1 2 /* @(#)e_sqrt.c 1.3 95/01/18 */ 3 /* 4 * ==================================================== 5 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. 6 * 7 * Developed at SunSoft, a Sun Microsystems, Inc. business. 8 * Permission to use, copy, modify, and distribute this 9 * software is freely granted, provided that this notice 10 * is preserved. 11 * ==================================================== 12 */ 13 14 #include "cdefs-compat.h" 15 //__FBSDID("$FreeBSD: src/lib/msun/src/e_sqrt.c,v 1.11 2008/03/02 01:47:58 das Exp $"); 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 <float.h> 88 #include <openlibm_math.h> 89 90 #include "math_private.h" 91 92 static const double one = 1.0, tiny=1.0e-300; 93 94 OLM_DLLEXPORT double 95 __ieee754_sqrt(double x) 96 { 97 double z; 98 int32_t sign = (int)0x80000000; 99 int32_t ix0,s0,q,m,t,i; 100 u_int32_t r,t1,s1,ix1,q1; 101 102 EXTRACT_WORDS(ix0,ix1,x); 103 104 /* take care of Inf and NaN */ 105 if((ix0&0x7ff00000)==0x7ff00000) { 106 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf 107 sqrt(-inf)=sNaN */ 108 } 109 /* take care of zero */ 110 if(ix0<=0) { 111 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ 112 else if(ix0<0) 113 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ 114 } 115 /* normalize x */ 116 m = (ix0>>20); 117 if(m==0) { /* subnormal x */ 118 while(ix0==0) { 119 m -= 21; 120 ix0 |= (ix1>>11); ix1 <<= 21; 121 } 122 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; 123 m -= i-1; 124 ix0 |= (ix1>>(32-i)); 125 ix1 <<= i; 126 } 127 m -= 1023; /* unbias exponent */ 128 ix0 = (ix0&0x000fffff)|0x00100000; 129 if(m&1){ /* odd m, double x to make it even */ 130 ix0 += ix0 + ((ix1&sign)>>31); 131 ix1 += ix1; 132 } 133 m >>= 1; /* m = [m/2] */ 134 135 /* generate sqrt(x) bit by bit */ 136 ix0 += ix0 + ((ix1&sign)>>31); 137 ix1 += ix1; 138 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ 139 r = 0x00200000; /* r = moving bit from right to left */ 140 141 while(r!=0) { 142 t = s0+r; 143 if(t<=ix0) { 144 s0 = t+r; 145 ix0 -= t; 146 q += r; 147 } 148 ix0 += ix0 + ((ix1&sign)>>31); 149 ix1 += ix1; 150 r>>=1; 151 } 152 153 r = sign; 154 while(r!=0) { 155 t1 = s1+r; 156 t = s0; 157 if((t<ix0)||((t==ix0)&&(t1<=ix1))) { 158 s1 = t1+r; 159 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; 160 ix0 -= t; 161 if (ix1 < t1) ix0 -= 1; 162 ix1 -= t1; 163 q1 += r; 164 } 165 ix0 += ix0 + ((ix1&sign)>>31); 166 ix1 += ix1; 167 r>>=1; 168 } 169 170 /* use floating add to find out rounding direction */ 171 if((ix0|ix1)!=0) { 172 z = one-tiny; /* trigger inexact flag */ 173 if (z>=one) { 174 z = one+tiny; 175 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;} 176 else if (z>one) { 177 if (q1==(u_int32_t)0xfffffffe) q+=1; 178 q1+=2; 179 } else 180 q1 += (q1&1); 181 } 182 } 183 ix0 = (q>>1)+0x3fe00000; 184 ix1 = q1>>1; 185 if ((q&1)==1) ix1 |= sign; 186 ix0 += (m <<20); 187 INSERT_WORDS(z,ix0,ix1); 188 return z; 189 } 190 191 #if (LDBL_MANT_DIG == 53) 192 __weak_reference(sqrt, sqrtl); 193 #endif 194 195 /* 196 Other methods (use floating-point arithmetic) 197 ------------- 198 (This is a copy of a drafted paper by Prof W. Kahan 199 and K.C. Ng, written in May, 1986) 200 201 Two algorithms are given here to implement sqrt(x) 202 (IEEE double precision arithmetic) in software. 203 Both supply sqrt(x) correctly rounded. The first algorithm (in 204 Section A) uses newton iterations and involves four divisions. 205 The second one uses reciproot iterations to avoid division, but 206 requires more multiplications. Both algorithms need the ability 207 to chop results of arithmetic operations instead of round them, 208 and the INEXACT flag to indicate when an arithmetic operation 209 is executed exactly with no roundoff error, all part of the 210 standard (IEEE 754-1985). The ability to perform shift, add, 211 subtract and logical AND operations upon 32-bit words is needed 212 too, though not part of the standard. 213 214 A. sqrt(x) by Newton Iteration 215 216 (1) Initial approximation 217 218 Let x0 and x1 be the leading and the trailing 32-bit words of 219 a floating point number x (in IEEE double format) respectively 220 221 1 11 52 ...widths 222 ------------------------------------------------------ 223 x: |s| e | f | 224 ------------------------------------------------------ 225 msb lsb msb lsb ...order 226 227 228 ------------------------ ------------------------ 229 x0: |s| e | f1 | x1: | f2 | 230 ------------------------ ------------------------ 231 232 By performing shifts and subtracts on x0 and x1 (both regarded 233 as integers), we obtain an 8-bit approximation of sqrt(x) as 234 follows. 235 236 k := (x0>>1) + 0x1ff80000; 237 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits 238 Here k is a 32-bit integer and T1[] is an integer array containing 239 correction terms. Now magically the floating value of y (y's 240 leading 32-bit word is y0, the value of its trailing word is 0) 241 approximates sqrt(x) to almost 8-bit. 242 243 Value of T1: 244 static int T1[32]= { 245 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, 246 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, 247 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, 248 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; 249 250 (2) Iterative refinement 251 252 Apply Heron's rule three times to y, we have y approximates 253 sqrt(x) to within 1 ulp (Unit in the Last Place): 254 255 y := (y+x/y)/2 ... almost 17 sig. bits 256 y := (y+x/y)/2 ... almost 35 sig. bits 257 y := y-(y-x/y)/2 ... within 1 ulp 258 259 260 Remark 1. 261 Another way to improve y to within 1 ulp is: 262 263 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) 264 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) 265 266 2 267 (x-y )*y 268 y := y + 2* ---------- ...within 1 ulp 269 2 270 3y + x 271 272 273 This formula has one division fewer than the one above; however, 274 it requires more multiplications and additions. Also x must be 275 scaled in advance to avoid spurious overflow in evaluating the 276 expression 3y*y+x. Hence it is not recommended uless division 277 is slow. If division is very slow, then one should use the 278 reciproot algorithm given in section B. 279 280 (3) Final adjustment 281 282 By twiddling y's last bit it is possible to force y to be 283 correctly rounded according to the prevailing rounding mode 284 as follows. Let r and i be copies of the rounding mode and 285 inexact flag before entering the square root program. Also we 286 use the expression y+-ulp for the next representable floating 287 numbers (up and down) of y. Note that y+-ulp = either fixed 288 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 289 mode. 290 291 I := FALSE; ... reset INEXACT flag I 292 R := RZ; ... set rounding mode to round-toward-zero 293 z := x/y; ... chopped quotient, possibly inexact 294 If(not I) then { ... if the quotient is exact 295 if(z=y) { 296 I := i; ... restore inexact flag 297 R := r; ... restore rounded mode 298 return sqrt(x):=y. 299 } else { 300 z := z - ulp; ... special rounding 301 } 302 } 303 i := TRUE; ... sqrt(x) is inexact 304 If (r=RN) then z=z+ulp ... rounded-to-nearest 305 If (r=RP) then { ... round-toward-+inf 306 y = y+ulp; z=z+ulp; 307 } 308 y := y+z; ... chopped sum 309 y0:=y0-0x00100000; ... y := y/2 is correctly rounded. 310 I := i; ... restore inexact flag 311 R := r; ... restore rounded mode 312 return sqrt(x):=y. 313 314 (4) Special cases 315 316 Square root of +inf, +-0, or NaN is itself; 317 Square root of a negative number is NaN with invalid signal. 318 319 320 B. sqrt(x) by Reciproot Iteration 321 322 (1) Initial approximation 323 324 Let x0 and x1 be the leading and the trailing 32-bit words of 325 a floating point number x (in IEEE double format) respectively 326 (see section A). By performing shifs and subtracts on x0 and y0, 327 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. 328 329 k := 0x5fe80000 - (x0>>1); 330 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits 331 332 Here k is a 32-bit integer and T2[] is an integer array 333 containing correction terms. Now magically the floating 334 value of y (y's leading 32-bit word is y0, the value of 335 its trailing word y1 is set to zero) approximates 1/sqrt(x) 336 to almost 7.8-bit. 337 338 Value of T2: 339 static int T2[64]= { 340 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, 341 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, 342 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, 343 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, 344 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, 345 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, 346 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, 347 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; 348 349 (2) Iterative refinement 350 351 Apply Reciproot iteration three times to y and multiply the 352 result by x to get an approximation z that matches sqrt(x) 353 to about 1 ulp. To be exact, we will have 354 -1ulp < sqrt(x)-z<1.0625ulp. 355 356 ... set rounding mode to Round-to-nearest 357 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) 358 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) 359 ... special arrangement for better accuracy 360 z := x*y ... 29 bits to sqrt(x), with z*y<1 361 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) 362 363 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that 364 (a) the term z*y in the final iteration is always less than 1; 365 (b) the error in the final result is biased upward so that 366 -1 ulp < sqrt(x) - z < 1.0625 ulp 367 instead of |sqrt(x)-z|<1.03125ulp. 368 369 (3) Final adjustment 370 371 By twiddling y's last bit it is possible to force y to be 372 correctly rounded according to the prevailing rounding mode 373 as follows. Let r and i be copies of the rounding mode and 374 inexact flag before entering the square root program. Also we 375 use the expression y+-ulp for the next representable floating 376 numbers (up and down) of y. Note that y+-ulp = either fixed 377 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 378 mode. 379 380 R := RZ; ... set rounding mode to round-toward-zero 381 switch(r) { 382 case RN: ... round-to-nearest 383 if(x<= z*(z-ulp)...chopped) z = z - ulp; else 384 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; 385 break; 386 case RZ:case RM: ... round-to-zero or round-to--inf 387 R:=RP; ... reset rounding mod to round-to-+inf 388 if(x<z*z ... rounded up) z = z - ulp; else 389 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; 390 break; 391 case RP: ... round-to-+inf 392 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else 393 if(x>z*z ...chopped) z = z+ulp; 394 break; 395 } 396 397 Remark 3. The above comparisons can be done in fixed point. For 398 example, to compare x and w=z*z chopped, it suffices to compare 399 x1 and w1 (the trailing parts of x and w), regarding them as 400 two's complement integers. 401 402 ...Is z an exact square root? 403 To determine whether z is an exact square root of x, let z1 be the 404 trailing part of z, and also let x0 and x1 be the leading and 405 trailing parts of x. 406 407 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 408 I := 1; ... Raise Inexact flag: z is not exact 409 else { 410 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 411 k := z1 >> 26; ... get z's 25-th and 26-th 412 fraction bits 413 I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); 414 } 415 R:= r ... restore rounded mode 416 return sqrt(x):=z. 417 418 If multiplication is cheaper then the foregoing red tape, the 419 Inexact flag can be evaluated by 420 421 I := i; 422 I := (z*z!=x) or I. 423 424 Note that z*z can overwrite I; this value must be sensed if it is 425 True. 426 427 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be 428 zero. 429 430 -------------------- 431 z1: | f2 | 432 -------------------- 433 bit 31 bit 0 434 435 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd 436 or even of logb(x) have the following relations: 437 438 ------------------------------------------------- 439 bit 27,26 of z1 bit 1,0 of x1 logb(x) 440 ------------------------------------------------- 441 00 00 odd and even 442 01 01 even 443 10 10 odd 444 10 00 even 445 11 01 even 446 ------------------------------------------------- 447 448 (4) Special cases (see (4) of Section A). 449 450 */ 451 452