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4acf9396 GCI |
1 | /* @(#)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 1.4 1994/03/03 17:04:21 jtc 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 <machine/endian.h> | |
89 | ||
90 | #if BYTE_ORDER == LITTLE_ENDIAN | |
91 | #define n0 1 | |
92 | #else | |
93 | #define n0 0 | |
94 | #endif | |
95 | ||
96 | #ifdef __STDC__ | |
97 | static const double one = 1.0, tiny=1.0e-300; | |
98 | #else | |
99 | static double one = 1.0, tiny=1.0e-300; | |
100 | #endif | |
101 | ||
102 | #ifdef __STDC__ | |
103 | double __ieee754_sqrt(double x) | |
104 | #else | |
105 | double __ieee754_sqrt(x) | |
106 | double x; | |
107 | #endif | |
108 | { | |
109 | double z; | |
110 | int sign = (int)0x80000000; | |
111 | unsigned r,t1,s1,ix1,q1; | |
112 | int ix0,s0,q,m,t,i; | |
113 | ||
114 | ix0 = *(n0+(int*)&x); /* high word of x */ | |
115 | ix1 = *((1-n0)+(int*)&x); /* low word of x */ | |
116 | ||
117 | /* take care of Inf and NaN */ | |
118 | if((ix0&0x7ff00000)==0x7ff00000) { | |
119 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf | |
120 | sqrt(-inf)=sNaN */ | |
121 | } | |
122 | /* take care of zero */ | |
123 | if(ix0<=0) { | |
124 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ | |
125 | else if(ix0<0) | |
126 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ | |
127 | } | |
128 | /* normalize x */ | |
129 | m = (ix0>>20); | |
130 | if(m==0) { /* subnormal x */ | |
131 | while(ix0==0) { | |
132 | m -= 21; | |
133 | ix0 |= (ix1>>11); ix1 <<= 21; | |
134 | } | |
135 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; | |
136 | m -= i-1; | |
137 | ix0 |= (ix1>>(32-i)); | |
138 | ix1 <<= i; | |
139 | } | |
140 | m -= 1023; /* unbias exponent */ | |
141 | ix0 = (ix0&0x000fffff)|0x00100000; | |
142 | if(m&1){ /* odd m, double x to make it even */ | |
143 | ix0 += ix0 + ((ix1&sign)>>31); | |
144 | ix1 += ix1; | |
145 | } | |
146 | m >>= 1; /* m = [m/2] */ | |
147 | ||
148 | /* generate sqrt(x) bit by bit */ | |
149 | ix0 += ix0 + ((ix1&sign)>>31); | |
150 | ix1 += ix1; | |
151 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ | |
152 | r = 0x00200000; /* r = moving bit from right to left */ | |
153 | ||
154 | while(r!=0) { | |
155 | t = s0+r; | |
156 | if(t<=ix0) { | |
157 | s0 = t+r; | |
158 | ix0 -= t; | |
159 | q += r; | |
160 | } | |
161 | ix0 += ix0 + ((ix1&sign)>>31); | |
162 | ix1 += ix1; | |
163 | r>>=1; | |
164 | } | |
165 | ||
166 | r = sign; | |
167 | while(r!=0) { | |
168 | t1 = s1+r; | |
169 | t = s0; | |
170 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) { | |
171 | s1 = t1+r; | |
172 | if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; | |
173 | ix0 -= t; | |
174 | if (ix1 < t1) ix0 -= 1; | |
175 | ix1 -= t1; | |
176 | q1 += r; | |
177 | } | |
178 | ix0 += ix0 + ((ix1&sign)>>31); | |
179 | ix1 += ix1; | |
180 | r>>=1; | |
181 | } | |
182 | ||
183 | /* use floating add to find out rounding direction */ | |
184 | if((ix0|ix1)!=0) { | |
185 | z = one-tiny; /* trigger inexact flag */ | |
186 | if (z>=one) { | |
187 | z = one+tiny; | |
188 | if (q1==(unsigned)0xffffffff) { q1=0; q += 1;} | |
189 | else if (z>one) { | |
190 | if (q1==(unsigned)0xfffffffe) q+=1; | |
191 | q1+=2; | |
192 | } else | |
193 | q1 += (q1&1); | |
194 | } | |
195 | } | |
196 | ix0 = (q>>1)+0x3fe00000; | |
197 | ix1 = q1>>1; | |
198 | if ((q&1)==1) ix1 |= sign; | |
199 | ix0 += (m <<20); | |
200 | *(n0+(int*)&z) = ix0; | |
201 | *((1-n0)+(int*)&z) = ix1; | |
202 | return z; | |
203 | } | |
204 | ||
205 | /* | |
206 | Other methods (use floating-point arithmetic) | |
207 | ------------- | |
208 | (This is a copy of a drafted paper by Prof W. Kahan | |
209 | and K.C. Ng, written in May, 1986) | |
210 | ||
211 | Two algorithms are given here to implement sqrt(x) | |
212 | (IEEE double precision arithmetic) in software. | |
213 | Both supply sqrt(x) correctly rounded. The first algorithm (in | |
214 | Section A) uses newton iterations and involves four divisions. | |
215 | The second one uses reciproot iterations to avoid division, but | |
216 | requires more multiplications. Both algorithms need the ability | |
217 | to chop results of arithmetic operations instead of round them, | |
218 | and the INEXACT flag to indicate when an arithmetic operation | |
219 | is executed exactly with no roundoff error, all part of the | |
220 | standard (IEEE 754-1985). The ability to perform shift, add, | |
221 | subtract and logical AND operations upon 32-bit words is needed | |
222 | too, though not part of the standard. | |
223 | ||
224 | A. sqrt(x) by Newton Iteration | |
225 | ||
226 | (1) Initial approximation | |
227 | ||
228 | Let x0 and x1 be the leading and the trailing 32-bit words of | |
229 | a floating point number x (in IEEE double format) respectively | |
230 | ||
231 | 1 11 52 ...widths | |
232 | ------------------------------------------------------ | |
233 | x: |s| e | f | | |
234 | ------------------------------------------------------ | |
235 | msb lsb msb lsb ...order | |
236 | ||
237 | ||
238 | ------------------------ ------------------------ | |
239 | x0: |s| e | f1 | x1: | f2 | | |
240 | ------------------------ ------------------------ | |
241 | ||
242 | By performing shifts and subtracts on x0 and x1 (both regarded | |
243 | as integers), we obtain an 8-bit approximation of sqrt(x) as | |
244 | follows. | |
245 | ||
246 | k := (x0>>1) + 0x1ff80000; | |
247 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits | |
248 | Here k is a 32-bit integer and T1[] is an integer array containing | |
249 | correction terms. Now magically the floating value of y (y's | |
250 | leading 32-bit word is y0, the value of its trailing word is 0) | |
251 | approximates sqrt(x) to almost 8-bit. | |
252 | ||
253 | Value of T1: | |
254 | static int T1[32]= { | |
255 | 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, | |
256 | 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, | |
257 | 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, | |
258 | 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; | |
259 | ||
260 | (2) Iterative refinement | |
261 | ||
262 | Apply Heron's rule three times to y, we have y approximates | |
263 | sqrt(x) to within 1 ulp (Unit in the Last Place): | |
264 | ||
265 | y := (y+x/y)/2 ... almost 17 sig. bits | |
266 | y := (y+x/y)/2 ... almost 35 sig. bits | |
267 | y := y-(y-x/y)/2 ... within 1 ulp | |
268 | ||
269 | ||
270 | Remark 1. | |
271 | Another way to improve y to within 1 ulp is: | |
272 | ||
273 | y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) | |
274 | y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) | |
275 | ||
276 | 2 | |
277 | (x-y )*y | |
278 | y := y + 2* ---------- ...within 1 ulp | |
279 | 2 | |
280 | 3y + x | |
281 | ||
282 | ||
283 | This formula has one division fewer than the one above; however, | |
284 | it requires more multiplications and additions. Also x must be | |
285 | scaled in advance to avoid spurious overflow in evaluating the | |
286 | expression 3y*y+x. Hence it is not recommended uless division | |
287 | is slow. If division is very slow, then one should use the | |
288 | reciproot algorithm given in section B. | |
289 | ||
290 | (3) Final adjustment | |
291 | ||
292 | By twiddling y's last bit it is possible to force y to be | |
293 | correctly rounded according to the prevailing rounding mode | |
294 | as follows. Let r and i be copies of the rounding mode and | |
295 | inexact flag before entering the square root program. Also we | |
296 | use the expression y+-ulp for the next representable floating | |
297 | numbers (up and down) of y. Note that y+-ulp = either fixed | |
298 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped | |
299 | mode. | |
300 | ||
301 | I := FALSE; ... reset INEXACT flag I | |
302 | R := RZ; ... set rounding mode to round-toward-zero | |
303 | z := x/y; ... chopped quotient, possibly inexact | |
304 | If(not I) then { ... if the quotient is exact | |
305 | if(z=y) { | |
306 | I := i; ... restore inexact flag | |
307 | R := r; ... restore rounded mode | |
308 | return sqrt(x):=y. | |
309 | } else { | |
310 | z := z - ulp; ... special rounding | |
311 | } | |
312 | } | |
313 | i := TRUE; ... sqrt(x) is inexact | |
314 | If (r=RN) then z=z+ulp ... rounded-to-nearest | |
315 | If (r=RP) then { ... round-toward-+inf | |
316 | y = y+ulp; z=z+ulp; | |
317 | } | |
318 | y := y+z; ... chopped sum | |
319 | y0:=y0-0x00100000; ... y := y/2 is correctly rounded. | |
320 | I := i; ... restore inexact flag | |
321 | R := r; ... restore rounded mode | |
322 | return sqrt(x):=y. | |
323 | ||
324 | (4) Special cases | |
325 | ||
326 | Square root of +inf, +-0, or NaN is itself; | |
327 | Square root of a negative number is NaN with invalid signal. | |
328 | ||
329 | ||
330 | B. sqrt(x) by Reciproot Iteration | |
331 | ||
332 | (1) Initial approximation | |
333 | ||
334 | Let x0 and x1 be the leading and the trailing 32-bit words of | |
335 | a floating point number x (in IEEE double format) respectively | |
336 | (see section A). By performing shifs and subtracts on x0 and y0, | |
337 | we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. | |
338 | ||
339 | k := 0x5fe80000 - (x0>>1); | |
340 | y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits | |
341 | ||
342 | Here k is a 32-bit integer and T2[] is an integer array | |
343 | containing correction terms. Now magically the floating | |
344 | value of y (y's leading 32-bit word is y0, the value of | |
345 | its trailing word y1 is set to zero) approximates 1/sqrt(x) | |
346 | to almost 7.8-bit. | |
347 | ||
348 | Value of T2: | |
349 | static int T2[64]= { | |
350 | 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, | |
351 | 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, | |
352 | 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, | |
353 | 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, | |
354 | 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, | |
355 | 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, | |
356 | 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, | |
357 | 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; | |
358 | ||
359 | (2) Iterative refinement | |
360 | ||
361 | Apply Reciproot iteration three times to y and multiply the | |
362 | result by x to get an approximation z that matches sqrt(x) | |
363 | to about 1 ulp. To be exact, we will have | |
364 | -1ulp < sqrt(x)-z<1.0625ulp. | |
365 | ||
366 | ... set rounding mode to Round-to-nearest | |
367 | y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) | |
368 | y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) | |
369 | ... special arrangement for better accuracy | |
370 | z := x*y ... 29 bits to sqrt(x), with z*y<1 | |
371 | z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) | |
372 | ||
373 | Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that | |
374 | (a) the term z*y in the final iteration is always less than 1; | |
375 | (b) the error in the final result is biased upward so that | |
376 | -1 ulp < sqrt(x) - z < 1.0625 ulp | |
377 | instead of |sqrt(x)-z|<1.03125ulp. | |
378 | ||
379 | (3) Final adjustment | |
380 | ||
381 | By twiddling y's last bit it is possible to force y to be | |
382 | correctly rounded according to the prevailing rounding mode | |
383 | as follows. Let r and i be copies of the rounding mode and | |
384 | inexact flag before entering the square root program. Also we | |
385 | use the expression y+-ulp for the next representable floating | |
386 | numbers (up and down) of y. Note that y+-ulp = either fixed | |
387 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped | |
388 | mode. | |
389 | ||
390 | R := RZ; ... set rounding mode to round-toward-zero | |
391 | switch(r) { | |
392 | case RN: ... round-to-nearest | |
393 | if(x<= z*(z-ulp)...chopped) z = z - ulp; else | |
394 | if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; | |
395 | break; | |
396 | case RZ:case RM: ... round-to-zero or round-to--inf | |
397 | R:=RP; ... reset rounding mod to round-to-+inf | |
398 | if(x<z*z ... rounded up) z = z - ulp; else | |
399 | if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; | |
400 | break; | |
401 | case RP: ... round-to-+inf | |
402 | if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else | |
403 | if(x>z*z ...chopped) z = z+ulp; | |
404 | break; | |
405 | } | |
406 | ||
407 | Remark 3. The above comparisons can be done in fixed point. For | |
408 | example, to compare x and w=z*z chopped, it suffices to compare | |
409 | x1 and w1 (the trailing parts of x and w), regarding them as | |
410 | two's complement integers. | |
411 | ||
412 | ...Is z an exact square root? | |
413 | To determine whether z is an exact square root of x, let z1 be the | |
414 | trailing part of z, and also let x0 and x1 be the leading and | |
415 | trailing parts of x. | |
416 | ||
417 | If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 | |
418 | I := 1; ... Raise Inexact flag: z is not exact | |
419 | else { | |
420 | j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 | |
421 | k := z1 >> 26; ... get z's 25-th and 26-th | |
422 | fraction bits | |
423 | I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); | |
424 | } | |
425 | R:= r ... restore rounded mode | |
426 | return sqrt(x):=z. | |
427 | ||
428 | If multiplication is cheaper then the foregoing red tape, the | |
429 | Inexact flag can be evaluated by | |
430 | ||
431 | I := i; | |
432 | I := (z*z!=x) or I. | |
433 | ||
434 | Note that z*z can overwrite I; this value must be sensed if it is | |
435 | True. | |
436 | ||
437 | Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be | |
438 | zero. | |
439 | ||
440 | -------------------- | |
441 | z1: | f2 | | |
442 | -------------------- | |
443 | bit 31 bit 0 | |
444 | ||
445 | Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd | |
446 | or even of logb(x) have the following relations: | |
447 | ||
448 | ------------------------------------------------- | |
449 | bit 27,26 of z1 bit 1,0 of x1 logb(x) | |
450 | ------------------------------------------------- | |
451 | 00 00 odd and even | |
452 | 01 01 even | |
453 | 10 10 odd | |
454 | 10 00 even | |
455 | 11 01 even | |
456 | ------------------------------------------------- | |
457 | ||
458 | (4) Special cases (see (4) of Section A). | |
459 | ||
460 | */ | |
461 |