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| 2 | /* @(#)e_sqrt.c 5.1 93/09/24 */
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| 3 | /*
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| 4 | * ====================================================
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| 5 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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| 6 | *
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| 7 | * Developed at SunPro, a Sun Microsystems, Inc. business.
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| 8 | * Permission to use, copy, modify, and distribute this
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| 9 | * software is freely granted, provided that this notice
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| 10 | * is preserved.
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| 11 | * ====================================================
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| 12 | */
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| 13 |
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| 14 | /* __ieee754_sqrt(x)
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| 15 | * Return correctly rounded sqrt.
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| 16 | * ------------------------------------------
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| 17 | * | Use the hardware sqrt if you have one |
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| 18 | * ------------------------------------------
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| 19 | * Method:
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| 20 | * Bit by bit method using integer arithmetic. (Slow, but portable)
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| 21 | * 1. Normalization
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| 22 | * Scale x to y in [1,4) with even powers of 2:
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| 23 | * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
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| 24 | * sqrt(x) = 2^k * sqrt(y)
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| 25 | * 2. Bit by bit computation
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| 26 | * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
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| 27 | * i 0
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| 28 | * i+1 2
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| 29 | * s = 2*q , and y = 2 * ( y - q ). (1)
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| 30 | * i i i i
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| 31 | *
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| 32 | * To compute q from q , one checks whether
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| 33 | * i+1 i
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| 34 | *
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| 35 | * -(i+1) 2
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| 36 | * (q + 2 ) <= y. (2)
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| 37 | * i
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| 38 | * -(i+1)
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| 39 | * If (2) is false, then q = q ; otherwise q = q + 2 .
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| 40 | * i+1 i i+1 i
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| 41 | *
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| 42 | * With some algebric manipulation, it is not difficult to see
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| 43 | * that (2) is equivalent to
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| 44 | * -(i+1)
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| 45 | * s + 2 <= y (3)
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| 46 | * i i
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| 47 | *
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| 48 | * The advantage of (3) is that s and y can be computed by
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| 49 | * i i
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| 50 | * the following recurrence formula:
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| 51 | * if (3) is false
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| 52 | *
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| 53 | * s = s , y = y ; (4)
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| 54 | * i+1 i i+1 i
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| 55 | *
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| 56 | * otherwise,
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| 57 | * -i -(i+1)
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| 58 | * s = s + 2 , y = y - s - 2 (5)
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| 59 | * i+1 i i+1 i i
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| 60 | *
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| 61 | * One may easily use induction to prove (4) and (5).
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| 62 | * Note. Since the left hand side of (3) contain only i+2 bits,
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| 63 | * it does not necessary to do a full (53-bit) comparison
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| 64 | * in (3).
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| 65 | * 3. Final rounding
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| 66 | * After generating the 53 bits result, we compute one more bit.
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| 67 | * Together with the remainder, we can decide whether the
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| 68 | * result is exact, bigger than 1/2ulp, or less than 1/2ulp
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| 69 | * (it will never equal to 1/2ulp).
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| 70 | * The rounding mode can be detected by checking whether
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| 71 | * huge + tiny is equal to huge, and whether huge - tiny is
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| 72 | * equal to huge for some floating point number "huge" and "tiny".
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| 73 | *
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| 74 | * Special cases:
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| 75 | * sqrt(+-0) = +-0 ... exact
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| 76 | * sqrt(inf) = inf
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| 77 | * sqrt(-ve) = NaN ... with invalid signal
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| 78 | * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
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| 79 | *
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| 80 | * Other methods : see the appended file at the end of the program below.
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| 81 | *---------------
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| 82 | */
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| 83 |
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| 84 | #include "fdlibm.h"
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| 85 |
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| 86 | #ifndef _DOUBLE_IS_32BITS
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| 87 |
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| 88 | #ifdef __STDC__
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| 89 | static const double one = 1.0, tiny=1.0e-300;
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| 90 | #else
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| 91 | static double one = 1.0, tiny=1.0e-300;
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| 92 | #endif
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| 93 |
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| 94 | #ifdef __STDC__
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| 95 | double __ieee754_sqrt(double x)
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| 96 | #else
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| 97 | double __ieee754_sqrt(x)
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| 98 | double x;
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| 99 | #endif
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| 100 | {
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| 101 | double z;
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| 102 | int32_t sign = (int)0x80000000;
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| 103 | uint32_t r,t1,s1,ix1,q1;
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| 104 | int32_t ix0,s0,q,m,t,i;
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| 105 |
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| 106 | EXTRACT_WORDS(ix0,ix1,x);
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| 107 |
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| 108 | /* take care of Inf and NaN */
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| 109 | if((ix0&0x7ff00000)==0x7ff00000) {
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| 110 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
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| 111 | sqrt(-inf)=sNaN */
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| 112 | }
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| 113 | /* take care of zero */
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| 114 | if(ix0<=0) {
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| 115 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
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| 116 | else if(ix0<0)
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| 117 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
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| 118 | }
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| 119 | /* normalize x */
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| 120 | m = (ix0>>20);
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| 121 | if(m==0) { /* subnormal x */
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| 122 | while(ix0==0) {
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| 123 | m -= 21;
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| 124 | ix0 |= (ix1>>11); ix1 <<= 21;
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| 125 | }
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| 126 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
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| 127 | m -= i-1;
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| 128 | ix0 |= (ix1>>(32-i));
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| 129 | ix1 <<= i;
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| 130 | }
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| 131 | m -= 1023; /* unbias exponent */
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| 132 | ix0 = (ix0&0x000fffff)|0x00100000;
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| 133 | if(m&1){ /* odd m, double x to make it even */
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| 134 | ix0 += ix0 + ((ix1&sign)>>31);
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| 135 | ix1 += ix1;
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| 136 | }
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| 137 | m >>= 1; /* m = [m/2] */
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| 138 |
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| 139 | /* generate sqrt(x) bit by bit */
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| 140 | ix0 += ix0 + ((ix1&sign)>>31);
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| 141 | ix1 += ix1;
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| 142 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
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| 143 | r = 0x00200000; /* r = moving bit from right to left */
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| 144 |
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| 145 | while(r!=0) {
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| 146 | t = s0+r;
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| 147 | if(t<=ix0) {
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| 148 | s0 = t+r;
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| 149 | ix0 -= t;
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| 150 | q += r;
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| 151 | }
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| 152 | ix0 += ix0 + ((ix1&sign)>>31);
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| 153 | ix1 += ix1;
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| 154 | r>>=1;
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| 155 | }
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| 156 |
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| 157 | r = sign;
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| 158 | while(r!=0) {
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| 159 | t1 = s1+r;
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| 160 | t = s0;
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| 161 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
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| 162 | s1 = t1+r;
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| 163 | if(((t1&sign)==(uint32_t)sign)&&(s1&sign)==0) s0 += 1;
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| 164 | ix0 -= t;
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| 165 | if (ix1 < t1) ix0 -= 1;
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| 166 | ix1 -= t1;
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| 167 | q1 += r;
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| 168 | }
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| 169 | ix0 += ix0 + ((ix1&sign)>>31);
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| 170 | ix1 += ix1;
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| 171 | r>>=1;
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| 172 | }
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| 173 |
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| 174 | /* use floating add to find out rounding direction */
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| 175 | if((ix0|ix1)!=0) {
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| 176 | z = one-tiny; /* trigger inexact flag */
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| 177 | if (z>=one) {
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| 178 | z = one+tiny;
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| 179 | if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;}
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| 180 | else if (z>one) {
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| 181 | if (q1==(uint32_t)0xfffffffe) q+=1;
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| 182 | q1+=2;
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| 183 | } else
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| 184 | q1 += (q1&1);
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| 185 | }
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| 186 | }
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| 187 | ix0 = (q>>1)+0x3fe00000;
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| 188 | ix1 = q1>>1;
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| 189 | if ((q&1)==1) ix1 |= sign;
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| 190 | ix0 += (m <<20);
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| 191 | INSERT_WORDS(z,ix0,ix1);
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| 192 | return z;
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| 193 | }
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| 194 |
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| 195 | #endif /* defined(_DOUBLE_IS_32BITS) */
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| 196 |
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| 197 | /*
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| 198 | Other methods (use floating-point arithmetic)
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| 199 | -------------
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| 200 | (This is a copy of a drafted paper by Prof W. Kahan
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| 201 | and K.C. Ng, written in May, 1986)
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| 202 |
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| 203 | Two algorithms are given here to implement sqrt(x)
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| 204 | (IEEE double precision arithmetic) in software.
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| 205 | Both supply sqrt(x) correctly rounded. The first algorithm (in
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| 206 | Section A) uses newton iterations and involves four divisions.
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| 207 | The second one uses reciproot iterations to avoid division, but
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| 208 | requires more multiplications. Both algorithms need the ability
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| 209 | to chop results of arithmetic operations instead of round them,
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| 210 | and the INEXACT flag to indicate when an arithmetic operation
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| 211 | is executed exactly with no roundoff error, all part of the
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| 212 | standard (IEEE 754-1985). The ability to perform shift, add,
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| 213 | subtract and logical AND operations upon 32-bit words is needed
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| 214 | too, though not part of the standard.
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| 215 |
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| 216 | A. sqrt(x) by Newton Iteration
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| 217 |
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| 218 | (1) Initial approximation
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| 219 |
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| 220 | Let x0 and x1 be the leading and the trailing 32-bit words of
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| 221 | a floating point number x (in IEEE double format) respectively
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| 222 |
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| 223 | 1 11 52 ...widths
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| 224 | ------------------------------------------------------
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| 225 | x: |s| e | f |
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| 226 | ------------------------------------------------------
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| 227 | msb lsb msb lsb ...order
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| 228 |
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| 229 |
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| 230 | ------------------------ ------------------------
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| 231 | x0: |s| e | f1 | x1: | f2 |
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| 232 | ------------------------ ------------------------
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| 233 |
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| 234 | By performing shifts and subtracts on x0 and x1 (both regarded
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| 235 | as integers), we obtain an 8-bit approximation of sqrt(x) as
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| 236 | follows.
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| 237 |
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| 238 | k := (x0>>1) + 0x1ff80000;
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| 239 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
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| 240 | Here k is a 32-bit integer and T1[] is an integer array containing
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| 241 | correction terms. Now magically the floating value of y (y's
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| 242 | leading 32-bit word is y0, the value of its trailing word is 0)
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| 243 | approximates sqrt(x) to almost 8-bit.
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