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