| 1 | /* |
| 2 | * solo.c: the number-placing puzzle most popularly known as `Sudoku'. |
| 3 | * |
| 4 | * TODO: |
| 5 | * |
| 6 | * - reports from users are that `Trivial'-mode puzzles are still |
| 7 | * rather hard compared to newspapers' easy ones, so some better |
| 8 | * low-end difficulty grading would be nice |
| 9 | * + it's possible that really easy puzzles always have |
| 10 | * _several_ things you can do, so don't make you hunt too |
| 11 | * hard for the one deduction you can currently make |
| 12 | * + it's also possible that easy puzzles require fewer |
| 13 | * cross-eliminations: perhaps there's a higher incidence of |
| 14 | * things you can deduce by looking only at (say) rows, |
| 15 | * rather than things you have to check both rows and columns |
| 16 | * for |
| 17 | * + but really, what I need to do is find some really easy |
| 18 | * puzzles and _play_ them, to see what's actually easy about |
| 19 | * them |
| 20 | * + while I'm revamping this area, filling in the _last_ |
| 21 | * number in a nearly-full row or column should certainly be |
| 22 | * permitted even at the lowest difficulty level. |
| 23 | * + also Owen noticed that `Basic' grids requiring numeric |
| 24 | * elimination are actually very hard, so I wonder if a |
| 25 | * difficulty gradation between that and positional- |
| 26 | * elimination-only might be in order |
| 27 | * + but it's not good to have _too_ many difficulty levels, or |
| 28 | * it'll take too long to randomly generate a given level. |
| 29 | * |
| 30 | * - it might still be nice to do some prioritisation on the |
| 31 | * removal of numbers from the grid |
| 32 | * + one possibility is to try to minimise the maximum number |
| 33 | * of filled squares in any block, which in particular ought |
| 34 | * to enforce never leaving a completely filled block in the |
| 35 | * puzzle as presented. |
| 36 | * |
| 37 | * - alternative interface modes |
| 38 | * + sudoku.com's Windows program has a palette of possible |
| 39 | * entries; you select a palette entry first and then click |
| 40 | * on the square you want it to go in, thus enabling |
| 41 | * mouse-only play. Useful for PDAs! I don't think it's |
| 42 | * actually incompatible with the current highlight-then-type |
| 43 | * approach: you _either_ highlight a palette entry and then |
| 44 | * click, _or_ you highlight a square and then type. At most |
| 45 | * one thing is ever highlighted at a time, so there's no way |
| 46 | * to confuse the two. |
| 47 | * + then again, I don't actually like sudoku.com's interface; |
| 48 | * it's too much like a paint package whereas I prefer to |
| 49 | * think of Solo as a text editor. |
| 50 | * + another PDA-friendly possibility is a drag interface: |
| 51 | * _drag_ numbers from the palette into the grid squares. |
| 52 | * Thought experiments suggest I'd prefer that to the |
| 53 | * sudoku.com approach, but I haven't actually tried it. |
| 54 | */ |
| 55 | |
| 56 | /* |
| 57 | * Solo puzzles need to be square overall (since each row and each |
| 58 | * column must contain one of every digit), but they need not be |
| 59 | * subdivided the same way internally. I am going to adopt a |
| 60 | * convention whereby I _always_ refer to `r' as the number of rows |
| 61 | * of _big_ divisions, and `c' as the number of columns of _big_ |
| 62 | * divisions. Thus, a 2c by 3r puzzle looks something like this: |
| 63 | * |
| 64 | * 4 5 1 | 2 6 3 |
| 65 | * 6 3 2 | 5 4 1 |
| 66 | * ------+------ (Of course, you can't subdivide it the other way |
| 67 | * 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the |
| 68 | * 3 2 6 | 4 1 5 top left would conflict with the 4 in the second |
| 69 | * ------+------ box down on the left-hand side.) |
| 70 | * 5 1 4 | 3 2 6 |
| 71 | * 2 6 3 | 1 5 4 |
| 72 | * |
| 73 | * The need for a strong naming convention should now be clear: |
| 74 | * each small box is two rows of digits by three columns, while the |
| 75 | * overall puzzle has three rows of small boxes by two columns. So |
| 76 | * I will (hopefully) consistently use `r' to denote the number of |
| 77 | * rows _of small boxes_ (here 3), which is also the number of |
| 78 | * columns of digits in each small box; and `c' vice versa (here |
| 79 | * 2). |
| 80 | * |
| 81 | * I'm also going to choose arbitrarily to list c first wherever |
| 82 | * possible: the above is a 2x3 puzzle, not a 3x2 one. |
| 83 | */ |
| 84 | |
| 85 | #include <stdio.h> |
| 86 | #include <stdlib.h> |
| 87 | #include <string.h> |
| 88 | #include <assert.h> |
| 89 | #include <ctype.h> |
| 90 | #include <math.h> |
| 91 | |
| 92 | #ifdef STANDALONE_SOLVER |
| 93 | #include <stdarg.h> |
| 94 | int solver_show_working, solver_recurse_depth; |
| 95 | #endif |
| 96 | |
| 97 | #include "puzzles.h" |
| 98 | |
| 99 | /* |
| 100 | * To save space, I store digits internally as unsigned char. This |
| 101 | * imposes a hard limit of 255 on the order of the puzzle. Since |
| 102 | * even a 5x5 takes unacceptably long to generate, I don't see this |
| 103 | * as a serious limitation unless something _really_ impressive |
| 104 | * happens in computing technology; but here's a typedef anyway for |
| 105 | * general good practice. |
| 106 | */ |
| 107 | typedef unsigned char digit; |
| 108 | #define ORDER_MAX 255 |
| 109 | |
| 110 | #define PREFERRED_TILE_SIZE 32 |
| 111 | #define TILE_SIZE (ds->tilesize) |
| 112 | #define BORDER (TILE_SIZE / 2) |
| 113 | #define GRIDEXTRA (TILE_SIZE / 32) |
| 114 | |
| 115 | #define FLASH_TIME 0.4F |
| 116 | |
| 117 | enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4, |
| 118 | SYMM_REF4D, SYMM_REF8 }; |
| 119 | |
| 120 | enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME, |
| 121 | DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE }; |
| 122 | |
| 123 | enum { |
| 124 | COL_BACKGROUND, |
| 125 | COL_XDIAGONALS, |
| 126 | COL_GRID, |
| 127 | COL_CLUE, |
| 128 | COL_USER, |
| 129 | COL_HIGHLIGHT, |
| 130 | COL_ERROR, |
| 131 | COL_PENCIL, |
| 132 | NCOLOURS |
| 133 | }; |
| 134 | |
| 135 | struct game_params { |
| 136 | /* |
| 137 | * For a square puzzle, `c' and `r' indicate the puzzle |
| 138 | * parameters as described above. |
| 139 | * |
| 140 | * A jigsaw-style puzzle is indicated by r==1, in which case c |
| 141 | * can be whatever it likes (there is no constraint on |
| 142 | * compositeness - a 7x7 jigsaw sudoku makes perfect sense). |
| 143 | */ |
| 144 | int c, r, symm, diff; |
| 145 | int xtype; /* require all digits in X-diagonals */ |
| 146 | }; |
| 147 | |
| 148 | struct block_structure { |
| 149 | int refcount; |
| 150 | |
| 151 | /* |
| 152 | * For text formatting, we do need c and r here. |
| 153 | */ |
| 154 | int c, r; |
| 155 | |
| 156 | /* |
| 157 | * For any square index, whichblock[i] gives its block index. |
| 158 | * |
| 159 | * For 0 <= b,i < cr, blocks[b][i] gives the index of the ith |
| 160 | * square in block b. |
| 161 | * |
| 162 | * whichblock and blocks are each dynamically allocated in |
| 163 | * their own right, but the subarrays in blocks are appended |
| 164 | * to the whichblock array, so shouldn't be freed |
| 165 | * individually. |
| 166 | */ |
| 167 | int *whichblock, **blocks; |
| 168 | |
| 169 | #ifdef STANDALONE_SOLVER |
| 170 | /* |
| 171 | * Textual descriptions of each block. For normal Sudoku these |
| 172 | * are of the form "(1,3)"; for jigsaw they are "starting at |
| 173 | * (5,7)". So the sensible usage in both cases is to say |
| 174 | * "elimination within block %s" with one of these strings. |
| 175 | * |
| 176 | * Only blocknames itself needs individually freeing; it's all |
| 177 | * one block. |
| 178 | */ |
| 179 | char **blocknames; |
| 180 | #endif |
| 181 | }; |
| 182 | |
| 183 | struct game_state { |
| 184 | /* |
| 185 | * For historical reasons, I use `cr' to denote the overall |
| 186 | * width/height of the puzzle. It was a natural notation when |
| 187 | * all puzzles were divided into blocks in a grid, but doesn't |
| 188 | * really make much sense given jigsaw puzzles. However, the |
| 189 | * obvious `n' is heavily used in the solver to describe the |
| 190 | * index of a number being placed, so `cr' will have to stay. |
| 191 | */ |
| 192 | int cr; |
| 193 | struct block_structure *blocks; |
| 194 | int xtype; |
| 195 | digit *grid; |
| 196 | unsigned char *pencil; /* c*r*c*r elements */ |
| 197 | unsigned char *immutable; /* marks which digits are clues */ |
| 198 | int completed, cheated; |
| 199 | }; |
| 200 | |
| 201 | static game_params *default_params(void) |
| 202 | { |
| 203 | game_params *ret = snew(game_params); |
| 204 | |
| 205 | ret->c = ret->r = 3; |
| 206 | ret->xtype = FALSE; |
| 207 | ret->symm = SYMM_ROT2; /* a plausible default */ |
| 208 | ret->diff = DIFF_BLOCK; /* so is this */ |
| 209 | |
| 210 | return ret; |
| 211 | } |
| 212 | |
| 213 | static void free_params(game_params *params) |
| 214 | { |
| 215 | sfree(params); |
| 216 | } |
| 217 | |
| 218 | static game_params *dup_params(game_params *params) |
| 219 | { |
| 220 | game_params *ret = snew(game_params); |
| 221 | *ret = *params; /* structure copy */ |
| 222 | return ret; |
| 223 | } |
| 224 | |
| 225 | static int game_fetch_preset(int i, char **name, game_params **params) |
| 226 | { |
| 227 | static struct { |
| 228 | char *title; |
| 229 | game_params params; |
| 230 | } presets[] = { |
| 231 | { "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK, FALSE } }, |
| 232 | { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, |
| 233 | { "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK, FALSE } }, |
| 234 | { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, |
| 235 | { "3x3 Basic X", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, TRUE } }, |
| 236 | { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT, FALSE } }, |
| 237 | { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET, FALSE } }, |
| 238 | { "3x3 Advanced X", { 3, 3, SYMM_ROT2, DIFF_SET, TRUE } }, |
| 239 | { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME, FALSE } }, |
| 240 | { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE, FALSE } }, |
| 241 | { "9 Jigsaw Basic", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, |
| 242 | { "9 Jigsaw Basic X", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, TRUE } }, |
| 243 | { "9 Jigsaw Advanced", { 9, 1, SYMM_ROT2, DIFF_SET, FALSE } }, |
| 244 | #ifndef SLOW_SYSTEM |
| 245 | { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, |
| 246 | { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, |
| 247 | #endif |
| 248 | }; |
| 249 | |
| 250 | if (i < 0 || i >= lenof(presets)) |
| 251 | return FALSE; |
| 252 | |
| 253 | *name = dupstr(presets[i].title); |
| 254 | *params = dup_params(&presets[i].params); |
| 255 | |
| 256 | return TRUE; |
| 257 | } |
| 258 | |
| 259 | static void decode_params(game_params *ret, char const *string) |
| 260 | { |
| 261 | int seen_r = FALSE; |
| 262 | |
| 263 | ret->c = ret->r = atoi(string); |
| 264 | ret->xtype = FALSE; |
| 265 | while (*string && isdigit((unsigned char)*string)) string++; |
| 266 | if (*string == 'x') { |
| 267 | string++; |
| 268 | ret->r = atoi(string); |
| 269 | seen_r = TRUE; |
| 270 | while (*string && isdigit((unsigned char)*string)) string++; |
| 271 | } |
| 272 | while (*string) { |
| 273 | if (*string == 'j') { |
| 274 | string++; |
| 275 | if (seen_r) |
| 276 | ret->c *= ret->r; |
| 277 | ret->r = 1; |
| 278 | } else if (*string == 'x') { |
| 279 | string++; |
| 280 | ret->xtype = TRUE; |
| 281 | } else if (*string == 'r' || *string == 'm' || *string == 'a') { |
| 282 | int sn, sc, sd; |
| 283 | sc = *string++; |
| 284 | if (sc == 'm' && *string == 'd') { |
| 285 | sd = TRUE; |
| 286 | string++; |
| 287 | } else { |
| 288 | sd = FALSE; |
| 289 | } |
| 290 | sn = atoi(string); |
| 291 | while (*string && isdigit((unsigned char)*string)) string++; |
| 292 | if (sc == 'm' && sn == 8) |
| 293 | ret->symm = SYMM_REF8; |
| 294 | if (sc == 'm' && sn == 4) |
| 295 | ret->symm = sd ? SYMM_REF4D : SYMM_REF4; |
| 296 | if (sc == 'm' && sn == 2) |
| 297 | ret->symm = sd ? SYMM_REF2D : SYMM_REF2; |
| 298 | if (sc == 'r' && sn == 4) |
| 299 | ret->symm = SYMM_ROT4; |
| 300 | if (sc == 'r' && sn == 2) |
| 301 | ret->symm = SYMM_ROT2; |
| 302 | if (sc == 'a') |
| 303 | ret->symm = SYMM_NONE; |
| 304 | } else if (*string == 'd') { |
| 305 | string++; |
| 306 | if (*string == 't') /* trivial */ |
| 307 | string++, ret->diff = DIFF_BLOCK; |
| 308 | else if (*string == 'b') /* basic */ |
| 309 | string++, ret->diff = DIFF_SIMPLE; |
| 310 | else if (*string == 'i') /* intermediate */ |
| 311 | string++, ret->diff = DIFF_INTERSECT; |
| 312 | else if (*string == 'a') /* advanced */ |
| 313 | string++, ret->diff = DIFF_SET; |
| 314 | else if (*string == 'e') /* extreme */ |
| 315 | string++, ret->diff = DIFF_EXTREME; |
| 316 | else if (*string == 'u') /* unreasonable */ |
| 317 | string++, ret->diff = DIFF_RECURSIVE; |
| 318 | } else |
| 319 | string++; /* eat unknown character */ |
| 320 | } |
| 321 | } |
| 322 | |
| 323 | static char *encode_params(game_params *params, int full) |
| 324 | { |
| 325 | char str[80]; |
| 326 | |
| 327 | if (params->r > 1) |
| 328 | sprintf(str, "%dx%d", params->c, params->r); |
| 329 | else |
| 330 | sprintf(str, "%dj", params->c); |
| 331 | if (params->xtype) |
| 332 | strcat(str, "x"); |
| 333 | |
| 334 | if (full) { |
| 335 | switch (params->symm) { |
| 336 | case SYMM_REF8: strcat(str, "m8"); break; |
| 337 | case SYMM_REF4: strcat(str, "m4"); break; |
| 338 | case SYMM_REF4D: strcat(str, "md4"); break; |
| 339 | case SYMM_REF2: strcat(str, "m2"); break; |
| 340 | case SYMM_REF2D: strcat(str, "md2"); break; |
| 341 | case SYMM_ROT4: strcat(str, "r4"); break; |
| 342 | /* case SYMM_ROT2: strcat(str, "r2"); break; [default] */ |
| 343 | case SYMM_NONE: strcat(str, "a"); break; |
| 344 | } |
| 345 | switch (params->diff) { |
| 346 | /* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */ |
| 347 | case DIFF_SIMPLE: strcat(str, "db"); break; |
| 348 | case DIFF_INTERSECT: strcat(str, "di"); break; |
| 349 | case DIFF_SET: strcat(str, "da"); break; |
| 350 | case DIFF_EXTREME: strcat(str, "de"); break; |
| 351 | case DIFF_RECURSIVE: strcat(str, "du"); break; |
| 352 | } |
| 353 | } |
| 354 | return dupstr(str); |
| 355 | } |
| 356 | |
| 357 | static config_item *game_configure(game_params *params) |
| 358 | { |
| 359 | config_item *ret; |
| 360 | char buf[80]; |
| 361 | |
| 362 | ret = snewn(7, config_item); |
| 363 | |
| 364 | ret[0].name = "Columns of sub-blocks"; |
| 365 | ret[0].type = C_STRING; |
| 366 | sprintf(buf, "%d", params->c); |
| 367 | ret[0].sval = dupstr(buf); |
| 368 | ret[0].ival = 0; |
| 369 | |
| 370 | ret[1].name = "Rows of sub-blocks"; |
| 371 | ret[1].type = C_STRING; |
| 372 | sprintf(buf, "%d", params->r); |
| 373 | ret[1].sval = dupstr(buf); |
| 374 | ret[1].ival = 0; |
| 375 | |
| 376 | ret[2].name = "\"X\" (require every number in each main diagonal)"; |
| 377 | ret[2].type = C_BOOLEAN; |
| 378 | ret[2].sval = NULL; |
| 379 | ret[2].ival = params->xtype; |
| 380 | |
| 381 | ret[3].name = "Jigsaw (irregularly shaped sub-blocks)"; |
| 382 | ret[3].type = C_BOOLEAN; |
| 383 | ret[3].sval = NULL; |
| 384 | ret[3].ival = (params->r == 1); |
| 385 | |
| 386 | ret[4].name = "Symmetry"; |
| 387 | ret[4].type = C_CHOICES; |
| 388 | ret[4].sval = ":None:2-way rotation:4-way rotation:2-way mirror:" |
| 389 | "2-way diagonal mirror:4-way mirror:4-way diagonal mirror:" |
| 390 | "8-way mirror"; |
| 391 | ret[4].ival = params->symm; |
| 392 | |
| 393 | ret[5].name = "Difficulty"; |
| 394 | ret[5].type = C_CHOICES; |
| 395 | ret[5].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable"; |
| 396 | ret[5].ival = params->diff; |
| 397 | |
| 398 | ret[6].name = NULL; |
| 399 | ret[6].type = C_END; |
| 400 | ret[6].sval = NULL; |
| 401 | ret[6].ival = 0; |
| 402 | |
| 403 | return ret; |
| 404 | } |
| 405 | |
| 406 | static game_params *custom_params(config_item *cfg) |
| 407 | { |
| 408 | game_params *ret = snew(game_params); |
| 409 | |
| 410 | ret->c = atoi(cfg[0].sval); |
| 411 | ret->r = atoi(cfg[1].sval); |
| 412 | ret->xtype = cfg[2].ival; |
| 413 | if (cfg[3].ival) { |
| 414 | ret->c *= ret->r; |
| 415 | ret->r = 1; |
| 416 | } |
| 417 | ret->symm = cfg[4].ival; |
| 418 | ret->diff = cfg[5].ival; |
| 419 | |
| 420 | return ret; |
| 421 | } |
| 422 | |
| 423 | static char *validate_params(game_params *params, int full) |
| 424 | { |
| 425 | if (params->c < 2) |
| 426 | return "Both dimensions must be at least 2"; |
| 427 | if (params->c > ORDER_MAX || params->r > ORDER_MAX) |
| 428 | return "Dimensions greater than "STR(ORDER_MAX)" are not supported"; |
| 429 | if ((params->c * params->r) > 35) |
| 430 | return "Unable to support more than 35 distinct symbols in a puzzle"; |
| 431 | return NULL; |
| 432 | } |
| 433 | |
| 434 | /* ---------------------------------------------------------------------- |
| 435 | * Solver. |
| 436 | * |
| 437 | * This solver is used for two purposes: |
| 438 | * + to check solubility of a grid as we gradually remove numbers |
| 439 | * from it |
| 440 | * + to solve an externally generated puzzle when the user selects |
| 441 | * `Solve'. |
| 442 | * |
| 443 | * It supports a variety of specific modes of reasoning. By |
| 444 | * enabling or disabling subsets of these modes we can arrange a |
| 445 | * range of difficulty levels. |
| 446 | */ |
| 447 | |
| 448 | /* |
| 449 | * Modes of reasoning currently supported: |
| 450 | * |
| 451 | * - Positional elimination: a number must go in a particular |
| 452 | * square because all the other empty squares in a given |
| 453 | * row/col/blk are ruled out. |
| 454 | * |
| 455 | * - Numeric elimination: a square must have a particular number |
| 456 | * in because all the other numbers that could go in it are |
| 457 | * ruled out. |
| 458 | * |
| 459 | * - Intersectional analysis: given two domains which overlap |
| 460 | * (hence one must be a block, and the other can be a row or |
| 461 | * col), if the possible locations for a particular number in |
| 462 | * one of the domains can be narrowed down to the overlap, then |
| 463 | * that number can be ruled out everywhere but the overlap in |
| 464 | * the other domain too. |
| 465 | * |
| 466 | * - Set elimination: if there is a subset of the empty squares |
| 467 | * within a domain such that the union of the possible numbers |
| 468 | * in that subset has the same size as the subset itself, then |
| 469 | * those numbers can be ruled out everywhere else in the domain. |
| 470 | * (For example, if there are five empty squares and the |
| 471 | * possible numbers in each are 12, 23, 13, 134 and 1345, then |
| 472 | * the first three empty squares form such a subset: the numbers |
| 473 | * 1, 2 and 3 _must_ be in those three squares in some |
| 474 | * permutation, and hence we can deduce none of them can be in |
| 475 | * the fourth or fifth squares.) |
| 476 | * + You can also see this the other way round, concentrating |
| 477 | * on numbers rather than squares: if there is a subset of |
| 478 | * the unplaced numbers within a domain such that the union |
| 479 | * of all their possible positions has the same size as the |
| 480 | * subset itself, then all other numbers can be ruled out for |
| 481 | * those positions. However, it turns out that this is |
| 482 | * exactly equivalent to the first formulation at all times: |
| 483 | * there is a 1-1 correspondence between suitable subsets of |
| 484 | * the unplaced numbers and suitable subsets of the unfilled |
| 485 | * places, found by taking the _complement_ of the union of |
| 486 | * the numbers' possible positions (or the spaces' possible |
| 487 | * contents). |
| 488 | * |
| 489 | * - Forcing chains (see comment for solver_forcing().) |
| 490 | * |
| 491 | * - Recursion. If all else fails, we pick one of the currently |
| 492 | * most constrained empty squares and take a random guess at its |
| 493 | * contents, then continue solving on that basis and see if we |
| 494 | * get any further. |
| 495 | */ |
| 496 | |
| 497 | struct solver_usage { |
| 498 | int cr; |
| 499 | struct block_structure *blocks; |
| 500 | /* |
| 501 | * We set up a cubic array, indexed by x, y and digit; each |
| 502 | * element of this array is TRUE or FALSE according to whether |
| 503 | * or not that digit _could_ in principle go in that position. |
| 504 | * |
| 505 | * The way to index this array is cube[(y*cr+x)*cr+n-1]; there |
| 506 | * are macros below to help with this. |
| 507 | */ |
| 508 | unsigned char *cube; |
| 509 | /* |
| 510 | * This is the grid in which we write down our final |
| 511 | * deductions. y-coordinates in here are _not_ transformed. |
| 512 | */ |
| 513 | digit *grid; |
| 514 | /* |
| 515 | * Now we keep track, at a slightly higher level, of what we |
| 516 | * have yet to work out, to prevent doing the same deduction |
| 517 | * many times. |
| 518 | */ |
| 519 | /* row[y*cr+n-1] TRUE if digit n has been placed in row y */ |
| 520 | unsigned char *row; |
| 521 | /* col[x*cr+n-1] TRUE if digit n has been placed in row x */ |
| 522 | unsigned char *col; |
| 523 | /* blk[i*cr+n-1] TRUE if digit n has been placed in block i */ |
| 524 | unsigned char *blk; |
| 525 | /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */ |
| 526 | unsigned char *diag; /* diag 0 is \, 1 is / */ |
| 527 | }; |
| 528 | #define cubepos2(xy,n) ((xy)*usage->cr+(n)-1) |
| 529 | #define cubepos(x,y,n) cubepos2((y)*usage->cr+(x),n) |
| 530 | #define cube(x,y,n) (usage->cube[cubepos(x,y,n)]) |
| 531 | #define cube2(xy,n) (usage->cube[cubepos2(xy,n)]) |
| 532 | |
| 533 | #define ondiag0(xy) ((xy) % (cr+1) == 0) |
| 534 | #define ondiag1(xy) ((xy) % (cr-1) == 0 && (xy) > 0 && (xy) < cr*cr-1) |
| 535 | #define diag0(i) ((i) * (cr+1)) |
| 536 | #define diag1(i) ((i+1) * (cr-1)) |
| 537 | |
| 538 | /* |
| 539 | * Function called when we are certain that a particular square has |
| 540 | * a particular number in it. The y-coordinate passed in here is |
| 541 | * transformed. |
| 542 | */ |
| 543 | static void solver_place(struct solver_usage *usage, int x, int y, int n) |
| 544 | { |
| 545 | int cr = usage->cr; |
| 546 | int sqindex = y*cr+x; |
| 547 | int i, bi; |
| 548 | |
| 549 | assert(cube(x,y,n)); |
| 550 | |
| 551 | /* |
| 552 | * Rule out all other numbers in this square. |
| 553 | */ |
| 554 | for (i = 1; i <= cr; i++) |
| 555 | if (i != n) |
| 556 | cube(x,y,i) = FALSE; |
| 557 | |
| 558 | /* |
| 559 | * Rule out this number in all other positions in the row. |
| 560 | */ |
| 561 | for (i = 0; i < cr; i++) |
| 562 | if (i != y) |
| 563 | cube(x,i,n) = FALSE; |
| 564 | |
| 565 | /* |
| 566 | * Rule out this number in all other positions in the column. |
| 567 | */ |
| 568 | for (i = 0; i < cr; i++) |
| 569 | if (i != x) |
| 570 | cube(i,y,n) = FALSE; |
| 571 | |
| 572 | /* |
| 573 | * Rule out this number in all other positions in the block. |
| 574 | */ |
| 575 | bi = usage->blocks->whichblock[sqindex]; |
| 576 | for (i = 0; i < cr; i++) { |
| 577 | int bp = usage->blocks->blocks[bi][i]; |
| 578 | if (bp != sqindex) |
| 579 | cube2(bp,n) = FALSE; |
| 580 | } |
| 581 | |
| 582 | /* |
| 583 | * Enter the number in the result grid. |
| 584 | */ |
| 585 | usage->grid[sqindex] = n; |
| 586 | |
| 587 | /* |
| 588 | * Cross out this number from the list of numbers left to place |
| 589 | * in its row, its column and its block. |
| 590 | */ |
| 591 | usage->row[y*cr+n-1] = usage->col[x*cr+n-1] = |
| 592 | usage->blk[bi*cr+n-1] = TRUE; |
| 593 | |
| 594 | if (usage->diag) { |
| 595 | if (ondiag0(sqindex)) { |
| 596 | for (i = 0; i < cr; i++) |
| 597 | if (diag0(i) != sqindex) |
| 598 | cube2(diag0(i),n) = FALSE; |
| 599 | usage->diag[n-1] = TRUE; |
| 600 | } |
| 601 | if (ondiag1(sqindex)) { |
| 602 | for (i = 0; i < cr; i++) |
| 603 | if (diag1(i) != sqindex) |
| 604 | cube2(diag1(i),n) = FALSE; |
| 605 | usage->diag[cr+n-1] = TRUE; |
| 606 | } |
| 607 | } |
| 608 | } |
| 609 | |
| 610 | static int solver_elim(struct solver_usage *usage, int *indices |
| 611 | #ifdef STANDALONE_SOLVER |
| 612 | , char *fmt, ... |
| 613 | #endif |
| 614 | ) |
| 615 | { |
| 616 | int cr = usage->cr; |
| 617 | int fpos, m, i; |
| 618 | |
| 619 | /* |
| 620 | * Count the number of set bits within this section of the |
| 621 | * cube. |
| 622 | */ |
| 623 | m = 0; |
| 624 | fpos = -1; |
| 625 | for (i = 0; i < cr; i++) |
| 626 | if (usage->cube[indices[i]]) { |
| 627 | fpos = indices[i]; |
| 628 | m++; |
| 629 | } |
| 630 | |
| 631 | if (m == 1) { |
| 632 | int x, y, n; |
| 633 | assert(fpos >= 0); |
| 634 | |
| 635 | n = 1 + fpos % cr; |
| 636 | x = fpos / cr; |
| 637 | y = x / cr; |
| 638 | x %= cr; |
| 639 | |
| 640 | if (!usage->grid[y*cr+x]) { |
| 641 | #ifdef STANDALONE_SOLVER |
| 642 | if (solver_show_working) { |
| 643 | va_list ap; |
| 644 | printf("%*s", solver_recurse_depth*4, ""); |
| 645 | va_start(ap, fmt); |
| 646 | vprintf(fmt, ap); |
| 647 | va_end(ap); |
| 648 | printf(":\n%*s placing %d at (%d,%d)\n", |
| 649 | solver_recurse_depth*4, "", n, 1+x, 1+y); |
| 650 | } |
| 651 | #endif |
| 652 | solver_place(usage, x, y, n); |
| 653 | return +1; |
| 654 | } |
| 655 | } else if (m == 0) { |
| 656 | #ifdef STANDALONE_SOLVER |
| 657 | if (solver_show_working) { |
| 658 | va_list ap; |
| 659 | printf("%*s", solver_recurse_depth*4, ""); |
| 660 | va_start(ap, fmt); |
| 661 | vprintf(fmt, ap); |
| 662 | va_end(ap); |
| 663 | printf(":\n%*s no possibilities available\n", |
| 664 | solver_recurse_depth*4, ""); |
| 665 | } |
| 666 | #endif |
| 667 | return -1; |
| 668 | } |
| 669 | |
| 670 | return 0; |
| 671 | } |
| 672 | |
| 673 | static int solver_intersect(struct solver_usage *usage, |
| 674 | int *indices1, int *indices2 |
| 675 | #ifdef STANDALONE_SOLVER |
| 676 | , char *fmt, ... |
| 677 | #endif |
| 678 | ) |
| 679 | { |
| 680 | int cr = usage->cr; |
| 681 | int ret, i, j; |
| 682 | |
| 683 | /* |
| 684 | * Loop over the first domain and see if there's any set bit |
| 685 | * not also in the second. |
| 686 | */ |
| 687 | for (i = j = 0; i < cr; i++) { |
| 688 | int p = indices1[i]; |
| 689 | while (j < cr && indices2[j] < p) |
| 690 | j++; |
| 691 | if (usage->cube[p]) { |
| 692 | if (j < cr && indices2[j] == p) |
| 693 | continue; /* both domains contain this index */ |
| 694 | else |
| 695 | return 0; /* there is, so we can't deduce */ |
| 696 | } |
| 697 | } |
| 698 | |
| 699 | /* |
| 700 | * We have determined that all set bits in the first domain are |
| 701 | * within its overlap with the second. So loop over the second |
| 702 | * domain and remove all set bits that aren't also in that |
| 703 | * overlap; return +1 iff we actually _did_ anything. |
| 704 | */ |
| 705 | ret = 0; |
| 706 | for (i = j = 0; i < cr; i++) { |
| 707 | int p = indices2[i]; |
| 708 | while (j < cr && indices1[j] < p) |
| 709 | j++; |
| 710 | if (usage->cube[p] && (j >= cr || indices1[j] != p)) { |
| 711 | #ifdef STANDALONE_SOLVER |
| 712 | if (solver_show_working) { |
| 713 | int px, py, pn; |
| 714 | |
| 715 | if (!ret) { |
| 716 | va_list ap; |
| 717 | printf("%*s", solver_recurse_depth*4, ""); |
| 718 | va_start(ap, fmt); |
| 719 | vprintf(fmt, ap); |
| 720 | va_end(ap); |
| 721 | printf(":\n"); |
| 722 | } |
| 723 | |
| 724 | pn = 1 + p % cr; |
| 725 | px = p / cr; |
| 726 | py = px / cr; |
| 727 | px %= cr; |
| 728 | |
| 729 | printf("%*s ruling out %d at (%d,%d)\n", |
| 730 | solver_recurse_depth*4, "", pn, 1+px, 1+py); |
| 731 | } |
| 732 | #endif |
| 733 | ret = +1; /* we did something */ |
| 734 | usage->cube[p] = 0; |
| 735 | } |
| 736 | } |
| 737 | |
| 738 | return ret; |
| 739 | } |
| 740 | |
| 741 | struct solver_scratch { |
| 742 | unsigned char *grid, *rowidx, *colidx, *set; |
| 743 | int *neighbours, *bfsqueue; |
| 744 | int *indexlist, *indexlist2; |
| 745 | #ifdef STANDALONE_SOLVER |
| 746 | int *bfsprev; |
| 747 | #endif |
| 748 | }; |
| 749 | |
| 750 | static int solver_set(struct solver_usage *usage, |
| 751 | struct solver_scratch *scratch, |
| 752 | int *indices |
| 753 | #ifdef STANDALONE_SOLVER |
| 754 | , char *fmt, ... |
| 755 | #endif |
| 756 | ) |
| 757 | { |
| 758 | int cr = usage->cr; |
| 759 | int i, j, n, count; |
| 760 | unsigned char *grid = scratch->grid; |
| 761 | unsigned char *rowidx = scratch->rowidx; |
| 762 | unsigned char *colidx = scratch->colidx; |
| 763 | unsigned char *set = scratch->set; |
| 764 | |
| 765 | /* |
| 766 | * We are passed a cr-by-cr matrix of booleans. Our first job |
| 767 | * is to winnow it by finding any definite placements - i.e. |
| 768 | * any row with a solitary 1 - and discarding that row and the |
| 769 | * column containing the 1. |
| 770 | */ |
| 771 | memset(rowidx, TRUE, cr); |
| 772 | memset(colidx, TRUE, cr); |
| 773 | for (i = 0; i < cr; i++) { |
| 774 | int count = 0, first = -1; |
| 775 | for (j = 0; j < cr; j++) |
| 776 | if (usage->cube[indices[i*cr+j]]) |
| 777 | first = j, count++; |
| 778 | |
| 779 | /* |
| 780 | * If count == 0, then there's a row with no 1s at all and |
| 781 | * the puzzle is internally inconsistent. However, we ought |
| 782 | * to have caught this already during the simpler reasoning |
| 783 | * methods, so we can safely fail an assertion if we reach |
| 784 | * this point here. |
| 785 | */ |
| 786 | assert(count > 0); |
| 787 | if (count == 1) |
| 788 | rowidx[i] = colidx[first] = FALSE; |
| 789 | } |
| 790 | |
| 791 | /* |
| 792 | * Convert each of rowidx/colidx from a list of 0s and 1s to a |
| 793 | * list of the indices of the 1s. |
| 794 | */ |
| 795 | for (i = j = 0; i < cr; i++) |
| 796 | if (rowidx[i]) |
| 797 | rowidx[j++] = i; |
| 798 | n = j; |
| 799 | for (i = j = 0; i < cr; i++) |
| 800 | if (colidx[i]) |
| 801 | colidx[j++] = i; |
| 802 | assert(n == j); |
| 803 | |
| 804 | /* |
| 805 | * And create the smaller matrix. |
| 806 | */ |
| 807 | for (i = 0; i < n; i++) |
| 808 | for (j = 0; j < n; j++) |
| 809 | grid[i*cr+j] = usage->cube[indices[rowidx[i]*cr+colidx[j]]]; |
| 810 | |
| 811 | /* |
| 812 | * Having done that, we now have a matrix in which every row |
| 813 | * has at least two 1s in. Now we search to see if we can find |
| 814 | * a rectangle of zeroes (in the set-theoretic sense of |
| 815 | * `rectangle', i.e. a subset of rows crossed with a subset of |
| 816 | * columns) whose width and height add up to n. |
| 817 | */ |
| 818 | |
| 819 | memset(set, 0, n); |
| 820 | count = 0; |
| 821 | while (1) { |
| 822 | /* |
| 823 | * We have a candidate set. If its size is <=1 or >=n-1 |
| 824 | * then we move on immediately. |
| 825 | */ |
| 826 | if (count > 1 && count < n-1) { |
| 827 | /* |
| 828 | * The number of rows we need is n-count. See if we can |
| 829 | * find that many rows which each have a zero in all |
| 830 | * the positions listed in `set'. |
| 831 | */ |
| 832 | int rows = 0; |
| 833 | for (i = 0; i < n; i++) { |
| 834 | int ok = TRUE; |
| 835 | for (j = 0; j < n; j++) |
| 836 | if (set[j] && grid[i*cr+j]) { |
| 837 | ok = FALSE; |
| 838 | break; |
| 839 | } |
| 840 | if (ok) |
| 841 | rows++; |
| 842 | } |
| 843 | |
| 844 | /* |
| 845 | * We expect never to be able to get _more_ than |
| 846 | * n-count suitable rows: this would imply that (for |
| 847 | * example) there are four numbers which between them |
| 848 | * have at most three possible positions, and hence it |
| 849 | * indicates a faulty deduction before this point or |
| 850 | * even a bogus clue. |
| 851 | */ |
| 852 | if (rows > n - count) { |
| 853 | #ifdef STANDALONE_SOLVER |
| 854 | if (solver_show_working) { |
| 855 | va_list ap; |
| 856 | printf("%*s", solver_recurse_depth*4, |
| 857 | ""); |
| 858 | va_start(ap, fmt); |
| 859 | vprintf(fmt, ap); |
| 860 | va_end(ap); |
| 861 | printf(":\n%*s contradiction reached\n", |
| 862 | solver_recurse_depth*4, ""); |
| 863 | } |
| 864 | #endif |
| 865 | return -1; |
| 866 | } |
| 867 | |
| 868 | if (rows >= n - count) { |
| 869 | int progress = FALSE; |
| 870 | |
| 871 | /* |
| 872 | * We've got one! Now, for each row which _doesn't_ |
| 873 | * satisfy the criterion, eliminate all its set |
| 874 | * bits in the positions _not_ listed in `set'. |
| 875 | * Return +1 (meaning progress has been made) if we |
| 876 | * successfully eliminated anything at all. |
| 877 | * |
| 878 | * This involves referring back through |
| 879 | * rowidx/colidx in order to work out which actual |
| 880 | * positions in the cube to meddle with. |
| 881 | */ |
| 882 | for (i = 0; i < n; i++) { |
| 883 | int ok = TRUE; |
| 884 | for (j = 0; j < n; j++) |
| 885 | if (set[j] && grid[i*cr+j]) { |
| 886 | ok = FALSE; |
| 887 | break; |
| 888 | } |
| 889 | if (!ok) { |
| 890 | for (j = 0; j < n; j++) |
| 891 | if (!set[j] && grid[i*cr+j]) { |
| 892 | int fpos = indices[rowidx[i]*cr+colidx[j]]; |
| 893 | #ifdef STANDALONE_SOLVER |
| 894 | if (solver_show_working) { |
| 895 | int px, py, pn; |
| 896 | |
| 897 | if (!progress) { |
| 898 | va_list ap; |
| 899 | printf("%*s", solver_recurse_depth*4, |
| 900 | ""); |
| 901 | va_start(ap, fmt); |
| 902 | vprintf(fmt, ap); |
| 903 | va_end(ap); |
| 904 | printf(":\n"); |
| 905 | } |
| 906 | |
| 907 | pn = 1 + fpos % cr; |
| 908 | px = fpos / cr; |
| 909 | py = px / cr; |
| 910 | px %= cr; |
| 911 | |
| 912 | printf("%*s ruling out %d at (%d,%d)\n", |
| 913 | solver_recurse_depth*4, "", |
| 914 | pn, 1+px, 1+py); |
| 915 | } |
| 916 | #endif |
| 917 | progress = TRUE; |
| 918 | usage->cube[fpos] = FALSE; |
| 919 | } |
| 920 | } |
| 921 | } |
| 922 | |
| 923 | if (progress) { |
| 924 | return +1; |
| 925 | } |
| 926 | } |
| 927 | } |
| 928 | |
| 929 | /* |
| 930 | * Binary increment: change the rightmost 0 to a 1, and |
| 931 | * change all 1s to the right of it to 0s. |
| 932 | */ |
| 933 | i = n; |
| 934 | while (i > 0 && set[i-1]) |
| 935 | set[--i] = 0, count--; |
| 936 | if (i > 0) |
| 937 | set[--i] = 1, count++; |
| 938 | else |
| 939 | break; /* done */ |
| 940 | } |
| 941 | |
| 942 | return 0; |
| 943 | } |
| 944 | |
| 945 | /* |
| 946 | * Look for forcing chains. A forcing chain is a path of |
| 947 | * pairwise-exclusive squares (i.e. each pair of adjacent squares |
| 948 | * in the path are in the same row, column or block) with the |
| 949 | * following properties: |
| 950 | * |
| 951 | * (a) Each square on the path has precisely two possible numbers. |
| 952 | * |
| 953 | * (b) Each pair of squares which are adjacent on the path share |
| 954 | * at least one possible number in common. |
| 955 | * |
| 956 | * (c) Each square in the middle of the path shares _both_ of its |
| 957 | * numbers with at least one of its neighbours (not the same |
| 958 | * one with both neighbours). |
| 959 | * |
| 960 | * These together imply that at least one of the possible number |
| 961 | * choices at one end of the path forces _all_ the rest of the |
| 962 | * numbers along the path. In order to make real use of this, we |
| 963 | * need further properties: |
| 964 | * |
| 965 | * (c) Ruling out some number N from the square at one end of the |
| 966 | * path forces the square at the other end to take the same |
| 967 | * number N. |
| 968 | * |
| 969 | * (d) The two end squares are both in line with some third |
| 970 | * square. |
| 971 | * |
| 972 | * (e) That third square currently has N as a possibility. |
| 973 | * |
| 974 | * If we can find all of that lot, we can deduce that at least one |
| 975 | * of the two ends of the forcing chain has number N, and that |
| 976 | * therefore the mutually adjacent third square does not. |
| 977 | * |
| 978 | * To find forcing chains, we're going to start a bfs at each |
| 979 | * suitable square, once for each of its two possible numbers. |
| 980 | */ |
| 981 | static int solver_forcing(struct solver_usage *usage, |
| 982 | struct solver_scratch *scratch) |
| 983 | { |
| 984 | int cr = usage->cr; |
| 985 | int *bfsqueue = scratch->bfsqueue; |
| 986 | #ifdef STANDALONE_SOLVER |
| 987 | int *bfsprev = scratch->bfsprev; |
| 988 | #endif |
| 989 | unsigned char *number = scratch->grid; |
| 990 | int *neighbours = scratch->neighbours; |
| 991 | int x, y; |
| 992 | |
| 993 | for (y = 0; y < cr; y++) |
| 994 | for (x = 0; x < cr; x++) { |
| 995 | int count, t, n; |
| 996 | |
| 997 | /* |
| 998 | * If this square doesn't have exactly two candidate |
| 999 | * numbers, don't try it. |
| 1000 | * |
| 1001 | * In this loop we also sum the candidate numbers, |
| 1002 | * which is a nasty hack to allow us to quickly find |
| 1003 | * `the other one' (since we will shortly know there |
| 1004 | * are exactly two). |
| 1005 | */ |
| 1006 | for (count = t = 0, n = 1; n <= cr; n++) |
| 1007 | if (cube(x, y, n)) |
| 1008 | count++, t += n; |
| 1009 | if (count != 2) |
| 1010 | continue; |
| 1011 | |
| 1012 | /* |
| 1013 | * Now attempt a bfs for each candidate. |
| 1014 | */ |
| 1015 | for (n = 1; n <= cr; n++) |
| 1016 | if (cube(x, y, n)) { |
| 1017 | int orign, currn, head, tail; |
| 1018 | |
| 1019 | /* |
| 1020 | * Begin a bfs. |
| 1021 | */ |
| 1022 | orign = n; |
| 1023 | |
| 1024 | memset(number, cr+1, cr*cr); |
| 1025 | head = tail = 0; |
| 1026 | bfsqueue[tail++] = y*cr+x; |
| 1027 | #ifdef STANDALONE_SOLVER |
| 1028 | bfsprev[y*cr+x] = -1; |
| 1029 | #endif |
| 1030 | number[y*cr+x] = t - n; |
| 1031 | |
| 1032 | while (head < tail) { |
| 1033 | int xx, yy, nneighbours, xt, yt, i; |
| 1034 | |
| 1035 | xx = bfsqueue[head++]; |
| 1036 | yy = xx / cr; |
| 1037 | xx %= cr; |
| 1038 | |
| 1039 | currn = number[yy*cr+xx]; |
| 1040 | |
| 1041 | /* |
| 1042 | * Find neighbours of yy,xx. |
| 1043 | */ |
| 1044 | nneighbours = 0; |
| 1045 | for (yt = 0; yt < cr; yt++) |
| 1046 | neighbours[nneighbours++] = yt*cr+xx; |
| 1047 | for (xt = 0; xt < cr; xt++) |
| 1048 | neighbours[nneighbours++] = yy*cr+xt; |
| 1049 | xt = usage->blocks->whichblock[yy*cr+xx]; |
| 1050 | for (yt = 0; yt < cr; yt++) |
| 1051 | neighbours[nneighbours++] = usage->blocks->blocks[xt][yt]; |
| 1052 | if (usage->diag) { |
| 1053 | int sqindex = yy*cr+xx; |
| 1054 | if (ondiag0(sqindex)) { |
| 1055 | for (i = 0; i < cr; i++) |
| 1056 | neighbours[nneighbours++] = diag0(i); |
| 1057 | } |
| 1058 | if (ondiag1(sqindex)) { |
| 1059 | for (i = 0; i < cr; i++) |
| 1060 | neighbours[nneighbours++] = diag1(i); |
| 1061 | } |
| 1062 | } |
| 1063 | |
| 1064 | /* |
| 1065 | * Try visiting each of those neighbours. |
| 1066 | */ |
| 1067 | for (i = 0; i < nneighbours; i++) { |
| 1068 | int cc, tt, nn; |
| 1069 | |
| 1070 | xt = neighbours[i] % cr; |
| 1071 | yt = neighbours[i] / cr; |
| 1072 | |
| 1073 | /* |
| 1074 | * We need this square to not be |
| 1075 | * already visited, and to include |
| 1076 | * currn as a possible number. |
| 1077 | */ |
| 1078 | if (number[yt*cr+xt] <= cr) |
| 1079 | continue; |
| 1080 | if (!cube(xt, yt, currn)) |
| 1081 | continue; |
| 1082 | |
| 1083 | /* |
| 1084 | * Don't visit _this_ square a second |
| 1085 | * time! |
| 1086 | */ |
| 1087 | if (xt == xx && yt == yy) |
| 1088 | continue; |
| 1089 | |
| 1090 | /* |
| 1091 | * To continue with the bfs, we need |
| 1092 | * this square to have exactly two |
| 1093 | * possible numbers. |
| 1094 | */ |
| 1095 | for (cc = tt = 0, nn = 1; nn <= cr; nn++) |
| 1096 | if (cube(xt, yt, nn)) |
| 1097 | cc++, tt += nn; |
| 1098 | if (cc == 2) { |
| 1099 | bfsqueue[tail++] = yt*cr+xt; |
| 1100 | #ifdef STANDALONE_SOLVER |
| 1101 | bfsprev[yt*cr+xt] = yy*cr+xx; |
| 1102 | #endif |
| 1103 | number[yt*cr+xt] = tt - currn; |
| 1104 | } |
| 1105 | |
| 1106 | /* |
| 1107 | * One other possibility is that this |
| 1108 | * might be the square in which we can |
| 1109 | * make a real deduction: if it's |
| 1110 | * adjacent to x,y, and currn is equal |
| 1111 | * to the original number we ruled out. |
| 1112 | */ |
| 1113 | if (currn == orign && |
| 1114 | (xt == x || yt == y || |
| 1115 | (usage->blocks->whichblock[yt*cr+xt] == usage->blocks->whichblock[y*cr+x]) || |
| 1116 | (usage->diag && ((ondiag0(yt*cr+xt) && ondiag0(y*cr+x)) || |
| 1117 | (ondiag1(yt*cr+xt) && ondiag1(y*cr+x)))))) { |
| 1118 | #ifdef STANDALONE_SOLVER |
| 1119 | if (solver_show_working) { |
| 1120 | char *sep = ""; |
| 1121 | int xl, yl; |
| 1122 | printf("%*sforcing chain, %d at ends of ", |
| 1123 | solver_recurse_depth*4, "", orign); |
| 1124 | xl = xx; |
| 1125 | yl = yy; |
| 1126 | while (1) { |
| 1127 | printf("%s(%d,%d)", sep, 1+xl, |
| 1128 | 1+yl); |
| 1129 | xl = bfsprev[yl*cr+xl]; |
| 1130 | if (xl < 0) |
| 1131 | break; |
| 1132 | yl = xl / cr; |
| 1133 | xl %= cr; |
| 1134 | sep = "-"; |
| 1135 | } |
| 1136 | printf("\n%*s ruling out %d at (%d,%d)\n", |
| 1137 | solver_recurse_depth*4, "", |
| 1138 | orign, 1+xt, 1+yt); |
| 1139 | } |
| 1140 | #endif |
| 1141 | cube(xt, yt, orign) = FALSE; |
| 1142 | return 1; |
| 1143 | } |
| 1144 | } |
| 1145 | } |
| 1146 | } |
| 1147 | } |
| 1148 | |
| 1149 | return 0; |
| 1150 | } |
| 1151 | |
| 1152 | static struct solver_scratch *solver_new_scratch(struct solver_usage *usage) |
| 1153 | { |
| 1154 | struct solver_scratch *scratch = snew(struct solver_scratch); |
| 1155 | int cr = usage->cr; |
| 1156 | scratch->grid = snewn(cr*cr, unsigned char); |
| 1157 | scratch->rowidx = snewn(cr, unsigned char); |
| 1158 | scratch->colidx = snewn(cr, unsigned char); |
| 1159 | scratch->set = snewn(cr, unsigned char); |
| 1160 | scratch->neighbours = snewn(5*cr, int); |
| 1161 | scratch->bfsqueue = snewn(cr*cr, int); |
| 1162 | #ifdef STANDALONE_SOLVER |
| 1163 | scratch->bfsprev = snewn(cr*cr, int); |
| 1164 | #endif |
| 1165 | scratch->indexlist = snewn(cr*cr, int); /* used for set elimination */ |
| 1166 | scratch->indexlist2 = snewn(cr, int); /* only used for intersect() */ |
| 1167 | return scratch; |
| 1168 | } |
| 1169 | |
| 1170 | static void solver_free_scratch(struct solver_scratch *scratch) |
| 1171 | { |
| 1172 | #ifdef STANDALONE_SOLVER |
| 1173 | sfree(scratch->bfsprev); |
| 1174 | #endif |
| 1175 | sfree(scratch->bfsqueue); |
| 1176 | sfree(scratch->neighbours); |
| 1177 | sfree(scratch->set); |
| 1178 | sfree(scratch->colidx); |
| 1179 | sfree(scratch->rowidx); |
| 1180 | sfree(scratch->grid); |
| 1181 | sfree(scratch->indexlist); |
| 1182 | sfree(scratch->indexlist2); |
| 1183 | sfree(scratch); |
| 1184 | } |
| 1185 | |
| 1186 | static int solver(int cr, struct block_structure *blocks, int xtype, |
| 1187 | digit *grid, int maxdiff) |
| 1188 | { |
| 1189 | struct solver_usage *usage; |
| 1190 | struct solver_scratch *scratch; |
| 1191 | int x, y, b, i, n, ret; |
| 1192 | int diff = DIFF_BLOCK; |
| 1193 | |
| 1194 | /* |
| 1195 | * Set up a usage structure as a clean slate (everything |
| 1196 | * possible). |
| 1197 | */ |
| 1198 | usage = snew(struct solver_usage); |
| 1199 | usage->cr = cr; |
| 1200 | usage->blocks = blocks; |
| 1201 | usage->cube = snewn(cr*cr*cr, unsigned char); |
| 1202 | usage->grid = grid; /* write straight back to the input */ |
| 1203 | memset(usage->cube, TRUE, cr*cr*cr); |
| 1204 | |
| 1205 | usage->row = snewn(cr * cr, unsigned char); |
| 1206 | usage->col = snewn(cr * cr, unsigned char); |
| 1207 | usage->blk = snewn(cr * cr, unsigned char); |
| 1208 | memset(usage->row, FALSE, cr * cr); |
| 1209 | memset(usage->col, FALSE, cr * cr); |
| 1210 | memset(usage->blk, FALSE, cr * cr); |
| 1211 | |
| 1212 | if (xtype) { |
| 1213 | usage->diag = snewn(cr * 2, unsigned char); |
| 1214 | memset(usage->diag, FALSE, cr * 2); |
| 1215 | } else |
| 1216 | usage->diag = NULL; |
| 1217 | |
| 1218 | scratch = solver_new_scratch(usage); |
| 1219 | |
| 1220 | /* |
| 1221 | * Place all the clue numbers we are given. |
| 1222 | */ |
| 1223 | for (x = 0; x < cr; x++) |
| 1224 | for (y = 0; y < cr; y++) |
| 1225 | if (grid[y*cr+x]) |
| 1226 | solver_place(usage, x, y, grid[y*cr+x]); |
| 1227 | |
| 1228 | /* |
| 1229 | * Now loop over the grid repeatedly trying all permitted modes |
| 1230 | * of reasoning. The loop terminates if we complete an |
| 1231 | * iteration without making any progress; we then return |
| 1232 | * failure or success depending on whether the grid is full or |
| 1233 | * not. |
| 1234 | */ |
| 1235 | while (1) { |
| 1236 | /* |
| 1237 | * I'd like to write `continue;' inside each of the |
| 1238 | * following loops, so that the solver returns here after |
| 1239 | * making some progress. However, I can't specify that I |
| 1240 | * want to continue an outer loop rather than the innermost |
| 1241 | * one, so I'm apologetically resorting to a goto. |
| 1242 | */ |
| 1243 | cont: |
| 1244 | |
| 1245 | /* |
| 1246 | * Blockwise positional elimination. |
| 1247 | */ |
| 1248 | for (b = 0; b < cr; b++) |
| 1249 | for (n = 1; n <= cr; n++) |
| 1250 | if (!usage->blk[b*cr+n-1]) { |
| 1251 | for (i = 0; i < cr; i++) |
| 1252 | scratch->indexlist[i] = cubepos2(usage->blocks->blocks[b][i],n); |
| 1253 | ret = solver_elim(usage, scratch->indexlist |
| 1254 | #ifdef STANDALONE_SOLVER |
| 1255 | , "positional elimination," |
| 1256 | " %d in block %s", n, |
| 1257 | usage->blocks->blocknames[b] |
| 1258 | #endif |
| 1259 | ); |
| 1260 | if (ret < 0) { |
| 1261 | diff = DIFF_IMPOSSIBLE; |
| 1262 | goto got_result; |
| 1263 | } else if (ret > 0) { |
| 1264 | diff = max(diff, DIFF_BLOCK); |
| 1265 | goto cont; |
| 1266 | } |
| 1267 | } |
| 1268 | |
| 1269 | if (maxdiff <= DIFF_BLOCK) |
| 1270 | break; |
| 1271 | |
| 1272 | /* |
| 1273 | * Row-wise positional elimination. |
| 1274 | */ |
| 1275 | for (y = 0; y < cr; y++) |
| 1276 | for (n = 1; n <= cr; n++) |
| 1277 | if (!usage->row[y*cr+n-1]) { |
| 1278 | for (x = 0; x < cr; x++) |
| 1279 | scratch->indexlist[x] = cubepos(x, y, n); |
| 1280 | ret = solver_elim(usage, scratch->indexlist |
| 1281 | #ifdef STANDALONE_SOLVER |
| 1282 | , "positional elimination," |
| 1283 | " %d in row %d", n, 1+y |
| 1284 | #endif |
| 1285 | ); |
| 1286 | if (ret < 0) { |
| 1287 | diff = DIFF_IMPOSSIBLE; |
| 1288 | goto got_result; |
| 1289 | } else if (ret > 0) { |
| 1290 | diff = max(diff, DIFF_SIMPLE); |
| 1291 | goto cont; |
| 1292 | } |
| 1293 | } |
| 1294 | /* |
| 1295 | * Column-wise positional elimination. |
| 1296 | */ |
| 1297 | for (x = 0; x < cr; x++) |
| 1298 | for (n = 1; n <= cr; n++) |
| 1299 | if (!usage->col[x*cr+n-1]) { |
| 1300 | for (y = 0; y < cr; y++) |
| 1301 | scratch->indexlist[y] = cubepos(x, y, n); |
| 1302 | ret = solver_elim(usage, scratch->indexlist |
| 1303 | #ifdef STANDALONE_SOLVER |
| 1304 | , "positional elimination," |
| 1305 | " %d in column %d", n, 1+x |
| 1306 | #endif |
| 1307 | ); |
| 1308 | if (ret < 0) { |
| 1309 | diff = DIFF_IMPOSSIBLE; |
| 1310 | goto got_result; |
| 1311 | } else if (ret > 0) { |
| 1312 | diff = max(diff, DIFF_SIMPLE); |
| 1313 | goto cont; |
| 1314 | } |
| 1315 | } |
| 1316 | |
| 1317 | /* |
| 1318 | * X-diagonal positional elimination. |
| 1319 | */ |
| 1320 | if (usage->diag) { |
| 1321 | for (n = 1; n <= cr; n++) |
| 1322 | if (!usage->diag[n-1]) { |
| 1323 | for (i = 0; i < cr; i++) |
| 1324 | scratch->indexlist[i] = cubepos2(diag0(i), n); |
| 1325 | ret = solver_elim(usage, scratch->indexlist |
| 1326 | #ifdef STANDALONE_SOLVER |
| 1327 | , "positional elimination," |
| 1328 | " %d in \\-diagonal", n |
| 1329 | #endif |
| 1330 | ); |
| 1331 | if (ret < 0) { |
| 1332 | diff = DIFF_IMPOSSIBLE; |
| 1333 | goto got_result; |
| 1334 | } else if (ret > 0) { |
| 1335 | diff = max(diff, DIFF_SIMPLE); |
| 1336 | goto cont; |
| 1337 | } |
| 1338 | } |
| 1339 | for (n = 1; n <= cr; n++) |
| 1340 | if (!usage->diag[cr+n-1]) { |
| 1341 | for (i = 0; i < cr; i++) |
| 1342 | scratch->indexlist[i] = cubepos2(diag1(i), n); |
| 1343 | ret = solver_elim(usage, scratch->indexlist |
| 1344 | #ifdef STANDALONE_SOLVER |
| 1345 | , "positional elimination," |
| 1346 | " %d in /-diagonal", n |
| 1347 | #endif |
| 1348 | ); |
| 1349 | if (ret < 0) { |
| 1350 | diff = DIFF_IMPOSSIBLE; |
| 1351 | goto got_result; |
| 1352 | } else if (ret > 0) { |
| 1353 | diff = max(diff, DIFF_SIMPLE); |
| 1354 | goto cont; |
| 1355 | } |
| 1356 | } |
| 1357 | } |
| 1358 | |
| 1359 | /* |
| 1360 | * Numeric elimination. |
| 1361 | */ |
| 1362 | for (x = 0; x < cr; x++) |
| 1363 | for (y = 0; y < cr; y++) |
| 1364 | if (!usage->grid[y*cr+x]) { |
| 1365 | for (n = 1; n <= cr; n++) |
| 1366 | scratch->indexlist[n-1] = cubepos(x, y, n); |
| 1367 | ret = solver_elim(usage, scratch->indexlist |
| 1368 | #ifdef STANDALONE_SOLVER |
| 1369 | , "numeric elimination at (%d,%d)", |
| 1370 | 1+x, 1+y |
| 1371 | #endif |
| 1372 | ); |
| 1373 | if (ret < 0) { |
| 1374 | diff = DIFF_IMPOSSIBLE; |
| 1375 | goto got_result; |
| 1376 | } else if (ret > 0) { |
| 1377 | diff = max(diff, DIFF_SIMPLE); |
| 1378 | goto cont; |
| 1379 | } |
| 1380 | } |
| 1381 | |
| 1382 | if (maxdiff <= DIFF_SIMPLE) |
| 1383 | break; |
| 1384 | |
| 1385 | /* |
| 1386 | * Intersectional analysis, rows vs blocks. |
| 1387 | */ |
| 1388 | for (y = 0; y < cr; y++) |
| 1389 | for (b = 0; b < cr; b++) |
| 1390 | for (n = 1; n <= cr; n++) { |
| 1391 | if (usage->row[y*cr+n-1] || |
| 1392 | usage->blk[b*cr+n-1]) |
| 1393 | continue; |
| 1394 | for (i = 0; i < cr; i++) { |
| 1395 | scratch->indexlist[i] = cubepos(i, y, n); |
| 1396 | scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n); |
| 1397 | } |
| 1398 | /* |
| 1399 | * solver_intersect() never returns -1. |
| 1400 | */ |
| 1401 | if (solver_intersect(usage, scratch->indexlist, |
| 1402 | scratch->indexlist2 |
| 1403 | #ifdef STANDALONE_SOLVER |
| 1404 | , "intersectional analysis," |
| 1405 | " %d in row %d vs block %s", |
| 1406 | n, 1+y, usage->blocks->blocknames[b] |
| 1407 | #endif |
| 1408 | ) || |
| 1409 | solver_intersect(usage, scratch->indexlist2, |
| 1410 | scratch->indexlist |
| 1411 | #ifdef STANDALONE_SOLVER |
| 1412 | , "intersectional analysis," |
| 1413 | " %d in block %s vs row %d", |
| 1414 | n, usage->blocks->blocknames[b], 1+y |
| 1415 | #endif |
| 1416 | )) { |
| 1417 | diff = max(diff, DIFF_INTERSECT); |
| 1418 | goto cont; |
| 1419 | } |
| 1420 | } |
| 1421 | |
| 1422 | /* |
| 1423 | * Intersectional analysis, columns vs blocks. |
| 1424 | */ |
| 1425 | for (x = 0; x < cr; x++) |
| 1426 | for (b = 0; b < cr; b++) |
| 1427 | for (n = 1; n <= cr; n++) { |
| 1428 | if (usage->col[x*cr+n-1] || |
| 1429 | usage->blk[b*cr+n-1]) |
| 1430 | continue; |
| 1431 | for (i = 0; i < cr; i++) { |
| 1432 | scratch->indexlist[i] = cubepos(x, i, n); |
| 1433 | scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n); |
| 1434 | } |
| 1435 | if (solver_intersect(usage, scratch->indexlist, |
| 1436 | scratch->indexlist2 |
| 1437 | #ifdef STANDALONE_SOLVER |
| 1438 | , "intersectional analysis," |
| 1439 | " %d in column %d vs block %s", |
| 1440 | n, 1+x, usage->blocks->blocknames[b] |
| 1441 | #endif |
| 1442 | ) || |
| 1443 | solver_intersect(usage, scratch->indexlist2, |
| 1444 | scratch->indexlist |
| 1445 | #ifdef STANDALONE_SOLVER |
| 1446 | , "intersectional analysis," |
| 1447 | " %d in block %s vs column %d", |
| 1448 | n, usage->blocks->blocknames[b], 1+x |
| 1449 | #endif |
| 1450 | )) { |
| 1451 | diff = max(diff, DIFF_INTERSECT); |
| 1452 | goto cont; |
| 1453 | } |
| 1454 | } |
| 1455 | |
| 1456 | if (usage->diag) { |
| 1457 | /* |
| 1458 | * Intersectional analysis, \-diagonal vs blocks. |
| 1459 | */ |
| 1460 | for (b = 0; b < cr; b++) |
| 1461 | for (n = 1; n <= cr; n++) { |
| 1462 | if (usage->diag[n-1] || |
| 1463 | usage->blk[b*cr+n-1]) |
| 1464 | continue; |
| 1465 | for (i = 0; i < cr; i++) { |
| 1466 | scratch->indexlist[i] = cubepos2(diag0(i), n); |
| 1467 | scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n); |
| 1468 | } |
| 1469 | if (solver_intersect(usage, scratch->indexlist, |
| 1470 | scratch->indexlist2 |
| 1471 | #ifdef STANDALONE_SOLVER |
| 1472 | , "intersectional analysis," |
| 1473 | " %d in \\-diagonal vs block %s", |
| 1474 | n, 1+x, usage->blocks->blocknames[b] |
| 1475 | #endif |
| 1476 | ) || |
| 1477 | solver_intersect(usage, scratch->indexlist2, |
| 1478 | scratch->indexlist |
| 1479 | #ifdef STANDALONE_SOLVER |
| 1480 | , "intersectional analysis," |
| 1481 | " %d in block %s vs \\-diagonal", |
| 1482 | n, usage->blocks->blocknames[b], 1+x |
| 1483 | #endif |
| 1484 | )) { |
| 1485 | diff = max(diff, DIFF_INTERSECT); |
| 1486 | goto cont; |
| 1487 | } |
| 1488 | } |
| 1489 | |
| 1490 | /* |
| 1491 | * Intersectional analysis, /-diagonal vs blocks. |
| 1492 | */ |
| 1493 | for (b = 0; b < cr; b++) |
| 1494 | for (n = 1; n <= cr; n++) { |
| 1495 | if (usage->diag[cr+n-1] || |
| 1496 | usage->blk[b*cr+n-1]) |
| 1497 | continue; |
| 1498 | for (i = 0; i < cr; i++) { |
| 1499 | scratch->indexlist[i] = cubepos2(diag1(i), n); |
| 1500 | scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n); |
| 1501 | } |
| 1502 | if (solver_intersect(usage, scratch->indexlist, |
| 1503 | scratch->indexlist2 |
| 1504 | #ifdef STANDALONE_SOLVER |
| 1505 | , "intersectional analysis," |
| 1506 | " %d in /-diagonal vs block %s", |
| 1507 | n, 1+x, usage->blocks->blocknames[b] |
| 1508 | #endif |
| 1509 | ) || |
| 1510 | solver_intersect(usage, scratch->indexlist2, |
| 1511 | scratch->indexlist |
| 1512 | #ifdef STANDALONE_SOLVER |
| 1513 | , "intersectional analysis," |
| 1514 | " %d in block %s vs /-diagonal", |
| 1515 | n, usage->blocks->blocknames[b], 1+x |
| 1516 | #endif |
| 1517 | )) { |
| 1518 | diff = max(diff, DIFF_INTERSECT); |
| 1519 | goto cont; |
| 1520 | } |
| 1521 | } |
| 1522 | } |
| 1523 | |
| 1524 | if (maxdiff <= DIFF_INTERSECT) |
| 1525 | break; |
| 1526 | |
| 1527 | /* |
| 1528 | * Blockwise set elimination. |
| 1529 | */ |
| 1530 | for (b = 0; b < cr; b++) { |
| 1531 | for (i = 0; i < cr; i++) |
| 1532 | for (n = 1; n <= cr; n++) |
| 1533 | scratch->indexlist[i*cr+n-1] = cubepos2(usage->blocks->blocks[b][i], n); |
| 1534 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1535 | #ifdef STANDALONE_SOLVER |
| 1536 | , "set elimination, block %s", |
| 1537 | usage->blocks->blocknames[b] |
| 1538 | #endif |
| 1539 | ); |
| 1540 | if (ret < 0) { |
| 1541 | diff = DIFF_IMPOSSIBLE; |
| 1542 | goto got_result; |
| 1543 | } else if (ret > 0) { |
| 1544 | diff = max(diff, DIFF_SET); |
| 1545 | goto cont; |
| 1546 | } |
| 1547 | } |
| 1548 | |
| 1549 | /* |
| 1550 | * Row-wise set elimination. |
| 1551 | */ |
| 1552 | for (y = 0; y < cr; y++) { |
| 1553 | for (x = 0; x < cr; x++) |
| 1554 | for (n = 1; n <= cr; n++) |
| 1555 | scratch->indexlist[x*cr+n-1] = cubepos(x, y, n); |
| 1556 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1557 | #ifdef STANDALONE_SOLVER |
| 1558 | , "set elimination, row %d", 1+y |
| 1559 | #endif |
| 1560 | ); |
| 1561 | if (ret < 0) { |
| 1562 | diff = DIFF_IMPOSSIBLE; |
| 1563 | goto got_result; |
| 1564 | } else if (ret > 0) { |
| 1565 | diff = max(diff, DIFF_SET); |
| 1566 | goto cont; |
| 1567 | } |
| 1568 | } |
| 1569 | |
| 1570 | /* |
| 1571 | * Column-wise set elimination. |
| 1572 | */ |
| 1573 | for (x = 0; x < cr; x++) { |
| 1574 | for (y = 0; y < cr; y++) |
| 1575 | for (n = 1; n <= cr; n++) |
| 1576 | scratch->indexlist[y*cr+n-1] = cubepos(x, y, n); |
| 1577 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1578 | #ifdef STANDALONE_SOLVER |
| 1579 | , "set elimination, column %d", 1+x |
| 1580 | #endif |
| 1581 | ); |
| 1582 | if (ret < 0) { |
| 1583 | diff = DIFF_IMPOSSIBLE; |
| 1584 | goto got_result; |
| 1585 | } else if (ret > 0) { |
| 1586 | diff = max(diff, DIFF_SET); |
| 1587 | goto cont; |
| 1588 | } |
| 1589 | } |
| 1590 | |
| 1591 | if (usage->diag) { |
| 1592 | /* |
| 1593 | * \-diagonal set elimination. |
| 1594 | */ |
| 1595 | for (i = 0; i < cr; i++) |
| 1596 | for (n = 1; n <= cr; n++) |
| 1597 | scratch->indexlist[i*cr+n-1] = cubepos2(diag0(i), n); |
| 1598 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1599 | #ifdef STANDALONE_SOLVER |
| 1600 | , "set elimination, \\-diagonal" |
| 1601 | #endif |
| 1602 | ); |
| 1603 | if (ret < 0) { |
| 1604 | diff = DIFF_IMPOSSIBLE; |
| 1605 | goto got_result; |
| 1606 | } else if (ret > 0) { |
| 1607 | diff = max(diff, DIFF_SET); |
| 1608 | goto cont; |
| 1609 | } |
| 1610 | |
| 1611 | /* |
| 1612 | * /-diagonal set elimination. |
| 1613 | */ |
| 1614 | for (i = 0; i < cr; i++) |
| 1615 | for (n = 1; n <= cr; n++) |
| 1616 | scratch->indexlist[i*cr+n-1] = cubepos2(diag1(i), n); |
| 1617 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1618 | #ifdef STANDALONE_SOLVER |
| 1619 | , "set elimination, \\-diagonal" |
| 1620 | #endif |
| 1621 | ); |
| 1622 | if (ret < 0) { |
| 1623 | diff = DIFF_IMPOSSIBLE; |
| 1624 | goto got_result; |
| 1625 | } else if (ret > 0) { |
| 1626 | diff = max(diff, DIFF_SET); |
| 1627 | goto cont; |
| 1628 | } |
| 1629 | } |
| 1630 | |
| 1631 | if (maxdiff <= DIFF_SET) |
| 1632 | break; |
| 1633 | |
| 1634 | /* |
| 1635 | * Row-vs-column set elimination on a single number. |
| 1636 | */ |
| 1637 | for (n = 1; n <= cr; n++) { |
| 1638 | for (y = 0; y < cr; y++) |
| 1639 | for (x = 0; x < cr; x++) |
| 1640 | scratch->indexlist[y*cr+x] = cubepos(x, y, n); |
| 1641 | ret = solver_set(usage, scratch, scratch->indexlist |
| 1642 | #ifdef STANDALONE_SOLVER |
| 1643 | , "positional set elimination, number %d", n |
| 1644 | #endif |
| 1645 | ); |
| 1646 | if (ret < 0) { |
| 1647 | diff = DIFF_IMPOSSIBLE; |
| 1648 | goto got_result; |
| 1649 | } else if (ret > 0) { |
| 1650 | diff = max(diff, DIFF_EXTREME); |
| 1651 | goto cont; |
| 1652 | } |
| 1653 | } |
| 1654 | |
| 1655 | /* |
| 1656 | * Forcing chains. |
| 1657 | */ |
| 1658 | if (solver_forcing(usage, scratch)) { |
| 1659 | diff = max(diff, DIFF_EXTREME); |
| 1660 | goto cont; |
| 1661 | } |
| 1662 | |
| 1663 | /* |
| 1664 | * If we reach here, we have made no deductions in this |
| 1665 | * iteration, so the algorithm terminates. |
| 1666 | */ |
| 1667 | break; |
| 1668 | } |
| 1669 | |
| 1670 | /* |
| 1671 | * Last chance: if we haven't fully solved the puzzle yet, try |
| 1672 | * recursing based on guesses for a particular square. We pick |
| 1673 | * one of the most constrained empty squares we can find, which |
| 1674 | * has the effect of pruning the search tree as much as |
| 1675 | * possible. |
| 1676 | */ |
| 1677 | if (maxdiff >= DIFF_RECURSIVE) { |
| 1678 | int best, bestcount; |
| 1679 | |
| 1680 | best = -1; |
| 1681 | bestcount = cr+1; |
| 1682 | |
| 1683 | for (y = 0; y < cr; y++) |
| 1684 | for (x = 0; x < cr; x++) |
| 1685 | if (!grid[y*cr+x]) { |
| 1686 | int count; |
| 1687 | |
| 1688 | /* |
| 1689 | * An unfilled square. Count the number of |
| 1690 | * possible digits in it. |
| 1691 | */ |
| 1692 | count = 0; |
| 1693 | for (n = 1; n <= cr; n++) |
| 1694 | if (cube(x,y,n)) |
| 1695 | count++; |
| 1696 | |
| 1697 | /* |
| 1698 | * We should have found any impossibilities |
| 1699 | * already, so this can safely be an assert. |
| 1700 | */ |
| 1701 | assert(count > 1); |
| 1702 | |
| 1703 | if (count < bestcount) { |
| 1704 | bestcount = count; |
| 1705 | best = y*cr+x; |
| 1706 | } |
| 1707 | } |
| 1708 | |
| 1709 | if (best != -1) { |
| 1710 | int i, j; |
| 1711 | digit *list, *ingrid, *outgrid; |
| 1712 | |
| 1713 | diff = DIFF_IMPOSSIBLE; /* no solution found yet */ |
| 1714 | |
| 1715 | /* |
| 1716 | * Attempt recursion. |
| 1717 | */ |
| 1718 | y = best / cr; |
| 1719 | x = best % cr; |
| 1720 | |
| 1721 | list = snewn(cr, digit); |
| 1722 | ingrid = snewn(cr * cr, digit); |
| 1723 | outgrid = snewn(cr * cr, digit); |
| 1724 | memcpy(ingrid, grid, cr * cr); |
| 1725 | |
| 1726 | /* Make a list of the possible digits. */ |
| 1727 | for (j = 0, n = 1; n <= cr; n++) |
| 1728 | if (cube(x,y,n)) |
| 1729 | list[j++] = n; |
| 1730 | |
| 1731 | #ifdef STANDALONE_SOLVER |
| 1732 | if (solver_show_working) { |
| 1733 | char *sep = ""; |
| 1734 | printf("%*srecursing on (%d,%d) [", |
| 1735 | solver_recurse_depth*4, "", x + 1, y + 1); |
| 1736 | for (i = 0; i < j; i++) { |
| 1737 | printf("%s%d", sep, list[i]); |
| 1738 | sep = " or "; |
| 1739 | } |
| 1740 | printf("]\n"); |
| 1741 | } |
| 1742 | #endif |
| 1743 | |
| 1744 | /* |
| 1745 | * And step along the list, recursing back into the |
| 1746 | * main solver at every stage. |
| 1747 | */ |
| 1748 | for (i = 0; i < j; i++) { |
| 1749 | int ret; |
| 1750 | |
| 1751 | memcpy(outgrid, ingrid, cr * cr); |
| 1752 | outgrid[y*cr+x] = list[i]; |
| 1753 | |
| 1754 | #ifdef STANDALONE_SOLVER |
| 1755 | if (solver_show_working) |
| 1756 | printf("%*sguessing %d at (%d,%d)\n", |
| 1757 | solver_recurse_depth*4, "", list[i], x + 1, y + 1); |
| 1758 | solver_recurse_depth++; |
| 1759 | #endif |
| 1760 | |
| 1761 | ret = solver(cr, blocks, xtype, outgrid, maxdiff); |
| 1762 | |
| 1763 | #ifdef STANDALONE_SOLVER |
| 1764 | solver_recurse_depth--; |
| 1765 | if (solver_show_working) { |
| 1766 | printf("%*sretracting %d at (%d,%d)\n", |
| 1767 | solver_recurse_depth*4, "", list[i], x + 1, y + 1); |
| 1768 | } |
| 1769 | #endif |
| 1770 | |
| 1771 | /* |
| 1772 | * If we have our first solution, copy it into the |
| 1773 | * grid we will return. |
| 1774 | */ |
| 1775 | if (diff == DIFF_IMPOSSIBLE && ret != DIFF_IMPOSSIBLE) |
| 1776 | memcpy(grid, outgrid, cr*cr); |
| 1777 | |
| 1778 | if (ret == DIFF_AMBIGUOUS) |
| 1779 | diff = DIFF_AMBIGUOUS; |
| 1780 | else if (ret == DIFF_IMPOSSIBLE) |
| 1781 | /* do not change our return value */; |
| 1782 | else { |
| 1783 | /* the recursion turned up exactly one solution */ |
| 1784 | if (diff == DIFF_IMPOSSIBLE) |
| 1785 | diff = DIFF_RECURSIVE; |
| 1786 | else |
| 1787 | diff = DIFF_AMBIGUOUS; |
| 1788 | } |
| 1789 | |
| 1790 | /* |
| 1791 | * As soon as we've found more than one solution, |
| 1792 | * give up immediately. |
| 1793 | */ |
| 1794 | if (diff == DIFF_AMBIGUOUS) |
| 1795 | break; |
| 1796 | } |
| 1797 | |
| 1798 | sfree(outgrid); |
| 1799 | sfree(ingrid); |
| 1800 | sfree(list); |
| 1801 | } |
| 1802 | |
| 1803 | } else { |
| 1804 | /* |
| 1805 | * We're forbidden to use recursion, so we just see whether |
| 1806 | * our grid is fully solved, and return DIFF_IMPOSSIBLE |
| 1807 | * otherwise. |
| 1808 | */ |
| 1809 | for (y = 0; y < cr; y++) |
| 1810 | for (x = 0; x < cr; x++) |
| 1811 | if (!grid[y*cr+x]) |
| 1812 | diff = DIFF_IMPOSSIBLE; |
| 1813 | } |
| 1814 | |
| 1815 | got_result:; |
| 1816 | |
| 1817 | #ifdef STANDALONE_SOLVER |
| 1818 | if (solver_show_working) |
| 1819 | printf("%*s%s found\n", |
| 1820 | solver_recurse_depth*4, "", |
| 1821 | diff == DIFF_IMPOSSIBLE ? "no solution" : |
| 1822 | diff == DIFF_AMBIGUOUS ? "multiple solutions" : |
| 1823 | "one solution"); |
| 1824 | #endif |
| 1825 | |
| 1826 | sfree(usage->cube); |
| 1827 | sfree(usage->row); |
| 1828 | sfree(usage->col); |
| 1829 | sfree(usage->blk); |
| 1830 | sfree(usage); |
| 1831 | |
| 1832 | solver_free_scratch(scratch); |
| 1833 | |
| 1834 | return diff; |
| 1835 | } |
| 1836 | |
| 1837 | /* ---------------------------------------------------------------------- |
| 1838 | * End of solver code. |
| 1839 | */ |
| 1840 | |
| 1841 | /* ---------------------------------------------------------------------- |
| 1842 | * Solo filled-grid generator. |
| 1843 | * |
| 1844 | * This grid generator works by essentially trying to solve a grid |
| 1845 | * starting from no clues, and not worrying that there's more than |
| 1846 | * one possible solution. Unfortunately, it isn't computationally |
| 1847 | * feasible to do this by calling the above solver with an empty |
| 1848 | * grid, because that one needs to allocate a lot of scratch space |
| 1849 | * at every recursion level. Instead, I have a much simpler |
| 1850 | * algorithm which I shamelessly copied from a Python solver |
| 1851 | * written by Andrew Wilkinson (which is GPLed, but I've reused |
| 1852 | * only ideas and no code). It mostly just does the obvious |
| 1853 | * recursive thing: pick an empty square, put one of the possible |
| 1854 | * digits in it, recurse until all squares are filled, backtrack |
| 1855 | * and change some choices if necessary. |
| 1856 | * |
| 1857 | * The clever bit is that every time it chooses which square to |
| 1858 | * fill in next, it does so by counting the number of _possible_ |
| 1859 | * numbers that can go in each square, and it prioritises so that |
| 1860 | * it picks a square with the _lowest_ number of possibilities. The |
| 1861 | * idea is that filling in lots of the obvious bits (particularly |
| 1862 | * any squares with only one possibility) will cut down on the list |
| 1863 | * of possibilities for other squares and hence reduce the enormous |
| 1864 | * search space as much as possible as early as possible. |
| 1865 | */ |
| 1866 | |
| 1867 | /* |
| 1868 | * Internal data structure used in gridgen to keep track of |
| 1869 | * progress. |
| 1870 | */ |
| 1871 | struct gridgen_coord { int x, y, r; }; |
| 1872 | struct gridgen_usage { |
| 1873 | int cr; |
| 1874 | struct block_structure *blocks; |
| 1875 | /* grid is a copy of the input grid, modified as we go along */ |
| 1876 | digit *grid; |
| 1877 | /* row[y*cr+n-1] TRUE if digit n has been placed in row y */ |
| 1878 | unsigned char *row; |
| 1879 | /* col[x*cr+n-1] TRUE if digit n has been placed in row x */ |
| 1880 | unsigned char *col; |
| 1881 | /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */ |
| 1882 | unsigned char *blk; |
| 1883 | /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */ |
| 1884 | unsigned char *diag; |
| 1885 | /* This lists all the empty spaces remaining in the grid. */ |
| 1886 | struct gridgen_coord *spaces; |
| 1887 | int nspaces; |
| 1888 | /* If we need randomisation in the solve, this is our random state. */ |
| 1889 | random_state *rs; |
| 1890 | }; |
| 1891 | |
| 1892 | static void gridgen_place(struct gridgen_usage *usage, int x, int y, digit n, |
| 1893 | int placing) |
| 1894 | { |
| 1895 | int cr = usage->cr; |
| 1896 | usage->row[y*cr+n-1] = usage->col[x*cr+n-1] = |
| 1897 | usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n-1] = placing; |
| 1898 | if (usage->diag) { |
| 1899 | if (ondiag0(y*cr+x)) |
| 1900 | usage->diag[n-1] = placing; |
| 1901 | if (ondiag1(y*cr+x)) |
| 1902 | usage->diag[cr+n-1] = placing; |
| 1903 | } |
| 1904 | usage->grid[y*cr+x] = placing ? n : 0; |
| 1905 | } |
| 1906 | |
| 1907 | /* |
| 1908 | * The real recursive step in the generating function. |
| 1909 | * |
| 1910 | * Return values: 1 means solution found, 0 means no solution |
| 1911 | * found on this branch. |
| 1912 | */ |
| 1913 | static int gridgen_real(struct gridgen_usage *usage, digit *grid, int *steps) |
| 1914 | { |
| 1915 | int cr = usage->cr; |
| 1916 | int i, j, n, sx, sy, bestm, bestr, ret; |
| 1917 | int *digits; |
| 1918 | |
| 1919 | /* |
| 1920 | * Firstly, check for completion! If there are no spaces left |
| 1921 | * in the grid, we have a solution. |
| 1922 | */ |
| 1923 | if (usage->nspaces == 0) |
| 1924 | return TRUE; |
| 1925 | |
| 1926 | /* |
| 1927 | * Next, abandon generation if we went over our steps limit. |
| 1928 | */ |
| 1929 | if (*steps <= 0) |
| 1930 | return FALSE; |
| 1931 | (*steps)--; |
| 1932 | |
| 1933 | /* |
| 1934 | * Otherwise, there must be at least one space. Find the most |
| 1935 | * constrained space, using the `r' field as a tie-breaker. |
| 1936 | */ |
| 1937 | bestm = cr+1; /* so that any space will beat it */ |
| 1938 | bestr = 0; |
| 1939 | i = sx = sy = -1; |
| 1940 | for (j = 0; j < usage->nspaces; j++) { |
| 1941 | int x = usage->spaces[j].x, y = usage->spaces[j].y; |
| 1942 | int m; |
| 1943 | |
| 1944 | /* |
| 1945 | * Find the number of digits that could go in this space. |
| 1946 | */ |
| 1947 | m = 0; |
| 1948 | for (n = 0; n < cr; n++) |
| 1949 | if (!usage->row[y*cr+n] && !usage->col[x*cr+n] && |
| 1950 | !usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n] && |
| 1951 | (!usage->diag || ((!ondiag0(y*cr+x) || !usage->diag[n]) && |
| 1952 | (!ondiag1(y*cr+x) || !usage->diag[cr+n])))) |
| 1953 | m++; |
| 1954 | |
| 1955 | if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) { |
| 1956 | bestm = m; |
| 1957 | bestr = usage->spaces[j].r; |
| 1958 | sx = x; |
| 1959 | sy = y; |
| 1960 | i = j; |
| 1961 | } |
| 1962 | } |
| 1963 | |
| 1964 | /* |
| 1965 | * Swap that square into the final place in the spaces array, |
| 1966 | * so that decrementing nspaces will remove it from the list. |
| 1967 | */ |
| 1968 | if (i != usage->nspaces-1) { |
| 1969 | struct gridgen_coord t; |
| 1970 | t = usage->spaces[usage->nspaces-1]; |
| 1971 | usage->spaces[usage->nspaces-1] = usage->spaces[i]; |
| 1972 | usage->spaces[i] = t; |
| 1973 | } |
| 1974 | |
| 1975 | /* |
| 1976 | * Now we've decided which square to start our recursion at, |
| 1977 | * simply go through all possible values, shuffling them |
| 1978 | * randomly first if necessary. |
| 1979 | */ |
| 1980 | digits = snewn(bestm, int); |
| 1981 | j = 0; |
| 1982 | for (n = 0; n < cr; n++) |
| 1983 | if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] && |
| 1984 | !usage->blk[usage->blocks->whichblock[sy*cr+sx]*cr+n] && |
| 1985 | (!usage->diag || ((!ondiag0(sy*cr+sx) || !usage->diag[n]) && |
| 1986 | (!ondiag1(sy*cr+sx) || !usage->diag[cr+n])))) { |
| 1987 | digits[j++] = n+1; |
| 1988 | } |
| 1989 | |
| 1990 | if (usage->rs) |
| 1991 | shuffle(digits, j, sizeof(*digits), usage->rs); |
| 1992 | |
| 1993 | /* And finally, go through the digit list and actually recurse. */ |
| 1994 | ret = FALSE; |
| 1995 | for (i = 0; i < j; i++) { |
| 1996 | n = digits[i]; |
| 1997 | |
| 1998 | /* Update the usage structure to reflect the placing of this digit. */ |
| 1999 | gridgen_place(usage, sx, sy, n, TRUE); |
| 2000 | usage->nspaces--; |
| 2001 | |
| 2002 | /* Call the solver recursively. Stop when we find a solution. */ |
| 2003 | if (gridgen_real(usage, grid, steps)) { |
| 2004 | ret = TRUE; |
| 2005 | break; |
| 2006 | } |
| 2007 | |
| 2008 | /* Revert the usage structure. */ |
| 2009 | gridgen_place(usage, sx, sy, n, FALSE); |
| 2010 | usage->nspaces++; |
| 2011 | } |
| 2012 | |
| 2013 | sfree(digits); |
| 2014 | return ret; |
| 2015 | } |
| 2016 | |
| 2017 | /* |
| 2018 | * Entry point to generator. You give it parameters and a starting |
| 2019 | * grid, which is simply an array of cr*cr digits. |
| 2020 | */ |
| 2021 | static int gridgen(int cr, struct block_structure *blocks, int xtype, |
| 2022 | digit *grid, random_state *rs, int maxsteps) |
| 2023 | { |
| 2024 | struct gridgen_usage *usage; |
| 2025 | int x, y, ret; |
| 2026 | |
| 2027 | /* |
| 2028 | * Clear the grid to start with. |
| 2029 | */ |
| 2030 | memset(grid, 0, cr*cr); |
| 2031 | |
| 2032 | /* |
| 2033 | * Create a gridgen_usage structure. |
| 2034 | */ |
| 2035 | usage = snew(struct gridgen_usage); |
| 2036 | |
| 2037 | usage->cr = cr; |
| 2038 | usage->blocks = blocks; |
| 2039 | |
| 2040 | usage->grid = grid; |
| 2041 | |
| 2042 | usage->row = snewn(cr * cr, unsigned char); |
| 2043 | usage->col = snewn(cr * cr, unsigned char); |
| 2044 | usage->blk = snewn(cr * cr, unsigned char); |
| 2045 | memset(usage->row, FALSE, cr * cr); |
| 2046 | memset(usage->col, FALSE, cr * cr); |
| 2047 | memset(usage->blk, FALSE, cr * cr); |
| 2048 | |
| 2049 | if (xtype) { |
| 2050 | usage->diag = snewn(2 * cr, unsigned char); |
| 2051 | memset(usage->diag, FALSE, 2 * cr); |
| 2052 | } else { |
| 2053 | usage->diag = NULL; |
| 2054 | } |
| 2055 | |
| 2056 | /* |
| 2057 | * Begin by filling in the whole top row with randomly chosen |
| 2058 | * numbers. This cannot introduce any bias or restriction on |
| 2059 | * the available grids, since we already know those numbers |
| 2060 | * are all distinct so all we're doing is choosing their |
| 2061 | * labels. |
| 2062 | */ |
| 2063 | for (x = 0; x < cr; x++) |
| 2064 | grid[x] = x+1; |
| 2065 | shuffle(grid, cr, sizeof(*grid), rs); |
| 2066 | for (x = 0; x < cr; x++) |
| 2067 | gridgen_place(usage, x, 0, grid[x], TRUE); |
| 2068 | |
| 2069 | usage->spaces = snewn(cr * cr, struct gridgen_coord); |
| 2070 | usage->nspaces = 0; |
| 2071 | |
| 2072 | usage->rs = rs; |
| 2073 | |
| 2074 | /* |
| 2075 | * Initialise the list of grid spaces, taking care to leave |
| 2076 | * out the row I've already filled in above. |
| 2077 | */ |
| 2078 | for (y = 1; y < cr; y++) { |
| 2079 | for (x = 0; x < cr; x++) { |
| 2080 | usage->spaces[usage->nspaces].x = x; |
| 2081 | usage->spaces[usage->nspaces].y = y; |
| 2082 | usage->spaces[usage->nspaces].r = random_bits(rs, 31); |
| 2083 | usage->nspaces++; |
| 2084 | } |
| 2085 | } |
| 2086 | |
| 2087 | /* |
| 2088 | * Run the real generator function. |
| 2089 | */ |
| 2090 | ret = gridgen_real(usage, grid, &maxsteps); |
| 2091 | |
| 2092 | /* |
| 2093 | * Clean up the usage structure now we have our answer. |
| 2094 | */ |
| 2095 | sfree(usage->spaces); |
| 2096 | sfree(usage->blk); |
| 2097 | sfree(usage->col); |
| 2098 | sfree(usage->row); |
| 2099 | sfree(usage); |
| 2100 | |
| 2101 | return ret; |
| 2102 | } |
| 2103 | |
| 2104 | /* ---------------------------------------------------------------------- |
| 2105 | * End of grid generator code. |
| 2106 | */ |
| 2107 | |
| 2108 | /* |
| 2109 | * Check whether a grid contains a valid complete puzzle. |
| 2110 | */ |
| 2111 | static int check_valid(int cr, struct block_structure *blocks, int xtype, |
| 2112 | digit *grid) |
| 2113 | { |
| 2114 | unsigned char *used; |
| 2115 | int x, y, i, j, n; |
| 2116 | |
| 2117 | used = snewn(cr, unsigned char); |
| 2118 | |
| 2119 | /* |
| 2120 | * Check that each row contains precisely one of everything. |
| 2121 | */ |
| 2122 | for (y = 0; y < cr; y++) { |
| 2123 | memset(used, FALSE, cr); |
| 2124 | for (x = 0; x < cr; x++) |
| 2125 | if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr) |
| 2126 | used[grid[y*cr+x]-1] = TRUE; |
| 2127 | for (n = 0; n < cr; n++) |
| 2128 | if (!used[n]) { |
| 2129 | sfree(used); |
| 2130 | return FALSE; |
| 2131 | } |
| 2132 | } |
| 2133 | |
| 2134 | /* |
| 2135 | * Check that each column contains precisely one of everything. |
| 2136 | */ |
| 2137 | for (x = 0; x < cr; x++) { |
| 2138 | memset(used, FALSE, cr); |
| 2139 | for (y = 0; y < cr; y++) |
| 2140 | if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr) |
| 2141 | used[grid[y*cr+x]-1] = TRUE; |
| 2142 | for (n = 0; n < cr; n++) |
| 2143 | if (!used[n]) { |
| 2144 | sfree(used); |
| 2145 | return FALSE; |
| 2146 | } |
| 2147 | } |
| 2148 | |
| 2149 | /* |
| 2150 | * Check that each block contains precisely one of everything. |
| 2151 | */ |
| 2152 | for (i = 0; i < cr; i++) { |
| 2153 | memset(used, FALSE, cr); |
| 2154 | for (j = 0; j < cr; j++) |
| 2155 | if (grid[blocks->blocks[i][j]] > 0 && |
| 2156 | grid[blocks->blocks[i][j]] <= cr) |
| 2157 | used[grid[blocks->blocks[i][j]]-1] = TRUE; |
| 2158 | for (n = 0; n < cr; n++) |
| 2159 | if (!used[n]) { |
| 2160 | sfree(used); |
| 2161 | return FALSE; |
| 2162 | } |
| 2163 | } |
| 2164 | |
| 2165 | /* |
| 2166 | * Check that each diagonal contains precisely one of everything. |
| 2167 | */ |
| 2168 | if (xtype) { |
| 2169 | memset(used, FALSE, cr); |
| 2170 | for (i = 0; i < cr; i++) |
| 2171 | if (grid[diag0(i)] > 0 && grid[diag0(i)] <= cr) |
| 2172 | used[grid[diag0(i)]-1] = TRUE; |
| 2173 | for (n = 0; n < cr; n++) |
| 2174 | if (!used[n]) { |
| 2175 | sfree(used); |
| 2176 | return FALSE; |
| 2177 | } |
| 2178 | for (i = 0; i < cr; i++) |
| 2179 | if (grid[diag1(i)] > 0 && grid[diag1(i)] <= cr) |
| 2180 | used[grid[diag1(i)]-1] = TRUE; |
| 2181 | for (n = 0; n < cr; n++) |
| 2182 | if (!used[n]) { |
| 2183 | sfree(used); |
| 2184 | return FALSE; |
| 2185 | } |
| 2186 | } |
| 2187 | |
| 2188 | sfree(used); |
| 2189 | return TRUE; |
| 2190 | } |
| 2191 | |
| 2192 | static int symmetries(game_params *params, int x, int y, int *output, int s) |
| 2193 | { |
| 2194 | int c = params->c, r = params->r, cr = c*r; |
| 2195 | int i = 0; |
| 2196 | |
| 2197 | #define ADD(x,y) (*output++ = (x), *output++ = (y), i++) |
| 2198 | |
| 2199 | ADD(x, y); |
| 2200 | |
| 2201 | switch (s) { |
| 2202 | case SYMM_NONE: |
| 2203 | break; /* just x,y is all we need */ |
| 2204 | case SYMM_ROT2: |
| 2205 | ADD(cr - 1 - x, cr - 1 - y); |
| 2206 | break; |
| 2207 | case SYMM_ROT4: |
| 2208 | ADD(cr - 1 - y, x); |
| 2209 | ADD(y, cr - 1 - x); |
| 2210 | ADD(cr - 1 - x, cr - 1 - y); |
| 2211 | break; |
| 2212 | case SYMM_REF2: |
| 2213 | ADD(cr - 1 - x, y); |
| 2214 | break; |
| 2215 | case SYMM_REF2D: |
| 2216 | ADD(y, x); |
| 2217 | break; |
| 2218 | case SYMM_REF4: |
| 2219 | ADD(cr - 1 - x, y); |
| 2220 | ADD(x, cr - 1 - y); |
| 2221 | ADD(cr - 1 - x, cr - 1 - y); |
| 2222 | break; |
| 2223 | case SYMM_REF4D: |
| 2224 | ADD(y, x); |
| 2225 | ADD(cr - 1 - x, cr - 1 - y); |
| 2226 | ADD(cr - 1 - y, cr - 1 - x); |
| 2227 | break; |
| 2228 | case SYMM_REF8: |
| 2229 | ADD(cr - 1 - x, y); |
| 2230 | ADD(x, cr - 1 - y); |
| 2231 | ADD(cr - 1 - x, cr - 1 - y); |
| 2232 | ADD(y, x); |
| 2233 | ADD(y, cr - 1 - x); |
| 2234 | ADD(cr - 1 - y, x); |
| 2235 | ADD(cr - 1 - y, cr - 1 - x); |
| 2236 | break; |
| 2237 | } |
| 2238 | |
| 2239 | #undef ADD |
| 2240 | |
| 2241 | return i; |
| 2242 | } |
| 2243 | |
| 2244 | static char *encode_solve_move(int cr, digit *grid) |
| 2245 | { |
| 2246 | int i, len; |
| 2247 | char *ret, *p, *sep; |
| 2248 | |
| 2249 | /* |
| 2250 | * It's surprisingly easy to work out _exactly_ how long this |
| 2251 | * string needs to be. To decimal-encode all the numbers from 1 |
| 2252 | * to n: |
| 2253 | * |
| 2254 | * - every number has a units digit; total is n. |
| 2255 | * - all numbers above 9 have a tens digit; total is max(n-9,0). |
| 2256 | * - all numbers above 99 have a hundreds digit; total is max(n-99,0). |
| 2257 | * - and so on. |
| 2258 | */ |
| 2259 | len = 0; |
| 2260 | for (i = 1; i <= cr; i *= 10) |
| 2261 | len += max(cr - i + 1, 0); |
| 2262 | len += cr; /* don't forget the commas */ |
| 2263 | len *= cr; /* there are cr rows of these */ |
| 2264 | |
| 2265 | /* |
| 2266 | * Now len is one bigger than the total size of the |
| 2267 | * comma-separated numbers (because we counted an |
| 2268 | * additional leading comma). We need to have a leading S |
| 2269 | * and a trailing NUL, so we're off by one in total. |
| 2270 | */ |
| 2271 | len++; |
| 2272 | |
| 2273 | ret = snewn(len, char); |
| 2274 | p = ret; |
| 2275 | *p++ = 'S'; |
| 2276 | sep = ""; |
| 2277 | for (i = 0; i < cr*cr; i++) { |
| 2278 | p += sprintf(p, "%s%d", sep, grid[i]); |
| 2279 | sep = ","; |
| 2280 | } |
| 2281 | *p++ = '\0'; |
| 2282 | assert(p - ret == len); |
| 2283 | |
| 2284 | return ret; |
| 2285 | } |
| 2286 | |
| 2287 | static char *new_game_desc(game_params *params, random_state *rs, |
| 2288 | char **aux, int interactive) |
| 2289 | { |
| 2290 | int c = params->c, r = params->r, cr = c*r; |
| 2291 | int area = cr*cr; |
| 2292 | struct block_structure *blocks; |
| 2293 | digit *grid, *grid2; |
| 2294 | struct xy { int x, y; } *locs; |
| 2295 | int nlocs; |
| 2296 | char *desc; |
| 2297 | int coords[16], ncoords; |
| 2298 | int maxdiff; |
| 2299 | int x, y, i, j; |
| 2300 | |
| 2301 | /* |
| 2302 | * Adjust the maximum difficulty level to be consistent with |
| 2303 | * the puzzle size: all 2x2 puzzles appear to be Trivial |
| 2304 | * (DIFF_BLOCK) so we cannot hold out for even a Basic |
| 2305 | * (DIFF_SIMPLE) one. |
| 2306 | */ |
| 2307 | maxdiff = params->diff; |
| 2308 | if (c == 2 && r == 2) |
| 2309 | maxdiff = DIFF_BLOCK; |
| 2310 | |
| 2311 | grid = snewn(area, digit); |
| 2312 | locs = snewn(area, struct xy); |
| 2313 | grid2 = snewn(area, digit); |
| 2314 | |
| 2315 | blocks = snew(struct block_structure); |
| 2316 | blocks->c = params->c; blocks->r = params->r; |
| 2317 | blocks->whichblock = snewn(area*2, int); |
| 2318 | blocks->blocks = snewn(cr, int *); |
| 2319 | for (i = 0; i < cr; i++) |
| 2320 | blocks->blocks[i] = blocks->whichblock + area + i*cr; |
| 2321 | #ifdef STANDALONE_SOLVER |
| 2322 | assert(!"This should never happen, so we don't need to create blocknames"); |
| 2323 | #endif |
| 2324 | |
| 2325 | /* |
| 2326 | * Loop until we get a grid of the required difficulty. This is |
| 2327 | * nasty, but it seems to be unpleasantly hard to generate |
| 2328 | * difficult grids otherwise. |
| 2329 | */ |
| 2330 | while (1) { |
| 2331 | /* |
| 2332 | * Generate a random solved state, starting by |
| 2333 | * constructing the block structure. |
| 2334 | */ |
| 2335 | if (r == 1) { /* jigsaw mode */ |
| 2336 | int *dsf = divvy_rectangle(cr, cr, cr, rs); |
| 2337 | int nb = 0; |
| 2338 | |
| 2339 | for (i = 0; i < area; i++) |
| 2340 | blocks->whichblock[i] = -1; |
| 2341 | for (i = 0; i < area; i++) { |
| 2342 | int j = dsf_canonify(dsf, i); |
| 2343 | if (blocks->whichblock[j] < 0) |
| 2344 | blocks->whichblock[j] = nb++; |
| 2345 | blocks->whichblock[i] = blocks->whichblock[j]; |
| 2346 | } |
| 2347 | assert(nb == cr); |
| 2348 | |
| 2349 | sfree(dsf); |
| 2350 | } else { /* basic Sudoku mode */ |
| 2351 | for (y = 0; y < cr; y++) |
| 2352 | for (x = 0; x < cr; x++) |
| 2353 | blocks->whichblock[y*cr+x] = (y/c) * c + (x/r); |
| 2354 | } |
| 2355 | for (i = 0; i < cr; i++) |
| 2356 | blocks->blocks[i][cr-1] = 0; |
| 2357 | for (i = 0; i < area; i++) { |
| 2358 | int b = blocks->whichblock[i]; |
| 2359 | j = blocks->blocks[b][cr-1]++; |
| 2360 | assert(j < cr); |
| 2361 | blocks->blocks[b][j] = i; |
| 2362 | } |
| 2363 | |
| 2364 | if (!gridgen(cr, blocks, params->xtype, grid, rs, area*area)) |
| 2365 | continue; |
| 2366 | assert(check_valid(cr, blocks, params->xtype, grid)); |
| 2367 | |
| 2368 | /* |
| 2369 | * Save the solved grid in aux. |
| 2370 | */ |
| 2371 | { |
| 2372 | /* |
| 2373 | * We might already have written *aux the last time we |
| 2374 | * went round this loop, in which case we should free |
| 2375 | * the old aux before overwriting it with the new one. |
| 2376 | */ |
| 2377 | if (*aux) { |
| 2378 | sfree(*aux); |
| 2379 | } |
| 2380 | |
| 2381 | *aux = encode_solve_move(cr, grid); |
| 2382 | } |
| 2383 | |
| 2384 | /* |
| 2385 | * Now we have a solved grid, start removing things from it |
| 2386 | * while preserving solubility. |
| 2387 | */ |
| 2388 | |
| 2389 | /* |
| 2390 | * Find the set of equivalence classes of squares permitted |
| 2391 | * by the selected symmetry. We do this by enumerating all |
| 2392 | * the grid squares which have no symmetric companion |
| 2393 | * sorting lower than themselves. |
| 2394 | */ |
| 2395 | nlocs = 0; |
| 2396 | for (y = 0; y < cr; y++) |
| 2397 | for (x = 0; x < cr; x++) { |
| 2398 | int i = y*cr+x; |
| 2399 | int j; |
| 2400 | |
| 2401 | ncoords = symmetries(params, x, y, coords, params->symm); |
| 2402 | for (j = 0; j < ncoords; j++) |
| 2403 | if (coords[2*j+1]*cr+coords[2*j] < i) |
| 2404 | break; |
| 2405 | if (j == ncoords) { |
| 2406 | locs[nlocs].x = x; |
| 2407 | locs[nlocs].y = y; |
| 2408 | nlocs++; |
| 2409 | } |
| 2410 | } |
| 2411 | |
| 2412 | /* |
| 2413 | * Now shuffle that list. |
| 2414 | */ |
| 2415 | shuffle(locs, nlocs, sizeof(*locs), rs); |
| 2416 | |
| 2417 | /* |
| 2418 | * Now loop over the shuffled list and, for each element, |
| 2419 | * see whether removing that element (and its reflections) |
| 2420 | * from the grid will still leave the grid soluble. |
| 2421 | */ |
| 2422 | for (i = 0; i < nlocs; i++) { |
| 2423 | int ret; |
| 2424 | |
| 2425 | x = locs[i].x; |
| 2426 | y = locs[i].y; |
| 2427 | |
| 2428 | memcpy(grid2, grid, area); |
| 2429 | ncoords = symmetries(params, x, y, coords, params->symm); |
| 2430 | for (j = 0; j < ncoords; j++) |
| 2431 | grid2[coords[2*j+1]*cr+coords[2*j]] = 0; |
| 2432 | |
| 2433 | ret = solver(cr, blocks, params->xtype, grid2, maxdiff); |
| 2434 | if (ret <= maxdiff) { |
| 2435 | for (j = 0; j < ncoords; j++) |
| 2436 | grid[coords[2*j+1]*cr+coords[2*j]] = 0; |
| 2437 | } |
| 2438 | } |
| 2439 | |
| 2440 | memcpy(grid2, grid, area); |
| 2441 | |
| 2442 | if (solver(cr, blocks, params->xtype, grid2, maxdiff) == maxdiff) |
| 2443 | break; /* found one! */ |
| 2444 | } |
| 2445 | |
| 2446 | sfree(grid2); |
| 2447 | sfree(locs); |
| 2448 | |
| 2449 | /* |
| 2450 | * Now we have the grid as it will be presented to the user. |
| 2451 | * Encode it in a game desc. |
| 2452 | */ |
| 2453 | { |
| 2454 | char *p; |
| 2455 | int run, i; |
| 2456 | |
| 2457 | desc = snewn(7 * area, char); |
| 2458 | p = desc; |
| 2459 | run = 0; |
| 2460 | for (i = 0; i <= area; i++) { |
| 2461 | int n = (i < area ? grid[i] : -1); |
| 2462 | |
| 2463 | if (!n) |
| 2464 | run++; |
| 2465 | else { |
| 2466 | if (run) { |
| 2467 | while (run > 0) { |
| 2468 | int c = 'a' - 1 + run; |
| 2469 | if (run > 26) |
| 2470 | c = 'z'; |
| 2471 | *p++ = c; |
| 2472 | run -= c - ('a' - 1); |
| 2473 | } |
| 2474 | } else { |
| 2475 | /* |
| 2476 | * If there's a number in the very top left or |
| 2477 | * bottom right, there's no point putting an |
| 2478 | * unnecessary _ before or after it. |
| 2479 | */ |
| 2480 | if (p > desc && n > 0) |
| 2481 | *p++ = '_'; |
| 2482 | } |
| 2483 | if (n > 0) |
| 2484 | p += sprintf(p, "%d", n); |
| 2485 | run = 0; |
| 2486 | } |
| 2487 | } |
| 2488 | |
| 2489 | if (r == 1) { |
| 2490 | int currrun = 0; |
| 2491 | |
| 2492 | *p++ = ','; |
| 2493 | |
| 2494 | /* |
| 2495 | * Encode the block structure. We do this by encoding |
| 2496 | * the pattern of dividing lines: first we iterate |
| 2497 | * over the cr*(cr-1) internal vertical grid lines in |
| 2498 | * ordinary reading order, then over the cr*(cr-1) |
| 2499 | * internal horizontal ones in transposed reading |
| 2500 | * order. |
| 2501 | * |
| 2502 | * We encode the number of non-lines between the |
| 2503 | * lines; _ means zero (two adjacent divisions), a |
| 2504 | * means 1, ..., y means 25, and z means 25 non-lines |
| 2505 | * _and no following line_ (so that za means 26, zb 27 |
| 2506 | * etc). |
| 2507 | */ |
| 2508 | for (i = 0; i <= 2*cr*(cr-1); i++) { |
| 2509 | int p0, p1, edge; |
| 2510 | |
| 2511 | if (i == 2*cr*(cr-1)) { |
| 2512 | edge = TRUE; /* terminating virtual edge */ |
| 2513 | } else { |
| 2514 | if (i < cr*(cr-1)) { |
| 2515 | y = i/(cr-1); |
| 2516 | x = i%(cr-1); |
| 2517 | p0 = y*cr+x; |
| 2518 | p1 = y*cr+x+1; |
| 2519 | } else { |
| 2520 | x = i/(cr-1) - cr; |
| 2521 | y = i%(cr-1); |
| 2522 | p0 = y*cr+x; |
| 2523 | p1 = (y+1)*cr+x; |
| 2524 | } |
| 2525 | edge = (blocks->whichblock[p0] != blocks->whichblock[p1]); |
| 2526 | } |
| 2527 | |
| 2528 | if (edge) { |
| 2529 | while (currrun > 25) |
| 2530 | *p++ = 'z', currrun -= 25; |
| 2531 | if (currrun) |
| 2532 | *p++ = 'a'-1 + currrun; |
| 2533 | else |
| 2534 | *p++ = '_'; |
| 2535 | currrun = 0; |
| 2536 | } else |
| 2537 | currrun++; |
| 2538 | } |
| 2539 | } |
| 2540 | |
| 2541 | assert(p - desc < 7 * area); |
| 2542 | *p++ = '\0'; |
| 2543 | desc = sresize(desc, p - desc, char); |
| 2544 | } |
| 2545 | |
| 2546 | sfree(grid); |
| 2547 | |
| 2548 | return desc; |
| 2549 | } |
| 2550 | |
| 2551 | static char *validate_desc(game_params *params, char *desc) |
| 2552 | { |
| 2553 | int cr = params->c * params->r, area = cr*cr; |
| 2554 | int squares = 0; |
| 2555 | int *dsf; |
| 2556 | |
| 2557 | while (*desc && *desc != ',') { |
| 2558 | int n = *desc++; |
| 2559 | if (n >= 'a' && n <= 'z') { |
| 2560 | squares += n - 'a' + 1; |
| 2561 | } else if (n == '_') { |
| 2562 | /* do nothing */; |
| 2563 | } else if (n > '0' && n <= '9') { |
| 2564 | int val = atoi(desc-1); |
| 2565 | if (val < 1 || val > params->c * params->r) |
| 2566 | return "Out-of-range number in game description"; |
| 2567 | squares++; |
| 2568 | while (*desc >= '0' && *desc <= '9') |
| 2569 | desc++; |
| 2570 | } else |
| 2571 | return "Invalid character in game description"; |
| 2572 | } |
| 2573 | |
| 2574 | if (squares < area) |
| 2575 | return "Not enough data to fill grid"; |
| 2576 | |
| 2577 | if (squares > area) |
| 2578 | return "Too much data to fit in grid"; |
| 2579 | |
| 2580 | if (params->r == 1) { |
| 2581 | int pos; |
| 2582 | |
| 2583 | /* |
| 2584 | * Now we expect a suffix giving the jigsaw block |
| 2585 | * structure. Parse it and validate that it divides the |
| 2586 | * grid into the right number of regions which are the |
| 2587 | * right size. |
| 2588 | */ |
| 2589 | if (*desc != ',') |
| 2590 | return "Expected jigsaw block structure in game description"; |
| 2591 | pos = 0; |
| 2592 | |
| 2593 | dsf = snew_dsf(area); |
| 2594 | desc++; |
| 2595 | |
| 2596 | while (*desc) { |
| 2597 | int c, adv; |
| 2598 | |
| 2599 | if (*desc == '_') |
| 2600 | c = 0; |
| 2601 | else if (*desc >= 'a' && *desc <= 'z') |
| 2602 | c = *desc - 'a' + 1; |
| 2603 | else { |
| 2604 | sfree(dsf); |
| 2605 | return "Invalid character in game description"; |
| 2606 | } |
| 2607 | desc++; |
| 2608 | |
| 2609 | adv = (c != 25); /* 'z' is a special case */ |
| 2610 | |
| 2611 | while (c-- > 0) { |
| 2612 | int p0, p1; |
| 2613 | |
| 2614 | /* |
| 2615 | * Non-edge; merge the two dsf classes on either |
| 2616 | * side of it. |
| 2617 | */ |
| 2618 | if (pos >= 2*cr*(cr-1)) { |
| 2619 | sfree(dsf); |
| 2620 | return "Too much data in block structure specification"; |
| 2621 | } else if (pos < cr*(cr-1)) { |
| 2622 | int y = pos/(cr-1); |
| 2623 | int x = pos%(cr-1); |
| 2624 | p0 = y*cr+x; |
| 2625 | p1 = y*cr+x+1; |
| 2626 | } else { |
| 2627 | int x = pos/(cr-1) - cr; |
| 2628 | int y = pos%(cr-1); |
| 2629 | p0 = y*cr+x; |
| 2630 | p1 = (y+1)*cr+x; |
| 2631 | } |
| 2632 | dsf_merge(dsf, p0, p1); |
| 2633 | |
| 2634 | pos++; |
| 2635 | } |
| 2636 | if (adv) |
| 2637 | pos++; |
| 2638 | } |
| 2639 | |
| 2640 | /* |
| 2641 | * When desc is exhausted, we expect to have gone exactly |
| 2642 | * one space _past_ the end of the grid, due to the dummy |
| 2643 | * edge at the end. |
| 2644 | */ |
| 2645 | if (pos != 2*cr*(cr-1)+1) { |
| 2646 | sfree(dsf); |
| 2647 | return "Not enough data in block structure specification"; |
| 2648 | } |
| 2649 | |
| 2650 | /* |
| 2651 | * Now we've got our dsf. Verify that it matches |
| 2652 | * expectations. |
| 2653 | */ |
| 2654 | { |
| 2655 | int *canons, *counts; |
| 2656 | int i, j, c, ncanons = 0; |
| 2657 | |
| 2658 | canons = snewn(cr, int); |
| 2659 | counts = snewn(cr, int); |
| 2660 | |
| 2661 | for (i = 0; i < area; i++) { |
| 2662 | j = dsf_canonify(dsf, i); |
| 2663 | |
| 2664 | for (c = 0; c < ncanons; c++) |
| 2665 | if (canons[c] == j) { |
| 2666 | counts[c]++; |
| 2667 | if (counts[c] > cr) { |
| 2668 | sfree(dsf); |
| 2669 | sfree(canons); |
| 2670 | sfree(counts); |
| 2671 | return "A jigsaw block is too big"; |
| 2672 | } |
| 2673 | break; |
| 2674 | } |
| 2675 | |
| 2676 | if (c == ncanons) { |
| 2677 | if (ncanons >= cr) { |
| 2678 | sfree(dsf); |
| 2679 | sfree(canons); |
| 2680 | sfree(counts); |
| 2681 | return "Too many distinct jigsaw blocks"; |
| 2682 | } |
| 2683 | canons[ncanons] = j; |
| 2684 | counts[ncanons] = 1; |
| 2685 | ncanons++; |
| 2686 | } |
| 2687 | } |
| 2688 | |
| 2689 | /* |
| 2690 | * If we've managed to get through that loop without |
| 2691 | * tripping either of the error conditions, then we |
| 2692 | * must have partitioned the entire grid into at most |
| 2693 | * cr blocks of at most cr squares each; therefore we |
| 2694 | * must have _exactly_ cr blocks of _exactly_ cr |
| 2695 | * squares each. I'll verify that by assertion just in |
| 2696 | * case something has gone horribly wrong, but it |
| 2697 | * shouldn't have been able to happen by duff input, |
| 2698 | * only by a bug in the above code. |
| 2699 | */ |
| 2700 | assert(ncanons == cr); |
| 2701 | for (c = 0; c < ncanons; c++) |
| 2702 | assert(counts[c] == cr); |
| 2703 | |
| 2704 | sfree(canons); |
| 2705 | sfree(counts); |
| 2706 | } |
| 2707 | |
| 2708 | sfree(dsf); |
| 2709 | } else { |
| 2710 | if (*desc) |
| 2711 | return "Unexpected jigsaw block structure in game description"; |
| 2712 | } |
| 2713 | |
| 2714 | return NULL; |
| 2715 | } |
| 2716 | |
| 2717 | static game_state *new_game(midend *me, game_params *params, char *desc) |
| 2718 | { |
| 2719 | game_state *state = snew(game_state); |
| 2720 | int c = params->c, r = params->r, cr = c*r, area = cr * cr; |
| 2721 | int i; |
| 2722 | |
| 2723 | state->cr = cr; |
| 2724 | state->xtype = params->xtype; |
| 2725 | |
| 2726 | state->grid = snewn(area, digit); |
| 2727 | state->pencil = snewn(area * cr, unsigned char); |
| 2728 | memset(state->pencil, 0, area * cr); |
| 2729 | state->immutable = snewn(area, unsigned char); |
| 2730 | memset(state->immutable, FALSE, area); |
| 2731 | |
| 2732 | state->blocks = snew(struct block_structure); |
| 2733 | state->blocks->c = c; state->blocks->r = r; |
| 2734 | state->blocks->refcount = 1; |
| 2735 | state->blocks->whichblock = snewn(area*2, int); |
| 2736 | state->blocks->blocks = snewn(cr, int *); |
| 2737 | for (i = 0; i < cr; i++) |
| 2738 | state->blocks->blocks[i] = state->blocks->whichblock + area + i*cr; |
| 2739 | #ifdef STANDALONE_SOLVER |
| 2740 | state->blocks->blocknames = (char **)smalloc(cr*(sizeof(char *)+80)); |
| 2741 | #endif |
| 2742 | |
| 2743 | state->completed = state->cheated = FALSE; |
| 2744 | |
| 2745 | i = 0; |
| 2746 | while (*desc && *desc != ',') { |
| 2747 | int n = *desc++; |
| 2748 | if (n >= 'a' && n <= 'z') { |
| 2749 | int run = n - 'a' + 1; |
| 2750 | assert(i + run <= area); |
| 2751 | while (run-- > 0) |
| 2752 | state->grid[i++] = 0; |
| 2753 | } else if (n == '_') { |
| 2754 | /* do nothing */; |
| 2755 | } else if (n > '0' && n <= '9') { |
| 2756 | assert(i < area); |
| 2757 | state->immutable[i] = TRUE; |
| 2758 | state->grid[i++] = atoi(desc-1); |
| 2759 | while (*desc >= '0' && *desc <= '9') |
| 2760 | desc++; |
| 2761 | } else { |
| 2762 | assert(!"We can't get here"); |
| 2763 | } |
| 2764 | } |
| 2765 | assert(i == area); |
| 2766 | |
| 2767 | if (r == 1) { |
| 2768 | int pos = 0; |
| 2769 | int *dsf; |
| 2770 | int nb; |
| 2771 | |
| 2772 | assert(*desc == ','); |
| 2773 | |
| 2774 | dsf = snew_dsf(area); |
| 2775 | desc++; |
| 2776 | |
| 2777 | while (*desc) { |
| 2778 | int c, adv; |
| 2779 | |
| 2780 | if (*desc == '_') |
| 2781 | c = 0; |
| 2782 | else { |
| 2783 | assert(*desc >= 'a' && *desc <= 'z'); |
| 2784 | c = *desc - 'a' + 1; |
| 2785 | } |
| 2786 | desc++; |
| 2787 | |
| 2788 | adv = (c != 25); /* 'z' is a special case */ |
| 2789 | |
| 2790 | while (c-- > 0) { |
| 2791 | int p0, p1; |
| 2792 | |
| 2793 | /* |
| 2794 | * Non-edge; merge the two dsf classes on either |
| 2795 | * side of it. |
| 2796 | */ |
| 2797 | assert(pos < 2*cr*(cr-1)); |
| 2798 | if (pos < cr*(cr-1)) { |
| 2799 | int y = pos/(cr-1); |
| 2800 | int x = pos%(cr-1); |
| 2801 | p0 = y*cr+x; |
| 2802 | p1 = y*cr+x+1; |
| 2803 | } else { |
| 2804 | int x = pos/(cr-1) - cr; |
| 2805 | int y = pos%(cr-1); |
| 2806 | p0 = y*cr+x; |
| 2807 | p1 = (y+1)*cr+x; |
| 2808 | } |
| 2809 | dsf_merge(dsf, p0, p1); |
| 2810 | |
| 2811 | pos++; |
| 2812 | } |
| 2813 | if (adv) |
| 2814 | pos++; |
| 2815 | } |
| 2816 | |
| 2817 | /* |
| 2818 | * When desc is exhausted, we expect to have gone exactly |
| 2819 | * one space _past_ the end of the grid, due to the dummy |
| 2820 | * edge at the end. |
| 2821 | */ |
| 2822 | assert(pos == 2*cr*(cr-1)+1); |
| 2823 | |
| 2824 | /* |
| 2825 | * Now we've got our dsf. Translate it into a block |
| 2826 | * structure. |
| 2827 | */ |
| 2828 | nb = 0; |
| 2829 | for (i = 0; i < area; i++) |
| 2830 | state->blocks->whichblock[i] = -1; |
| 2831 | for (i = 0; i < area; i++) { |
| 2832 | int j = dsf_canonify(dsf, i); |
| 2833 | if (state->blocks->whichblock[j] < 0) |
| 2834 | state->blocks->whichblock[j] = nb++; |
| 2835 | state->blocks->whichblock[i] = state->blocks->whichblock[j]; |
| 2836 | } |
| 2837 | assert(nb == cr); |
| 2838 | |
| 2839 | sfree(dsf); |
| 2840 | } else { |
| 2841 | int x, y; |
| 2842 | |
| 2843 | assert(!*desc); |
| 2844 | |
| 2845 | for (y = 0; y < cr; y++) |
| 2846 | for (x = 0; x < cr; x++) |
| 2847 | state->blocks->whichblock[y*cr+x] = (y/c) * c + (x/r); |
| 2848 | } |
| 2849 | |
| 2850 | /* |
| 2851 | * Having sorted out whichblock[], set up the block index arrays. |
| 2852 | */ |
| 2853 | for (i = 0; i < cr; i++) |
| 2854 | state->blocks->blocks[i][cr-1] = 0; |
| 2855 | for (i = 0; i < area; i++) { |
| 2856 | int b = state->blocks->whichblock[i]; |
| 2857 | int j = state->blocks->blocks[b][cr-1]++; |
| 2858 | assert(j < cr); |
| 2859 | state->blocks->blocks[b][j] = i; |
| 2860 | } |
| 2861 | |
| 2862 | #ifdef STANDALONE_SOLVER |
| 2863 | /* |
| 2864 | * Set up the block names for solver diagnostic output. |
| 2865 | */ |
| 2866 | { |
| 2867 | char *p = (char *)(state->blocks->blocknames + cr); |
| 2868 | |
| 2869 | if (r == 1) { |
| 2870 | for (i = 0; i < cr; i++) |
| 2871 | state->blocks->blocknames[i] = NULL; |
| 2872 | |
| 2873 | for (i = 0; i < area; i++) { |
| 2874 | int j = state->blocks->whichblock[i]; |
| 2875 | if (!state->blocks->blocknames[j]) { |
| 2876 | state->blocks->blocknames[j] = p; |
| 2877 | p += 1 + sprintf(p, "starting at (%d,%d)", |
| 2878 | 1 + i%cr, 1 + i/cr); |
| 2879 | } |
| 2880 | } |
| 2881 | } else { |
| 2882 | int bx, by; |
| 2883 | for (by = 0; by < r; by++) |
| 2884 | for (bx = 0; bx < c; bx++) { |
| 2885 | state->blocks->blocknames[by*c+bx] = p; |
| 2886 | p += 1 + sprintf(p, "(%d,%d)", bx+1, by+1); |
| 2887 | } |
| 2888 | } |
| 2889 | assert(p - (char *)state->blocks->blocknames < cr*(sizeof(char *)+80)); |
| 2890 | for (i = 0; i < cr; i++) |
| 2891 | assert(state->blocks->blocknames[i]); |
| 2892 | } |
| 2893 | #endif |
| 2894 | |
| 2895 | return state; |
| 2896 | } |
| 2897 | |
| 2898 | static game_state *dup_game(game_state *state) |
| 2899 | { |
| 2900 | game_state *ret = snew(game_state); |
| 2901 | int cr = state->cr, area = cr * cr; |
| 2902 | |
| 2903 | ret->cr = state->cr; |
| 2904 | ret->xtype = state->xtype; |
| 2905 | |
| 2906 | ret->blocks = state->blocks; |
| 2907 | ret->blocks->refcount++; |
| 2908 | |
| 2909 | ret->grid = snewn(area, digit); |
| 2910 | memcpy(ret->grid, state->grid, area); |
| 2911 | |
| 2912 | ret->pencil = snewn(area * cr, unsigned char); |
| 2913 | memcpy(ret->pencil, state->pencil, area * cr); |
| 2914 | |
| 2915 | ret->immutable = snewn(area, unsigned char); |
| 2916 | memcpy(ret->immutable, state->immutable, area); |
| 2917 | |
| 2918 | ret->completed = state->completed; |
| 2919 | ret->cheated = state->cheated; |
| 2920 | |
| 2921 | return ret; |
| 2922 | } |
| 2923 | |
| 2924 | static void free_game(game_state *state) |
| 2925 | { |
| 2926 | if (--state->blocks->refcount == 0) { |
| 2927 | sfree(state->blocks->whichblock); |
| 2928 | sfree(state->blocks->blocks); |
| 2929 | #ifdef STANDALONE_SOLVER |
| 2930 | sfree(state->blocks->blocknames); |
| 2931 | #endif |
| 2932 | sfree(state->blocks); |
| 2933 | } |
| 2934 | sfree(state->immutable); |
| 2935 | sfree(state->pencil); |
| 2936 | sfree(state->grid); |
| 2937 | sfree(state); |
| 2938 | } |
| 2939 | |
| 2940 | static char *solve_game(game_state *state, game_state *currstate, |
| 2941 | char *ai, char **error) |
| 2942 | { |
| 2943 | int cr = state->cr; |
| 2944 | char *ret; |
| 2945 | digit *grid; |
| 2946 | int solve_ret; |
| 2947 | |
| 2948 | /* |
| 2949 | * If we already have the solution in ai, save ourselves some |
| 2950 | * time. |
| 2951 | */ |
| 2952 | if (ai) |
| 2953 | return dupstr(ai); |
| 2954 | |
| 2955 | grid = snewn(cr*cr, digit); |
| 2956 | memcpy(grid, state->grid, cr*cr); |
| 2957 | solve_ret = solver(cr, state->blocks, state->xtype, grid, DIFF_RECURSIVE); |
| 2958 | |
| 2959 | *error = NULL; |
| 2960 | |
| 2961 | if (solve_ret == DIFF_IMPOSSIBLE) |
| 2962 | *error = "No solution exists for this puzzle"; |
| 2963 | else if (solve_ret == DIFF_AMBIGUOUS) |
| 2964 | *error = "Multiple solutions exist for this puzzle"; |
| 2965 | |
| 2966 | if (*error) { |
| 2967 | sfree(grid); |
| 2968 | return NULL; |
| 2969 | } |
| 2970 | |
| 2971 | ret = encode_solve_move(cr, grid); |
| 2972 | |
| 2973 | sfree(grid); |
| 2974 | |
| 2975 | return ret; |
| 2976 | } |
| 2977 | |
| 2978 | static char *grid_text_format(int cr, struct block_structure *blocks, |
| 2979 | int xtype, digit *grid) |
| 2980 | { |
| 2981 | int vmod, hmod; |
| 2982 | int x, y; |
| 2983 | int totallen, linelen, nlines; |
| 2984 | char *ret, *p, ch; |
| 2985 | |
| 2986 | /* |
| 2987 | * For non-jigsaw Sudoku, we format in the way we always have, |
| 2988 | * by having the digits unevenly spaced so that the dividing |
| 2989 | * lines can fit in: |
| 2990 | * |
| 2991 | * . . | . . |
| 2992 | * . . | . . |
| 2993 | * ----+---- |
| 2994 | * . . | . . |
| 2995 | * . . | . . |
| 2996 | * |
| 2997 | * For jigsaw puzzles, however, we must leave space between |
| 2998 | * _all_ pairs of digits for an optional dividing line, so we |
| 2999 | * have to move to the rather ugly |
| 3000 | * |
| 3001 | * . . . . |
| 3002 | * ------+------ |
| 3003 | * . . | . . |
| 3004 | * +---+ |
| 3005 | * . . | . | . |
| 3006 | * ------+ | |
| 3007 | * . . . | . |
| 3008 | * |
| 3009 | * We deal with both cases using the same formatting code; we |
| 3010 | * simply invent a vmod value such that there's a vertical |
| 3011 | * dividing line before column i iff i is divisible by vmod |
| 3012 | * (so it's r in the first case and 1 in the second), and hmod |
| 3013 | * likewise for horizontal dividing lines. |
| 3014 | */ |
| 3015 | |
| 3016 | if (blocks->r != 1) { |
| 3017 | vmod = blocks->r; |
| 3018 | hmod = blocks->c; |
| 3019 | } else { |
| 3020 | vmod = hmod = 1; |
| 3021 | } |
| 3022 | |
| 3023 | /* |
| 3024 | * Line length: we have cr digits, each with a space after it, |
| 3025 | * and (cr-1)/vmod dividing lines, each with a space after it. |
| 3026 | * The final space is replaced by a newline, but that doesn't |
| 3027 | * affect the length. |
| 3028 | */ |
| 3029 | linelen = 2*(cr + (cr-1)/vmod); |
| 3030 | |
| 3031 | /* |
| 3032 | * Number of lines: we have cr rows of digits, and (cr-1)/hmod |
| 3033 | * dividing rows. |
| 3034 | */ |
| 3035 | nlines = cr + (cr-1)/hmod; |
| 3036 | |
| 3037 | /* |
| 3038 | * Allocate the space. |
| 3039 | */ |
| 3040 | totallen = linelen * nlines; |
| 3041 | ret = snewn(totallen+1, char); /* leave room for terminating NUL */ |
| 3042 | |
| 3043 | /* |
| 3044 | * Write the text. |
| 3045 | */ |
| 3046 | p = ret; |
| 3047 | for (y = 0; y < cr; y++) { |
| 3048 | /* |
| 3049 | * Row of digits. |
| 3050 | */ |
| 3051 | for (x = 0; x < cr; x++) { |
| 3052 | /* |
| 3053 | * Digit. |
| 3054 | */ |
| 3055 | digit d = grid[y*cr+x]; |
| 3056 | |
| 3057 | if (d == 0) { |
| 3058 | /* |
| 3059 | * Empty space: we usually write a dot, but we'll |
| 3060 | * highlight spaces on the X-diagonals (in X mode) |
| 3061 | * by using underscores instead. |
| 3062 | */ |
| 3063 | if (xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x))) |
| 3064 | ch = '_'; |
| 3065 | else |
| 3066 | ch = '.'; |
| 3067 | } else if (d <= 9) { |
| 3068 | ch = '0' + d; |
| 3069 | } else { |
| 3070 | ch = 'a' + d-10; |
| 3071 | } |
| 3072 | |
| 3073 | *p++ = ch; |
| 3074 | if (x == cr-1) { |
| 3075 | *p++ = '\n'; |
| 3076 | continue; |
| 3077 | } |
| 3078 | *p++ = ' '; |
| 3079 | |
| 3080 | if ((x+1) % vmod) |
| 3081 | continue; |
| 3082 | |
| 3083 | /* |
| 3084 | * Optional dividing line. |
| 3085 | */ |
| 3086 | if (blocks->whichblock[y*cr+x] != blocks->whichblock[y*cr+x+1]) |
| 3087 | ch = '|'; |
| 3088 | else |
| 3089 | ch = ' '; |
| 3090 | *p++ = ch; |
| 3091 | *p++ = ' '; |
| 3092 | } |
| 3093 | if (y == cr-1 || (y+1) % hmod) |
| 3094 | continue; |
| 3095 | |
| 3096 | /* |
| 3097 | * Dividing row. |
| 3098 | */ |
| 3099 | for (x = 0; x < cr; x++) { |
| 3100 | int dwid; |
| 3101 | int tl, tr, bl, br; |
| 3102 | |
| 3103 | /* |
| 3104 | * Division between two squares. This varies |
| 3105 | * complicatedly in length. |
| 3106 | */ |
| 3107 | dwid = 2; /* digit and its following space */ |
| 3108 | if (x == cr-1) |
| 3109 | dwid--; /* no following space at end of line */ |
| 3110 | if (x > 0 && x % vmod == 0) |
| 3111 | dwid++; /* preceding space after a divider */ |
| 3112 | |
| 3113 | if (blocks->whichblock[y*cr+x] != blocks->whichblock[(y+1)*cr+x]) |
| 3114 | ch = '-'; |
| 3115 | else |
| 3116 | ch = ' '; |
| 3117 | |
| 3118 | while (dwid-- > 0) |
| 3119 | *p++ = ch; |
| 3120 | |
| 3121 | if (x == cr-1) { |
| 3122 | *p++ = '\n'; |
| 3123 | break; |
| 3124 | } |
| 3125 | |
| 3126 | if ((x+1) % vmod) |
| 3127 | continue; |
| 3128 | |
| 3129 | /* |
| 3130 | * Corner square. This is: |
| 3131 | * - a space if all four surrounding squares are in |
| 3132 | * the same block |
| 3133 | * - a vertical line if the two left ones are in one |
| 3134 | * block and the two right in another |
| 3135 | * - a horizontal line if the two top ones are in one |
| 3136 | * block and the two bottom in another |
| 3137 | * - a plus sign in all other cases. (If we had a |
| 3138 | * richer character set available we could break |
| 3139 | * this case up further by doing fun things with |
| 3140 | * line-drawing T-pieces.) |
| 3141 | */ |
| 3142 | tl = blocks->whichblock[y*cr+x]; |
| 3143 | tr = blocks->whichblock[y*cr+x+1]; |
| 3144 | bl = blocks->whichblock[(y+1)*cr+x]; |
| 3145 | br = blocks->whichblock[(y+1)*cr+x+1]; |
| 3146 | |
| 3147 | if (tl == tr && tr == bl && bl == br) |
| 3148 | ch = ' '; |
| 3149 | else if (tl == bl && tr == br) |
| 3150 | ch = '|'; |
| 3151 | else if (tl == tr && bl == br) |
| 3152 | ch = '-'; |
| 3153 | else |
| 3154 | ch = '+'; |
| 3155 | |
| 3156 | *p++ = ch; |
| 3157 | } |
| 3158 | } |
| 3159 | |
| 3160 | assert(p - ret == totallen); |
| 3161 | *p = '\0'; |
| 3162 | return ret; |
| 3163 | } |
| 3164 | |
| 3165 | static int game_can_format_as_text_now(game_params *params) |
| 3166 | { |
| 3167 | return TRUE; |
| 3168 | } |
| 3169 | |
| 3170 | static char *game_text_format(game_state *state) |
| 3171 | { |
| 3172 | return grid_text_format(state->cr, state->blocks, state->xtype, |
| 3173 | state->grid); |
| 3174 | } |
| 3175 | |
| 3176 | struct game_ui { |
| 3177 | /* |
| 3178 | * These are the coordinates of the currently highlighted |
| 3179 | * square on the grid, or -1,-1 if there isn't one. When there |
| 3180 | * is, pressing a valid number or letter key or Space will |
| 3181 | * enter that number or letter in the grid. |
| 3182 | */ |
| 3183 | int hx, hy; |
| 3184 | /* |
| 3185 | * This indicates whether the current highlight is a |
| 3186 | * pencil-mark one or a real one. |
| 3187 | */ |
| 3188 | int hpencil; |
| 3189 | }; |
| 3190 | |
| 3191 | static game_ui *new_ui(game_state *state) |
| 3192 | { |
| 3193 | game_ui *ui = snew(game_ui); |
| 3194 | |
| 3195 | ui->hx = ui->hy = -1; |
| 3196 | ui->hpencil = 0; |
| 3197 | |
| 3198 | return ui; |
| 3199 | } |
| 3200 | |
| 3201 | static void free_ui(game_ui *ui) |
| 3202 | { |
| 3203 | sfree(ui); |
| 3204 | } |
| 3205 | |
| 3206 | static char *encode_ui(game_ui *ui) |
| 3207 | { |
| 3208 | return NULL; |
| 3209 | } |
| 3210 | |
| 3211 | static void decode_ui(game_ui *ui, char *encoding) |
| 3212 | { |
| 3213 | } |
| 3214 | |
| 3215 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
| 3216 | game_state *newstate) |
| 3217 | { |
| 3218 | int cr = newstate->cr; |
| 3219 | /* |
| 3220 | * We prevent pencil-mode highlighting of a filled square. So |
| 3221 | * if the user has just filled in a square which we had a |
| 3222 | * pencil-mode highlight in (by Undo, or by Redo, or by Solve), |
| 3223 | * then we cancel the highlight. |
| 3224 | */ |
| 3225 | if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil && |
| 3226 | newstate->grid[ui->hy * cr + ui->hx] != 0) { |
| 3227 | ui->hx = ui->hy = -1; |
| 3228 | } |
| 3229 | } |
| 3230 | |
| 3231 | struct game_drawstate { |
| 3232 | int started; |
| 3233 | int cr, xtype; |
| 3234 | int tilesize; |
| 3235 | digit *grid; |
| 3236 | unsigned char *pencil; |
| 3237 | unsigned char *hl; |
| 3238 | /* This is scratch space used within a single call to game_redraw. */ |
| 3239 | int *entered_items; |
| 3240 | }; |
| 3241 | |
| 3242 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
| 3243 | int x, int y, int button) |
| 3244 | { |
| 3245 | int cr = state->cr; |
| 3246 | int tx, ty; |
| 3247 | char buf[80]; |
| 3248 | |
| 3249 | button &= ~MOD_MASK; |
| 3250 | |
| 3251 | tx = (x + TILE_SIZE - BORDER) / TILE_SIZE - 1; |
| 3252 | ty = (y + TILE_SIZE - BORDER) / TILE_SIZE - 1; |
| 3253 | |
| 3254 | if (tx >= 0 && tx < cr && ty >= 0 && ty < cr) { |
| 3255 | if (button == LEFT_BUTTON) { |
| 3256 | if (state->immutable[ty*cr+tx]) { |
| 3257 | ui->hx = ui->hy = -1; |
| 3258 | } else if (tx == ui->hx && ty == ui->hy && ui->hpencil == 0) { |
| 3259 | ui->hx = ui->hy = -1; |
| 3260 | } else { |
| 3261 | ui->hx = tx; |
| 3262 | ui->hy = ty; |
| 3263 | ui->hpencil = 0; |
| 3264 | } |
| 3265 | return ""; /* UI activity occurred */ |
| 3266 | } |
| 3267 | if (button == RIGHT_BUTTON) { |
| 3268 | /* |
| 3269 | * Pencil-mode highlighting for non filled squares. |
| 3270 | */ |
| 3271 | if (state->grid[ty*cr+tx] == 0) { |
| 3272 | if (tx == ui->hx && ty == ui->hy && ui->hpencil) { |
| 3273 | ui->hx = ui->hy = -1; |
| 3274 | } else { |
| 3275 | ui->hpencil = 1; |
| 3276 | ui->hx = tx; |
| 3277 | ui->hy = ty; |
| 3278 | } |
| 3279 | } else { |
| 3280 | ui->hx = ui->hy = -1; |
| 3281 | } |
| 3282 | return ""; /* UI activity occurred */ |
| 3283 | } |
| 3284 | } |
| 3285 | |
| 3286 | if (ui->hx != -1 && ui->hy != -1 && |
| 3287 | ((button >= '1' && button <= '9' && button - '0' <= cr) || |
| 3288 | (button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) || |
| 3289 | (button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) || |
| 3290 | button == ' ' || button == '\010' || button == '\177')) { |
| 3291 | int n = button - '0'; |
| 3292 | if (button >= 'A' && button <= 'Z') |
| 3293 | n = button - 'A' + 10; |
| 3294 | if (button >= 'a' && button <= 'z') |
| 3295 | n = button - 'a' + 10; |
| 3296 | if (button == ' ' || button == '\010' || button == '\177') |
| 3297 | n = 0; |
| 3298 | |
| 3299 | /* |
| 3300 | * Can't overwrite this square. In principle this shouldn't |
| 3301 | * happen anyway because we should never have even been |
| 3302 | * able to highlight the square, but it never hurts to be |
| 3303 | * careful. |
| 3304 | */ |
| 3305 | if (state->immutable[ui->hy*cr+ui->hx]) |
| 3306 | return NULL; |
| 3307 | |
| 3308 | /* |
| 3309 | * Can't make pencil marks in a filled square. In principle |
| 3310 | * this shouldn't happen anyway because we should never |
| 3311 | * have even been able to pencil-highlight the square, but |
| 3312 | * it never hurts to be careful. |
| 3313 | */ |
| 3314 | if (ui->hpencil && state->grid[ui->hy*cr+ui->hx]) |
| 3315 | return NULL; |
| 3316 | |
| 3317 | sprintf(buf, "%c%d,%d,%d", |
| 3318 | (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n); |
| 3319 | |
| 3320 | ui->hx = ui->hy = -1; |
| 3321 | |
| 3322 | return dupstr(buf); |
| 3323 | } |
| 3324 | |
| 3325 | return NULL; |
| 3326 | } |
| 3327 | |
| 3328 | static game_state *execute_move(game_state *from, char *move) |
| 3329 | { |
| 3330 | int cr = from->cr; |
| 3331 | game_state *ret; |
| 3332 | int x, y, n; |
| 3333 | |
| 3334 | if (move[0] == 'S') { |
| 3335 | char *p; |
| 3336 | |
| 3337 | ret = dup_game(from); |
| 3338 | ret->completed = ret->cheated = TRUE; |
| 3339 | |
| 3340 | p = move+1; |
| 3341 | for (n = 0; n < cr*cr; n++) { |
| 3342 | ret->grid[n] = atoi(p); |
| 3343 | |
| 3344 | if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) { |
| 3345 | free_game(ret); |
| 3346 | return NULL; |
| 3347 | } |
| 3348 | |
| 3349 | while (*p && isdigit((unsigned char)*p)) p++; |
| 3350 | if (*p == ',') p++; |
| 3351 | } |
| 3352 | |
| 3353 | return ret; |
| 3354 | } else if ((move[0] == 'P' || move[0] == 'R') && |
| 3355 | sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 && |
| 3356 | x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) { |
| 3357 | |
| 3358 | ret = dup_game(from); |
| 3359 | if (move[0] == 'P' && n > 0) { |
| 3360 | int index = (y*cr+x) * cr + (n-1); |
| 3361 | ret->pencil[index] = !ret->pencil[index]; |
| 3362 | } else { |
| 3363 | ret->grid[y*cr+x] = n; |
| 3364 | memset(ret->pencil + (y*cr+x)*cr, 0, cr); |
| 3365 | |
| 3366 | /* |
| 3367 | * We've made a real change to the grid. Check to see |
| 3368 | * if the game has been completed. |
| 3369 | */ |
| 3370 | if (!ret->completed && check_valid(cr, ret->blocks, ret->xtype, |
| 3371 | ret->grid)) { |
| 3372 | ret->completed = TRUE; |
| 3373 | } |
| 3374 | } |
| 3375 | return ret; |
| 3376 | } else |
| 3377 | return NULL; /* couldn't parse move string */ |
| 3378 | } |
| 3379 | |
| 3380 | /* ---------------------------------------------------------------------- |
| 3381 | * Drawing routines. |
| 3382 | */ |
| 3383 | |
| 3384 | #define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1) |
| 3385 | #define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) ) |
| 3386 | |
| 3387 | static void game_compute_size(game_params *params, int tilesize, |
| 3388 | int *x, int *y) |
| 3389 | { |
| 3390 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
| 3391 | struct { int tilesize; } ads, *ds = &ads; |
| 3392 | ads.tilesize = tilesize; |
| 3393 | |
| 3394 | *x = SIZE(params->c * params->r); |
| 3395 | *y = SIZE(params->c * params->r); |
| 3396 | } |
| 3397 | |
| 3398 | static void game_set_size(drawing *dr, game_drawstate *ds, |
| 3399 | game_params *params, int tilesize) |
| 3400 | { |
| 3401 | ds->tilesize = tilesize; |
| 3402 | } |
| 3403 | |
| 3404 | static float *game_colours(frontend *fe, int *ncolours) |
| 3405 | { |
| 3406 | float *ret = snewn(3 * NCOLOURS, float); |
| 3407 | |
| 3408 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
| 3409 | |
| 3410 | ret[COL_XDIAGONALS * 3 + 0] = 0.9F * ret[COL_BACKGROUND * 3 + 0]; |
| 3411 | ret[COL_XDIAGONALS * 3 + 1] = 0.9F * ret[COL_BACKGROUND * 3 + 1]; |
| 3412 | ret[COL_XDIAGONALS * 3 + 2] = 0.9F * ret[COL_BACKGROUND * 3 + 2]; |
| 3413 | |
| 3414 | ret[COL_GRID * 3 + 0] = 0.0F; |
| 3415 | ret[COL_GRID * 3 + 1] = 0.0F; |
| 3416 | ret[COL_GRID * 3 + 2] = 0.0F; |
| 3417 | |
| 3418 | ret[COL_CLUE * 3 + 0] = 0.0F; |
| 3419 | ret[COL_CLUE * 3 + 1] = 0.0F; |
| 3420 | ret[COL_CLUE * 3 + 2] = 0.0F; |
| 3421 | |
| 3422 | ret[COL_USER * 3 + 0] = 0.0F; |
| 3423 | ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1]; |
| 3424 | ret[COL_USER * 3 + 2] = 0.0F; |
| 3425 | |
| 3426 | ret[COL_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0]; |
| 3427 | ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1]; |
| 3428 | ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * ret[COL_BACKGROUND * 3 + 2]; |
| 3429 | |
| 3430 | ret[COL_ERROR * 3 + 0] = 1.0F; |
| 3431 | ret[COL_ERROR * 3 + 1] = 0.0F; |
| 3432 | ret[COL_ERROR * 3 + 2] = 0.0F; |
| 3433 | |
| 3434 | ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0]; |
| 3435 | ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1]; |
| 3436 | ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2]; |
| 3437 | |
| 3438 | *ncolours = NCOLOURS; |
| 3439 | return ret; |
| 3440 | } |
| 3441 | |
| 3442 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
| 3443 | { |
| 3444 | struct game_drawstate *ds = snew(struct game_drawstate); |
| 3445 | int cr = state->cr; |
| 3446 | |
| 3447 | ds->started = FALSE; |
| 3448 | ds->cr = cr; |
| 3449 | ds->xtype = state->xtype; |
| 3450 | ds->grid = snewn(cr*cr, digit); |
| 3451 | memset(ds->grid, cr+2, cr*cr); |
| 3452 | ds->pencil = snewn(cr*cr*cr, digit); |
| 3453 | memset(ds->pencil, 0, cr*cr*cr); |
| 3454 | ds->hl = snewn(cr*cr, unsigned char); |
| 3455 | memset(ds->hl, 0, cr*cr); |
| 3456 | ds->entered_items = snewn(cr*cr, int); |
| 3457 | ds->tilesize = 0; /* not decided yet */ |
| 3458 | return ds; |
| 3459 | } |
| 3460 | |
| 3461 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
| 3462 | { |
| 3463 | sfree(ds->hl); |
| 3464 | sfree(ds->pencil); |
| 3465 | sfree(ds->grid); |
| 3466 | sfree(ds->entered_items); |
| 3467 | sfree(ds); |
| 3468 | } |
| 3469 | |
| 3470 | static void draw_number(drawing *dr, game_drawstate *ds, game_state *state, |
| 3471 | int x, int y, int hl) |
| 3472 | { |
| 3473 | int cr = state->cr; |
| 3474 | int tx, ty; |
| 3475 | int cx, cy, cw, ch; |
| 3476 | char str[2]; |
| 3477 | |
| 3478 | if (ds->grid[y*cr+x] == state->grid[y*cr+x] && |
| 3479 | ds->hl[y*cr+x] == hl && |
| 3480 | !memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr)) |
| 3481 | return; /* no change required */ |
| 3482 | |
| 3483 | tx = BORDER + x * TILE_SIZE + 1 + GRIDEXTRA; |
| 3484 | ty = BORDER + y * TILE_SIZE + 1 + GRIDEXTRA; |
| 3485 | |
| 3486 | cx = tx; |
| 3487 | cy = ty; |
| 3488 | cw = TILE_SIZE-1-2*GRIDEXTRA; |
| 3489 | ch = TILE_SIZE-1-2*GRIDEXTRA; |
| 3490 | |
| 3491 | if (x > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x-1]) |
| 3492 | cx -= GRIDEXTRA, cw += GRIDEXTRA; |
| 3493 | if (x+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x+1]) |
| 3494 | cw += GRIDEXTRA; |
| 3495 | if (y > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y-1)*cr+x]) |
| 3496 | cy -= GRIDEXTRA, ch += GRIDEXTRA; |
| 3497 | if (y+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y+1)*cr+x]) |
| 3498 | ch += GRIDEXTRA; |
| 3499 | |
| 3500 | clip(dr, cx, cy, cw, ch); |
| 3501 | |
| 3502 | /* background needs erasing */ |
| 3503 | draw_rect(dr, cx, cy, cw, ch, |
| 3504 | ((hl & 15) == 1 ? COL_HIGHLIGHT : |
| 3505 | (ds->xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x))) ? COL_XDIAGONALS : |
| 3506 | COL_BACKGROUND)); |
| 3507 | |
| 3508 | /* |
| 3509 | * Draw the corners of thick lines in corner-adjacent squares, |
| 3510 | * which jut into this square by one pixel. |
| 3511 | */ |
| 3512 | if (x > 0 && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x-1]) |
| 3513 | draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); |
| 3514 | if (x+1 < cr && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x+1]) |
| 3515 | draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); |
| 3516 | if (x > 0 && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x-1]) |
| 3517 | draw_rect(dr, tx-GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); |
| 3518 | if (x+1 < cr && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x+1]) |
| 3519 | draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); |
| 3520 | |
| 3521 | /* pencil-mode highlight */ |
| 3522 | if ((hl & 15) == 2) { |
| 3523 | int coords[6]; |
| 3524 | coords[0] = cx; |
| 3525 | coords[1] = cy; |
| 3526 | coords[2] = cx+cw/2; |
| 3527 | coords[3] = cy; |
| 3528 | coords[4] = cx; |
| 3529 | coords[5] = cy+ch/2; |
| 3530 | draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT); |
| 3531 | } |
| 3532 | |
| 3533 | /* new number needs drawing? */ |
| 3534 | if (state->grid[y*cr+x]) { |
| 3535 | str[1] = '\0'; |
| 3536 | str[0] = state->grid[y*cr+x] + '0'; |
| 3537 | if (str[0] > '9') |
| 3538 | str[0] += 'a' - ('9'+1); |
| 3539 | draw_text(dr, tx + TILE_SIZE/2, ty + TILE_SIZE/2, |
| 3540 | FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE, |
| 3541 | state->immutable[y*cr+x] ? COL_CLUE : (hl & 16) ? COL_ERROR : COL_USER, str); |
| 3542 | } else { |
| 3543 | int i, j, npencil; |
| 3544 | int pw, ph, pmax, fontsize; |
| 3545 | |
| 3546 | /* count the pencil marks required */ |
| 3547 | for (i = npencil = 0; i < cr; i++) |
| 3548 | if (state->pencil[(y*cr+x)*cr+i]) |
| 3549 | npencil++; |
| 3550 | |
| 3551 | /* |
| 3552 | * It's not sensible to arrange pencil marks in the same |
| 3553 | * layout as the squares within a block, because this leads |
| 3554 | * to the font being too small. Instead, we arrange pencil |
| 3555 | * marks in the nearest thing we can to a square layout, |
| 3556 | * and we adjust the square layout depending on the number |
| 3557 | * of pencil marks in the square. |
| 3558 | */ |
| 3559 | for (pw = 1; pw * pw < npencil; pw++); |
| 3560 | if (pw < 3) pw = 3; /* otherwise it just looks _silly_ */ |
| 3561 | ph = (npencil + pw - 1) / pw; |
| 3562 | if (ph < 2) ph = 2; /* likewise */ |
| 3563 | pmax = max(pw, ph); |
| 3564 | fontsize = TILE_SIZE/(pmax*(11-pmax)/8); |
| 3565 | |
| 3566 | for (i = j = 0; i < cr; i++) |
| 3567 | if (state->pencil[(y*cr+x)*cr+i]) { |
| 3568 | int dx = j % pw, dy = j / pw; |
| 3569 | |
| 3570 | str[1] = '\0'; |
| 3571 | str[0] = i + '1'; |
| 3572 | if (str[0] > '9') |
| 3573 | str[0] += 'a' - ('9'+1); |
| 3574 | draw_text(dr, tx + (4*dx+3) * TILE_SIZE / (4*pw+2), |
| 3575 | ty + (4*dy+3) * TILE_SIZE / (4*ph+2), |
| 3576 | FONT_VARIABLE, fontsize, |
| 3577 | ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str); |
| 3578 | j++; |
| 3579 | } |
| 3580 | } |
| 3581 | |
| 3582 | unclip(dr); |
| 3583 | |
| 3584 | draw_update(dr, cx, cy, cw, ch); |
| 3585 | |
| 3586 | ds->grid[y*cr+x] = state->grid[y*cr+x]; |
| 3587 | memcpy(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr); |
| 3588 | ds->hl[y*cr+x] = hl; |
| 3589 | } |
| 3590 | |
| 3591 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
| 3592 | game_state *state, int dir, game_ui *ui, |
| 3593 | float animtime, float flashtime) |
| 3594 | { |
| 3595 | int cr = state->cr; |
| 3596 | int x, y; |
| 3597 | |
| 3598 | if (!ds->started) { |
| 3599 | /* |
| 3600 | * The initial contents of the window are not guaranteed |
| 3601 | * and can vary with front ends. To be on the safe side, |
| 3602 | * all games should start by drawing a big |
| 3603 | * background-colour rectangle covering the whole window. |
| 3604 | */ |
| 3605 | draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND); |
| 3606 | |
| 3607 | /* |
| 3608 | * Draw the grid. We draw it as a big thick rectangle of |
| 3609 | * COL_GRID initially; individual calls to draw_number() |
| 3610 | * will poke the right-shaped holes in it. |
| 3611 | */ |
| 3612 | draw_rect(dr, BORDER-GRIDEXTRA, BORDER-GRIDEXTRA, |
| 3613 | cr*TILE_SIZE+1+2*GRIDEXTRA, cr*TILE_SIZE+1+2*GRIDEXTRA, |
| 3614 | COL_GRID); |
| 3615 | } |
| 3616 | |
| 3617 | /* |
| 3618 | * This array is used to keep track of rows, columns and boxes |
| 3619 | * which contain a number more than once. |
| 3620 | */ |
| 3621 | for (x = 0; x < cr * cr; x++) |
| 3622 | ds->entered_items[x] = 0; |
| 3623 | for (x = 0; x < cr; x++) |
| 3624 | for (y = 0; y < cr; y++) { |
| 3625 | digit d = state->grid[y*cr+x]; |
| 3626 | if (d) { |
| 3627 | int box = state->blocks->whichblock[y*cr+x]; |
| 3628 | ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1; |
| 3629 | ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4; |
| 3630 | ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16; |
| 3631 | if (ds->xtype) { |
| 3632 | if (ondiag0(y*cr+x)) |
| 3633 | ds->entered_items[d-1] |= ((ds->entered_items[d-1] & 64) << 1) | 64; |
| 3634 | if (ondiag1(y*cr+x)) |
| 3635 | ds->entered_items[cr+d-1] |= ((ds->entered_items[cr+d-1] & 64) << 1) | 64; |
| 3636 | } |
| 3637 | } |
| 3638 | } |
| 3639 | |
| 3640 | /* |
| 3641 | * Draw any numbers which need redrawing. |
| 3642 | */ |
| 3643 | for (x = 0; x < cr; x++) { |
| 3644 | for (y = 0; y < cr; y++) { |
| 3645 | int highlight = 0; |
| 3646 | digit d = state->grid[y*cr+x]; |
| 3647 | |
| 3648 | if (flashtime > 0 && |
| 3649 | (flashtime <= FLASH_TIME/3 || |
| 3650 | flashtime >= FLASH_TIME*2/3)) |
| 3651 | highlight = 1; |
| 3652 | |
| 3653 | /* Highlight active input areas. */ |
| 3654 | if (x == ui->hx && y == ui->hy) |
| 3655 | highlight = ui->hpencil ? 2 : 1; |
| 3656 | |
| 3657 | /* Mark obvious errors (ie, numbers which occur more than once |
| 3658 | * in a single row, column, or box). */ |
| 3659 | if (d && ((ds->entered_items[x*cr+d-1] & 2) || |
| 3660 | (ds->entered_items[y*cr+d-1] & 8) || |
| 3661 | (ds->entered_items[state->blocks->whichblock[y*cr+x]*cr+d-1] & 32) || |
| 3662 | (ds->xtype && ((ondiag0(y*cr+x) && (ds->entered_items[d-1] & 128)) || |
| 3663 | (ondiag1(y*cr+x) && (ds->entered_items[cr+d-1] & 128)))))) |
| 3664 | highlight |= 16; |
| 3665 | |
| 3666 | draw_number(dr, ds, state, x, y, highlight); |
| 3667 | } |
| 3668 | } |
| 3669 | |
| 3670 | /* |
| 3671 | * Update the _entire_ grid if necessary. |
| 3672 | */ |
| 3673 | if (!ds->started) { |
| 3674 | draw_update(dr, 0, 0, SIZE(cr), SIZE(cr)); |
| 3675 | ds->started = TRUE; |
| 3676 | } |
| 3677 | } |
| 3678 | |
| 3679 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
| 3680 | int dir, game_ui *ui) |
| 3681 | { |
| 3682 | return 0.0F; |
| 3683 | } |
| 3684 | |
| 3685 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
| 3686 | int dir, game_ui *ui) |
| 3687 | { |
| 3688 | if (!oldstate->completed && newstate->completed && |
| 3689 | !oldstate->cheated && !newstate->cheated) |
| 3690 | return FLASH_TIME; |
| 3691 | return 0.0F; |
| 3692 | } |
| 3693 | |
| 3694 | static int game_timing_state(game_state *state, game_ui *ui) |
| 3695 | { |
| 3696 | return TRUE; |
| 3697 | } |
| 3698 | |
| 3699 | static void game_print_size(game_params *params, float *x, float *y) |
| 3700 | { |
| 3701 | int pw, ph; |
| 3702 | |
| 3703 | /* |
| 3704 | * I'll use 9mm squares by default. They should be quite big |
| 3705 | * for this game, because players will want to jot down no end |
| 3706 | * of pencil marks in the squares. |
| 3707 | */ |
| 3708 | game_compute_size(params, 900, &pw, &ph); |
| 3709 | *x = pw / 100.0; |
| 3710 | *y = ph / 100.0; |
| 3711 | } |
| 3712 | |
| 3713 | static void game_print(drawing *dr, game_state *state, int tilesize) |
| 3714 | { |
| 3715 | int cr = state->cr; |
| 3716 | int ink = print_mono_colour(dr, 0); |
| 3717 | int x, y; |
| 3718 | |
| 3719 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
| 3720 | game_drawstate ads, *ds = &ads; |
| 3721 | game_set_size(dr, ds, NULL, tilesize); |
| 3722 | |
| 3723 | /* |
| 3724 | * Border. |
| 3725 | */ |
| 3726 | print_line_width(dr, 3 * TILE_SIZE / 40); |
| 3727 | draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink); |
| 3728 | |
| 3729 | /* |
| 3730 | * Highlight X-diagonal squares. |
| 3731 | */ |
| 3732 | if (state->xtype) { |
| 3733 | int i; |
| 3734 | int xhighlight = print_grey_colour(dr, 0.90F); |
| 3735 | |
| 3736 | for (i = 0; i < cr; i++) |
| 3737 | draw_rect(dr, BORDER + i*TILE_SIZE, BORDER + i*TILE_SIZE, |
| 3738 | TILE_SIZE, TILE_SIZE, xhighlight); |
| 3739 | for (i = 0; i < cr; i++) |
| 3740 | if (i*2 != cr-1) /* avoid redoing centre square, just for fun */ |
| 3741 | draw_rect(dr, BORDER + i*TILE_SIZE, |
| 3742 | BORDER + (cr-1-i)*TILE_SIZE, |
| 3743 | TILE_SIZE, TILE_SIZE, xhighlight); |
| 3744 | } |
| 3745 | |
| 3746 | /* |
| 3747 | * Main grid. |
| 3748 | */ |
| 3749 | for (x = 1; x < cr; x++) { |
| 3750 | print_line_width(dr, TILE_SIZE / 40); |
| 3751 | draw_line(dr, BORDER+x*TILE_SIZE, BORDER, |
| 3752 | BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink); |
| 3753 | } |
| 3754 | for (y = 1; y < cr; y++) { |
| 3755 | print_line_width(dr, TILE_SIZE / 40); |
| 3756 | draw_line(dr, BORDER, BORDER+y*TILE_SIZE, |
| 3757 | BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink); |
| 3758 | } |
| 3759 | |
| 3760 | /* |
| 3761 | * Thick lines between cells. In order to do this using the |
| 3762 | * line-drawing rather than rectangle-drawing API (so as to |
| 3763 | * get line thicknesses to scale correctly) and yet have |
| 3764 | * correctly mitred joins between lines, we must do this by |
| 3765 | * tracing the boundary of each sub-block and drawing it in |
| 3766 | * one go as a single polygon. |
| 3767 | */ |
| 3768 | { |
| 3769 | int *coords; |
| 3770 | int bi, i, n; |
| 3771 | int x, y, dx, dy, sx, sy, sdx, sdy; |
| 3772 | |
| 3773 | print_line_width(dr, 3 * TILE_SIZE / 40); |
| 3774 | |
| 3775 | /* |
| 3776 | * Maximum perimeter of a k-omino is 2k+2. (Proof: start |
| 3777 | * with k unconnected squares, with total perimeter 4k. |
| 3778 | * Now repeatedly join two disconnected components |
| 3779 | * together into a larger one; every time you do so you |
| 3780 | * remove at least two unit edges, and you require k-1 of |
| 3781 | * these operations to create a single connected piece, so |
| 3782 | * you must have at most 4k-2(k-1) = 2k+2 unit edges left |
| 3783 | * afterwards.) |
| 3784 | */ |
| 3785 | coords = snewn(4*cr+4, int); /* 2k+2 points, 2 coords per point */ |
| 3786 | |
| 3787 | /* |
| 3788 | * Iterate over all the blocks. |
| 3789 | */ |
| 3790 | for (bi = 0; bi < cr; bi++) { |
| 3791 | |
| 3792 | /* |
| 3793 | * For each block, find a starting square within it |
| 3794 | * which has a boundary at the left. |
| 3795 | */ |
| 3796 | for (i = 0; i < cr; i++) { |
| 3797 | int j = state->blocks->blocks[bi][i]; |
| 3798 | if (j % cr == 0 || state->blocks->whichblock[j-1] != bi) |
| 3799 | break; |
| 3800 | } |
| 3801 | assert(i < cr); /* every block must have _some_ leftmost square */ |
| 3802 | x = state->blocks->blocks[bi][i] % cr; |
| 3803 | y = state->blocks->blocks[bi][i] / cr; |
| 3804 | dx = -1; |
| 3805 | dy = 0; |
| 3806 | |
| 3807 | /* |
| 3808 | * Now begin tracing round the perimeter. At all |
| 3809 | * times, (x,y) describes some square within the |
| 3810 | * block, and (x+dx,y+dy) is some adjacent square |
| 3811 | * outside it; so the edge between those two squares |
| 3812 | * is always an edge of the block. |
| 3813 | */ |
| 3814 | sx = x, sy = y, sdx = dx, sdy = dy; /* save starting position */ |
| 3815 | n = 0; |
| 3816 | do { |
| 3817 | int cx, cy, tx, ty, nin; |
| 3818 | |
| 3819 | /* |
| 3820 | * To begin with, record the point at one end of |
| 3821 | * the edge. To do this, we translate (x,y) down |
| 3822 | * and right by half a unit (so they're describing |
| 3823 | * a point in the _centre_ of the square) and then |
| 3824 | * translate back again in a manner rotated by dy |
| 3825 | * and dx. |
| 3826 | */ |
| 3827 | assert(n < 2*cr+2); |
| 3828 | cx = ((2*x+1) + dy + dx) / 2; |
| 3829 | cy = ((2*y+1) - dx + dy) / 2; |
| 3830 | coords[2*n+0] = BORDER + cx * TILE_SIZE; |
| 3831 | coords[2*n+1] = BORDER + cy * TILE_SIZE; |
| 3832 | n++; |
| 3833 | |
| 3834 | /* |
| 3835 | * Now advance to the next edge, by looking at the |
| 3836 | * two squares beyond it. If they're both outside |
| 3837 | * the block, we turn right (by leaving x,y the |
| 3838 | * same and rotating dx,dy clockwise); if they're |
| 3839 | * both inside, we turn left (by rotating dx,dy |
| 3840 | * anticlockwise and contriving to leave x+dx,y+dy |
| 3841 | * unchanged); if one of each, we go straight on |
| 3842 | * (and may enforce by assertion that they're one |
| 3843 | * of each the _right_ way round). |
| 3844 | */ |
| 3845 | nin = 0; |
| 3846 | tx = x - dy + dx; |
| 3847 | ty = y + dx + dy; |
| 3848 | nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr && |
| 3849 | state->blocks->whichblock[ty*cr+tx] == bi); |
| 3850 | tx = x - dy; |
| 3851 | ty = y + dx; |
| 3852 | nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr && |
| 3853 | state->blocks->whichblock[ty*cr+tx] == bi); |
| 3854 | if (nin == 0) { |
| 3855 | /* |
| 3856 | * Turn right. |
| 3857 | */ |
| 3858 | int tmp; |
| 3859 | tmp = dx; |
| 3860 | dx = -dy; |
| 3861 | dy = tmp; |
| 3862 | } else if (nin == 2) { |
| 3863 | /* |
| 3864 | * Turn left. |
| 3865 | */ |
| 3866 | int tmp; |
| 3867 | |
| 3868 | x += dx; |
| 3869 | y += dy; |
| 3870 | |
| 3871 | tmp = dx; |
| 3872 | dx = dy; |
| 3873 | dy = -tmp; |
| 3874 | |
| 3875 | x -= dx; |
| 3876 | y -= dy; |
| 3877 | } else { |
| 3878 | /* |
| 3879 | * Go straight on. |
| 3880 | */ |
| 3881 | x -= dy; |
| 3882 | y += dx; |
| 3883 | } |
| 3884 | |
| 3885 | /* |
| 3886 | * Now enforce by assertion that we ended up |
| 3887 | * somewhere sensible. |
| 3888 | */ |
| 3889 | assert(x >= 0 && x < cr && y >= 0 && y < cr && |
| 3890 | state->blocks->whichblock[y*cr+x] == bi); |
| 3891 | assert(x+dx < 0 || x+dx >= cr || y+dy < 0 || y+dy >= cr || |
| 3892 | state->blocks->whichblock[(y+dy)*cr+(x+dx)] != bi); |
| 3893 | |
| 3894 | } while (x != sx || y != sy || dx != sdx || dy != sdy); |
| 3895 | |
| 3896 | /* |
| 3897 | * That's our polygon; now draw it. |
| 3898 | */ |
| 3899 | draw_polygon(dr, coords, n, -1, ink); |
| 3900 | } |
| 3901 | |
| 3902 | sfree(coords); |
| 3903 | } |
| 3904 | |
| 3905 | /* |
| 3906 | * Numbers. |
| 3907 | */ |
| 3908 | for (y = 0; y < cr; y++) |
| 3909 | for (x = 0; x < cr; x++) |
| 3910 | if (state->grid[y*cr+x]) { |
| 3911 | char str[2]; |
| 3912 | str[1] = '\0'; |
| 3913 | str[0] = state->grid[y*cr+x] + '0'; |
| 3914 | if (str[0] > '9') |
| 3915 | str[0] += 'a' - ('9'+1); |
| 3916 | draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2, |
| 3917 | BORDER + y*TILE_SIZE + TILE_SIZE/2, |
| 3918 | FONT_VARIABLE, TILE_SIZE/2, |
| 3919 | ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str); |
| 3920 | } |
| 3921 | } |
| 3922 | |
| 3923 | #ifdef COMBINED |
| 3924 | #define thegame solo |
| 3925 | #endif |
| 3926 | |
| 3927 | const struct game thegame = { |
| 3928 | "Solo", "games.solo", "solo", |
| 3929 | default_params, |
| 3930 | game_fetch_preset, |
| 3931 | decode_params, |
| 3932 | encode_params, |
| 3933 | free_params, |
| 3934 | dup_params, |
| 3935 | TRUE, game_configure, custom_params, |
| 3936 | validate_params, |
| 3937 | new_game_desc, |
| 3938 | validate_desc, |
| 3939 | new_game, |
| 3940 | dup_game, |
| 3941 | free_game, |
| 3942 | TRUE, solve_game, |
| 3943 | TRUE, game_can_format_as_text_now, game_text_format, |
| 3944 | new_ui, |
| 3945 | free_ui, |
| 3946 | encode_ui, |
| 3947 | decode_ui, |
| 3948 | game_changed_state, |
| 3949 | interpret_move, |
| 3950 | execute_move, |
| 3951 | PREFERRED_TILE_SIZE, game_compute_size, game_set_size, |
| 3952 | game_colours, |
| 3953 | game_new_drawstate, |
| 3954 | game_free_drawstate, |
| 3955 | game_redraw, |
| 3956 | game_anim_length, |
| 3957 | game_flash_length, |
| 3958 | TRUE, FALSE, game_print_size, game_print, |
| 3959 | FALSE, /* wants_statusbar */ |
| 3960 | FALSE, game_timing_state, |
| 3961 | REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */ |
| 3962 | }; |
| 3963 | |
| 3964 | #ifdef STANDALONE_SOLVER |
| 3965 | |
| 3966 | int main(int argc, char **argv) |
| 3967 | { |
| 3968 | game_params *p; |
| 3969 | game_state *s; |
| 3970 | char *id = NULL, *desc, *err; |
| 3971 | int grade = FALSE; |
| 3972 | int ret; |
| 3973 | |
| 3974 | while (--argc > 0) { |
| 3975 | char *p = *++argv; |
| 3976 | if (!strcmp(p, "-v")) { |
| 3977 | solver_show_working = TRUE; |
| 3978 | } else if (!strcmp(p, "-g")) { |
| 3979 | grade = TRUE; |
| 3980 | } else if (*p == '-') { |
| 3981 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
| 3982 | return 1; |
| 3983 | } else { |
| 3984 | id = p; |
| 3985 | } |
| 3986 | } |
| 3987 | |
| 3988 | if (!id) { |
| 3989 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
| 3990 | return 1; |
| 3991 | } |
| 3992 | |
| 3993 | desc = strchr(id, ':'); |
| 3994 | if (!desc) { |
| 3995 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
| 3996 | return 1; |
| 3997 | } |
| 3998 | *desc++ = '\0'; |
| 3999 | |
| 4000 | p = default_params(); |
| 4001 | decode_params(p, id); |
| 4002 | err = validate_desc(p, desc); |
| 4003 | if (err) { |
| 4004 | fprintf(stderr, "%s: %s\n", argv[0], err); |
| 4005 | return 1; |
| 4006 | } |
| 4007 | s = new_game(NULL, p, desc); |
| 4008 | |
| 4009 | ret = solver(s->cr, s->blocks, s->xtype, s->grid, DIFF_RECURSIVE); |
| 4010 | if (grade) { |
| 4011 | printf("Difficulty rating: %s\n", |
| 4012 | ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)": |
| 4013 | ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)": |
| 4014 | ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)": |
| 4015 | ret==DIFF_SET ? "Advanced (set elimination required)": |
| 4016 | ret==DIFF_EXTREME ? "Extreme (complex non-recursive techniques required)": |
| 4017 | ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)": |
| 4018 | ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)": |
| 4019 | ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)": |
| 4020 | "INTERNAL ERROR: unrecognised difficulty code"); |
| 4021 | } else { |
| 4022 | printf("%s\n", grid_text_format(s->cr, s->blocks, s->xtype, s->grid)); |
| 4023 | } |
| 4024 | |
| 4025 | return 0; |
| 4026 | } |
| 4027 | |
| 4028 | #endif |