| 1 | #include <assert.h> |
| 2 | #include <string.h> |
| 3 | #include <stdarg.h> |
| 4 | |
| 5 | #include "puzzles.h" |
| 6 | #include "tree234.h" |
| 7 | #include "maxflow.h" |
| 8 | |
| 9 | #ifdef STANDALONE_LATIN_TEST |
| 10 | #define STANDALONE_SOLVER |
| 11 | #endif |
| 12 | |
| 13 | #include "latin.h" |
| 14 | |
| 15 | /* -------------------------------------------------------- |
| 16 | * Solver. |
| 17 | */ |
| 18 | |
| 19 | #ifdef STANDALONE_SOLVER |
| 20 | int solver_show_working, solver_recurse_depth; |
| 21 | #endif |
| 22 | |
| 23 | /* |
| 24 | * Function called when we are certain that a particular square has |
| 25 | * a particular number in it. The y-coordinate passed in here is |
| 26 | * transformed. |
| 27 | */ |
| 28 | void latin_solver_place(struct latin_solver *solver, int x, int y, int n) |
| 29 | { |
| 30 | int i, o = solver->o; |
| 31 | |
| 32 | assert(n <= o); |
| 33 | assert(cube(x,y,n)); |
| 34 | |
| 35 | /* |
| 36 | * Rule out all other numbers in this square. |
| 37 | */ |
| 38 | for (i = 1; i <= o; i++) |
| 39 | if (i != n) |
| 40 | cube(x,y,i) = FALSE; |
| 41 | |
| 42 | /* |
| 43 | * Rule out this number in all other positions in the row. |
| 44 | */ |
| 45 | for (i = 0; i < o; i++) |
| 46 | if (i != y) |
| 47 | cube(x,i,n) = FALSE; |
| 48 | |
| 49 | /* |
| 50 | * Rule out this number in all other positions in the column. |
| 51 | */ |
| 52 | for (i = 0; i < o; i++) |
| 53 | if (i != x) |
| 54 | cube(i,y,n) = FALSE; |
| 55 | |
| 56 | /* |
| 57 | * Enter the number in the result grid. |
| 58 | */ |
| 59 | solver->grid[YUNTRANS(y)*o+x] = n; |
| 60 | |
| 61 | /* |
| 62 | * Cross out this number from the list of numbers left to place |
| 63 | * in its row, its column and its block. |
| 64 | */ |
| 65 | solver->row[y*o+n-1] = solver->col[x*o+n-1] = TRUE; |
| 66 | } |
| 67 | |
| 68 | int latin_solver_elim(struct latin_solver *solver, int start, int step |
| 69 | #ifdef STANDALONE_SOLVER |
| 70 | , char *fmt, ... |
| 71 | #endif |
| 72 | ) |
| 73 | { |
| 74 | int o = solver->o; |
| 75 | int fpos, m, i; |
| 76 | |
| 77 | /* |
| 78 | * Count the number of set bits within this section of the |
| 79 | * cube. |
| 80 | */ |
| 81 | m = 0; |
| 82 | fpos = -1; |
| 83 | for (i = 0; i < o; i++) |
| 84 | if (solver->cube[start+i*step]) { |
| 85 | fpos = start+i*step; |
| 86 | m++; |
| 87 | } |
| 88 | |
| 89 | if (m == 1) { |
| 90 | int x, y, n; |
| 91 | assert(fpos >= 0); |
| 92 | |
| 93 | n = 1 + fpos % o; |
| 94 | y = fpos / o; |
| 95 | x = y / o; |
| 96 | y %= o; |
| 97 | |
| 98 | if (!solver->grid[YUNTRANS(y)*o+x]) { |
| 99 | #ifdef STANDALONE_SOLVER |
| 100 | if (solver_show_working) { |
| 101 | va_list ap; |
| 102 | printf("%*s", solver_recurse_depth*4, ""); |
| 103 | va_start(ap, fmt); |
| 104 | vprintf(fmt, ap); |
| 105 | va_end(ap); |
| 106 | printf(":\n%*s placing %d at (%d,%d)\n", |
| 107 | solver_recurse_depth*4, "", n, x+1, YUNTRANS(y)+1); |
| 108 | } |
| 109 | #endif |
| 110 | latin_solver_place(solver, x, y, n); |
| 111 | return +1; |
| 112 | } |
| 113 | } else if (m == 0) { |
| 114 | #ifdef STANDALONE_SOLVER |
| 115 | if (solver_show_working) { |
| 116 | va_list ap; |
| 117 | printf("%*s", solver_recurse_depth*4, ""); |
| 118 | va_start(ap, fmt); |
| 119 | vprintf(fmt, ap); |
| 120 | va_end(ap); |
| 121 | printf(":\n%*s no possibilities available\n", |
| 122 | solver_recurse_depth*4, ""); |
| 123 | } |
| 124 | #endif |
| 125 | return -1; |
| 126 | } |
| 127 | |
| 128 | return 0; |
| 129 | } |
| 130 | |
| 131 | struct latin_solver_scratch { |
| 132 | unsigned char *grid, *rowidx, *colidx, *set; |
| 133 | int *neighbours, *bfsqueue; |
| 134 | #ifdef STANDALONE_SOLVER |
| 135 | int *bfsprev; |
| 136 | #endif |
| 137 | }; |
| 138 | |
| 139 | int latin_solver_set(struct latin_solver *solver, |
| 140 | struct latin_solver_scratch *scratch, |
| 141 | int start, int step1, int step2 |
| 142 | #ifdef STANDALONE_SOLVER |
| 143 | , char *fmt, ... |
| 144 | #endif |
| 145 | ) |
| 146 | { |
| 147 | int o = solver->o; |
| 148 | int i, j, n, count; |
| 149 | unsigned char *grid = scratch->grid; |
| 150 | unsigned char *rowidx = scratch->rowidx; |
| 151 | unsigned char *colidx = scratch->colidx; |
| 152 | unsigned char *set = scratch->set; |
| 153 | |
| 154 | /* |
| 155 | * We are passed a o-by-o matrix of booleans. Our first job |
| 156 | * is to winnow it by finding any definite placements - i.e. |
| 157 | * any row with a solitary 1 - and discarding that row and the |
| 158 | * column containing the 1. |
| 159 | */ |
| 160 | memset(rowidx, TRUE, o); |
| 161 | memset(colidx, TRUE, o); |
| 162 | for (i = 0; i < o; i++) { |
| 163 | int count = 0, first = -1; |
| 164 | for (j = 0; j < o; j++) |
| 165 | if (solver->cube[start+i*step1+j*step2]) |
| 166 | first = j, count++; |
| 167 | |
| 168 | if (count == 0) return -1; |
| 169 | if (count == 1) |
| 170 | rowidx[i] = colidx[first] = FALSE; |
| 171 | } |
| 172 | |
| 173 | /* |
| 174 | * Convert each of rowidx/colidx from a list of 0s and 1s to a |
| 175 | * list of the indices of the 1s. |
| 176 | */ |
| 177 | for (i = j = 0; i < o; i++) |
| 178 | if (rowidx[i]) |
| 179 | rowidx[j++] = i; |
| 180 | n = j; |
| 181 | for (i = j = 0; i < o; i++) |
| 182 | if (colidx[i]) |
| 183 | colidx[j++] = i; |
| 184 | assert(n == j); |
| 185 | |
| 186 | /* |
| 187 | * And create the smaller matrix. |
| 188 | */ |
| 189 | for (i = 0; i < n; i++) |
| 190 | for (j = 0; j < n; j++) |
| 191 | grid[i*o+j] = solver->cube[start+rowidx[i]*step1+colidx[j]*step2]; |
| 192 | |
| 193 | /* |
| 194 | * Having done that, we now have a matrix in which every row |
| 195 | * has at least two 1s in. Now we search to see if we can find |
| 196 | * a rectangle of zeroes (in the set-theoretic sense of |
| 197 | * `rectangle', i.e. a subset of rows crossed with a subset of |
| 198 | * columns) whose width and height add up to n. |
| 199 | */ |
| 200 | |
| 201 | memset(set, 0, n); |
| 202 | count = 0; |
| 203 | while (1) { |
| 204 | /* |
| 205 | * We have a candidate set. If its size is <=1 or >=n-1 |
| 206 | * then we move on immediately. |
| 207 | */ |
| 208 | if (count > 1 && count < n-1) { |
| 209 | /* |
| 210 | * The number of rows we need is n-count. See if we can |
| 211 | * find that many rows which each have a zero in all |
| 212 | * the positions listed in `set'. |
| 213 | */ |
| 214 | int rows = 0; |
| 215 | for (i = 0; i < n; i++) { |
| 216 | int ok = TRUE; |
| 217 | for (j = 0; j < n; j++) |
| 218 | if (set[j] && grid[i*o+j]) { |
| 219 | ok = FALSE; |
| 220 | break; |
| 221 | } |
| 222 | if (ok) |
| 223 | rows++; |
| 224 | } |
| 225 | |
| 226 | /* |
| 227 | * We expect never to be able to get _more_ than |
| 228 | * n-count suitable rows: this would imply that (for |
| 229 | * example) there are four numbers which between them |
| 230 | * have at most three possible positions, and hence it |
| 231 | * indicates a faulty deduction before this point or |
| 232 | * even a bogus clue. |
| 233 | */ |
| 234 | if (rows > n - count) { |
| 235 | #ifdef STANDALONE_SOLVER |
| 236 | if (solver_show_working) { |
| 237 | va_list ap; |
| 238 | printf("%*s", solver_recurse_depth*4, |
| 239 | ""); |
| 240 | va_start(ap, fmt); |
| 241 | vprintf(fmt, ap); |
| 242 | va_end(ap); |
| 243 | printf(":\n%*s contradiction reached\n", |
| 244 | solver_recurse_depth*4, ""); |
| 245 | } |
| 246 | #endif |
| 247 | return -1; |
| 248 | } |
| 249 | |
| 250 | if (rows >= n - count) { |
| 251 | int progress = FALSE; |
| 252 | |
| 253 | /* |
| 254 | * We've got one! Now, for each row which _doesn't_ |
| 255 | * satisfy the criterion, eliminate all its set |
| 256 | * bits in the positions _not_ listed in `set'. |
| 257 | * Return +1 (meaning progress has been made) if we |
| 258 | * successfully eliminated anything at all. |
| 259 | * |
| 260 | * This involves referring back through |
| 261 | * rowidx/colidx in order to work out which actual |
| 262 | * positions in the cube to meddle with. |
| 263 | */ |
| 264 | for (i = 0; i < n; i++) { |
| 265 | int ok = TRUE; |
| 266 | for (j = 0; j < n; j++) |
| 267 | if (set[j] && grid[i*o+j]) { |
| 268 | ok = FALSE; |
| 269 | break; |
| 270 | } |
| 271 | if (!ok) { |
| 272 | for (j = 0; j < n; j++) |
| 273 | if (!set[j] && grid[i*o+j]) { |
| 274 | int fpos = (start+rowidx[i]*step1+ |
| 275 | colidx[j]*step2); |
| 276 | #ifdef STANDALONE_SOLVER |
| 277 | if (solver_show_working) { |
| 278 | int px, py, pn; |
| 279 | |
| 280 | if (!progress) { |
| 281 | va_list ap; |
| 282 | printf("%*s", solver_recurse_depth*4, |
| 283 | ""); |
| 284 | va_start(ap, fmt); |
| 285 | vprintf(fmt, ap); |
| 286 | va_end(ap); |
| 287 | printf(":\n"); |
| 288 | } |
| 289 | |
| 290 | pn = 1 + fpos % o; |
| 291 | py = fpos / o; |
| 292 | px = py / o; |
| 293 | py %= o; |
| 294 | |
| 295 | printf("%*s ruling out %d at (%d,%d)\n", |
| 296 | solver_recurse_depth*4, "", |
| 297 | pn, px+1, YUNTRANS(py)+1); |
| 298 | } |
| 299 | #endif |
| 300 | progress = TRUE; |
| 301 | solver->cube[fpos] = FALSE; |
| 302 | } |
| 303 | } |
| 304 | } |
| 305 | |
| 306 | if (progress) { |
| 307 | return +1; |
| 308 | } |
| 309 | } |
| 310 | } |
| 311 | |
| 312 | /* |
| 313 | * Binary increment: change the rightmost 0 to a 1, and |
| 314 | * change all 1s to the right of it to 0s. |
| 315 | */ |
| 316 | i = n; |
| 317 | while (i > 0 && set[i-1]) |
| 318 | set[--i] = 0, count--; |
| 319 | if (i > 0) |
| 320 | set[--i] = 1, count++; |
| 321 | else |
| 322 | break; /* done */ |
| 323 | } |
| 324 | |
| 325 | return 0; |
| 326 | } |
| 327 | |
| 328 | /* |
| 329 | * Look for forcing chains. A forcing chain is a path of |
| 330 | * pairwise-exclusive squares (i.e. each pair of adjacent squares |
| 331 | * in the path are in the same row, column or block) with the |
| 332 | * following properties: |
| 333 | * |
| 334 | * (a) Each square on the path has precisely two possible numbers. |
| 335 | * |
| 336 | * (b) Each pair of squares which are adjacent on the path share |
| 337 | * at least one possible number in common. |
| 338 | * |
| 339 | * (c) Each square in the middle of the path shares _both_ of its |
| 340 | * numbers with at least one of its neighbours (not the same |
| 341 | * one with both neighbours). |
| 342 | * |
| 343 | * These together imply that at least one of the possible number |
| 344 | * choices at one end of the path forces _all_ the rest of the |
| 345 | * numbers along the path. In order to make real use of this, we |
| 346 | * need further properties: |
| 347 | * |
| 348 | * (c) Ruling out some number N from the square at one end |
| 349 | * of the path forces the square at the other end to |
| 350 | * take number N. |
| 351 | * |
| 352 | * (d) The two end squares are both in line with some third |
| 353 | * square. |
| 354 | * |
| 355 | * (e) That third square currently has N as a possibility. |
| 356 | * |
| 357 | * If we can find all of that lot, we can deduce that at least one |
| 358 | * of the two ends of the forcing chain has number N, and that |
| 359 | * therefore the mutually adjacent third square does not. |
| 360 | * |
| 361 | * To find forcing chains, we're going to start a bfs at each |
| 362 | * suitable square, once for each of its two possible numbers. |
| 363 | */ |
| 364 | int latin_solver_forcing(struct latin_solver *solver, |
| 365 | struct latin_solver_scratch *scratch) |
| 366 | { |
| 367 | int o = solver->o; |
| 368 | int *bfsqueue = scratch->bfsqueue; |
| 369 | #ifdef STANDALONE_SOLVER |
| 370 | int *bfsprev = scratch->bfsprev; |
| 371 | #endif |
| 372 | unsigned char *number = scratch->grid; |
| 373 | int *neighbours = scratch->neighbours; |
| 374 | int x, y; |
| 375 | |
| 376 | for (y = 0; y < o; y++) |
| 377 | for (x = 0; x < o; x++) { |
| 378 | int count, t, n; |
| 379 | |
| 380 | /* |
| 381 | * If this square doesn't have exactly two candidate |
| 382 | * numbers, don't try it. |
| 383 | * |
| 384 | * In this loop we also sum the candidate numbers, |
| 385 | * which is a nasty hack to allow us to quickly find |
| 386 | * `the other one' (since we will shortly know there |
| 387 | * are exactly two). |
| 388 | */ |
| 389 | for (count = t = 0, n = 1; n <= o; n++) |
| 390 | if (cube(x, y, n)) |
| 391 | count++, t += n; |
| 392 | if (count != 2) |
| 393 | continue; |
| 394 | |
| 395 | /* |
| 396 | * Now attempt a bfs for each candidate. |
| 397 | */ |
| 398 | for (n = 1; n <= o; n++) |
| 399 | if (cube(x, y, n)) { |
| 400 | int orign, currn, head, tail; |
| 401 | |
| 402 | /* |
| 403 | * Begin a bfs. |
| 404 | */ |
| 405 | orign = n; |
| 406 | |
| 407 | memset(number, o+1, o*o); |
| 408 | head = tail = 0; |
| 409 | bfsqueue[tail++] = y*o+x; |
| 410 | #ifdef STANDALONE_SOLVER |
| 411 | bfsprev[y*o+x] = -1; |
| 412 | #endif |
| 413 | number[y*o+x] = t - n; |
| 414 | |
| 415 | while (head < tail) { |
| 416 | int xx, yy, nneighbours, xt, yt, i; |
| 417 | |
| 418 | xx = bfsqueue[head++]; |
| 419 | yy = xx / o; |
| 420 | xx %= o; |
| 421 | |
| 422 | currn = number[yy*o+xx]; |
| 423 | |
| 424 | /* |
| 425 | * Find neighbours of yy,xx. |
| 426 | */ |
| 427 | nneighbours = 0; |
| 428 | for (yt = 0; yt < o; yt++) |
| 429 | neighbours[nneighbours++] = yt*o+xx; |
| 430 | for (xt = 0; xt < o; xt++) |
| 431 | neighbours[nneighbours++] = yy*o+xt; |
| 432 | |
| 433 | /* |
| 434 | * Try visiting each of those neighbours. |
| 435 | */ |
| 436 | for (i = 0; i < nneighbours; i++) { |
| 437 | int cc, tt, nn; |
| 438 | |
| 439 | xt = neighbours[i] % o; |
| 440 | yt = neighbours[i] / o; |
| 441 | |
| 442 | /* |
| 443 | * We need this square to not be |
| 444 | * already visited, and to include |
| 445 | * currn as a possible number. |
| 446 | */ |
| 447 | if (number[yt*o+xt] <= o) |
| 448 | continue; |
| 449 | if (!cube(xt, yt, currn)) |
| 450 | continue; |
| 451 | |
| 452 | /* |
| 453 | * Don't visit _this_ square a second |
| 454 | * time! |
| 455 | */ |
| 456 | if (xt == xx && yt == yy) |
| 457 | continue; |
| 458 | |
| 459 | /* |
| 460 | * To continue with the bfs, we need |
| 461 | * this square to have exactly two |
| 462 | * possible numbers. |
| 463 | */ |
| 464 | for (cc = tt = 0, nn = 1; nn <= o; nn++) |
| 465 | if (cube(xt, yt, nn)) |
| 466 | cc++, tt += nn; |
| 467 | if (cc == 2) { |
| 468 | bfsqueue[tail++] = yt*o+xt; |
| 469 | #ifdef STANDALONE_SOLVER |
| 470 | bfsprev[yt*o+xt] = yy*o+xx; |
| 471 | #endif |
| 472 | number[yt*o+xt] = tt - currn; |
| 473 | } |
| 474 | |
| 475 | /* |
| 476 | * One other possibility is that this |
| 477 | * might be the square in which we can |
| 478 | * make a real deduction: if it's |
| 479 | * adjacent to x,y, and currn is equal |
| 480 | * to the original number we ruled out. |
| 481 | */ |
| 482 | if (currn == orign && |
| 483 | (xt == x || yt == y)) { |
| 484 | #ifdef STANDALONE_SOLVER |
| 485 | if (solver_show_working) { |
| 486 | char *sep = ""; |
| 487 | int xl, yl; |
| 488 | printf("%*sforcing chain, %d at ends of ", |
| 489 | solver_recurse_depth*4, "", orign); |
| 490 | xl = xx; |
| 491 | yl = yy; |
| 492 | while (1) { |
| 493 | printf("%s(%d,%d)", sep, xl+1, |
| 494 | YUNTRANS(yl)+1); |
| 495 | xl = bfsprev[yl*o+xl]; |
| 496 | if (xl < 0) |
| 497 | break; |
| 498 | yl = xl / o; |
| 499 | xl %= o; |
| 500 | sep = "-"; |
| 501 | } |
| 502 | printf("\n%*s ruling out %d at (%d,%d)\n", |
| 503 | solver_recurse_depth*4, "", |
| 504 | orign, xt+1, YUNTRANS(yt)+1); |
| 505 | } |
| 506 | #endif |
| 507 | cube(xt, yt, orign) = FALSE; |
| 508 | return 1; |
| 509 | } |
| 510 | } |
| 511 | } |
| 512 | } |
| 513 | } |
| 514 | |
| 515 | return 0; |
| 516 | } |
| 517 | |
| 518 | struct latin_solver_scratch *latin_solver_new_scratch(struct latin_solver *solver) |
| 519 | { |
| 520 | struct latin_solver_scratch *scratch = snew(struct latin_solver_scratch); |
| 521 | int o = solver->o; |
| 522 | scratch->grid = snewn(o*o, unsigned char); |
| 523 | scratch->rowidx = snewn(o, unsigned char); |
| 524 | scratch->colidx = snewn(o, unsigned char); |
| 525 | scratch->set = snewn(o, unsigned char); |
| 526 | scratch->neighbours = snewn(3*o, int); |
| 527 | scratch->bfsqueue = snewn(o*o, int); |
| 528 | #ifdef STANDALONE_SOLVER |
| 529 | scratch->bfsprev = snewn(o*o, int); |
| 530 | #endif |
| 531 | return scratch; |
| 532 | } |
| 533 | |
| 534 | void latin_solver_free_scratch(struct latin_solver_scratch *scratch) |
| 535 | { |
| 536 | #ifdef STANDALONE_SOLVER |
| 537 | sfree(scratch->bfsprev); |
| 538 | #endif |
| 539 | sfree(scratch->bfsqueue); |
| 540 | sfree(scratch->neighbours); |
| 541 | sfree(scratch->set); |
| 542 | sfree(scratch->colidx); |
| 543 | sfree(scratch->rowidx); |
| 544 | sfree(scratch->grid); |
| 545 | sfree(scratch); |
| 546 | } |
| 547 | |
| 548 | void latin_solver_alloc(struct latin_solver *solver, digit *grid, int o) |
| 549 | { |
| 550 | int x, y; |
| 551 | |
| 552 | solver->o = o; |
| 553 | solver->cube = snewn(o*o*o, unsigned char); |
| 554 | solver->grid = grid; /* write straight back to the input */ |
| 555 | memset(solver->cube, TRUE, o*o*o); |
| 556 | |
| 557 | solver->row = snewn(o*o, unsigned char); |
| 558 | solver->col = snewn(o*o, unsigned char); |
| 559 | memset(solver->row, FALSE, o*o); |
| 560 | memset(solver->col, FALSE, o*o); |
| 561 | |
| 562 | for (x = 0; x < o; x++) |
| 563 | for (y = 0; y < o; y++) |
| 564 | if (grid[y*o+x]) |
| 565 | latin_solver_place(solver, x, YTRANS(y), grid[y*o+x]); |
| 566 | } |
| 567 | |
| 568 | void latin_solver_free(struct latin_solver *solver) |
| 569 | { |
| 570 | sfree(solver->cube); |
| 571 | sfree(solver->row); |
| 572 | sfree(solver->col); |
| 573 | } |
| 574 | |
| 575 | int latin_solver_diff_simple(struct latin_solver *solver) |
| 576 | { |
| 577 | int x, y, n, ret, o = solver->o; |
| 578 | /* |
| 579 | * Row-wise positional elimination. |
| 580 | */ |
| 581 | for (y = 0; y < o; y++) |
| 582 | for (n = 1; n <= o; n++) |
| 583 | if (!solver->row[y*o+n-1]) { |
| 584 | ret = latin_solver_elim(solver, cubepos(0,y,n), o*o |
| 585 | #ifdef STANDALONE_SOLVER |
| 586 | , "positional elimination," |
| 587 | " %d in row %d", n, YUNTRANS(y)+1 |
| 588 | #endif |
| 589 | ); |
| 590 | if (ret != 0) return ret; |
| 591 | } |
| 592 | /* |
| 593 | * Column-wise positional elimination. |
| 594 | */ |
| 595 | for (x = 0; x < o; x++) |
| 596 | for (n = 1; n <= o; n++) |
| 597 | if (!solver->col[x*o+n-1]) { |
| 598 | ret = latin_solver_elim(solver, cubepos(x,0,n), o |
| 599 | #ifdef STANDALONE_SOLVER |
| 600 | , "positional elimination," |
| 601 | " %d in column %d", n, x+1 |
| 602 | #endif |
| 603 | ); |
| 604 | if (ret != 0) return ret; |
| 605 | } |
| 606 | |
| 607 | /* |
| 608 | * Numeric elimination. |
| 609 | */ |
| 610 | for (x = 0; x < o; x++) |
| 611 | for (y = 0; y < o; y++) |
| 612 | if (!solver->grid[YUNTRANS(y)*o+x]) { |
| 613 | ret = latin_solver_elim(solver, cubepos(x,y,1), 1 |
| 614 | #ifdef STANDALONE_SOLVER |
| 615 | , "numeric elimination at (%d,%d)", |
| 616 | x+1, YUNTRANS(y)+1 |
| 617 | #endif |
| 618 | ); |
| 619 | if (ret != 0) return ret; |
| 620 | } |
| 621 | return 0; |
| 622 | } |
| 623 | |
| 624 | int latin_solver_diff_set(struct latin_solver *solver, |
| 625 | struct latin_solver_scratch *scratch, |
| 626 | int extreme) |
| 627 | { |
| 628 | int x, y, n, ret, o = solver->o; |
| 629 | |
| 630 | if (!extreme) { |
| 631 | /* |
| 632 | * Row-wise set elimination. |
| 633 | */ |
| 634 | for (y = 0; y < o; y++) { |
| 635 | ret = latin_solver_set(solver, scratch, cubepos(0,y,1), o*o, 1 |
| 636 | #ifdef STANDALONE_SOLVER |
| 637 | , "set elimination, row %d", YUNTRANS(y)+1 |
| 638 | #endif |
| 639 | ); |
| 640 | if (ret != 0) return ret; |
| 641 | } |
| 642 | /* |
| 643 | * Column-wise set elimination. |
| 644 | */ |
| 645 | for (x = 0; x < o; x++) { |
| 646 | ret = latin_solver_set(solver, scratch, cubepos(x,0,1), o, 1 |
| 647 | #ifdef STANDALONE_SOLVER |
| 648 | , "set elimination, column %d", x+1 |
| 649 | #endif |
| 650 | ); |
| 651 | if (ret != 0) return ret; |
| 652 | } |
| 653 | } else { |
| 654 | /* |
| 655 | * Row-vs-column set elimination on a single number |
| 656 | * (much tricker for a human to do!) |
| 657 | */ |
| 658 | for (n = 1; n <= o; n++) { |
| 659 | ret = latin_solver_set(solver, scratch, cubepos(0,0,n), o*o, o |
| 660 | #ifdef STANDALONE_SOLVER |
| 661 | , "positional set elimination, number %d", n |
| 662 | #endif |
| 663 | ); |
| 664 | if (ret != 0) return ret; |
| 665 | } |
| 666 | } |
| 667 | return 0; |
| 668 | } |
| 669 | |
| 670 | /* |
| 671 | * Returns: |
| 672 | * 0 for 'didn't do anything' implying it was already solved. |
| 673 | * -1 for 'impossible' (no solution) |
| 674 | * 1 for 'single solution' |
| 675 | * >1 for 'multiple solutions' (you don't get to know how many, and |
| 676 | * the first such solution found will be set. |
| 677 | * |
| 678 | * and this function may well assert if given an impossible board. |
| 679 | */ |
| 680 | static int latin_solver_recurse |
| 681 | (struct latin_solver *solver, int diff_simple, int diff_set_0, |
| 682 | int diff_set_1, int diff_forcing, int diff_recursive, |
| 683 | usersolver_t const *usersolvers, void *ctx, |
| 684 | ctxnew_t ctxnew, ctxfree_t ctxfree) |
| 685 | { |
| 686 | int best, bestcount; |
| 687 | int o = solver->o, x, y, n; |
| 688 | |
| 689 | best = -1; |
| 690 | bestcount = o+1; |
| 691 | |
| 692 | for (y = 0; y < o; y++) |
| 693 | for (x = 0; x < o; x++) |
| 694 | if (!solver->grid[y*o+x]) { |
| 695 | int count; |
| 696 | |
| 697 | /* |
| 698 | * An unfilled square. Count the number of |
| 699 | * possible digits in it. |
| 700 | */ |
| 701 | count = 0; |
| 702 | for (n = 1; n <= o; n++) |
| 703 | if (cube(x,YTRANS(y),n)) |
| 704 | count++; |
| 705 | |
| 706 | /* |
| 707 | * We should have found any impossibilities |
| 708 | * already, so this can safely be an assert. |
| 709 | */ |
| 710 | assert(count > 1); |
| 711 | |
| 712 | if (count < bestcount) { |
| 713 | bestcount = count; |
| 714 | best = y*o+x; |
| 715 | } |
| 716 | } |
| 717 | |
| 718 | if (best == -1) |
| 719 | /* we were complete already. */ |
| 720 | return 0; |
| 721 | else { |
| 722 | int i, j; |
| 723 | digit *list, *ingrid, *outgrid; |
| 724 | int diff = diff_impossible; /* no solution found yet */ |
| 725 | |
| 726 | /* |
| 727 | * Attempt recursion. |
| 728 | */ |
| 729 | y = best / o; |
| 730 | x = best % o; |
| 731 | |
| 732 | list = snewn(o, digit); |
| 733 | ingrid = snewn(o*o, digit); |
| 734 | outgrid = snewn(o*o, digit); |
| 735 | memcpy(ingrid, solver->grid, o*o); |
| 736 | |
| 737 | /* Make a list of the possible digits. */ |
| 738 | for (j = 0, n = 1; n <= o; n++) |
| 739 | if (cube(x,YTRANS(y),n)) |
| 740 | list[j++] = n; |
| 741 | |
| 742 | #ifdef STANDALONE_SOLVER |
| 743 | if (solver_show_working) { |
| 744 | char *sep = ""; |
| 745 | printf("%*srecursing on (%d,%d) [", |
| 746 | solver_recurse_depth*4, "", x+1, y+1); |
| 747 | for (i = 0; i < j; i++) { |
| 748 | printf("%s%d", sep, list[i]); |
| 749 | sep = " or "; |
| 750 | } |
| 751 | printf("]\n"); |
| 752 | } |
| 753 | #endif |
| 754 | |
| 755 | /* |
| 756 | * And step along the list, recursing back into the |
| 757 | * main solver at every stage. |
| 758 | */ |
| 759 | for (i = 0; i < j; i++) { |
| 760 | int ret; |
| 761 | void *newctx; |
| 762 | |
| 763 | memcpy(outgrid, ingrid, o*o); |
| 764 | outgrid[y*o+x] = list[i]; |
| 765 | |
| 766 | #ifdef STANDALONE_SOLVER |
| 767 | if (solver_show_working) |
| 768 | printf("%*sguessing %d at (%d,%d)\n", |
| 769 | solver_recurse_depth*4, "", list[i], x+1, y+1); |
| 770 | solver_recurse_depth++; |
| 771 | #endif |
| 772 | |
| 773 | if (ctxnew) { |
| 774 | newctx = ctxnew(ctx); |
| 775 | } else { |
| 776 | newctx = ctx; |
| 777 | } |
| 778 | ret = latin_solver(outgrid, o, diff_recursive, |
| 779 | diff_simple, diff_set_0, diff_set_1, |
| 780 | diff_forcing, diff_recursive, |
| 781 | usersolvers, newctx, ctxnew, ctxfree); |
| 782 | if (ctxnew) |
| 783 | ctxfree(newctx); |
| 784 | |
| 785 | #ifdef STANDALONE_SOLVER |
| 786 | solver_recurse_depth--; |
| 787 | if (solver_show_working) { |
| 788 | printf("%*sretracting %d at (%d,%d)\n", |
| 789 | solver_recurse_depth*4, "", list[i], x+1, y+1); |
| 790 | } |
| 791 | #endif |
| 792 | /* we recurse as deep as we can, so we should never find |
| 793 | * find ourselves giving up on a puzzle without declaring it |
| 794 | * impossible. */ |
| 795 | assert(ret != diff_unfinished); |
| 796 | |
| 797 | /* |
| 798 | * If we have our first solution, copy it into the |
| 799 | * grid we will return. |
| 800 | */ |
| 801 | if (diff == diff_impossible && ret != diff_impossible) |
| 802 | memcpy(solver->grid, outgrid, o*o); |
| 803 | |
| 804 | if (ret == diff_ambiguous) |
| 805 | diff = diff_ambiguous; |
| 806 | else if (ret == diff_impossible) |
| 807 | /* do not change our return value */; |
| 808 | else { |
| 809 | /* the recursion turned up exactly one solution */ |
| 810 | if (diff == diff_impossible) |
| 811 | diff = diff_recursive; |
| 812 | else |
| 813 | diff = diff_ambiguous; |
| 814 | } |
| 815 | |
| 816 | /* |
| 817 | * As soon as we've found more than one solution, |
| 818 | * give up immediately. |
| 819 | */ |
| 820 | if (diff == diff_ambiguous) |
| 821 | break; |
| 822 | } |
| 823 | |
| 824 | sfree(outgrid); |
| 825 | sfree(ingrid); |
| 826 | sfree(list); |
| 827 | |
| 828 | if (diff == diff_impossible) |
| 829 | return -1; |
| 830 | else if (diff == diff_ambiguous) |
| 831 | return 2; |
| 832 | else { |
| 833 | assert(diff == diff_recursive); |
| 834 | return 1; |
| 835 | } |
| 836 | } |
| 837 | } |
| 838 | |
| 839 | int latin_solver_main(struct latin_solver *solver, int maxdiff, |
| 840 | int diff_simple, int diff_set_0, int diff_set_1, |
| 841 | int diff_forcing, int diff_recursive, |
| 842 | usersolver_t const *usersolvers, void *ctx, |
| 843 | ctxnew_t ctxnew, ctxfree_t ctxfree) |
| 844 | { |
| 845 | struct latin_solver_scratch *scratch = latin_solver_new_scratch(solver); |
| 846 | int ret, diff = diff_simple; |
| 847 | |
| 848 | assert(maxdiff <= diff_recursive); |
| 849 | /* |
| 850 | * Now loop over the grid repeatedly trying all permitted modes |
| 851 | * of reasoning. The loop terminates if we complete an |
| 852 | * iteration without making any progress; we then return |
| 853 | * failure or success depending on whether the grid is full or |
| 854 | * not. |
| 855 | */ |
| 856 | while (1) { |
| 857 | int i; |
| 858 | |
| 859 | cont: |
| 860 | |
| 861 | latin_solver_debug(solver->cube, solver->o); |
| 862 | |
| 863 | for (i = 0; i <= maxdiff; i++) { |
| 864 | if (usersolvers[i]) |
| 865 | ret = usersolvers[i](solver, ctx); |
| 866 | else |
| 867 | ret = 0; |
| 868 | if (ret == 0 && i == diff_simple) |
| 869 | ret = latin_solver_diff_simple(solver); |
| 870 | if (ret == 0 && i == diff_set_0) |
| 871 | ret = latin_solver_diff_set(solver, scratch, 0); |
| 872 | if (ret == 0 && i == diff_set_1) |
| 873 | ret = latin_solver_diff_set(solver, scratch, 1); |
| 874 | if (ret == 0 && i == diff_forcing) |
| 875 | ret = latin_solver_forcing(solver, scratch); |
| 876 | |
| 877 | if (ret < 0) { |
| 878 | diff = diff_impossible; |
| 879 | goto got_result; |
| 880 | } else if (ret > 0) { |
| 881 | diff = max(diff, i); |
| 882 | goto cont; |
| 883 | } |
| 884 | } |
| 885 | |
| 886 | /* |
| 887 | * If we reach here, we have made no deductions in this |
| 888 | * iteration, so the algorithm terminates. |
| 889 | */ |
| 890 | break; |
| 891 | } |
| 892 | |
| 893 | /* |
| 894 | * Last chance: if we haven't fully solved the puzzle yet, try |
| 895 | * recursing based on guesses for a particular square. We pick |
| 896 | * one of the most constrained empty squares we can find, which |
| 897 | * has the effect of pruning the search tree as much as |
| 898 | * possible. |
| 899 | */ |
| 900 | if (maxdiff == diff_recursive) { |
| 901 | int nsol = latin_solver_recurse(solver, |
| 902 | diff_simple, diff_set_0, diff_set_1, |
| 903 | diff_forcing, diff_recursive, |
| 904 | usersolvers, ctx, ctxnew, ctxfree); |
| 905 | if (nsol < 0) diff = diff_impossible; |
| 906 | else if (nsol == 1) diff = diff_recursive; |
| 907 | else if (nsol > 1) diff = diff_ambiguous; |
| 908 | /* if nsol == 0 then we were complete anyway |
| 909 | * (and thus don't need to change diff) */ |
| 910 | } else { |
| 911 | /* |
| 912 | * We're forbidden to use recursion, so we just see whether |
| 913 | * our grid is fully solved, and return diff_unfinished |
| 914 | * otherwise. |
| 915 | */ |
| 916 | int x, y, o = solver->o; |
| 917 | |
| 918 | for (y = 0; y < o; y++) |
| 919 | for (x = 0; x < o; x++) |
| 920 | if (!solver->grid[y*o+x]) |
| 921 | diff = diff_unfinished; |
| 922 | } |
| 923 | |
| 924 | got_result: |
| 925 | |
| 926 | #ifdef STANDALONE_SOLVER |
| 927 | if (solver_show_working) |
| 928 | printf("%*s%s found\n", |
| 929 | solver_recurse_depth*4, "", |
| 930 | diff == diff_impossible ? "no solution (impossible)" : |
| 931 | diff == diff_unfinished ? "no solution (unfinished)" : |
| 932 | diff == diff_ambiguous ? "multiple solutions" : |
| 933 | "one solution"); |
| 934 | #endif |
| 935 | |
| 936 | latin_solver_free_scratch(scratch); |
| 937 | |
| 938 | return diff; |
| 939 | } |
| 940 | |
| 941 | int latin_solver(digit *grid, int o, int maxdiff, |
| 942 | int diff_simple, int diff_set_0, int diff_set_1, |
| 943 | int diff_forcing, int diff_recursive, |
| 944 | usersolver_t const *usersolvers, void *ctx, |
| 945 | ctxnew_t ctxnew, ctxfree_t ctxfree) |
| 946 | { |
| 947 | struct latin_solver solver; |
| 948 | int diff; |
| 949 | |
| 950 | latin_solver_alloc(&solver, grid, o); |
| 951 | diff = latin_solver_main(&solver, maxdiff, |
| 952 | diff_simple, diff_set_0, diff_set_1, |
| 953 | diff_forcing, diff_recursive, |
| 954 | usersolvers, ctx, ctxnew, ctxfree); |
| 955 | latin_solver_free(&solver); |
| 956 | return diff; |
| 957 | } |
| 958 | |
| 959 | void latin_solver_debug(unsigned char *cube, int o) |
| 960 | { |
| 961 | #ifdef STANDALONE_SOLVER |
| 962 | if (solver_show_working > 1) { |
| 963 | struct latin_solver ls, *solver = &ls; |
| 964 | char *dbg; |
| 965 | int x, y, i, c = 0; |
| 966 | |
| 967 | ls.cube = cube; ls.o = o; /* for cube() to work */ |
| 968 | |
| 969 | dbg = snewn(3*o*o*o, char); |
| 970 | for (y = 0; y < o; y++) { |
| 971 | for (x = 0; x < o; x++) { |
| 972 | for (i = 1; i <= o; i++) { |
| 973 | if (cube(x,y,i)) |
| 974 | dbg[c++] = i + '0'; |
| 975 | else |
| 976 | dbg[c++] = '.'; |
| 977 | } |
| 978 | dbg[c++] = ' '; |
| 979 | } |
| 980 | dbg[c++] = '\n'; |
| 981 | } |
| 982 | dbg[c++] = '\n'; |
| 983 | dbg[c++] = '\0'; |
| 984 | |
| 985 | printf("%s", dbg); |
| 986 | sfree(dbg); |
| 987 | } |
| 988 | #endif |
| 989 | } |
| 990 | |
| 991 | void latin_debug(digit *sq, int o) |
| 992 | { |
| 993 | #ifdef STANDALONE_SOLVER |
| 994 | if (solver_show_working) { |
| 995 | int x, y; |
| 996 | |
| 997 | for (y = 0; y < o; y++) { |
| 998 | for (x = 0; x < o; x++) { |
| 999 | printf("%2d ", sq[y*o+x]); |
| 1000 | } |
| 1001 | printf("\n"); |
| 1002 | } |
| 1003 | printf("\n"); |
| 1004 | } |
| 1005 | #endif |
| 1006 | } |
| 1007 | |
| 1008 | /* -------------------------------------------------------- |
| 1009 | * Generation. |
| 1010 | */ |
| 1011 | |
| 1012 | digit *latin_generate(int o, random_state *rs) |
| 1013 | { |
| 1014 | digit *sq; |
| 1015 | int *edges, *backedges, *capacity, *flow; |
| 1016 | void *scratch; |
| 1017 | int ne, scratchsize; |
| 1018 | int i, j, k; |
| 1019 | digit *row, *col, *numinv, *num; |
| 1020 | |
| 1021 | /* |
| 1022 | * To efficiently generate a latin square in such a way that |
| 1023 | * all possible squares are possible outputs from the function, |
| 1024 | * we make use of a theorem which states that any r x n latin |
| 1025 | * rectangle, with r < n, can be extended into an (r+1) x n |
| 1026 | * latin rectangle. In other words, we can reliably generate a |
| 1027 | * latin square row by row, by at every stage writing down any |
| 1028 | * row at all which doesn't conflict with previous rows, and |
| 1029 | * the theorem guarantees that we will never have to backtrack. |
| 1030 | * |
| 1031 | * To find a viable row at each stage, we can make use of the |
| 1032 | * support functions in maxflow.c. |
| 1033 | */ |
| 1034 | |
| 1035 | sq = snewn(o*o, digit); |
| 1036 | |
| 1037 | /* |
| 1038 | * In case this method of generation introduces a really subtle |
| 1039 | * top-to-bottom directional bias, we'll generate the rows in |
| 1040 | * random order. |
| 1041 | */ |
| 1042 | row = snewn(o, digit); |
| 1043 | col = snewn(o, digit); |
| 1044 | numinv = snewn(o, digit); |
| 1045 | num = snewn(o, digit); |
| 1046 | for (i = 0; i < o; i++) |
| 1047 | row[i] = i; |
| 1048 | shuffle(row, i, sizeof(*row), rs); |
| 1049 | |
| 1050 | /* |
| 1051 | * Set up the infrastructure for the maxflow algorithm. |
| 1052 | */ |
| 1053 | scratchsize = maxflow_scratch_size(o * 2 + 2); |
| 1054 | scratch = smalloc(scratchsize); |
| 1055 | backedges = snewn(o*o + 2*o, int); |
| 1056 | edges = snewn((o*o + 2*o) * 2, int); |
| 1057 | capacity = snewn(o*o + 2*o, int); |
| 1058 | flow = snewn(o*o + 2*o, int); |
| 1059 | /* Set up the edge array, and the initial capacities. */ |
| 1060 | ne = 0; |
| 1061 | for (i = 0; i < o; i++) { |
| 1062 | /* Each LHS vertex is connected to all RHS vertices. */ |
| 1063 | for (j = 0; j < o; j++) { |
| 1064 | edges[ne*2] = i; |
| 1065 | edges[ne*2+1] = j+o; |
| 1066 | /* capacity for this edge is set later on */ |
| 1067 | ne++; |
| 1068 | } |
| 1069 | } |
| 1070 | for (i = 0; i < o; i++) { |
| 1071 | /* Each RHS vertex is connected to the distinguished sink vertex. */ |
| 1072 | edges[ne*2] = i+o; |
| 1073 | edges[ne*2+1] = o*2+1; |
| 1074 | capacity[ne] = 1; |
| 1075 | ne++; |
| 1076 | } |
| 1077 | for (i = 0; i < o; i++) { |
| 1078 | /* And the distinguished source vertex connects to each LHS vertex. */ |
| 1079 | edges[ne*2] = o*2; |
| 1080 | edges[ne*2+1] = i; |
| 1081 | capacity[ne] = 1; |
| 1082 | ne++; |
| 1083 | } |
| 1084 | assert(ne == o*o + 2*o); |
| 1085 | /* Now set up backedges. */ |
| 1086 | maxflow_setup_backedges(ne, edges, backedges); |
| 1087 | |
| 1088 | /* |
| 1089 | * Now generate each row of the latin square. |
| 1090 | */ |
| 1091 | for (i = 0; i < o; i++) { |
| 1092 | /* |
| 1093 | * To prevent maxflow from behaving deterministically, we |
| 1094 | * separately permute the columns and the digits for the |
| 1095 | * purposes of the algorithm, differently for every row. |
| 1096 | */ |
| 1097 | for (j = 0; j < o; j++) |
| 1098 | col[j] = num[j] = j; |
| 1099 | shuffle(col, j, sizeof(*col), rs); |
| 1100 | shuffle(num, j, sizeof(*num), rs); |
| 1101 | /* We need the num permutation in both forward and inverse forms. */ |
| 1102 | for (j = 0; j < o; j++) |
| 1103 | numinv[num[j]] = j; |
| 1104 | |
| 1105 | /* |
| 1106 | * Set up the capacities for the maxflow run, by examining |
| 1107 | * the existing latin square. |
| 1108 | */ |
| 1109 | for (j = 0; j < o*o; j++) |
| 1110 | capacity[j] = 1; |
| 1111 | for (j = 0; j < i; j++) |
| 1112 | for (k = 0; k < o; k++) { |
| 1113 | int n = num[sq[row[j]*o + col[k]] - 1]; |
| 1114 | capacity[k*o+n] = 0; |
| 1115 | } |
| 1116 | |
| 1117 | /* |
| 1118 | * Run maxflow. |
| 1119 | */ |
| 1120 | j = maxflow_with_scratch(scratch, o*2+2, 2*o, 2*o+1, ne, |
| 1121 | edges, backedges, capacity, flow, NULL); |
| 1122 | assert(j == o); /* by the above theorem, this must have succeeded */ |
| 1123 | |
| 1124 | /* |
| 1125 | * And examine the flow array to pick out the new row of |
| 1126 | * the latin square. |
| 1127 | */ |
| 1128 | for (j = 0; j < o; j++) { |
| 1129 | for (k = 0; k < o; k++) { |
| 1130 | if (flow[j*o+k]) |
| 1131 | break; |
| 1132 | } |
| 1133 | assert(k < o); |
| 1134 | sq[row[i]*o + col[j]] = numinv[k] + 1; |
| 1135 | } |
| 1136 | } |
| 1137 | |
| 1138 | /* |
| 1139 | * Done. Free our internal workspaces... |
| 1140 | */ |
| 1141 | sfree(flow); |
| 1142 | sfree(capacity); |
| 1143 | sfree(edges); |
| 1144 | sfree(backedges); |
| 1145 | sfree(scratch); |
| 1146 | sfree(numinv); |
| 1147 | sfree(num); |
| 1148 | sfree(col); |
| 1149 | sfree(row); |
| 1150 | |
| 1151 | /* |
| 1152 | * ... and return our completed latin square. |
| 1153 | */ |
| 1154 | return sq; |
| 1155 | } |
| 1156 | |
| 1157 | /* -------------------------------------------------------- |
| 1158 | * Checking. |
| 1159 | */ |
| 1160 | |
| 1161 | typedef struct lcparams { |
| 1162 | digit elt; |
| 1163 | int count; |
| 1164 | } lcparams; |
| 1165 | |
| 1166 | static int latin_check_cmp(void *v1, void *v2) |
| 1167 | { |
| 1168 | lcparams *lc1 = (lcparams *)v1; |
| 1169 | lcparams *lc2 = (lcparams *)v2; |
| 1170 | |
| 1171 | if (lc1->elt < lc2->elt) return -1; |
| 1172 | if (lc1->elt > lc2->elt) return 1; |
| 1173 | return 0; |
| 1174 | } |
| 1175 | |
| 1176 | #define ELT(sq,x,y) (sq[((y)*order)+(x)]) |
| 1177 | |
| 1178 | /* returns non-zero if sq is not a latin square. */ |
| 1179 | int latin_check(digit *sq, int order) |
| 1180 | { |
| 1181 | tree234 *dict = newtree234(latin_check_cmp); |
| 1182 | int c, r; |
| 1183 | int ret = 0; |
| 1184 | lcparams *lcp, lc, *aret; |
| 1185 | |
| 1186 | /* Use a tree234 as a simple hash table, go through the square |
| 1187 | * adding elements as we go or incrementing their counts. */ |
| 1188 | for (c = 0; c < order; c++) { |
| 1189 | for (r = 0; r < order; r++) { |
| 1190 | lc.elt = ELT(sq, c, r); lc.count = 0; |
| 1191 | lcp = find234(dict, &lc, NULL); |
| 1192 | if (!lcp) { |
| 1193 | lcp = snew(lcparams); |
| 1194 | lcp->elt = ELT(sq, c, r); |
| 1195 | lcp->count = 1; |
| 1196 | aret = add234(dict, lcp); |
| 1197 | assert(aret == lcp); |
| 1198 | } else { |
| 1199 | lcp->count++; |
| 1200 | } |
| 1201 | } |
| 1202 | } |
| 1203 | |
| 1204 | /* There should be precisely 'order' letters in the alphabet, |
| 1205 | * each occurring 'order' times (making the OxO tree) */ |
| 1206 | if (count234(dict) != order) ret = 1; |
| 1207 | else { |
| 1208 | for (c = 0; (lcp = index234(dict, c)) != NULL; c++) { |
| 1209 | if (lcp->count != order) ret = 1; |
| 1210 | } |
| 1211 | } |
| 1212 | for (c = 0; (lcp = index234(dict, c)) != NULL; c++) |
| 1213 | sfree(lcp); |
| 1214 | freetree234(dict); |
| 1215 | |
| 1216 | return ret; |
| 1217 | } |
| 1218 | |
| 1219 | |
| 1220 | /* -------------------------------------------------------- |
| 1221 | * Testing (and printing). |
| 1222 | */ |
| 1223 | |
| 1224 | #ifdef STANDALONE_LATIN_TEST |
| 1225 | |
| 1226 | #include <stdio.h> |
| 1227 | #include <time.h> |
| 1228 | |
| 1229 | const char *quis; |
| 1230 | |
| 1231 | static void latin_print(digit *sq, int order) |
| 1232 | { |
| 1233 | int x, y; |
| 1234 | |
| 1235 | for (y = 0; y < order; y++) { |
| 1236 | for (x = 0; x < order; x++) { |
| 1237 | printf("%2u ", ELT(sq, x, y)); |
| 1238 | } |
| 1239 | printf("\n"); |
| 1240 | } |
| 1241 | printf("\n"); |
| 1242 | } |
| 1243 | |
| 1244 | static void gen(int order, random_state *rs, int debug) |
| 1245 | { |
| 1246 | digit *sq; |
| 1247 | |
| 1248 | solver_show_working = debug; |
| 1249 | |
| 1250 | sq = latin_generate(order, rs); |
| 1251 | latin_print(sq, order); |
| 1252 | if (latin_check(sq, order)) { |
| 1253 | fprintf(stderr, "Square is not a latin square!"); |
| 1254 | exit(1); |
| 1255 | } |
| 1256 | |
| 1257 | sfree(sq); |
| 1258 | } |
| 1259 | |
| 1260 | void test_soak(int order, random_state *rs) |
| 1261 | { |
| 1262 | digit *sq; |
| 1263 | int n = 0; |
| 1264 | time_t tt_start, tt_now, tt_last; |
| 1265 | |
| 1266 | solver_show_working = 0; |
| 1267 | tt_now = tt_start = time(NULL); |
| 1268 | |
| 1269 | while(1) { |
| 1270 | sq = latin_generate(order, rs); |
| 1271 | sfree(sq); |
| 1272 | n++; |
| 1273 | |
| 1274 | tt_last = time(NULL); |
| 1275 | if (tt_last > tt_now) { |
| 1276 | tt_now = tt_last; |
| 1277 | printf("%d total, %3.1f/s\n", n, |
| 1278 | (double)n / (double)(tt_now - tt_start)); |
| 1279 | } |
| 1280 | } |
| 1281 | } |
| 1282 | |
| 1283 | void usage_exit(const char *msg) |
| 1284 | { |
| 1285 | if (msg) |
| 1286 | fprintf(stderr, "%s: %s\n", quis, msg); |
| 1287 | fprintf(stderr, "Usage: %s [--seed SEED] --soak <params> | [game_id [game_id ...]]\n", quis); |
| 1288 | exit(1); |
| 1289 | } |
| 1290 | |
| 1291 | int main(int argc, char *argv[]) |
| 1292 | { |
| 1293 | int i, soak = 0; |
| 1294 | random_state *rs; |
| 1295 | time_t seed = time(NULL); |
| 1296 | |
| 1297 | quis = argv[0]; |
| 1298 | while (--argc > 0) { |
| 1299 | const char *p = *++argv; |
| 1300 | if (!strcmp(p, "--soak")) |
| 1301 | soak = 1; |
| 1302 | else if (!strcmp(p, "--seed")) { |
| 1303 | if (argc == 0) |
| 1304 | usage_exit("--seed needs an argument"); |
| 1305 | seed = (time_t)atoi(*++argv); |
| 1306 | argc--; |
| 1307 | } else if (*p == '-') |
| 1308 | usage_exit("unrecognised option"); |
| 1309 | else |
| 1310 | break; /* finished options */ |
| 1311 | } |
| 1312 | |
| 1313 | rs = random_new((void*)&seed, sizeof(time_t)); |
| 1314 | |
| 1315 | if (soak == 1) { |
| 1316 | if (argc != 1) usage_exit("only one argument for --soak"); |
| 1317 | test_soak(atoi(*argv), rs); |
| 1318 | } else { |
| 1319 | if (argc > 0) { |
| 1320 | for (i = 0; i < argc; i++) { |
| 1321 | gen(atoi(*argv++), rs, 1); |
| 1322 | } |
| 1323 | } else { |
| 1324 | while (1) { |
| 1325 | i = random_upto(rs, 20) + 1; |
| 1326 | gen(i, rs, 0); |
| 1327 | } |
| 1328 | } |
| 1329 | } |
| 1330 | random_free(rs); |
| 1331 | return 0; |
| 1332 | } |
| 1333 | |
| 1334 | #endif |
| 1335 | |
| 1336 | /* vim: set shiftwidth=4 tabstop=8: */ |