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[/] [or1k/] [trunk/] [newlib/] [newlib/] [libm/] [math/] [e_sqrt.c] - Blame information for rev 1765

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1 39 lampret
 
2
/* @(#)e_sqrt.c 5.1 93/09/24 */
3
/*
4
 * ====================================================
5
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6
 *
7
 * Developed at SunPro, a Sun Microsystems, Inc. business.
8
 * Permission to use, copy, modify, and distribute this
9
 * software is freely granted, provided that this notice
10
 * is preserved.
11
 * ====================================================
12
 */
13
 
14
/* __ieee754_sqrt(x)
15
 * Return correctly rounded sqrt.
16
 *           ------------------------------------------
17
 *           |  Use the hardware sqrt if you have one |
18
 *           ------------------------------------------
19
 * Method:
20
 *   Bit by bit method using integer arithmetic. (Slow, but portable)
21
 *   1. Normalization
22
 *      Scale x to y in [1,4) with even powers of 2:
23
 *      find an integer k such that  1 <= (y=x*2^(2k)) < 4, then
24
 *              sqrt(x) = 2^k * sqrt(y)
25
 *   2. Bit by bit computation
26
 *      Let q  = sqrt(y) truncated to i bit after binary point (q = 1),
27
 *           i                                                   0
28
 *                                     i+1         2
29
 *          s  = 2*q , and      y  =  2   * ( y - q  ).         (1)
30
 *           i      i            i                 i
31
 *
32
 *      To compute q    from q , one checks whether
33
 *                  i+1       i
34
 *
35
 *                            -(i+1) 2
36
 *                      (q + 2      ) <= y.                     (2)
37
 *                        i
38
 *                                                            -(i+1)
39
 *      If (2) is false, then q   = q ; otherwise q   = q  + 2      .
40
 *                             i+1   i             i+1   i
41
 *
42
 *      With some algebric manipulation, it is not difficult to see
43
 *      that (2) is equivalent to
44
 *                             -(i+1)
45
 *                      s  +  2       <= y                      (3)
46
 *                       i                i
47
 *
48
 *      The advantage of (3) is that s  and y  can be computed by
49
 *                                    i      i
50
 *      the following recurrence formula:
51
 *          if (3) is false
52
 *
53
 *          s     =  s  ,       y    = y   ;                    (4)
54
 *           i+1      i          i+1    i
55
 *
56
 *          otherwise,
57
 *                         -i                     -(i+1)
58
 *          s     =  s  + 2  ,  y    = y  -  s  - 2             (5)
59
 *           i+1      i          i+1    i     i
60
 *
61
 *      One may easily use induction to prove (4) and (5).
62
 *      Note. Since the left hand side of (3) contain only i+2 bits,
63
 *            it does not necessary to do a full (53-bit) comparison
64
 *            in (3).
65
 *   3. Final rounding
66
 *      After generating the 53 bits result, we compute one more bit.
67
 *      Together with the remainder, we can decide whether the
68
 *      result is exact, bigger than 1/2ulp, or less than 1/2ulp
69
 *      (it will never equal to 1/2ulp).
70
 *      The rounding mode can be detected by checking whether
71
 *      huge + tiny is equal to huge, and whether huge - tiny is
72
 *      equal to huge for some floating point number "huge" and "tiny".
73
 *
74
 * Special cases:
75
 *      sqrt(+-0) = +-0         ... exact
76
 *      sqrt(inf) = inf
77
 *      sqrt(-ve) = NaN         ... with invalid signal
78
 *      sqrt(NaN) = NaN         ... with invalid signal for signaling NaN
79
 *
80
 * Other methods : see the appended file at the end of the program below.
81
 *---------------
82
 */
83
 
84
#include "fdlibm.h"
85
 
86
#ifndef _DOUBLE_IS_32BITS
87
 
88
#ifdef __STDC__
89
static  const double    one     = 1.0, tiny=1.0e-300;
90
#else
91
static  double  one     = 1.0, tiny=1.0e-300;
92
#endif
93
 
94
#ifdef __STDC__
95
        double __ieee754_sqrt(double x)
96
#else
97
        double __ieee754_sqrt(x)
98
        double x;
99
#endif
100
{
101
        double z;
102
        __int32_t sign = (int)0x80000000;
103
        __uint32_t r,t1,s1,ix1,q1;
104
        __int32_t ix0,s0,q,m,t,i;
105
 
106
        EXTRACT_WORDS(ix0,ix1,x);
107
 
108
    /* take care of Inf and NaN */
109
        if((ix0&0x7ff00000)==0x7ff00000) {
110
            return x*x+x;               /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
111
                                           sqrt(-inf)=sNaN */
112
        }
113
    /* take care of zero */
114
        if(ix0<=0) {
115
            if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
116
            else if(ix0<0)
117
                return (x-x)/(x-x);             /* sqrt(-ve) = sNaN */
118
        }
119
    /* normalize x */
120
        m = (ix0>>20);
121
        if(m==0) {                               /* subnormal x */
122
            while(ix0==0) {
123
                m -= 21;
124
                ix0 |= (ix1>>11); ix1 <<= 21;
125
            }
126
            for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
127
            m -= i-1;
128
            ix0 |= (ix1>>(32-i));
129
            ix1 <<= i;
130
        }
131
        m -= 1023;      /* unbias exponent */
132
        ix0 = (ix0&0x000fffff)|0x00100000;
133
        if(m&1){        /* odd m, double x to make it even */
134
            ix0 += ix0 + ((ix1&sign)>>31);
135
            ix1 += ix1;
136
        }
137
        m >>= 1;        /* m = [m/2] */
138
 
139
    /* generate sqrt(x) bit by bit */
140
        ix0 += ix0 + ((ix1&sign)>>31);
141
        ix1 += ix1;
142
        q = q1 = s0 = s1 = 0;    /* [q,q1] = sqrt(x) */
143
        r = 0x00200000;         /* r = moving bit from right to left */
144
 
145
        while(r!=0) {
146
            t = s0+r;
147
            if(t<=ix0) {
148
                s0   = t+r;
149
                ix0 -= t;
150
                q   += r;
151
            }
152
            ix0 += ix0 + ((ix1&sign)>>31);
153
            ix1 += ix1;
154
            r>>=1;
155
        }
156
 
157
        r = sign;
158
        while(r!=0) {
159
            t1 = s1+r;
160
            t  = s0;
161
            if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
162
                s1  = t1+r;
163
                if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
164
                ix0 -= t;
165
                if (ix1 < t1) ix0 -= 1;
166
                ix1 -= t1;
167
                q1  += r;
168
            }
169
            ix0 += ix0 + ((ix1&sign)>>31);
170
            ix1 += ix1;
171
            r>>=1;
172
        }
173
 
174
    /* use floating add to find out rounding direction */
175
        if((ix0|ix1)!=0) {
176
            z = one-tiny; /* trigger inexact flag */
177
            if (z>=one) {
178
                z = one+tiny;
179
                if (q1==(__uint32_t)0xffffffff) { q1=0; q += 1;}
180
                else if (z>one) {
181
                    if (q1==(__uint32_t)0xfffffffe) q+=1;
182
                    q1+=2;
183
                } else
184
                    q1 += (q1&1);
185
            }
186
        }
187
        ix0 = (q>>1)+0x3fe00000;
188
        ix1 =  q1>>1;
189
        if ((q&1)==1) ix1 |= sign;
190
        ix0 += (m <<20);
191
        INSERT_WORDS(z,ix0,ix1);
192
        return z;
193
}
194
 
195
#endif /* defined(_DOUBLE_IS_32BITS) */
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
 */

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