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Viet-Trung Luu96b05c12016-01-11 11:26:36 -08001/* origin: FreeBSD /usr/src/lib/msun/src/e_sqrt.c */
2/*
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
9 * is preserved.
10 * ====================================================
11 */
12/* sqrt(x)
13 * Return correctly rounded sqrt.
14 * ------------------------------------------
15 * | Use the hardware sqrt if you have one |
16 * ------------------------------------------
17 * Method:
18 * Bit by bit method using integer arithmetic. (Slow, but portable)
19 * 1. Normalization
20 * Scale x to y in [1,4) with even powers of 2:
21 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
22 * sqrt(x) = 2^k * sqrt(y)
23 * 2. Bit by bit computation
24 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
25 * i 0
26 * i+1 2
27 * s = 2*q , and y = 2 * ( y - q ). (1)
28 * i i i i
29 *
30 * To compute q from q , one checks whether
31 * i+1 i
32 *
33 * -(i+1) 2
34 * (q + 2 ) <= y. (2)
35 * i
36 * -(i+1)
37 * If (2) is false, then q = q ; otherwise q = q + 2 .
38 * i+1 i i+1 i
39 *
40 * With some algebric manipulation, it is not difficult to see
41 * that (2) is equivalent to
42 * -(i+1)
43 * s + 2 <= y (3)
44 * i i
45 *
46 * The advantage of (3) is that s and y can be computed by
47 * i i
48 * the following recurrence formula:
49 * if (3) is false
50 *
51 * s = s , y = y ; (4)
52 * i+1 i i+1 i
53 *
54 * otherwise,
55 * -i -(i+1)
56 * s = s + 2 , y = y - s - 2 (5)
57 * i+1 i i+1 i i
58 *
59 * One may easily use induction to prove (4) and (5).
60 * Note. Since the left hand side of (3) contain only i+2 bits,
61 * it does not necessary to do a full (53-bit) comparison
62 * in (3).
63 * 3. Final rounding
64 * After generating the 53 bits result, we compute one more bit.
65 * Together with the remainder, we can decide whether the
66 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
67 * (it will never equal to 1/2ulp).
68 * The rounding mode can be detected by checking whether
69 * huge + tiny is equal to huge, and whether huge - tiny is
70 * equal to huge for some floating point number "huge" and "tiny".
71 *
72 * Special cases:
73 * sqrt(+-0) = +-0 ... exact
74 * sqrt(inf) = inf
75 * sqrt(-ve) = NaN ... with invalid signal
76 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
77 */
78
79#include "libm.h"
80
81static const double tiny = 1.0e-300;
82
George Kulakowski17e3b042016-02-18 15:59:50 -080083double sqrt(double x) {
84 double z;
85 int32_t sign = (int)0x80000000;
86 int32_t ix0, s0, q, m, t, i;
87 uint32_t r, t1, s1, ix1, q1;
Viet-Trung Luu96b05c12016-01-11 11:26:36 -080088
George Kulakowski17e3b042016-02-18 15:59:50 -080089 EXTRACT_WORDS(ix0, ix1, x);
Viet-Trung Luu96b05c12016-01-11 11:26:36 -080090
George Kulakowski17e3b042016-02-18 15:59:50 -080091 /* take care of Inf and NaN */
92 if ((ix0 & 0x7ff00000) == 0x7ff00000) {
93 return x * x + x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf, sqrt(-inf)=sNaN */
94 }
95 /* take care of zero */
96 if (ix0 <= 0) {
97 if (((ix0 & ~sign) | ix1) == 0)
98 return x; /* sqrt(+-0) = +-0 */
99 if (ix0 < 0)
100 return (x - x) / (x - x); /* sqrt(-ve) = sNaN */
101 }
102 /* normalize x */
103 m = ix0 >> 20;
104 if (m == 0) { /* subnormal x */
105 while (ix0 == 0) {
106 m -= 21;
107 ix0 |= (ix1 >> 11);
108 ix1 <<= 21;
109 }
110 for (i = 0; (ix0 & 0x00100000) == 0; i++)
111 ix0 <<= 1;
112 m -= i - 1;
113 ix0 |= ix1 >> (32 - i);
114 ix1 <<= i;
115 }
116 m -= 1023; /* unbias exponent */
117 ix0 = (ix0 & 0x000fffff) | 0x00100000;
118 if (m & 1) { /* odd m, double x to make it even */
119 ix0 += ix0 + ((ix1 & sign) >> 31);
120 ix1 += ix1;
121 }
122 m >>= 1; /* m = [m/2] */
Viet-Trung Luu96b05c12016-01-11 11:26:36 -0800123
George Kulakowski17e3b042016-02-18 15:59:50 -0800124 /* generate sqrt(x) bit by bit */
125 ix0 += ix0 + ((ix1 & sign) >> 31);
126 ix1 += ix1;
127 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
128 r = 0x00200000; /* r = moving bit from right to left */
Viet-Trung Luu96b05c12016-01-11 11:26:36 -0800129
George Kulakowski17e3b042016-02-18 15:59:50 -0800130 while (r != 0) {
131 t = s0 + r;
132 if (t <= ix0) {
133 s0 = t + r;
134 ix0 -= t;
135 q += r;
136 }
137 ix0 += ix0 + ((ix1 & sign) >> 31);
138 ix1 += ix1;
139 r >>= 1;
140 }
Viet-Trung Luu96b05c12016-01-11 11:26:36 -0800141
George Kulakowski17e3b042016-02-18 15:59:50 -0800142 r = sign;
143 while (r != 0) {
144 t1 = s1 + r;
145 t = s0;
146 if (t < ix0 || (t == ix0 && t1 <= ix1)) {
147 s1 = t1 + r;
148 if ((t1 & sign) == sign && (s1 & sign) == 0)
149 s0++;
150 ix0 -= t;
151 if (ix1 < t1)
152 ix0--;
153 ix1 -= t1;
154 q1 += r;
155 }
156 ix0 += ix0 + ((ix1 & sign) >> 31);
157 ix1 += ix1;
158 r >>= 1;
159 }
Viet-Trung Luu96b05c12016-01-11 11:26:36 -0800160
George Kulakowski17e3b042016-02-18 15:59:50 -0800161 /* use floating add to find out rounding direction */
162 if ((ix0 | ix1) != 0) {
163 z = 1.0 - tiny; /* raise inexact flag */
164 if (z >= 1.0) {
165 z = 1.0 + tiny;
166 if (q1 == (uint32_t)0xffffffff) {
167 q1 = 0;
168 q++;
169 } else if (z > 1.0) {
170 if (q1 == (uint32_t)0xfffffffe)
171 q++;
172 q1 += 2;
173 } else
174 q1 += q1 & 1;
175 }
176 }
177 ix0 = (q >> 1) + 0x3fe00000;
178 ix1 = q1 >> 1;
179 if (q & 1)
180 ix1 |= sign;
181 ix0 += m << 20;
182 INSERT_WORDS(z, ix0, ix1);
183 return z;
Viet-Trung Luu96b05c12016-01-11 11:26:36 -0800184}