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make sure we have a large enough buffer to hold the output of the print statement
[AROS.git] / compiler / stdc / math / e_sqrt.c
blob219783cc7b729b0e34ef3d95e165e8c7cb5d5998
1
2 /* @(#)e_sqrt.c 1.3 95/01/18 */
3 /*
4 * ====================================================
5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 *
7 * Developed at SunSoft, 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 #ifndef lint
15 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_sqrt.c,v 1.11 2008/03/02 01:47:58 das Exp $";
16 #endif
17
18 /* __ieee754_sqrt(x)
19 * Return correctly rounded sqrt.
20 * ------------------------------------------
21 * | Use the hardware sqrt if you have one |
22 * ------------------------------------------
23 * Method:
24 * Bit by bit method using integer arithmetic. (Slow, but portable)
25 * 1. Normalization
26 * Scale x to y in [1,4) with even powers of 2:
27 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
28 * sqrt(x) = 2^k * sqrt(y)
29 * 2. Bit by bit computation
30 * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
31 * i 0
32 * i+1 2
33 * s = 2*q , and y = 2 * ( y - q ). (1)
34 * i i i i
35 *
36 * To compute q from q , one checks whether
37 * i+1 i
38 *
39 * -(i+1) 2
40 * (q + 2 ) <= y. (2)
41 * i
42 * -(i+1)
43 * If (2) is false, then q = q ; otherwise q = q + 2 .
44 * i+1 i i+1 i
45 *
46 * With some algebric manipulation, it is not difficult to see
47 * that (2) is equivalent to
48 * -(i+1)
49 * s + 2 <= y (3)
50 * i i
51 *
52 * The advantage of (3) is that s and y can be computed by
53 * i i
54 * the following recurrence formula:
55 * if (3) is false
56 *
57 * s = s , y = y ; (4)
58 * i+1 i i+1 i
59 *
60 * otherwise,
61 * -i -(i+1)
62 * s = s + 2 , y = y - s - 2 (5)
63 * i+1 i i+1 i i
64 *
65 * One may easily use induction to prove (4) and (5).
66 * Note. Since the left hand side of (3) contain only i+2 bits,
67 * it does not necessary to do a full (53-bit) comparison
68 * in (3).
69 * 3. Final rounding
70 * After generating the 53 bits result, we compute one more bit.
71 * Together with the remainder, we can decide whether the
72 * result is exact, bigger than 1/2ulp, or less than 1/2ulp
73 * (it will never equal to 1/2ulp).
74 * The rounding mode can be detected by checking whether
75 * huge + tiny is equal to huge, and whether huge - tiny is
76 * equal to huge for some floating point number "huge" and "tiny".
77 *
78 * Special cases:
79 * sqrt(+-0) = +-0 ... exact
80 * sqrt(inf) = inf
81 * sqrt(-ve) = NaN ... with invalid signal
82 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
83 *
84 * Other methods : see the appended file at the end of the program below.
85 *---------------
86 */
87
88 #include <float.h>
89 #include "math.h"
90 #include "math_private.h"
91
92 static const double one = 1.0, tiny=1.0e-300;
93
94 double
95 __ieee754_sqrt(double x)
96 {
97 double z;
98 int32_t sign = (int)0x80000000;
99 int32_t ix0,s0,q,m,t,i;
100 uint32_t r,t1,s1,ix1,q1;
101
102 EXTRACT_WORDS(ix0,ix1,x);
103
104 /* take care of Inf and NaN */
105 if((ix0&0x7ff00000)==0x7ff00000) {
106 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
107 sqrt(-inf)=sNaN */
108 }
109 /* take care of zero */
110 if(ix0<=0) {
111 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
112 else if(ix0<0)
113 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
114 }
115 /* normalize x */
116 m = (ix0>>20);
117 if(m==0) { /* subnormal x */
118 while(ix0==0) {
119 m -= 21;
120 ix0 |= (ix1>>11); ix1 <<= 21;
121 }
122 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
123 m -= i-1;
124 ix0 |= (ix1>>(32-i));
125 ix1 <<= i;
126 }
127 m -= 1023; /* unbias exponent */
128 ix0 = (ix0&0x000fffff)|0x00100000;
129 if(m&1){ /* odd m, double x to make it even */
130 ix0 += ix0 + ((ix1&sign)>>31);
131 ix1 += ix1;
132 }
133 m >>= 1; /* m = [m/2] */
134
135 /* generate sqrt(x) bit by bit */
136 ix0 += ix0 + ((ix1&sign)>>31);
137 ix1 += ix1;
138 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
139 r = 0x00200000; /* r = moving bit from right to left */
140
141 while(r!=0) {
142 t = s0+r;
143 if(t<=ix0) {
144 s0 = t+r;
145 ix0 -= t;
146 q += r;
147 }
148 ix0 += ix0 + ((ix1&sign)>>31);
149 ix1 += ix1;
150 r>>=1;
151 }
152
153 r = sign;
154 while(r!=0) {
155 t1 = s1+r;
156 t = s0;
157 if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
158 s1 = t1+r;
159 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
160 ix0 -= t;
161 if (ix1 < t1) ix0 -= 1;
162 ix1 -= t1;
163 q1 += r;
164 }
165 ix0 += ix0 + ((ix1&sign)>>31);
166 ix1 += ix1;
167 r>>=1;
168 }
169
170 /* use floating add to find out rounding direction */
171 if((ix0|ix1)!=0) {
172 z = one-tiny; /* trigger inexact flag */
173 if (z>=one) {
174 z = one+tiny;
175 if (q1==(uint32_t)0xffffffff) { q1=0; q += 1;}
176 else if (z>one) {
177 if (q1==(uint32_t)0xfffffffe) q+=1;
178 q1+=2;
179 } else
180 q1 += (q1&1);
181 }
182 }
183 ix0 = (q>>1)+0x3fe00000;
184 ix1 = q1>>1;
185 if ((q&1)==1) ix1 |= sign;
186 ix0 += (m <<20);
187 INSERT_WORDS(z,ix0,ix1);
188 return z;
189 }
190
191 #if (LDBL_MANT_DIG == 53)
192 AROS_MAKE_ASM_SYM(typeof(sqrtl), sqrtl, AROS_CSYM_FROM_ASM_NAME(sqrtl), AROS_CSYM_FROM_ASM_NAME(sqrt));
193 AROS_EXPORT_ASM_SYM(AROS_CSYM_FROM_ASM_NAME(sqrtl));
194 #endif
195
196 /*
197 Other methods (use floating-point arithmetic)
198 -------------
199 (This is a copy of a drafted paper by Prof W. Kahan
200 and K.C. Ng, written in May, 1986)
201
202 Two algorithms are given here to implement sqrt(x)
203 (IEEE double precision arithmetic) in software.
204 Both supply sqrt(x) correctly rounded. The first algorithm (in
205 Section A) uses newton iterations and involves four divisions.
206 The second one uses reciproot iterations to avoid division, but
207 requires more multiplications. Both algorithms need the ability
208 to chop results of arithmetic operations instead of round them,
209 and the INEXACT flag to indicate when an arithmetic operation
210 is executed exactly with no roundoff error, all part of the
211 standard (IEEE 754-1985). The ability to perform shift, add,
212 subtract and logical AND operations upon 32-bit words is needed
213 too, though not part of the standard.
214
215 A. sqrt(x) by Newton Iteration
216
217 (1) Initial approximation
218
219 Let x0 and x1 be the leading and the trailing 32-bit words of
220 a floating point number x (in IEEE double format) respectively
221
222 1 11 52 ...widths
223 ------------------------------------------------------
224 x: |s| e | f |
225 ------------------------------------------------------
226 msb lsb msb lsb ...order
227
228
229 ------------------------ ------------------------
230 x0: |s| e | f1 | x1: | f2 |
231 ------------------------ ------------------------
232
233 By performing shifts and subtracts on x0 and x1 (both regarded
234 as integers), we obtain an 8-bit approximation of sqrt(x) as
235 follows.
236
237 k := (x0>>1) + 0x1ff80000;
238 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
239 Here k is a 32-bit integer and T1[] is an integer array containing
240 correction terms. Now magically the floating value of y (y's
241 leading 32-bit word is y0, the value of its trailing word is 0)
242 approximates sqrt(x) to almost 8-bit.
243
244 Value of T1:
245 static int T1[32]= {
246 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
247 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
248 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
249 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
250
251 (2) Iterative refinement
252
253 Apply Heron's rule three times to y, we have y approximates
254 sqrt(x) to within 1 ulp (Unit in the Last Place):
255
256 y := (y+x/y)/2 ... almost 17 sig. bits
257 y := (y+x/y)/2 ... almost 35 sig. bits
258 y := y-(y-x/y)/2 ... within 1 ulp
259
260
261 Remark 1.
262 Another way to improve y to within 1 ulp is:
263
264 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
265 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
266
267 2
268 (x-y )*y
269 y := y + 2* ---------- ...within 1 ulp
270 2
271 3y + x
272
273
274 This formula has one division fewer than the one above; however,
275 it requires more multiplications and additions. Also x must be
276 scaled in advance to avoid spurious overflow in evaluating the
277 expression 3y*y+x. Hence it is not recommended uless division
278 is slow. If division is very slow, then one should use the
279 reciproot algorithm given in section B.
280
281 (3) Final adjustment
282
283 By twiddling y's last bit it is possible to force y to be
284 correctly rounded according to the prevailing rounding mode
285 as follows. Let r and i be copies of the rounding mode and
286 inexact flag before entering the square root program. Also we
287 use the expression y+-ulp for the next representable floating
288 numbers (up and down) of y. Note that y+-ulp = either fixed
289 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
290 mode.
291
292 I := FALSE; ... reset INEXACT flag I
293 R := RZ; ... set rounding mode to round-toward-zero
294 z := x/y; ... chopped quotient, possibly inexact
295 If(not I) then { ... if the quotient is exact
296 if(z=y) {
297 I := i; ... restore inexact flag
298 R := r; ... restore rounded mode
299 return sqrt(x):=y.
300 } else {
301 z := z - ulp; ... special rounding
302 }
303 }
304 i := TRUE; ... sqrt(x) is inexact
305 If (r=RN) then z=z+ulp ... rounded-to-nearest
306 If (r=RP) then { ... round-toward-+inf
307 y = y+ulp; z=z+ulp;
308 }
309 y := y+z; ... chopped sum
310 y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
311 I := i; ... restore inexact flag
312 R := r; ... restore rounded mode
313 return sqrt(x):=y.
314
315 (4) Special cases
316
317 Square root of +inf, +-0, or NaN is itself;
318 Square root of a negative number is NaN with invalid signal.
319
320
321 B. sqrt(x) by Reciproot Iteration
322
323 (1) Initial approximation
324
325 Let x0 and x1 be the leading and the trailing 32-bit words of
326 a floating point number x (in IEEE double format) respectively
327 (see section A). By performing shifs and subtracts on x0 and y0,
328 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
329
330 k := 0x5fe80000 - (x0>>1);
331 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
332
333 Here k is a 32-bit integer and T2[] is an integer array
334 containing correction terms. Now magically the floating
335 value of y (y's leading 32-bit word is y0, the value of
336 its trailing word y1 is set to zero) approximates 1/sqrt(x)
337 to almost 7.8-bit.
338
339 Value of T2:
340 static int T2[64]= {
341 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
342 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
343 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
344 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
345 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
346 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
347 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
348 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
349
350 (2) Iterative refinement
351
352 Apply Reciproot iteration three times to y and multiply the
353 result by x to get an approximation z that matches sqrt(x)
354 to about 1 ulp. To be exact, we will have
355 -1ulp < sqrt(x)-z<1.0625ulp.
356
357 ... set rounding mode to Round-to-nearest
358 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
359 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
360 ... special arrangement for better accuracy
361 z := x*y ... 29 bits to sqrt(x), with z*y<1
362 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
363
364 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
365 (a) the term z*y in the final iteration is always less than 1;
366 (b) the error in the final result is biased upward so that
367 -1 ulp < sqrt(x) - z < 1.0625 ulp
368 instead of |sqrt(x)-z|<1.03125ulp.
369
370 (3) Final adjustment
371
372 By twiddling y's last bit it is possible to force y to be
373 correctly rounded according to the prevailing rounding mode
374 as follows. Let r and i be copies of the rounding mode and
375 inexact flag before entering the square root program. Also we
376 use the expression y+-ulp for the next representable floating
377 numbers (up and down) of y. Note that y+-ulp = either fixed
378 point y+-1, or multiply y by nextafter(1,+-inf) in chopped
379 mode.
380
381 R := RZ; ... set rounding mode to round-toward-zero
382 switch(r) {
383 case RN: ... round-to-nearest
384 if(x<= z*(z-ulp)...chopped) z = z - ulp; else
385 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
386 break;
387 case RZ:case RM: ... round-to-zero or round-to--inf
388 R:=RP; ... reset rounding mod to round-to-+inf
389 if(x<z*z ... rounded up) z = z - ulp; else
390 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
391 break;
392 case RP: ... round-to-+inf
393 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
394 if(x>z*z ...chopped) z = z+ulp;
395 break;
396 }
397
398 Remark 3. The above comparisons can be done in fixed point. For
399 example, to compare x and w=z*z chopped, it suffices to compare
400 x1 and w1 (the trailing parts of x and w), regarding them as
401 two's complement integers.
402
403 ...Is z an exact square root?
404 To determine whether z is an exact square root of x, let z1 be the
405 trailing part of z, and also let x0 and x1 be the leading and
406 trailing parts of x.
407
408 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
409 I := 1; ... Raise Inexact flag: z is not exact
410 else {
411 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
412 k := z1 >> 26; ... get z's 25-th and 26-th
413 fraction bits
414 I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
415 }
416 R:= r ... restore rounded mode
417 return sqrt(x):=z.
418
419 If multiplication is cheaper then the foregoing red tape, the
420 Inexact flag can be evaluated by
421
422 I := i;
423 I := (z*z!=x) or I.
424
425 Note that z*z can overwrite I; this value must be sensed if it is
426 True.
427
428 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
429 zero.
430
431 --------------------
432 z1: | f2 |
433 --------------------
434 bit 31 bit 0
435
436 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
437 or even of logb(x) have the following relations:
438
439 -------------------------------------------------
440 bit 27,26 of z1 bit 1,0 of x1 logb(x)
441 -------------------------------------------------
442 00 00 odd and even
443 01 01 even
444 10 10 odd
445 10 00 even
446 11 01 even
447 -------------------------------------------------
448
449 (4) Special cases (see (4) of Section A).
450
451 */
Mirror of AROS' svn.