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 attribute_hidden __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. Other methods (use floating-point arithmetic)
  174. -------------
  175. (This is a copy of a drafted paper by Prof W. Kahan
  176. and K.C. Ng, written in May, 1986)
  177. Two algorithms are given here to implement sqrt(x)
  178. (IEEE double precision arithmetic) in software.
  179. Both supply sqrt(x) correctly rounded. The first algorithm (in
  180. Section A) uses newton iterations and involves four divisions.
  181. The second one uses reciproot iterations to avoid division, but
  182. requires more multiplications. Both algorithms need the ability
  183. to chop results of arithmetic operations instead of round them,
  184. and the INEXACT flag to indicate when an arithmetic operation
  185. is executed exactly with no roundoff error, all part of the
  186. standard (IEEE 754-1985). The ability to perform shift, add,
  187. subtract and logical AND operations upon 32-bit words is needed
  188. too, though not part of the standard.
  189. A. sqrt(x) by Newton Iteration
  190. (1) Initial approximation
  191. Let x0 and x1 be the leading and the trailing 32-bit words of
  192. a floating point number x (in IEEE double format) respectively
  193. 1 11 52 ...widths
  194. ------------------------------------------------------
  195. x: |s| e | f |
  196. ------------------------------------------------------
  197. msb lsb msb lsb ...order
  198. ------------------------ ------------------------
  199. x0: |s| e | f1 | x1: | f2 |
  200. ------------------------ ------------------------
  201. By performing shifts and subtracts on x0 and x1 (both regarded
  202. as integers), we obtain an 8-bit approximation of sqrt(x) as
  203. follows.
  204. k := (x0>>1) + 0x1ff80000;
  205. y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
  206. Here k is a 32-bit integer and T1[] is an integer array containing
  207. correction terms. Now magically the floating value of y (y's
  208. leading 32-bit word is y0, the value of its trailing word is 0)
  209. approximates sqrt(x) to almost 8-bit.
  210. Value of T1:
  211. static int T1[32]= {
  212. 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
  213. 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
  214. 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
  215. 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
  216. (2) Iterative refinement
  217. Apply Heron's rule three times to y, we have y approximates
  218. sqrt(x) to within 1 ulp (Unit in the Last Place):
  219. y := (y+x/y)/2 ... almost 17 sig. bits
  220. y := (y+x/y)/2 ... almost 35 sig. bits
  221. y := y-(y-x/y)/2 ... within 1 ulp
  222. Remark 1.
  223. Another way to improve y to within 1 ulp is:
  224. y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
  225. y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
  226. 2
  227. (x-y )*y
  228. y := y + 2* ---------- ...within 1 ulp
  229. 2
  230. 3y + x
  231. This formula has one division fewer than the one above; however,
  232. it requires more multiplications and additions. Also x must be
  233. scaled in advance to avoid spurious overflow in evaluating the
  234. expression 3y*y+x. Hence it is not recommended uless division
  235. is slow. If division is very slow, then one should use the
  236. reciproot algorithm given in section B.
  237. (3) Final adjustment
  238. By twiddling y's last bit it is possible to force y to be
  239. correctly rounded according to the prevailing rounding mode
  240. as follows. Let r and i be copies of the rounding mode and
  241. inexact flag before entering the square root program. Also we
  242. use the expression y+-ulp for the next representable floating
  243. numbers (up and down) of y. Note that y+-ulp = either fixed
  244. point y+-1, or multiply y by nextafter(1,+-inf) in chopped
  245. mode.
  246. I := FALSE; ... reset INEXACT flag I
  247. R := RZ; ... set rounding mode to round-toward-zero
  248. z := x/y; ... chopped quotient, possibly inexact
  249. If(not I) then { ... if the quotient is exact
  250. if(z=y) {
  251. I := i; ... restore inexact flag
  252. R := r; ... restore rounded mode
  253. return sqrt(x):=y.
  254. } else {
  255. z := z - ulp; ... special rounding
  256. }
  257. }
  258. i := TRUE; ... sqrt(x) is inexact
  259. If (r=RN) then z=z+ulp ... rounded-to-nearest
  260. If (r=RP) then { ... round-toward-+inf
  261. y = y+ulp; z=z+ulp;
  262. }
  263. y := y+z; ... chopped sum
  264. y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
  265. I := i; ... restore inexact flag
  266. R := r; ... restore rounded mode
  267. return sqrt(x):=y.
  268. (4) Special cases
  269. Square root of +inf, +-0, or NaN is itself;
  270. Square root of a negative number is NaN with invalid signal.
  271. B. sqrt(x) by Reciproot Iteration
  272. (1) Initial approximation
  273. Let x0 and x1 be the leading and the trailing 32-bit words of
  274. a floating point number x (in IEEE double format) respectively
  275. (see section A). By performing shifs and subtracts on x0 and y0,
  276. we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
  277. k := 0x5fe80000 - (x0>>1);
  278. y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
  279. Here k is a 32-bit integer and T2[] is an integer array
  280. containing correction terms. Now magically the floating
  281. value of y (y's leading 32-bit word is y0, the value of
  282. its trailing word y1 is set to zero) approximates 1/sqrt(x)
  283. to almost 7.8-bit.
  284. Value of T2:
  285. static int T2[64]= {
  286. 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
  287. 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
  288. 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
  289. 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
  290. 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
  291. 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
  292. 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
  293. 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
  294. (2) Iterative refinement
  295. Apply Reciproot iteration three times to y and multiply the
  296. result by x to get an approximation z that matches sqrt(x)
  297. to about 1 ulp. To be exact, we will have
  298. -1ulp < sqrt(x)-z<1.0625ulp.
  299. ... set rounding mode to Round-to-nearest
  300. y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
  301. y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
  302. ... special arrangement for better accuracy
  303. z := x*y ... 29 bits to sqrt(x), with z*y<1
  304. z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
  305. Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
  306. (a) the term z*y in the final iteration is always less than 1;
  307. (b) the error in the final result is biased upward so that
  308. -1 ulp < sqrt(x) - z < 1.0625 ulp
  309. instead of |sqrt(x)-z|<1.03125ulp.
  310. (3) Final adjustment
  311. By twiddling y's last bit it is possible to force y to be
  312. correctly rounded according to the prevailing rounding mode
  313. as follows. Let r and i be copies of the rounding mode and
  314. inexact flag before entering the square root program. Also we
  315. use the expression y+-ulp for the next representable floating
  316. numbers (up and down) of y. Note that y+-ulp = either fixed
  317. point y+-1, or multiply y by nextafter(1,+-inf) in chopped
  318. mode.
  319. R := RZ; ... set rounding mode to round-toward-zero
  320. switch(r) {
  321. case RN: ... round-to-nearest
  322. if(x<= z*(z-ulp)...chopped) z = z - ulp; else
  323. if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
  324. break;
  325. case RZ:case RM: ... round-to-zero or round-to--inf
  326. R:=RP; ... reset rounding mod to round-to-+inf
  327. if(x<z*z ... rounded up) z = z - ulp; else
  328. if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
  329. break;
  330. case RP: ... round-to-+inf
  331. if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
  332. if(x>z*z ...chopped) z = z+ulp;
  333. break;
  334. }
  335. Remark 3. The above comparisons can be done in fixed point. For
  336. example, to compare x and w=z*z chopped, it suffices to compare
  337. x1 and w1 (the trailing parts of x and w), regarding them as
  338. two's complement integers.
  339. ...Is z an exact square root?
  340. To determine whether z is an exact square root of x, let z1 be the
  341. trailing part of z, and also let x0 and x1 be the leading and
  342. trailing parts of x.
  343. If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
  344. I := 1; ... Raise Inexact flag: z is not exact
  345. else {
  346. j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
  347. k := z1 >> 26; ... get z's 25-th and 26-th
  348. fraction bits
  349. I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
  350. }
  351. R:= r ... restore rounded mode
  352. return sqrt(x):=z.
  353. If multiplication is cheaper then the foregoing red tape, the
  354. Inexact flag can be evaluated by
  355. I := i;
  356. I := (z*z!=x) or I.
  357. Note that z*z can overwrite I; this value must be sensed if it is
  358. True.
  359. Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
  360. zero.
  361. --------------------
  362. z1: | f2 |
  363. --------------------
  364. bit 31 bit 0
  365. Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
  366. or even of logb(x) have the following relations:
  367. -------------------------------------------------
  368. bit 27,26 of z1 bit 1,0 of x1 logb(x)
  369. -------------------------------------------------
  370. 00 00 odd and even
  371. 01 01 even
  372. 10 10 odd
  373. 10 00 even
  374. 11 01 even
  375. -------------------------------------------------
  376. (4) Special cases (see (4) of Section A).
  377. */