1 /* derived from /netlib/fdlibm */ 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 /* __ieee754_sqrt(x) 15 * Return correctly rounded sqrt. 16 * ------------------------------------------ 17 * | Use the hardware sqrt if you have one | 18 * ------------------------------------------ 19 * Method: 20 * Bit by bit method using integer arithmetic. (Slow, but portable) 21 * 1. Normalization 22 * Scale x to y in [1,4) with even powers of 2: 23 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then 24 * sqrt(x) = 2^k * sqrt(y) 25 * 2. Bit by bit computation 26 * Let q = sqrt(y) truncated to i bit after binary point (q = 1), 27 * i 0 28 * i+1 2 29 * s = 2*q , and y = 2 * ( y - q ). (1) 30 * i i i i 31 * 32 * To compute q from q , one checks whether 33 * i+1 i 34 * 35 * -(i+1) 2 36 * (q + 2 ) <= y. (2) 37 * i 38 * -(i+1) 39 * If (2) is false, then q = q ; otherwise q = q + 2 . 40 * i+1 i i+1 i 41 * 42 * With some algebric manipulation, it is not difficult to see 43 * that (2) is equivalent to 44 * -(i+1) 45 * s + 2 <= y (3) 46 * i i 47 * 48 * The advantage of (3) is that s and y can be computed by 49 * i i 50 * the following recurrence formula: 51 * if (3) is false 52 * 53 * s = s , y = y ; (4) 54 * i+1 i i+1 i 55 * 56 * otherwise, 57 * -i -(i+1) 58 * s = s + 2 , y = y - s - 2 (5) 59 * i+1 i i+1 i i 60 * 61 * One may easily use induction to prove (4) and (5). 62 * Note. Since the left hand side of (3) contain only i+2 bits, 63 * it does not necessary to do a full (53-bit) comparison 64 * in (3). 65 * 3. Final rounding 66 * After generating the 53 bits result, we compute one more bit. 67 * Together with the remainder, we can decide whether the 68 * result is exact, bigger than 1/2ulp, or less than 1/2ulp 69 * (it will never equal to 1/2ulp). 70 * The rounding mode can be detected by checking whether 71 * Huge + tiny is equal to Huge, and whether Huge - tiny is 72 * equal to Huge for some floating point number "Huge" and "tiny". 73 * 74 * Special cases: 75 * sqrt(+-0) = +-0 ... exact 76 * sqrt(inf) = inf 77 * sqrt(-ve) = NaN ... with invalid signal 78 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN 79 * 80 * Other methods : see the appended file at the end of the program below. 81 *--------------- 82 */ 83 84 #include "fdlibm.h" 85 86 static const double one = 1.0, tiny=1.0e-300; 87 __ieee754_sqrt(double x)88 double __ieee754_sqrt(double x) 89 { 90 double z; 91 int sign = (int)0x80000000; 92 unsigned r,t1,s1,ix1,q1; 93 int ix0,s0,q,m,t,i; 94 95 ix0 = __HI(x); /* high word of x */ 96 ix1 = __LO(x); /* low word of x */ 97 98 /* take care of Inf and NaN */ 99 if((ix0&0x7ff00000)==0x7ff00000) { 100 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf 101 sqrt(-inf)=sNaN */ 102 } 103 /* take care of zero */ 104 if(ix0<=0) { 105 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ 106 else if(ix0<0) 107 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ 108 } 109 /* normalize x */ 110 m = (ix0>>20); 111 if(m==0) { /* subnormal x */ 112 while(ix0==0) { 113 m -= 21; 114 ix0 |= (ix1>>11); ix1 <<= 21; 115 } 116 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; 117 m -= i-1; 118 ix0 |= (ix1>>(32-i)); 119 ix1 <<= i; 120 } 121 m -= 1023; /* unbias exponent */ 122 ix0 = (ix0&0x000fffff)|0x00100000; 123 if(m&1){ /* odd m, double x to make it even */ 124 ix0 += ix0 + ((ix1&sign)>>31); 125 ix1 += ix1; 126 } 127 m >>= 1; /* m = [m/2] */ 128 129 /* generate sqrt(x) bit by bit */ 130 ix0 += ix0 + ((ix1&sign)>>31); 131 ix1 += ix1; 132 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ 133 r = 0x00200000; /* r = moving bit from right to left */ 134 135 while(r!=0) { 136 t = s0+r; 137 if(t<=ix0) { 138 s0 = t+r; 139 ix0 -= t; 140 q += r; 141 } 142 ix0 += ix0 + ((ix1&sign)>>31); 143 ix1 += ix1; 144 r>>=1; 145 } 146 147 r = sign; 148 while(r!=0) { 149 t1 = s1+r; 150 t = s0; 151 if((t<ix0)||((t==ix0)&&(t1<=ix1))) { 152 s1 = t1+r; 153 if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1; 154 ix0 -= t; 155 if (ix1 < t1) ix0 -= 1; 156 ix1 -= t1; 157 q1 += r; 158 } 159 ix0 += ix0 + ((ix1&sign)>>31); 160 ix1 += ix1; 161 r>>=1; 162 } 163 164 /* use floating add to find out rounding direction */ 165 if((ix0|ix1)!=0) { 166 z = one-tiny; /* trigger inexact flag */ 167 if (z>=one) { 168 z = one+tiny; 169 if (q1==(unsigned)0xffffffff) { q1=0; q += 1;} 170 else if (z>one) { 171 if (q1==(unsigned)0xfffffffe) q+=1; 172 q1+=2; 173 } else 174 q1 += (q1&1); 175 } 176 } 177 ix0 = (q>>1)+0x3fe00000; 178 ix1 = q1>>1; 179 if ((q&1)==1) ix1 |= sign; 180 ix0 += (m <<20); 181 __HI(z) = ix0; 182 __LO(z) = ix1; 183 return z; 184 } 185 186 /* 187 Other methods (use floating-point arithmetic) 188 ------------- 189 (This is a copy of a drafted paper by Prof W. Kahan 190 and K.C. Ng, written in May, 1986) 191 192 Two algorithms are given here to implement sqrt(x) 193 (IEEE double precision arithmetic) in software. 194 Both supply sqrt(x) correctly rounded. The first algorithm (in 195 Section A) uses newton iterations and involves four divisions. 196 The second one uses reciproot iterations to avoid division, but 197 requires more multiplications. Both algorithms need the ability 198 to chop results of arithmetic operations instead of round them, 199 and the INEXACT flag to indicate when an arithmetic operation 200 is executed exactly with no roundoff error, all part of the 201 standard (IEEE 754-1985). The ability to perform shift, add, 202 subtract and logical AND operations upon 32-bit words is needed 203 too, though not part of the standard. 204 205 A. sqrt(x) by Newton Iteration 206 207 (1) Initial approximation 208 209 Let x0 and x1 be the leading and the trailing 32-bit words of 210 a floating point number x (in IEEE double format) respectively 211 212 1 11 52 ...widths 213 ------------------------------------------------------ 214 x: |s| e | f | 215 ------------------------------------------------------ 216 msb lsb msb lsb ...order 217 218 219 ------------------------ ------------------------ 220 x0: |s| e | f1 | x1: | f2 | 221 ------------------------ ------------------------ 222 223 By performing shifts and subtracts on x0 and x1 (both regarded 224 as integers), we obtain an 8-bit approximation of sqrt(x) as 225 follows. 226 227 k := (x0>>1) + 0x1ff80000; 228 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits 229 Here k is a 32-bit integer and T1[] is an integer array containing 230 correction terms. Now magically the floating value of y (y's 231 leading 32-bit word is y0, the value of its trailing word is 0) 232 approximates sqrt(x) to almost 8-bit. 233 234 Value of T1: 235 static int T1[32]= { 236 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, 237 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, 238 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, 239 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; 240 241 (2) Iterative refinement 242 243 Apply Heron's rule three times to y, we have y approximates 244 sqrt(x) to within 1 ulp (Unit in the Last Place): 245 246 y := (y+x/y)/2 ... almost 17 sig. bits 247 y := (y+x/y)/2 ... almost 35 sig. bits 248 y := y-(y-x/y)/2 ... within 1 ulp 249 250 251 Remark 1. 252 Another way to improve y to within 1 ulp is: 253 254 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) 255 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) 256 257 2 258 (x-y )*y 259 y := y + 2* ---------- ...within 1 ulp 260 2 261 3y + x 262 263 264 This formula has one division fewer than the one above; however, 265 it requires more multiplications and additions. Also x must be 266 scaled in advance to avoid spurious overflow in evaluating the 267 expression 3y*y+x. Hence it is not recommended uless division 268 is slow. If division is very slow, then one should use the 269 reciproot algorithm given in section B. 270 271 (3) Final adjustment 272 273 By twiddling y's last bit it is possible to force y to be 274 correctly rounded according to the prevailing rounding mode 275 as follows. Let r and i be copies of the rounding mode and 276 inexact flag before entering the square root program. Also we 277 use the expression y+-ulp for the next representable floating 278 numbers (up and down) of y. Note that y+-ulp = either fixed 279 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 280 mode. 281 282 I := FALSE; ... reset INEXACT flag I 283 R := RZ; ... set rounding mode to round-toward-zero 284 z := x/y; ... chopped quotient, possibly inexact 285 If(not I) then { ... if the quotient is exact 286 if(z=y) { 287 I := i; ... restore inexact flag 288 R := r; ... restore rounded mode 289 return sqrt(x):=y. 290 } else { 291 z := z - ulp; ... special rounding 292 } 293 } 294 i := TRUE; ... sqrt(x) is inexact 295 If (r=RN) then z=z+ulp ... rounded-to-nearest 296 If (r=RP) then { ... round-toward-+inf 297 y = y+ulp; z=z+ulp; 298 } 299 y := y+z; ... chopped sum 300 y0:=y0-0x00100000; ... y := y/2 is correctly rounded. 301 I := i; ... restore inexact flag 302 R := r; ... restore rounded mode 303 return sqrt(x):=y. 304 305 (4) Special cases 306 307 Square root of +inf, +-0, or NaN is itself; 308 Square root of a negative number is NaN with invalid signal. 309 310 311 B. sqrt(x) by Reciproot Iteration 312 313 (1) Initial approximation 314 315 Let x0 and x1 be the leading and the trailing 32-bit words of 316 a floating point number x (in IEEE double format) respectively 317 (see section A). By performing shifs and subtracts on x0 and y0, 318 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. 319 320 k := 0x5fe80000 - (x0>>1); 321 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits 322 323 Here k is a 32-bit integer and T2[] is an integer array 324 containing correction terms. Now magically the floating 325 value of y (y's leading 32-bit word is y0, the value of 326 its trailing word y1 is set to zero) approximates 1/sqrt(x) 327 to almost 7.8-bit. 328 329 Value of T2: 330 static int T2[64]= { 331 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, 332 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, 333 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, 334 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, 335 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, 336 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, 337 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, 338 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; 339 340 (2) Iterative refinement 341 342 Apply Reciproot iteration three times to y and multiply the 343 result by x to get an approximation z that matches sqrt(x) 344 to about 1 ulp. To be exact, we will have 345 -1ulp < sqrt(x)-z<1.0625ulp. 346 347 ... set rounding mode to Round-to-nearest 348 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) 349 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) 350 ... special arrangement for better accuracy 351 z := x*y ... 29 bits to sqrt(x), with z*y<1 352 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) 353 354 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that 355 (a) the term z*y in the final iteration is always less than 1; 356 (b) the error in the final result is biased upward so that 357 -1 ulp < sqrt(x) - z < 1.0625 ulp 358 instead of |sqrt(x)-z|<1.03125ulp. 359 360 (3) Final adjustment 361 362 By twiddling y's last bit it is possible to force y to be 363 correctly rounded according to the prevailing rounding mode 364 as follows. Let r and i be copies of the rounding mode and 365 inexact flag before entering the square root program. Also we 366 use the expression y+-ulp for the next representable floating 367 numbers (up and down) of y. Note that y+-ulp = either fixed 368 point y+-1, or multiply y by nextafter(1,+-inf) in chopped 369 mode. 370 371 R := RZ; ... set rounding mode to round-toward-zero 372 switch(r) { 373 case RN: ... round-to-nearest 374 if(x<= z*(z-ulp)...chopped) z = z - ulp; else 375 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; 376 break; 377 case RZ:case RM: ... round-to-zero or round-to--inf 378 R:=RP; ... reset rounding mod to round-to-+inf 379 if(x<z*z ... rounded up) z = z - ulp; else 380 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; 381 break; 382 case RP: ... round-to-+inf 383 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else 384 if(x>z*z ...chopped) z = z+ulp; 385 break; 386 } 387 388 Remark 3. The above comparisons can be done in fixed point. For 389 example, to compare x and w=z*z chopped, it suffices to compare 390 x1 and w1 (the trailing parts of x and w), regarding them as 391 two's complement integers. 392 393 ...Is z an exact square root? 394 To determine whether z is an exact square root of x, let z1 be the 395 trailing part of z, and also let x0 and x1 be the leading and 396 trailing parts of x. 397 398 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 399 I := 1; ... Raise Inexact flag: z is not exact 400 else { 401 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 402 k := z1 >> 26; ... get z's 25-th and 26-th 403 fraction bits 404 I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); 405 } 406 R:= r ... restore rounded mode 407 return sqrt(x):=z. 408 409 If multiplication is cheaper then the foregoing red tape, the 410 Inexact flag can be evaluated by 411 412 I := i; 413 I := (z*z!=x) or I. 414 415 Note that z*z can overwrite I; this value must be sensed if it is 416 True. 417 418 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be 419 zero. 420 421 -------------------- 422 z1: | f2 | 423 -------------------- 424 bit 31 bit 0 425 426 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd 427 or even of logb(x) have the following relations: 428 429 ------------------------------------------------- 430 bit 27,26 of z1 bit 1,0 of x1 logb(x) 431 ------------------------------------------------- 432 00 00 odd and even 433 01 01 even 434 10 10 odd 435 10 00 even 436 11 01 even 437 ------------------------------------------------- 438 439 (4) Special cases (see (4) of Section A). 440 441 */ 442 443