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