1/* $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */ 2 3/*- 4 * SPDX-License-Identifier: BSD-3-Clause 5 * 6 * Copyright (c) 1992, 1993 7 * The Regents of the University of California. All rights reserved. 8 * 9 * This software was developed by the Computer Systems Engineering group 10 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 11 * contributed to Berkeley. 12 * 13 * All advertising materials mentioning features or use of this software 14 * must display the following acknowledgement: 15 * This product includes software developed by the University of 16 * California, Lawrence Berkeley Laboratory. 17 * 18 * Redistribution and use in source and binary forms, with or without 19 * modification, are permitted provided that the following conditions 20 * are met: 21 * 1. Redistributions of source code must retain the above copyright 22 * notice, this list of conditions and the following disclaimer. 23 * 2. Redistributions in binary form must reproduce the above copyright 24 * notice, this list of conditions and the following disclaimer in the 25 * documentation and/or other materials provided with the distribution. 26 * 3. Neither the name of the University nor the names of its contributors 27 * may be used to endorse or promote products derived from this software 28 * without specific prior written permission. 29 * 30 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 31 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 32 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 33 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 34 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 35 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 36 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 37 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 38 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 39 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 40 * SUCH DAMAGE. 41 */ 42 43/* 44 * Perform an FPU square root (return sqrt(x)). 45 */ 46 47#include <sys/types.h> 48#include <sys/systm.h> 49 50#include <machine/fpu.h> 51 52#include <powerpc/fpu/fpu_arith.h> 53#include <powerpc/fpu/fpu_emu.h> 54 55/* 56 * Our task is to calculate the square root of a floating point number x0. 57 * This number x normally has the form: 58 * 59 * exp 60 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 61 * 62 * This can be left as it stands, or the mantissa can be doubled and the 63 * exponent decremented: 64 * 65 * exp-1 66 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 67 * 68 * If the exponent `exp' is even, the square root of the number is best 69 * handled using the first form, and is by definition equal to: 70 * 71 * exp/2 72 * sqrt(x) = sqrt(mant) * 2 73 * 74 * If exp is odd, on the other hand, it is convenient to use the second 75 * form, giving: 76 * 77 * (exp-1)/2 78 * sqrt(x) = sqrt(2 * mant) * 2 79 * 80 * In the first case, we have 81 * 82 * 1 <= mant < 2 83 * 84 * and therefore 85 * 86 * sqrt(1) <= sqrt(mant) < sqrt(2) 87 * 88 * while in the second case we have 89 * 90 * 2 <= 2*mant < 4 91 * 92 * and therefore 93 * 94 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 95 * 96 * so that in any case, we are sure that 97 * 98 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 99 * 100 * or 101 * 102 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 103 * 104 * This root is therefore a properly formed mantissa for a floating 105 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 106 * as above. This leaves us with the problem of finding the square root 107 * of a fixed-point number in the range [1..4). 108 * 109 * Though it may not be instantly obvious, the following square root 110 * algorithm works for any integer x of an even number of bits, provided 111 * that no overflows occur: 112 * 113 * let q = 0 114 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 115 * x *= 2 -- multiply by radix, for next digit 116 * if x >= 2q + 2^k then -- if adding 2^k does not 117 * x -= 2q + 2^k -- exceed the correct root, 118 * q += 2^k -- add 2^k and adjust x 119 * fi 120 * done 121 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 122 * 123 * If NBITS is odd (so that k is initially even), we can just add another 124 * zero bit at the top of x. Doing so means that q is not going to acquire 125 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 126 * final value in x is not needed, or can be off by a factor of 2, this is 127 * equivalant to moving the `x *= 2' step to the bottom of the loop: 128 * 129 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 130 * 131 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 132 * (Since the algorithm is destructive on x, we will call x's initial 133 * value, for which q is some power of two times its square root, x0.) 134 * 135 * If we insert a loop invariant y = 2q, we can then rewrite this using 136 * C notation as: 137 * 138 * q = y = 0; x = x0; 139 * for (k = NBITS; --k >= 0;) { 140 * #if (NBITS is even) 141 * x *= 2; 142 * #endif 143 * t = y + (1 << k); 144 * if (x >= t) { 145 * x -= t; 146 * q += 1 << k; 147 * y += 1 << (k + 1); 148 * } 149 * #if (NBITS is odd) 150 * x *= 2; 151 * #endif 152 * } 153 * 154 * If x0 is fixed point, rather than an integer, we can simply alter the 155 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 156 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 157 * 158 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 159 * integers, which adds some complication. But note that q is built one 160 * bit at a time, from the top down, and is not used itself in the loop 161 * (we use 2q as held in y instead). This means we can build our answer 162 * in an integer, one word at a time, which saves a bit of work. Also, 163 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 164 * `new' bits in y and we can set them with an `or' operation rather than 165 * a full-blown multiword add. 166 * 167 * We are almost done, except for one snag. We must prove that none of our 168 * intermediate calculations can overflow. We know that x0 is in [1..4) 169 * and therefore the square root in q will be in [1..2), but what about x, 170 * y, and t? 171 * 172 * We know that y = 2q at the beginning of each loop. (The relation only 173 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 174 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 175 * Furthermore, we can prove with a bit of work that x never exceeds y by 176 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 177 * an exercise to the reader, mostly because I have become tired of working 178 * on this comment.) 179 * 180 * If our floating point mantissas (which are of the form 1.frac) occupy 181 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 182 * In fact, we want even one more bit (for a carry, to avoid compares), or 183 * three extra. There is a comment in fpu_emu.h reminding maintainers of 184 * this, so we have some justification in assuming it. 185 */ 186struct fpn * 187fpu_sqrt(struct fpemu *fe) 188{ 189 struct fpn *x = &fe->fe_f1; 190 u_int bit, q, tt; 191 u_int x0, x1, x2, x3; 192 u_int y0, y1, y2, y3; 193 u_int d0, d1, d2, d3; 194 int e; 195 FPU_DECL_CARRY; 196 197 /* 198 * Take care of special cases first. In order: 199 * 200 * sqrt(NaN) = NaN 201 * sqrt(+0) = +0 202 * sqrt(-0) = -0 203 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 204 * sqrt(+Inf) = +Inf 205 * 206 * Then all that remains are numbers with mantissas in [1..2). 207 */ 208 DPRINTF(FPE_REG, ("fpu_sqer:\n")); 209 DUMPFPN(FPE_REG, x); 210 DPRINTF(FPE_REG, ("=>\n")); 211 if (ISNAN(x)) { 212 fe->fe_cx |= FPSCR_VXSNAN; 213 DUMPFPN(FPE_REG, x); 214 return (x); 215 } 216 if (ISZERO(x)) { 217 fe->fe_cx |= FPSCR_ZX; 218 x->fp_class = FPC_INF; 219 DUMPFPN(FPE_REG, x); 220 return (x); 221 } 222 if (x->fp_sign) { 223 fe->fe_cx |= FPSCR_VXSQRT; 224 return (fpu_newnan(fe)); 225 } 226 if (ISINF(x)) { 227 DUMPFPN(FPE_REG, x); 228 return (x); 229 } 230 231 /* 232 * Calculate result exponent. As noted above, this may involve 233 * doubling the mantissa. We will also need to double x each 234 * time around the loop, so we define a macro for this here, and 235 * we break out the multiword mantissa. 236 */ 237#ifdef FPU_SHL1_BY_ADD 238#define DOUBLE_X { \ 239 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 240 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 241} 242#else 243#define DOUBLE_X { \ 244 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 245 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 246} 247#endif 248#if (FP_NMANT & 1) != 0 249# define ODD_DOUBLE DOUBLE_X 250# define EVEN_DOUBLE /* nothing */ 251#else 252# define ODD_DOUBLE /* nothing */ 253# define EVEN_DOUBLE DOUBLE_X 254#endif 255 x0 = x->fp_mant[0]; 256 x1 = x->fp_mant[1]; 257 x2 = x->fp_mant[2]; 258 x3 = x->fp_mant[3]; 259 e = x->fp_exp; 260 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 261 DOUBLE_X; 262 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 263 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 264 265 /* 266 * Now calculate the mantissa root. Since x is now in [1..4), 267 * we know that the first trip around the loop will definitely 268 * set the top bit in q, so we can do that manually and start 269 * the loop at the next bit down instead. We must be sure to 270 * double x correctly while doing the `known q=1.0'. 271 * 272 * We do this one mantissa-word at a time, as noted above, to 273 * save work. To avoid `(1U << 31) << 1', we also do the top bit 274 * outside of each per-word loop. 275 * 276 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 277 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 278 * is always a `new' one, this means that three of the `t?'s are 279 * just the corresponding `y?'; we use `#define's here for this. 280 * The variable `tt' holds the actual `t?' variable. 281 */ 282 283 /* calculate q0 */ 284#define t0 tt 285 bit = FP_1; 286 EVEN_DOUBLE; 287 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 288 q = bit; 289 x0 -= bit; 290 y0 = bit << 1; 291 /* } */ 292 ODD_DOUBLE; 293 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 294 EVEN_DOUBLE; 295 t0 = y0 | bit; /* t = y + bit */ 296 if (x0 >= t0) { /* if x >= t then */ 297 x0 -= t0; /* x -= t */ 298 q |= bit; /* q += bit */ 299 y0 |= bit << 1; /* y += bit << 1 */ 300 } 301 ODD_DOUBLE; 302 } 303 x->fp_mant[0] = q; 304#undef t0 305 306 /* calculate q1. note (y0&1)==0. */ 307#define t0 y0 308#define t1 tt 309 q = 0; 310 y1 = 0; 311 bit = 1 << 31; 312 EVEN_DOUBLE; 313 t1 = bit; 314 FPU_SUBS(d1, x1, t1); 315 FPU_SUBC(d0, x0, t0); /* d = x - t */ 316 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 317 x0 = d0, x1 = d1; /* x -= t */ 318 q = bit; /* q += bit */ 319 y0 |= 1; /* y += bit << 1 */ 320 } 321 ODD_DOUBLE; 322 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 323 EVEN_DOUBLE; /* as before */ 324 t1 = y1 | bit; 325 FPU_SUBS(d1, x1, t1); 326 FPU_SUBC(d0, x0, t0); 327 if ((int)d0 >= 0) { 328 x0 = d0, x1 = d1; 329 q |= bit; 330 y1 |= bit << 1; 331 } 332 ODD_DOUBLE; 333 } 334 x->fp_mant[1] = q; 335#undef t1 336 337 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 338#define t1 y1 339#define t2 tt 340 q = 0; 341 y2 = 0; 342 bit = 1 << 31; 343 EVEN_DOUBLE; 344 t2 = bit; 345 FPU_SUBS(d2, x2, t2); 346 FPU_SUBCS(d1, x1, t1); 347 FPU_SUBC(d0, x0, t0); 348 if ((int)d0 >= 0) { 349 x0 = d0, x1 = d1, x2 = d2; 350 q = bit; 351 y1 |= 1; /* now t1, y1 are set in concrete */ 352 } 353 ODD_DOUBLE; 354 while ((bit >>= 1) != 0) { 355 EVEN_DOUBLE; 356 t2 = y2 | bit; 357 FPU_SUBS(d2, x2, t2); 358 FPU_SUBCS(d1, x1, t1); 359 FPU_SUBC(d0, x0, t0); 360 if ((int)d0 >= 0) { 361 x0 = d0, x1 = d1, x2 = d2; 362 q |= bit; 363 y2 |= bit << 1; 364 } 365 ODD_DOUBLE; 366 } 367 x->fp_mant[2] = q; 368#undef t2 369 370 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 371#define t2 y2 372#define t3 tt 373 q = 0; 374 y3 = 0; 375 bit = 1 << 31; 376 EVEN_DOUBLE; 377 t3 = bit; 378 FPU_SUBS(d3, x3, t3); 379 FPU_SUBCS(d2, x2, t2); 380 FPU_SUBCS(d1, x1, t1); 381 FPU_SUBC(d0, x0, t0); 382 if ((int)d0 >= 0) { 383 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 384 q = bit; 385 y2 |= 1; 386 } 387 ODD_DOUBLE; 388 while ((bit >>= 1) != 0) { 389 EVEN_DOUBLE; 390 t3 = y3 | bit; 391 FPU_SUBS(d3, x3, t3); 392 FPU_SUBCS(d2, x2, t2); 393 FPU_SUBCS(d1, x1, t1); 394 FPU_SUBC(d0, x0, t0); 395 if ((int)d0 >= 0) { 396 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 397 q |= bit; 398 y3 |= bit << 1; 399 } 400 ODD_DOUBLE; 401 } 402 x->fp_mant[3] = q; 403 404 /* 405 * The result, which includes guard and round bits, is exact iff 406 * x is now zero; any nonzero bits in x represent sticky bits. 407 */ 408 x->fp_sticky = x0 | x1 | x2 | x3; 409 DUMPFPN(FPE_REG, x); 410 return (x); 411} 412