e_sqrt.c 14 KB

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  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. #if defined(LIBM_SCCS) && !defined(lint)
  13. static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $";
  14. #endif
  15. /* __ieee754_sqrt(x)
  16. * Return correctly rounded sqrt.
  17. * ------------------------------------------
  18. * | Use the hardware sqrt if you have one |
  19. * ------------------------------------------
  20. * Method:
  21. * Bit by bit method using integer arithmetic. (Slow, but portable)
  22. * 1. Normalization
  23. * Scale x to y in [1,4) with even powers of 2:
  24. * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
  25. * sqrt(x) = 2^k * sqrt(y)
  26. * 2. Bit by bit computation
  27. * Let q = sqrt(y) truncated to i bit after binary point (q = 1),
  28. * i 0
  29. * i+1 2
  30. * s = 2*q , and y = 2 * ( y - q ). (1)
  31. * i i i i
  32. *
  33. * To compute q from q , one checks whether
  34. * i+1 i
  35. *
  36. * -(i+1) 2
  37. * (q + 2 ) <= y. (2)
  38. * i
  39. * -(i+1)
  40. * If (2) is false, then q = q ; otherwise q = q + 2 .
  41. * i+1 i i+1 i
  42. *
  43. * With some algebric manipulation, it is not difficult to see
  44. * that (2) is equivalent to
  45. * -(i+1)
  46. * s + 2 <= y (3)
  47. * i i
  48. *
  49. * The advantage of (3) is that s and y can be computed by
  50. * i i
  51. * the following recurrence formula:
  52. * if (3) is false
  53. *
  54. * s = s , y = y ; (4)
  55. * i+1 i i+1 i
  56. *
  57. * otherwise,
  58. * -i -(i+1)
  59. * s = s + 2 , y = y - s - 2 (5)
  60. * i+1 i i+1 i i
  61. *
  62. * One may easily use induction to prove (4) and (5).
  63. * Note. Since the left hand side of (3) contain only i+2 bits,
  64. * it does not necessary to do a full (53-bit) comparison
  65. * in (3).
  66. * 3. Final rounding
  67. * After generating the 53 bits result, we compute one more bit.
  68. * Together with the remainder, we can decide whether the
  69. * result is exact, bigger than 1/2ulp, or less than 1/2ulp
  70. * (it will never equal to 1/2ulp).
  71. * The rounding mode can be detected by checking whether
  72. * huge + tiny is equal to huge, and whether huge - tiny is
  73. * equal to huge for some floating point number "huge" and "tiny".
  74. *
  75. * Special cases:
  76. * sqrt(+-0) = +-0 ... exact
  77. * sqrt(inf) = inf
  78. * sqrt(-ve) = NaN ... with invalid signal
  79. * sqrt(NaN) = NaN ... with invalid signal for signaling NaN
  80. *
  81. * Other methods : see the appended file at the end of the program below.
  82. *---------------
  83. */
  84. #include "math.h"
  85. #include "math_private.h"
  86. #ifdef __STDC__
  87. static const double one = 1.0, tiny=1.0e-300;
  88. #else
  89. static double one = 1.0, tiny=1.0e-300;
  90. #endif
  91. #ifdef __STDC__
  92. double __ieee754_sqrt(double x)
  93. #else
  94. double __ieee754_sqrt(x)
  95. double x;
  96. #endif
  97. {
  98. double z;
  99. int32_t sign = (int)0x80000000;
  100. int32_t ix0,s0,q,m,t,i;
  101. u_int32_t r,t1,s1,ix1,q1;
  102. EXTRACT_WORDS(ix0,ix1,x);
  103. /* take care of Inf and NaN */
  104. if((ix0&0x7ff00000)==0x7ff00000) {
  105. return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
  106. sqrt(-inf)=sNaN */
  107. }
  108. /* take care of zero */
  109. if(ix0<=0) {
  110. if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
  111. else if(ix0<0)
  112. return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
  113. }
  114. /* normalize x */
  115. m = (ix0>>20);
  116. if(m==0) { /* subnormal x */
  117. while(ix0==0) {
  118. m -= 21;
  119. ix0 |= (ix1>>11); ix1 <<= 21;
  120. }
  121. for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
  122. m -= i-1;
  123. ix0 |= (ix1>>(32-i));
  124. ix1 <<= i;
  125. }
  126. m -= 1023; /* unbias exponent */
  127. ix0 = (ix0&0x000fffff)|0x00100000;
  128. if(m&1){ /* odd m, double x to make it even */
  129. ix0 += ix0 + ((ix1&sign)>>31);
  130. ix1 += ix1;
  131. }
  132. m >>= 1; /* m = [m/2] */
  133. /* generate sqrt(x) bit by bit */
  134. ix0 += ix0 + ((ix1&sign)>>31);
  135. ix1 += ix1;
  136. q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
  137. r = 0x00200000; /* r = moving bit from right to left */
  138. while(r!=0) {
  139. t = s0+r;
  140. if(t<=ix0) {
  141. s0 = t+r;
  142. ix0 -= t;
  143. q += r;
  144. }
  145. ix0 += ix0 + ((ix1&sign)>>31);
  146. ix1 += ix1;
  147. r>>=1;
  148. }
  149. r = sign;
  150. while(r!=0) {
  151. t1 = s1+r;
  152. t = s0;
  153. if((t<ix0)||((t==ix0)&&(t1<=ix1))) {
  154. s1 = t1+r;
  155. if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
  156. ix0 -= t;
  157. if (ix1 < t1) ix0 -= 1;
  158. ix1 -= t1;
  159. q1 += r;
  160. }
  161. ix0 += ix0 + ((ix1&sign)>>31);
  162. ix1 += ix1;
  163. r>>=1;
  164. }
  165. /* use floating add to find out rounding direction */
  166. if((ix0|ix1)!=0) {
  167. z = one-tiny; /* trigger inexact flag */
  168. if (z>=one) {
  169. z = one+tiny;
  170. if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;}
  171. else if (z>one) {
  172. if (q1==(u_int32_t)0xfffffffe) q+=1;
  173. q1+=2;
  174. } else
  175. q1 += (q1&1);
  176. }
  177. }
  178. ix0 = (q>>1)+0x3fe00000;
  179. ix1 = q1>>1;
  180. if ((q&1)==1) ix1 |= sign;
  181. ix0 += (m <<20);
  182. INSERT_WORDS(z,ix0,ix1);
  183. return z;
  184. }
  185. /*
  186. Other methods (use floating-point arithmetic)
  187. -------------
  188. (This is a copy of a drafted paper by Prof W. Kahan
  189. and K.C. Ng, written in May, 1986)
  190. Two algorithms are given here to implement sqrt(x)
  191. (IEEE double precision arithmetic) in software.
  192. Both supply sqrt(x) correctly rounded. The first algorithm (in
  193. Section A) uses newton iterations and involves four divisions.
  194. The second one uses reciproot iterations to avoid division, but
  195. requires more multiplications. Both algorithms need the ability
  196. to chop results of arithmetic operations instead of round them,
  197. and the INEXACT flag to indicate when an arithmetic operation
  198. is executed exactly with no roundoff error, all part of the
  199. standard (IEEE 754-1985). The ability to perform shift, add,
  200. subtract and logical AND operations upon 32-bit words is needed
  201. too, though not part of the standard.
  202. A. sqrt(x) by Newton Iteration
  203. (1) Initial approximation
  204. Let x0 and x1 be the leading and the trailing 32-bit words of
  205. a floating point number x (in IEEE double format) respectively
  206. 1 11 52 ...widths
  207. ------------------------------------------------------
  208. x: |s| e | f |
  209. ------------------------------------------------------
  210. msb lsb msb lsb ...order
  211. ------------------------ ------------------------
  212. x0: |s| e | f1 | x1: | f2 |
  213. ------------------------ ------------------------
  214. By performing shifts and subtracts on x0 and x1 (both regarded
  215. as integers), we obtain an 8-bit approximation of sqrt(x) as
  216. follows.
  217. k := (x0>>1) + 0x1ff80000;
  218. y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
  219. Here k is a 32-bit integer and T1[] is an integer array containing
  220. correction terms. Now magically the floating value of y (y's
  221. leading 32-bit word is y0, the value of its trailing word is 0)
  222. approximates sqrt(x) to almost 8-bit.
  223. Value of T1:
  224. static int T1[32]= {
  225. 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
  226. 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
  227. 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
  228. 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
  229. (2) Iterative refinement
  230. Apply Heron's rule three times to y, we have y approximates
  231. sqrt(x) to within 1 ulp (Unit in the Last Place):
  232. y := (y+x/y)/2 ... almost 17 sig. bits
  233. y := (y+x/y)/2 ... almost 35 sig. bits
  234. y := y-(y-x/y)/2 ... within 1 ulp
  235. Remark 1.
  236. Another way to improve y to within 1 ulp is:
  237. y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
  238. y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
  239. 2
  240. (x-y )*y
  241. y := y + 2* ---------- ...within 1 ulp
  242. 2
  243. 3y + x
  244. This formula has one division fewer than the one above; however,
  245. it requires more multiplications and additions. Also x must be
  246. scaled in advance to avoid spurious overflow in evaluating the
  247. expression 3y*y+x. Hence it is not recommended uless division
  248. is slow. If division is very slow, then one should use the
  249. reciproot algorithm given in section B.
  250. (3) Final adjustment
  251. By twiddling y's last bit it is possible to force y to be
  252. correctly rounded according to the prevailing rounding mode
  253. as follows. Let r and i be copies of the rounding mode and
  254. inexact flag before entering the square root program. Also we
  255. use the expression y+-ulp for the next representable floating
  256. numbers (up and down) of y. Note that y+-ulp = either fixed
  257. point y+-1, or multiply y by nextafter(1,+-inf) in chopped
  258. mode.
  259. I := FALSE; ... reset INEXACT flag I
  260. R := RZ; ... set rounding mode to round-toward-zero
  261. z := x/y; ... chopped quotient, possibly inexact
  262. If(not I) then { ... if the quotient is exact
  263. if(z=y) {
  264. I := i; ... restore inexact flag
  265. R := r; ... restore rounded mode
  266. return sqrt(x):=y.
  267. } else {
  268. z := z - ulp; ... special rounding
  269. }
  270. }
  271. i := TRUE; ... sqrt(x) is inexact
  272. If (r=RN) then z=z+ulp ... rounded-to-nearest
  273. If (r=RP) then { ... round-toward-+inf
  274. y = y+ulp; z=z+ulp;
  275. }
  276. y := y+z; ... chopped sum
  277. y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
  278. I := i; ... restore inexact flag
  279. R := r; ... restore rounded mode
  280. return sqrt(x):=y.
  281. (4) Special cases
  282. Square root of +inf, +-0, or NaN is itself;
  283. Square root of a negative number is NaN with invalid signal.
  284. B. sqrt(x) by Reciproot Iteration
  285. (1) Initial approximation
  286. Let x0 and x1 be the leading and the trailing 32-bit words of
  287. a floating point number x (in IEEE double format) respectively
  288. (see section A). By performing shifs and subtracts on x0 and y0,
  289. we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
  290. k := 0x5fe80000 - (x0>>1);
  291. y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
  292. Here k is a 32-bit integer and T2[] is an integer array
  293. containing correction terms. Now magically the floating
  294. value of y (y's leading 32-bit word is y0, the value of
  295. its trailing word y1 is set to zero) approximates 1/sqrt(x)
  296. to almost 7.8-bit.
  297. Value of T2:
  298. static int T2[64]= {
  299. 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
  300. 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
  301. 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
  302. 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
  303. 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
  304. 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
  305. 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
  306. 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
  307. (2) Iterative refinement
  308. Apply Reciproot iteration three times to y and multiply the
  309. result by x to get an approximation z that matches sqrt(x)
  310. to about 1 ulp. To be exact, we will have
  311. -1ulp < sqrt(x)-z<1.0625ulp.
  312. ... set rounding mode to Round-to-nearest
  313. y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
  314. y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
  315. ... special arrangement for better accuracy
  316. z := x*y ... 29 bits to sqrt(x), with z*y<1
  317. z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
  318. Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
  319. (a) the term z*y in the final iteration is always less than 1;
  320. (b) the error in the final result is biased upward so that
  321. -1 ulp < sqrt(x) - z < 1.0625 ulp
  322. instead of |sqrt(x)-z|<1.03125ulp.
  323. (3) Final adjustment
  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. R := RZ; ... set rounding mode to round-toward-zero
  333. switch(r) {
  334. case RN: ... round-to-nearest
  335. if(x<= z*(z-ulp)...chopped) z = z - ulp; else
  336. if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
  337. break;
  338. case RZ:case RM: ... round-to-zero or round-to--inf
  339. R:=RP; ... reset rounding mod to round-to-+inf
  340. if(x<z*z ... rounded up) z = z - ulp; else
  341. if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
  342. break;
  343. case RP: ... round-to-+inf
  344. if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
  345. if(x>z*z ...chopped) z = z+ulp;
  346. break;
  347. }
  348. Remark 3. The above comparisons can be done in fixed point. For
  349. example, to compare x and w=z*z chopped, it suffices to compare
  350. x1 and w1 (the trailing parts of x and w), regarding them as
  351. two's complement integers.
  352. ...Is z an exact square root?
  353. To determine whether z is an exact square root of x, let z1 be the
  354. trailing part of z, and also let x0 and x1 be the leading and
  355. trailing parts of x.
  356. If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
  357. I := 1; ... Raise Inexact flag: z is not exact
  358. else {
  359. j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
  360. k := z1 >> 26; ... get z's 25-th and 26-th
  361. fraction bits
  362. I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
  363. }
  364. R:= r ... restore rounded mode
  365. return sqrt(x):=z.
  366. If multiplication is cheaper then the foregoing red tape, the
  367. Inexact flag can be evaluated by
  368. I := i;
  369. I := (z*z!=x) or I.
  370. Note that z*z can overwrite I; this value must be sensed if it is
  371. True.
  372. Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
  373. zero.
  374. --------------------
  375. z1: | f2 |
  376. --------------------
  377. bit 31 bit 0
  378. Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
  379. or even of logb(x) have the following relations:
  380. -------------------------------------------------
  381. bit 27,26 of z1 bit 1,0 of x1 logb(x)
  382. -------------------------------------------------
  383. 00 00 odd and even
  384. 01 01 even
  385. 10 10 odd
  386. 10 00 even
  387. 11 01 even
  388. -------------------------------------------------
  389. (4) Special cases (see (4) of Section A).
  390. */