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 13 #ifndef lint 14 static char rcsid[] = "$FreeBSD$"; 15 #endif 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 "math.h" 88 #include "math_private.h" 89 90 #ifdef __STDC__ 91 static const double one = 1.0, tiny=1.0e-300; 92 #else 93 static double one = 1.0, tiny=1.0e-300; 94 #endif 95 96 #ifdef __STDC__ 97 double __generic___ieee754_sqrt(double x) 98 #else 99 double __generic___ieee754_sqrt(x) 100 double x; 101 #endif 102 { 103 double z; 104 int32_t sign = (int)0x80000000; 105 int32_t ix0,s0,q,m,t,i; 106 u_int32_t r,t1,s1,ix1,q1; 107 108 EXTRACT_WORDS(ix0,ix1,x); 109 110 /* take care of Inf and NaN */ 111 if((ix0&0x7ff00000)==0x7ff00000) { 112 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf 113 sqrt(-inf)=sNaN */ 114 } 115 /* take care of zero */ 116 if(ix0<=0) { 117 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ 118 else if(ix0<0) 119 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ 120 } 121 /* normalize x */ 122 m = (ix0>>20); 123 if(m==0) { /* subnormal x */ 124 while(ix0==0) { 125 m -= 21; 126 ix0 |= (ix1>>11); ix1 <<= 21; 127 } 128 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; 129 m -= i-1; 130 ix0 |= (ix1>>(32-i)); 131 ix1 <<= i; 132 } 133 m -= 1023; /* unbias exponent */ 134 ix0 = (ix0&0x000fffff)|0x00100000; 135 if(m&1){ /* odd m, double x to make it even */ 136 ix0 += ix0 + ((ix1&sign)>>31); 137 ix1 += ix1; 138 } 139 m >>= 1; /* m = [m/2] */ 140 141 /* generate sqrt(x) bit by bit */ 142 ix0 += ix0 + ((ix1&sign)>>31); 143 ix1 += ix1; 144 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ 145 r = 0x00200000; /* r = moving bit from right to left */ 146 147 while(r!=0) { 148 t = s0+r; 149 if(t<=ix0) { 150 s0 = t+r; 151 ix0 -= t; 152 q += r; 153 } 154 ix0 += ix0 + ((ix1&sign)>>31); 155 ix1 += ix1; 156 r>>=1; 157 } 158 159 r = sign; 160 while(r!=0) { 161 t1 = s1+r; 162 t = s0; 163 if((t<ix0)||((t==ix0)&&(t1<=ix1))) { 164 s1 = t1+r; 165 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; 166 ix0 -= t; 167 if (ix1 < t1) ix0 -= 1; 168 ix1 -= t1; 169 q1 += r; 170 } 171 ix0 += ix0 + ((ix1&sign)>>31); 172 ix1 += ix1; 173 r>>=1; 174 } 175 176 /* use floating add to find out rounding direction */ 177 if((ix0|ix1)!=0) { 178 z = one-tiny; /* trigger inexact flag */ 179 if (z>=one) { 180 z = one+tiny; 181 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;} 182 else if (z>one) { 183 if (q1==(u_int32_t)0xfffffffe) q+=1; 184 q1+=2; 185 } else 186 q1 += (q1&1); 187 } 188 } 189 ix0 = (q>>1)+0x3fe00000; 190 ix1 = q1>>1; 191 if ((q&1)==1) ix1 |= sign; 192 ix0 += (m <<20); 193 INSERT_WORDS(z,ix0,ix1); 194 return z; 195 } 196 197 /* 198 Other methods (use floating-point arithmetic) 199 ------------- 200 (This is a copy of a drafted paper by Prof W. Kahan 201 and K.C. Ng, written in May, 1986) 202 203 Two algorithms are given here to implement sqrt(x) 204 (IEEE double precision arithmetic) in software. 205 Both supply sqrt(x) correctly rounded. The first algorithm (in 206 Section A) uses newton iterations and involves four divisions. 207 The second one uses reciproot iterations to avoid division, but 208 requires more multiplications. Both algorithms need the ability 209 to chop results of arithmetic operations instead of round them, 210 and the INEXACT flag to indicate when an arithmetic operation 211 is executed exactly with no roundoff error, all part of the 212 standard (IEEE 754-1985). The ability to perform shift, add, 213 subtract and logical AND operations upon 32-bit words is needed 214 too, though not part of the standard. 215 216 A. sqrt(x) by Newton Iteration 217 218 (1) Initial approximation 219 220 Let x0 and x1 be the leading and the trailing 32-bit words of 221 a floating point number x (in IEEE double format) respectively 222 223 1 11 52 ...widths 224 ------------------------------------------------------ 225 x: |s| e | f | 226 ------------------------------------------------------ 227 msb lsb msb lsb ...order 228 229 230 ------------------------ ------------------------ 231 x0: |s| e | f1 | x1: | f2 | 232 ------------------------ ------------------------ 233 234 By performing shifts and subtracts on x0 and x1 (both regarded 235 as integers), we obtain an 8-bit approximation of sqrt(x) as 236 follows. 237 238 k := (x0>>1) + 0x1ff80000; 239 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits 240 Here k is a 32-bit integer and T1[] is an integer array containing 241 correction terms. Now magically the floating value of y (y's 242 leading 32-bit word is y0, the value of its trailing word is 0) 243 approximates sqrt(x) to almost 8-bit. 244 245 Value of T1: 246 static int T1[32]= { 247 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, 248 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, 249 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, 250 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; 251 252 (2) Iterative refinement 253 254 Apply Heron's rule three times to y, we have y approximates 255 sqrt(x) to within 1 ulp (Unit in the Last Place): 256 257 y := (y+x/y)/2 ... almost 17 sig. bits 258 y := (y+x/y)/2 ... almost 35 sig. bits 259 y := y-(y-x/y)/2 ... within 1 ulp 260 261 262 Remark 1. 263 Another way to improve y to within 1 ulp is: 264 265 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) 266 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) 267 268 2 269 (x-y )*y 270 y := y + 2* ---------- ...within 1 ulp 271 2 272 3y + x 273 274 275 This formula has one division fewer than the one above; however, 276 it requires more multiplications and additions. Also x must be 277 scaled in advance to avoid spurious overflow in evaluating the 278 expression 3y*y+x. Hence it is not recommended uless division 279 is slow. If division is very slow, then one should use the 280 reciproot algorithm given in section B. 281 282 (3) Final adjustment 283 284 By twiddling y's last bit it is possible to force y to be 285 correctly rounded according to the prevailing rounding mode 286 as follows. Let r and i be copies of the rounding mode and 287 inexact flag before entering the square root program. Also we 288 use the expression y+-ulp for the next representable floating 289 numbers (up and down) of y. Note that y+-ulp = either fixed 290 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 291 mode. 292 293 I := FALSE; ... reset INEXACT flag I 294 R := RZ; ... set rounding mode to round-toward-zero 295 z := x/y; ... chopped quotient, possibly inexact 296 If(not I) then { ... if the quotient is exact 297 if(z=y) { 298 I := i; ... restore inexact flag 299 R := r; ... restore rounded mode 300 return sqrt(x):=y. 301 } else { 302 z := z - ulp; ... special rounding 303 } 304 } 305 i := TRUE; ... sqrt(x) is inexact 306 If (r=RN) then z=z+ulp ... rounded-to-nearest 307 If (r=RP) then { ... round-toward-+inf 308 y = y+ulp; z=z+ulp; 309 } 310 y := y+z; ... chopped sum 311 y0:=y0-0x00100000; ... y := y/2 is correctly rounded. 312 I := i; ... restore inexact flag 313 R := r; ... restore rounded mode 314 return sqrt(x):=y. 315 316 (4) Special cases 317 318 Square root of +inf, +-0, or NaN is itself; 319 Square root of a negative number is NaN with invalid signal. 320 321 322 B. sqrt(x) by Reciproot Iteration 323 324 (1) Initial approximation 325 326 Let x0 and x1 be the leading and the trailing 32-bit words of 327 a floating point number x (in IEEE double format) respectively 328 (see section A). By performing shifs and subtracts on x0 and y0, 329 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. 330 331 k := 0x5fe80000 - (x0>>1); 332 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits 333 334 Here k is a 32-bit integer and T2[] is an integer array 335 containing correction terms. Now magically the floating 336 value of y (y's leading 32-bit word is y0, the value of 337 its trailing word y1 is set to zero) approximates 1/sqrt(x) 338 to almost 7.8-bit. 339 340 Value of T2: 341 static int T2[64]= { 342 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, 343 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, 344 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, 345 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, 346 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, 347 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, 348 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, 349 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; 350 351 (2) Iterative refinement 352 353 Apply Reciproot iteration three times to y and multiply the 354 result by x to get an approximation z that matches sqrt(x) 355 to about 1 ulp. To be exact, we will have 356 -1ulp < sqrt(x)-z<1.0625ulp. 357 358 ... set rounding mode to Round-to-nearest 359 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) 360 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) 361 ... special arrangement for better accuracy 362 z := x*y ... 29 bits to sqrt(x), with z*y<1 363 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) 364 365 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that 366 (a) the term z*y in the final iteration is always less than 1; 367 (b) the error in the final result is biased upward so that 368 -1 ulp < sqrt(x) - z < 1.0625 ulp 369 instead of |sqrt(x)-z|<1.03125ulp. 370 371 (3) Final adjustment 372 373 By twiddling y's last bit it is possible to force y to be 374 correctly rounded according to the prevailing rounding mode 375 as follows. Let r and i be copies of the rounding mode and 376 inexact flag before entering the square root program. Also we 377 use the expression y+-ulp for the next representable floating 378 numbers (up and down) of y. Note that y+-ulp = either fixed 379 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 380 mode. 381 382 R := RZ; ... set rounding mode to round-toward-zero 383 switch(r) { 384 case RN: ... round-to-nearest 385 if(x<= z*(z-ulp)...chopped) z = z - ulp; else 386 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; 387 break; 388 case RZ:case RM: ... round-to-zero or round-to--inf 389 R:=RP; ... reset rounding mod to round-to-+inf 390 if(x<z*z ... rounded up) z = z - ulp; else 391 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; 392 break; 393 case RP: ... round-to-+inf 394 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else 395 if(x>z*z ...chopped) z = z+ulp; 396 break; 397 } 398 399 Remark 3. The above comparisons can be done in fixed point. For 400 example, to compare x and w=z*z chopped, it suffices to compare 401 x1 and w1 (the trailing parts of x and w), regarding them as 402 two's complement integers. 403 404 ...Is z an exact square root? 405 To determine whether z is an exact square root of x, let z1 be the 406 trailing part of z, and also let x0 and x1 be the leading and 407 trailing parts of x. 408 409 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 410 I := 1; ... Raise Inexact flag: z is not exact 411 else { 412 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 413 k := z1 >> 26; ... get z's 25-th and 26-th 414 fraction bits 415 I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); 416 } 417 R:= r ... restore rounded mode 418 return sqrt(x):=z. 419 420 If multiplication is cheaper then the foregoing red tape, the 421 Inexact flag can be evaluated by 422 423 I := i; 424 I := (z*z!=x) or I. 425 426 Note that z*z can overwrite I; this value must be sensed if it is 427 True. 428 429 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be 430 zero. 431 432 -------------------- 433 z1: | f2 | 434 -------------------- 435 bit 31 bit 0 436 437 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd 438 or even of logb(x) have the following relations: 439 440 ------------------------------------------------- 441 bit 27,26 of z1 bit 1,0 of x1 logb(x) 442 ------------------------------------------------- 443 00 00 odd and even 444 01 01 even 445 10 10 odd 446 10 00 even 447 11 01 even 448 ------------------------------------------------- 449 450 (4) Special cases (see (4) of Section A). 451 452 */ 453 454