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2aa62f2b | 1 | /** @file\r |
2 | Compute the logrithm of x.\r | |
3 | \r | |
4 | Copyright (c) 2010 - 2011, Intel Corporation. All rights reserved.<BR>\r | |
5 | This program and the accompanying materials are licensed and made available under\r | |
6 | the terms and conditions of the BSD License that accompanies this distribution.\r | |
7 | The full text of the license may be found at\r | |
8 | http://opensource.org/licenses/bsd-license.\r | |
9 | \r | |
10 | THE PROGRAM IS DISTRIBUTED UNDER THE BSD LICENSE ON AN "AS IS" BASIS,\r | |
11 | WITHOUT WARRANTIES OR REPRESENTATIONS OF ANY KIND, EITHER EXPRESS OR IMPLIED.\r | |
12 | \r | |
13 | * ====================================================\r | |
14 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.\r | |
15 | *\r | |
16 | * Developed at SunPro, a Sun Microsystems, Inc. business.\r | |
17 | * Permission to use, copy, modify, and distribute this\r | |
18 | * software is freely granted, provided that this notice\r | |
19 | * is preserved.\r | |
20 | * ====================================================\r | |
21 | \r | |
22 | e_sqrt.c 5.1 93/09/24\r | |
23 | NetBSD: e_sqrt.c,v 1.12 2002/05/26 22:01:52 wiz Exp\r | |
24 | **/\r | |
25 | #include <LibConfig.h>\r | |
26 | #include <sys/EfiCdefs.h>\r | |
27 | \r | |
28 | #include <errno.h>\r | |
29 | #include "math.h"\r | |
30 | #include "math_private.h"\r | |
31 | \r | |
32 | #if defined(_MSC_VER) /* Handle Microsoft VC++ compiler specifics. */\r | |
33 | // potential divide by 0 -- near line 129, (x-x)/(x-x) is on purpose\r | |
34 | #pragma warning ( disable : 4723 )\r | |
35 | #endif\r | |
36 | \r | |
37 | /* __ieee754_sqrt(x)\r | |
38 | * Return correctly rounded sqrt.\r | |
39 | * ------------------------------------------\r | |
40 | * | Use the hardware sqrt if you have one |\r | |
41 | * ------------------------------------------\r | |
42 | * Method:\r | |
43 | * Bit by bit method using integer arithmetic. (Slow, but portable)\r | |
44 | * 1. Normalization\r | |
45 | * Scale x to y in [1,4) with even powers of 2:\r | |
46 | * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then\r | |
47 | * sqrt(x) = 2^k * sqrt(y)\r | |
48 | * 2. Bit by bit computation\r | |
49 | * Let q = sqrt(y) truncated to i bit after binary point (q = 1),\r | |
50 | * i 0\r | |
51 | * i+1 2\r | |
52 | * s = 2*q , and y = 2 * ( y - q ). (1)\r | |
53 | * i i i i\r | |
54 | *\r | |
55 | * To compute q from q , one checks whether\r | |
56 | * i+1 i\r | |
57 | *\r | |
58 | * -(i+1) 2\r | |
59 | * (q + 2 ) <= y. (2)\r | |
60 | * i\r | |
61 | * -(i+1)\r | |
62 | * If (2) is false, then q = q ; otherwise q = q + 2 .\r | |
63 | * i+1 i i+1 i\r | |
64 | *\r | |
65 | * With some algebric manipulation, it is not difficult to see\r | |
66 | * that (2) is equivalent to\r | |
67 | * -(i+1)\r | |
68 | * s + 2 <= y (3)\r | |
69 | * i i\r | |
70 | *\r | |
71 | * The advantage of (3) is that s and y can be computed by\r | |
72 | * i i\r | |
73 | * the following recurrence formula:\r | |
74 | * if (3) is false\r | |
75 | *\r | |
76 | * s = s , y = y ; (4)\r | |
77 | * i+1 i i+1 i\r | |
78 | *\r | |
79 | * otherwise,\r | |
80 | * -i -(i+1)\r | |
81 | * s = s + 2 , y = y - s - 2 (5)\r | |
82 | * i+1 i i+1 i i\r | |
83 | *\r | |
84 | * One may easily use induction to prove (4) and (5).\r | |
85 | * Note. Since the left hand side of (3) contain only i+2 bits,\r | |
86 | * it does not necessary to do a full (53-bit) comparison\r | |
87 | * in (3).\r | |
88 | * 3. Final rounding\r | |
89 | * After generating the 53 bits result, we compute one more bit.\r | |
90 | * Together with the remainder, we can decide whether the\r | |
91 | * result is exact, bigger than 1/2ulp, or less than 1/2ulp\r | |
92 | * (it will never equal to 1/2ulp).\r | |
93 | * The rounding mode can be detected by checking whether\r | |
94 | * huge + tiny is equal to huge, and whether huge - tiny is\r | |
95 | * equal to huge for some floating point number "huge" and "tiny".\r | |
96 | *\r | |
97 | * Special cases:\r | |
98 | * sqrt(+-0) = +-0 ... exact\r | |
99 | * sqrt(inf) = inf\r | |
100 | * sqrt(-ve) = NaN ... with invalid signal\r | |
101 | * sqrt(NaN) = NaN ... with invalid signal for signaling NaN\r | |
102 | *\r | |
103 | * Other methods : see the appended file at the end of the program below.\r | |
104 | *---------------\r | |
105 | */\r | |
106 | \r | |
107 | static const double one = 1.0, tiny=1.0e-300;\r | |
108 | \r | |
109 | double\r | |
110 | __ieee754_sqrt(double x)\r | |
111 | {\r | |
112 | double z;\r | |
113 | int32_t sign = (int)0x80000000;\r | |
114 | int32_t ix0,s0,q,m,t,i;\r | |
115 | u_int32_t r,t1,s1,ix1,q1;\r | |
116 | \r | |
117 | EXTRACT_WORDS(ix0,ix1,x);\r | |
118 | \r | |
119 | /* take care of Inf and NaN */\r | |
120 | if((ix0&0x7ff00000)==0x7ff00000) {\r | |
121 | return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf\r | |
122 | sqrt(-inf)=sNaN */\r | |
123 | }\r | |
124 | /* take care of zero */\r | |
125 | if(ix0<=0) {\r | |
126 | if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */\r | |
127 | else if(ix0<0) {\r | |
128 | errno = EDOM;\r | |
129 | return (x-x)/(x-x); /* sqrt(-ve) = sNaN */\r | |
130 | }\r | |
131 | }\r | |
132 | /* normalize x */\r | |
133 | m = (ix0>>20);\r | |
134 | if(m==0) { /* subnormal x */\r | |
135 | while(ix0==0) {\r | |
136 | m -= 21;\r | |
137 | ix0 |= (ix1>>11); ix1 <<= 21;\r | |
138 | }\r | |
139 | for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;\r | |
140 | m -= i-1;\r | |
141 | ix0 |= (ix1>>(32-i));\r | |
142 | ix1 <<= i;\r | |
143 | }\r | |
144 | m -= 1023; /* unbias exponent */\r | |
145 | ix0 = (ix0&0x000fffff)|0x00100000;\r | |
146 | if(m&1){ /* odd m, double x to make it even */\r | |
147 | ix0 += ix0 + ((ix1&sign)>>31);\r | |
148 | ix1 += ix1;\r | |
149 | }\r | |
150 | m >>= 1; /* m = [m/2] */\r | |
151 | \r | |
152 | /* generate sqrt(x) bit by bit */\r | |
153 | ix0 += ix0 + ((ix1&sign)>>31);\r | |
154 | ix1 += ix1;\r | |
155 | q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */\r | |
156 | r = 0x00200000; /* r = moving bit from right to left */\r | |
157 | \r | |
158 | while(r!=0) {\r | |
159 | t = s0+r;\r | |
160 | if(t<=ix0) {\r | |
161 | s0 = t+r;\r | |
162 | ix0 -= t;\r | |
163 | q += r;\r | |
164 | }\r | |
165 | ix0 += ix0 + ((ix1&sign)>>31);\r | |
166 | ix1 += ix1;\r | |
167 | r>>=1;\r | |
168 | }\r | |
169 | \r | |
170 | r = sign;\r | |
171 | while(r!=0) {\r | |
172 | t1 = s1+r;\r | |
173 | t = s0;\r | |
174 | if((t<ix0)||((t==ix0)&&(t1<=ix1))) {\r | |
175 | s1 = t1+r;\r | |
176 | if(((t1&sign)==(u_int32_t)sign)&&(s1&sign)==0) s0 += 1;\r | |
177 | ix0 -= t;\r | |
178 | if (ix1 < t1) ix0 -= 1;\r | |
179 | ix1 -= t1;\r | |
180 | q1 += r;\r | |
181 | }\r | |
182 | ix0 += ix0 + ((ix1&sign)>>31);\r | |
183 | ix1 += ix1;\r | |
184 | r>>=1;\r | |
185 | }\r | |
186 | \r | |
187 | /* use floating add to find out rounding direction */\r | |
188 | if((ix0|ix1)!=0) {\r | |
189 | z = one-tiny; /* trigger inexact flag */\r | |
190 | if (z>=one) {\r | |
191 | z = one+tiny;\r | |
192 | if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}\r | |
193 | else if (z>one) {\r | |
194 | if (q1==(u_int32_t)0xfffffffe) q+=1;\r | |
195 | q1+=2;\r | |
196 | } else\r | |
197 | q1 += (q1&1);\r | |
198 | }\r | |
199 | }\r | |
200 | ix0 = (q>>1)+0x3fe00000;\r | |
201 | ix1 = q1>>1;\r | |
202 | if ((q&1)==1) ix1 |= sign;\r | |
203 | ix0 += (m <<20);\r | |
204 | INSERT_WORDS(z,ix0,ix1);\r | |
205 | return z;\r | |
206 | }\r | |
207 | \r | |
208 | /*\r | |
209 | Other methods (use floating-point arithmetic)\r | |
210 | -------------\r | |
211 | (This is a copy of a drafted paper by Prof W. Kahan\r | |
212 | and K.C. Ng, written in May, 1986)\r | |
213 | \r | |
214 | Two algorithms are given here to implement sqrt(x)\r | |
215 | (IEEE double precision arithmetic) in software.\r | |
216 | Both supply sqrt(x) correctly rounded. The first algorithm (in\r | |
217 | Section A) uses newton iterations and involves four divisions.\r | |
218 | The second one uses reciproot iterations to avoid division, but\r | |
219 | requires more multiplications. Both algorithms need the ability\r | |
220 | to chop results of arithmetic operations instead of round them,\r | |
221 | and the INEXACT flag to indicate when an arithmetic operation\r | |
222 | is executed exactly with no roundoff error, all part of the\r | |
223 | standard (IEEE 754-1985). The ability to perform shift, add,\r | |
224 | subtract and logical AND operations upon 32-bit words is needed\r | |
225 | too, though not part of the standard.\r | |
226 | \r | |
227 | A. sqrt(x) by Newton Iteration\r | |
228 | \r | |
229 | (1) Initial approximation\r | |
230 | \r | |
231 | Let x0 and x1 be the leading and the trailing 32-bit words of\r | |
232 | a floating point number x (in IEEE double format) respectively\r | |
233 | \r | |
234 | 1 11 52 ...widths\r | |
235 | ------------------------------------------------------\r | |
236 | x: |s| e | f |\r | |
237 | ------------------------------------------------------\r | |
238 | msb lsb msb lsb ...order\r | |
239 | \r | |
240 | \r | |
241 | ------------------------ ------------------------\r | |
242 | x0: |s| e | f1 | x1: | f2 |\r | |
243 | ------------------------ ------------------------\r | |
244 | \r | |
245 | By performing shifts and subtracts on x0 and x1 (both regarded\r | |
246 | as integers), we obtain an 8-bit approximation of sqrt(x) as\r | |
247 | follows.\r | |
248 | \r | |
249 | k := (x0>>1) + 0x1ff80000;\r | |
250 | y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits\r | |
251 | Here k is a 32-bit integer and T1[] is an integer array containing\r | |
252 | correction terms. Now magically the floating value of y (y's\r | |
253 | leading 32-bit word is y0, the value of its trailing word is 0)\r | |
254 | approximates sqrt(x) to almost 8-bit.\r | |
255 | \r | |
256 | Value of T1:\r | |
257 | static int T1[32]= {\r | |
258 | 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,\r | |
259 | 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,\r | |
260 | 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,\r | |
261 | 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};\r | |
262 | \r | |
263 | (2) Iterative refinement\r | |
264 | \r | |
265 | Apply Heron's rule three times to y, we have y approximates\r | |
266 | sqrt(x) to within 1 ulp (Unit in the Last Place):\r | |
267 | \r | |
268 | y := (y+x/y)/2 ... almost 17 sig. bits\r | |
269 | y := (y+x/y)/2 ... almost 35 sig. bits\r | |
270 | y := y-(y-x/y)/2 ... within 1 ulp\r | |
271 | \r | |
272 | \r | |
273 | Remark 1.\r | |
274 | Another way to improve y to within 1 ulp is:\r | |
275 | \r | |
276 | y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)\r | |
277 | y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)\r | |
278 | \r | |
279 | 2\r | |
280 | (x-y )*y\r | |
281 | y := y + 2* ---------- ...within 1 ulp\r | |
282 | 2\r | |
283 | 3y + x\r | |
284 | \r | |
285 | \r | |
286 | This formula has one division fewer than the one above; however,\r | |
287 | it requires more multiplications and additions. Also x must be\r | |
288 | scaled in advance to avoid spurious overflow in evaluating the\r | |
289 | expression 3y*y+x. Hence it is not recommended uless division\r | |
290 | is slow. If division is very slow, then one should use the\r | |
291 | reciproot algorithm given in section B.\r | |
292 | \r | |
293 | (3) Final adjustment\r | |
294 | \r | |
295 | By twiddling y's last bit it is possible to force y to be\r | |
296 | correctly rounded according to the prevailing rounding mode\r | |
297 | as follows. Let r and i be copies of the rounding mode and\r | |
298 | inexact flag before entering the square root program. Also we\r | |
299 | use the expression y+-ulp for the next representable floating\r | |
300 | numbers (up and down) of y. Note that y+-ulp = either fixed\r | |
301 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped\r | |
302 | mode.\r | |
303 | \r | |
304 | I := FALSE; ... reset INEXACT flag I\r | |
305 | R := RZ; ... set rounding mode to round-toward-zero\r | |
306 | z := x/y; ... chopped quotient, possibly inexact\r | |
307 | If(not I) then { ... if the quotient is exact\r | |
308 | if(z=y) {\r | |
309 | I := i; ... restore inexact flag\r | |
310 | R := r; ... restore rounded mode\r | |
311 | return sqrt(x):=y.\r | |
312 | } else {\r | |
313 | z := z - ulp; ... special rounding\r | |
314 | }\r | |
315 | }\r | |
316 | i := TRUE; ... sqrt(x) is inexact\r | |
317 | If (r=RN) then z=z+ulp ... rounded-to-nearest\r | |
318 | If (r=RP) then { ... round-toward-+inf\r | |
319 | y = y+ulp; z=z+ulp;\r | |
320 | }\r | |
321 | y := y+z; ... chopped sum\r | |
322 | y0:=y0-0x00100000; ... y := y/2 is correctly rounded.\r | |
323 | I := i; ... restore inexact flag\r | |
324 | R := r; ... restore rounded mode\r | |
325 | return sqrt(x):=y.\r | |
326 | \r | |
327 | (4) Special cases\r | |
328 | \r | |
329 | Square root of +inf, +-0, or NaN is itself;\r | |
330 | Square root of a negative number is NaN with invalid signal.\r | |
331 | \r | |
332 | \r | |
333 | B. sqrt(x) by Reciproot Iteration\r | |
334 | \r | |
335 | (1) Initial approximation\r | |
336 | \r | |
337 | Let x0 and x1 be the leading and the trailing 32-bit words of\r | |
338 | a floating point number x (in IEEE double format) respectively\r | |
339 | (see section A). By performing shifs and subtracts on x0 and y0,\r | |
340 | we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.\r | |
341 | \r | |
342 | k := 0x5fe80000 - (x0>>1);\r | |
343 | y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits\r | |
344 | \r | |
345 | Here k is a 32-bit integer and T2[] is an integer array\r | |
346 | containing correction terms. Now magically the floating\r | |
347 | value of y (y's leading 32-bit word is y0, the value of\r | |
348 | its trailing word y1 is set to zero) approximates 1/sqrt(x)\r | |
349 | to almost 7.8-bit.\r | |
350 | \r | |
351 | Value of T2:\r | |
352 | static int T2[64]= {\r | |
353 | 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,\r | |
354 | 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,\r | |
355 | 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,\r | |
356 | 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,\r | |
357 | 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,\r | |
358 | 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,\r | |
359 | 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,\r | |
360 | 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};\r | |
361 | \r | |
362 | (2) Iterative refinement\r | |
363 | \r | |
364 | Apply Reciproot iteration three times to y and multiply the\r | |
365 | result by x to get an approximation z that matches sqrt(x)\r | |
366 | to about 1 ulp. To be exact, we will have\r | |
367 | -1ulp < sqrt(x)-z<1.0625ulp.\r | |
368 | \r | |
369 | ... set rounding mode to Round-to-nearest\r | |
370 | y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)\r | |
371 | y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)\r | |
372 | ... special arrangement for better accuracy\r | |
373 | z := x*y ... 29 bits to sqrt(x), with z*y<1\r | |
374 | z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)\r | |
375 | \r | |
376 | Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that\r | |
377 | (a) the term z*y in the final iteration is always less than 1;\r | |
378 | (b) the error in the final result is biased upward so that\r | |
379 | -1 ulp < sqrt(x) - z < 1.0625 ulp\r | |
380 | instead of |sqrt(x)-z|<1.03125ulp.\r | |
381 | \r | |
382 | (3) Final adjustment\r | |
383 | \r | |
384 | By twiddling y's last bit it is possible to force y to be\r | |
385 | correctly rounded according to the prevailing rounding mode\r | |
386 | as follows. Let r and i be copies of the rounding mode and\r | |
387 | inexact flag before entering the square root program. Also we\r | |
388 | use the expression y+-ulp for the next representable floating\r | |
389 | numbers (up and down) of y. Note that y+-ulp = either fixed\r | |
390 | point y+-1, or multiply y by nextafter(1,+-inf) in chopped\r | |
391 | mode.\r | |
392 | \r | |
393 | R := RZ; ... set rounding mode to round-toward-zero\r | |
394 | switch(r) {\r | |
395 | case RN: ... round-to-nearest\r | |
396 | if(x<= z*(z-ulp)...chopped) z = z - ulp; else\r | |
397 | if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;\r | |
398 | break;\r | |
399 | case RZ:case RM: ... round-to-zero or round-to--inf\r | |
400 | R:=RP; ... reset rounding mod to round-to-+inf\r | |
401 | if(x<z*z ... rounded up) z = z - ulp; else\r | |
402 | if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;\r | |
403 | break;\r | |
404 | case RP: ... round-to-+inf\r | |
405 | if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else\r | |
406 | if(x>z*z ...chopped) z = z+ulp;\r | |
407 | break;\r | |
408 | }\r | |
409 | \r | |
410 | Remark 3. The above comparisons can be done in fixed point. For\r | |
411 | example, to compare x and w=z*z chopped, it suffices to compare\r | |
412 | x1 and w1 (the trailing parts of x and w), regarding them as\r | |
413 | two's complement integers.\r | |
414 | \r | |
415 | ...Is z an exact square root?\r | |
416 | To determine whether z is an exact square root of x, let z1 be the\r | |
417 | trailing part of z, and also let x0 and x1 be the leading and\r | |
418 | trailing parts of x.\r | |
419 | \r | |
420 | If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0\r | |
421 | I := 1; ... Raise Inexact flag: z is not exact\r | |
422 | else {\r | |
423 | j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2\r | |
424 | k := z1 >> 26; ... get z's 25-th and 26-th\r | |
425 | fraction bits\r | |
426 | I := i or (k&j) or ((k&(j+j+1))!=(x1&3));\r | |
427 | }\r | |
428 | R:= r ... restore rounded mode\r | |
429 | return sqrt(x):=z.\r | |
430 | \r | |
431 | If multiplication is cheaper than the foregoing red tape, the\r | |
432 | Inexact flag can be evaluated by\r | |
433 | \r | |
434 | I := i;\r | |
435 | I := (z*z!=x) or I.\r | |
436 | \r | |
437 | Note that z*z can overwrite I; this value must be sensed if it is\r | |
438 | True.\r | |
439 | \r | |
440 | Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be\r | |
441 | zero.\r | |
442 | \r | |
443 | --------------------\r | |
444 | z1: | f2 |\r | |
445 | --------------------\r | |
446 | bit 31 bit 0\r | |
447 | \r | |
448 | Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd\r | |
449 | or even of logb(x) have the following relations:\r | |
450 | \r | |
451 | -------------------------------------------------\r | |
452 | bit 27,26 of z1 bit 1,0 of x1 logb(x)\r | |
453 | -------------------------------------------------\r | |
454 | 00 00 odd and even\r | |
455 | 01 01 even\r | |
456 | 10 10 odd\r | |
457 | 10 00 even\r | |
458 | 11 01 even\r | |
459 | -------------------------------------------------\r | |
460 | \r | |
461 | (4) Special cases (see (4) of Section A).\r | |
462 | \r | |
463 | */\r | |
464 | \r |