1/* $NetBSD: fpu_sqrt.c,v 1.5 2005/12/11 12:17:52 christos Exp $ */ 2 3/* 4 * Copyright (c) 1992, 1993 5 * The Regents of the University of California. All rights reserved. 6 * 7 * This software was developed by the Computer Systems Engineering group 8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 9 * contributed to Berkeley. 10 * 11 * All advertising materials mentioning features or use of this software 12 * must display the following acknowledgement: 13 * This product includes software developed by the University of 14 * California, Lawrence Berkeley Laboratory. 15 * 16 * Redistribution and use in source and binary forms, with or without 17 * modification, are permitted provided that the following conditions 18 * are met: 19 * 1. Redistributions of source code must retain the above copyright 20 * notice, this list of conditions and the following disclaimer. 21 * 2. Redistributions in binary form must reproduce the above copyright 22 * notice, this list of conditions and the following disclaimer in the 23 * documentation and/or other materials provided with the distribution. 24 * 3. Neither the name of the University nor the names of its contributors 25 * may be used to endorse or promote products derived from this software 26 * without specific prior written permission. 27 * 28 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 29 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 30 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 31 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 32 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 33 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 34 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 35 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 36 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 37 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 38 * SUCH DAMAGE. 39 * 40 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93 41 */ 42 43/* 44 * Perform an FPU square root (return sqrt(x)). 45 */ 46 47#include <sys/cdefs.h> 48__KERNEL_RCSID(0, "$NetBSD: fpu_sqrt.c,v 1.5 2005/12/11 12:17:52 christos Exp $"); 49 50#include <sys/types.h> 51 52#include <machine/reg.h> 53 54#include "fpu_arith.h" 55#include "fpu_emulate.h" 56 57/* 58 * Our task is to calculate the square root of a floating point number x0. 59 * This number x normally has the form: 60 * 61 * exp 62 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 63 * 64 * This can be left as it stands, or the mantissa can be doubled and the 65 * exponent decremented: 66 * 67 * exp-1 68 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 69 * 70 * If the exponent `exp' is even, the square root of the number is best 71 * handled using the first form, and is by definition equal to: 72 * 73 * exp/2 74 * sqrt(x) = sqrt(mant) * 2 75 * 76 * If exp is odd, on the other hand, it is convenient to use the second 77 * form, giving: 78 * 79 * (exp-1)/2 80 * sqrt(x) = sqrt(2 * mant) * 2 81 * 82 * In the first case, we have 83 * 84 * 1 <= mant < 2 85 * 86 * and therefore 87 * 88 * sqrt(1) <= sqrt(mant) < sqrt(2) 89 * 90 * while in the second case we have 91 * 92 * 2 <= 2*mant < 4 93 * 94 * and therefore 95 * 96 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 97 * 98 * so that in any case, we are sure that 99 * 100 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 101 * 102 * or 103 * 104 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 105 * 106 * This root is therefore a properly formed mantissa for a floating 107 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 108 * as above. This leaves us with the problem of finding the square root 109 * of a fixed-point number in the range [1..4). 110 * 111 * Though it may not be instantly obvious, the following square root 112 * algorithm works for any integer x of an even number of bits, provided 113 * that no overflows occur: 114 * 115 * let q = 0 116 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 117 * x *= 2 -- multiply by radix, for next digit 118 * if x >= 2q + 2^k then -- if adding 2^k does not 119 * x -= 2q + 2^k -- exceed the correct root, 120 * q += 2^k -- add 2^k and adjust x 121 * fi 122 * done 123 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 124 * 125 * If NBITS is odd (so that k is initially even), we can just add another 126 * zero bit at the top of x. Doing so means that q is not going to acquire 127 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 128 * final value in x is not needed, or can be off by a factor of 2, this is 129 * equivalant to moving the `x *= 2' step to the bottom of the loop: 130 * 131 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 132 * 133 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 134 * (Since the algorithm is destructive on x, we will call x's initial 135 * value, for which q is some power of two times its square root, x0.) 136 * 137 * If we insert a loop invariant y = 2q, we can then rewrite this using 138 * C notation as: 139 * 140 * q = y = 0; x = x0; 141 * for (k = NBITS; --k >= 0;) { 142 * #if (NBITS is even) 143 * x *= 2; 144 * #endif 145 * t = y + (1 << k); 146 * if (x >= t) { 147 * x -= t; 148 * q += 1 << k; 149 * y += 1 << (k + 1); 150 * } 151 * #if (NBITS is odd) 152 * x *= 2; 153 * #endif 154 * } 155 * 156 * If x0 is fixed point, rather than an integer, we can simply alter the 157 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 158 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 159 * 160 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 161 * integers, which adds some complication. But note that q is built one 162 * bit at a time, from the top down, and is not used itself in the loop 163 * (we use 2q as held in y instead). This means we can build our answer 164 * in an integer, one word at a time, which saves a bit of work. Also, 165 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 166 * `new' bits in y and we can set them with an `or' operation rather than 167 * a full-blown multiword add. 168 * 169 * We are almost done, except for one snag. We must prove that none of our 170 * intermediate calculations can overflow. We know that x0 is in [1..4) 171 * and therefore the square root in q will be in [1..2), but what about x, 172 * y, and t? 173 * 174 * We know that y = 2q at the beginning of each loop. (The relation only 175 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 176 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 177 * Furthermore, we can prove with a bit of work that x never exceeds y by 178 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 179 * an exercise to the reader, mostly because I have become tired of working 180 * on this comment.) 181 * 182 * If our floating point mantissas (which are of the form 1.frac) occupy 183 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 184 * In fact, we want even one more bit (for a carry, to avoid compares), or 185 * three extra. There is a comment in fpu_emu.h reminding maintainers of 186 * this, so we have some justification in assuming it. 187 */ 188struct fpn * 189fpu_sqrt(struct fpemu *fe) 190{ 191 register struct fpn *x = &fe->fe_f2; 192 register u_int bit, q, tt; 193 register u_int x0, x1, x2; 194 register u_int y0, y1, y2; 195 register u_int d0, d1, d2; 196 register int e; 197 FPU_DECL_CARRY 198 199 /* 200 * Take care of special cases first. In order: 201 * 202 * sqrt(NaN) = NaN 203 * sqrt(+0) = +0 204 * sqrt(-0) = -0 205 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 206 * sqrt(+Inf) = +Inf 207 * 208 * Then all that remains are numbers with mantissas in [1..2). 209 */ 210 if (ISNAN(x) || ISZERO(x)) 211 return (x); 212 if (x->fp_sign) 213 return (fpu_newnan(fe)); 214 if (ISINF(x)) 215 return (x); 216 217 /* 218 * Calculate result exponent. As noted above, this may involve 219 * doubling the mantissa. We will also need to double x each 220 * time around the loop, so we define a macro for this here, and 221 * we break out the multiword mantissa. 222 */ 223#ifdef FPU_SHL1_BY_ADD 224#define DOUBLE_X { \ 225 FPU_ADDS(x2, x2, x2); \ 226 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 227} 228#else 229#define DOUBLE_X { \ 230 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 231 x2 <<= 1; \ 232} 233#endif 234#if (FP_NMANT & 1) != 0 235# define ODD_DOUBLE DOUBLE_X 236# define EVEN_DOUBLE /* nothing */ 237#else 238# define ODD_DOUBLE /* nothing */ 239# define EVEN_DOUBLE DOUBLE_X 240#endif 241 x0 = x->fp_mant[0]; 242 x1 = x->fp_mant[1]; 243 x2 = x->fp_mant[2]; 244 e = x->fp_exp; 245 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 246 DOUBLE_X; 247 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 248 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 249 250 /* 251 * Now calculate the mantissa root. Since x is now in [1..4), 252 * we know that the first trip around the loop will definitely 253 * set the top bit in q, so we can do that manually and start 254 * the loop at the next bit down instead. We must be sure to 255 * double x correctly while doing the `known q=1.0'. 256 * 257 * We do this one mantissa-word at a time, as noted above, to 258 * save work. To avoid `(1 << 31) << 1', we also do the top bit 259 * outside of each per-word loop. 260 * 261 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 262 * t2 = y2, t? |= bit' for the appropriate word. Since the bit 263 * is always a `new' one, this means that three of the `t?'s are 264 * just the corresponding `y?'; we use `#define's here for this. 265 * The variable `tt' holds the actual `t?' variable. 266 */ 267 268 /* calculate q0 */ 269#define t0 tt 270 bit = FP_1; 271 EVEN_DOUBLE; 272 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 273 q = bit; 274 x0 -= bit; 275 y0 = bit << 1; 276 /* } */ 277 ODD_DOUBLE; 278 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 279 EVEN_DOUBLE; 280 t0 = y0 | bit; /* t = y + bit */ 281 if (x0 >= t0) { /* if x >= t then */ 282 x0 -= t0; /* x -= t */ 283 q |= bit; /* q += bit */ 284 y0 |= bit << 1; /* y += bit << 1 */ 285 } 286 ODD_DOUBLE; 287 } 288 x->fp_mant[0] = q; 289#undef t0 290 291 /* calculate q1. note (y0&1)==0. */ 292#define t0 y0 293#define t1 tt 294 q = 0; 295 y1 = 0; 296 bit = 1 << 31; 297 EVEN_DOUBLE; 298 t1 = bit; 299 FPU_SUBS(d1, x1, t1); 300 FPU_SUBC(d0, x0, t0); /* d = x - t */ 301 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 302 x0 = d0, x1 = d1; /* x -= t */ 303 q = bit; /* q += bit */ 304 y0 |= 1; /* y += bit << 1 */ 305 } 306 ODD_DOUBLE; 307 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 308 EVEN_DOUBLE; /* as before */ 309 t1 = y1 | bit; 310 FPU_SUBS(d1, x1, t1); 311 FPU_SUBC(d0, x0, t0); 312 if ((int)d0 >= 0) { 313 x0 = d0, x1 = d1; 314 q |= bit; 315 y1 |= bit << 1; 316 } 317 ODD_DOUBLE; 318 } 319 x->fp_mant[1] = q; 320#undef t1 321 322 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 323#define t1 y1 324#define t2 tt 325 q = 0; 326 y2 = 0; 327 bit = 1 << 31; 328 EVEN_DOUBLE; 329 t2 = bit; 330 FPU_SUBS(d2, x2, t2); 331 FPU_SUBCS(d1, x1, t1); 332 FPU_SUBC(d0, x0, t0); 333 if ((int)d0 >= 0) { 334 x0 = d0, x1 = d1, x2 = d2; 335 q |= bit; 336 y1 |= 1; /* now t1, y1 are set in concrete */ 337 } 338 ODD_DOUBLE; 339 while ((bit >>= 1) != 0) { 340 EVEN_DOUBLE; 341 t2 = y2 | bit; 342 FPU_SUBS(d2, x2, t2); 343 FPU_SUBCS(d1, x1, t1); 344 FPU_SUBC(d0, x0, t0); 345 if ((int)d0 >= 0) { 346 x0 = d0, x1 = d1, x2 = d2; 347 q |= bit; 348 y2 |= bit << 1; 349 } 350 ODD_DOUBLE; 351 } 352 x->fp_mant[2] = q; 353#undef t2 354 355 /* 356 * The result, which includes guard and round bits, is exact iff 357 * x is now zero; any nonzero bits in x represent sticky bits. 358 */ 359 x->fp_sticky = x0 | x1 | x2; 360 return (x); 361} 362