e_sqrt.c 14 KB

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  1. /*
  2. * ====================================================
  3. * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
  4. *
  5. * Developed at SunPro, a Sun Microsystems, Inc. business.
  6. * Permission to use, copy, modify, and distribute this
  7. * software is freely granted, provided that this notice
  8. * is preserved.
  9. * ====================================================
  10. */
  11. /* __ieee754_sqrt(x)
  12. * Return correctly rounded sqrt.
  13. * ------------------------------------------
  14. * | Use the hardware sqrt if you have one |
  15. * ------------------------------------------
  16. * Method:
  17. * Bit by bit method using integer arithmetic. (Slow, but portable)
  18. * 1. Normalization
  19. * Scale x to y in [1,4) with even powers of 2:
  20. * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
  21. * sqrt(x) = 2^k * sqrt(y)
  22. * 2. Bit by bit computation
  23. * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
  24. * i 0
  25. * i+1 2
  26. * s = 2*q , and y = 2 * ( y - q ). (1)
  27. * i i i i
  28. *
  29. * To compute q from q , one checks whether
  30. * i+1 i
  31. *
  32. * -(i+1) 2
  33. * (q + 2 ) <= y. (2)
  34. * i
  35. * -(i+1)
  36. * If (2) is false, then q = q ; otherwise q = q + 2 .
  37. * i+1 i i+1 i
  38. *
  39. * With some algebric manipulation, it is not difficult to see
  40. * that (2) is equivalent to
  41. * -(i+1)
  42. * s + 2 <= y (3)
  43. * i i
  44. *
  45. * The advantage of (3) is that s and y can be computed by
  46. * i i
  47. * the following recurrence formula:
  48. * if (3) is false
  49. *
  50. * s = s , y = y ; (4)
  51. * i+1 i i+1 i
  52. *
  53. * otherwise,
  54. * -i -(i+1)
  55. * s = s + 2 , y = y - s - 2 (5)
  56. * i+1 i i+1 i i
  57. *
  58. * One may easily use induction to prove (4) and (5).
  59. * Note. Since the left hand side of (3) contain only i+2 bits,
  60. * it does not necessary to do a full (53-bit) comparison
  61. * in (3).
  62. * 3. Final rounding
  63. * After generating the 53 bits result, we compute one more bit.
  64. * Together with the remainder, we can decide whether the
  65. * result is exact, bigger than 1/2ulp, or less than 1/2ulp
  66. * (it will never equal to 1/2ulp).
  67. * The rounding mode can be detected by checking whether
  68. * huge + tiny is equal to huge, and whether huge - tiny is
  69. * equal to huge for some floating point number "huge" and "tiny".
  70. *
  71. * Special cases:
  72. * sqrt(+-0) = +-0 ... exact
  73. * sqrt(inf) = inf
  74. * sqrt(-ve) = NaN ... with invalid signal
  75. * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
  76. *
  77. * Other methods : see the appended file at the end of the program below.
  78. *---------------
  79. */
  80. #include "math.h"
  81. #include "math_private.h"
  82. static const double one = 1.0, tiny = 1.0e-300;
  83. double __ieee754_sqrt(double x)
  84. {
  85. double z;
  86. int32_t sign = (int)0x80000000;
  87. int32_t ix0,s0,q,m,t,i;
  88. u_int32_t r,t1,s1,ix1,q1;
  89. EXTRACT_WORDS(ix0,ix1,x);
  90. /* take care of Inf and NaN */
  91. if((ix0&0x7ff00000)==0x7ff00000) {
  92. return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
  93. sqrt(-inf)=sNaN */
  94. }
  95. /* take care of zero */
  96. if(ix0<=0) {
  97. if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
  98. else if(ix0<0)
  99. return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
  100. }
  101. /* normalize x */
  102. m = (ix0>>20);
  103. if(m==0) { /* subnormal x */
  104. while(ix0==0) {
  105. m -= 21;
  106. ix0 |= (ix1>>11); ix1 <<= 21;
  107. }
  108. for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
  109. m -= i-1;
  110. ix0 |= (ix1>>(32-i));
  111. ix1 <<= i;
  112. }
  113. m -= 1023; /* unbias exponent */
  114. ix0 = (ix0&0x000fffff)|0x00100000;
  115. if(m&1){ /* odd m, double x to make it even */
  116. ix0 += ix0 + ((ix1&sign)>>31);
  117. ix1 += ix1;
  118. }
  119. m >>= 1; /* m = [m/2] */
  120. /* generate sqrt(x) bit by bit */
  121. ix0 += ix0 + ((ix1&sign)>>31);
  122. ix1 += ix1;
  123. q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
  124. r = 0x00200000; /* r = moving bit from right to left */
  125. while(r!=0) {
  126. t = s0+r;
  127. if(t<=ix0) {
  128. s0 = t+r;
  129. ix0 -= t;
  130. q += r;
  131. }
  132. ix0 += ix0 + ((ix1&sign)>>31);
  133. ix1 += ix1;
  134. r>>=1;
  135. }
  136. r = sign;
  137. while(r!=0) {
  138. t1 = s1+r;
  139. t = s0;
  140. if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
  141. s1 = t1+r;
  142. if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
  143. ix0 -= t;
  144. if (ix1 < t1) ix0 -= 1;
  145. ix1 -= t1;
  146. q1 += r;
  147. }
  148. ix0 += ix0 + ((ix1&sign)>>31);
  149. ix1 += ix1;
  150. r>>=1;
  151. }
  152. /* use floating add to find out rounding direction */
  153. if((ix0|ix1)!=0) {
  154. z = one-tiny; /* trigger inexact flag */
  155. if (z>=one) {
  156. z = one+tiny;
  157. if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
  158. else if (z>one) {
  159. if (q1==(u_int32_t)0xfffffffe) q+=1;
  160. q1+=2;
  161. } else
  162. q1 += (q1&1);
  163. }
  164. }
  165. ix0 = (q>>1)+0x3fe00000;
  166. ix1 = q1>>1;
  167. if ((q&1)==1) ix1 |= sign;
  168. ix0 += (m <<20);
  169. INSERT_WORDS(z,ix0,ix1);
  170. return z;
  171. }
  172. /*
  173. * wrapper sqrt(x)
  174. */
  175. #ifndef _IEEE_LIBM
  176. double sqrt(double x)
  177. {
  178. double z = __ieee754_sqrt(x);
  179. if (_LIB_VERSION == _IEEE_ || isnan(x))
  180. return z;
  181. if (x < 0.0)
  182. return __kernel_standard(x, x, 26); /* sqrt(negative) */
  183. return z;
  184. }
  185. #else
  186. strong_alias(__ieee754_sqrt, sqrt)
  187. #endif
  188. libm_hidden_def(sqrt)
  189. /*
  190. Other methods (use floating-point arithmetic)
  191. -------------
  192. (This is a copy of a drafted paper by Prof W. Kahan
  193. and K.C. Ng, written in May, 1986)
  194. Two algorithms are given here to implement sqrt(x)
  195. (IEEE double precision arithmetic) in software.
  196. Both supply sqrt(x) correctly rounded. The first algorithm (in
  197. Section A) uses newton iterations and involves four divisions.
  198. The second one uses reciproot iterations to avoid division, but
  199. requires more multiplications. Both algorithms need the ability
  200. to chop results of arithmetic operations instead of round them,
  201. and the INEXACT flag to indicate when an arithmetic operation
  202. is executed exactly with no roundoff error, all part of the
  203. standard (IEEE 754-1985). The ability to perform shift, add,
  204. subtract and logical AND operations upon 32-bit words is needed
  205. too, though not part of the standard.
  206. A. sqrt(x) by Newton Iteration
  207. (1) Initial approximation
  208. Let x0 and x1 be the leading and the trailing 32-bit words of
  209. a floating point number x (in IEEE double format) respectively
  210. 1 11 52 ...widths
  211. ------------------------------------------------------
  212. x: |s| e | f |
  213. ------------------------------------------------------
  214. msb lsb msb lsb ...order
  215. ------------------------ ------------------------
  216. x0: |s| e | f1 | x1: | f2 |
  217. ------------------------ ------------------------
  218. By performing shifts and subtracts on x0 and x1 (both regarded
  219. as integers), we obtain an 8-bit approximation of sqrt(x) as
  220. follows.
  221. k := (x0>>1) + 0x1ff80000;
  222. y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
  223. Here k is a 32-bit integer and T1[] is an integer array containing
  224. correction terms. Now magically the floating value of y (y's
  225. leading 32-bit word is y0, the value of its trailing word is 0)
  226. approximates sqrt(x) to almost 8-bit.
  227. Value of T1:
  228. static int T1[32]= {
  229. 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
  230. 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
  231. 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
  232. 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
  233. (2) Iterative refinement
  234. Apply Heron's rule three times to y, we have y approximates
  235. sqrt(x) to within 1 ulp (Unit in the Last Place):
  236. y := (y+x/y)/2 ... almost 17 sig. bits
  237. y := (y+x/y)/2 ... almost 35 sig. bits
  238. y := y-(y-x/y)/2 ... within 1 ulp
  239. Remark 1.
  240. Another way to improve y to within 1 ulp is:
  241. y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
  242. y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
  243. 2
  244. (x-y )*y
  245. y := y + 2* ---------- ...within 1 ulp
  246. 2
  247. 3y + x
  248. This formula has one division fewer than the one above; however,
  249. it requires more multiplications and additions. Also x must be
  250. scaled in advance to avoid spurious overflow in evaluating the
  251. expression 3y*y+x. Hence it is not recommended uless division
  252. is slow. If division is very slow, then one should use the
  253. reciproot algorithm given in section B.
  254. (3) Final adjustment
  255. By twiddling y's last bit it is possible to force y to be
  256. correctly rounded according to the prevailing rounding mode
  257. as follows. Let r and i be copies of the rounding mode and
  258. inexact flag before entering the square root program. Also we
  259. use the expression y+-ulp for the next representable floating
  260. numbers (up and down) of y. Note that y+-ulp = either fixed
  261. point y+-1, or multiply y by nextafter(1,+-inf) in chopped
  262. mode.
  263. I := FALSE; ... reset INEXACT flag I
  264. R := RZ; ... set rounding mode to round-toward-zero
  265. z := x/y; ... chopped quotient, possibly inexact
  266. If(not I) then { ... if the quotient is exact
  267. if(z=y) {
  268. I := i; ... restore inexact flag
  269. R := r; ... restore rounded mode
  270. return sqrt(x):=y.
  271. } else {
  272. z := z - ulp; ... special rounding
  273. }
  274. }
  275. i := TRUE; ... sqrt(x) is inexact
  276. If (r=RN) then z=z+ulp ... rounded-to-nearest
  277. If (r=RP) then { ... round-toward-+inf
  278. y = y+ulp; z=z+ulp;
  279. }
  280. y := y+z; ... chopped sum
  281. y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
  282. I := i; ... restore inexact flag
  283. R := r; ... restore rounded mode
  284. return sqrt(x):=y.
  285. (4) Special cases
  286. Square root of +inf, +-0, or NaN is itself;
  287. Square root of a negative number is NaN with invalid signal.
  288. B. sqrt(x) by Reciproot Iteration
  289. (1) Initial approximation
  290. Let x0 and x1 be the leading and the trailing 32-bit words of
  291. a floating point number x (in IEEE double format) respectively
  292. (see section A). By performing shifs and subtracts on x0 and y0,
  293. we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
  294. k := 0x5fe80000 - (x0>>1);
  295. y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
  296. Here k is a 32-bit integer and T2[] is an integer array
  297. containing correction terms. Now magically the floating
  298. value of y (y's leading 32-bit word is y0, the value of
  299. its trailing word y1 is set to zero) approximates 1/sqrt(x)
  300. to almost 7.8-bit.
  301. Value of T2:
  302. static int T2[64]= {
  303. 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
  304. 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
  305. 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
  306. 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
  307. 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
  308. 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
  309. 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
  310. 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
  311. (2) Iterative refinement
  312. Apply Reciproot iteration three times to y and multiply the
  313. result by x to get an approximation z that matches sqrt(x)
  314. to about 1 ulp. To be exact, we will have
  315. -1ulp < sqrt(x)-z<1.0625ulp.
  316. ... set rounding mode to Round-to-nearest
  317. y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
  318. y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
  319. ... special arrangement for better accuracy
  320. z := x*y ... 29 bits to sqrt(x), with z*y<1
  321. z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
  322. Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
  323. (a) the term z*y in the final iteration is always less than 1;
  324. (b) the error in the final result is biased upward so that
  325. -1 ulp < sqrt(x) - z < 1.0625 ulp
  326. instead of |sqrt(x)-z|<1.03125ulp.
  327. (3) Final adjustment
  328. By twiddling y's last bit it is possible to force y to be
  329. correctly rounded according to the prevailing rounding mode
  330. as follows. Let r and i be copies of the rounding mode and
  331. inexact flag before entering the square root program. Also we
  332. use the expression y+-ulp for the next representable floating
  333. numbers (up and down) of y. Note that y+-ulp = either fixed
  334. point y+-1, or multiply y by nextafter(1,+-inf) in chopped
  335. mode.
  336. R := RZ; ... set rounding mode to round-toward-zero
  337. switch(r) {
  338. case RN: ... round-to-nearest
  339. if(x<= z*(z-ulp)...chopped) z = z - ulp; else
  340. if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
  341. break;
  342. case RZ:case RM: ... round-to-zero or round-to--inf
  343. R:=RP; ... reset rounding mod to round-to-+inf
  344. if(x<z*z ... rounded up) z = z - ulp; else
  345. if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
  346. break;
  347. case RP: ... round-to-+inf
  348. if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
  349. if(x>z*z ...chopped) z = z+ulp;
  350. break;
  351. }
  352. Remark 3. The above comparisons can be done in fixed point. For
  353. example, to compare x and w=z*z chopped, it suffices to compare
  354. x1 and w1 (the trailing parts of x and w), regarding them as
  355. two's complement integers.
  356. ...Is z an exact square root?
  357. To determine whether z is an exact square root of x, let z1 be the
  358. trailing part of z, and also let x0 and x1 be the leading and
  359. trailing parts of x.
  360. If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
  361. I := 1; ... Raise Inexact flag: z is not exact
  362. else {
  363. j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
  364. k := z1 >> 26; ... get z's 25-th and 26-th
  365. fraction bits
  366. I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
  367. }
  368. R:= r ... restore rounded mode
  369. return sqrt(x):=z.
  370. If multiplication is cheaper then the foregoing red tape, the
  371. Inexact flag can be evaluated by
  372. I := i;
  373. I := (z*z!=x) or I.
  374. Note that z*z can overwrite I; this value must be sensed if it is
  375. True.
  376. Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
  377. zero.
  378. --------------------
  379. z1: | f2 |
  380. --------------------
  381. bit 31 bit 0
  382. Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
  383. or even of logb(x) have the following relations:
  384. -------------------------------------------------
  385. bit 27,26 of z1 bit 1,0 of x1 logb(x)
  386. -------------------------------------------------
  387. 00 00 odd and even
  388. 01 01 even
  389. 10 10 odd
  390. 10 00 even
  391. 11 01 even
  392. -------------------------------------------------
  393. (4) Special cases (see (4) of Section A).
  394. */