source:trunk/sys/libm/e_sqrt.c@1

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1
2/* @(#)e_sqrt.c 5.1 93/09/24 */
3/*
4 * ====================================================
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 <libm/fdlibm.h>
85
86#ifdef __STDC__
87static  const double    one     = 1.0, tiny=1.0e-300;
88#else
89static  double  one     = 1.0, tiny=1.0e-300;
90#endif
91
92#ifdef __STDC__
93        double __ieee754_sqrt(double x)
94#else
95        double __ieee754_sqrt(x)
96        double x;
97#endif
98{
99        int     n0;
100
101        double z;
102        int     sign = (int)0x80000000;
103        unsigned r,t1,s1,ix1,q1;
104        int ix0,s0,q,m,t,i;
105
106        n0 = ((*(int*)&one)>>29)^1;             /* index of high word */
107        ix0 = *(n0+(int*)&x);                   /* high word of x */
108        ix1 = *((1-n0)+(int*)&x);               /* low word of 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)== (unsigned)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==(unsigned)0xffffffff) { q1=0; q += 1;}
182                else if (z>one) {
183                    if (q1==(unsigned)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        *(n0+(int*)&z) = ix0;
194        *((1-n0)+(int*)&z) = ix1;
195        return z;
196}
197
198/*
199Other methods  (use floating-point arithmetic)
200-------------
201(This is a copy of a drafted paper by Prof W. Kahan
202and K.C. Ng, written in May, 1986)
203
204        Two algorithms are given here to implement sqrt(x)
205        (IEEE double precision arithmetic) in software.
206        Both supply sqrt(x) correctly rounded. The first algorithm (in
207        Section A) uses newton iterations and involves four divisions.
208        The second one uses reciproot iterations to avoid division, but
209        requires more multiplications. Both algorithms need the ability
210        to chop results of arithmetic operations instead of round them,
211        and the INEXACT flag to indicate when an arithmetic operation
212        is executed exactly with no roundoff error, all part of the
213        standard (IEEE 754-1985). The ability to perform shift, add,
214        subtract and logical AND operations upon 32-bit words is needed
215        too, though not part of the standard.
216
217A.  sqrt(x) by Newton Iteration
218
219   (1)  Initial approximation
220
221        Let x0 and x1 be the leading and the trailing 32-bit words of
222        a floating point number x (in IEEE double format) respectively
223
224            1    11                  52                           ...widths
225           ------------------------------------------------------
226        x: |s|    e     |             f                         |
227           ------------------------------------------------------
228              msb    lsb  msb                                 lsb ...order
229
230
231             ------------------------        ------------------------
232        x0:  |s|   e    |    f1     |    x1: |          f2           |
233             ------------------------        ------------------------
234
235        By performing shifts and subtracts on x0 and x1 (both regarded
236        as integers), we obtain an 8-bit approximation of sqrt(x) as
237        follows.
238
239                k  := (x0>>1) + 0x1ff80000;
240                y0 := k - T1[31&(k>>15)].       ... y ~ sqrt(x) to 8 bits
241        Here k is a 32-bit integer and T1[] is an integer array containing
242        correction terms. Now magically the floating value of y (y's
243        leading 32-bit word is y0, the value of its trailing word is 0)
244        approximates sqrt(x) to almost 8-bit.
245
246        Value of T1:
247        static int T1= {
248        0,      1024,   3062,   5746,   9193,   13348,  18162,  23592,
249        29598,  36145,  43202,  50740,  58733,  67158,  75992,  85215,
250        83599,  71378,  60428,  50647,  41945,  34246,  27478,  21581,
251        16499,  12183,  8588,   5674,   3403,   1742,   661,    130,};
252
253    (2) Iterative refinement
254
255        Apply Heron's rule three times to y, we have y approximates
256        sqrt(x) to within 1 ulp (Unit in the Last Place):
257
258                y := (y+x/y)/2          ... almost 17 sig. bits
259                y := (y+x/y)/2          ... almost 35 sig. bits
260                y := y-(y-x/y)/2        ... within 1 ulp
261
262
263        Remark 1.
264            Another way to improve y to within 1 ulp is:
265
266                y := (y+x/y)            ... almost 17 sig. bits to 2*sqrt(x)
267                y := y - 0x00100006     ... almost 18 sig. bits to sqrt(x)
268
269                                2
270                            (x-y )*y
271                y := y + 2* ----------  ...within 1 ulp
272                               2
273                             3y  + x
274
275
276        This formula has one division fewer than the one above; however,
277        it requires more multiplications and additions. Also x must be
278        scaled in advance to avoid spurious overflow in evaluating the
279        expression 3y*y+x. Hence it is not recommended uless division
280        is slow. If division is very slow, then one should use the
281        reciproot algorithm given in section B.
282
284
285        By twiddling y's last bit it is possible to force y to be
286        correctly rounded according to the prevailing rounding mode
287        as follows. Let r and i be copies of the rounding mode and
288        inexact flag before entering the square root program. Also we
289        use the expression y+-ulp for the next representable floating
290        numbers (up and down) of y. Note that y+-ulp = either fixed
291        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
292        mode.
293
294                I := FALSE;     ... reset INEXACT flag I
295                R := RZ;        ... set rounding mode to round-toward-zero
296                z := x/y;       ... chopped quotient, possibly inexact
297                If(not I) then {        ... if the quotient is exact
298                    if(z=y) {
299                        I := i;  ... restore inexact flag
300                        R := r;  ... restore rounded mode
301                        return sqrt(x):=y.
302                    } else {
303                        z := z - ulp;   ... special rounding
304                    }
305                }
306                i := TRUE;              ... sqrt(x) is inexact
307                If (r=RN) then z=z+ulp  ... rounded-to-nearest
308                If (r=RP) then {        ... round-toward-+inf
309                    y = y+ulp; z=z+ulp;
310                }
311                y := y+z;               ... chopped sum
312                y0:=y0-0x00100000;      ... y := y/2 is correctly rounded.
313                I := i;                 ... restore inexact flag
314                R := r;                 ... restore rounded mode
315                return sqrt(x):=y.
316
317    (4) Special cases
318
319        Square root of +inf, +-0, or NaN is itself;
320        Square root of a negative number is NaN with invalid signal.
321
322
323B.  sqrt(x) by Reciproot Iteration
324
325   (1)  Initial approximation
326
327        Let x0 and x1 be the leading and the trailing 32-bit words of
328        a floating point number x (in IEEE double format) respectively
329        (see section A). By performing shifs and subtracts on x0 and y0,
330        we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
331
332            k := 0x5fe80000 - (x0>>1);
333            y0:= k - T2[63&(k>>14)].    ... y ~ 1/sqrt(x) to 7.8 bits
334
335        Here k is a 32-bit integer and T2[] is an integer array
336        containing correction terms. Now magically the floating
337        value of y (y's leading 32-bit word is y0, the value of
338        its trailing word y1 is set to zero) approximates 1/sqrt(x)
339        to almost 7.8-bit.
340
341        Value of T2:
342        static int T2= {
343        0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
344        0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
345        0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
346        0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
347        0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
349        0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
350        0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
351
352    (2) Iterative refinement
353
354        Apply Reciproot iteration three times to y and multiply the
355        result by x to get an approximation z that matches sqrt(x)
356        to about 1 ulp. To be exact, we will have
357                -1ulp < sqrt(x)-z<1.0625ulp.
358
359        ... set rounding mode to Round-to-nearest
360           y := y*(1.5-0.5*x*y*y)       ... almost 15 sig. bits to 1/sqrt(x)
361           y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
362        ... special arrangement for better accuracy
363           z := x*y                     ... 29 bits to sqrt(x), with z*y<1
364           z := z + 0.5*z*(1-z*y)       ... about 1 ulp to sqrt(x)
365
366        Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
367        (a) the term z*y in the final iteration is always less than 1;
368        (b) the error in the final result is biased upward so that
369                -1 ulp < sqrt(x) - z < 1.0625 ulp
371
373
374        By twiddling y's last bit it is possible to force y to be
375        correctly rounded according to the prevailing rounding mode
376        as follows. Let r and i be copies of the rounding mode and
377        inexact flag before entering the square root program. Also we
378        use the expression y+-ulp for the next representable floating
379        numbers (up and down) of y. Note that y+-ulp = either fixed
380        point y+-1, or multiply y by nextafter(1,+-inf) in chopped
381        mode.
382
383        R := RZ;                ... set rounding mode to round-toward-zero
384        switch(r) {
385            case RN:            ... round-to-nearest
386               if(x<= z*(z-ulp)...chopped) z = z - ulp; else
387               if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
388               break;
389            case RZ:case RM:    ... round-to-zero or round-to--inf
390               R:=RP;           ... reset rounding mod to round-to-+inf
391               if(x<z*z ... rounded up) z = z - ulp; else
392               if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
393               break;
394            case RP:            ... round-to-+inf
395               if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
396               if(x>z*z ...chopped) z = z+ulp;
397               break;
398        }
399
400        Remark 3. The above comparisons can be done in fixed point. For
401        example, to compare x and w=z*z chopped, it suffices to compare
402        x1 and w1 (the trailing parts of x and w), regarding them as
403        two's complement integers.
404
405        ...Is z an exact square root?
406        To determine whether z is an exact square root of x, let z1 be the
407        trailing part of z, and also let x0 and x1 be the leading and
408        trailing parts of x.
409
410        If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
411            I := 1;             ... Raise Inexact flag: z is not exact
412        else {
413            j := 1 - [(x0>>20)&1]       ... j = logb(x) mod 2
414            k := z1 >> 26;              ... get z's 25-th and 26-th
415                                            fraction bits
416            I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
417        }
418        R:= r           ... restore rounded mode
419        return sqrt(x):=z.
420
421        If multiplication is cheaper then the foregoing red tape, the
422        Inexact flag can be evaluated by
423
424            I := i;
425            I := (z*z!=x) or I.
426
427        Note that z*z can overwrite I; this value must be sensed if it is
428        True.
429
430        Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
431        zero.
432
433                    --------------------
434                z1: |        f2        |
435                    --------------------
436                bit 31             bit 0
437
438        Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
439        or even of logb(x) have the following relations:
440
441        -------------------------------------------------
442        bit 27,26 of z1         bit 1,0 of x1   logb(x)
443        -------------------------------------------------
444        00                      00              odd and even
445        01                      01              even
446        10                      10              odd
447        10                      00              even
448        11                      01              even
449        -------------------------------------------------
450
451    (4) Special cases (see (4) of Section A).
452
453 */
454
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