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