src/video/e_sqrt.h
author Sam Lantinga <slouken@libsdl.org>
Mon, 21 Sep 2009 08:58:51 +0000
branchSDL-1.2
changeset 4214 4250beeb5ad1
parent 1424 7a610f25c12f
child 1662 782fd950bd46
child 1895 c121d94672cb
permissions -rw-r--r--
Oh yeah, we have GLX support too.
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/* @(#)e_sqrt.c 5.1 93/09/24 */
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/*
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 * ====================================================
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 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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 *
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 * Developed at SunPro, a Sun Microsystems, Inc. business.
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 * Permission to use, copy, modify, and distribute this
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 * software is freely granted, provided that this notice
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 * is preserved.
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 * ====================================================
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 */
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#if defined(LIBM_SCCS) && !defined(lint)
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static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
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#endif
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/* __ieee754_sqrt(x)
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 * Return correctly rounded sqrt.
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 *           ------------------------------------------
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 *	     |  Use the hardware sqrt if you have one |
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 *           ------------------------------------------
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 * Method:
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 *   Bit by bit method using integer arithmetic. (Slow, but portable)
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 *   1. Normalization
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 *	Scale x to y in [1,4) with even powers of 2:
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 *	find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
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 *		sqrt(x) = 2^k * sqrt(y)
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 *   2. Bit by bit computation
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 *	Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
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 *	     i							 0
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 *                                     i+1         2
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 *	    s  = 2*q , and	y  =  2   * ( y - q  ).		(1)
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 *	     i      i            i                 i
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 *
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 *	To compute q    from q , one checks whether
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 *		    i+1       i
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 *
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 *			      -(i+1) 2
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 *			(q + 2      ) <= y.			(2)
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 *     			  i
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 *							      -(i+1)
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 *	If (2) is false, then q   = q ; otherwise q   = q  + 2      .
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 *		 	       i+1   i             i+1   i
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 *
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 *	With some algebric manipulation, it is not difficult to see
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 *	that (2) is equivalent to
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 *                             -(i+1)
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 *			s  +  2       <= y			(3)
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 *			 i                i
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 *
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 *	The advantage of (3) is that s  and y  can be computed by
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 *				      i      i
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 *	the following recurrence formula:
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 *	    if (3) is false
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 *
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 *	    s     =  s  ,	y    = y   ;			(4)
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 *	     i+1      i		 i+1    i
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 *
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 *	    otherwise,
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 *                         -i                     -(i+1)
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 *	    s	  =  s  + 2  ,  y    = y  -  s  - 2  		(5)
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 *           i+1      i          i+1    i     i
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 *
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 *	One may easily use induction to prove (4) and (5).
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 *	Note. Since the left hand side of (3) contain only i+2 bits,
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 *	      it does not necessary to do a full (53-bit) comparison
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 *	      in (3).
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 *   3. Final rounding
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 *	After generating the 53 bits result, we compute one more bit.
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 *	Together with the remainder, we can decide whether the
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 *	result is exact, bigger than 1/2ulp, or less than 1/2ulp
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 *	(it will never equal to 1/2ulp).
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 *	The rounding mode can be detected by checking whether
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 *	huge + tiny is equal to huge, and whether huge - tiny is
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 *	equal to huge for some floating point number "huge" and "tiny".
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 *
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 * Special cases:
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 *	sqrt(+-0) = +-0 	... exact
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 *	sqrt(inf) = inf
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 *	sqrt(-ve) = NaN		... with invalid signal
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 *	sqrt(NaN) = NaN		... with invalid signal for signaling NaN
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 *
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 * Other methods : see the appended file at the end of the program below.
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 *---------------
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 */
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/*#include "math.h"*/
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#include "math_private.h"
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#ifdef __STDC__
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	double SDL_NAME(copysign)(double x, double y)
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#else
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	double SDL_NAME(copysign)(x,y)
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	double x,y;
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#endif
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{
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	u_int32_t hx,hy;
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	GET_HIGH_WORD(hx,x);
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	GET_HIGH_WORD(hy,y);
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	SET_HIGH_WORD(x,(hx&0x7fffffff)|(hy&0x80000000));
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        return x;
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}
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#ifdef __STDC__
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	double SDL_NAME(scalbn) (double x, int n)
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#else
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	double SDL_NAME(scalbn) (x,n)
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	double x; int n;
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#endif
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{
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	int32_t k,hx,lx;
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	EXTRACT_WORDS(hx,lx,x);
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        k = (hx&0x7ff00000)>>20;		/* extract exponent */
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        if (k==0) {				/* 0 or subnormal x */
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            if ((lx|(hx&0x7fffffff))==0) return x; /* +-0 */
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	    x *= two54;
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	    GET_HIGH_WORD(hx,x);
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	    k = ((hx&0x7ff00000)>>20) - 54;
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            if (n< -50000) return tiny*x; 	/*underflow*/
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	    }
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        if (k==0x7ff) return x+x;		/* NaN or Inf */
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        k = k+n;
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        if (k >  0x7fe) return huge*SDL_NAME(copysign)(huge,x); /* overflow  */
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        if (k > 0) 				/* normal result */
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	    {SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); return x;}
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        if (k <= -54) {
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            if (n > 50000) 	/* in case integer overflow in n+k */
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		return huge*SDL_NAME(copysign)(huge,x);	/*overflow*/
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	    else return tiny*SDL_NAME(copysign)(tiny,x); 	/*underflow*/
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	}
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        k += 54;				/* subnormal result */
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	SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20));
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        return x*twom54;
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}
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#ifdef __STDC__
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	double __ieee754_sqrt(double x)
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#else
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	double __ieee754_sqrt(x)
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	double x;
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#endif
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{
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	double z;
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	int32_t sign = (int)0x80000000;
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	int32_t ix0,s0,q,m,t,i;
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	u_int32_t r,t1,s1,ix1,q1;
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	EXTRACT_WORDS(ix0,ix1,x);
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    /* take care of Inf and NaN */
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	if((ix0&0x7ff00000)==0x7ff00000) {
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	    return x*x+x;		/* sqrt(NaN)=NaN, sqrt(+inf)=+inf
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					   sqrt(-inf)=sNaN */
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	}
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    /* take care of zero */
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	if(ix0<=0) {
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	    if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
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	    else if(ix0<0)
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		return (x-x)/(x-x);		/* sqrt(-ve) = sNaN */
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	}
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    /* normalize x */
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	m = (ix0>>20);
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	if(m==0) {				/* subnormal x */
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	    while(ix0==0) {
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		m -= 21;
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		ix0 |= (ix1>>11); ix1 <<= 21;
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	    }
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	    for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
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	    m -= i-1;
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	    ix0 |= (ix1>>(32-i));
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	    ix1 <<= i;
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	}
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	m -= 1023;	/* unbias exponent */
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	ix0 = (ix0&0x000fffff)|0x00100000;
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	if(m&1){	/* odd m, double x to make it even */
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	    ix0 += ix0 + ((ix1&sign)>>31);
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	    ix1 += ix1;
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	}
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	m >>= 1;	/* m = [m/2] */
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    /* generate sqrt(x) bit by bit */
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	ix0 += ix0 + ((ix1&sign)>>31);
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	ix1 += ix1;
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	q = q1 = s0 = s1 = 0;	/* [q,q1] = sqrt(x) */
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	r = 0x00200000;		/* r = moving bit from right to left */
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	while(r!=0) {
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	    t = s0+r;
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	    if(t<=ix0) {
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		s0   = t+r;
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		ix0 -= t;
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		q   += r;
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	    }
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	    ix0 += ix0 + ((ix1&sign)>>31);
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	    ix1 += ix1;
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	    r>>=1;
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	}
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	r = sign;
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	while(r!=0) {
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	    t1 = s1+r;
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	    t  = s0;
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	    if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
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		s1  = t1+r;
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		if(((int32_t)(t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
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		ix0 -= t;
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		if (ix1 < t1) ix0 -= 1;
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		ix1 -= t1;
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		q1  += r;
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	    }
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	    ix0 += ix0 + ((ix1&sign)>>31);
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	    ix1 += ix1;
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	    r>>=1;
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	}
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    /* use floating add to find out rounding direction */
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	if((ix0|ix1)!=0) {
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	    z = one-tiny; /* trigger inexact flag */
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	    if (z>=one) {
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	        z = one+tiny;
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	        if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
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		else if (z>one) {
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		    if (q1==(u_int32_t)0xfffffffe) q+=1;
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		    q1+=2;
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		} else
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	            q1 += (q1&1);
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	    }
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	}
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	ix0 = (q>>1)+0x3fe00000;
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	ix1 =  q1>>1;
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	if ((q&1)==1) ix1 |= sign;
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	ix0 += (m <<20);
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	INSERT_WORDS(z,ix0,ix1);
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	return z;
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}
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/*
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Other methods  (use floating-point arithmetic)
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-------------
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(This is a copy of a drafted paper by Prof W. Kahan
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and K.C. Ng, written in May, 1986)
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	Two algorithms are given here to implement sqrt(x)
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	(IEEE double precision arithmetic) in software.
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	Both supply sqrt(x) correctly rounded. The first algorithm (in
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	Section A) uses newton iterations and involves four divisions.
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	The second one uses reciproot iterations to avoid division, but
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	requires more multiplications. Both algorithms need the ability
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	to chop results of arithmetic operations instead of round them,
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	and the INEXACT flag to indicate when an arithmetic operation
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	is executed exactly with no roundoff error, all part of the
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	standard (IEEE 754-1985). The ability to perform shift, add,
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	subtract and logical AND operations upon 32-bit words is needed
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	too, though not part of the standard.
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A.  sqrt(x) by Newton Iteration
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   (1)	Initial approximation
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	Let x0 and x1 be the leading and the trailing 32-bit words of
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	a floating point number x (in IEEE double format) respectively
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	    1    11		     52				  ...widths
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	   ------------------------------------------------------
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	x: |s|	  e     |	      f				|
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	   ------------------------------------------------------
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	      msb    lsb  msb				      lsb ...order
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	     ------------------------  	     ------------------------
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	x0:  |s|   e    |    f1     |	 x1: |          f2           |
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	     ------------------------  	     ------------------------
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	By performing shifts and subtracts on x0 and x1 (both regarded
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	as integers), we obtain an 8-bit approximation of sqrt(x) as
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	follows.
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		k  := (x0>>1) + 0x1ff80000;
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		y0 := k - T1[31&(k>>15)].	... y ~ sqrt(x) to 8 bits
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	Here k is a 32-bit integer and T1[] is an integer array containing
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	correction terms. Now magically the floating value of y (y's
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	leading 32-bit word is y0, the value of its trailing word is 0)
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	approximates sqrt(x) to almost 8-bit.
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	Value of T1:
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	static int T1[32]= {
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	0,	1024,	3062,	5746,	9193,	13348,	18162,	23592,
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	29598,	36145,	43202,	50740,	58733,	67158,	75992,	85215,
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	83599,	71378,	60428,	50647,	41945,	34246,	27478,	21581,
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	16499,	12183,	8588,	5674,	3403,	1742,	661,	130,};
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    (2)	Iterative refinement
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	Apply Heron's rule three times to y, we have y approximates
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	sqrt(x) to within 1 ulp (Unit in the Last Place):
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		y := (y+x/y)/2		... almost 17 sig. bits
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		y := (y+x/y)/2		... almost 35 sig. bits
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		y := y-(y-x/y)/2	... within 1 ulp
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	Remark 1.
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	    Another way to improve y to within 1 ulp is:
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		y := (y+x/y)		... almost 17 sig. bits to 2*sqrt(x)
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		y := y - 0x00100006	... almost 18 sig. bits to sqrt(x)
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				2
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			    (x-y )*y
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		y := y + 2* ----------	...within 1 ulp
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			       2
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			     3y  + x
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	This formula has one division fewer than the one above; however,
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	it requires more multiplications and additions. Also x must be
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	scaled in advance to avoid spurious overflow in evaluating the
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	expression 3y*y+x. Hence it is not recommended uless division
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	is slow. If division is very slow, then one should use the
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	reciproot algorithm given in section B.
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    (3) Final adjustment
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	By twiddling y's last bit it is possible to force y to be
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	correctly rounded according to the prevailing rounding mode
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	as follows. Let r and i be copies of the rounding mode and
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	inexact flag before entering the square root program. Also we
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	use the expression y+-ulp for the next representable floating
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	numbers (up and down) of y. Note that y+-ulp = either fixed
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	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
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	mode.
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   332
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		I := FALSE;	... reset INEXACT flag I
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		R := RZ;	... set rounding mode to round-toward-zero
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		z := x/y;	... chopped quotient, possibly inexact
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   336
		If(not I) then {	... if the quotient is exact
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   337
		    if(z=y) {
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		        I := i;	 ... restore inexact flag
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		        R := r;  ... restore rounded mode
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		        return sqrt(x):=y.
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   341
		    } else {
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			z := z - ulp;	... special rounding
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   343
		    }
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   344
		}
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		i := TRUE;		... sqrt(x) is inexact
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   346
		If (r=RN) then z=z+ulp	... rounded-to-nearest
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   347
		If (r=RP) then {	... round-toward-+inf
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		    y = y+ulp; z=z+ulp;
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   349
		}
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		y := y+z;		... chopped sum
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		y0:=y0-0x00100000;	... y := y/2 is correctly rounded.
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	        I := i;	 		... restore inexact flag
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	        R := r;  		... restore rounded mode
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	        return sqrt(x):=y.
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   355
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   356
    (4)	Special cases
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   358
	Square root of +inf, +-0, or NaN is itself;
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	Square root of a negative number is NaN with invalid signal.
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   360
slouken@1330
   361
slouken@1330
   362
B.  sqrt(x) by Reciproot Iteration
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   363
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   364
   (1)	Initial approximation
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   365
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   366
	Let x0 and x1 be the leading and the trailing 32-bit words of
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   367
	a floating point number x (in IEEE double format) respectively
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	(see section A). By performing shifs and subtracts on x0 and y0,
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   369
	we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
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   370
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   371
	    k := 0x5fe80000 - (x0>>1);
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   372
	    y0:= k - T2[63&(k>>14)].	... y ~ 1/sqrt(x) to 7.8 bits
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   373
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   374
	Here k is a 32-bit integer and T2[] is an integer array
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   375
	containing correction terms. Now magically the floating
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   376
	value of y (y's leading 32-bit word is y0, the value of
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   377
	its trailing word y1 is set to zero) approximates 1/sqrt(x)
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   378
	to almost 7.8-bit.
slouken@1330
   379
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   380
	Value of T2:
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	static int T2[64]= {
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   382
	0x1500,	0x2ef8,	0x4d67,	0x6b02,	0x87be,	0xa395,	0xbe7a,	0xd866,
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   383
	0xf14a,	0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
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   384
	0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
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   385
	0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
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   386
	0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
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   387
	0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
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   388
	0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
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   389
	0x1527f,0x1334a,0x11051,0xe951,	0xbe01,	0x8e0d,	0x5924,	0x1edd,};
slouken@1330
   390
slouken@1330
   391
    (2)	Iterative refinement
slouken@1330
   392
slouken@1330
   393
	Apply Reciproot iteration three times to y and multiply the
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   394
	result by x to get an approximation z that matches sqrt(x)
slouken@1330
   395
	to about 1 ulp. To be exact, we will have
slouken@1330
   396
		-1ulp < sqrt(x)-z<1.0625ulp.
slouken@1330
   397
slouken@1330
   398
	... set rounding mode to Round-to-nearest
slouken@1330
   399
	   y := y*(1.5-0.5*x*y*y)	... almost 15 sig. bits to 1/sqrt(x)
slouken@1330
   400
	   y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
slouken@1330
   401
	... special arrangement for better accuracy
slouken@1330
   402
	   z := x*y			... 29 bits to sqrt(x), with z*y<1
slouken@1330
   403
	   z := z + 0.5*z*(1-z*y)	... about 1 ulp to sqrt(x)
slouken@1330
   404
slouken@1330
   405
	Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
slouken@1330
   406
	(a) the term z*y in the final iteration is always less than 1;
slouken@1330
   407
	(b) the error in the final result is biased upward so that
slouken@1330
   408
		-1 ulp < sqrt(x) - z < 1.0625 ulp
slouken@1330
   409
	    instead of |sqrt(x)-z|<1.03125ulp.
slouken@1330
   410
slouken@1330
   411
    (3)	Final adjustment
slouken@1330
   412
slouken@1330
   413
	By twiddling y's last bit it is possible to force y to be
slouken@1330
   414
	correctly rounded according to the prevailing rounding mode
slouken@1330
   415
	as follows. Let r and i be copies of the rounding mode and
slouken@1330
   416
	inexact flag before entering the square root program. Also we
slouken@1330
   417
	use the expression y+-ulp for the next representable floating
slouken@1330
   418
	numbers (up and down) of y. Note that y+-ulp = either fixed
slouken@1330
   419
	point y+-1, or multiply y by nextafter(1,+-inf) in chopped
slouken@1330
   420
	mode.
slouken@1330
   421
slouken@1330
   422
	R := RZ;		... set rounding mode to round-toward-zero
slouken@1330
   423
	switch(r) {
slouken@1330
   424
	    case RN:		... round-to-nearest
slouken@1330
   425
	       if(x<= z*(z-ulp)...chopped) z = z - ulp; else
slouken@1330
   426
	       if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
slouken@1330
   427
	       break;
slouken@1330
   428
	    case RZ:case RM:	... round-to-zero or round-to--inf
slouken@1330
   429
	       R:=RP;		... reset rounding mod to round-to-+inf
slouken@1330
   430
	       if(x<z*z ... rounded up) z = z - ulp; else
slouken@1330
   431
	       if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
slouken@1330
   432
	       break;
slouken@1330
   433
	    case RP:		... round-to-+inf
slouken@1330
   434
	       if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
slouken@1330
   435
	       if(x>z*z ...chopped) z = z+ulp;
slouken@1330
   436
	       break;
slouken@1330
   437
	}
slouken@1330
   438
slouken@1330
   439
	Remark 3. The above comparisons can be done in fixed point. For
slouken@1330
   440
	example, to compare x and w=z*z chopped, it suffices to compare
slouken@1330
   441
	x1 and w1 (the trailing parts of x and w), regarding them as
slouken@1330
   442
	two's complement integers.
slouken@1330
   443
slouken@1330
   444
	...Is z an exact square root?
slouken@1330
   445
	To determine whether z is an exact square root of x, let z1 be the
slouken@1330
   446
	trailing part of z, and also let x0 and x1 be the leading and
slouken@1330
   447
	trailing parts of x.
slouken@1330
   448
slouken@1330
   449
	If ((z1&0x03ffffff)!=0)	... not exact if trailing 26 bits of z!=0
slouken@1330
   450
	    I := 1;		... Raise Inexact flag: z is not exact
slouken@1330
   451
	else {
slouken@1330
   452
	    j := 1 - [(x0>>20)&1]	... j = logb(x) mod 2
slouken@1330
   453
	    k := z1 >> 26;		... get z's 25-th and 26-th
slouken@1330
   454
					    fraction bits
slouken@1330
   455
	    I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
slouken@1330
   456
	}
slouken@1330
   457
	R:= r		... restore rounded mode
slouken@1330
   458
	return sqrt(x):=z.
slouken@1330
   459
slouken@1330
   460
	If multiplication is cheaper then the foregoing red tape, the
slouken@1330
   461
	Inexact flag can be evaluated by
slouken@1330
   462
slouken@1330
   463
	    I := i;
slouken@1330
   464
	    I := (z*z!=x) or I.
slouken@1330
   465
slouken@1330
   466
	Note that z*z can overwrite I; this value must be sensed if it is
slouken@1330
   467
	True.
slouken@1330
   468
slouken@1330
   469
	Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
slouken@1330
   470
	zero.
slouken@1330
   471
slouken@1330
   472
		    --------------------
slouken@1330
   473
		z1: |        f2        |
slouken@1330
   474
		    --------------------
slouken@1330
   475
		bit 31		   bit 0
slouken@1330
   476
slouken@1330
   477
	Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
slouken@1330
   478
	or even of logb(x) have the following relations:
slouken@1330
   479
slouken@1330
   480
	-------------------------------------------------
slouken@1330
   481
	bit 27,26 of z1		bit 1,0 of x1	logb(x)
slouken@1330
   482
	-------------------------------------------------
slouken@1330
   483
	00			00		odd and even
slouken@1330
   484
	01			01		even
slouken@1330
   485
	10			10		odd
slouken@1330
   486
	10			00		even
slouken@1330
   487
	11			01		even
slouken@1330
   488
	-------------------------------------------------
slouken@1330
   489
slouken@1330
   490
    (4)	Special cases (see (4) of Section A).
slouken@1330
   491
slouken@1330
   492
 */
slouken@1330
   493