| 1 | /* |
| 2 | * loopy.c: |
| 3 | * |
| 4 | * An implementation of the Nikoli game 'Loop the loop'. |
| 5 | * (c) Mike Pinna, 2005, 2006 |
| 6 | * Substantially rewritten to allowing for more general types of grid. |
| 7 | * (c) Lambros Lambrou 2008 |
| 8 | * |
| 9 | * vim: set shiftwidth=4 :set textwidth=80: |
| 10 | */ |
| 11 | |
| 12 | /* |
| 13 | * Possible future solver enhancements: |
| 14 | * |
| 15 | * - There's an interesting deductive technique which makes use |
| 16 | * of topology rather than just graph theory. Each _face_ in |
| 17 | * the grid is either inside or outside the loop; you can tell |
| 18 | * that two faces are on the same side of the loop if they're |
| 19 | * separated by a LINE_NO (or, more generally, by a path |
| 20 | * crossing no LINE_UNKNOWNs and an even number of LINE_YESes), |
| 21 | * and on the opposite side of the loop if they're separated by |
| 22 | * a LINE_YES (or an odd number of LINE_YESes and no |
| 23 | * LINE_UNKNOWNs). Oh, and any face separated from the outside |
| 24 | * of the grid by a LINE_YES or a LINE_NO is on the inside or |
| 25 | * outside respectively. So if you can track this for all |
| 26 | * faces, you figure out the state of the line between a pair |
| 27 | * once their relative insideness is known. |
| 28 | * + The way I envisage this working is simply to keep an edsf |
| 29 | * of all _faces_, which indicates whether they're on |
| 30 | * opposite sides of the loop from one another. We also |
| 31 | * include a special entry in the edsf for the infinite |
| 32 | * exterior "face". |
| 33 | * + So, the simple way to do this is to just go through the |
| 34 | * edges: every time we see an edge in a state other than |
| 35 | * LINE_UNKNOWN which separates two faces that aren't in the |
| 36 | * same edsf class, we can rectify that by merging the |
| 37 | * classes. Then, conversely, an edge in LINE_UNKNOWN state |
| 38 | * which separates two faces that _are_ in the same edsf |
| 39 | * class can immediately have its state determined. |
| 40 | * + But you can go one better, if you're prepared to loop |
| 41 | * over all _pairs_ of edges. Suppose we have edges A and B, |
| 42 | * which respectively separate faces A1,A2 and B1,B2. |
| 43 | * Suppose that A,B are in the same edge-edsf class and that |
| 44 | * A1,B1 (wlog) are in the same face-edsf class; then we can |
| 45 | * immediately place A2,B2 into the same face-edsf class (as |
| 46 | * each other, not as A1 and A2) one way round or the other. |
| 47 | * And conversely again, if A1,B1 are in the same face-edsf |
| 48 | * class and so are A2,B2, then we can put A,B into the same |
| 49 | * face-edsf class. |
| 50 | * * Of course, this deduction requires a quadratic-time |
| 51 | * loop over all pairs of edges in the grid, so it should |
| 52 | * be reserved until there's nothing easier left to be |
| 53 | * done. |
| 54 | * |
| 55 | * - The generalised grid support has made me (SGT) notice a |
| 56 | * possible extension to the loop-avoidance code. When you have |
| 57 | * a path of connected edges such that no other edges at all |
| 58 | * are incident on any vertex in the middle of the path - or, |
| 59 | * alternatively, such that any such edges are already known to |
| 60 | * be LINE_NO - then you know those edges are either all |
| 61 | * LINE_YES or all LINE_NO. Hence you can mentally merge the |
| 62 | * entire path into a single long curly edge for the purposes |
| 63 | * of loop avoidance, and look directly at whether or not the |
| 64 | * extreme endpoints of the path are connected by some other |
| 65 | * route. I find this coming up fairly often when I play on the |
| 66 | * octagonal grid setting, so it might be worth implementing in |
| 67 | * the solver. |
| 68 | * |
| 69 | * - (Just a speed optimisation.) Consider some todo list queue where every |
| 70 | * time we modify something we mark it for consideration by other bits of |
| 71 | * the solver, to save iteration over things that have already been done. |
| 72 | */ |
| 73 | |
| 74 | #include <stdio.h> |
| 75 | #include <stdlib.h> |
| 76 | #include <stddef.h> |
| 77 | #include <string.h> |
| 78 | #include <assert.h> |
| 79 | #include <ctype.h> |
| 80 | #include <math.h> |
| 81 | |
| 82 | #include "puzzles.h" |
| 83 | #include "tree234.h" |
| 84 | #include "grid.h" |
| 85 | |
| 86 | /* Debugging options */ |
| 87 | |
| 88 | /* |
| 89 | #define DEBUG_CACHES |
| 90 | #define SHOW_WORKING |
| 91 | #define DEBUG_DLINES |
| 92 | */ |
| 93 | |
| 94 | /* ---------------------------------------------------------------------- |
| 95 | * Struct, enum and function declarations |
| 96 | */ |
| 97 | |
| 98 | enum { |
| 99 | COL_BACKGROUND, |
| 100 | COL_FOREGROUND, |
| 101 | COL_LINEUNKNOWN, |
| 102 | COL_HIGHLIGHT, |
| 103 | COL_MISTAKE, |
| 104 | COL_SATISFIED, |
| 105 | COL_FAINT, |
| 106 | NCOLOURS |
| 107 | }; |
| 108 | |
| 109 | struct game_state { |
| 110 | grid *game_grid; /* ref-counted (internally) */ |
| 111 | |
| 112 | /* Put -1 in a face that doesn't get a clue */ |
| 113 | signed char *clues; |
| 114 | |
| 115 | /* Array of line states, to store whether each line is |
| 116 | * YES, NO or UNKNOWN */ |
| 117 | char *lines; |
| 118 | |
| 119 | unsigned char *line_errors; |
| 120 | |
| 121 | int solved; |
| 122 | int cheated; |
| 123 | |
| 124 | /* Used in game_text_format(), so that it knows what type of |
| 125 | * grid it's trying to render as ASCII text. */ |
| 126 | int grid_type; |
| 127 | }; |
| 128 | |
| 129 | enum solver_status { |
| 130 | SOLVER_SOLVED, /* This is the only solution the solver could find */ |
| 131 | SOLVER_MISTAKE, /* This is definitely not a solution */ |
| 132 | SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */ |
| 133 | SOLVER_INCOMPLETE /* This may be a partial solution */ |
| 134 | }; |
| 135 | |
| 136 | /* ------ Solver state ------ */ |
| 137 | typedef struct solver_state { |
| 138 | game_state *state; |
| 139 | enum solver_status solver_status; |
| 140 | /* NB looplen is the number of dots that are joined together at a point, ie a |
| 141 | * looplen of 1 means there are no lines to a particular dot */ |
| 142 | int *looplen; |
| 143 | |
| 144 | /* Difficulty level of solver. Used by solver functions that want to |
| 145 | * vary their behaviour depending on the requested difficulty level. */ |
| 146 | int diff; |
| 147 | |
| 148 | /* caches */ |
| 149 | char *dot_yes_count; |
| 150 | char *dot_no_count; |
| 151 | char *face_yes_count; |
| 152 | char *face_no_count; |
| 153 | char *dot_solved, *face_solved; |
| 154 | int *dotdsf; |
| 155 | |
| 156 | /* Information for Normal level deductions: |
| 157 | * For each dline, store a bitmask for whether we know: |
| 158 | * (bit 0) at least one is YES |
| 159 | * (bit 1) at most one is YES */ |
| 160 | char *dlines; |
| 161 | |
| 162 | /* Hard level information */ |
| 163 | int *linedsf; |
| 164 | } solver_state; |
| 165 | |
| 166 | /* |
| 167 | * Difficulty levels. I do some macro ickery here to ensure that my |
| 168 | * enum and the various forms of my name list always match up. |
| 169 | */ |
| 170 | |
| 171 | #define DIFFLIST(A) \ |
| 172 | A(EASY,Easy,e) \ |
| 173 | A(NORMAL,Normal,n) \ |
| 174 | A(TRICKY,Tricky,t) \ |
| 175 | A(HARD,Hard,h) |
| 176 | #define ENUM(upper,title,lower) DIFF_ ## upper, |
| 177 | #define TITLE(upper,title,lower) #title, |
| 178 | #define ENCODE(upper,title,lower) #lower |
| 179 | #define CONFIG(upper,title,lower) ":" #title |
| 180 | enum { DIFFLIST(ENUM) DIFF_MAX }; |
| 181 | static char const *const diffnames[] = { DIFFLIST(TITLE) }; |
| 182 | static char const diffchars[] = DIFFLIST(ENCODE); |
| 183 | #define DIFFCONFIG DIFFLIST(CONFIG) |
| 184 | |
| 185 | /* |
| 186 | * Solver routines, sorted roughly in order of computational cost. |
| 187 | * The solver will run the faster deductions first, and slower deductions are |
| 188 | * only invoked when the faster deductions are unable to make progress. |
| 189 | * Each function is associated with a difficulty level, so that the generated |
| 190 | * puzzles are solvable by applying only the functions with the chosen |
| 191 | * difficulty level or lower. |
| 192 | */ |
| 193 | #define SOLVERLIST(A) \ |
| 194 | A(trivial_deductions, DIFF_EASY) \ |
| 195 | A(dline_deductions, DIFF_NORMAL) \ |
| 196 | A(linedsf_deductions, DIFF_HARD) \ |
| 197 | A(loop_deductions, DIFF_EASY) |
| 198 | #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *); |
| 199 | #define SOLVER_FN(fn,diff) &fn, |
| 200 | #define SOLVER_DIFF(fn,diff) diff, |
| 201 | SOLVERLIST(SOLVER_FN_DECL) |
| 202 | static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) }; |
| 203 | static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) }; |
| 204 | static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs); |
| 205 | |
| 206 | struct game_params { |
| 207 | int w, h; |
| 208 | int diff; |
| 209 | int type; |
| 210 | }; |
| 211 | |
| 212 | /* line_drawstate is the same as line_state, but with the extra ERROR |
| 213 | * possibility. The drawing code copies line_state to line_drawstate, |
| 214 | * except in the case that the line is an error. */ |
| 215 | enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO }; |
| 216 | enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN, |
| 217 | DS_LINE_NO, DS_LINE_ERROR }; |
| 218 | |
| 219 | #define OPP(line_state) \ |
| 220 | (2 - line_state) |
| 221 | |
| 222 | |
| 223 | struct game_drawstate { |
| 224 | int started; |
| 225 | int tilesize; |
| 226 | int flashing; |
| 227 | int *textx, *texty; |
| 228 | char *lines; |
| 229 | char *clue_error; |
| 230 | char *clue_satisfied; |
| 231 | }; |
| 232 | |
| 233 | static char *validate_desc(game_params *params, char *desc); |
| 234 | static int dot_order(const game_state* state, int i, char line_type); |
| 235 | static int face_order(const game_state* state, int i, char line_type); |
| 236 | static solver_state *solve_game_rec(const solver_state *sstate); |
| 237 | |
| 238 | #ifdef DEBUG_CACHES |
| 239 | static void check_caches(const solver_state* sstate); |
| 240 | #else |
| 241 | #define check_caches(s) |
| 242 | #endif |
| 243 | |
| 244 | /* ------- List of grid generators ------- */ |
| 245 | #define GRIDLIST(A) \ |
| 246 | A(Squares,GRID_SQUARE,3,3) \ |
| 247 | A(Triangular,GRID_TRIANGULAR,3,3) \ |
| 248 | A(Honeycomb,GRID_HONEYCOMB,3,3) \ |
| 249 | A(Snub-Square,GRID_SNUBSQUARE,3,3) \ |
| 250 | A(Cairo,GRID_CAIRO,3,4) \ |
| 251 | A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \ |
| 252 | A(Octagonal,GRID_OCTAGONAL,3,3) \ |
| 253 | A(Kites,GRID_KITE,3,3) \ |
| 254 | A(Floret,GRID_FLORET,1,2) \ |
| 255 | A(Dodecagonal,GRID_DODECAGONAL,2,2) \ |
| 256 | A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \ |
| 257 | A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \ |
| 258 | A(Penrose (rhombs),GRID_PENROSE_P3,3,3) |
| 259 | |
| 260 | #define GRID_NAME(title,type,amin,omin) #title, |
| 261 | #define GRID_CONFIG(title,type,amin,omin) ":" #title |
| 262 | #define GRID_TYPE(title,type,amin,omin) type, |
| 263 | #define GRID_SIZES(title,type,amin,omin) \ |
| 264 | {amin, omin, \ |
| 265 | "Width and height for this grid type must both be at least " #amin, \ |
| 266 | "At least one of width and height for this grid type must be at least " #omin,}, |
| 267 | static char const *const gridnames[] = { GRIDLIST(GRID_NAME) }; |
| 268 | #define GRID_CONFIGS GRIDLIST(GRID_CONFIG) |
| 269 | static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) }; |
| 270 | #define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0])) |
| 271 | static const struct { |
| 272 | int amin, omin; |
| 273 | char *aerr, *oerr; |
| 274 | } grid_size_limits[] = { GRIDLIST(GRID_SIZES) }; |
| 275 | |
| 276 | /* Generates a (dynamically allocated) new grid, according to the |
| 277 | * type and size requested in params. Does nothing if the grid is already |
| 278 | * generated. */ |
| 279 | static grid *loopy_generate_grid(game_params *params, char *grid_desc) |
| 280 | { |
| 281 | return grid_new(grid_types[params->type], params->w, params->h, grid_desc); |
| 282 | } |
| 283 | |
| 284 | /* ---------------------------------------------------------------------- |
| 285 | * Preprocessor magic |
| 286 | */ |
| 287 | |
| 288 | /* General constants */ |
| 289 | #define PREFERRED_TILE_SIZE 32 |
| 290 | #define BORDER(tilesize) ((tilesize) / 2) |
| 291 | #define FLASH_TIME 0.5F |
| 292 | |
| 293 | #define BIT_SET(field, bit) ((field) & (1<<(bit))) |
| 294 | |
| 295 | #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \ |
| 296 | ((field) |= (1<<(bit)), TRUE)) |
| 297 | |
| 298 | #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \ |
| 299 | ((field) &= ~(1<<(bit)), TRUE) : FALSE) |
| 300 | |
| 301 | #define CLUE2CHAR(c) \ |
| 302 | ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A') |
| 303 | |
| 304 | /* ---------------------------------------------------------------------- |
| 305 | * General struct manipulation and other straightforward code |
| 306 | */ |
| 307 | |
| 308 | static game_state *dup_game(game_state *state) |
| 309 | { |
| 310 | game_state *ret = snew(game_state); |
| 311 | |
| 312 | ret->game_grid = state->game_grid; |
| 313 | ret->game_grid->refcount++; |
| 314 | |
| 315 | ret->solved = state->solved; |
| 316 | ret->cheated = state->cheated; |
| 317 | |
| 318 | ret->clues = snewn(state->game_grid->num_faces, signed char); |
| 319 | memcpy(ret->clues, state->clues, state->game_grid->num_faces); |
| 320 | |
| 321 | ret->lines = snewn(state->game_grid->num_edges, char); |
| 322 | memcpy(ret->lines, state->lines, state->game_grid->num_edges); |
| 323 | |
| 324 | ret->line_errors = snewn(state->game_grid->num_edges, unsigned char); |
| 325 | memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges); |
| 326 | |
| 327 | ret->grid_type = state->grid_type; |
| 328 | return ret; |
| 329 | } |
| 330 | |
| 331 | static void free_game(game_state *state) |
| 332 | { |
| 333 | if (state) { |
| 334 | grid_free(state->game_grid); |
| 335 | sfree(state->clues); |
| 336 | sfree(state->lines); |
| 337 | sfree(state->line_errors); |
| 338 | sfree(state); |
| 339 | } |
| 340 | } |
| 341 | |
| 342 | static solver_state *new_solver_state(game_state *state, int diff) { |
| 343 | int i; |
| 344 | int num_dots = state->game_grid->num_dots; |
| 345 | int num_faces = state->game_grid->num_faces; |
| 346 | int num_edges = state->game_grid->num_edges; |
| 347 | solver_state *ret = snew(solver_state); |
| 348 | |
| 349 | ret->state = dup_game(state); |
| 350 | |
| 351 | ret->solver_status = SOLVER_INCOMPLETE; |
| 352 | ret->diff = diff; |
| 353 | |
| 354 | ret->dotdsf = snew_dsf(num_dots); |
| 355 | ret->looplen = snewn(num_dots, int); |
| 356 | |
| 357 | for (i = 0; i < num_dots; i++) { |
| 358 | ret->looplen[i] = 1; |
| 359 | } |
| 360 | |
| 361 | ret->dot_solved = snewn(num_dots, char); |
| 362 | ret->face_solved = snewn(num_faces, char); |
| 363 | memset(ret->dot_solved, FALSE, num_dots); |
| 364 | memset(ret->face_solved, FALSE, num_faces); |
| 365 | |
| 366 | ret->dot_yes_count = snewn(num_dots, char); |
| 367 | memset(ret->dot_yes_count, 0, num_dots); |
| 368 | ret->dot_no_count = snewn(num_dots, char); |
| 369 | memset(ret->dot_no_count, 0, num_dots); |
| 370 | ret->face_yes_count = snewn(num_faces, char); |
| 371 | memset(ret->face_yes_count, 0, num_faces); |
| 372 | ret->face_no_count = snewn(num_faces, char); |
| 373 | memset(ret->face_no_count, 0, num_faces); |
| 374 | |
| 375 | if (diff < DIFF_NORMAL) { |
| 376 | ret->dlines = NULL; |
| 377 | } else { |
| 378 | ret->dlines = snewn(2*num_edges, char); |
| 379 | memset(ret->dlines, 0, 2*num_edges); |
| 380 | } |
| 381 | |
| 382 | if (diff < DIFF_HARD) { |
| 383 | ret->linedsf = NULL; |
| 384 | } else { |
| 385 | ret->linedsf = snew_dsf(state->game_grid->num_edges); |
| 386 | } |
| 387 | |
| 388 | return ret; |
| 389 | } |
| 390 | |
| 391 | static void free_solver_state(solver_state *sstate) { |
| 392 | if (sstate) { |
| 393 | free_game(sstate->state); |
| 394 | sfree(sstate->dotdsf); |
| 395 | sfree(sstate->looplen); |
| 396 | sfree(sstate->dot_solved); |
| 397 | sfree(sstate->face_solved); |
| 398 | sfree(sstate->dot_yes_count); |
| 399 | sfree(sstate->dot_no_count); |
| 400 | sfree(sstate->face_yes_count); |
| 401 | sfree(sstate->face_no_count); |
| 402 | |
| 403 | /* OK, because sfree(NULL) is a no-op */ |
| 404 | sfree(sstate->dlines); |
| 405 | sfree(sstate->linedsf); |
| 406 | |
| 407 | sfree(sstate); |
| 408 | } |
| 409 | } |
| 410 | |
| 411 | static solver_state *dup_solver_state(const solver_state *sstate) { |
| 412 | game_state *state = sstate->state; |
| 413 | int num_dots = state->game_grid->num_dots; |
| 414 | int num_faces = state->game_grid->num_faces; |
| 415 | int num_edges = state->game_grid->num_edges; |
| 416 | solver_state *ret = snew(solver_state); |
| 417 | |
| 418 | ret->state = state = dup_game(sstate->state); |
| 419 | |
| 420 | ret->solver_status = sstate->solver_status; |
| 421 | ret->diff = sstate->diff; |
| 422 | |
| 423 | ret->dotdsf = snewn(num_dots, int); |
| 424 | ret->looplen = snewn(num_dots, int); |
| 425 | memcpy(ret->dotdsf, sstate->dotdsf, |
| 426 | num_dots * sizeof(int)); |
| 427 | memcpy(ret->looplen, sstate->looplen, |
| 428 | num_dots * sizeof(int)); |
| 429 | |
| 430 | ret->dot_solved = snewn(num_dots, char); |
| 431 | ret->face_solved = snewn(num_faces, char); |
| 432 | memcpy(ret->dot_solved, sstate->dot_solved, num_dots); |
| 433 | memcpy(ret->face_solved, sstate->face_solved, num_faces); |
| 434 | |
| 435 | ret->dot_yes_count = snewn(num_dots, char); |
| 436 | memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots); |
| 437 | ret->dot_no_count = snewn(num_dots, char); |
| 438 | memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots); |
| 439 | |
| 440 | ret->face_yes_count = snewn(num_faces, char); |
| 441 | memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces); |
| 442 | ret->face_no_count = snewn(num_faces, char); |
| 443 | memcpy(ret->face_no_count, sstate->face_no_count, num_faces); |
| 444 | |
| 445 | if (sstate->dlines) { |
| 446 | ret->dlines = snewn(2*num_edges, char); |
| 447 | memcpy(ret->dlines, sstate->dlines, |
| 448 | 2*num_edges); |
| 449 | } else { |
| 450 | ret->dlines = NULL; |
| 451 | } |
| 452 | |
| 453 | if (sstate->linedsf) { |
| 454 | ret->linedsf = snewn(num_edges, int); |
| 455 | memcpy(ret->linedsf, sstate->linedsf, |
| 456 | num_edges * sizeof(int)); |
| 457 | } else { |
| 458 | ret->linedsf = NULL; |
| 459 | } |
| 460 | |
| 461 | return ret; |
| 462 | } |
| 463 | |
| 464 | static game_params *default_params(void) |
| 465 | { |
| 466 | game_params *ret = snew(game_params); |
| 467 | |
| 468 | #ifdef SLOW_SYSTEM |
| 469 | ret->h = 7; |
| 470 | ret->w = 7; |
| 471 | #else |
| 472 | ret->h = 10; |
| 473 | ret->w = 10; |
| 474 | #endif |
| 475 | ret->diff = DIFF_EASY; |
| 476 | ret->type = 0; |
| 477 | |
| 478 | return ret; |
| 479 | } |
| 480 | |
| 481 | static game_params *dup_params(game_params *params) |
| 482 | { |
| 483 | game_params *ret = snew(game_params); |
| 484 | |
| 485 | *ret = *params; /* structure copy */ |
| 486 | return ret; |
| 487 | } |
| 488 | |
| 489 | static const game_params presets[] = { |
| 490 | #ifdef SMALL_SCREEN |
| 491 | { 7, 7, DIFF_EASY, 0 }, |
| 492 | { 7, 7, DIFF_NORMAL, 0 }, |
| 493 | { 7, 7, DIFF_HARD, 0 }, |
| 494 | { 7, 7, DIFF_HARD, 1 }, |
| 495 | { 7, 7, DIFF_HARD, 2 }, |
| 496 | { 5, 5, DIFF_HARD, 3 }, |
| 497 | { 7, 7, DIFF_HARD, 4 }, |
| 498 | { 5, 4, DIFF_HARD, 5 }, |
| 499 | { 5, 5, DIFF_HARD, 6 }, |
| 500 | { 5, 5, DIFF_HARD, 7 }, |
| 501 | { 3, 3, DIFF_HARD, 8 }, |
| 502 | { 3, 3, DIFF_HARD, 9 }, |
| 503 | { 3, 3, DIFF_HARD, 10 }, |
| 504 | { 6, 6, DIFF_HARD, 11 }, |
| 505 | { 6, 6, DIFF_HARD, 12 }, |
| 506 | #else |
| 507 | { 7, 7, DIFF_EASY, 0 }, |
| 508 | { 10, 10, DIFF_EASY, 0 }, |
| 509 | { 7, 7, DIFF_NORMAL, 0 }, |
| 510 | { 10, 10, DIFF_NORMAL, 0 }, |
| 511 | { 7, 7, DIFF_HARD, 0 }, |
| 512 | { 10, 10, DIFF_HARD, 0 }, |
| 513 | { 10, 10, DIFF_HARD, 1 }, |
| 514 | { 12, 10, DIFF_HARD, 2 }, |
| 515 | { 7, 7, DIFF_HARD, 3 }, |
| 516 | { 9, 9, DIFF_HARD, 4 }, |
| 517 | { 5, 4, DIFF_HARD, 5 }, |
| 518 | { 7, 7, DIFF_HARD, 6 }, |
| 519 | { 5, 5, DIFF_HARD, 7 }, |
| 520 | { 5, 5, DIFF_HARD, 8 }, |
| 521 | { 5, 4, DIFF_HARD, 9 }, |
| 522 | { 5, 4, DIFF_HARD, 10 }, |
| 523 | { 10, 10, DIFF_HARD, 11 }, |
| 524 | { 10, 10, DIFF_HARD, 12 } |
| 525 | #endif |
| 526 | }; |
| 527 | |
| 528 | static int game_fetch_preset(int i, char **name, game_params **params) |
| 529 | { |
| 530 | game_params *tmppar; |
| 531 | char buf[80]; |
| 532 | |
| 533 | if (i < 0 || i >= lenof(presets)) |
| 534 | return FALSE; |
| 535 | |
| 536 | tmppar = snew(game_params); |
| 537 | *tmppar = presets[i]; |
| 538 | *params = tmppar; |
| 539 | sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w, |
| 540 | gridnames[tmppar->type], diffnames[tmppar->diff]); |
| 541 | *name = dupstr(buf); |
| 542 | |
| 543 | return TRUE; |
| 544 | } |
| 545 | |
| 546 | static void free_params(game_params *params) |
| 547 | { |
| 548 | sfree(params); |
| 549 | } |
| 550 | |
| 551 | static void decode_params(game_params *params, char const *string) |
| 552 | { |
| 553 | params->h = params->w = atoi(string); |
| 554 | params->diff = DIFF_EASY; |
| 555 | while (*string && isdigit((unsigned char)*string)) string++; |
| 556 | if (*string == 'x') { |
| 557 | string++; |
| 558 | params->h = atoi(string); |
| 559 | while (*string && isdigit((unsigned char)*string)) string++; |
| 560 | } |
| 561 | if (*string == 't') { |
| 562 | string++; |
| 563 | params->type = atoi(string); |
| 564 | while (*string && isdigit((unsigned char)*string)) string++; |
| 565 | } |
| 566 | if (*string == 'd') { |
| 567 | int i; |
| 568 | string++; |
| 569 | for (i = 0; i < DIFF_MAX; i++) |
| 570 | if (*string == diffchars[i]) |
| 571 | params->diff = i; |
| 572 | if (*string) string++; |
| 573 | } |
| 574 | } |
| 575 | |
| 576 | static char *encode_params(game_params *params, int full) |
| 577 | { |
| 578 | char str[80]; |
| 579 | sprintf(str, "%dx%dt%d", params->w, params->h, params->type); |
| 580 | if (full) |
| 581 | sprintf(str + strlen(str), "d%c", diffchars[params->diff]); |
| 582 | return dupstr(str); |
| 583 | } |
| 584 | |
| 585 | static config_item *game_configure(game_params *params) |
| 586 | { |
| 587 | config_item *ret; |
| 588 | char buf[80]; |
| 589 | |
| 590 | ret = snewn(5, config_item); |
| 591 | |
| 592 | ret[0].name = "Width"; |
| 593 | ret[0].type = C_STRING; |
| 594 | sprintf(buf, "%d", params->w); |
| 595 | ret[0].sval = dupstr(buf); |
| 596 | ret[0].ival = 0; |
| 597 | |
| 598 | ret[1].name = "Height"; |
| 599 | ret[1].type = C_STRING; |
| 600 | sprintf(buf, "%d", params->h); |
| 601 | ret[1].sval = dupstr(buf); |
| 602 | ret[1].ival = 0; |
| 603 | |
| 604 | ret[2].name = "Grid type"; |
| 605 | ret[2].type = C_CHOICES; |
| 606 | ret[2].sval = GRID_CONFIGS; |
| 607 | ret[2].ival = params->type; |
| 608 | |
| 609 | ret[3].name = "Difficulty"; |
| 610 | ret[3].type = C_CHOICES; |
| 611 | ret[3].sval = DIFFCONFIG; |
| 612 | ret[3].ival = params->diff; |
| 613 | |
| 614 | ret[4].name = NULL; |
| 615 | ret[4].type = C_END; |
| 616 | ret[4].sval = NULL; |
| 617 | ret[4].ival = 0; |
| 618 | |
| 619 | return ret; |
| 620 | } |
| 621 | |
| 622 | static game_params *custom_params(config_item *cfg) |
| 623 | { |
| 624 | game_params *ret = snew(game_params); |
| 625 | |
| 626 | ret->w = atoi(cfg[0].sval); |
| 627 | ret->h = atoi(cfg[1].sval); |
| 628 | ret->type = cfg[2].ival; |
| 629 | ret->diff = cfg[3].ival; |
| 630 | |
| 631 | return ret; |
| 632 | } |
| 633 | |
| 634 | static char *validate_params(game_params *params, int full) |
| 635 | { |
| 636 | if (params->type < 0 || params->type >= NUM_GRID_TYPES) |
| 637 | return "Illegal grid type"; |
| 638 | if (params->w < grid_size_limits[params->type].amin || |
| 639 | params->h < grid_size_limits[params->type].amin) |
| 640 | return grid_size_limits[params->type].aerr; |
| 641 | if (params->w < grid_size_limits[params->type].omin && |
| 642 | params->h < grid_size_limits[params->type].omin) |
| 643 | return grid_size_limits[params->type].oerr; |
| 644 | |
| 645 | /* |
| 646 | * This shouldn't be able to happen at all, since decode_params |
| 647 | * and custom_params will never generate anything that isn't |
| 648 | * within range. |
| 649 | */ |
| 650 | assert(params->diff < DIFF_MAX); |
| 651 | |
| 652 | return NULL; |
| 653 | } |
| 654 | |
| 655 | /* Returns a newly allocated string describing the current puzzle */ |
| 656 | static char *state_to_text(const game_state *state) |
| 657 | { |
| 658 | grid *g = state->game_grid; |
| 659 | char *retval; |
| 660 | int num_faces = g->num_faces; |
| 661 | char *description = snewn(num_faces + 1, char); |
| 662 | char *dp = description; |
| 663 | int empty_count = 0; |
| 664 | int i; |
| 665 | |
| 666 | for (i = 0; i < num_faces; i++) { |
| 667 | if (state->clues[i] < 0) { |
| 668 | if (empty_count > 25) { |
| 669 | dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); |
| 670 | empty_count = 0; |
| 671 | } |
| 672 | empty_count++; |
| 673 | } else { |
| 674 | if (empty_count) { |
| 675 | dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); |
| 676 | empty_count = 0; |
| 677 | } |
| 678 | dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i])); |
| 679 | } |
| 680 | } |
| 681 | |
| 682 | if (empty_count) |
| 683 | dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); |
| 684 | |
| 685 | retval = dupstr(description); |
| 686 | sfree(description); |
| 687 | |
| 688 | return retval; |
| 689 | } |
| 690 | |
| 691 | #define GRID_DESC_SEP '_' |
| 692 | |
| 693 | /* Splits up a (optional) grid_desc from the game desc. Returns the |
| 694 | * grid_desc (which needs freeing) and updates the desc pointer to |
| 695 | * start of real desc, or returns NULL if no desc. */ |
| 696 | static char *extract_grid_desc(char **desc) |
| 697 | { |
| 698 | char *sep = strchr(*desc, GRID_DESC_SEP), *gd; |
| 699 | int gd_len; |
| 700 | |
| 701 | if (!sep) return NULL; |
| 702 | |
| 703 | gd_len = sep - (*desc); |
| 704 | gd = snewn(gd_len+1, char); |
| 705 | memcpy(gd, *desc, gd_len); |
| 706 | gd[gd_len] = '\0'; |
| 707 | |
| 708 | *desc = sep+1; |
| 709 | |
| 710 | return gd; |
| 711 | } |
| 712 | |
| 713 | /* We require that the params pass the test in validate_params and that the |
| 714 | * description fills the entire game area */ |
| 715 | static char *validate_desc(game_params *params, char *desc) |
| 716 | { |
| 717 | int count = 0; |
| 718 | grid *g; |
| 719 | char *grid_desc, *ret; |
| 720 | |
| 721 | /* It's pretty inefficient to do this just for validation. All we need to |
| 722 | * know is the precise number of faces. */ |
| 723 | grid_desc = extract_grid_desc(&desc); |
| 724 | ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc); |
| 725 | if (ret) return ret; |
| 726 | |
| 727 | g = loopy_generate_grid(params, grid_desc); |
| 728 | if (grid_desc) sfree(grid_desc); |
| 729 | |
| 730 | for (; *desc; ++desc) { |
| 731 | if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) { |
| 732 | count++; |
| 733 | continue; |
| 734 | } |
| 735 | if (*desc >= 'a') { |
| 736 | count += *desc - 'a' + 1; |
| 737 | continue; |
| 738 | } |
| 739 | return "Unknown character in description"; |
| 740 | } |
| 741 | |
| 742 | if (count < g->num_faces) |
| 743 | return "Description too short for board size"; |
| 744 | if (count > g->num_faces) |
| 745 | return "Description too long for board size"; |
| 746 | |
| 747 | grid_free(g); |
| 748 | |
| 749 | return NULL; |
| 750 | } |
| 751 | |
| 752 | /* Sums the lengths of the numbers in range [0,n) */ |
| 753 | /* See equivalent function in solo.c for justification of this. */ |
| 754 | static int len_0_to_n(int n) |
| 755 | { |
| 756 | int len = 1; /* Counting 0 as a bit of a special case */ |
| 757 | int i; |
| 758 | |
| 759 | for (i = 1; i < n; i *= 10) { |
| 760 | len += max(n - i, 0); |
| 761 | } |
| 762 | |
| 763 | return len; |
| 764 | } |
| 765 | |
| 766 | static char *encode_solve_move(const game_state *state) |
| 767 | { |
| 768 | int len; |
| 769 | char *ret, *p; |
| 770 | int i; |
| 771 | int num_edges = state->game_grid->num_edges; |
| 772 | |
| 773 | /* This is going to return a string representing the moves needed to set |
| 774 | * every line in a grid to be the same as the ones in 'state'. The exact |
| 775 | * length of this string is predictable. */ |
| 776 | |
| 777 | len = 1; /* Count the 'S' prefix */ |
| 778 | /* Numbers in all lines */ |
| 779 | len += len_0_to_n(num_edges); |
| 780 | /* For each line we also have a letter */ |
| 781 | len += num_edges; |
| 782 | |
| 783 | ret = snewn(len + 1, char); |
| 784 | p = ret; |
| 785 | |
| 786 | p += sprintf(p, "S"); |
| 787 | |
| 788 | for (i = 0; i < num_edges; i++) { |
| 789 | switch (state->lines[i]) { |
| 790 | case LINE_YES: |
| 791 | p += sprintf(p, "%dy", i); |
| 792 | break; |
| 793 | case LINE_NO: |
| 794 | p += sprintf(p, "%dn", i); |
| 795 | break; |
| 796 | } |
| 797 | } |
| 798 | |
| 799 | /* No point in doing sums like that if they're going to be wrong */ |
| 800 | assert(strlen(ret) <= (size_t)len); |
| 801 | return ret; |
| 802 | } |
| 803 | |
| 804 | static game_ui *new_ui(game_state *state) |
| 805 | { |
| 806 | return NULL; |
| 807 | } |
| 808 | |
| 809 | static void free_ui(game_ui *ui) |
| 810 | { |
| 811 | } |
| 812 | |
| 813 | static char *encode_ui(game_ui *ui) |
| 814 | { |
| 815 | return NULL; |
| 816 | } |
| 817 | |
| 818 | static void decode_ui(game_ui *ui, char *encoding) |
| 819 | { |
| 820 | } |
| 821 | |
| 822 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
| 823 | game_state *newstate) |
| 824 | { |
| 825 | } |
| 826 | |
| 827 | static void game_compute_size(game_params *params, int tilesize, |
| 828 | int *x, int *y) |
| 829 | { |
| 830 | int grid_width, grid_height, rendered_width, rendered_height; |
| 831 | int g_tilesize; |
| 832 | |
| 833 | grid_compute_size(grid_types[params->type], params->w, params->h, |
| 834 | &g_tilesize, &grid_width, &grid_height); |
| 835 | |
| 836 | /* multiply first to minimise rounding error on integer division */ |
| 837 | rendered_width = grid_width * tilesize / g_tilesize; |
| 838 | rendered_height = grid_height * tilesize / g_tilesize; |
| 839 | *x = rendered_width + 2 * BORDER(tilesize) + 1; |
| 840 | *y = rendered_height + 2 * BORDER(tilesize) + 1; |
| 841 | } |
| 842 | |
| 843 | static void game_set_size(drawing *dr, game_drawstate *ds, |
| 844 | game_params *params, int tilesize) |
| 845 | { |
| 846 | ds->tilesize = tilesize; |
| 847 | } |
| 848 | |
| 849 | static float *game_colours(frontend *fe, int *ncolours) |
| 850 | { |
| 851 | float *ret = snewn(4 * NCOLOURS, float); |
| 852 | |
| 853 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
| 854 | |
| 855 | ret[COL_FOREGROUND * 3 + 0] = 0.0F; |
| 856 | ret[COL_FOREGROUND * 3 + 1] = 0.0F; |
| 857 | ret[COL_FOREGROUND * 3 + 2] = 0.0F; |
| 858 | |
| 859 | /* |
| 860 | * We want COL_LINEUNKNOWN to be a yellow which is a bit darker |
| 861 | * than the background. (I previously set it to 0.8,0.8,0, but |
| 862 | * found that this went badly with the 0.8,0.8,0.8 favoured as a |
| 863 | * background by the Java frontend.) |
| 864 | */ |
| 865 | ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F; |
| 866 | ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F; |
| 867 | ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F; |
| 868 | |
| 869 | ret[COL_HIGHLIGHT * 3 + 0] = 1.0F; |
| 870 | ret[COL_HIGHLIGHT * 3 + 1] = 1.0F; |
| 871 | ret[COL_HIGHLIGHT * 3 + 2] = 1.0F; |
| 872 | |
| 873 | ret[COL_MISTAKE * 3 + 0] = 1.0F; |
| 874 | ret[COL_MISTAKE * 3 + 1] = 0.0F; |
| 875 | ret[COL_MISTAKE * 3 + 2] = 0.0F; |
| 876 | |
| 877 | ret[COL_SATISFIED * 3 + 0] = 0.0F; |
| 878 | ret[COL_SATISFIED * 3 + 1] = 0.0F; |
| 879 | ret[COL_SATISFIED * 3 + 2] = 0.0F; |
| 880 | |
| 881 | /* We want the faint lines to be a bit darker than the background. |
| 882 | * Except if the background is pretty dark already; then it ought to be a |
| 883 | * bit lighter. Oy vey. |
| 884 | */ |
| 885 | ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F; |
| 886 | ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F; |
| 887 | ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F; |
| 888 | |
| 889 | *ncolours = NCOLOURS; |
| 890 | return ret; |
| 891 | } |
| 892 | |
| 893 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
| 894 | { |
| 895 | struct game_drawstate *ds = snew(struct game_drawstate); |
| 896 | int num_faces = state->game_grid->num_faces; |
| 897 | int num_edges = state->game_grid->num_edges; |
| 898 | int i; |
| 899 | |
| 900 | ds->tilesize = 0; |
| 901 | ds->started = 0; |
| 902 | ds->lines = snewn(num_edges, char); |
| 903 | ds->clue_error = snewn(num_faces, char); |
| 904 | ds->clue_satisfied = snewn(num_faces, char); |
| 905 | ds->textx = snewn(num_faces, int); |
| 906 | ds->texty = snewn(num_faces, int); |
| 907 | ds->flashing = 0; |
| 908 | |
| 909 | memset(ds->lines, LINE_UNKNOWN, num_edges); |
| 910 | memset(ds->clue_error, 0, num_faces); |
| 911 | memset(ds->clue_satisfied, 0, num_faces); |
| 912 | for (i = 0; i < num_faces; i++) |
| 913 | ds->textx[i] = ds->texty[i] = -1; |
| 914 | |
| 915 | return ds; |
| 916 | } |
| 917 | |
| 918 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
| 919 | { |
| 920 | sfree(ds->textx); |
| 921 | sfree(ds->texty); |
| 922 | sfree(ds->clue_error); |
| 923 | sfree(ds->clue_satisfied); |
| 924 | sfree(ds->lines); |
| 925 | sfree(ds); |
| 926 | } |
| 927 | |
| 928 | static int game_timing_state(game_state *state, game_ui *ui) |
| 929 | { |
| 930 | return TRUE; |
| 931 | } |
| 932 | |
| 933 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
| 934 | int dir, game_ui *ui) |
| 935 | { |
| 936 | return 0.0F; |
| 937 | } |
| 938 | |
| 939 | static int game_can_format_as_text_now(game_params *params) |
| 940 | { |
| 941 | if (params->type != 0) |
| 942 | return FALSE; |
| 943 | return TRUE; |
| 944 | } |
| 945 | |
| 946 | static char *game_text_format(game_state *state) |
| 947 | { |
| 948 | int w, h, W, H; |
| 949 | int x, y, i; |
| 950 | int cell_size; |
| 951 | char *ret; |
| 952 | grid *g = state->game_grid; |
| 953 | grid_face *f; |
| 954 | |
| 955 | assert(state->grid_type == 0); |
| 956 | |
| 957 | /* Work out the basic size unit */ |
| 958 | f = g->faces; /* first face */ |
| 959 | assert(f->order == 4); |
| 960 | /* The dots are ordered clockwise, so the two opposite |
| 961 | * corners are guaranteed to span the square */ |
| 962 | cell_size = abs(f->dots[0]->x - f->dots[2]->x); |
| 963 | |
| 964 | w = (g->highest_x - g->lowest_x) / cell_size; |
| 965 | h = (g->highest_y - g->lowest_y) / cell_size; |
| 966 | |
| 967 | /* Create a blank "canvas" to "draw" on */ |
| 968 | W = 2 * w + 2; |
| 969 | H = 2 * h + 1; |
| 970 | ret = snewn(W * H + 1, char); |
| 971 | for (y = 0; y < H; y++) { |
| 972 | for (x = 0; x < W-1; x++) { |
| 973 | ret[y*W + x] = ' '; |
| 974 | } |
| 975 | ret[y*W + W-1] = '\n'; |
| 976 | } |
| 977 | ret[H*W] = '\0'; |
| 978 | |
| 979 | /* Fill in edge info */ |
| 980 | for (i = 0; i < g->num_edges; i++) { |
| 981 | grid_edge *e = g->edges + i; |
| 982 | /* Cell coordinates, from (0,0) to (w-1,h-1) */ |
| 983 | int x1 = (e->dot1->x - g->lowest_x) / cell_size; |
| 984 | int x2 = (e->dot2->x - g->lowest_x) / cell_size; |
| 985 | int y1 = (e->dot1->y - g->lowest_y) / cell_size; |
| 986 | int y2 = (e->dot2->y - g->lowest_y) / cell_size; |
| 987 | /* Midpoint, in canvas coordinates (canvas coordinates are just twice |
| 988 | * cell coordinates) */ |
| 989 | x = x1 + x2; |
| 990 | y = y1 + y2; |
| 991 | switch (state->lines[i]) { |
| 992 | case LINE_YES: |
| 993 | ret[y*W + x] = (y1 == y2) ? '-' : '|'; |
| 994 | break; |
| 995 | case LINE_NO: |
| 996 | ret[y*W + x] = 'x'; |
| 997 | break; |
| 998 | case LINE_UNKNOWN: |
| 999 | break; /* already a space */ |
| 1000 | default: |
| 1001 | assert(!"Illegal line state"); |
| 1002 | } |
| 1003 | } |
| 1004 | |
| 1005 | /* Fill in clues */ |
| 1006 | for (i = 0; i < g->num_faces; i++) { |
| 1007 | int x1, x2, y1, y2; |
| 1008 | |
| 1009 | f = g->faces + i; |
| 1010 | assert(f->order == 4); |
| 1011 | /* Cell coordinates, from (0,0) to (w-1,h-1) */ |
| 1012 | x1 = (f->dots[0]->x - g->lowest_x) / cell_size; |
| 1013 | x2 = (f->dots[2]->x - g->lowest_x) / cell_size; |
| 1014 | y1 = (f->dots[0]->y - g->lowest_y) / cell_size; |
| 1015 | y2 = (f->dots[2]->y - g->lowest_y) / cell_size; |
| 1016 | /* Midpoint, in canvas coordinates */ |
| 1017 | x = x1 + x2; |
| 1018 | y = y1 + y2; |
| 1019 | ret[y*W + x] = CLUE2CHAR(state->clues[i]); |
| 1020 | } |
| 1021 | return ret; |
| 1022 | } |
| 1023 | |
| 1024 | /* ---------------------------------------------------------------------- |
| 1025 | * Debug code |
| 1026 | */ |
| 1027 | |
| 1028 | #ifdef DEBUG_CACHES |
| 1029 | static void check_caches(const solver_state* sstate) |
| 1030 | { |
| 1031 | int i; |
| 1032 | const game_state *state = sstate->state; |
| 1033 | const grid *g = state->game_grid; |
| 1034 | |
| 1035 | for (i = 0; i < g->num_dots; i++) { |
| 1036 | assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]); |
| 1037 | assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]); |
| 1038 | } |
| 1039 | |
| 1040 | for (i = 0; i < g->num_faces; i++) { |
| 1041 | assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]); |
| 1042 | assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]); |
| 1043 | } |
| 1044 | } |
| 1045 | |
| 1046 | #if 0 |
| 1047 | #define check_caches(s) \ |
| 1048 | do { \ |
| 1049 | fprintf(stderr, "check_caches at line %d\n", __LINE__); \ |
| 1050 | check_caches(s); \ |
| 1051 | } while (0) |
| 1052 | #endif |
| 1053 | #endif /* DEBUG_CACHES */ |
| 1054 | |
| 1055 | /* ---------------------------------------------------------------------- |
| 1056 | * Solver utility functions |
| 1057 | */ |
| 1058 | |
| 1059 | /* Sets the line (with index i) to the new state 'line_new', and updates |
| 1060 | * the cached counts of any affected faces and dots. |
| 1061 | * Returns TRUE if this actually changed the line's state. */ |
| 1062 | static int solver_set_line(solver_state *sstate, int i, |
| 1063 | enum line_state line_new |
| 1064 | #ifdef SHOW_WORKING |
| 1065 | , const char *reason |
| 1066 | #endif |
| 1067 | ) |
| 1068 | { |
| 1069 | game_state *state = sstate->state; |
| 1070 | grid *g; |
| 1071 | grid_edge *e; |
| 1072 | |
| 1073 | assert(line_new != LINE_UNKNOWN); |
| 1074 | |
| 1075 | check_caches(sstate); |
| 1076 | |
| 1077 | if (state->lines[i] == line_new) { |
| 1078 | return FALSE; /* nothing changed */ |
| 1079 | } |
| 1080 | state->lines[i] = line_new; |
| 1081 | |
| 1082 | #ifdef SHOW_WORKING |
| 1083 | fprintf(stderr, "solver: set line [%d] to %s (%s)\n", |
| 1084 | i, line_new == LINE_YES ? "YES" : "NO", |
| 1085 | reason); |
| 1086 | #endif |
| 1087 | |
| 1088 | g = state->game_grid; |
| 1089 | e = g->edges + i; |
| 1090 | |
| 1091 | /* Update the cache for both dots and both faces affected by this. */ |
| 1092 | if (line_new == LINE_YES) { |
| 1093 | sstate->dot_yes_count[e->dot1 - g->dots]++; |
| 1094 | sstate->dot_yes_count[e->dot2 - g->dots]++; |
| 1095 | if (e->face1) { |
| 1096 | sstate->face_yes_count[e->face1 - g->faces]++; |
| 1097 | } |
| 1098 | if (e->face2) { |
| 1099 | sstate->face_yes_count[e->face2 - g->faces]++; |
| 1100 | } |
| 1101 | } else { |
| 1102 | sstate->dot_no_count[e->dot1 - g->dots]++; |
| 1103 | sstate->dot_no_count[e->dot2 - g->dots]++; |
| 1104 | if (e->face1) { |
| 1105 | sstate->face_no_count[e->face1 - g->faces]++; |
| 1106 | } |
| 1107 | if (e->face2) { |
| 1108 | sstate->face_no_count[e->face2 - g->faces]++; |
| 1109 | } |
| 1110 | } |
| 1111 | |
| 1112 | check_caches(sstate); |
| 1113 | return TRUE; |
| 1114 | } |
| 1115 | |
| 1116 | #ifdef SHOW_WORKING |
| 1117 | #define solver_set_line(a, b, c) \ |
| 1118 | solver_set_line(a, b, c, __FUNCTION__) |
| 1119 | #endif |
| 1120 | |
| 1121 | /* |
| 1122 | * Merge two dots due to the existence of an edge between them. |
| 1123 | * Updates the dsf tracking equivalence classes, and keeps track of |
| 1124 | * the length of path each dot is currently a part of. |
| 1125 | * Returns TRUE if the dots were already linked, ie if they are part of a |
| 1126 | * closed loop, and false otherwise. |
| 1127 | */ |
| 1128 | static int merge_dots(solver_state *sstate, int edge_index) |
| 1129 | { |
| 1130 | int i, j, len; |
| 1131 | grid *g = sstate->state->game_grid; |
| 1132 | grid_edge *e = g->edges + edge_index; |
| 1133 | |
| 1134 | i = e->dot1 - g->dots; |
| 1135 | j = e->dot2 - g->dots; |
| 1136 | |
| 1137 | i = dsf_canonify(sstate->dotdsf, i); |
| 1138 | j = dsf_canonify(sstate->dotdsf, j); |
| 1139 | |
| 1140 | if (i == j) { |
| 1141 | return TRUE; |
| 1142 | } else { |
| 1143 | len = sstate->looplen[i] + sstate->looplen[j]; |
| 1144 | dsf_merge(sstate->dotdsf, i, j); |
| 1145 | i = dsf_canonify(sstate->dotdsf, i); |
| 1146 | sstate->looplen[i] = len; |
| 1147 | return FALSE; |
| 1148 | } |
| 1149 | } |
| 1150 | |
| 1151 | /* Merge two lines because the solver has deduced that they must be either |
| 1152 | * identical or opposite. Returns TRUE if this is new information, otherwise |
| 1153 | * FALSE. */ |
| 1154 | static int merge_lines(solver_state *sstate, int i, int j, int inverse |
| 1155 | #ifdef SHOW_WORKING |
| 1156 | , const char *reason |
| 1157 | #endif |
| 1158 | ) |
| 1159 | { |
| 1160 | int inv_tmp; |
| 1161 | |
| 1162 | assert(i < sstate->state->game_grid->num_edges); |
| 1163 | assert(j < sstate->state->game_grid->num_edges); |
| 1164 | |
| 1165 | i = edsf_canonify(sstate->linedsf, i, &inv_tmp); |
| 1166 | inverse ^= inv_tmp; |
| 1167 | j = edsf_canonify(sstate->linedsf, j, &inv_tmp); |
| 1168 | inverse ^= inv_tmp; |
| 1169 | |
| 1170 | edsf_merge(sstate->linedsf, i, j, inverse); |
| 1171 | |
| 1172 | #ifdef SHOW_WORKING |
| 1173 | if (i != j) { |
| 1174 | fprintf(stderr, "%s [%d] [%d] %s(%s)\n", |
| 1175 | __FUNCTION__, i, j, |
| 1176 | inverse ? "inverse " : "", reason); |
| 1177 | } |
| 1178 | #endif |
| 1179 | return (i != j); |
| 1180 | } |
| 1181 | |
| 1182 | #ifdef SHOW_WORKING |
| 1183 | #define merge_lines(a, b, c, d) \ |
| 1184 | merge_lines(a, b, c, d, __FUNCTION__) |
| 1185 | #endif |
| 1186 | |
| 1187 | /* Count the number of lines of a particular type currently going into the |
| 1188 | * given dot. */ |
| 1189 | static int dot_order(const game_state* state, int dot, char line_type) |
| 1190 | { |
| 1191 | int n = 0; |
| 1192 | grid *g = state->game_grid; |
| 1193 | grid_dot *d = g->dots + dot; |
| 1194 | int i; |
| 1195 | |
| 1196 | for (i = 0; i < d->order; i++) { |
| 1197 | grid_edge *e = d->edges[i]; |
| 1198 | if (state->lines[e - g->edges] == line_type) |
| 1199 | ++n; |
| 1200 | } |
| 1201 | return n; |
| 1202 | } |
| 1203 | |
| 1204 | /* Count the number of lines of a particular type currently surrounding the |
| 1205 | * given face */ |
| 1206 | static int face_order(const game_state* state, int face, char line_type) |
| 1207 | { |
| 1208 | int n = 0; |
| 1209 | grid *g = state->game_grid; |
| 1210 | grid_face *f = g->faces + face; |
| 1211 | int i; |
| 1212 | |
| 1213 | for (i = 0; i < f->order; i++) { |
| 1214 | grid_edge *e = f->edges[i]; |
| 1215 | if (state->lines[e - g->edges] == line_type) |
| 1216 | ++n; |
| 1217 | } |
| 1218 | return n; |
| 1219 | } |
| 1220 | |
| 1221 | /* Set all lines bordering a dot of type old_type to type new_type |
| 1222 | * Return value tells caller whether this function actually did anything */ |
| 1223 | static int dot_setall(solver_state *sstate, int dot, |
| 1224 | char old_type, char new_type) |
| 1225 | { |
| 1226 | int retval = FALSE, r; |
| 1227 | game_state *state = sstate->state; |
| 1228 | grid *g; |
| 1229 | grid_dot *d; |
| 1230 | int i; |
| 1231 | |
| 1232 | if (old_type == new_type) |
| 1233 | return FALSE; |
| 1234 | |
| 1235 | g = state->game_grid; |
| 1236 | d = g->dots + dot; |
| 1237 | |
| 1238 | for (i = 0; i < d->order; i++) { |
| 1239 | int line_index = d->edges[i] - g->edges; |
| 1240 | if (state->lines[line_index] == old_type) { |
| 1241 | r = solver_set_line(sstate, line_index, new_type); |
| 1242 | assert(r == TRUE); |
| 1243 | retval = TRUE; |
| 1244 | } |
| 1245 | } |
| 1246 | return retval; |
| 1247 | } |
| 1248 | |
| 1249 | /* Set all lines bordering a face of type old_type to type new_type */ |
| 1250 | static int face_setall(solver_state *sstate, int face, |
| 1251 | char old_type, char new_type) |
| 1252 | { |
| 1253 | int retval = FALSE, r; |
| 1254 | game_state *state = sstate->state; |
| 1255 | grid *g; |
| 1256 | grid_face *f; |
| 1257 | int i; |
| 1258 | |
| 1259 | if (old_type == new_type) |
| 1260 | return FALSE; |
| 1261 | |
| 1262 | g = state->game_grid; |
| 1263 | f = g->faces + face; |
| 1264 | |
| 1265 | for (i = 0; i < f->order; i++) { |
| 1266 | int line_index = f->edges[i] - g->edges; |
| 1267 | if (state->lines[line_index] == old_type) { |
| 1268 | r = solver_set_line(sstate, line_index, new_type); |
| 1269 | assert(r == TRUE); |
| 1270 | retval = TRUE; |
| 1271 | } |
| 1272 | } |
| 1273 | return retval; |
| 1274 | } |
| 1275 | |
| 1276 | /* ---------------------------------------------------------------------- |
| 1277 | * Loop generation and clue removal |
| 1278 | */ |
| 1279 | |
| 1280 | /* We're going to store lists of current candidate faces for colouring black |
| 1281 | * or white. |
| 1282 | * Each face gets a 'score', which tells us how adding that face right |
| 1283 | * now would affect the curliness of the solution loop. We're trying to |
| 1284 | * maximise that quantity so will bias our random selection of faces to |
| 1285 | * colour those with high scores */ |
| 1286 | struct face_score { |
| 1287 | int white_score; |
| 1288 | int black_score; |
| 1289 | unsigned long random; |
| 1290 | /* No need to store a grid_face* here. The 'face_scores' array will |
| 1291 | * be a list of 'face_score' objects, one for each face of the grid, so |
| 1292 | * the position (index) within the 'face_scores' array will determine |
| 1293 | * which face corresponds to a particular face_score. |
| 1294 | * Having a single 'face_scores' array for all faces simplifies memory |
| 1295 | * management, and probably improves performance, because we don't have to |
| 1296 | * malloc/free each individual face_score, and we don't have to maintain |
| 1297 | * a mapping from grid_face* pointers to face_score* pointers. |
| 1298 | */ |
| 1299 | }; |
| 1300 | |
| 1301 | static int generic_sort_cmpfn(void *v1, void *v2, size_t offset) |
| 1302 | { |
| 1303 | struct face_score *f1 = v1; |
| 1304 | struct face_score *f2 = v2; |
| 1305 | int r; |
| 1306 | |
| 1307 | r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset); |
| 1308 | if (r) { |
| 1309 | return r; |
| 1310 | } |
| 1311 | |
| 1312 | if (f1->random < f2->random) |
| 1313 | return -1; |
| 1314 | else if (f1->random > f2->random) |
| 1315 | return 1; |
| 1316 | |
| 1317 | /* |
| 1318 | * It's _just_ possible that two faces might have been given |
| 1319 | * the same random value. In that situation, fall back to |
| 1320 | * comparing based on the positions within the face_scores list. |
| 1321 | * This introduces a tiny directional bias, but not a significant one. |
| 1322 | */ |
| 1323 | return f1 - f2; |
| 1324 | } |
| 1325 | |
| 1326 | static int white_sort_cmpfn(void *v1, void *v2) |
| 1327 | { |
| 1328 | return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score)); |
| 1329 | } |
| 1330 | |
| 1331 | static int black_sort_cmpfn(void *v1, void *v2) |
| 1332 | { |
| 1333 | return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score)); |
| 1334 | } |
| 1335 | |
| 1336 | enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK }; |
| 1337 | |
| 1338 | /* face should be of type grid_face* here. */ |
| 1339 | #define FACE_COLOUR(face) \ |
| 1340 | ( (face) == NULL ? FACE_BLACK : \ |
| 1341 | board[(face) - g->faces] ) |
| 1342 | |
| 1343 | /* 'board' is an array of these enums, indicating which faces are |
| 1344 | * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK. |
| 1345 | * Returns whether it's legal to colour the given face with this colour. */ |
| 1346 | static int can_colour_face(grid *g, char* board, int face_index, |
| 1347 | enum face_colour colour) |
| 1348 | { |
| 1349 | int i, j; |
| 1350 | grid_face *test_face = g->faces + face_index; |
| 1351 | grid_face *starting_face, *current_face; |
| 1352 | grid_dot *starting_dot; |
| 1353 | int transitions; |
| 1354 | int current_state, s; /* booleans: equal or not-equal to 'colour' */ |
| 1355 | int found_same_coloured_neighbour = FALSE; |
| 1356 | assert(board[face_index] != colour); |
| 1357 | |
| 1358 | /* Can only consider a face for colouring if it's adjacent to a face |
| 1359 | * with the same colour. */ |
| 1360 | for (i = 0; i < test_face->order; i++) { |
| 1361 | grid_edge *e = test_face->edges[i]; |
| 1362 | grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1; |
| 1363 | if (FACE_COLOUR(f) == colour) { |
| 1364 | found_same_coloured_neighbour = TRUE; |
| 1365 | break; |
| 1366 | } |
| 1367 | } |
| 1368 | if (!found_same_coloured_neighbour) |
| 1369 | return FALSE; |
| 1370 | |
| 1371 | /* Need to avoid creating a loop of faces of this colour around some |
| 1372 | * differently-coloured faces. |
| 1373 | * Also need to avoid meeting a same-coloured face at a corner, with |
| 1374 | * other-coloured faces in between. Here's a simple test that (I believe) |
| 1375 | * takes care of both these conditions: |
| 1376 | * |
| 1377 | * Take the circular path formed by this face's edges, and inflate it |
| 1378 | * slightly outwards. Imagine walking around this path and consider |
| 1379 | * the faces that you visit in sequence. This will include all faces |
| 1380 | * touching the given face, either along an edge or just at a corner. |
| 1381 | * Count the number of 'colour'/not-'colour' transitions you encounter, as |
| 1382 | * you walk along the complete loop. This will obviously turn out to be |
| 1383 | * an even number. |
| 1384 | * If 0, we're either in the middle of an "island" of this colour (should |
| 1385 | * be impossible as we're not supposed to create black or white loops), |
| 1386 | * or we're about to start a new island - also not allowed. |
| 1387 | * If 4 or greater, there are too many separate coloured regions touching |
| 1388 | * this face, and colouring it would create a loop or a corner-violation. |
| 1389 | * The only allowed case is when the count is exactly 2. */ |
| 1390 | |
| 1391 | /* i points to a dot around the test face. |
| 1392 | * j points to a face around the i^th dot. |
| 1393 | * The current face will always be: |
| 1394 | * test_face->dots[i]->faces[j] |
| 1395 | * We assume dots go clockwise around the test face, |
| 1396 | * and faces go clockwise around dots. */ |
| 1397 | |
| 1398 | /* |
| 1399 | * The end condition is slightly fiddly. In sufficiently strange |
| 1400 | * degenerate grids, our test face may be adjacent to the same |
| 1401 | * other face multiple times (typically if it's the exterior |
| 1402 | * face). Consider this, in particular: |
| 1403 | * |
| 1404 | * +--+ |
| 1405 | * | | |
| 1406 | * +--+--+ |
| 1407 | * | | | |
| 1408 | * +--+--+ |
| 1409 | * |
| 1410 | * The bottom left face there is adjacent to the exterior face |
| 1411 | * twice, so we can't just terminate our iteration when we reach |
| 1412 | * the same _face_ we started at. Furthermore, we can't |
| 1413 | * condition on having the same (i,j) pair either, because |
| 1414 | * several (i,j) pairs identify the bottom left contiguity with |
| 1415 | * the exterior face! We canonicalise the (i,j) pair by taking |
| 1416 | * one step around before we set the termination tracking. |
| 1417 | */ |
| 1418 | |
| 1419 | i = j = 0; |
| 1420 | current_face = test_face->dots[0]->faces[0]; |
| 1421 | if (current_face == test_face) { |
| 1422 | j = 1; |
| 1423 | current_face = test_face->dots[0]->faces[1]; |
| 1424 | } |
| 1425 | transitions = 0; |
| 1426 | current_state = (FACE_COLOUR(current_face) == colour); |
| 1427 | starting_dot = NULL; |
| 1428 | starting_face = NULL; |
| 1429 | while (TRUE) { |
| 1430 | /* Advance to next face. |
| 1431 | * Need to loop here because it might take several goes to |
| 1432 | * find it. */ |
| 1433 | while (TRUE) { |
| 1434 | j++; |
| 1435 | if (j == test_face->dots[i]->order) |
| 1436 | j = 0; |
| 1437 | |
| 1438 | if (test_face->dots[i]->faces[j] == test_face) { |
| 1439 | /* Advance to next dot round test_face, then |
| 1440 | * find current_face around new dot |
| 1441 | * and advance to the next face clockwise */ |
| 1442 | i++; |
| 1443 | if (i == test_face->order) |
| 1444 | i = 0; |
| 1445 | for (j = 0; j < test_face->dots[i]->order; j++) { |
| 1446 | if (test_face->dots[i]->faces[j] == current_face) |
| 1447 | break; |
| 1448 | } |
| 1449 | /* Must actually find current_face around new dot, |
| 1450 | * or else something's wrong with the grid. */ |
| 1451 | assert(j != test_face->dots[i]->order); |
| 1452 | /* Found, so advance to next face and try again */ |
| 1453 | } else { |
| 1454 | break; |
| 1455 | } |
| 1456 | } |
| 1457 | /* (i,j) are now advanced to next face */ |
| 1458 | current_face = test_face->dots[i]->faces[j]; |
| 1459 | s = (FACE_COLOUR(current_face) == colour); |
| 1460 | if (!starting_dot) { |
| 1461 | starting_dot = test_face->dots[i]; |
| 1462 | starting_face = current_face; |
| 1463 | current_state = s; |
| 1464 | } else { |
| 1465 | if (s != current_state) { |
| 1466 | ++transitions; |
| 1467 | current_state = s; |
| 1468 | if (transitions > 2) |
| 1469 | break; |
| 1470 | } |
| 1471 | if (test_face->dots[i] == starting_dot && |
| 1472 | current_face == starting_face) |
| 1473 | break; |
| 1474 | } |
| 1475 | } |
| 1476 | |
| 1477 | return (transitions == 2) ? TRUE : FALSE; |
| 1478 | } |
| 1479 | |
| 1480 | /* Count the number of neighbours of 'face', having colour 'colour' */ |
| 1481 | static int face_num_neighbours(grid *g, char *board, grid_face *face, |
| 1482 | enum face_colour colour) |
| 1483 | { |
| 1484 | int colour_count = 0; |
| 1485 | int i; |
| 1486 | grid_face *f; |
| 1487 | grid_edge *e; |
| 1488 | for (i = 0; i < face->order; i++) { |
| 1489 | e = face->edges[i]; |
| 1490 | f = (e->face1 == face) ? e->face2 : e->face1; |
| 1491 | if (FACE_COLOUR(f) == colour) |
| 1492 | ++colour_count; |
| 1493 | } |
| 1494 | return colour_count; |
| 1495 | } |
| 1496 | |
| 1497 | /* The 'score' of a face reflects its current desirability for selection |
| 1498 | * as the next face to colour white or black. We want to encourage moving |
| 1499 | * into grey areas and increasing loopiness, so we give scores according to |
| 1500 | * how many of the face's neighbours are currently coloured the same as the |
| 1501 | * proposed colour. */ |
| 1502 | static int face_score(grid *g, char *board, grid_face *face, |
| 1503 | enum face_colour colour) |
| 1504 | { |
| 1505 | /* Simple formula: score = 0 - num. same-coloured neighbours, |
| 1506 | * so a higher score means fewer same-coloured neighbours. */ |
| 1507 | return -face_num_neighbours(g, board, face, colour); |
| 1508 | } |
| 1509 | |
| 1510 | /* Generate a new complete set of clues for the given game_state. |
| 1511 | * The method is to generate a WHITE/BLACK colouring of all the faces, |
| 1512 | * such that the WHITE faces will define the inside of the path, and the |
| 1513 | * BLACK faces define the outside. |
| 1514 | * To do this, we initially colour all faces GREY. The infinite space outside |
| 1515 | * the grid is coloured BLACK, and we choose a random face to colour WHITE. |
| 1516 | * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY |
| 1517 | * faces, until the grid is filled with BLACK/WHITE. As we grow the regions, |
| 1518 | * we avoid creating loops of a single colour, to preserve the topological |
| 1519 | * shape of the WHITE and BLACK regions. |
| 1520 | * We also try to make the boundary as loopy and twisty as possible, to avoid |
| 1521 | * generating paths that are uninteresting. |
| 1522 | * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY |
| 1523 | * face that can be coloured with that colour (without violating the |
| 1524 | * topological shape of that region). It's not obvious, but I think this |
| 1525 | * algorithm is guaranteed to terminate without leaving any GREY faces behind. |
| 1526 | * Indeed, if there are any GREY faces at all, both the WHITE and BLACK |
| 1527 | * regions can be grown. |
| 1528 | * This is checked using assert()ions, and I haven't seen any failures yet. |
| 1529 | * |
| 1530 | * Hand-wavy proof: imagine what can go wrong... |
| 1531 | * |
| 1532 | * Could the white faces get completely cut off by the black faces, and still |
| 1533 | * leave some grey faces remaining? |
| 1534 | * No, because then the black faces would form a loop around both the white |
| 1535 | * faces and the grey faces, which is disallowed because we continually |
| 1536 | * maintain the correct topological shape of the black region. |
| 1537 | * Similarly, the black faces can never get cut off by the white faces. That |
| 1538 | * means both the WHITE and BLACK regions always have some room to grow into |
| 1539 | * the GREY regions. |
| 1540 | * Could it be that we can't colour some GREY face, because there are too many |
| 1541 | * WHITE/BLACK transitions as we walk round the face? (see the |
| 1542 | * can_colour_face() function for details) |
| 1543 | * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk |
| 1544 | * around the face. The two WHITE faces would be connected by a WHITE path, |
| 1545 | * and the BLACK faces would be connected by a BLACK path. These paths would |
| 1546 | * have to cross, which is impossible. |
| 1547 | * Another thing that could go wrong: perhaps we can't find any GREY face to |
| 1548 | * colour WHITE, because it would create a loop-violation or a corner-violation |
| 1549 | * with the other WHITE faces? |
| 1550 | * This is a little bit tricky to prove impossible. Imagine you have such a |
| 1551 | * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop |
| 1552 | * or corner violation). |
| 1553 | * That would cut all the non-white area into two blobs. One of those blobs |
| 1554 | * must be free of BLACK faces (because the BLACK stuff is a connected blob). |
| 1555 | * So we have a connected GREY area, completely surrounded by WHITE |
| 1556 | * (including the GREY face we've tentatively coloured WHITE). |
| 1557 | * A well-known result in graph theory says that you can always find a GREY |
| 1558 | * face whose removal leaves the remaining GREY area connected. And it says |
| 1559 | * there are at least two such faces, so we can always choose the one that |
| 1560 | * isn't the "tentative" GREY face. Colouring that face WHITE leaves |
| 1561 | * everything nice and connected, including that "tentative" GREY face which |
| 1562 | * acts as a gateway to the rest of the non-WHITE grid. |
| 1563 | */ |
| 1564 | static void add_full_clues(game_state *state, random_state *rs) |
| 1565 | { |
| 1566 | signed char *clues = state->clues; |
| 1567 | char *board; |
| 1568 | grid *g = state->game_grid; |
| 1569 | int i, j; |
| 1570 | int num_faces = g->num_faces; |
| 1571 | struct face_score *face_scores; /* Array of face_score objects */ |
| 1572 | struct face_score *fs; /* Points somewhere in the above list */ |
| 1573 | struct grid_face *cur_face; |
| 1574 | tree234 *lightable_faces_sorted; |
| 1575 | tree234 *darkable_faces_sorted; |
| 1576 | int *face_list; |
| 1577 | int do_random_pass; |
| 1578 | |
| 1579 | board = snewn(num_faces, char); |
| 1580 | |
| 1581 | /* Make a board */ |
| 1582 | memset(board, FACE_GREY, num_faces); |
| 1583 | |
| 1584 | /* Create and initialise the list of face_scores */ |
| 1585 | face_scores = snewn(num_faces, struct face_score); |
| 1586 | for (i = 0; i < num_faces; i++) { |
| 1587 | face_scores[i].random = random_bits(rs, 31); |
| 1588 | face_scores[i].black_score = face_scores[i].white_score = 0; |
| 1589 | } |
| 1590 | |
| 1591 | /* Colour a random, finite face white. The infinite face is implicitly |
| 1592 | * coloured black. Together, they will seed the random growth process |
| 1593 | * for the black and white areas. */ |
| 1594 | i = random_upto(rs, num_faces); |
| 1595 | board[i] = FACE_WHITE; |
| 1596 | |
| 1597 | /* We need a way of favouring faces that will increase our loopiness. |
| 1598 | * We do this by maintaining a list of all candidate faces sorted by |
| 1599 | * their score and choose randomly from that with appropriate skew. |
| 1600 | * In order to avoid consistently biasing towards particular faces, we |
| 1601 | * need the sort order _within_ each group of scores to be completely |
| 1602 | * random. But it would be abusing the hospitality of the tree234 data |
| 1603 | * structure if our comparison function were nondeterministic :-). So with |
| 1604 | * each face we associate a random number that does not change during a |
| 1605 | * particular run of the generator, and use that as a secondary sort key. |
| 1606 | * Yes, this means we will be biased towards particular random faces in |
| 1607 | * any one run but that doesn't actually matter. */ |
| 1608 | |
| 1609 | lightable_faces_sorted = newtree234(white_sort_cmpfn); |
| 1610 | darkable_faces_sorted = newtree234(black_sort_cmpfn); |
| 1611 | |
| 1612 | /* Initialise the lists of lightable and darkable faces. This is |
| 1613 | * slightly different from the code inside the while-loop, because we need |
| 1614 | * to check every face of the board (the grid structure does not keep a |
| 1615 | * list of the infinite face's neighbours). */ |
| 1616 | for (i = 0; i < num_faces; i++) { |
| 1617 | grid_face *f = g->faces + i; |
| 1618 | struct face_score *fs = face_scores + i; |
| 1619 | if (board[i] != FACE_GREY) continue; |
| 1620 | /* We need the full colourability check here, it's not enough simply |
| 1621 | * to check neighbourhood. On some grids, a neighbour of the infinite |
| 1622 | * face is not necessarily darkable. */ |
| 1623 | if (can_colour_face(g, board, i, FACE_BLACK)) { |
| 1624 | fs->black_score = face_score(g, board, f, FACE_BLACK); |
| 1625 | add234(darkable_faces_sorted, fs); |
| 1626 | } |
| 1627 | if (can_colour_face(g, board, i, FACE_WHITE)) { |
| 1628 | fs->white_score = face_score(g, board, f, FACE_WHITE); |
| 1629 | add234(lightable_faces_sorted, fs); |
| 1630 | } |
| 1631 | } |
| 1632 | |
| 1633 | /* Colour faces one at a time until no more faces are colourable. */ |
| 1634 | while (TRUE) |
| 1635 | { |
| 1636 | enum face_colour colour; |
| 1637 | struct face_score *fs_white, *fs_black; |
| 1638 | int c_lightable = count234(lightable_faces_sorted); |
| 1639 | int c_darkable = count234(darkable_faces_sorted); |
| 1640 | if (c_lightable == 0 && c_darkable == 0) { |
| 1641 | /* No more faces we can use at all. */ |
| 1642 | break; |
| 1643 | } |
| 1644 | assert(c_lightable != 0 && c_darkable != 0); |
| 1645 | |
| 1646 | fs_white = (struct face_score *)index234(lightable_faces_sorted, 0); |
| 1647 | fs_black = (struct face_score *)index234(darkable_faces_sorted, 0); |
| 1648 | |
| 1649 | /* Choose a colour, and colour the best available face |
| 1650 | * with that colour. */ |
| 1651 | colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK; |
| 1652 | |
| 1653 | if (colour == FACE_WHITE) |
| 1654 | fs = fs_white; |
| 1655 | else |
| 1656 | fs = fs_black; |
| 1657 | assert(fs); |
| 1658 | i = fs - face_scores; |
| 1659 | assert(board[i] == FACE_GREY); |
| 1660 | board[i] = colour; |
| 1661 | |
| 1662 | /* Remove this newly-coloured face from the lists. These lists should |
| 1663 | * only contain grey faces. */ |
| 1664 | del234(lightable_faces_sorted, fs); |
| 1665 | del234(darkable_faces_sorted, fs); |
| 1666 | |
| 1667 | /* Remember which face we've just coloured */ |
| 1668 | cur_face = g->faces + i; |
| 1669 | |
| 1670 | /* The face we've just coloured potentially affects the colourability |
| 1671 | * and the scores of any neighbouring faces (touching at a corner or |
| 1672 | * edge). So the search needs to be conducted around all faces |
| 1673 | * touching the one we've just lit. Iterate over its corners, then |
| 1674 | * over each corner's faces. For each such face, we remove it from |
| 1675 | * the lists, recalculate any scores, then add it back to the lists |
| 1676 | * (depending on whether it is lightable, darkable or both). */ |
| 1677 | for (i = 0; i < cur_face->order; i++) { |
| 1678 | grid_dot *d = cur_face->dots[i]; |
| 1679 | for (j = 0; j < d->order; j++) { |
| 1680 | grid_face *f = d->faces[j]; |
| 1681 | int fi; /* face index of f */ |
| 1682 | |
| 1683 | if (f == NULL) |
| 1684 | continue; |
| 1685 | if (f == cur_face) |
| 1686 | continue; |
| 1687 | |
| 1688 | /* If the face is already coloured, it won't be on our |
| 1689 | * lightable/darkable lists anyway, so we can skip it without |
| 1690 | * bothering with the removal step. */ |
| 1691 | if (FACE_COLOUR(f) != FACE_GREY) continue; |
| 1692 | |
| 1693 | /* Find the face index and face_score* corresponding to f */ |
| 1694 | fi = f - g->faces; |
| 1695 | fs = face_scores + fi; |
| 1696 | |
| 1697 | /* Remove from lightable list if it's in there. We do this, |
| 1698 | * even if it is still lightable, because the score might |
| 1699 | * be different, and we need to remove-then-add to maintain |
| 1700 | * correct sort order. */ |
| 1701 | del234(lightable_faces_sorted, fs); |
| 1702 | if (can_colour_face(g, board, fi, FACE_WHITE)) { |
| 1703 | fs->white_score = face_score(g, board, f, FACE_WHITE); |
| 1704 | add234(lightable_faces_sorted, fs); |
| 1705 | } |
| 1706 | /* Do the same for darkable list. */ |
| 1707 | del234(darkable_faces_sorted, fs); |
| 1708 | if (can_colour_face(g, board, fi, FACE_BLACK)) { |
| 1709 | fs->black_score = face_score(g, board, f, FACE_BLACK); |
| 1710 | add234(darkable_faces_sorted, fs); |
| 1711 | } |
| 1712 | } |
| 1713 | } |
| 1714 | } |
| 1715 | |
| 1716 | /* Clean up */ |
| 1717 | freetree234(lightable_faces_sorted); |
| 1718 | freetree234(darkable_faces_sorted); |
| 1719 | sfree(face_scores); |
| 1720 | |
| 1721 | /* The next step requires a shuffled list of all faces */ |
| 1722 | face_list = snewn(num_faces, int); |
| 1723 | for (i = 0; i < num_faces; ++i) { |
| 1724 | face_list[i] = i; |
| 1725 | } |
| 1726 | shuffle(face_list, num_faces, sizeof(int), rs); |
| 1727 | |
| 1728 | /* The above loop-generation algorithm can often leave large clumps |
| 1729 | * of faces of one colour. In extreme cases, the resulting path can be |
| 1730 | * degenerate and not very satisfying to solve. |
| 1731 | * This next step alleviates this problem: |
| 1732 | * Go through the shuffled list, and flip the colour of any face we can |
| 1733 | * legally flip, and which is adjacent to only one face of the opposite |
| 1734 | * colour - this tends to grow 'tendrils' into any clumps. |
| 1735 | * Repeat until we can find no more faces to flip. This will |
| 1736 | * eventually terminate, because each flip increases the loop's |
| 1737 | * perimeter, which cannot increase for ever. |
| 1738 | * The resulting path will have maximal loopiness (in the sense that it |
| 1739 | * cannot be improved "locally". Unfortunately, this allows a player to |
| 1740 | * make some illicit deductions. To combat this (and make the path more |
| 1741 | * interesting), we do one final pass making random flips. */ |
| 1742 | |
| 1743 | /* Set to TRUE for final pass */ |
| 1744 | do_random_pass = FALSE; |
| 1745 | |
| 1746 | while (TRUE) { |
| 1747 | /* Remember whether a flip occurred during this pass */ |
| 1748 | int flipped = FALSE; |
| 1749 | |
| 1750 | for (i = 0; i < num_faces; ++i) { |
| 1751 | int j = face_list[i]; |
| 1752 | enum face_colour opp = |
| 1753 | (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE; |
| 1754 | if (can_colour_face(g, board, j, opp)) { |
| 1755 | grid_face *face = g->faces +j; |
| 1756 | if (do_random_pass) { |
| 1757 | /* final random pass */ |
| 1758 | if (!random_upto(rs, 10)) |
| 1759 | board[j] = opp; |
| 1760 | } else { |
| 1761 | /* normal pass - flip when neighbour count is 1 */ |
| 1762 | if (face_num_neighbours(g, board, face, opp) == 1) { |
| 1763 | board[j] = opp; |
| 1764 | flipped = TRUE; |
| 1765 | } |
| 1766 | } |
| 1767 | } |
| 1768 | } |
| 1769 | |
| 1770 | if (do_random_pass) break; |
| 1771 | if (!flipped) do_random_pass = TRUE; |
| 1772 | } |
| 1773 | |
| 1774 | sfree(face_list); |
| 1775 | |
| 1776 | /* Fill out all the clues by initialising to 0, then iterating over |
| 1777 | * all edges and incrementing each clue as we find edges that border |
| 1778 | * between BLACK/WHITE faces. While we're at it, we verify that the |
| 1779 | * algorithm does work, and there aren't any GREY faces still there. */ |
| 1780 | memset(clues, 0, num_faces); |
| 1781 | for (i = 0; i < g->num_edges; i++) { |
| 1782 | grid_edge *e = g->edges + i; |
| 1783 | grid_face *f1 = e->face1; |
| 1784 | grid_face *f2 = e->face2; |
| 1785 | enum face_colour c1 = FACE_COLOUR(f1); |
| 1786 | enum face_colour c2 = FACE_COLOUR(f2); |
| 1787 | assert(c1 != FACE_GREY); |
| 1788 | assert(c2 != FACE_GREY); |
| 1789 | if (c1 != c2) { |
| 1790 | if (f1) clues[f1 - g->faces]++; |
| 1791 | if (f2) clues[f2 - g->faces]++; |
| 1792 | } |
| 1793 | } |
| 1794 | |
| 1795 | sfree(board); |
| 1796 | } |
| 1797 | |
| 1798 | |
| 1799 | static int game_has_unique_soln(const game_state *state, int diff) |
| 1800 | { |
| 1801 | int ret; |
| 1802 | solver_state *sstate_new; |
| 1803 | solver_state *sstate = new_solver_state((game_state *)state, diff); |
| 1804 | |
| 1805 | sstate_new = solve_game_rec(sstate); |
| 1806 | |
| 1807 | assert(sstate_new->solver_status != SOLVER_MISTAKE); |
| 1808 | ret = (sstate_new->solver_status == SOLVER_SOLVED); |
| 1809 | |
| 1810 | free_solver_state(sstate_new); |
| 1811 | free_solver_state(sstate); |
| 1812 | |
| 1813 | return ret; |
| 1814 | } |
| 1815 | |
| 1816 | |
| 1817 | /* Remove clues one at a time at random. */ |
| 1818 | static game_state *remove_clues(game_state *state, random_state *rs, |
| 1819 | int diff) |
| 1820 | { |
| 1821 | int *face_list; |
| 1822 | int num_faces = state->game_grid->num_faces; |
| 1823 | game_state *ret = dup_game(state), *saved_ret; |
| 1824 | int n; |
| 1825 | |
| 1826 | /* We need to remove some clues. We'll do this by forming a list of all |
| 1827 | * available clues, shuffling it, then going along one at a |
| 1828 | * time clearing each clue in turn for which doing so doesn't render the |
| 1829 | * board unsolvable. */ |
| 1830 | face_list = snewn(num_faces, int); |
| 1831 | for (n = 0; n < num_faces; ++n) { |
| 1832 | face_list[n] = n; |
| 1833 | } |
| 1834 | |
| 1835 | shuffle(face_list, num_faces, sizeof(int), rs); |
| 1836 | |
| 1837 | for (n = 0; n < num_faces; ++n) { |
| 1838 | saved_ret = dup_game(ret); |
| 1839 | ret->clues[face_list[n]] = -1; |
| 1840 | |
| 1841 | if (game_has_unique_soln(ret, diff)) { |
| 1842 | free_game(saved_ret); |
| 1843 | } else { |
| 1844 | free_game(ret); |
| 1845 | ret = saved_ret; |
| 1846 | } |
| 1847 | } |
| 1848 | sfree(face_list); |
| 1849 | |
| 1850 | return ret; |
| 1851 | } |
| 1852 | |
| 1853 | |
| 1854 | static char *new_game_desc(game_params *params, random_state *rs, |
| 1855 | char **aux, int interactive) |
| 1856 | { |
| 1857 | /* solution and description both use run-length encoding in obvious ways */ |
| 1858 | char *retval, *game_desc, *grid_desc; |
| 1859 | grid *g; |
| 1860 | game_state *state = snew(game_state); |
| 1861 | game_state *state_new; |
| 1862 | |
| 1863 | grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs); |
| 1864 | state->game_grid = g = loopy_generate_grid(params, grid_desc); |
| 1865 | |
| 1866 | state->clues = snewn(g->num_faces, signed char); |
| 1867 | state->lines = snewn(g->num_edges, char); |
| 1868 | state->line_errors = snewn(g->num_edges, unsigned char); |
| 1869 | |
| 1870 | state->grid_type = params->type; |
| 1871 | |
| 1872 | newboard_please: |
| 1873 | |
| 1874 | memset(state->lines, LINE_UNKNOWN, g->num_edges); |
| 1875 | memset(state->line_errors, 0, g->num_edges); |
| 1876 | |
| 1877 | state->solved = state->cheated = FALSE; |
| 1878 | |
| 1879 | /* Get a new random solvable board with all its clues filled in. Yes, this |
| 1880 | * can loop for ever if the params are suitably unfavourable, but |
| 1881 | * preventing games smaller than 4x4 seems to stop this happening */ |
| 1882 | do { |
| 1883 | add_full_clues(state, rs); |
| 1884 | } while (!game_has_unique_soln(state, params->diff)); |
| 1885 | |
| 1886 | state_new = remove_clues(state, rs, params->diff); |
| 1887 | free_game(state); |
| 1888 | state = state_new; |
| 1889 | |
| 1890 | |
| 1891 | if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) { |
| 1892 | #ifdef SHOW_WORKING |
| 1893 | fprintf(stderr, "Rejecting board, it is too easy\n"); |
| 1894 | #endif |
| 1895 | goto newboard_please; |
| 1896 | } |
| 1897 | |
| 1898 | game_desc = state_to_text(state); |
| 1899 | |
| 1900 | free_game(state); |
| 1901 | |
| 1902 | if (grid_desc) { |
| 1903 | retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char); |
| 1904 | sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc); |
| 1905 | sfree(grid_desc); |
| 1906 | sfree(game_desc); |
| 1907 | } else { |
| 1908 | retval = game_desc; |
| 1909 | } |
| 1910 | |
| 1911 | assert(!validate_desc(params, retval)); |
| 1912 | |
| 1913 | return retval; |
| 1914 | } |
| 1915 | |
| 1916 | static game_state *new_game(midend *me, game_params *params, char *desc) |
| 1917 | { |
| 1918 | int i; |
| 1919 | game_state *state = snew(game_state); |
| 1920 | int empties_to_make = 0; |
| 1921 | int n,n2; |
| 1922 | const char *dp; |
| 1923 | char *grid_desc; |
| 1924 | grid *g; |
| 1925 | int num_faces, num_edges; |
| 1926 | |
| 1927 | grid_desc = extract_grid_desc(&desc); |
| 1928 | state->game_grid = g = loopy_generate_grid(params, grid_desc); |
| 1929 | if (grid_desc) sfree(grid_desc); |
| 1930 | |
| 1931 | dp = desc; |
| 1932 | |
| 1933 | num_faces = g->num_faces; |
| 1934 | num_edges = g->num_edges; |
| 1935 | |
| 1936 | state->clues = snewn(num_faces, signed char); |
| 1937 | state->lines = snewn(num_edges, char); |
| 1938 | state->line_errors = snewn(num_edges, unsigned char); |
| 1939 | |
| 1940 | state->solved = state->cheated = FALSE; |
| 1941 | |
| 1942 | state->grid_type = params->type; |
| 1943 | |
| 1944 | for (i = 0; i < num_faces; i++) { |
| 1945 | if (empties_to_make) { |
| 1946 | empties_to_make--; |
| 1947 | state->clues[i] = -1; |
| 1948 | continue; |
| 1949 | } |
| 1950 | |
| 1951 | assert(*dp); |
| 1952 | n = *dp - '0'; |
| 1953 | n2 = *dp - 'A' + 10; |
| 1954 | if (n >= 0 && n < 10) { |
| 1955 | state->clues[i] = n; |
| 1956 | } else if (n2 >= 10 && n2 < 36) { |
| 1957 | state->clues[i] = n2; |
| 1958 | } else { |
| 1959 | n = *dp - 'a' + 1; |
| 1960 | assert(n > 0); |
| 1961 | state->clues[i] = -1; |
| 1962 | empties_to_make = n - 1; |
| 1963 | } |
| 1964 | ++dp; |
| 1965 | } |
| 1966 | |
| 1967 | memset(state->lines, LINE_UNKNOWN, num_edges); |
| 1968 | memset(state->line_errors, 0, num_edges); |
| 1969 | return state; |
| 1970 | } |
| 1971 | |
| 1972 | /* Calculates the line_errors data, and checks if the current state is a |
| 1973 | * solution */ |
| 1974 | static int check_completion(game_state *state) |
| 1975 | { |
| 1976 | grid *g = state->game_grid; |
| 1977 | int *dsf; |
| 1978 | int num_faces = g->num_faces; |
| 1979 | int i; |
| 1980 | int infinite_area, finite_area; |
| 1981 | int loops_found = 0; |
| 1982 | int found_edge_not_in_loop = FALSE; |
| 1983 | |
| 1984 | memset(state->line_errors, 0, g->num_edges); |
| 1985 | |
| 1986 | /* LL implementation of SGT's idea: |
| 1987 | * A loop will partition the grid into an inside and an outside. |
| 1988 | * If there is more than one loop, the grid will be partitioned into |
| 1989 | * even more distinct regions. We can therefore track equivalence of |
| 1990 | * faces, by saying that two faces are equivalent when there is a non-YES |
| 1991 | * edge between them. |
| 1992 | * We could keep track of the number of connected components, by counting |
| 1993 | * the number of dsf-merges that aren't no-ops. |
| 1994 | * But we're only interested in 3 separate cases: |
| 1995 | * no loops, one loop, more than one loop. |
| 1996 | * |
| 1997 | * No loops: all faces are equivalent to the infinite face. |
| 1998 | * One loop: only two equivalence classes - finite and infinite. |
| 1999 | * >= 2 loops: there are 2 distinct finite regions. |
| 2000 | * |
| 2001 | * So we simply make two passes through all the edges. |
| 2002 | * In the first pass, we dsf-merge the two faces bordering each non-YES |
| 2003 | * edge. |
| 2004 | * In the second pass, we look for YES-edges bordering: |
| 2005 | * a) two non-equivalent faces. |
| 2006 | * b) two non-equivalent faces, and one of them is part of a different |
| 2007 | * finite area from the first finite area we've seen. |
| 2008 | * |
| 2009 | * An occurrence of a) means there is at least one loop. |
| 2010 | * An occurrence of b) means there is more than one loop. |
| 2011 | * Edges satisfying a) are marked as errors. |
| 2012 | * |
| 2013 | * While we're at it, we set a flag if we find a YES edge that is not |
| 2014 | * part of a loop. |
| 2015 | * This information will help decide, if there's a single loop, whether it |
| 2016 | * is a candidate for being a solution (that is, all YES edges are part of |
| 2017 | * this loop). |
| 2018 | * |
| 2019 | * If there is a candidate loop, we then go through all clues and check |
| 2020 | * they are all satisfied. If so, we have found a solution and we can |
| 2021 | * unmark all line_errors. |
| 2022 | */ |
| 2023 | |
| 2024 | /* Infinite face is at the end - its index is num_faces. |
| 2025 | * This macro is just to make this obvious! */ |
| 2026 | #define INF_FACE num_faces |
| 2027 | dsf = snewn(num_faces + 1, int); |
| 2028 | dsf_init(dsf, num_faces + 1); |
| 2029 | |
| 2030 | /* First pass */ |
| 2031 | for (i = 0; i < g->num_edges; i++) { |
| 2032 | grid_edge *e = g->edges + i; |
| 2033 | int f1 = e->face1 ? e->face1 - g->faces : INF_FACE; |
| 2034 | int f2 = e->face2 ? e->face2 - g->faces : INF_FACE; |
| 2035 | if (state->lines[i] != LINE_YES) |
| 2036 | dsf_merge(dsf, f1, f2); |
| 2037 | } |
| 2038 | |
| 2039 | /* Second pass */ |
| 2040 | infinite_area = dsf_canonify(dsf, INF_FACE); |
| 2041 | finite_area = -1; |
| 2042 | for (i = 0; i < g->num_edges; i++) { |
| 2043 | grid_edge *e = g->edges + i; |
| 2044 | int f1 = e->face1 ? e->face1 - g->faces : INF_FACE; |
| 2045 | int can1 = dsf_canonify(dsf, f1); |
| 2046 | int f2 = e->face2 ? e->face2 - g->faces : INF_FACE; |
| 2047 | int can2 = dsf_canonify(dsf, f2); |
| 2048 | if (state->lines[i] != LINE_YES) continue; |
| 2049 | |
| 2050 | if (can1 == can2) { |
| 2051 | /* Faces are equivalent, so this edge not part of a loop */ |
| 2052 | found_edge_not_in_loop = TRUE; |
| 2053 | continue; |
| 2054 | } |
| 2055 | state->line_errors[i] = TRUE; |
| 2056 | if (loops_found == 0) loops_found = 1; |
| 2057 | |
| 2058 | /* Don't bother with further checks if we've already found 2 loops */ |
| 2059 | if (loops_found == 2) continue; |
| 2060 | |
| 2061 | if (finite_area == -1) { |
| 2062 | /* Found our first finite area */ |
| 2063 | if (can1 != infinite_area) |
| 2064 | finite_area = can1; |
| 2065 | else |
| 2066 | finite_area = can2; |
| 2067 | } |
| 2068 | |
| 2069 | /* Have we found a second area? */ |
| 2070 | if (finite_area != -1) { |
| 2071 | if (can1 != infinite_area && can1 != finite_area) { |
| 2072 | loops_found = 2; |
| 2073 | continue; |
| 2074 | } |
| 2075 | if (can2 != infinite_area && can2 != finite_area) { |
| 2076 | loops_found = 2; |
| 2077 | } |
| 2078 | } |
| 2079 | } |
| 2080 | |
| 2081 | /* |
| 2082 | printf("loops_found = %d\n", loops_found); |
| 2083 | printf("found_edge_not_in_loop = %s\n", |
| 2084 | found_edge_not_in_loop ? "TRUE" : "FALSE"); |
| 2085 | */ |
| 2086 | |
| 2087 | sfree(dsf); /* No longer need the dsf */ |
| 2088 | |
| 2089 | /* Have we found a candidate loop? */ |
| 2090 | if (loops_found == 1 && !found_edge_not_in_loop) { |
| 2091 | /* Yes, so check all clues are satisfied */ |
| 2092 | int found_clue_violation = FALSE; |
| 2093 | for (i = 0; i < num_faces; i++) { |
| 2094 | int c = state->clues[i]; |
| 2095 | if (c >= 0) { |
| 2096 | if (face_order(state, i, LINE_YES) != c) { |
| 2097 | found_clue_violation = TRUE; |
| 2098 | break; |
| 2099 | } |
| 2100 | } |
| 2101 | } |
| 2102 | |
| 2103 | if (!found_clue_violation) { |
| 2104 | /* The loop is good */ |
| 2105 | memset(state->line_errors, 0, g->num_edges); |
| 2106 | return TRUE; /* No need to bother checking for dot violations */ |
| 2107 | } |
| 2108 | } |
| 2109 | |
| 2110 | /* Check for dot violations */ |
| 2111 | for (i = 0; i < g->num_dots; i++) { |
| 2112 | int yes = dot_order(state, i, LINE_YES); |
| 2113 | int unknown = dot_order(state, i, LINE_UNKNOWN); |
| 2114 | if ((yes == 1 && unknown == 0) || (yes >= 3)) { |
| 2115 | /* violation, so mark all YES edges as errors */ |
| 2116 | grid_dot *d = g->dots + i; |
| 2117 | int j; |
| 2118 | for (j = 0; j < d->order; j++) { |
| 2119 | int e = d->edges[j] - g->edges; |
| 2120 | if (state->lines[e] == LINE_YES) |
| 2121 | state->line_errors[e] = TRUE; |
| 2122 | } |
| 2123 | } |
| 2124 | } |
| 2125 | return FALSE; |
| 2126 | } |
| 2127 | |
| 2128 | /* ---------------------------------------------------------------------- |
| 2129 | * Solver logic |
| 2130 | * |
| 2131 | * Our solver modes operate as follows. Each mode also uses the modes above it. |
| 2132 | * |
| 2133 | * Easy Mode |
| 2134 | * Just implement the rules of the game. |
| 2135 | * |
| 2136 | * Normal and Tricky Modes |
| 2137 | * For each (adjacent) pair of lines through each dot we store a bit for |
| 2138 | * whether at least one of them is on and whether at most one is on. (If we |
| 2139 | * know both or neither is on that's already stored more directly.) |
| 2140 | * |
| 2141 | * Advanced Mode |
| 2142 | * Use edsf data structure to make equivalence classes of lines that are |
| 2143 | * known identical to or opposite to one another. |
| 2144 | */ |
| 2145 | |
| 2146 | |
| 2147 | /* DLines: |
| 2148 | * For general grids, we consider "dlines" to be pairs of lines joined |
| 2149 | * at a dot. The lines must be adjacent around the dot, so we can think of |
| 2150 | * a dline as being a dot+face combination. Or, a dot+edge combination where |
| 2151 | * the second edge is taken to be the next clockwise edge from the dot. |
| 2152 | * Original loopy code didn't have this extra restriction of the lines being |
| 2153 | * adjacent. From my tests with square grids, this extra restriction seems to |
| 2154 | * take little, if anything, away from the quality of the puzzles. |
| 2155 | * A dline can be uniquely identified by an edge/dot combination, given that |
| 2156 | * a dline-pair always goes clockwise around its common dot. The edge/dot |
| 2157 | * combination can be represented by an edge/bool combination - if bool is |
| 2158 | * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is |
| 2159 | * exactly twice the number of edges in the grid - although the dlines |
| 2160 | * spanning the infinite face are not all that useful to the solver. |
| 2161 | * Note that, by convention, a dline goes clockwise around its common dot, |
| 2162 | * which means the dline goes anti-clockwise around its common face. |
| 2163 | */ |
| 2164 | |
| 2165 | /* Helper functions for obtaining an index into an array of dlines, given |
| 2166 | * various information. We assume the grid layout conventions about how |
| 2167 | * the various lists are interleaved - see grid_make_consistent() for |
| 2168 | * details. */ |
| 2169 | |
| 2170 | /* i points to the first edge of the dline pair, reading clockwise around |
| 2171 | * the dot. */ |
| 2172 | static int dline_index_from_dot(grid *g, grid_dot *d, int i) |
| 2173 | { |
| 2174 | grid_edge *e = d->edges[i]; |
| 2175 | int ret; |
| 2176 | #ifdef DEBUG_DLINES |
| 2177 | grid_edge *e2; |
| 2178 | int i2 = i+1; |
| 2179 | if (i2 == d->order) i2 = 0; |
| 2180 | e2 = d->edges[i2]; |
| 2181 | #endif |
| 2182 | ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0); |
| 2183 | #ifdef DEBUG_DLINES |
| 2184 | printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n", |
| 2185 | (int)(d - g->dots), i, (int)(e - g->edges), |
| 2186 | (int)(e2 - g->edges), ret); |
| 2187 | #endif |
| 2188 | return ret; |
| 2189 | } |
| 2190 | /* i points to the second edge of the dline pair, reading clockwise around |
| 2191 | * the face. That is, the edges of the dline, starting at edge{i}, read |
| 2192 | * anti-clockwise around the face. By layout conventions, the common dot |
| 2193 | * of the dline will be f->dots[i] */ |
| 2194 | static int dline_index_from_face(grid *g, grid_face *f, int i) |
| 2195 | { |
| 2196 | grid_edge *e = f->edges[i]; |
| 2197 | grid_dot *d = f->dots[i]; |
| 2198 | int ret; |
| 2199 | #ifdef DEBUG_DLINES |
| 2200 | grid_edge *e2; |
| 2201 | int i2 = i - 1; |
| 2202 | if (i2 < 0) i2 += f->order; |
| 2203 | e2 = f->edges[i2]; |
| 2204 | #endif |
| 2205 | ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0); |
| 2206 | #ifdef DEBUG_DLINES |
| 2207 | printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n", |
| 2208 | (int)(f - g->faces), i, (int)(e - g->edges), |
| 2209 | (int)(e2 - g->edges), ret); |
| 2210 | #endif |
| 2211 | return ret; |
| 2212 | } |
| 2213 | static int is_atleastone(const char *dline_array, int index) |
| 2214 | { |
| 2215 | return BIT_SET(dline_array[index], 0); |
| 2216 | } |
| 2217 | static int set_atleastone(char *dline_array, int index) |
| 2218 | { |
| 2219 | return SET_BIT(dline_array[index], 0); |
| 2220 | } |
| 2221 | static int is_atmostone(const char *dline_array, int index) |
| 2222 | { |
| 2223 | return BIT_SET(dline_array[index], 1); |
| 2224 | } |
| 2225 | static int set_atmostone(char *dline_array, int index) |
| 2226 | { |
| 2227 | return SET_BIT(dline_array[index], 1); |
| 2228 | } |
| 2229 | |
| 2230 | static void array_setall(char *array, char from, char to, int len) |
| 2231 | { |
| 2232 | char *p = array, *p_old = p; |
| 2233 | int len_remaining = len; |
| 2234 | |
| 2235 | while ((p = memchr(p, from, len_remaining))) { |
| 2236 | *p = to; |
| 2237 | len_remaining -= p - p_old; |
| 2238 | p_old = p; |
| 2239 | } |
| 2240 | } |
| 2241 | |
| 2242 | /* Helper, called when doing dline dot deductions, in the case where we |
| 2243 | * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between |
| 2244 | * them (because of dline atmostone/atleastone). |
| 2245 | * On entry, edge points to the first of these two UNKNOWNs. This function |
| 2246 | * will find the opposite UNKNOWNS (if they are adjacent to one another) |
| 2247 | * and set their corresponding dline to atleastone. (Setting atmostone |
| 2248 | * already happens in earlier dline deductions) */ |
| 2249 | static int dline_set_opp_atleastone(solver_state *sstate, |
| 2250 | grid_dot *d, int edge) |
| 2251 | { |
| 2252 | game_state *state = sstate->state; |
| 2253 | grid *g = state->game_grid; |
| 2254 | int N = d->order; |
| 2255 | int opp, opp2; |
| 2256 | for (opp = 0; opp < N; opp++) { |
| 2257 | int opp_dline_index; |
| 2258 | if (opp == edge || opp == edge+1 || opp == edge-1) |
| 2259 | continue; |
| 2260 | if (opp == 0 && edge == N-1) |
| 2261 | continue; |
| 2262 | if (opp == N-1 && edge == 0) |
| 2263 | continue; |
| 2264 | opp2 = opp + 1; |
| 2265 | if (opp2 == N) opp2 = 0; |
| 2266 | /* Check if opp, opp2 point to LINE_UNKNOWNs */ |
| 2267 | if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN) |
| 2268 | continue; |
| 2269 | if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN) |
| 2270 | continue; |
| 2271 | /* Found opposite UNKNOWNS and they're next to each other */ |
| 2272 | opp_dline_index = dline_index_from_dot(g, d, opp); |
| 2273 | return set_atleastone(sstate->dlines, opp_dline_index); |
| 2274 | } |
| 2275 | return FALSE; |
| 2276 | } |
| 2277 | |
| 2278 | |
| 2279 | /* Set pairs of lines around this face which are known to be identical, to |
| 2280 | * the given line_state */ |
| 2281 | static int face_setall_identical(solver_state *sstate, int face_index, |
| 2282 | enum line_state line_new) |
| 2283 | { |
| 2284 | /* can[dir] contains the canonical line associated with the line in |
| 2285 | * direction dir from the square in question. Similarly inv[dir] is |
| 2286 | * whether or not the line in question is inverse to its canonical |
| 2287 | * element. */ |
| 2288 | int retval = FALSE; |
| 2289 | game_state *state = sstate->state; |
| 2290 | grid *g = state->game_grid; |
| 2291 | grid_face *f = g->faces + face_index; |
| 2292 | int N = f->order; |
| 2293 | int i, j; |
| 2294 | int can1, can2, inv1, inv2; |
| 2295 | |
| 2296 | for (i = 0; i < N; i++) { |
| 2297 | int line1_index = f->edges[i] - g->edges; |
| 2298 | if (state->lines[line1_index] != LINE_UNKNOWN) |
| 2299 | continue; |
| 2300 | for (j = i + 1; j < N; j++) { |
| 2301 | int line2_index = f->edges[j] - g->edges; |
| 2302 | if (state->lines[line2_index] != LINE_UNKNOWN) |
| 2303 | continue; |
| 2304 | |
| 2305 | /* Found two UNKNOWNS */ |
| 2306 | can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1); |
| 2307 | can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2); |
| 2308 | if (can1 == can2 && inv1 == inv2) { |
| 2309 | solver_set_line(sstate, line1_index, line_new); |
| 2310 | solver_set_line(sstate, line2_index, line_new); |
| 2311 | } |
| 2312 | } |
| 2313 | } |
| 2314 | return retval; |
| 2315 | } |
| 2316 | |
| 2317 | /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and |
| 2318 | * return the edge indices into e. */ |
| 2319 | static void find_unknowns(game_state *state, |
| 2320 | grid_edge **edge_list, /* Edge list to search (from a face or a dot) */ |
| 2321 | int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */ |
| 2322 | int *e /* Returned edge indices */) |
| 2323 | { |
| 2324 | int c = 0; |
| 2325 | grid *g = state->game_grid; |
| 2326 | while (c < expected_count) { |
| 2327 | int line_index = *edge_list - g->edges; |
| 2328 | if (state->lines[line_index] == LINE_UNKNOWN) { |
| 2329 | e[c] = line_index; |
| 2330 | c++; |
| 2331 | } |
| 2332 | ++edge_list; |
| 2333 | } |
| 2334 | } |
| 2335 | |
| 2336 | /* If we have a list of edges, and we know whether the number of YESs should |
| 2337 | * be odd or even, and there are only a few UNKNOWNs, we can do some simple |
| 2338 | * linedsf deductions. This can be used for both face and dot deductions. |
| 2339 | * Returns the difficulty level of the next solver that should be used, |
| 2340 | * or DIFF_MAX if no progress was made. */ |
| 2341 | static int parity_deductions(solver_state *sstate, |
| 2342 | grid_edge **edge_list, /* Edge list (from a face or a dot) */ |
| 2343 | int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */ |
| 2344 | int unknown_count) |
| 2345 | { |
| 2346 | game_state *state = sstate->state; |
| 2347 | int diff = DIFF_MAX; |
| 2348 | int *linedsf = sstate->linedsf; |
| 2349 | |
| 2350 | if (unknown_count == 2) { |
| 2351 | /* Lines are known alike/opposite, depending on inv. */ |
| 2352 | int e[2]; |
| 2353 | find_unknowns(state, edge_list, 2, e); |
| 2354 | if (merge_lines(sstate, e[0], e[1], total_parity)) |
| 2355 | diff = min(diff, DIFF_HARD); |
| 2356 | } else if (unknown_count == 3) { |
| 2357 | int e[3]; |
| 2358 | int can[3]; /* canonical edges */ |
| 2359 | int inv[3]; /* whether can[x] is inverse to e[x] */ |
| 2360 | find_unknowns(state, edge_list, 3, e); |
| 2361 | can[0] = edsf_canonify(linedsf, e[0], inv); |
| 2362 | can[1] = edsf_canonify(linedsf, e[1], inv+1); |
| 2363 | can[2] = edsf_canonify(linedsf, e[2], inv+2); |
| 2364 | if (can[0] == can[1]) { |
| 2365 | if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ? |
| 2366 | LINE_YES : LINE_NO)) |
| 2367 | diff = min(diff, DIFF_EASY); |
| 2368 | } |
| 2369 | if (can[0] == can[2]) { |
| 2370 | if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ? |
| 2371 | LINE_YES : LINE_NO)) |
| 2372 | diff = min(diff, DIFF_EASY); |
| 2373 | } |
| 2374 | if (can[1] == can[2]) { |
| 2375 | if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ? |
| 2376 | LINE_YES : LINE_NO)) |
| 2377 | diff = min(diff, DIFF_EASY); |
| 2378 | } |
| 2379 | } else if (unknown_count == 4) { |
| 2380 | int e[4]; |
| 2381 | int can[4]; /* canonical edges */ |
| 2382 | int inv[4]; /* whether can[x] is inverse to e[x] */ |
| 2383 | find_unknowns(state, edge_list, 4, e); |
| 2384 | can[0] = edsf_canonify(linedsf, e[0], inv); |
| 2385 | can[1] = edsf_canonify(linedsf, e[1], inv+1); |
| 2386 | can[2] = edsf_canonify(linedsf, e[2], inv+2); |
| 2387 | can[3] = edsf_canonify(linedsf, e[3], inv+3); |
| 2388 | if (can[0] == can[1]) { |
| 2389 | if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1])) |
| 2390 | diff = min(diff, DIFF_HARD); |
| 2391 | } else if (can[0] == can[2]) { |
| 2392 | if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2])) |
| 2393 | diff = min(diff, DIFF_HARD); |
| 2394 | } else if (can[0] == can[3]) { |
| 2395 | if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3])) |
| 2396 | diff = min(diff, DIFF_HARD); |
| 2397 | } else if (can[1] == can[2]) { |
| 2398 | if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2])) |
| 2399 | diff = min(diff, DIFF_HARD); |
| 2400 | } else if (can[1] == can[3]) { |
| 2401 | if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3])) |
| 2402 | diff = min(diff, DIFF_HARD); |
| 2403 | } else if (can[2] == can[3]) { |
| 2404 | if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3])) |
| 2405 | diff = min(diff, DIFF_HARD); |
| 2406 | } |
| 2407 | } |
| 2408 | return diff; |
| 2409 | } |
| 2410 | |
| 2411 | |
| 2412 | /* |
| 2413 | * These are the main solver functions. |
| 2414 | * |
| 2415 | * Their return values are diff values corresponding to the lowest mode solver |
| 2416 | * that would notice the work that they have done. For example if the normal |
| 2417 | * mode solver adds actual lines or crosses, it will return DIFF_EASY as the |
| 2418 | * easy mode solver might be able to make progress using that. It doesn't make |
| 2419 | * sense for one of them to return a diff value higher than that of the |
| 2420 | * function itself. |
| 2421 | * |
| 2422 | * Each function returns the lowest value it can, as early as possible, in |
| 2423 | * order to try and pass as much work as possible back to the lower level |
| 2424 | * solvers which progress more quickly. |
| 2425 | */ |
| 2426 | |
| 2427 | /* PROPOSED NEW DESIGN: |
| 2428 | * We have a work queue consisting of 'events' notifying us that something has |
| 2429 | * happened that a particular solver mode might be interested in. For example |
| 2430 | * the hard mode solver might do something that helps the normal mode solver at |
| 2431 | * dot [x,y] in which case it will enqueue an event recording this fact. Then |
| 2432 | * we pull events off the work queue, and hand each in turn to the solver that |
| 2433 | * is interested in them. If a solver reports that it failed we pass the same |
| 2434 | * event on to progressively more advanced solvers and the loop detector. Once |
| 2435 | * we've exhausted an event, or it has helped us progress, we drop it and |
| 2436 | * continue to the next one. The events are sorted first in order of solver |
| 2437 | * complexity (easy first) then order of insertion (oldest first). |
| 2438 | * Once we run out of events we loop over each permitted solver in turn |
| 2439 | * (easiest first) until either a deduction is made (and an event therefore |
| 2440 | * emerges) or no further deductions can be made (in which case we've failed). |
| 2441 | * |
| 2442 | * QUESTIONS: |
| 2443 | * * How do we 'loop over' a solver when both dots and squares are concerned. |
| 2444 | * Answer: first all squares then all dots. |
| 2445 | */ |
| 2446 | |
| 2447 | static int trivial_deductions(solver_state *sstate) |
| 2448 | { |
| 2449 | int i, current_yes, current_no; |
| 2450 | game_state *state = sstate->state; |
| 2451 | grid *g = state->game_grid; |
| 2452 | int diff = DIFF_MAX; |
| 2453 | |
| 2454 | /* Per-face deductions */ |
| 2455 | for (i = 0; i < g->num_faces; i++) { |
| 2456 | grid_face *f = g->faces + i; |
| 2457 | |
| 2458 | if (sstate->face_solved[i]) |
| 2459 | continue; |
| 2460 | |
| 2461 | current_yes = sstate->face_yes_count[i]; |
| 2462 | current_no = sstate->face_no_count[i]; |
| 2463 | |
| 2464 | if (current_yes + current_no == f->order) { |
| 2465 | sstate->face_solved[i] = TRUE; |
| 2466 | continue; |
| 2467 | } |
| 2468 | |
| 2469 | if (state->clues[i] < 0) |
| 2470 | continue; |
| 2471 | |
| 2472 | /* |
| 2473 | * This code checks whether the numeric clue on a face is so |
| 2474 | * large as to permit all its remaining LINE_UNKNOWNs to be |
| 2475 | * filled in as LINE_YES, or alternatively so small as to |
| 2476 | * permit them all to be filled in as LINE_NO. |
| 2477 | */ |
| 2478 | |
| 2479 | if (state->clues[i] < current_yes) { |
| 2480 | sstate->solver_status = SOLVER_MISTAKE; |
| 2481 | return DIFF_EASY; |
| 2482 | } |
| 2483 | if (state->clues[i] == current_yes) { |
| 2484 | if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO)) |
| 2485 | diff = min(diff, DIFF_EASY); |
| 2486 | sstate->face_solved[i] = TRUE; |
| 2487 | continue; |
| 2488 | } |
| 2489 | |
| 2490 | if (f->order - state->clues[i] < current_no) { |
| 2491 | sstate->solver_status = SOLVER_MISTAKE; |
| 2492 | return DIFF_EASY; |
| 2493 | } |
| 2494 | if (f->order - state->clues[i] == current_no) { |
| 2495 | if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES)) |
| 2496 | diff = min(diff, DIFF_EASY); |
| 2497 | sstate->face_solved[i] = TRUE; |
| 2498 | continue; |
| 2499 | } |
| 2500 | |
| 2501 | if (f->order - state->clues[i] == current_no + 1 && |
| 2502 | f->order - current_yes - current_no > 2) { |
| 2503 | /* |
| 2504 | * One small refinement to the above: we also look for any |
| 2505 | * adjacent pair of LINE_UNKNOWNs around the face with |
| 2506 | * some LINE_YES incident on it from elsewhere. If we find |
| 2507 | * one, then we know that pair of LINE_UNKNOWNs can't |
| 2508 | * _both_ be LINE_YES, and hence that pushes us one line |
| 2509 | * closer to being able to determine all the rest. |
| 2510 | */ |
| 2511 | int j, k, e1, e2, e, d; |
| 2512 | |
| 2513 | for (j = 0; j < f->order; j++) { |
| 2514 | e1 = f->edges[j] - g->edges; |
| 2515 | e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges; |
| 2516 | |
| 2517 | if (g->edges[e1].dot1 == g->edges[e2].dot1 || |
| 2518 | g->edges[e1].dot1 == g->edges[e2].dot2) { |
| 2519 | d = g->edges[e1].dot1 - g->dots; |
| 2520 | } else { |
| 2521 | assert(g->edges[e1].dot2 == g->edges[e2].dot1 || |
| 2522 | g->edges[e1].dot2 == g->edges[e2].dot2); |
| 2523 | d = g->edges[e1].dot2 - g->dots; |
| 2524 | } |
| 2525 | |
| 2526 | if (state->lines[e1] == LINE_UNKNOWN && |
| 2527 | state->lines[e2] == LINE_UNKNOWN) { |
| 2528 | for (k = 0; k < g->dots[d].order; k++) { |
| 2529 | int e = g->dots[d].edges[k] - g->edges; |
| 2530 | if (state->lines[e] == LINE_YES) |
| 2531 | goto found; /* multi-level break */ |
| 2532 | } |
| 2533 | } |
| 2534 | } |
| 2535 | continue; |
| 2536 | |
| 2537 | found: |
| 2538 | /* |
| 2539 | * If we get here, we've found such a pair of edges, and |
| 2540 | * they're e1 and e2. |
| 2541 | */ |
| 2542 | for (j = 0; j < f->order; j++) { |
| 2543 | e = f->edges[j] - g->edges; |
| 2544 | if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) { |
| 2545 | int r = solver_set_line(sstate, e, LINE_YES); |
| 2546 | assert(r); |
| 2547 | diff = min(diff, DIFF_EASY); |
| 2548 | } |
| 2549 | } |
| 2550 | } |
| 2551 | } |
| 2552 | |
| 2553 | check_caches(sstate); |
| 2554 | |
| 2555 | /* Per-dot deductions */ |
| 2556 | for (i = 0; i < g->num_dots; i++) { |
| 2557 | grid_dot *d = g->dots + i; |
| 2558 | int yes, no, unknown; |
| 2559 | |
| 2560 | if (sstate->dot_solved[i]) |
| 2561 | continue; |
| 2562 | |
| 2563 | yes = sstate->dot_yes_count[i]; |
| 2564 | no = sstate->dot_no_count[i]; |
| 2565 | unknown = d->order - yes - no; |
| 2566 | |
| 2567 | if (yes == 0) { |
| 2568 | if (unknown == 0) { |
| 2569 | sstate->dot_solved[i] = TRUE; |
| 2570 | } else if (unknown == 1) { |
| 2571 | dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO); |
| 2572 | diff = min(diff, DIFF_EASY); |
| 2573 | sstate->dot_solved[i] = TRUE; |
| 2574 | } |
| 2575 | } else if (yes == 1) { |
| 2576 | if (unknown == 0) { |
| 2577 | sstate->solver_status = SOLVER_MISTAKE; |
| 2578 | return DIFF_EASY; |
| 2579 | } else if (unknown == 1) { |
| 2580 | dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES); |
| 2581 | diff = min(diff, DIFF_EASY); |
| 2582 | } |
| 2583 | } else if (yes == 2) { |
| 2584 | if (unknown > 0) { |
| 2585 | dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO); |
| 2586 | diff = min(diff, DIFF_EASY); |
| 2587 | } |
| 2588 | sstate->dot_solved[i] = TRUE; |
| 2589 | } else { |
| 2590 | sstate->solver_status = SOLVER_MISTAKE; |
| 2591 | return DIFF_EASY; |
| 2592 | } |
| 2593 | } |
| 2594 | |
| 2595 | check_caches(sstate); |
| 2596 | |
| 2597 | return diff; |
| 2598 | } |
| 2599 | |
| 2600 | static int dline_deductions(solver_state *sstate) |
| 2601 | { |
| 2602 | game_state *state = sstate->state; |
| 2603 | grid *g = state->game_grid; |
| 2604 | char *dlines = sstate->dlines; |
| 2605 | int i; |
| 2606 | int diff = DIFF_MAX; |
| 2607 | |
| 2608 | /* ------ Face deductions ------ */ |
| 2609 | |
| 2610 | /* Given a set of dline atmostone/atleastone constraints, need to figure |
| 2611 | * out if we can deduce any further info. For more general faces than |
| 2612 | * squares, this turns out to be a tricky problem. |
| 2613 | * The approach taken here is to define (per face) NxN matrices: |
| 2614 | * "maxs" and "mins". |
| 2615 | * The entries maxs(j,k) and mins(j,k) define the upper and lower limits |
| 2616 | * for the possible number of edges that are YES between positions j and k |
| 2617 | * going clockwise around the face. Can think of j and k as marking dots |
| 2618 | * around the face (recall the labelling scheme: edge0 joins dot0 to dot1, |
| 2619 | * edge1 joins dot1 to dot2 etc). |
| 2620 | * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing |
| 2621 | * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j} |
| 2622 | * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to |
| 2623 | * the dline atmostone/atleastone status for edges j and j+1. |
| 2624 | * |
| 2625 | * Then we calculate the remaining entries recursively. We definitely |
| 2626 | * know that |
| 2627 | * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k. |
| 2628 | * This is because any valid placement of YESs between j and k must give |
| 2629 | * a valid placement between j and u, and also between u and k. |
| 2630 | * I believe it's sufficient to use just the two values of u: |
| 2631 | * j+1 and j+2. Seems to work well in practice - the bounds we compute |
| 2632 | * are rigorous, even if they might not be best-possible. |
| 2633 | * |
| 2634 | * Once we have maxs and mins calculated, we can make inferences about |
| 2635 | * each dline{j,j+1} by looking at the possible complementary edge-counts |
| 2636 | * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue. |
| 2637 | * As well as dlines, we can make similar inferences about single edges. |
| 2638 | * For example, consider a pentagon with clue 3, and we know at most one |
| 2639 | * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES. |
| 2640 | * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so |
| 2641 | * that final edge would have to be YES to make the count up to 3. |
| 2642 | */ |
| 2643 | |
| 2644 | /* Much quicker to allocate arrays on the stack than the heap, so |
| 2645 | * define the largest possible face size, and base our array allocations |
| 2646 | * on that. We check this with an assertion, in case someone decides to |
| 2647 | * make a grid which has larger faces than this. Note, this algorithm |
| 2648 | * could get quite expensive if there are many large faces. */ |
| 2649 | #define MAX_FACE_SIZE 12 |
| 2650 | |
| 2651 | for (i = 0; i < g->num_faces; i++) { |
| 2652 | int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE]; |
| 2653 | int mins[MAX_FACE_SIZE][MAX_FACE_SIZE]; |
| 2654 | grid_face *f = g->faces + i; |
| 2655 | int N = f->order; |
| 2656 | int j,m; |
| 2657 | int clue = state->clues[i]; |
| 2658 | assert(N <= MAX_FACE_SIZE); |
| 2659 | if (sstate->face_solved[i]) |
| 2660 | continue; |
| 2661 | if (clue < 0) continue; |
| 2662 | |
| 2663 | /* Calculate the (j,j+1) entries */ |
| 2664 | for (j = 0; j < N; j++) { |
| 2665 | int edge_index = f->edges[j] - g->edges; |
| 2666 | int dline_index; |
| 2667 | enum line_state line1 = state->lines[edge_index]; |
| 2668 | enum line_state line2; |
| 2669 | int tmp; |
| 2670 | int k = j + 1; |
| 2671 | if (k >= N) k = 0; |
| 2672 | maxs[j][k] = (line1 == LINE_NO) ? 0 : 1; |
| 2673 | mins[j][k] = (line1 == LINE_YES) ? 1 : 0; |
| 2674 | /* Calculate the (j,j+2) entries */ |
| 2675 | dline_index = dline_index_from_face(g, f, k); |
| 2676 | edge_index = f->edges[k] - g->edges; |
| 2677 | line2 = state->lines[edge_index]; |
| 2678 | k++; |
| 2679 | if (k >= N) k = 0; |
| 2680 | |
| 2681 | /* max */ |
| 2682 | tmp = 2; |
| 2683 | if (line1 == LINE_NO) tmp--; |
| 2684 | if (line2 == LINE_NO) tmp--; |
| 2685 | if (tmp == 2 && is_atmostone(dlines, dline_index)) |
| 2686 | tmp = 1; |
| 2687 | maxs[j][k] = tmp; |
| 2688 | |
| 2689 | /* min */ |
| 2690 | tmp = 0; |
| 2691 | if (line1 == LINE_YES) tmp++; |
| 2692 | if (line2 == LINE_YES) tmp++; |
| 2693 | if (tmp == 0 && is_atleastone(dlines, dline_index)) |
| 2694 | tmp = 1; |
| 2695 | mins[j][k] = tmp; |
| 2696 | } |
| 2697 | |
| 2698 | /* Calculate the (j,j+m) entries for m between 3 and N-1 */ |
| 2699 | for (m = 3; m < N; m++) { |
| 2700 | for (j = 0; j < N; j++) { |
| 2701 | int k = j + m; |
| 2702 | int u = j + 1; |
| 2703 | int v = j + 2; |
| 2704 | int tmp; |
| 2705 | if (k >= N) k -= N; |
| 2706 | if (u >= N) u -= N; |
| 2707 | if (v >= N) v -= N; |
| 2708 | maxs[j][k] = maxs[j][u] + maxs[u][k]; |
| 2709 | mins[j][k] = mins[j][u] + mins[u][k]; |
| 2710 | tmp = maxs[j][v] + maxs[v][k]; |
| 2711 | maxs[j][k] = min(maxs[j][k], tmp); |
| 2712 | tmp = mins[j][v] + mins[v][k]; |
| 2713 | mins[j][k] = max(mins[j][k], tmp); |
| 2714 | } |
| 2715 | } |
| 2716 | |
| 2717 | /* See if we can make any deductions */ |
| 2718 | for (j = 0; j < N; j++) { |
| 2719 | int k; |
| 2720 | grid_edge *e = f->edges[j]; |
| 2721 | int line_index = e - g->edges; |
| 2722 | int dline_index; |
| 2723 | |
| 2724 | if (state->lines[line_index] != LINE_UNKNOWN) |
| 2725 | continue; |
| 2726 | k = j + 1; |
| 2727 | if (k >= N) k = 0; |
| 2728 | |
| 2729 | /* minimum YESs in the complement of this edge */ |
| 2730 | if (mins[k][j] > clue) { |
| 2731 | sstate->solver_status = SOLVER_MISTAKE; |
| 2732 | return DIFF_EASY; |
| 2733 | } |
| 2734 | if (mins[k][j] == clue) { |
| 2735 | /* setting this edge to YES would make at least |
| 2736 | * (clue+1) edges - contradiction */ |
| 2737 | solver_set_line(sstate, line_index, LINE_NO); |
| 2738 | diff = min(diff, DIFF_EASY); |
| 2739 | } |
| 2740 | if (maxs[k][j] < clue - 1) { |
| 2741 | sstate->solver_status = SOLVER_MISTAKE; |
| 2742 | return DIFF_EASY; |
| 2743 | } |
| 2744 | if (maxs[k][j] == clue - 1) { |
| 2745 | /* Only way to satisfy the clue is to set edge{j} as YES */ |
| 2746 | solver_set_line(sstate, line_index, LINE_YES); |
| 2747 | diff = min(diff, DIFF_EASY); |
| 2748 | } |
| 2749 | |
| 2750 | /* More advanced deduction that allows propagation along diagonal |
| 2751 | * chains of faces connected by dots, for example, 3-2-...-2-3 |
| 2752 | * in square grids. */ |
| 2753 | if (sstate->diff >= DIFF_TRICKY) { |
| 2754 | /* Now see if we can make dline deduction for edges{j,j+1} */ |
| 2755 | e = f->edges[k]; |
| 2756 | if (state->lines[e - g->edges] != LINE_UNKNOWN) |
| 2757 | /* Only worth doing this for an UNKNOWN,UNKNOWN pair. |
| 2758 | * Dlines where one of the edges is known, are handled in the |
| 2759 | * dot-deductions */ |
| 2760 | continue; |
| 2761 | |
| 2762 | dline_index = dline_index_from_face(g, f, k); |
| 2763 | k++; |
| 2764 | if (k >= N) k = 0; |
| 2765 | |
| 2766 | /* minimum YESs in the complement of this dline */ |
| 2767 | if (mins[k][j] > clue - 2) { |
| 2768 | /* Adding 2 YESs would break the clue */ |
| 2769 | if (set_atmostone(dlines, dline_index)) |
| 2770 | diff = min(diff, DIFF_NORMAL); |
| 2771 | } |
| 2772 | /* maximum YESs in the complement of this dline */ |
| 2773 | if (maxs[k][j] < clue) { |
| 2774 | /* Adding 2 NOs would mean not enough YESs */ |
| 2775 | if (set_atleastone(dlines, dline_index)) |
| 2776 | diff = min(diff, DIFF_NORMAL); |
| 2777 | } |
| 2778 | } |
| 2779 | } |
| 2780 | } |
| 2781 | |
| 2782 | if (diff < DIFF_NORMAL) |
| 2783 | return diff; |
| 2784 | |
| 2785 | /* ------ Dot deductions ------ */ |
| 2786 | |
| 2787 | for (i = 0; i < g->num_dots; i++) { |
| 2788 | grid_dot *d = g->dots + i; |
| 2789 | int N = d->order; |
| 2790 | int yes, no, unknown; |
| 2791 | int j; |
| 2792 | if (sstate->dot_solved[i]) |
| 2793 | continue; |
| 2794 | yes = sstate->dot_yes_count[i]; |
| 2795 | no = sstate->dot_no_count[i]; |
| 2796 | unknown = N - yes - no; |
| 2797 | |
| 2798 | for (j = 0; j < N; j++) { |
| 2799 | int k; |
| 2800 | int dline_index; |
| 2801 | int line1_index, line2_index; |
| 2802 | enum line_state line1, line2; |
| 2803 | k = j + 1; |
| 2804 | if (k >= N) k = 0; |
| 2805 | dline_index = dline_index_from_dot(g, d, j); |
| 2806 | line1_index = d->edges[j] - g->edges; |
| 2807 | line2_index = d->edges[k] - g->edges; |
| 2808 | line1 = state->lines[line1_index]; |
| 2809 | line2 = state->lines[line2_index]; |
| 2810 | |
| 2811 | /* Infer dline state from line state */ |
| 2812 | if (line1 == LINE_NO || line2 == LINE_NO) { |
| 2813 | if (set_atmostone(dlines, dline_index)) |
| 2814 | diff = min(diff, DIFF_NORMAL); |
| 2815 | } |
| 2816 | if (line1 == LINE_YES || line2 == LINE_YES) { |
| 2817 | if (set_atleastone(dlines, dline_index)) |
| 2818 | diff = min(diff, DIFF_NORMAL); |
| 2819 | } |
| 2820 | /* Infer line state from dline state */ |
| 2821 | if (is_atmostone(dlines, dline_index)) { |
| 2822 | if (line1 == LINE_YES && line2 == LINE_UNKNOWN) { |
| 2823 | solver_set_line(sstate, line2_index, LINE_NO); |
| 2824 | diff = min(diff, DIFF_EASY); |
| 2825 | } |
| 2826 | if (line2 == LINE_YES && line1 == LINE_UNKNOWN) { |
| 2827 | solver_set_line(sstate, line1_index, LINE_NO); |
| 2828 | diff = min(diff, DIFF_EASY); |
| 2829 | } |
| 2830 | } |
| 2831 | if (is_atleastone(dlines, dline_index)) { |
| 2832 | if (line1 == LINE_NO && line2 == LINE_UNKNOWN) { |
| 2833 | solver_set_line(sstate, line2_index, LINE_YES); |
| 2834 | diff = min(diff, DIFF_EASY); |
| 2835 | } |
| 2836 | if (line2 == LINE_NO && line1 == LINE_UNKNOWN) { |
| 2837 | solver_set_line(sstate, line1_index, LINE_YES); |
| 2838 | diff = min(diff, DIFF_EASY); |
| 2839 | } |
| 2840 | } |
| 2841 | /* Deductions that depend on the numbers of lines. |
| 2842 | * Only bother if both lines are UNKNOWN, otherwise the |
| 2843 | * easy-mode solver (or deductions above) would have taken |
| 2844 | * care of it. */ |
| 2845 | if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN) |
| 2846 | continue; |
| 2847 | |
| 2848 | if (yes == 0 && unknown == 2) { |
| 2849 | /* Both these unknowns must be identical. If we know |
| 2850 | * atmostone or atleastone, we can make progress. */ |
| 2851 | if (is_atmostone(dlines, dline_index)) { |
| 2852 | solver_set_line(sstate, line1_index, LINE_NO); |
| 2853 | solver_set_line(sstate, line2_index, LINE_NO); |
| 2854 | diff = min(diff, DIFF_EASY); |
| 2855 | } |
| 2856 | if (is_atleastone(dlines, dline_index)) { |
| 2857 | solver_set_line(sstate, line1_index, LINE_YES); |
| 2858 | solver_set_line(sstate, line2_index, LINE_YES); |
| 2859 | diff = min(diff, DIFF_EASY); |
| 2860 | } |
| 2861 | } |
| 2862 | if (yes == 1) { |
| 2863 | if (set_atmostone(dlines, dline_index)) |
| 2864 | diff = min(diff, DIFF_NORMAL); |
| 2865 | if (unknown == 2) { |
| 2866 | if (set_atleastone(dlines, dline_index)) |
| 2867 | diff = min(diff, DIFF_NORMAL); |
| 2868 | } |
| 2869 | } |
| 2870 | |
| 2871 | /* More advanced deduction that allows propagation along diagonal |
| 2872 | * chains of faces connected by dots, for example: 3-2-...-2-3 |
| 2873 | * in square grids. */ |
| 2874 | if (sstate->diff >= DIFF_TRICKY) { |
| 2875 | /* If we have atleastone set for this dline, infer |
| 2876 | * atmostone for each "opposite" dline (that is, each |
| 2877 | * dline without edges in common with this one). |
| 2878 | * Again, this test is only worth doing if both these |
| 2879 | * lines are UNKNOWN. For if one of these lines were YES, |
| 2880 | * the (yes == 1) test above would kick in instead. */ |
| 2881 | if (is_atleastone(dlines, dline_index)) { |
| 2882 | int opp; |
| 2883 | for (opp = 0; opp < N; opp++) { |
| 2884 | int opp_dline_index; |
| 2885 | if (opp == j || opp == j+1 || opp == j-1) |
| 2886 | continue; |
| 2887 | if (j == 0 && opp == N-1) |
| 2888 | continue; |
| 2889 | if (j == N-1 && opp == 0) |
| 2890 | continue; |
| 2891 | opp_dline_index = dline_index_from_dot(g, d, opp); |
| 2892 | if (set_atmostone(dlines, opp_dline_index)) |
| 2893 | diff = min(diff, DIFF_NORMAL); |
| 2894 | } |
| 2895 | if (yes == 0 && is_atmostone(dlines, dline_index)) { |
| 2896 | /* This dline has *exactly* one YES and there are no |
| 2897 | * other YESs. This allows more deductions. */ |
| 2898 | if (unknown == 3) { |
| 2899 | /* Third unknown must be YES */ |
| 2900 | for (opp = 0; opp < N; opp++) { |
| 2901 | int opp_index; |
| 2902 | if (opp == j || opp == k) |
| 2903 | continue; |
| 2904 | opp_index = d->edges[opp] - g->edges; |
| 2905 | if (state->lines[opp_index] == LINE_UNKNOWN) { |
| 2906 | solver_set_line(sstate, opp_index, |
| 2907 | LINE_YES); |
| 2908 | diff = min(diff, DIFF_EASY); |
| 2909 | } |
| 2910 | } |
| 2911 | } else if (unknown == 4) { |
| 2912 | /* Exactly one of opposite UNKNOWNS is YES. We've |
| 2913 | * already set atmostone, so set atleastone as |
| 2914 | * well. |
| 2915 | */ |
| 2916 | if (dline_set_opp_atleastone(sstate, d, j)) |
| 2917 | diff = min(diff, DIFF_NORMAL); |
| 2918 | } |
| 2919 | } |
| 2920 | } |
| 2921 | } |
| 2922 | } |
| 2923 | } |
| 2924 | return diff; |
| 2925 | } |
| 2926 | |
| 2927 | static int linedsf_deductions(solver_state *sstate) |
| 2928 | { |
| 2929 | game_state *state = sstate->state; |
| 2930 | grid *g = state->game_grid; |
| 2931 | char *dlines = sstate->dlines; |
| 2932 | int i; |
| 2933 | int diff = DIFF_MAX; |
| 2934 | int diff_tmp; |
| 2935 | |
| 2936 | /* ------ Face deductions ------ */ |
| 2937 | |
| 2938 | /* A fully-general linedsf deduction seems overly complicated |
| 2939 | * (I suspect the problem is NP-complete, though in practice it might just |
| 2940 | * be doable because faces are limited in size). |
| 2941 | * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are |
| 2942 | * known to be identical. If setting them both to YES (or NO) would break |
| 2943 | * the clue, set them to NO (or YES). */ |
| 2944 | |
| 2945 | for (i = 0; i < g->num_faces; i++) { |
| 2946 | int N, yes, no, unknown; |
| 2947 | int clue; |
| 2948 | |
| 2949 | if (sstate->face_solved[i]) |
| 2950 | continue; |
| 2951 | clue = state->clues[i]; |
| 2952 | if (clue < 0) |
| 2953 | continue; |
| 2954 | |
| 2955 | N = g->faces[i].order; |
| 2956 | yes = sstate->face_yes_count[i]; |
| 2957 | if (yes + 1 == clue) { |
| 2958 | if (face_setall_identical(sstate, i, LINE_NO)) |
| 2959 | diff = min(diff, DIFF_EASY); |
| 2960 | } |
| 2961 | no = sstate->face_no_count[i]; |
| 2962 | if (no + 1 == N - clue) { |
| 2963 | if (face_setall_identical(sstate, i, LINE_YES)) |
| 2964 | diff = min(diff, DIFF_EASY); |
| 2965 | } |
| 2966 | |
| 2967 | /* Reload YES count, it might have changed */ |
| 2968 | yes = sstate->face_yes_count[i]; |
| 2969 | unknown = N - no - yes; |
| 2970 | |
| 2971 | /* Deductions with small number of LINE_UNKNOWNs, based on overall |
| 2972 | * parity of lines. */ |
| 2973 | diff_tmp = parity_deductions(sstate, g->faces[i].edges, |
| 2974 | (clue - yes) % 2, unknown); |
| 2975 | diff = min(diff, diff_tmp); |
| 2976 | } |
| 2977 | |
| 2978 | /* ------ Dot deductions ------ */ |
| 2979 | for (i = 0; i < g->num_dots; i++) { |
| 2980 | grid_dot *d = g->dots + i; |
| 2981 | int N = d->order; |
| 2982 | int j; |
| 2983 | int yes, no, unknown; |
| 2984 | /* Go through dlines, and do any dline<->linedsf deductions wherever |
| 2985 | * we find two UNKNOWNS. */ |
| 2986 | for (j = 0; j < N; j++) { |
| 2987 | int dline_index = dline_index_from_dot(g, d, j); |
| 2988 | int line1_index; |
| 2989 | int line2_index; |
| 2990 | int can1, can2, inv1, inv2; |
| 2991 | int j2; |
| 2992 | line1_index = d->edges[j] - g->edges; |
| 2993 | if (state->lines[line1_index] != LINE_UNKNOWN) |
| 2994 | continue; |
| 2995 | j2 = j + 1; |
| 2996 | if (j2 == N) j2 = 0; |
| 2997 | line2_index = d->edges[j2] - g->edges; |
| 2998 | if (state->lines[line2_index] != LINE_UNKNOWN) |
| 2999 | continue; |
| 3000 | /* Infer dline flags from linedsf */ |
| 3001 | can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1); |
| 3002 | can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2); |
| 3003 | if (can1 == can2 && inv1 != inv2) { |
| 3004 | /* These are opposites, so set dline atmostone/atleastone */ |
| 3005 | if (set_atmostone(dlines, dline_index)) |
| 3006 | diff = min(diff, DIFF_NORMAL); |
| 3007 | if (set_atleastone(dlines, dline_index)) |
| 3008 | diff = min(diff, DIFF_NORMAL); |
| 3009 | continue; |
| 3010 | } |
| 3011 | /* Infer linedsf from dline flags */ |
| 3012 | if (is_atmostone(dlines, dline_index) |
| 3013 | && is_atleastone(dlines, dline_index)) { |
| 3014 | if (merge_lines(sstate, line1_index, line2_index, 1)) |
| 3015 | diff = min(diff, DIFF_HARD); |
| 3016 | } |
| 3017 | } |
| 3018 | |
| 3019 | /* Deductions with small number of LINE_UNKNOWNs, based on overall |
| 3020 | * parity of lines. */ |
| 3021 | yes = sstate->dot_yes_count[i]; |
| 3022 | no = sstate->dot_no_count[i]; |
| 3023 | unknown = N - yes - no; |
| 3024 | diff_tmp = parity_deductions(sstate, d->edges, |
| 3025 | yes % 2, unknown); |
| 3026 | diff = min(diff, diff_tmp); |
| 3027 | } |
| 3028 | |
| 3029 | /* ------ Edge dsf deductions ------ */ |
| 3030 | |
| 3031 | /* If the state of a line is known, deduce the state of its canonical line |
| 3032 | * too, and vice versa. */ |
| 3033 | for (i = 0; i < g->num_edges; i++) { |
| 3034 | int can, inv; |
| 3035 | enum line_state s; |
| 3036 | can = edsf_canonify(sstate->linedsf, i, &inv); |
| 3037 | if (can == i) |
| 3038 | continue; |
| 3039 | s = sstate->state->lines[can]; |
| 3040 | if (s != LINE_UNKNOWN) { |
| 3041 | if (solver_set_line(sstate, i, inv ? OPP(s) : s)) |
| 3042 | diff = min(diff, DIFF_EASY); |
| 3043 | } else { |
| 3044 | s = sstate->state->lines[i]; |
| 3045 | if (s != LINE_UNKNOWN) { |
| 3046 | if (solver_set_line(sstate, can, inv ? OPP(s) : s)) |
| 3047 | diff = min(diff, DIFF_EASY); |
| 3048 | } |
| 3049 | } |
| 3050 | } |
| 3051 | |
| 3052 | return diff; |
| 3053 | } |
| 3054 | |
| 3055 | static int loop_deductions(solver_state *sstate) |
| 3056 | { |
| 3057 | int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0; |
| 3058 | game_state *state = sstate->state; |
| 3059 | grid *g = state->game_grid; |
| 3060 | int shortest_chainlen = g->num_dots; |
| 3061 | int loop_found = FALSE; |
| 3062 | int dots_connected; |
| 3063 | int progress = FALSE; |
| 3064 | int i; |
| 3065 | |
| 3066 | /* |
| 3067 | * Go through the grid and update for all the new edges. |
| 3068 | * Since merge_dots() is idempotent, the simplest way to |
| 3069 | * do this is just to update for _all_ the edges. |
| 3070 | * Also, while we're here, we count the edges. |
| 3071 | */ |
| 3072 | for (i = 0; i < g->num_edges; i++) { |
| 3073 | if (state->lines[i] == LINE_YES) { |
| 3074 | loop_found |= merge_dots(sstate, i); |
| 3075 | edgecount++; |
| 3076 | } |
| 3077 | } |
| 3078 | |
| 3079 | /* |
| 3080 | * Count the clues, count the satisfied clues, and count the |
| 3081 | * satisfied-minus-one clues. |
| 3082 | */ |
| 3083 | for (i = 0; i < g->num_faces; i++) { |
| 3084 | int c = state->clues[i]; |
| 3085 | if (c >= 0) { |
| 3086 | int o = sstate->face_yes_count[i]; |
| 3087 | if (o == c) |
| 3088 | satclues++; |
| 3089 | else if (o == c-1) |
| 3090 | sm1clues++; |
| 3091 | clues++; |
| 3092 | } |
| 3093 | } |
| 3094 | |
| 3095 | for (i = 0; i < g->num_dots; ++i) { |
| 3096 | dots_connected = |
| 3097 | sstate->looplen[dsf_canonify(sstate->dotdsf, i)]; |
| 3098 | if (dots_connected > 1) |
| 3099 | shortest_chainlen = min(shortest_chainlen, dots_connected); |
| 3100 | } |
| 3101 | |
| 3102 | assert(sstate->solver_status == SOLVER_INCOMPLETE); |
| 3103 | |
| 3104 | if (satclues == clues && shortest_chainlen == edgecount) { |
| 3105 | sstate->solver_status = SOLVER_SOLVED; |
| 3106 | /* This discovery clearly counts as progress, even if we haven't |
| 3107 | * just added any lines or anything */ |
| 3108 | progress = TRUE; |
| 3109 | goto finished_loop_deductionsing; |
| 3110 | } |
| 3111 | |
| 3112 | /* |
| 3113 | * Now go through looking for LINE_UNKNOWN edges which |
| 3114 | * connect two dots that are already in the same |
| 3115 | * equivalence class. If we find one, test to see if the |
| 3116 | * loop it would create is a solution. |
| 3117 | */ |
| 3118 | for (i = 0; i < g->num_edges; i++) { |
| 3119 | grid_edge *e = g->edges + i; |
| 3120 | int d1 = e->dot1 - g->dots; |
| 3121 | int d2 = e->dot2 - g->dots; |
| 3122 | int eqclass, val; |
| 3123 | if (state->lines[i] != LINE_UNKNOWN) |
| 3124 | continue; |
| 3125 | |
| 3126 | eqclass = dsf_canonify(sstate->dotdsf, d1); |
| 3127 | if (eqclass != dsf_canonify(sstate->dotdsf, d2)) |
| 3128 | continue; |
| 3129 | |
| 3130 | val = LINE_NO; /* loop is bad until proven otherwise */ |
| 3131 | |
| 3132 | /* |
| 3133 | * This edge would form a loop. Next |
| 3134 | * question: how long would the loop be? |
| 3135 | * Would it equal the total number of edges |
| 3136 | * (plus the one we'd be adding if we added |
| 3137 | * it)? |
| 3138 | */ |
| 3139 | if (sstate->looplen[eqclass] == edgecount + 1) { |
| 3140 | int sm1_nearby; |
| 3141 | |
| 3142 | /* |
| 3143 | * This edge would form a loop which |
| 3144 | * took in all the edges in the entire |
| 3145 | * grid. So now we need to work out |
| 3146 | * whether it would be a valid solution |
| 3147 | * to the puzzle, which means we have to |
| 3148 | * check if it satisfies all the clues. |
| 3149 | * This means that every clue must be |
| 3150 | * either satisfied or satisfied-minus- |
| 3151 | * 1, and also that the number of |
| 3152 | * satisfied-minus-1 clues must be at |
| 3153 | * most two and they must lie on either |
| 3154 | * side of this edge. |
| 3155 | */ |
| 3156 | sm1_nearby = 0; |
| 3157 | if (e->face1) { |
| 3158 | int f = e->face1 - g->faces; |
| 3159 | int c = state->clues[f]; |
| 3160 | if (c >= 0 && sstate->face_yes_count[f] == c - 1) |
| 3161 | sm1_nearby++; |
| 3162 | } |
| 3163 | if (e->face2) { |
| 3164 | int f = e->face2 - g->faces; |
| 3165 | int c = state->clues[f]; |
| 3166 | if (c >= 0 && sstate->face_yes_count[f] == c - 1) |
| 3167 | sm1_nearby++; |
| 3168 | } |
| 3169 | if (sm1clues == sm1_nearby && |
| 3170 | sm1clues + satclues == clues) { |
| 3171 | val = LINE_YES; /* loop is good! */ |
| 3172 | } |
| 3173 | } |
| 3174 | |
| 3175 | /* |
| 3176 | * Right. Now we know that adding this edge |
| 3177 | * would form a loop, and we know whether |
| 3178 | * that loop would be a viable solution or |
| 3179 | * not. |
| 3180 | * |
| 3181 | * If adding this edge produces a solution, |
| 3182 | * then we know we've found _a_ solution but |
| 3183 | * we don't know that it's _the_ solution - |
| 3184 | * if it were provably the solution then |
| 3185 | * we'd have deduced this edge some time ago |
| 3186 | * without the need to do loop detection. So |
| 3187 | * in this state we return SOLVER_AMBIGUOUS, |
| 3188 | * which has the effect that hitting Solve |
| 3189 | * on a user-provided puzzle will fill in a |
| 3190 | * solution but using the solver to |
| 3191 | * construct new puzzles won't consider this |
| 3192 | * a reasonable deduction for the user to |
| 3193 | * make. |
| 3194 | */ |
| 3195 | progress = solver_set_line(sstate, i, val); |
| 3196 | assert(progress == TRUE); |
| 3197 | if (val == LINE_YES) { |
| 3198 | sstate->solver_status = SOLVER_AMBIGUOUS; |
| 3199 | goto finished_loop_deductionsing; |
| 3200 | } |
| 3201 | } |
| 3202 | |
| 3203 | finished_loop_deductionsing: |
| 3204 | return progress ? DIFF_EASY : DIFF_MAX; |
| 3205 | } |
| 3206 | |
| 3207 | /* This will return a dynamically allocated solver_state containing the (more) |
| 3208 | * solved grid */ |
| 3209 | static solver_state *solve_game_rec(const solver_state *sstate_start) |
| 3210 | { |
| 3211 | solver_state *sstate; |
| 3212 | |
| 3213 | /* Index of the solver we should call next. */ |
| 3214 | int i = 0; |
| 3215 | |
| 3216 | /* As a speed-optimisation, we avoid re-running solvers that we know |
| 3217 | * won't make any progress. This happens when a high-difficulty |
| 3218 | * solver makes a deduction that can only help other high-difficulty |
| 3219 | * solvers. |
| 3220 | * For example: if a new 'dline' flag is set by dline_deductions, the |
| 3221 | * trivial_deductions solver cannot do anything with this information. |
| 3222 | * If we've already run the trivial_deductions solver (because it's |
| 3223 | * earlier in the list), there's no point running it again. |
| 3224 | * |
| 3225 | * Therefore: if a solver is earlier in the list than "threshold_index", |
| 3226 | * we don't bother running it if it's difficulty level is less than |
| 3227 | * "threshold_diff". |
| 3228 | */ |
| 3229 | int threshold_diff = 0; |
| 3230 | int threshold_index = 0; |
| 3231 | |
| 3232 | sstate = dup_solver_state(sstate_start); |
| 3233 | |
| 3234 | check_caches(sstate); |
| 3235 | |
| 3236 | while (i < NUM_SOLVERS) { |
| 3237 | if (sstate->solver_status == SOLVER_MISTAKE) |
| 3238 | return sstate; |
| 3239 | if (sstate->solver_status == SOLVER_SOLVED || |
| 3240 | sstate->solver_status == SOLVER_AMBIGUOUS) { |
| 3241 | /* solver finished */ |
| 3242 | break; |
| 3243 | } |
| 3244 | |
| 3245 | if ((solver_diffs[i] >= threshold_diff || i >= threshold_index) |
| 3246 | && solver_diffs[i] <= sstate->diff) { |
| 3247 | /* current_solver is eligible, so use it */ |
| 3248 | int next_diff = solver_fns[i](sstate); |
| 3249 | if (next_diff != DIFF_MAX) { |
| 3250 | /* solver made progress, so use new thresholds and |
| 3251 | * start again at top of list. */ |
| 3252 | threshold_diff = next_diff; |
| 3253 | threshold_index = i; |
| 3254 | i = 0; |
| 3255 | continue; |
| 3256 | } |
| 3257 | } |
| 3258 | /* current_solver is ineligible, or failed to make progress, so |
| 3259 | * go to the next solver in the list */ |
| 3260 | i++; |
| 3261 | } |
| 3262 | |
| 3263 | if (sstate->solver_status == SOLVER_SOLVED || |
| 3264 | sstate->solver_status == SOLVER_AMBIGUOUS) { |
| 3265 | /* s/LINE_UNKNOWN/LINE_NO/g */ |
| 3266 | array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO, |
| 3267 | sstate->state->game_grid->num_edges); |
| 3268 | return sstate; |
| 3269 | } |
| 3270 | |
| 3271 | return sstate; |
| 3272 | } |
| 3273 | |
| 3274 | static char *solve_game(game_state *state, game_state *currstate, |
| 3275 | char *aux, char **error) |
| 3276 | { |
| 3277 | char *soln = NULL; |
| 3278 | solver_state *sstate, *new_sstate; |
| 3279 | |
| 3280 | sstate = new_solver_state(state, DIFF_MAX); |
| 3281 | new_sstate = solve_game_rec(sstate); |
| 3282 | |
| 3283 | if (new_sstate->solver_status == SOLVER_SOLVED) { |
| 3284 | soln = encode_solve_move(new_sstate->state); |
| 3285 | } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) { |
| 3286 | soln = encode_solve_move(new_sstate->state); |
| 3287 | /**error = "Solver found ambiguous solutions"; */ |
| 3288 | } else { |
| 3289 | soln = encode_solve_move(new_sstate->state); |
| 3290 | /**error = "Solver failed"; */ |
| 3291 | } |
| 3292 | |
| 3293 | free_solver_state(new_sstate); |
| 3294 | free_solver_state(sstate); |
| 3295 | |
| 3296 | return soln; |
| 3297 | } |
| 3298 | |
| 3299 | /* ---------------------------------------------------------------------- |
| 3300 | * Drawing and mouse-handling |
| 3301 | */ |
| 3302 | |
| 3303 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
| 3304 | int x, int y, int button) |
| 3305 | { |
| 3306 | grid *g = state->game_grid; |
| 3307 | grid_edge *e; |
| 3308 | int i; |
| 3309 | char *ret, buf[80]; |
| 3310 | char button_char = ' '; |
| 3311 | enum line_state old_state; |
| 3312 | |
| 3313 | button &= ~MOD_MASK; |
| 3314 | |
| 3315 | /* Convert mouse-click (x,y) to grid coordinates */ |
| 3316 | x -= BORDER(ds->tilesize); |
| 3317 | y -= BORDER(ds->tilesize); |
| 3318 | x = x * g->tilesize / ds->tilesize; |
| 3319 | y = y * g->tilesize / ds->tilesize; |
| 3320 | x += g->lowest_x; |
| 3321 | y += g->lowest_y; |
| 3322 | |
| 3323 | e = grid_nearest_edge(g, x, y); |
| 3324 | if (e == NULL) |
| 3325 | return NULL; |
| 3326 | |
| 3327 | i = e - g->edges; |
| 3328 | |
| 3329 | /* I think it's only possible to play this game with mouse clicks, sorry */ |
| 3330 | /* Maybe will add mouse drag support some time */ |
| 3331 | old_state = state->lines[i]; |
| 3332 | |
| 3333 | switch (button) { |
| 3334 | case LEFT_BUTTON: |
| 3335 | switch (old_state) { |
| 3336 | case LINE_UNKNOWN: |
| 3337 | button_char = 'y'; |
| 3338 | break; |
| 3339 | case LINE_YES: |
| 3340 | #ifdef STYLUS_BASED |
| 3341 | button_char = 'n'; |
| 3342 | break; |
| 3343 | #endif |
| 3344 | case LINE_NO: |
| 3345 | button_char = 'u'; |
| 3346 | break; |
| 3347 | } |
| 3348 | break; |
| 3349 | case MIDDLE_BUTTON: |
| 3350 | button_char = 'u'; |
| 3351 | break; |
| 3352 | case RIGHT_BUTTON: |
| 3353 | switch (old_state) { |
| 3354 | case LINE_UNKNOWN: |
| 3355 | button_char = 'n'; |
| 3356 | break; |
| 3357 | case LINE_NO: |
| 3358 | #ifdef STYLUS_BASED |
| 3359 | button_char = 'y'; |
| 3360 | break; |
| 3361 | #endif |
| 3362 | case LINE_YES: |
| 3363 | button_char = 'u'; |
| 3364 | break; |
| 3365 | } |
| 3366 | break; |
| 3367 | default: |
| 3368 | return NULL; |
| 3369 | } |
| 3370 | |
| 3371 | |
| 3372 | sprintf(buf, "%d%c", i, (int)button_char); |
| 3373 | ret = dupstr(buf); |
| 3374 | |
| 3375 | return ret; |
| 3376 | } |
| 3377 | |
| 3378 | static game_state *execute_move(game_state *state, char *move) |
| 3379 | { |
| 3380 | int i; |
| 3381 | game_state *newstate = dup_game(state); |
| 3382 | |
| 3383 | if (move[0] == 'S') { |
| 3384 | move++; |
| 3385 | newstate->cheated = TRUE; |
| 3386 | } |
| 3387 | |
| 3388 | while (*move) { |
| 3389 | i = atoi(move); |
| 3390 | if (i < 0 || i >= newstate->game_grid->num_edges) |
| 3391 | goto fail; |
| 3392 | move += strspn(move, "1234567890"); |
| 3393 | switch (*(move++)) { |
| 3394 | case 'y': |
| 3395 | newstate->lines[i] = LINE_YES; |
| 3396 | break; |
| 3397 | case 'n': |
| 3398 | newstate->lines[i] = LINE_NO; |
| 3399 | break; |
| 3400 | case 'u': |
| 3401 | newstate->lines[i] = LINE_UNKNOWN; |
| 3402 | break; |
| 3403 | default: |
| 3404 | goto fail; |
| 3405 | } |
| 3406 | } |
| 3407 | |
| 3408 | /* |
| 3409 | * Check for completion. |
| 3410 | */ |
| 3411 | if (check_completion(newstate)) |
| 3412 | newstate->solved = TRUE; |
| 3413 | |
| 3414 | return newstate; |
| 3415 | |
| 3416 | fail: |
| 3417 | free_game(newstate); |
| 3418 | return NULL; |
| 3419 | } |
| 3420 | |
| 3421 | /* ---------------------------------------------------------------------- |
| 3422 | * Drawing routines. |
| 3423 | */ |
| 3424 | |
| 3425 | /* Convert from grid coordinates to screen coordinates */ |
| 3426 | static void grid_to_screen(const game_drawstate *ds, const grid *g, |
| 3427 | int grid_x, int grid_y, int *x, int *y) |
| 3428 | { |
| 3429 | *x = grid_x - g->lowest_x; |
| 3430 | *y = grid_y - g->lowest_y; |
| 3431 | *x = *x * ds->tilesize / g->tilesize; |
| 3432 | *y = *y * ds->tilesize / g->tilesize; |
| 3433 | *x += BORDER(ds->tilesize); |
| 3434 | *y += BORDER(ds->tilesize); |
| 3435 | } |
| 3436 | |
| 3437 | /* Returns (into x,y) position of centre of face for rendering the text clue. |
| 3438 | */ |
| 3439 | static void face_text_pos(const game_drawstate *ds, const grid *g, |
| 3440 | grid_face *f, int *xret, int *yret) |
| 3441 | { |
| 3442 | int faceindex = f - g->faces; |
| 3443 | |
| 3444 | /* |
| 3445 | * Return the cached position for this face, if we've already |
| 3446 | * worked it out. |
| 3447 | */ |
| 3448 | if (ds->textx[faceindex] >= 0) { |
| 3449 | *xret = ds->textx[faceindex]; |
| 3450 | *yret = ds->texty[faceindex]; |
| 3451 | return; |
| 3452 | } |
| 3453 | |
| 3454 | /* |
| 3455 | * Otherwise, use the incentre computed by grid.c and convert it |
| 3456 | * to screen coordinates. |
| 3457 | */ |
| 3458 | grid_find_incentre(f); |
| 3459 | grid_to_screen(ds, g, f->ix, f->iy, |
| 3460 | &ds->textx[faceindex], &ds->texty[faceindex]); |
| 3461 | |
| 3462 | *xret = ds->textx[faceindex]; |
| 3463 | *yret = ds->texty[faceindex]; |
| 3464 | } |
| 3465 | |
| 3466 | static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f, |
| 3467 | int *x, int *y, int *w, int *h) |
| 3468 | { |
| 3469 | int xx, yy; |
| 3470 | face_text_pos(ds, g, f, &xx, &yy); |
| 3471 | |
| 3472 | /* There seems to be a certain amount of trial-and-error involved |
| 3473 | * in working out the correct bounding-box for the text. */ |
| 3474 | |
| 3475 | *x = xx - ds->tilesize/4 - 1; |
| 3476 | *y = yy - ds->tilesize/4 - 3; |
| 3477 | *w = ds->tilesize/2 + 2; |
| 3478 | *h = ds->tilesize/2 + 5; |
| 3479 | } |
| 3480 | |
| 3481 | static void game_redraw_clue(drawing *dr, game_drawstate *ds, |
| 3482 | game_state *state, int i) |
| 3483 | { |
| 3484 | grid *g = state->game_grid; |
| 3485 | grid_face *f = g->faces + i; |
| 3486 | int x, y; |
| 3487 | char c[3]; |
| 3488 | |
| 3489 | if (state->clues[i] < 10) { |
| 3490 | c[0] = CLUE2CHAR(state->clues[i]); |
| 3491 | c[1] = '\0'; |
| 3492 | } else { |
| 3493 | sprintf(c, "%d", state->clues[i]); |
| 3494 | } |
| 3495 | |
| 3496 | face_text_pos(ds, g, f, &x, &y); |
| 3497 | draw_text(dr, x, y, |
| 3498 | FONT_VARIABLE, ds->tilesize/2, |
| 3499 | ALIGN_VCENTRE | ALIGN_HCENTRE, |
| 3500 | ds->clue_error[i] ? COL_MISTAKE : |
| 3501 | ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c); |
| 3502 | } |
| 3503 | |
| 3504 | static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e, |
| 3505 | int *x, int *y, int *w, int *h) |
| 3506 | { |
| 3507 | int x1 = e->dot1->x; |
| 3508 | int y1 = e->dot1->y; |
| 3509 | int x2 = e->dot2->x; |
| 3510 | int y2 = e->dot2->y; |
| 3511 | int xmin, xmax, ymin, ymax; |
| 3512 | |
| 3513 | grid_to_screen(ds, g, x1, y1, &x1, &y1); |
| 3514 | grid_to_screen(ds, g, x2, y2, &x2, &y2); |
| 3515 | /* Allow extra margin for dots, and thickness of lines */ |
| 3516 | xmin = min(x1, x2) - 2; |
| 3517 | xmax = max(x1, x2) + 2; |
| 3518 | ymin = min(y1, y2) - 2; |
| 3519 | ymax = max(y1, y2) + 2; |
| 3520 | |
| 3521 | *x = xmin; |
| 3522 | *y = ymin; |
| 3523 | *w = xmax - xmin + 1; |
| 3524 | *h = ymax - ymin + 1; |
| 3525 | } |
| 3526 | |
| 3527 | static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d, |
| 3528 | int *x, int *y, int *w, int *h) |
| 3529 | { |
| 3530 | int x1, y1; |
| 3531 | |
| 3532 | grid_to_screen(ds, g, d->x, d->y, &x1, &y1); |
| 3533 | |
| 3534 | *x = x1 - 2; |
| 3535 | *y = y1 - 2; |
| 3536 | *w = 5; |
| 3537 | *h = 5; |
| 3538 | } |
| 3539 | |
| 3540 | static const int loopy_line_redraw_phases[] = { |
| 3541 | COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE |
| 3542 | }; |
| 3543 | #define NPHASES lenof(loopy_line_redraw_phases) |
| 3544 | |
| 3545 | static void game_redraw_line(drawing *dr, game_drawstate *ds, |
| 3546 | game_state *state, int i, int phase) |
| 3547 | { |
| 3548 | grid *g = state->game_grid; |
| 3549 | grid_edge *e = g->edges + i; |
| 3550 | int x1, x2, y1, y2; |
| 3551 | int line_colour; |
| 3552 | |
| 3553 | if (state->line_errors[i]) |
| 3554 | line_colour = COL_MISTAKE; |
| 3555 | else if (state->lines[i] == LINE_UNKNOWN) |
| 3556 | line_colour = COL_LINEUNKNOWN; |
| 3557 | else if (state->lines[i] == LINE_NO) |
| 3558 | line_colour = COL_FAINT; |
| 3559 | else if (ds->flashing) |
| 3560 | line_colour = COL_HIGHLIGHT; |
| 3561 | else |
| 3562 | line_colour = COL_FOREGROUND; |
| 3563 | if (line_colour != loopy_line_redraw_phases[phase]) |
| 3564 | return; |
| 3565 | |
| 3566 | /* Convert from grid to screen coordinates */ |
| 3567 | grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1); |
| 3568 | grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2); |
| 3569 | |
| 3570 | if (line_colour == COL_FAINT) { |
| 3571 | static int draw_faint_lines = -1; |
| 3572 | if (draw_faint_lines < 0) { |
| 3573 | char *env = getenv("LOOPY_FAINT_LINES"); |
| 3574 | draw_faint_lines = (!env || (env[0] == 'y' || |
| 3575 | env[0] == 'Y')); |
| 3576 | } |
| 3577 | if (draw_faint_lines) |
| 3578 | draw_line(dr, x1, y1, x2, y2, line_colour); |
| 3579 | } else { |
| 3580 | draw_thick_line(dr, 3.0, |
| 3581 | x1 + 0.5, y1 + 0.5, |
| 3582 | x2 + 0.5, y2 + 0.5, |
| 3583 | line_colour); |
| 3584 | } |
| 3585 | } |
| 3586 | |
| 3587 | static void game_redraw_dot(drawing *dr, game_drawstate *ds, |
| 3588 | game_state *state, int i) |
| 3589 | { |
| 3590 | grid *g = state->game_grid; |
| 3591 | grid_dot *d = g->dots + i; |
| 3592 | int x, y; |
| 3593 | |
| 3594 | grid_to_screen(ds, g, d->x, d->y, &x, &y); |
| 3595 | draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND); |
| 3596 | } |
| 3597 | |
| 3598 | static int boxes_intersect(int x0, int y0, int w0, int h0, |
| 3599 | int x1, int y1, int w1, int h1) |
| 3600 | { |
| 3601 | /* |
| 3602 | * Two intervals intersect iff neither is wholly on one side of |
| 3603 | * the other. Two boxes intersect iff their horizontal and |
| 3604 | * vertical intervals both intersect. |
| 3605 | */ |
| 3606 | return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0); |
| 3607 | } |
| 3608 | |
| 3609 | static void game_redraw_in_rect(drawing *dr, game_drawstate *ds, |
| 3610 | game_state *state, int x, int y, int w, int h) |
| 3611 | { |
| 3612 | grid *g = state->game_grid; |
| 3613 | int i, phase; |
| 3614 | int bx, by, bw, bh; |
| 3615 | |
| 3616 | clip(dr, x, y, w, h); |
| 3617 | draw_rect(dr, x, y, w, h, COL_BACKGROUND); |
| 3618 | |
| 3619 | for (i = 0; i < g->num_faces; i++) { |
| 3620 | if (state->clues[i] >= 0) { |
| 3621 | face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh); |
| 3622 | if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) |
| 3623 | game_redraw_clue(dr, ds, state, i); |
| 3624 | } |
| 3625 | } |
| 3626 | for (phase = 0; phase < NPHASES; phase++) { |
| 3627 | for (i = 0; i < g->num_edges; i++) { |
| 3628 | edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh); |
| 3629 | if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) |
| 3630 | game_redraw_line(dr, ds, state, i, phase); |
| 3631 | } |
| 3632 | } |
| 3633 | for (i = 0; i < g->num_dots; i++) { |
| 3634 | dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh); |
| 3635 | if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) |
| 3636 | game_redraw_dot(dr, ds, state, i); |
| 3637 | } |
| 3638 | |
| 3639 | unclip(dr); |
| 3640 | draw_update(dr, x, y, w, h); |
| 3641 | } |
| 3642 | |
| 3643 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
| 3644 | game_state *state, int dir, game_ui *ui, |
| 3645 | float animtime, float flashtime) |
| 3646 | { |
| 3647 | #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */ |
| 3648 | |
| 3649 | grid *g = state->game_grid; |
| 3650 | int border = BORDER(ds->tilesize); |
| 3651 | int i; |
| 3652 | int flash_changed; |
| 3653 | int redraw_everything = FALSE; |
| 3654 | |
| 3655 | int edges[REDRAW_OBJECTS_LIMIT], nedges = 0; |
| 3656 | int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0; |
| 3657 | |
| 3658 | /* Redrawing is somewhat involved. |
| 3659 | * |
| 3660 | * An update can theoretically affect an arbitrary number of edges |
| 3661 | * (consider, for example, completing or breaking a cycle which doesn't |
| 3662 | * satisfy all the clues -- we'll switch many edges between error and |
| 3663 | * normal states). On the other hand, redrawing the whole grid takes a |
| 3664 | * while, making the game feel sluggish, and many updates are actually |
| 3665 | * quite well localized. |
| 3666 | * |
| 3667 | * This redraw algorithm attempts to cope with both situations gracefully |
| 3668 | * and correctly. For localized changes, we set a clip rectangle, fill |
| 3669 | * it with background, and then redraw (a plausible but conservative |
| 3670 | * guess at) the objects which intersect the rectangle; if several |
| 3671 | * objects need redrawing, we'll do them individually. However, if lots |
| 3672 | * of objects are affected, we'll just redraw everything. |
| 3673 | * |
| 3674 | * The reason for all of this is that it's just not safe to do the redraw |
| 3675 | * piecemeal. If you try to draw an antialiased diagonal line over |
| 3676 | * itself, you get a slightly thicker antialiased diagonal line, which |
| 3677 | * looks rather ugly after a while. |
| 3678 | * |
| 3679 | * So, we take two passes over the grid. The first attempts to work out |
| 3680 | * what needs doing, and the second actually does it. |
| 3681 | */ |
| 3682 | |
| 3683 | if (!ds->started) |
| 3684 | redraw_everything = TRUE; |
| 3685 | else { |
| 3686 | |
| 3687 | /* First, trundle through the faces. */ |
| 3688 | for (i = 0; i < g->num_faces; i++) { |
| 3689 | grid_face *f = g->faces + i; |
| 3690 | int sides = f->order; |
| 3691 | int clue_mistake; |
| 3692 | int clue_satisfied; |
| 3693 | int n = state->clues[i]; |
| 3694 | if (n < 0) |
| 3695 | continue; |
| 3696 | |
| 3697 | clue_mistake = (face_order(state, i, LINE_YES) > n || |
| 3698 | face_order(state, i, LINE_NO ) > (sides-n)); |
| 3699 | clue_satisfied = (face_order(state, i, LINE_YES) == n && |
| 3700 | face_order(state, i, LINE_NO ) == (sides-n)); |
| 3701 | |
| 3702 | if (clue_mistake != ds->clue_error[i] || |
| 3703 | clue_satisfied != ds->clue_satisfied[i]) { |
| 3704 | ds->clue_error[i] = clue_mistake; |
| 3705 | ds->clue_satisfied[i] = clue_satisfied; |
| 3706 | if (nfaces == REDRAW_OBJECTS_LIMIT) |
| 3707 | redraw_everything = TRUE; |
| 3708 | else |
| 3709 | faces[nfaces++] = i; |
| 3710 | } |
| 3711 | } |
| 3712 | |
| 3713 | /* Work out what the flash state needs to be. */ |
| 3714 | if (flashtime > 0 && |
| 3715 | (flashtime <= FLASH_TIME/3 || |
| 3716 | flashtime >= FLASH_TIME*2/3)) { |
| 3717 | flash_changed = !ds->flashing; |
| 3718 | ds->flashing = TRUE; |
| 3719 | } else { |
| 3720 | flash_changed = ds->flashing; |
| 3721 | ds->flashing = FALSE; |
| 3722 | } |
| 3723 | |
| 3724 | /* Now, trundle through the edges. */ |
| 3725 | for (i = 0; i < g->num_edges; i++) { |
| 3726 | char new_ds = |
| 3727 | state->line_errors[i] ? DS_LINE_ERROR : state->lines[i]; |
| 3728 | if (new_ds != ds->lines[i] || |
| 3729 | (flash_changed && state->lines[i] == LINE_YES)) { |
| 3730 | ds->lines[i] = new_ds; |
| 3731 | if (nedges == REDRAW_OBJECTS_LIMIT) |
| 3732 | redraw_everything = TRUE; |
| 3733 | else |
| 3734 | edges[nedges++] = i; |
| 3735 | } |
| 3736 | } |
| 3737 | } |
| 3738 | |
| 3739 | /* Pass one is now done. Now we do the actual drawing. */ |
| 3740 | if (redraw_everything) { |
| 3741 | int grid_width = g->highest_x - g->lowest_x; |
| 3742 | int grid_height = g->highest_y - g->lowest_y; |
| 3743 | int w = grid_width * ds->tilesize / g->tilesize; |
| 3744 | int h = grid_height * ds->tilesize / g->tilesize; |
| 3745 | |
| 3746 | game_redraw_in_rect(dr, ds, state, |
| 3747 | 0, 0, w + 2*border + 1, h + 2*border + 1); |
| 3748 | } else { |
| 3749 | |
| 3750 | /* Right. Now we roll up our sleeves. */ |
| 3751 | |
| 3752 | for (i = 0; i < nfaces; i++) { |
| 3753 | grid_face *f = g->faces + faces[i]; |
| 3754 | int x, y, w, h; |
| 3755 | |
| 3756 | face_text_bbox(ds, g, f, &x, &y, &w, &h); |
| 3757 | game_redraw_in_rect(dr, ds, state, x, y, w, h); |
| 3758 | } |
| 3759 | |
| 3760 | for (i = 0; i < nedges; i++) { |
| 3761 | grid_edge *e = g->edges + edges[i]; |
| 3762 | int x, y, w, h; |
| 3763 | |
| 3764 | edge_bbox(ds, g, e, &x, &y, &w, &h); |
| 3765 | game_redraw_in_rect(dr, ds, state, x, y, w, h); |
| 3766 | } |
| 3767 | } |
| 3768 | |
| 3769 | ds->started = TRUE; |
| 3770 | } |
| 3771 | |
| 3772 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
| 3773 | int dir, game_ui *ui) |
| 3774 | { |
| 3775 | if (!oldstate->solved && newstate->solved && |
| 3776 | !oldstate->cheated && !newstate->cheated) { |
| 3777 | return FLASH_TIME; |
| 3778 | } |
| 3779 | |
| 3780 | return 0.0F; |
| 3781 | } |
| 3782 | |
| 3783 | static int game_status(game_state *state) |
| 3784 | { |
| 3785 | return state->solved ? +1 : 0; |
| 3786 | } |
| 3787 | |
| 3788 | static void game_print_size(game_params *params, float *x, float *y) |
| 3789 | { |
| 3790 | int pw, ph; |
| 3791 | |
| 3792 | /* |
| 3793 | * I'll use 7mm "squares" by default. |
| 3794 | */ |
| 3795 | game_compute_size(params, 700, &pw, &ph); |
| 3796 | *x = pw / 100.0F; |
| 3797 | *y = ph / 100.0F; |
| 3798 | } |
| 3799 | |
| 3800 | static void game_print(drawing *dr, game_state *state, int tilesize) |
| 3801 | { |
| 3802 | int ink = print_mono_colour(dr, 0); |
| 3803 | int i; |
| 3804 | game_drawstate ads, *ds = &ads; |
| 3805 | grid *g = state->game_grid; |
| 3806 | |
| 3807 | ds->tilesize = tilesize; |
| 3808 | ds->textx = snewn(g->num_faces, int); |
| 3809 | ds->texty = snewn(g->num_faces, int); |
| 3810 | for (i = 0; i < g->num_faces; i++) |
| 3811 | ds->textx[i] = ds->texty[i] = -1; |
| 3812 | |
| 3813 | for (i = 0; i < g->num_dots; i++) { |
| 3814 | int x, y; |
| 3815 | grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y); |
| 3816 | draw_circle(dr, x, y, ds->tilesize / 15, ink, ink); |
| 3817 | } |
| 3818 | |
| 3819 | /* |
| 3820 | * Clues. |
| 3821 | */ |
| 3822 | for (i = 0; i < g->num_faces; i++) { |
| 3823 | grid_face *f = g->faces + i; |
| 3824 | int clue = state->clues[i]; |
| 3825 | if (clue >= 0) { |
| 3826 | char c[2]; |
| 3827 | int x, y; |
| 3828 | c[0] = CLUE2CHAR(clue); |
| 3829 | c[1] = '\0'; |
| 3830 | face_text_pos(ds, g, f, &x, &y); |
| 3831 | draw_text(dr, x, y, |
| 3832 | FONT_VARIABLE, ds->tilesize / 2, |
| 3833 | ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c); |
| 3834 | } |
| 3835 | } |
| 3836 | |
| 3837 | /* |
| 3838 | * Lines. |
| 3839 | */ |
| 3840 | for (i = 0; i < g->num_edges; i++) { |
| 3841 | int thickness = (state->lines[i] == LINE_YES) ? 30 : 150; |
| 3842 | grid_edge *e = g->edges + i; |
| 3843 | int x1, y1, x2, y2; |
| 3844 | grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1); |
| 3845 | grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2); |
| 3846 | if (state->lines[i] == LINE_YES) |
| 3847 | { |
| 3848 | /* (dx, dy) points from (x1, y1) to (x2, y2). |
| 3849 | * The line is then "fattened" in a perpendicular |
| 3850 | * direction to create a thin rectangle. */ |
| 3851 | double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2)); |
| 3852 | double dx = (x2 - x1) / d; |
| 3853 | double dy = (y2 - y1) / d; |
| 3854 | int points[8]; |
| 3855 | |
| 3856 | dx = (dx * ds->tilesize) / thickness; |
| 3857 | dy = (dy * ds->tilesize) / thickness; |
| 3858 | points[0] = x1 + (int)dy; |
| 3859 | points[1] = y1 - (int)dx; |
| 3860 | points[2] = x1 - (int)dy; |
| 3861 | points[3] = y1 + (int)dx; |
| 3862 | points[4] = x2 - (int)dy; |
| 3863 | points[5] = y2 + (int)dx; |
| 3864 | points[6] = x2 + (int)dy; |
| 3865 | points[7] = y2 - (int)dx; |
| 3866 | draw_polygon(dr, points, 4, ink, ink); |
| 3867 | } |
| 3868 | else |
| 3869 | { |
| 3870 | /* Draw a dotted line */ |
| 3871 | int divisions = 6; |
| 3872 | int j; |
| 3873 | for (j = 1; j < divisions; j++) { |
| 3874 | /* Weighted average */ |
| 3875 | int x = (x1 * (divisions -j) + x2 * j) / divisions; |
| 3876 | int y = (y1 * (divisions -j) + y2 * j) / divisions; |
| 3877 | draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink); |
| 3878 | } |
| 3879 | } |
| 3880 | } |
| 3881 | |
| 3882 | sfree(ds->textx); |
| 3883 | sfree(ds->texty); |
| 3884 | } |
| 3885 | |
| 3886 | #ifdef COMBINED |
| 3887 | #define thegame loopy |
| 3888 | #endif |
| 3889 | |
| 3890 | const struct game thegame = { |
| 3891 | "Loopy", "games.loopy", "loopy", |
| 3892 | default_params, |
| 3893 | game_fetch_preset, |
| 3894 | decode_params, |
| 3895 | encode_params, |
| 3896 | free_params, |
| 3897 | dup_params, |
| 3898 | TRUE, game_configure, custom_params, |
| 3899 | validate_params, |
| 3900 | new_game_desc, |
| 3901 | validate_desc, |
| 3902 | new_game, |
| 3903 | dup_game, |
| 3904 | free_game, |
| 3905 | 1, solve_game, |
| 3906 | TRUE, game_can_format_as_text_now, game_text_format, |
| 3907 | new_ui, |
| 3908 | free_ui, |
| 3909 | encode_ui, |
| 3910 | decode_ui, |
| 3911 | game_changed_state, |
| 3912 | interpret_move, |
| 3913 | execute_move, |
| 3914 | PREFERRED_TILE_SIZE, game_compute_size, game_set_size, |
| 3915 | game_colours, |
| 3916 | game_new_drawstate, |
| 3917 | game_free_drawstate, |
| 3918 | game_redraw, |
| 3919 | game_anim_length, |
| 3920 | game_flash_length, |
| 3921 | game_status, |
| 3922 | TRUE, FALSE, game_print_size, game_print, |
| 3923 | FALSE /* wants_statusbar */, |
| 3924 | FALSE, game_timing_state, |
| 3925 | 0, /* mouse_priorities */ |
| 3926 | }; |
| 3927 | |
| 3928 | #ifdef STANDALONE_SOLVER |
| 3929 | |
| 3930 | /* |
| 3931 | * Half-hearted standalone solver. It can't output the solution to |
| 3932 | * anything but a square puzzle, and it can't log the deductions |
| 3933 | * it makes either. But it can solve square puzzles, and more |
| 3934 | * importantly it can use its solver to grade the difficulty of |
| 3935 | * any puzzle you give it. |
| 3936 | */ |
| 3937 | |
| 3938 | #include <stdarg.h> |
| 3939 | |
| 3940 | int main(int argc, char **argv) |
| 3941 | { |
| 3942 | game_params *p; |
| 3943 | game_state *s; |
| 3944 | char *id = NULL, *desc, *err; |
| 3945 | int grade = FALSE; |
| 3946 | int ret, diff; |
| 3947 | #if 0 /* verbose solver not supported here (yet) */ |
| 3948 | int really_verbose = FALSE; |
| 3949 | #endif |
| 3950 | |
| 3951 | while (--argc > 0) { |
| 3952 | char *p = *++argv; |
| 3953 | #if 0 /* verbose solver not supported here (yet) */ |
| 3954 | if (!strcmp(p, "-v")) { |
| 3955 | really_verbose = TRUE; |
| 3956 | } else |
| 3957 | #endif |
| 3958 | if (!strcmp(p, "-g")) { |
| 3959 | grade = TRUE; |
| 3960 | } else if (*p == '-') { |
| 3961 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
| 3962 | return 1; |
| 3963 | } else { |
| 3964 | id = p; |
| 3965 | } |
| 3966 | } |
| 3967 | |
| 3968 | if (!id) { |
| 3969 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
| 3970 | return 1; |
| 3971 | } |
| 3972 | |
| 3973 | desc = strchr(id, ':'); |
| 3974 | if (!desc) { |
| 3975 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
| 3976 | return 1; |
| 3977 | } |
| 3978 | *desc++ = '\0'; |
| 3979 | |
| 3980 | p = default_params(); |
| 3981 | decode_params(p, id); |
| 3982 | err = validate_desc(p, desc); |
| 3983 | if (err) { |
| 3984 | fprintf(stderr, "%s: %s\n", argv[0], err); |
| 3985 | return 1; |
| 3986 | } |
| 3987 | s = new_game(NULL, p, desc); |
| 3988 | |
| 3989 | /* |
| 3990 | * When solving an Easy puzzle, we don't want to bother the |
| 3991 | * user with Hard-level deductions. For this reason, we grade |
| 3992 | * the puzzle internally before doing anything else. |
| 3993 | */ |
| 3994 | ret = -1; /* placate optimiser */ |
| 3995 | for (diff = 0; diff < DIFF_MAX; diff++) { |
| 3996 | solver_state *sstate_new; |
| 3997 | solver_state *sstate = new_solver_state((game_state *)s, diff); |
| 3998 | |
| 3999 | sstate_new = solve_game_rec(sstate); |
| 4000 | |
| 4001 | if (sstate_new->solver_status == SOLVER_MISTAKE) |
| 4002 | ret = 0; |
| 4003 | else if (sstate_new->solver_status == SOLVER_SOLVED) |
| 4004 | ret = 1; |
| 4005 | else |
| 4006 | ret = 2; |
| 4007 | |
| 4008 | free_solver_state(sstate_new); |
| 4009 | free_solver_state(sstate); |
| 4010 | |
| 4011 | if (ret < 2) |
| 4012 | break; |
| 4013 | } |
| 4014 | |
| 4015 | if (diff == DIFF_MAX) { |
| 4016 | if (grade) |
| 4017 | printf("Difficulty rating: harder than Hard, or ambiguous\n"); |
| 4018 | else |
| 4019 | printf("Unable to find a unique solution\n"); |
| 4020 | } else { |
| 4021 | if (grade) { |
| 4022 | if (ret == 0) |
| 4023 | printf("Difficulty rating: impossible (no solution exists)\n"); |
| 4024 | else if (ret == 1) |
| 4025 | printf("Difficulty rating: %s\n", diffnames[diff]); |
| 4026 | } else { |
| 4027 | solver_state *sstate_new; |
| 4028 | solver_state *sstate = new_solver_state((game_state *)s, diff); |
| 4029 | |
| 4030 | /* If we supported a verbose solver, we'd set verbosity here */ |
| 4031 | |
| 4032 | sstate_new = solve_game_rec(sstate); |
| 4033 | |
| 4034 | if (sstate_new->solver_status == SOLVER_MISTAKE) |
| 4035 | printf("Puzzle is inconsistent\n"); |
| 4036 | else { |
| 4037 | assert(sstate_new->solver_status == SOLVER_SOLVED); |
| 4038 | if (s->grid_type == 0) { |
| 4039 | fputs(game_text_format(sstate_new->state), stdout); |
| 4040 | } else { |
| 4041 | printf("Unable to output non-square grids\n"); |
| 4042 | } |
| 4043 | } |
| 4044 | |
| 4045 | free_solver_state(sstate_new); |
| 4046 | free_solver_state(sstate); |
| 4047 | } |
| 4048 | } |
| 4049 | |
| 4050 | return 0; |
| 4051 | } |
| 4052 | |
| 4053 | #endif |
| 4054 | |
| 4055 | /* vim: set shiftwidth=4 tabstop=8: */ |