| [048a6e9] | 1 | /*
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| 2 | * ====================================================
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| 3 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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| 4 | *
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| 5 | * Developed at SunSoft, a Sun Microsystems, Inc. business.
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| 6 | * Permission to use, copy, modify, and distribute this
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| 7 | * software is freely granted, provided that this notice
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| 8 | * is preserved.
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| 9 | * ====================================================
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| 10 | */
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| 11 |
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| 12 | /** @addtogroup libmath
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| 13 | * @{
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| 14 | */
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| 15 | /** @file sqrt mathematical function
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| 16 | */
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| 17 |
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| 18 |
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| 19 | /* __ieee754_sqrt(x)
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| 20 | * Return correctly rounded sqrt.
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| 21 | * ------------------------------------------
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| 22 | * | Use the hardware sqrt if you have one |
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| 23 | * ------------------------------------------
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| 24 | * Method:
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| 25 | * Bit by bit method using integer arithmetic. (Slow, but portable)
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| 26 | * 1. Normalization
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| 27 | * Scale x to y in [1,4) with even powers of 2:
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| 28 | * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
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| 29 | * sqrt(x) = 2^k * sqrt(y)
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| 30 | * 2. Bit by bit computation
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| 31 | * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
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| 32 | * i 0
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| 33 | * i+1 2
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| 34 | * s = 2*q , and y = 2 * ( y - q ). (1)
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| 35 | * i i i i
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| 36 | *
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| 37 | * To compute q from q , one checks whether
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| 38 | * i+1 i
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| 39 | *
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| 40 | * -(i+1) 2
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| 41 | * (q + 2 ) <= y. (2)
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| 42 | * i
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| 43 | * -(i+1)
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| 44 | * If (2) is false, then q = q ; otherwise q = q + 2 .
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| 45 | * i+1 i i+1 i
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| 46 | *
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| 47 | * With some algebric manipulation, it is not difficult to see
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| 48 | * that (2) is equivalent to
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| 49 | * -(i+1)
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| 50 | * s + 2 <= y (3)
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| 51 | * i i
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| 52 | *
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| 53 | * The advantage of (3) is that s and y can be computed by
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| 54 | * i i
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| 55 | * the following recurrence formula:
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| 56 | * if (3) is false
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| 57 | *
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| 58 | * s = s , y = y ; (4)
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| 59 | * i+1 i i+1 i
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| 60 | *
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| 61 | * otherwise,
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| 62 | * -i -(i+1)
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| 63 | * s = s + 2 , y = y - s - 2 (5)
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| 64 | * i+1 i i+1 i i
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| 65 | *
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| 66 | * One may easily use induction to prove (4) and (5).
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| 67 | * Note. Since the left hand side of (3) contain only i+2 bits,
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| 68 | * it does not necessary to do a full (53-bit) comparison
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| 69 | * in (3).
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| 70 | * 3. Final rounding
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| 71 | * After generating the 53 bits result, we compute one more bit.
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| 72 | * Together with the remainder, we can decide whether the
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| 73 | * result is exact, bigger than 1/2ulp, or less than 1/2ulp
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| 74 | * (it will never equal to 1/2ulp).
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| 75 | * The rounding mode can be detected by checking whether
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| 76 | * huge + tiny is equal to huge, and whether huge - tiny is
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| 77 | * equal to huge for some floating point number "huge" and "tiny".
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| 78 | *
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| 79 | * Special cases:
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| 80 | * sqrt(+-0) = +-0 ... exact
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| 81 | * sqrt(inf) = inf
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| 82 | * sqrt(-ve) = NaN ... with invalid signal
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| 83 | * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
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| 84 | *
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| 85 | * Other methods : see the appended file at the end of the program below.
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| 86 | *---------------
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| 87 | */
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| 88 |
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| 89 | #include <math.h>
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| 90 | #include <stdint.h>
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| 91 |
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| 92 | #include "internal.h"
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| 93 |
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| [ca113cf] | 94 | static const double one = 1.0, tiny=1.0e-300;
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| [048a6e9] | 95 |
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| 96 | double sqrt(double x)
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| 97 | {
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| 98 | double z;
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| 99 | int32_t sign = (int)0x80000000;
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| 100 | int32_t ix0,s0,q,m,t,i;
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| 101 | uint32_t r,t1,s1,ix1,q1;
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| 102 |
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| [ca113cf] | 103 | EXTRACT_WORDS(ix0,ix1,x);
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| [048a6e9] | 104 |
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| [ca113cf] | 105 | /* take care of Inf and NaN */
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| 106 | if((ix0&0x7ff00000)==0x7ff00000) {
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| 107 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
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| 108 | sqrt(-inf)=sNaN */
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| 109 | }
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| 110 | /* take care of zero */
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| 111 | if(ix0<=0) {
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| 112 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
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| 113 | else if(ix0<0)
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| 114 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
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| [048a6e9] | 115 | }
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| [ca113cf] | 116 | /* normalize x */
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| 117 | m = (ix0>>20);
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| 118 | if(m==0) { /* subnormal x */
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| 119 | while(ix0==0) {
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| 120 | m -= 21;
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| 121 | ix0 |= (ix1>>11); ix1 <<= 21;
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| 122 | }
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| 123 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
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| 124 | m -= i-1;
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| 125 | ix0 |= (ix1>>(32-i));
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| 126 | ix1 <<= i;
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| [048a6e9] | 127 | }
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| [ca113cf] | 128 | m -= 1023; /* unbias exponent */
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| [048a6e9] | 129 | ix0 = (ix0&0x000fffff)|0x00100000;
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| [ca113cf] | 130 | if(m&1){ /* odd m, double x to make it even */
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| 131 | ix0 += ix0 + ((ix1&sign)>>31);
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| 132 | ix1 += ix1;
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| [048a6e9] | 133 | }
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| [ca113cf] | 134 | m >>= 1; /* m = [m/2] */
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| [048a6e9] | 135 |
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| [ca113cf] | 136 | /* generate sqrt(x) bit by bit */
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| [048a6e9] | 137 | ix0 += ix0 + ((ix1&sign)>>31);
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| 138 | ix1 += ix1;
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| [ca113cf] | 139 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
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| 140 | r = 0x00200000; /* r = moving bit from right to left */
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| [048a6e9] | 141 |
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| [ca113cf] | 142 | while(r!=0) {
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| 143 | t = s0+r;
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| 144 | if(t<=ix0) {
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| 145 | s0 = t+r;
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| 146 | ix0 -= t;
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| 147 | q += r;
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| 148 | }
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| 149 | ix0 += ix0 + ((ix1&sign)>>31);
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| 150 | ix1 += ix1;
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| 151 | r>>=1;
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| [048a6e9] | 152 | }
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| 153 |
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| 154 | r = sign;
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| [ca113cf] | 155 | while(r!=0) {
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| 156 | t1 = s1+r;
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| 157 | t = s0;
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| 158 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
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| 159 | s1 = t1+r;
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| 160 | if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
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| 161 | ix0 -= t;
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| 162 | if (ix1 < t1) ix0 -= 1;
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| 163 | ix1 -= t1;
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| 164 | q1 += r;
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| 165 | }
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| 166 | ix0 += ix0 + ((ix1&sign)>>31);
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| 167 | ix1 += ix1;
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| 168 | r>>=1;
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| [048a6e9] | 169 | }
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| 170 |
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| [ca113cf] | 171 | /* use floating add to find out rounding direction */
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| 172 | if((ix0|ix1)!=0) {
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| 173 | z = one-tiny; /* trigger inexact flag */
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| 174 | if (z>=one) {
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| 175 | z = one+tiny;
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| 176 | if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;}
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| 177 | else if (z>one) {
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| 178 | if (q1==(uint32_t)0xfffffffe) q+=1;
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| 179 | q1+=2;
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| 180 | } else
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| 181 | q1 += (q1&1);
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| 182 | }
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| [048a6e9] | 183 | }
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| [ca113cf] | 184 | ix0 = (q>>1)+0x3fe00000;
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| 185 | ix1 = q1>>1;
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| 186 | if ((q&1)==1) ix1 |= sign;
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| 187 | ix0 += (m <<20);
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| 188 | INSERT_WORDS(z,ix0,ix1);
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| [048a6e9] | 189 | return z;
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| 190 | }
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| 191 |
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| 192 | /*
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| 193 | Other methods (use floating-point arithmetic)
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| 194 | -------------
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| 195 | (This is a copy of a drafted paper by Prof W. Kahan
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| 196 | and K.C. Ng, written in May, 1986)
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| 197 | Two algorithms are given here to implement sqrt(x)
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| 198 | (IEEE double precision arithmetic) in software.
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| 199 | Both supply sqrt(x) correctly rounded. The first algorithm (in
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| 200 | Section A) uses newton iterations and involves four divisions.
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| 201 | The second one uses reciproot iterations to avoid division, but
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| 202 | requires more multiplications. Both algorithms need the ability
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| 203 | to chop results of arithmetic operations instead of round them,
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| 204 | and the INEXACT flag to indicate when an arithmetic operation
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| 205 | is executed exactly with no roundoff error, all part of the
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| 206 | standard (IEEE 754-1985). The ability to perform shift, add,
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| 207 | subtract and logical AND operations upon 32-bit words is needed
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| 208 | too, though not part of the standard.
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| 209 | A. sqrt(x) by Newton Iteration
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| 210 | (1) Initial approximation
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| 211 | Let x0 and x1 be the leading and the trailing 32-bit words of
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| 212 | a floating point number x (in IEEE double format) respectively
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| 213 | 1 11 52 ...widths
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| 214 | ------------------------------------------------------
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| 215 | x: |s| e | f |
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| 216 | ------------------------------------------------------
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| 217 | msb lsb msb lsb ...order
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| 218 |
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| 219 | ------------------------ ------------------------
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| 220 | x0: |s| e | f1 | x1: | f2 |
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| 221 | ------------------------ ------------------------
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| 222 | By performing shifts and subtracts on x0 and x1 (both regarded
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| 223 | as integers), we obtain an 8-bit approximation of sqrt(x) as
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| 224 | follows.
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| 225 | k := (x0>>1) + 0x1ff80000;
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| 226 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
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| 227 | Here k is a 32-bit integer and T1[] is an integer array containing
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| 228 | correction terms. Now magically the floating value of y (y's
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| 229 | leading 32-bit word is y0, the value of its trailing word is 0)
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| 230 | approximates sqrt(x) to almost 8-bit.
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| 231 | Value of T1:
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| 232 | static int T1[32]= {
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| 233 | 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
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| 234 | 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
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| 235 | 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
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| 236 | 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
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| 237 | (2) Iterative refinement
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| 238 | Apply Heron's rule three times to y, we have y approximates
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| 239 | sqrt(x) to within 1 ulp (Unit in the Last Place):
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| 240 | y := (y+x/y)/2 ... almost 17 sig. bits
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| 241 | y := (y+x/y)/2 ... almost 35 sig. bits
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| 242 | y := y-(y-x/y)/2 ... within 1 ulp
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| 243 | Remark 1.
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| 244 | Another way to improve y to within 1 ulp is:
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| 245 | y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
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| 246 | y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
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| 247 | 2
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| 248 | (x-y )*y
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| 249 | y := y + 2* ---------- ...within 1 ulp
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| 250 | 2
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| 251 | 3y + x
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| 252 | This formula has one division fewer than the one above; however,
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| 253 | it requires more multiplications and additions. Also x must be
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| 254 | scaled in advance to avoid spurious overflow in evaluating the
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| 255 | expression 3y*y+x. Hence it is not recommended uless division
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| 256 | is slow. If division is very slow, then one should use the
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| 257 | reciproot algorithm given in section B.
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| 258 | (3) Final adjustment
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| 259 | By twiddling y's last bit it is possible to force y to be
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| 260 | correctly rounded according to the prevailing rounding mode
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| 261 | as follows. Let r and i be copies of the rounding mode and
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| 262 | inexact flag before entering the square root program. Also we
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| 263 | use the expression y+-ulp for the next representable floating
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| 264 | numbers (up and down) of y. Note that y+-ulp = either fixed
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| 265 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped
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| 266 | mode.
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| 267 | I := FALSE; ... reset INEXACT flag I
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| 268 | R := RZ; ... set rounding mode to round-toward-zero
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| 269 | z := x/y; ... chopped quotient, possibly inexact
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| 270 | If(not I) then { ... if the quotient is exact
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| 271 | if(z=y) {
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| 272 | I := i; ... restore inexact flag
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| 273 | R := r; ... restore rounded mode
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| 274 | return sqrt(x):=y.
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| 275 | } else {
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| 276 | z := z - ulp; ... special rounding
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| 277 | }
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| 278 | }
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| 279 | i := TRUE; ... sqrt(x) is inexact
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| 280 | If (r=RN) then z=z+ulp ... rounded-to-nearest
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| 281 | If (r=RP) then { ... round-toward-+inf
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| 282 | y = y+ulp; z=z+ulp;
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| 283 | }
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| 284 | y := y+z; ... chopped sum
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| 285 | y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
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| 286 | I := i; ... restore inexact flag
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| 287 | R := r; ... restore rounded mode
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| 288 | return sqrt(x):=y.
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| 289 |
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| 290 | (4) Special cases
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| 291 | Square root of +inf, +-0, or NaN is itself;
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| 292 | Square root of a negative number is NaN with invalid signal.
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| 293 | B. sqrt(x) by Reciproot Iteration
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| 294 | (1) Initial approximation
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| 295 | Let x0 and x1 be the leading and the trailing 32-bit words of
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| 296 | a floating point number x (in IEEE double format) respectively
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| 297 | (see section A). By performing shifs and subtracts on x0 and y0,
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| 298 | we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
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| 299 | k := 0x5fe80000 - (x0>>1);
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| 300 | y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
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| 301 | Here k is a 32-bit integer and T2[] is an integer array
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| 302 | containing correction terms. Now magically the floating
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| 303 | value of y (y's leading 32-bit word is y0, the value of
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| 304 | its trailing word y1 is set to zero) approximates 1/sqrt(x)
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| 305 | to almost 7.8-bit.
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| 306 | Value of T2:
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| 307 | static int T2[64]= {
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| 308 | 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
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| 309 | 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
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| 310 | 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
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| 311 | 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
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| 312 | 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
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| 313 | 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
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| 314 | 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
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| 315 | 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
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| 316 | (2) Iterative refinement
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| 317 | Apply Reciproot iteration three times to y and multiply the
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| 318 | result by x to get an approximation z that matches sqrt(x)
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| 319 | to about 1 ulp. To be exact, we will have
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| 320 | -1ulp < sqrt(x)-z<1.0625ulp.
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| 321 |
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| 322 | ... set rounding mode to Round-to-nearest
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| 323 | y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
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| 324 | y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
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| 325 | ... special arrangement for better accuracy
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| 326 | z := x*y ... 29 bits to sqrt(x), with z*y<1
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| 327 | z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
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| 328 | Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
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| 329 | (a) the term z*y in the final iteration is always less than 1;
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| 330 | (b) the error in the final result is biased upward so that
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| 331 | -1 ulp < sqrt(x) - z < 1.0625 ulp
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| 332 | instead of |sqrt(x)-z|<1.03125ulp.
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| 333 | (3) Final adjustment
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| 334 | By twiddling y's last bit it is possible to force y to be
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| 335 | correctly rounded according to the prevailing rounding mode
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| 336 | as follows. Let r and i be copies of the rounding mode and
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| 337 | inexact flag before entering the square root program. Also we
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| 338 | use the expression y+-ulp for the next representable floating
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| 339 | numbers (up and down) of y. Note that y+-ulp = either fixed
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| 340 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped
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| 341 | mode.
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| 342 | R := RZ; ... set rounding mode to round-toward-zero
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| 343 | switch(r) {
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| 344 | case RN: ... round-to-nearest
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| 345 | if(x<= z*(z-ulp)...chopped) z = z - ulp; else
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| 346 | if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
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| 347 | break;
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| 348 | case RZ:case RM: ... round-to-zero or round-to--inf
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| 349 | R:=RP; ... reset rounding mod to round-to-+inf
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| 350 | if(x<z*z ... rounded up) z = z - ulp; else
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| 351 | if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
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| 352 | break;
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| 353 | case RP: ... round-to-+inf
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| 354 | if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
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| 355 | if(x>z*z ...chopped) z = z+ulp;
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| 356 | break;
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| 357 | }
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| 358 | Remark 3. The above comparisons can be done in fixed point. For
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| 359 | example, to compare x and w=z*z chopped, it suffices to compare
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| 360 | x1 and w1 (the trailing parts of x and w), regarding them as
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| 361 | two's complement integers.
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| 362 | ...Is z an exact square root?
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| 363 | To determine whether z is an exact square root of x, let z1 be the
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| 364 | trailing part of z, and also let x0 and x1 be the leading and
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| 365 | trailing parts of x.
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| 366 | If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
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| 367 | I := 1; ... Raise Inexact flag: z is not exact
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| 368 | else {
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| 369 | j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
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| 370 | k := z1 >> 26; ... get z's 25-th and 26-th
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| 371 | fraction bits
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| 372 | I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
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| 373 | }
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| 374 | R:= r ... restore rounded mode
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| 375 | return sqrt(x):=z.
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| 376 | If multiplication is cheaper then the foregoing red tape, the
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| 377 | Inexact flag can be evaluated by
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| 378 | I := i;
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| 379 | I := (z*z!=x) or I.
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| 380 | Note that z*z can overwrite I; this value must be sensed if it is
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| 381 | True.
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| 382 | Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
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| 383 | zero.
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| 384 | --------------------
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| 385 | z1: | f2 |
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| 386 | --------------------
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| 387 | bit 31 bit 0
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| 388 | Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
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| 389 | or even of logb(x) have the following relations:
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| 390 | -------------------------------------------------
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| 391 | bit 27,26 of z1 bit 1,0 of x1 logb(x)
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| 392 | -------------------------------------------------
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| 393 | 00 00 odd and even
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| 394 | 01 01 even
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| 395 | 10 10 odd
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| 396 | 10 00 even
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| 397 | 11 01 even
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| 398 | -------------------------------------------------
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| 399 | (4) Special cases (see (4) of Section A).
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| 400 | */
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| 401 |
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| 402 | /** @}
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| 403 | */
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