e14743d1 |
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 | #if defined(LIBM_SCCS) && !defined(lint) |
14 | static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 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 "math_private.h" |
89 | |
90 | #ifdef __STDC__ |
91 | double SDL_NAME(copysign)(double x, double y) |
92 | #else |
93 | double SDL_NAME(copysign)(x,y) |
94 | double x,y; |
95 | #endif |
96 | { |
97 | u_int32_t hx,hy; |
98 | GET_HIGH_WORD(hx,x); |
99 | GET_HIGH_WORD(hy,y); |
100 | SET_HIGH_WORD(x,(hx&0x7fffffff)|(hy&0x80000000)); |
101 | return x; |
102 | } |
103 | |
104 | #ifdef __STDC__ |
105 | double SDL_NAME(scalbn) (double x, int n) |
106 | #else |
107 | double SDL_NAME(scalbn) (x,n) |
108 | double x; int n; |
109 | #endif |
110 | { |
111 | int32_t k,hx,lx; |
112 | EXTRACT_WORDS(hx,lx,x); |
113 | k = (hx&0x7ff00000)>>20; /* extract exponent */ |
114 | if (k==0) { /* 0 or subnormal x */ |
115 | if ((lx|(hx&0x7fffffff))==0) return x; /* +-0 */ |
116 | x *= two54; |
117 | GET_HIGH_WORD(hx,x); |
118 | k = ((hx&0x7ff00000)>>20) - 54; |
119 | if (n< -50000) return tiny*x; /*underflow*/ |
120 | } |
121 | if (k==0x7ff) return x+x; /* NaN or Inf */ |
122 | k = k+n; |
123 | if (k > 0x7fe) return huge*SDL_NAME(copysign)(huge,x); /* overflow */ |
124 | if (k > 0) /* normal result */ |
125 | {SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); return x;} |
126 | if (k <= -54) { |
127 | if (n > 50000) /* in case integer overflow in n+k */ |
128 | return huge*SDL_NAME(copysign)(huge,x); /*overflow*/ |
129 | else return tiny*SDL_NAME(copysign)(tiny,x); /*underflow*/ |
130 | } |
131 | k += 54; /* subnormal result */ |
132 | SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); |
133 | return x*twom54; |
134 | } |
135 | |
136 | #ifdef __STDC__ |
137 | double __ieee754_sqrt(double x) |
138 | #else |
139 | double __ieee754_sqrt(x) |
140 | double x; |
141 | #endif |
142 | { |
143 | double z; |
144 | int32_t sign = (int)0x80000000; |
145 | int32_t ix0,s0,q,m,t,i; |
146 | u_int32_t r,t1,s1,ix1,q1; |
147 | |
148 | EXTRACT_WORDS(ix0,ix1,x); |
149 | |
150 | /* take care of Inf and NaN */ |
151 | if((ix0&0x7ff00000)==0x7ff00000) { |
152 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf |
153 | sqrt(-inf)=sNaN */ |
154 | } |
155 | /* take care of zero */ |
156 | if(ix0<=0) { |
157 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ |
158 | else if(ix0<0) |
159 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ |
160 | } |
161 | /* normalize x */ |
162 | m = (ix0>>20); |
163 | if(m==0) { /* subnormal x */ |
164 | while(ix0==0) { |
165 | m -= 21; |
166 | ix0 |= (ix1>>11); ix1 <<= 21; |
167 | } |
168 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; |
169 | m -= i-1; |
170 | ix0 |= (ix1>>(32-i)); |
171 | ix1 <<= i; |
172 | } |
173 | m -= 1023; /* unbias exponent */ |
174 | ix0 = (ix0&0x000fffff)|0x00100000; |
175 | if(m&1){ /* odd m, double x to make it even */ |
176 | ix0 += ix0 + ((ix1&sign)>>31); |
177 | ix1 += ix1; |
178 | } |
179 | m >>= 1; /* m = [m/2] */ |
180 | |
181 | /* generate sqrt(x) bit by bit */ |
182 | ix0 += ix0 + ((ix1&sign)>>31); |
183 | ix1 += ix1; |
184 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ |
185 | r = 0x00200000; /* r = moving bit from right to left */ |
186 | |
187 | while(r!=0) { |
188 | t = s0+r; |
189 | if(t<=ix0) { |
190 | s0 = t+r; |
191 | ix0 -= t; |
192 | q += r; |
193 | } |
194 | ix0 += ix0 + ((ix1&sign)>>31); |
195 | ix1 += ix1; |
196 | r>>=1; |
197 | } |
198 | |
199 | r = sign; |
200 | while(r!=0) { |
201 | t1 = s1+r; |
202 | t = s0; |
203 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) { |
204 | s1 = t1+r; |
205 | if(((int32_t)(t1&sign)==sign)&&(s1&sign)==0) s0 += 1; |
206 | ix0 -= t; |
207 | if (ix1 < t1) ix0 -= 1; |
208 | ix1 -= t1; |
209 | q1 += r; |
210 | } |
211 | ix0 += ix0 + ((ix1&sign)>>31); |
212 | ix1 += ix1; |
213 | r>>=1; |
214 | } |
215 | |
216 | /* use floating add to find out rounding direction */ |
217 | if((ix0|ix1)!=0) { |
218 | z = one-tiny; /* trigger inexact flag */ |
219 | if (z>=one) { |
220 | z = one+tiny; |
221 | if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;} |
222 | else if (z>one) { |
223 | if (q1==(u_int32_t)0xfffffffe) q+=1; |
224 | q1+=2; |
225 | } else |
226 | q1 += (q1&1); |
227 | } |
228 | } |
229 | ix0 = (q>>1)+0x3fe00000; |
230 | ix1 = q1>>1; |
231 | if ((q&1)==1) ix1 |= sign; |
232 | ix0 += (m <<20); |
233 | INSERT_WORDS(z,ix0,ix1); |
234 | return z; |
235 | } |
236 | |
237 | /* |
238 | Other methods (use floating-point arithmetic) |
239 | ------------- |
240 | (This is a copy of a drafted paper by Prof W. Kahan |
241 | and K.C. Ng, written in May, 1986) |
242 | |
243 | Two algorithms are given here to implement sqrt(x) |
244 | (IEEE double precision arithmetic) in software. |
245 | Both supply sqrt(x) correctly rounded. The first algorithm (in |
246 | Section A) uses newton iterations and involves four divisions. |
247 | The second one uses reciproot iterations to avoid division, but |
248 | requires more multiplications. Both algorithms need the ability |
249 | to chop results of arithmetic operations instead of round them, |
250 | and the INEXACT flag to indicate when an arithmetic operation |
251 | is executed exactly with no roundoff error, all part of the |
252 | standard (IEEE 754-1985). The ability to perform shift, add, |
253 | subtract and logical AND operations upon 32-bit words is needed |
254 | too, though not part of the standard. |
255 | |
256 | A. sqrt(x) by Newton Iteration |
257 | |
258 | (1) Initial approximation |
259 | |
260 | Let x0 and x1 be the leading and the trailing 32-bit words of |
261 | a floating point number x (in IEEE double format) respectively |
262 | |
263 | 1 11 52 ...widths |
264 | ------------------------------------------------------ |
265 | x: |s| e | f | |
266 | ------------------------------------------------------ |
267 | msb lsb msb lsb ...order |
268 | |
269 | |
270 | ------------------------ ------------------------ |
271 | x0: |s| e | f1 | x1: | f2 | |
272 | ------------------------ ------------------------ |
273 | |
274 | By performing shifts and subtracts on x0 and x1 (both regarded |
275 | as integers), we obtain an 8-bit approximation of sqrt(x) as |
276 | follows. |
277 | |
278 | k := (x0>>1) + 0x1ff80000; |
279 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits |
280 | Here k is a 32-bit integer and T1[] is an integer array containing |
281 | correction terms. Now magically the floating value of y (y's |
282 | leading 32-bit word is y0, the value of its trailing word is 0) |
283 | approximates sqrt(x) to almost 8-bit. |
284 | |
285 | Value of T1: |
286 | static int T1[32]= { |
287 | 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, |
288 | 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, |
289 | 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, |
290 | 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; |
291 | |
292 | (2) Iterative refinement |
293 | |
294 | Apply Heron's rule three times to y, we have y approximates |
295 | sqrt(x) to within 1 ulp (Unit in the Last Place): |
296 | |
297 | y := (y+x/y)/2 ... almost 17 sig. bits |
298 | y := (y+x/y)/2 ... almost 35 sig. bits |
299 | y := y-(y-x/y)/2 ... within 1 ulp |
300 | |
301 | |
302 | Remark 1. |
303 | Another way to improve y to within 1 ulp is: |
304 | |
305 | y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) |
306 | y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) |
307 | |
308 | 2 |
309 | (x-y )*y |
310 | y := y + 2* ---------- ...within 1 ulp |
311 | 2 |
312 | 3y + x |
313 | |
314 | |
315 | This formula has one division fewer than the one above; however, |
316 | it requires more multiplications and additions. Also x must be |
317 | scaled in advance to avoid spurious overflow in evaluating the |
318 | expression 3y*y+x. Hence it is not recommended uless division |
319 | is slow. If division is very slow, then one should use the |
320 | reciproot algorithm given in section B. |
321 | |
322 | (3) Final adjustment |
323 | |
324 | By twiddling y's last bit it is possible to force y to be |
325 | correctly rounded according to the prevailing rounding mode |
326 | as follows. Let r and i be copies of the rounding mode and |
327 | inexact flag before entering the square root program. Also we |
328 | use the expression y+-ulp for the next representable floating |
329 | numbers (up and down) of y. Note that y+-ulp = either fixed |
330 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
331 | mode. |
332 | |
333 | I := FALSE; ... reset INEXACT flag I |
334 | R := RZ; ... set rounding mode to round-toward-zero |
335 | z := x/y; ... chopped quotient, possibly inexact |
336 | If(not I) then { ... if the quotient is exact |
337 | if(z=y) { |
338 | I := i; ... restore inexact flag |
339 | R := r; ... restore rounded mode |
340 | return sqrt(x):=y. |
341 | } else { |
342 | z := z - ulp; ... special rounding |
343 | } |
344 | } |
345 | i := TRUE; ... sqrt(x) is inexact |
346 | If (r=RN) then z=z+ulp ... rounded-to-nearest |
347 | If (r=RP) then { ... round-toward-+inf |
348 | y = y+ulp; z=z+ulp; |
349 | } |
350 | y := y+z; ... chopped sum |
351 | y0:=y0-0x00100000; ... y := y/2 is correctly rounded. |
352 | I := i; ... restore inexact flag |
353 | R := r; ... restore rounded mode |
354 | return sqrt(x):=y. |
355 | |
356 | (4) Special cases |
357 | |
358 | Square root of +inf, +-0, or NaN is itself; |
359 | Square root of a negative number is NaN with invalid signal. |
360 | |
361 | |
362 | B. sqrt(x) by Reciproot Iteration |
363 | |
364 | (1) Initial approximation |
365 | |
366 | Let x0 and x1 be the leading and the trailing 32-bit words of |
367 | a floating point number x (in IEEE double format) respectively |
368 | (see section A). By performing shifs and subtracts on x0 and y0, |
369 | we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. |
370 | |
371 | k := 0x5fe80000 - (x0>>1); |
372 | y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits |
373 | |
374 | Here k is a 32-bit integer and T2[] is an integer array |
375 | containing correction terms. Now magically the floating |
376 | value of y (y's leading 32-bit word is y0, the value of |
377 | its trailing word y1 is set to zero) approximates 1/sqrt(x) |
378 | to almost 7.8-bit. |
379 | |
380 | Value of T2: |
381 | static int T2[64]= { |
382 | 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, |
383 | 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, |
384 | 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, |
385 | 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, |
386 | 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, |
387 | 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, |
388 | 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, |
389 | 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; |
390 | |
391 | (2) Iterative refinement |
392 | |
393 | Apply Reciproot iteration three times to y and multiply the |
394 | result by x to get an approximation z that matches sqrt(x) |
395 | to about 1 ulp. To be exact, we will have |
396 | -1ulp < sqrt(x)-z<1.0625ulp. |
397 | |
398 | ... set rounding mode to Round-to-nearest |
399 | y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) |
400 | y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) |
401 | ... special arrangement for better accuracy |
402 | z := x*y ... 29 bits to sqrt(x), with z*y<1 |
403 | z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) |
404 | |
405 | Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that |
406 | (a) the term z*y in the final iteration is always less than 1; |
407 | (b) the error in the final result is biased upward so that |
408 | -1 ulp < sqrt(x) - z < 1.0625 ulp |
409 | instead of |sqrt(x)-z|<1.03125ulp. |
410 | |
411 | (3) Final adjustment |
412 | |
413 | By twiddling y's last bit it is possible to force y to be |
414 | correctly rounded according to the prevailing rounding mode |
415 | as follows. Let r and i be copies of the rounding mode and |
416 | inexact flag before entering the square root program. Also we |
417 | use the expression y+-ulp for the next representable floating |
418 | numbers (up and down) of y. Note that y+-ulp = either fixed |
419 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
420 | mode. |
421 | |
422 | R := RZ; ... set rounding mode to round-toward-zero |
423 | switch(r) { |
424 | case RN: ... round-to-nearest |
425 | if(x<= z*(z-ulp)...chopped) z = z - ulp; else |
426 | if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; |
427 | break; |
428 | case RZ:case RM: ... round-to-zero or round-to--inf |
429 | R:=RP; ... reset rounding mod to round-to-+inf |
430 | if(x<z*z ... rounded up) z = z - ulp; else |
431 | if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; |
432 | break; |
433 | case RP: ... round-to-+inf |
434 | if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else |
435 | if(x>z*z ...chopped) z = z+ulp; |
436 | break; |
437 | } |
438 | |
439 | Remark 3. The above comparisons can be done in fixed point. For |
440 | example, to compare x and w=z*z chopped, it suffices to compare |
441 | x1 and w1 (the trailing parts of x and w), regarding them as |
442 | two's complement integers. |
443 | |
444 | ...Is z an exact square root? |
445 | To determine whether z is an exact square root of x, let z1 be the |
446 | trailing part of z, and also let x0 and x1 be the leading and |
447 | trailing parts of x. |
448 | |
449 | If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 |
450 | I := 1; ... Raise Inexact flag: z is not exact |
451 | else { |
452 | j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 |
453 | k := z1 >> 26; ... get z's 25-th and 26-th |
454 | fraction bits |
455 | I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); |
456 | } |
457 | R:= r ... restore rounded mode |
458 | return sqrt(x):=z. |
459 | |
460 | If multiplication is cheaper then the foregoing red tape, the |
461 | Inexact flag can be evaluated by |
462 | |
463 | I := i; |
464 | I := (z*z!=x) or I. |
465 | |
466 | Note that z*z can overwrite I; this value must be sensed if it is |
467 | True. |
468 | |
469 | Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be |
470 | zero. |
471 | |
472 | -------------------- |
473 | z1: | f2 | |
474 | -------------------- |
475 | bit 31 bit 0 |
476 | |
477 | Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd |
478 | or even of logb(x) have the following relations: |
479 | |
480 | ------------------------------------------------- |
481 | bit 27,26 of z1 bit 1,0 of x1 logb(x) |
482 | ------------------------------------------------- |
483 | 00 00 odd and even |
484 | 01 01 even |
485 | 10 10 odd |
486 | 10 00 even |
487 | 11 01 even |
488 | ------------------------------------------------- |
489 | |
490 | (4) Special cases (see (4) of Section A). |
491 | |
492 | */ |
493 | |