial approximation
       
   388 
       
   389 	Let x0 and x1 be the leading and the trailing 32-bit words of
       
   390 	a floating point number x (in IEEE double format) respectively
       
   391 	(see section A). By performing shifs and subtracts on x0 and y0,
       
   392 	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
       
   393 
       
   394 	    k := 0x5fe80000 - (x0>>1);
       
   395 	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
       
   396 
       
   397 	Here k is a 32-bit integer and T2[] is an integer array
       
   398 	containing correction terms. Now magically the floating
       
   399 	value of y (y's leading 32-bit word is y0, the value of
       
   400 	its trailing word y1 is set to zero) approximates 1/sqrt(x)
       
   401 	to almost 7.8-bit.
       
   402 
       
   403 	Value of T2:
       
   404 	static int T2[64]= {
       
   405 	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
       
   406 	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
       
   407 	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
       
   408 	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
       
   409 	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
       
   410 	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
       
   411 	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
       
   412 	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
       
   413 
       
   414     (2)	Iterative refinement
       
   415 
       
   416 	Apply Reciproot iteration three times to y and multiply the
       
   417 	result by x to get an approximation z that matches sqrt(x)
       
   418 	to about 1 ulp. To be exact, we will have
       
   419 		-1ulp < sqrt(x)-z<1.0625ulp.
       
   420 
       
   421 	... set rounding mode to Round-to-nearest
       
   422 	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
       
   423 	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
       
   424 	... special arrangement for better accuracy
       
   425 	   z := x*y			... 29 bits to sqrt(x), with z*y<1
       
   426 	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
       
   427 
       
   428 	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
       
   429 	(a) the term z*y in the final iteration is always less than 1;
       
   430 	(b) the error in the final result is biased upward so that
       
   431 		-1 ulp < sqrt(x) - z < 1.0625 ulp
       
   432 	    instead of |sqrt(x)-z|<1.03125ulp.
       
   433 
       
   434     (3)	Final adjustment
       
   435 
       
   436 	By twiddling y's last bit it is possible to force y to be
       
   437 	correctly rounded according to the prevailing rounding mode
       
   438 	as follows. Let r and i be copies of the rounding mode and
       
   439 	inexact flag before entering the square root program. Also we
       
   440 	use the expression y+-ulp for the next representable floating
       
   441 	numbers (up and down) of y. Note that y+-ulp = either fixed
       
   442 	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
       
   443 	mode.
       
   444 
       
   445 	R := RZ;		... set rounding mode to round-toward-zero
       
   446 	switch(r) {
       
   447 	    case RN:		... round-to-nearest
       
   448 	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
       
   449 	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
       
   450 	       break;
       
   451 	    case RZ:case RM:	... round-to-zero or round-to--inf
       
   452 	       R:=RP;		... reset rounding mod to round-to-+inf
       
   453 	       if(x<z*z ... rounded up) z = z - ulp; else
       
   454 	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
       
   455 	       break;
       
   456 	    case RP:		... round-to-+inf
       
   457 	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
       
   458 	       if(x>z*z ...chopped) z = z+ulp;
       
   459 	       break;
       
   460 	}
       
   461 
       
   462 	Remark 3. The above comparisons can be done in fixed point. For
       
   463 	example, to compare x and w=z*z chopped, it suffices to compare
       
   464 	x1 and w1 (the trailing parts of x and w), regarding them as
       
   465 	two's complement integers.
       
   466 
       
   467 	...Is z an exact square root?
       
   468 	To determine whether z is an exact square root of x, let z1 be the
       
   469 	trailing part of z, and also let x0 and x1 be the leading and
       
   470 	trailing parts of x.
       
   471 
       
   472 	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
       
   473 	    I := 1;		... Raise Inexact flag: z is not exact
       
   474 	else {
       
   475 	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
       
   476 	    k := z1 >> 26;		... get z's 25-th and 26-th
       
   477 					    fraction bits
       
   478 	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
       
   479 	}
       
   480 	R:= r		... restore rounded mode
       
   481 	return sqrt(x):=z.
       
   482 
       
   483 	If multiplication is cheaper then the foregoing red tape, the
       
   484 	Inexact flag can be evaluated by
       
   485 
       
   486 	    I := i;
       
   487 	    I := (z*z!=x) or I.
       
   488 
       
   489 	Note that z*z can overwrite I; this value must be sensed if it is
       
   490 	True.
       
   491 
       
   492 	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
       
   493 	zero.
       
   494 
       
   495 		    --------------------
       
   496 		z1: |        f2        |
       
   497 		    --------------------
       
   498 		bit 31		   bit 0
       
   499 
       
   500 	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
       
   501 	or even of logb(x) have the following relations:
       
   502 
       
   503 	-------------------------------------------------
       
   504 	bit 27,26 of z1		bit 1,0 of x1	logb(x)
       
   505 	-------------------------------------------------
       
   506 	00			00		odd and even
       
   507 	01			01		even
       
   508 	10			10		odd
       
   509 	10			00		even
       
   510 	11			01		even
       
   511 	-------------------------------------------------
       
   512 
       
   513     (4)	Special cases (see (4) of Section A).
       
   514 
       
   515  */
       
   516 /* vi: set ts=4 sw=4 expandtab: */