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/* @(#)e_sqrt.c 5.1 93/09/24 */
2
/*
3
 * ====================================================
4
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5
 *
6
 * Developed at SunPro, a Sun Microsystems, Inc. business.
7
 * Permission to use, copy, modify, and distribute this
8
 * software is freely granted, provided that this notice
9
 * is preserved.
10
 * ====================================================
11
 */
12
 
13
#ifndef lint
14
static char rcsid[] = "$\Id: e_sqrt.c,v 1.2.6.1 1997/02/23 11:03:09 joerg Exp $";
15
#endif
16
 
17
/* __ieee754_sqrt(x)
18
 * Return correctly rounded sqrt.
19
 *           ------------------------------------------
20
 *           |  Use the hardware sqrt if you have one |
21
 *           ------------------------------------------
22
 * Method:
23
 *   Bit by bit method using integer arithmetic. (Slow, but portable)
24
 *   1. Normalization
25
 *      Scale x to y in [1,4) with even powers of 2:
26
 *      find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
27
 *              sqrt(x) = 2^k * sqrt(y)
28
 *   2. Bit by bit computation
29
 *      Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
30
 *           i                                                   0
31
 *                                     i+1         2
32
 *          s  = 2*q , and      y  =  2   * ( y - q  ).         (1)
33
 *           i      i            i                 i
34
 *
35
 *      To compute q    from q , one checks whether
36
 *                  i+1       i
37
 *
38
 *                            -(i+1) 2
39
 *                      (q + 2      ) <= y.                     (2)
40
 *                        i
41
 *                                                            -(i+1)
42
 *      If (2) is false, then q   = q ; otherwise q   = q  + 2      .
43
 *                             i+1   i             i+1   i
44
 *
45
 *      With some algebric manipulation, it is not difficult to see
46
 *      that (2) is equivalent to
47
 *                             -(i+1)
48
 *                      s  +  2       <= y                      (3)
49
 *                       i                i
50
 *
51
 *      The advantage of (3) is that s  and y  can be computed by
52
 *                                    i      i
53
 *      the following recurrence formula:
54
 *          if (3) is false
55
 *
56
 *          s     =  s  ,       y    = y   ;                    (4)
57
 *           i+1      i          i+1    i
58
 *
59
 *          otherwise,
60
 *                         -i                     -(i+1)
61
 *          s     =  s  + 2  ,  y    = y  -  s  - 2             (5)
62
 *           i+1      i          i+1    i     i
63
 *
64
 *      One may easily use induction to prove (4) and (5).
65
 *      Note. Since the left hand side of (3) contain only i+2 bits,
66
 *            it does not necessary to do a full (53-bit) comparison
67
 *            in (3).
68
 *   3. Final rounding
69
 *      After generating the 53 bits result, we compute one more bit.
70
 *      Together with the remainder, we can decide whether the
71
 *      result is exact, bigger than 1/2ulp, or less than 1/2ulp
72
 *      (it will never equal to 1/2ulp).
73
 *      The rounding mode can be detected by checking whether
74
 *      huge + tiny is equal to huge, and whether huge - tiny is
75
 *      equal to huge for some floating point number "huge" and "tiny".
76
 *
77
 * Special cases:
78
 *      sqrt(+-0) = +-0         ... exact
79
 *      sqrt(inf) = inf
80
 *      sqrt(-ve) = NaN         ... with invalid signal
81
 *      sqrt(NaN) = NaN         ... with invalid signal for signaling NaN
82
 *
83
 * Other methods : see the appended file at the end of the program below.
84
 *---------------
85
 */
86
 
87
#include "math.h"
88
#include "math_private.h"
89
 
90
#ifdef __STDC__
91
static  const double    one     = 1.0, tiny=1.0e-300;
92
#else
93
static  double  one     = 1.0, tiny=1.0e-300;
94
#endif
95
 
96
#ifdef __STDC__
97
        double __generic___ieee754_sqrt(double x)
98
#else
99
        double __generic___ieee754_sqrt(x)
100
        double x;
101
#endif
102
{
103
        double z;
104
        int32_t sign = (int)0x80000000;
105
        int32_t ix0,s0,q,m,t,i;
106
        u_int32_t r,t1,s1,ix1,q1;
107
 
108
        EXTRACT_WORDS(ix0,ix1,x);
109
 
110
    /* take care of Inf and NaN */
111
        if((ix0&0x7ff00000)==0x7ff00000) {
112
            return x*x+x;               /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
113
                                           sqrt(-inf)=sNaN */
114
        }
115
    /* take care of zero */
116
        if(ix0<=0) {
117
            if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
118
            else if(ix0<0)
119
                return (x-x)/(x-x);             /* sqrt(-ve) = sNaN */
120
        }
121
    /* normalize x */
122
        m = (ix0>>20);
123
        if(m==0) {                              /* subnormal x */
124
            while(ix0==0) {
125
                m -= 21;
126
                ix0 |= (ix1>>11); ix1 <<= 21;
127
            }
128
            for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
129
            m -= i-1;
130
            ix0 |= (ix1>>(32-i));
131
            ix1 <<= i;
132
        }
133
        m -= 1023;      /* unbias exponent */
134
        ix0 = (ix0&0x000fffff)|0x00100000;
135
        if(m&1){        /* odd m, double x to make it even */
136
            ix0 += ix0 + ((ix1&sign)>>31);
137
            ix1 += ix1;
138
        }
139
        m >>= 1;        /* m = [m/2] */
140
 
141
    /* generate sqrt(x) bit by bit */
142
        ix0 += ix0 + ((ix1&sign)>>31);
143
        ix1 += ix1;
144
        q = q1 = s0 = s1 = 0;   /* [q,q1] = sqrt(x) */
145
        r = 0x00200000;         /* r = moving bit from right to left */
146
 
147
        while(r!=0) {
148
            t = s0+r;
149
            if(t<=ix0) {
150
                s0   = t+r;
151
                ix0 -= t;
152
                q   += r;
153
            }
154
            ix0 += ix0 + ((ix1&sign)>>31);
155
            ix1 += ix1;
156
            r>>=1;
157
        }
158
 
159
        r = sign;
160
        while(r!=0) {
161
            t1 = s1+r;
162
            t  = s0;
163
            if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
164
                s1  = t1+r;
165
                if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
166
                ix0 -= t;
167
                if (ix1 < t1) ix0 -= 1;
168
                ix1 -= t1;
169
                q1  += r;
170
            }
171
            ix0 += ix0 + ((ix1&sign)>>31);
172
            ix1 += ix1;
173
            r>>=1;
174
        }
175
 
176
    /* use floating add to find out rounding direction */
177
        if((ix0|ix1)!=0) {
178
            z = one-tiny; /* trigger inexact flag */
179
            if (z>=one) {
180
                z = one+tiny;
181
                if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
182
                else if (z>one) {
183
                    if (q1==(u_int32_t)0xfffffffe) q+=1;
184
                    q1+=2;
185
                } else
186
                    q1 += (q1&1);
187
            }
188
        }
189
        ix0 = (q>>1)+0x3fe00000;
190
        ix1 =  q1>>1;
191
        if ((q&1)==1) ix1 |= sign;
192
        ix0 += (m <<20);
193
        INSERT_WORDS(z,ix0,ix1);
194
        return z;
195
}
196
 
197
/*
198
Other methods  (use floating-point arithmetic)
199
-------------
200
(This is a copy of a drafted paper by Prof W. Kahan
201
and K.C. Ng, written in May, 1986)
202
 
203
        Two algorithms are given here to implement sqrt(x)
204
        (IEEE double precision arithmetic) in software.
205
        Both supply sqrt(x) correctly rounded. The first algorithm (in
206
        Section A) uses newton iterations and involves four divisions.
207
        The second one uses reciproot iterations to avoid division, but
208
        requires more multiplications. Both algorithms need the ability
209
        to chop results of arithmetic operations instead of round them,
210
        and the INEXACT flag to indicate when an arithmetic operation
211
        is executed exactly with no roundoff error, all part of the
212
        standard (IEEE 754-1985). The ability to perform shift, add,
213
        subtract and logical AND operations upon 32-bit words is needed
214
        too, though not part of the standard.
215
 
216
A.  sqrt(x) by Newton Iteration
217
 
218
   (1)  Initial approximation
219
 
220
        Let x0 and x1 be the leading and the trailing 32-bit words of
221
        a floating point number x (in IEEE double format) respectively
222
 
223
            1    11                  52                           ...widths
224
           ------------------------------------------------------
225
        x: |s|    e     |             f                         |
226
           ------------------------------------------------------
227
              msb    lsb  msb                                 lsb ...order
228
 
229
 
230
             ------------------------        ------------------------
231
        x0:  |s|   e    |    f1     |    x1: |          f2           |
232
             ------------------------        ------------------------
233
 
234
        By performing shifts and subtracts on x0 and x1 (both regarded
235
        as integers), we obtain an 8-bit approximation of sqrt(x) as
236
        follows.
237
 
238
                k  := (x0>>1) + 0x1ff80000;
239
                y0 := k - T1[31&(k>>15)].       ... y ~ sqrt(x) to 8 bits
240
        Here k is a 32-bit integer and T1[] is an integer array containing
241
        correction terms. Now magically the floating value of y (y's
242
        leading 32-bit word is y0, the value of its trailing word is 0)
243
        approximates sqrt(x) to almost 8-bit.
244
 
245
        Value of T1:
246
        static int T1[32]= {
247
        0,      1024,   3062,   5746,   9193,   13348,  18162,  23592,
248
        29598,  36145,  43202,  50740,  58733,  67158,  75992,  85215,
249
        83599,  71378,  60428,  50647,  41945,  34246,  27478,  21581,
250
        16499,  12183,  8588,   5674,   3403,   1742,   661,    130,};
251
 
252
    (2) Iterative refinement
253
 
254
        Apply Heron's rule three times to y, we have y approximates
255
        sqrt(x) to within 1 ulp (Unit in the Last Place):
256
 
257
                y := (y+x/y)/2          ... almost 17 sig. bits
258
                y := (y+x/y)/2          ... almost 35 sig. bits
259
                y := y-(y-x/y)/2        ... within 1 ulp
260
 
261
 
262
        Remark 1.
263
            Another way to improve y to within 1 ulp is:
264
 
265
                y := (y+x/y)            ... almost 17 sig. bits to 2*sqrt(x)
266
                y := y - 0x00100006     ... almost 18 sig. bits to sqrt(x)
267
 
268
                                2
269
                            (x-y )*y
270
                y := y + 2* ----------  ...within 1 ulp
271
                               2
272
                             3y  + x
273
 
274
 
275
        This formula has one division fewer than the one above; however,
276
        it requires more multiplications and additions. Also x must be
277
        scaled in advance to avoid spurious overflow in evaluating the
278
        expression 3y*y+x. Hence it is not recommended uless division
279
        is slow. If division is very slow, then one should use the
280
        reciproot algorithm given in section B.
281
 
282
    (3) Final adjustment
283
 
284
        By twiddling y's last bit it is possible to force y to be
285
        correctly rounded according to the prevailing rounding mode
286
        as follows. Let r and i be copies of the rounding mode and
287
        inexact flag before entering the square root program. Also we
288
        use the expression y+-ulp for the next representable floating
289
        numbers (up and down) of y. Note that y+-ulp = either fixed
290
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
291
        mode.
292
 
293
                I := FALSE;     ... reset INEXACT flag I
294
                R := RZ;        ... set rounding mode to round-toward-zero
295
                z := x/y;       ... chopped quotient, possibly inexact
296
                If(not I) then {        ... if the quotient is exact
297
                    if(z=y) {
298
                        I := i;  ... restore inexact flag
299
                        R := r;  ... restore rounded mode
300
                        return sqrt(x):=y.
301
                    } else {
302
                        z := z - ulp;   ... special rounding
303
                    }
304
                }
305
                i := TRUE;              ... sqrt(x) is inexact
306
                If (r=RN) then z=z+ulp  ... rounded-to-nearest
307
                If (r=RP) then {        ... round-toward-+inf
308
                    y = y+ulp; z=z+ulp;
309
                }
310
                y := y+z;               ... chopped sum
311
                y0:=y0-0x00100000;      ... y := y/2 is correctly rounded.
312
                I := i;                 ... restore inexact flag
313
                R := r;                 ... restore rounded mode
314
                return sqrt(x):=y.
315
 
316
    (4) Special cases
317
 
318
        Square root of +inf, +-0, or NaN is itself;
319
        Square root of a negative number is NaN with invalid signal.
320
 
321
 
322
B.  sqrt(x) by Reciproot Iteration
323
 
324
   (1)  Initial approximation
325
 
326
        Let x0 and x1 be the leading and the trailing 32-bit words of
327
        a floating point number x (in IEEE double format) respectively
328
        (see section A). By performing shifs and subtracts on x0 and y0,
329
        we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
330
 
331
            k := 0x5fe80000 - (x0>>1);
332
            y0:= k - T2[63&(k>>14)].    ... y ~ 1/sqrt(x) to 7.8 bits
333
 
334
        Here k is a 32-bit integer and T2[] is an integer array
335
        containing correction terms. Now magically the floating
336
        value of y (y's leading 32-bit word is y0, the value of
337
        its trailing word y1 is set to zero) approximates 1/sqrt(x)
338
        to almost 7.8-bit.
339
 
340
        Value of T2:
341
        static int T2[64]= {
342
        0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
343
        0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
344
        0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
345
        0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
346
        0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
347
        0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
348
        0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
349
        0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
350
 
351
    (2) Iterative refinement
352
 
353
        Apply Reciproot iteration three times to y and multiply the
354
        result by x to get an approximation z that matches sqrt(x)
355
        to about 1 ulp. To be exact, we will have
356
                -1ulp < sqrt(x)-z<1.0625ulp.
357
 
358
        ... set rounding mode to Round-to-nearest
359
           y := y*(1.5-0.5*x*y*y)       ... almost 15 sig. bits to 1/sqrt(x)
360
           y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
361
        ... special arrangement for better accuracy
362
           z := x*y                     ... 29 bits to sqrt(x), with z*y<1
363
           z := z + 0.5*z*(1-z*y)       ... about 1 ulp to sqrt(x)
364
 
365
        Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
366
        (a) the term z*y in the final iteration is always less than 1;
367
        (b) the error in the final result is biased upward so that
368
                -1 ulp < sqrt(x) - z < 1.0625 ulp
369
            instead of |sqrt(x)-z|<1.03125ulp.
370
 
371
    (3) Final adjustment
372
 
373
        By twiddling y's last bit it is possible to force y to be
374
        correctly rounded according to the prevailing rounding mode
375
        as follows. Let r and i be copies of the rounding mode and
376
        inexact flag before entering the square root program. Also we
377
        use the expression y+-ulp for the next representable floating
378
        numbers (up and down) of y. Note that y+-ulp = either fixed
379
        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
380
        mode.
381
 
382
        R := RZ;                ... set rounding mode to round-toward-zero
383
        switch(r) {
384
            case RN:            ... round-to-nearest
385
               if(x<= z*(z-ulp)...chopped) z = z - ulp; else
386
               if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
387
               break;
388
            case RZ:case RM:    ... round-to-zero or round-to--inf
389
               R:=RP;           ... reset rounding mod to round-to-+inf
390
               if(x<z*z ... rounded up) z = z - ulp; else
391
               if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
392
               break;
393
            case RP:            ... round-to-+inf
394
               if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
395
               if(x>z*z ...chopped) z = z+ulp;
396
               break;
397
        }
398
 
399
        Remark 3. The above comparisons can be done in fixed point. For
400
        example, to compare x and w=z*z chopped, it suffices to compare
401
        x1 and w1 (the trailing parts of x and w), regarding them as
402
        two's complement integers.
403
 
404
        ...Is z an exact square root?
405
        To determine whether z is an exact square root of x, let z1 be the
406
        trailing part of z, and also let x0 and x1 be the leading and
407
        trailing parts of x.
408
 
409
        If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
410
            I := 1;             ... Raise Inexact flag: z is not exact
411
        else {
412
            j := 1 - [(x0>>20)&1]       ... j = logb(x) mod 2
413
            k := z1 >> 26;              ... get z's 25-th and 26-th
414
                                            fraction bits
415
            I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
416
        }
417
        R:= r           ... restore rounded mode
418
        return sqrt(x):=z.
419
 
420
        If multiplication is cheaper then the foregoing red tape, the
421
        Inexact flag can be evaluated by
422
 
423
            I := i;
424
            I := (z*z!=x) or I.
425
 
426
        Note that z*z can overwrite I; this value must be sensed if it is
427
        True.
428
 
429
        Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
430
        zero.
431
 
432
                    --------------------
433
                z1: |        f2        |
434
                    --------------------
435
                bit 31             bit 0
436
 
437
        Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
438
        or even of logb(x) have the following relations:
439
 
440
        -------------------------------------------------
441
        bit 27,26 of z1         bit 1,0 of x1   logb(x)
442
        -------------------------------------------------
443
        00                      00              odd and even
444
        01                      01              even
445
        10                      10              odd
446
        10                      00              even
447
        11                      01              even
448
        -------------------------------------------------
449
 
450
    (4) Special cases (see (4) of Section A).
451
 
452
 */
453