| 1 | /* |
| 2 | * tents.c: Puzzle involving placing tents next to trees subject to |
| 3 | * some confusing conditions. |
| 4 | * |
| 5 | * TODO: |
| 6 | * |
| 7 | * - it might be nice to make setter-provided tent/nontent clues |
| 8 | * inviolable? |
| 9 | * * on the other hand, this would introduce considerable extra |
| 10 | * complexity and size into the game state; also inviolable |
| 11 | * clues would have to be marked as such somehow, in an |
| 12 | * intrusive and annoying manner. Since they're never |
| 13 | * generated by _my_ generator, I'm currently more inclined |
| 14 | * not to bother. |
| 15 | * |
| 16 | * - more difficult levels at the top end? |
| 17 | * * for example, sometimes we can deduce that two BLANKs in |
| 18 | * the same row are each adjacent to the same unattached tree |
| 19 | * and to nothing else, implying that they can't both be |
| 20 | * tents; this enables us to rule out some extra combinations |
| 21 | * in the row-based deduction loop, and hence deduce more |
| 22 | * from the number in that row than we could otherwise do. |
| 23 | * * that by itself doesn't seem worth implementing a new |
| 24 | * difficulty level for, but if I can find a few more things |
| 25 | * like that then it might become worthwhile. |
| 26 | * * I wonder if there's a sensible heuristic for where to |
| 27 | * guess which would make a recursive solver viable? |
| 28 | */ |
| 29 | |
| 30 | #include <stdio.h> |
| 31 | #include <stdlib.h> |
| 32 | #include <string.h> |
| 33 | #include <assert.h> |
| 34 | #include <ctype.h> |
| 35 | #include <math.h> |
| 36 | |
| 37 | #include "puzzles.h" |
| 38 | #include "maxflow.h" |
| 39 | |
| 40 | /* |
| 41 | * Design discussion |
| 42 | * ----------------- |
| 43 | * |
| 44 | * The rules of this puzzle as available on the WWW are poorly |
| 45 | * specified. The bits about tents having to be orthogonally |
| 46 | * adjacent to trees, tents not being even diagonally adjacent to |
| 47 | * one another, and the number of tents in each row and column |
| 48 | * being given are simple enough; the difficult bit is the |
| 49 | * tent-to-tree matching. |
| 50 | * |
| 51 | * Some sources use simplistic wordings such as `each tree is |
| 52 | * exactly connected to only one tent', which is extremely unclear: |
| 53 | * it's easy to read erroneously as `each tree is _orthogonally |
| 54 | * adjacent_ to exactly one tent', which is definitely incorrect. |
| 55 | * Even the most coherent sources I've found don't do a much better |
| 56 | * job of stating the rule. |
| 57 | * |
| 58 | * A more precise statement of the rule is that it must be possible |
| 59 | * to find a bijection f between tents and trees such that each |
| 60 | * tree T is orthogonally adjacent to the tent f(T), but that a |
| 61 | * tent is permitted to be adjacent to other trees in addition to |
| 62 | * its own. This slightly non-obvious criterion is what gives this |
| 63 | * puzzle most of its subtlety. |
| 64 | * |
| 65 | * However, there's a particularly subtle ambiguity left over. Is |
| 66 | * the bijection between tents and trees required to be _unique_? |
| 67 | * In other words, is that bijection conceptually something the |
| 68 | * player should be able to exhibit as part of the solution (even |
| 69 | * if they aren't actually required to do so)? Or is it sufficient |
| 70 | * to have a unique _placement_ of the tents which gives rise to at |
| 71 | * least one suitable bijection? |
| 72 | * |
| 73 | * The puzzle shown to the right of this .T. 2 *T* 2 |
| 74 | * paragraph illustrates the problem. There T.T 0 -> T-T 0 |
| 75 | * are two distinct bijections available. .T. 2 *T* 2 |
| 76 | * The answer to the above question will |
| 77 | * determine whether it's a valid puzzle. 202 202 |
| 78 | * |
| 79 | * This is an important question, because it affects both the |
| 80 | * player and the generator. Eventually I found all the instances |
| 81 | * of this puzzle I could Google up, solved them all by hand, and |
| 82 | * verified that in all cases the tree/tent matching was uniquely |
| 83 | * determined given the tree and tent positions. Therefore, the |
| 84 | * puzzle as implemented in this source file takes the following |
| 85 | * policy: |
| 86 | * |
| 87 | * - When checking a user-supplied solution for correctness, only |
| 88 | * verify that there exists _at least_ one matching. |
| 89 | * - When generating a puzzle, enforce that there must be |
| 90 | * _exactly_ one. |
| 91 | * |
| 92 | * Algorithmic implications |
| 93 | * ------------------------ |
| 94 | * |
| 95 | * Another way of phrasing the tree/tent matching criterion is to |
| 96 | * say that the bipartite adjacency graph between trees and tents |
| 97 | * has a perfect matching. That is, if you construct a graph which |
| 98 | * has a vertex per tree and a vertex per tent, and an edge between |
| 99 | * any tree and tent which are orthogonally adjacent, it is |
| 100 | * possible to find a set of N edges of that graph (where N is the |
| 101 | * number of trees and also the number of tents) which between them |
| 102 | * connect every tree to every tent. |
| 103 | * |
| 104 | * The most efficient known algorithms for finding such a matching |
| 105 | * given a graph, as far as I'm aware, are the Munkres assignment |
| 106 | * algorithm (also known as the Hungarian algorithm) and the |
| 107 | * Ford-Fulkerson algorithm (for finding optimal flows in |
| 108 | * networks). Each of these takes O(N^3) running time; so we're |
| 109 | * talking O(N^3) time to verify any candidate solution to this |
| 110 | * puzzle. That's just about OK if you're doing it once per mouse |
| 111 | * click (and in fact not even that, since the sensible thing to do |
| 112 | * is check all the _other_ puzzle criteria and only wade into this |
| 113 | * quagmire if none are violated); but if the solver had to keep |
| 114 | * doing N^3 work internally, then it would probably end up with |
| 115 | * more like N^5 or N^6 running time, and grid generation would |
| 116 | * become very clunky. |
| 117 | * |
| 118 | * Fortunately, I've been able to prove a very useful property of |
| 119 | * _unique_ perfect matchings, by adapting the proof of Hall's |
| 120 | * Marriage Theorem. For those unaware of Hall's Theorem, I'll |
| 121 | * recap it and its proof: it states that a bipartite graph |
| 122 | * contains a perfect matching iff every set of vertices on the |
| 123 | * left side of the graph have a neighbourhood _at least_ as big on |
| 124 | * the right. |
| 125 | * |
| 126 | * This condition is obviously satisfied if a perfect matching does |
| 127 | * exist; each left-side node has a distinct right-side node which |
| 128 | * is the one assigned to it by the matching, and thus any set of n |
| 129 | * left vertices must have a combined neighbourhood containing at |
| 130 | * least the n corresponding right vertices, and possibly others |
| 131 | * too. Alternatively, imagine if you had (say) three left-side |
| 132 | * nodes all of which were connected to only two right-side nodes |
| 133 | * between them: any perfect matching would have to assign one of |
| 134 | * those two right nodes to each of the three left nodes, and still |
| 135 | * give the three left nodes a different right node each. This is |
| 136 | * of course impossible. |
| 137 | * |
| 138 | * To prove the converse (that if every subset of left vertices |
| 139 | * satisfies the Hall condition then a perfect matching exists), |
| 140 | * consider trying to find a proper subset of the left vertices |
| 141 | * which _exactly_ satisfies the Hall condition: that is, its right |
| 142 | * neighbourhood is precisely the same size as it. If we can find |
| 143 | * such a subset, then we can split the bipartite graph into two |
| 144 | * smaller ones: one consisting of the left subset and its right |
| 145 | * neighbourhood, the other consisting of everything else. Edges |
| 146 | * from the left side of the former graph to the right side of the |
| 147 | * latter do not exist, by construction; edges from the right side |
| 148 | * of the former to the left of the latter cannot be part of any |
| 149 | * perfect matching because otherwise the left subset would not be |
| 150 | * left with enough distinct right vertices to connect to (this is |
| 151 | * exactly the same deduction used in Solo's set analysis). You can |
| 152 | * then prove (left as an exercise) that both these smaller graphs |
| 153 | * still satisfy the Hall condition, and therefore the proof will |
| 154 | * follow by induction. |
| 155 | * |
| 156 | * There's one other possibility, which is the case where _no_ |
| 157 | * proper subset of the left vertices has a right neighbourhood of |
| 158 | * exactly the same size. That is, every left subset has a strictly |
| 159 | * _larger_ right neighbourhood. In this situation, we can simply |
| 160 | * remove an _arbitrary_ edge from the graph. This cannot reduce |
| 161 | * the size of any left subset's right neighbourhood by more than |
| 162 | * one, so if all neighbourhoods were strictly bigger than they |
| 163 | * needed to be initially, they must now still be _at least as big_ |
| 164 | * as they need to be. So we can keep throwing out arbitrary edges |
| 165 | * until we find a set which exactly satisfies the Hall condition, |
| 166 | * and then proceed as above. [] |
| 167 | * |
| 168 | * That's Hall's theorem. I now build on this by examining the |
| 169 | * circumstances in which a bipartite graph can have a _unique_ |
| 170 | * perfect matching. It is clear that in the second case, where no |
| 171 | * left subset exactly satisfies the Hall condition and so we can |
| 172 | * remove an arbitrary edge, there cannot be a unique perfect |
| 173 | * matching: given one perfect matching, we choose our arbitrary |
| 174 | * removed edge to be one of those contained in it, and then we can |
| 175 | * still find a perfect matching in the remaining graph, which will |
| 176 | * be a distinct perfect matching in the original. |
| 177 | * |
| 178 | * So it is a necessary condition for a unique perfect matching |
| 179 | * that there must be at least one proper left subset which |
| 180 | * _exactly_ satisfies the Hall condition. But now consider the |
| 181 | * smaller graph constructed by taking that left subset and its |
| 182 | * neighbourhood: if the graph as a whole had a unique perfect |
| 183 | * matching, then so must this smaller one, which means we can find |
| 184 | * a proper left subset _again_, and so on. Repeating this process |
| 185 | * must eventually reduce us to a graph with only one left-side |
| 186 | * vertex (so there are no proper subsets at all); this vertex must |
| 187 | * be connected to only one right-side vertex, and hence must be so |
| 188 | * in the original graph as well (by construction). So we can |
| 189 | * discard this vertex pair from the graph, and any other edges |
| 190 | * that involved it (which will by construction be from other left |
| 191 | * vertices only), and the resulting smaller graph still has a |
| 192 | * unique perfect matching which means we can do the same thing |
| 193 | * again. |
| 194 | * |
| 195 | * In other words, given any bipartite graph with a unique perfect |
| 196 | * matching, we can find that matching by the following extremely |
| 197 | * simple algorithm: |
| 198 | * |
| 199 | * - Find a left-side vertex which is only connected to one |
| 200 | * right-side vertex. |
| 201 | * - Assign those vertices to one another, and therefore discard |
| 202 | * any other edges connecting to that right vertex. |
| 203 | * - Repeat until all vertices have been matched. |
| 204 | * |
| 205 | * This algorithm can be run in O(V+E) time (where V is the number |
| 206 | * of vertices and E is the number of edges in the graph), and the |
| 207 | * only way it can fail is if there is not a unique perfect |
| 208 | * matching (either because there is no matching at all, or because |
| 209 | * it isn't unique; but it can't distinguish those cases). |
| 210 | * |
| 211 | * Thus, the internal solver in this source file can be confident |
| 212 | * that if the tree/tent matching is uniquely determined by the |
| 213 | * tree and tent positions, it can find it using only this kind of |
| 214 | * obvious and simple operation: assign a tree to a tent if it |
| 215 | * cannot possibly belong to any other tent, and vice versa. If the |
| 216 | * solver were _only_ trying to determine the matching, even that |
| 217 | * `vice versa' wouldn't be required; but it can come in handy when |
| 218 | * not all the tents have been placed yet. I can therefore be |
| 219 | * reasonably confident that as long as my solver doesn't need to |
| 220 | * cope with grids that have a non-unique matching, it will also |
| 221 | * not need to do anything complicated like set analysis between |
| 222 | * trees and tents. |
| 223 | */ |
| 224 | |
| 225 | /* |
| 226 | * In standalone solver mode, `verbose' is a variable which can be |
| 227 | * set by command-line option; in debugging mode it's simply always |
| 228 | * true. |
| 229 | */ |
| 230 | #if defined STANDALONE_SOLVER |
| 231 | #define SOLVER_DIAGNOSTICS |
| 232 | int verbose = FALSE; |
| 233 | #elif defined SOLVER_DIAGNOSTICS |
| 234 | #define verbose TRUE |
| 235 | #endif |
| 236 | |
| 237 | /* |
| 238 | * Difficulty levels. I do some macro ickery here to ensure that my |
| 239 | * enum and the various forms of my name list always match up. |
| 240 | */ |
| 241 | #define DIFFLIST(A) \ |
| 242 | A(EASY,Easy,e) \ |
| 243 | A(TRICKY,Tricky,t) |
| 244 | #define ENUM(upper,title,lower) DIFF_ ## upper, |
| 245 | #define TITLE(upper,title,lower) #title, |
| 246 | #define ENCODE(upper,title,lower) #lower |
| 247 | #define CONFIG(upper,title,lower) ":" #title |
| 248 | enum { DIFFLIST(ENUM) DIFFCOUNT }; |
| 249 | static char const *const tents_diffnames[] = { DIFFLIST(TITLE) }; |
| 250 | static char const tents_diffchars[] = DIFFLIST(ENCODE); |
| 251 | #define DIFFCONFIG DIFFLIST(CONFIG) |
| 252 | |
| 253 | enum { |
| 254 | COL_BACKGROUND, |
| 255 | COL_GRID, |
| 256 | COL_GRASS, |
| 257 | COL_TREETRUNK, |
| 258 | COL_TREELEAF, |
| 259 | COL_TENT, |
| 260 | COL_ERROR, |
| 261 | COL_ERRTEXT, |
| 262 | COL_ERRTRUNK, |
| 263 | NCOLOURS |
| 264 | }; |
| 265 | |
| 266 | enum { BLANK, TREE, TENT, NONTENT, MAGIC }; |
| 267 | |
| 268 | struct game_params { |
| 269 | int w, h; |
| 270 | int diff; |
| 271 | }; |
| 272 | |
| 273 | struct numbers { |
| 274 | int refcount; |
| 275 | int *numbers; |
| 276 | }; |
| 277 | |
| 278 | struct game_state { |
| 279 | game_params p; |
| 280 | char *grid; |
| 281 | struct numbers *numbers; |
| 282 | int completed, used_solve; |
| 283 | }; |
| 284 | |
| 285 | static game_params *default_params(void) |
| 286 | { |
| 287 | game_params *ret = snew(game_params); |
| 288 | |
| 289 | ret->w = ret->h = 8; |
| 290 | ret->diff = DIFF_EASY; |
| 291 | |
| 292 | return ret; |
| 293 | } |
| 294 | |
| 295 | static const struct game_params tents_presets[] = { |
| 296 | {8, 8, DIFF_EASY}, |
| 297 | {8, 8, DIFF_TRICKY}, |
| 298 | {10, 10, DIFF_EASY}, |
| 299 | {10, 10, DIFF_TRICKY}, |
| 300 | {15, 15, DIFF_EASY}, |
| 301 | {15, 15, DIFF_TRICKY}, |
| 302 | }; |
| 303 | |
| 304 | static int game_fetch_preset(int i, char **name, game_params **params) |
| 305 | { |
| 306 | game_params *ret; |
| 307 | char str[80]; |
| 308 | |
| 309 | if (i < 0 || i >= lenof(tents_presets)) |
| 310 | return FALSE; |
| 311 | |
| 312 | ret = snew(game_params); |
| 313 | *ret = tents_presets[i]; |
| 314 | |
| 315 | sprintf(str, "%dx%d %s", ret->w, ret->h, tents_diffnames[ret->diff]); |
| 316 | |
| 317 | *name = dupstr(str); |
| 318 | *params = ret; |
| 319 | return TRUE; |
| 320 | } |
| 321 | |
| 322 | static void free_params(game_params *params) |
| 323 | { |
| 324 | sfree(params); |
| 325 | } |
| 326 | |
| 327 | static game_params *dup_params(game_params *params) |
| 328 | { |
| 329 | game_params *ret = snew(game_params); |
| 330 | *ret = *params; /* structure copy */ |
| 331 | return ret; |
| 332 | } |
| 333 | |
| 334 | static void decode_params(game_params *params, char const *string) |
| 335 | { |
| 336 | params->w = params->h = atoi(string); |
| 337 | while (*string && isdigit((unsigned char)*string)) string++; |
| 338 | if (*string == 'x') { |
| 339 | string++; |
| 340 | params->h = atoi(string); |
| 341 | while (*string && isdigit((unsigned char)*string)) string++; |
| 342 | } |
| 343 | if (*string == 'd') { |
| 344 | int i; |
| 345 | string++; |
| 346 | for (i = 0; i < DIFFCOUNT; i++) |
| 347 | if (*string == tents_diffchars[i]) |
| 348 | params->diff = i; |
| 349 | if (*string) string++; |
| 350 | } |
| 351 | } |
| 352 | |
| 353 | static char *encode_params(game_params *params, int full) |
| 354 | { |
| 355 | char buf[120]; |
| 356 | |
| 357 | sprintf(buf, "%dx%d", params->w, params->h); |
| 358 | if (full) |
| 359 | sprintf(buf + strlen(buf), "d%c", |
| 360 | tents_diffchars[params->diff]); |
| 361 | return dupstr(buf); |
| 362 | } |
| 363 | |
| 364 | static config_item *game_configure(game_params *params) |
| 365 | { |
| 366 | config_item *ret; |
| 367 | char buf[80]; |
| 368 | |
| 369 | ret = snewn(4, config_item); |
| 370 | |
| 371 | ret[0].name = "Width"; |
| 372 | ret[0].type = C_STRING; |
| 373 | sprintf(buf, "%d", params->w); |
| 374 | ret[0].sval = dupstr(buf); |
| 375 | ret[0].ival = 0; |
| 376 | |
| 377 | ret[1].name = "Height"; |
| 378 | ret[1].type = C_STRING; |
| 379 | sprintf(buf, "%d", params->h); |
| 380 | ret[1].sval = dupstr(buf); |
| 381 | ret[1].ival = 0; |
| 382 | |
| 383 | ret[2].name = "Difficulty"; |
| 384 | ret[2].type = C_CHOICES; |
| 385 | ret[2].sval = DIFFCONFIG; |
| 386 | ret[2].ival = params->diff; |
| 387 | |
| 388 | ret[3].name = NULL; |
| 389 | ret[3].type = C_END; |
| 390 | ret[3].sval = NULL; |
| 391 | ret[3].ival = 0; |
| 392 | |
| 393 | return ret; |
| 394 | } |
| 395 | |
| 396 | static game_params *custom_params(config_item *cfg) |
| 397 | { |
| 398 | game_params *ret = snew(game_params); |
| 399 | |
| 400 | ret->w = atoi(cfg[0].sval); |
| 401 | ret->h = atoi(cfg[1].sval); |
| 402 | ret->diff = cfg[2].ival; |
| 403 | |
| 404 | return ret; |
| 405 | } |
| 406 | |
| 407 | static char *validate_params(game_params *params, int full) |
| 408 | { |
| 409 | /* |
| 410 | * Generating anything under 4x4 runs into trouble of one kind |
| 411 | * or another. |
| 412 | */ |
| 413 | if (params->w < 4 || params->h < 4) |
| 414 | return "Width and height must both be at least four"; |
| 415 | return NULL; |
| 416 | } |
| 417 | |
| 418 | /* |
| 419 | * Scratch space for solver. |
| 420 | */ |
| 421 | enum { N, U, L, R, D, MAXDIR }; /* link directions */ |
| 422 | #define dx(d) ( ((d)==R) - ((d)==L) ) |
| 423 | #define dy(d) ( ((d)==D) - ((d)==U) ) |
| 424 | #define F(d) ( U + D - (d) ) |
| 425 | struct solver_scratch { |
| 426 | char *links; /* mapping between trees and tents */ |
| 427 | int *locs; |
| 428 | char *place, *mrows, *trows; |
| 429 | }; |
| 430 | |
| 431 | static struct solver_scratch *new_scratch(int w, int h) |
| 432 | { |
| 433 | struct solver_scratch *ret = snew(struct solver_scratch); |
| 434 | |
| 435 | ret->links = snewn(w*h, char); |
| 436 | ret->locs = snewn(max(w, h), int); |
| 437 | ret->place = snewn(max(w, h), char); |
| 438 | ret->mrows = snewn(3 * max(w, h), char); |
| 439 | ret->trows = snewn(3 * max(w, h), char); |
| 440 | |
| 441 | return ret; |
| 442 | } |
| 443 | |
| 444 | static void free_scratch(struct solver_scratch *sc) |
| 445 | { |
| 446 | sfree(sc->trows); |
| 447 | sfree(sc->mrows); |
| 448 | sfree(sc->place); |
| 449 | sfree(sc->locs); |
| 450 | sfree(sc->links); |
| 451 | sfree(sc); |
| 452 | } |
| 453 | |
| 454 | /* |
| 455 | * Solver. Returns 0 for impossibility, 1 for success, 2 for |
| 456 | * ambiguity or failure to converge. |
| 457 | */ |
| 458 | static int tents_solve(int w, int h, const char *grid, int *numbers, |
| 459 | char *soln, struct solver_scratch *sc, int diff) |
| 460 | { |
| 461 | int x, y, d, i, j; |
| 462 | char *mrow, *mrow1, *mrow2, *trow, *trow1, *trow2; |
| 463 | |
| 464 | /* |
| 465 | * Set up solver data. |
| 466 | */ |
| 467 | memset(sc->links, N, w*h); |
| 468 | |
| 469 | /* |
| 470 | * Set up solution array. |
| 471 | */ |
| 472 | memcpy(soln, grid, w*h); |
| 473 | |
| 474 | /* |
| 475 | * Main solver loop. |
| 476 | */ |
| 477 | while (1) { |
| 478 | int done_something = FALSE; |
| 479 | |
| 480 | /* |
| 481 | * Any tent which has only one unattached tree adjacent to |
| 482 | * it can be tied to that tree. |
| 483 | */ |
| 484 | for (y = 0; y < h; y++) |
| 485 | for (x = 0; x < w; x++) |
| 486 | if (soln[y*w+x] == TENT && !sc->links[y*w+x]) { |
| 487 | int linkd = 0; |
| 488 | |
| 489 | for (d = 1; d < MAXDIR; d++) { |
| 490 | int x2 = x + dx(d), y2 = y + dy(d); |
| 491 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
| 492 | soln[y2*w+x2] == TREE && |
| 493 | !sc->links[y2*w+x2]) { |
| 494 | if (linkd) |
| 495 | break; /* found more than one */ |
| 496 | else |
| 497 | linkd = d; |
| 498 | } |
| 499 | } |
| 500 | |
| 501 | if (d == MAXDIR && linkd == 0) { |
| 502 | #ifdef SOLVER_DIAGNOSTICS |
| 503 | if (verbose) |
| 504 | printf("tent at %d,%d cannot link to anything\n", |
| 505 | x, y); |
| 506 | #endif |
| 507 | return 0; /* no solution exists */ |
| 508 | } else if (d == MAXDIR) { |
| 509 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
| 510 | |
| 511 | #ifdef SOLVER_DIAGNOSTICS |
| 512 | if (verbose) |
| 513 | printf("tent at %d,%d can only link to tree at" |
| 514 | " %d,%d\n", x, y, x2, y2); |
| 515 | #endif |
| 516 | |
| 517 | sc->links[y*w+x] = linkd; |
| 518 | sc->links[y2*w+x2] = F(linkd); |
| 519 | done_something = TRUE; |
| 520 | } |
| 521 | } |
| 522 | |
| 523 | if (done_something) |
| 524 | continue; |
| 525 | if (diff < 0) |
| 526 | break; /* don't do anything else! */ |
| 527 | |
| 528 | /* |
| 529 | * Mark a blank square as NONTENT if it is not orthogonally |
| 530 | * adjacent to any unmatched tree. |
| 531 | */ |
| 532 | for (y = 0; y < h; y++) |
| 533 | for (x = 0; x < w; x++) |
| 534 | if (soln[y*w+x] == BLANK) { |
| 535 | int can_be_tent = FALSE; |
| 536 | |
| 537 | for (d = 1; d < MAXDIR; d++) { |
| 538 | int x2 = x + dx(d), y2 = y + dy(d); |
| 539 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
| 540 | soln[y2*w+x2] == TREE && |
| 541 | !sc->links[y2*w+x2]) |
| 542 | can_be_tent = TRUE; |
| 543 | } |
| 544 | |
| 545 | if (!can_be_tent) { |
| 546 | #ifdef SOLVER_DIAGNOSTICS |
| 547 | if (verbose) |
| 548 | printf("%d,%d cannot be a tent (no adjacent" |
| 549 | " unmatched tree)\n", x, y); |
| 550 | #endif |
| 551 | soln[y*w+x] = NONTENT; |
| 552 | done_something = TRUE; |
| 553 | } |
| 554 | } |
| 555 | |
| 556 | if (done_something) |
| 557 | continue; |
| 558 | |
| 559 | /* |
| 560 | * Mark a blank square as NONTENT if it is (perhaps |
| 561 | * diagonally) adjacent to any other tent. |
| 562 | */ |
| 563 | for (y = 0; y < h; y++) |
| 564 | for (x = 0; x < w; x++) |
| 565 | if (soln[y*w+x] == BLANK) { |
| 566 | int dx, dy, imposs = FALSE; |
| 567 | |
| 568 | for (dy = -1; dy <= +1; dy++) |
| 569 | for (dx = -1; dx <= +1; dx++) |
| 570 | if (dy || dx) { |
| 571 | int x2 = x + dx, y2 = y + dy; |
| 572 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
| 573 | soln[y2*w+x2] == TENT) |
| 574 | imposs = TRUE; |
| 575 | } |
| 576 | |
| 577 | if (imposs) { |
| 578 | #ifdef SOLVER_DIAGNOSTICS |
| 579 | if (verbose) |
| 580 | printf("%d,%d cannot be a tent (adjacent tent)\n", |
| 581 | x, y); |
| 582 | #endif |
| 583 | soln[y*w+x] = NONTENT; |
| 584 | done_something = TRUE; |
| 585 | } |
| 586 | } |
| 587 | |
| 588 | if (done_something) |
| 589 | continue; |
| 590 | |
| 591 | /* |
| 592 | * Any tree which has exactly one {unattached tent, BLANK} |
| 593 | * adjacent to it must have its tent in that square. |
| 594 | */ |
| 595 | for (y = 0; y < h; y++) |
| 596 | for (x = 0; x < w; x++) |
| 597 | if (soln[y*w+x] == TREE && !sc->links[y*w+x]) { |
| 598 | int linkd = 0, linkd2 = 0, nd = 0; |
| 599 | |
| 600 | for (d = 1; d < MAXDIR; d++) { |
| 601 | int x2 = x + dx(d), y2 = y + dy(d); |
| 602 | if (!(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h)) |
| 603 | continue; |
| 604 | if (soln[y2*w+x2] == BLANK || |
| 605 | (soln[y2*w+x2] == TENT && !sc->links[y2*w+x2])) { |
| 606 | if (linkd) |
| 607 | linkd2 = d; |
| 608 | else |
| 609 | linkd = d; |
| 610 | nd++; |
| 611 | } |
| 612 | } |
| 613 | |
| 614 | if (nd == 0) { |
| 615 | #ifdef SOLVER_DIAGNOSTICS |
| 616 | if (verbose) |
| 617 | printf("tree at %d,%d cannot link to anything\n", |
| 618 | x, y); |
| 619 | #endif |
| 620 | return 0; /* no solution exists */ |
| 621 | } else if (nd == 1) { |
| 622 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
| 623 | |
| 624 | #ifdef SOLVER_DIAGNOSTICS |
| 625 | if (verbose) |
| 626 | printf("tree at %d,%d can only link to tent at" |
| 627 | " %d,%d\n", x, y, x2, y2); |
| 628 | #endif |
| 629 | soln[y2*w+x2] = TENT; |
| 630 | sc->links[y*w+x] = linkd; |
| 631 | sc->links[y2*w+x2] = F(linkd); |
| 632 | done_something = TRUE; |
| 633 | } else if (nd == 2 && (!dx(linkd) != !dx(linkd2)) && |
| 634 | diff >= DIFF_TRICKY) { |
| 635 | /* |
| 636 | * If there are two possible places where |
| 637 | * this tree's tent can go, and they are |
| 638 | * diagonally separated rather than being |
| 639 | * on opposite sides of the tree, then the |
| 640 | * square (other than the tree square) |
| 641 | * which is adjacent to both of them must |
| 642 | * be a non-tent. |
| 643 | */ |
| 644 | int x2 = x + dx(linkd) + dx(linkd2); |
| 645 | int y2 = y + dy(linkd) + dy(linkd2); |
| 646 | assert(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h); |
| 647 | if (soln[y2*w+x2] == BLANK) { |
| 648 | #ifdef SOLVER_DIAGNOSTICS |
| 649 | if (verbose) |
| 650 | printf("possible tent locations for tree at" |
| 651 | " %d,%d rule out tent at %d,%d\n", |
| 652 | x, y, x2, y2); |
| 653 | #endif |
| 654 | soln[y2*w+x2] = NONTENT; |
| 655 | done_something = TRUE; |
| 656 | } |
| 657 | } |
| 658 | } |
| 659 | |
| 660 | if (done_something) |
| 661 | continue; |
| 662 | |
| 663 | /* |
| 664 | * If localised deductions about the trees and tents |
| 665 | * themselves haven't helped us, it's time to resort to the |
| 666 | * numbers round the grid edge. For each row and column, we |
| 667 | * go through all possible combinations of locations for |
| 668 | * the unplaced tents, rule out any which have adjacent |
| 669 | * tents, and spot any square which is given the same state |
| 670 | * by all remaining combinations. |
| 671 | */ |
| 672 | for (i = 0; i < w+h; i++) { |
| 673 | int start, step, len, start1, start2, n, k; |
| 674 | |
| 675 | if (i < w) { |
| 676 | /* |
| 677 | * This is the number for a column. |
| 678 | */ |
| 679 | start = i; |
| 680 | step = w; |
| 681 | len = h; |
| 682 | if (i > 0) |
| 683 | start1 = start - 1; |
| 684 | else |
| 685 | start1 = -1; |
| 686 | if (i+1 < w) |
| 687 | start2 = start + 1; |
| 688 | else |
| 689 | start2 = -1; |
| 690 | } else { |
| 691 | /* |
| 692 | * This is the number for a row. |
| 693 | */ |
| 694 | start = (i-w)*w; |
| 695 | step = 1; |
| 696 | len = w; |
| 697 | if (i > w) |
| 698 | start1 = start - w; |
| 699 | else |
| 700 | start1 = -1; |
| 701 | if (i+1 < w+h) |
| 702 | start2 = start + w; |
| 703 | else |
| 704 | start2 = -1; |
| 705 | } |
| 706 | |
| 707 | if (diff < DIFF_TRICKY) { |
| 708 | /* |
| 709 | * In Easy mode, we don't look at the effect of one |
| 710 | * row on the next (i.e. ruling out a square if all |
| 711 | * possibilities for an adjacent row place a tent |
| 712 | * next to it). |
| 713 | */ |
| 714 | start1 = start2 = -1; |
| 715 | } |
| 716 | |
| 717 | k = numbers[i]; |
| 718 | |
| 719 | /* |
| 720 | * Count and store the locations of the free squares, |
| 721 | * and also count the number of tents already placed. |
| 722 | */ |
| 723 | n = 0; |
| 724 | for (j = 0; j < len; j++) { |
| 725 | if (soln[start+j*step] == TENT) |
| 726 | k--; /* one fewer tent to place */ |
| 727 | else if (soln[start+j*step] == BLANK) |
| 728 | sc->locs[n++] = j; |
| 729 | } |
| 730 | |
| 731 | if (n == 0) |
| 732 | continue; /* nothing left to do here */ |
| 733 | |
| 734 | /* |
| 735 | * Now we know we're placing k tents in n squares. Set |
| 736 | * up the first possibility. |
| 737 | */ |
| 738 | for (j = 0; j < n; j++) |
| 739 | sc->place[j] = (j < k ? TENT : NONTENT); |
| 740 | |
| 741 | /* |
| 742 | * We're aiming to find squares in this row which are |
| 743 | * invariant over all valid possibilities. Thus, we |
| 744 | * maintain the current state of that invariance. We |
| 745 | * start everything off at MAGIC to indicate that it |
| 746 | * hasn't been set up yet. |
| 747 | */ |
| 748 | mrow = sc->mrows; |
| 749 | mrow1 = sc->mrows + len; |
| 750 | mrow2 = sc->mrows + 2*len; |
| 751 | trow = sc->trows; |
| 752 | trow1 = sc->trows + len; |
| 753 | trow2 = sc->trows + 2*len; |
| 754 | memset(mrow, MAGIC, 3*len); |
| 755 | |
| 756 | /* |
| 757 | * And iterate over all possibilities. |
| 758 | */ |
| 759 | while (1) { |
| 760 | int p, valid; |
| 761 | |
| 762 | /* |
| 763 | * See if this possibility is valid. The only way |
| 764 | * it can fail to be valid is if it contains two |
| 765 | * adjacent tents. (Other forms of invalidity, such |
| 766 | * as containing a tent adjacent to one already |
| 767 | * placed, will have been dealt with already by |
| 768 | * other parts of the solver.) |
| 769 | */ |
| 770 | valid = TRUE; |
| 771 | for (j = 0; j+1 < n; j++) |
| 772 | if (sc->place[j] == TENT && |
| 773 | sc->place[j+1] == TENT && |
| 774 | sc->locs[j+1] == sc->locs[j]+1) { |
| 775 | valid = FALSE; |
| 776 | break; |
| 777 | } |
| 778 | |
| 779 | if (valid) { |
| 780 | /* |
| 781 | * Merge this valid combination into mrow. |
| 782 | */ |
| 783 | memset(trow, MAGIC, len); |
| 784 | memset(trow+len, BLANK, 2*len); |
| 785 | for (j = 0; j < n; j++) { |
| 786 | trow[sc->locs[j]] = sc->place[j]; |
| 787 | if (sc->place[j] == TENT) { |
| 788 | int jj; |
| 789 | for (jj = sc->locs[j]-1; jj <= sc->locs[j]+1; jj++) |
| 790 | if (jj >= 0 && jj < len) |
| 791 | trow1[jj] = trow2[jj] = NONTENT; |
| 792 | } |
| 793 | } |
| 794 | |
| 795 | for (j = 0; j < 3*len; j++) { |
| 796 | if (trow[j] == MAGIC) |
| 797 | continue; |
| 798 | if (mrow[j] == MAGIC || mrow[j] == trow[j]) { |
| 799 | /* |
| 800 | * Either this is the first valid |
| 801 | * placement we've found at all, or |
| 802 | * this square's contents are |
| 803 | * consistent with every previous valid |
| 804 | * combination. |
| 805 | */ |
| 806 | mrow[j] = trow[j]; |
| 807 | } else { |
| 808 | /* |
| 809 | * This square's contents fail to match |
| 810 | * what they were in a different |
| 811 | * combination, so we cannot deduce |
| 812 | * anything about this square. |
| 813 | */ |
| 814 | mrow[j] = BLANK; |
| 815 | } |
| 816 | } |
| 817 | } |
| 818 | |
| 819 | /* |
| 820 | * Find the next combination of k choices from n. |
| 821 | * We do this by finding the rightmost tent which |
| 822 | * can be moved one place right, doing so, and |
| 823 | * shunting all tents to the right of that as far |
| 824 | * left as they can go. |
| 825 | */ |
| 826 | p = 0; |
| 827 | for (j = n-1; j > 0; j--) { |
| 828 | if (sc->place[j] == TENT) |
| 829 | p++; |
| 830 | if (sc->place[j] == NONTENT && sc->place[j-1] == TENT) { |
| 831 | sc->place[j-1] = NONTENT; |
| 832 | sc->place[j] = TENT; |
| 833 | while (p--) |
| 834 | sc->place[++j] = TENT; |
| 835 | while (++j < n) |
| 836 | sc->place[j] = NONTENT; |
| 837 | break; |
| 838 | } |
| 839 | } |
| 840 | if (j <= 0) |
| 841 | break; /* we've finished */ |
| 842 | } |
| 843 | |
| 844 | /* |
| 845 | * It's just possible that _no_ placement was valid, in |
| 846 | * which case we have an internally inconsistent |
| 847 | * puzzle. |
| 848 | */ |
| 849 | if (mrow[sc->locs[0]] == MAGIC) |
| 850 | return 0; /* inconsistent */ |
| 851 | |
| 852 | /* |
| 853 | * Now go through mrow and see if there's anything |
| 854 | * we've deduced which wasn't already mentioned in soln. |
| 855 | */ |
| 856 | for (j = 0; j < len; j++) { |
| 857 | int whichrow; |
| 858 | |
| 859 | for (whichrow = 0; whichrow < 3; whichrow++) { |
| 860 | char *mthis = mrow + whichrow * len; |
| 861 | int tstart = (whichrow == 0 ? start : |
| 862 | whichrow == 1 ? start1 : start2); |
| 863 | if (tstart >= 0 && |
| 864 | mthis[j] != MAGIC && mthis[j] != BLANK && |
| 865 | soln[tstart+j*step] == BLANK) { |
| 866 | int pos = tstart+j*step; |
| 867 | |
| 868 | #ifdef SOLVER_DIAGNOSTICS |
| 869 | if (verbose) |
| 870 | printf("%s %d forces %s at %d,%d\n", |
| 871 | step==1 ? "row" : "column", |
| 872 | step==1 ? start/w : start, |
| 873 | mthis[j] == TENT ? "tent" : "non-tent", |
| 874 | pos % w, pos / w); |
| 875 | #endif |
| 876 | soln[pos] = mthis[j]; |
| 877 | done_something = TRUE; |
| 878 | } |
| 879 | } |
| 880 | } |
| 881 | } |
| 882 | |
| 883 | if (done_something) |
| 884 | continue; |
| 885 | |
| 886 | if (!done_something) |
| 887 | break; |
| 888 | } |
| 889 | |
| 890 | /* |
| 891 | * The solver has nothing further it can do. Return 1 if both |
| 892 | * soln and sc->links are completely filled in, or 2 otherwise. |
| 893 | */ |
| 894 | for (y = 0; y < h; y++) |
| 895 | for (x = 0; x < w; x++) { |
| 896 | if (soln[y*w+x] == BLANK) |
| 897 | return 2; |
| 898 | if (soln[y*w+x] != NONTENT && sc->links[y*w+x] == 0) |
| 899 | return 2; |
| 900 | } |
| 901 | |
| 902 | return 1; |
| 903 | } |
| 904 | |
| 905 | static char *new_game_desc(game_params *params, random_state *rs, |
| 906 | char **aux, int interactive) |
| 907 | { |
| 908 | int w = params->w, h = params->h; |
| 909 | int ntrees = w * h / 5; |
| 910 | char *grid = snewn(w*h, char); |
| 911 | char *puzzle = snewn(w*h, char); |
| 912 | int *numbers = snewn(w+h, int); |
| 913 | char *soln = snewn(w*h, char); |
| 914 | int *temp = snewn(2*w*h, int); |
| 915 | int maxedges = ntrees*4 + w*h; |
| 916 | int *edges = snewn(2*maxedges, int); |
| 917 | int *capacity = snewn(maxedges, int); |
| 918 | int *flow = snewn(maxedges, int); |
| 919 | struct solver_scratch *sc = new_scratch(w, h); |
| 920 | char *ret, *p; |
| 921 | int i, j, nedges; |
| 922 | |
| 923 | /* |
| 924 | * Since this puzzle has many global deductions and doesn't |
| 925 | * permit limited clue sets, generating grids for this puzzle |
| 926 | * is hard enough that I see no better option than to simply |
| 927 | * generate a solution and see if it's unique and has the |
| 928 | * required difficulty. This turns out to be computationally |
| 929 | * plausible as well. |
| 930 | * |
| 931 | * We chose our tree count (hence also tent count) by dividing |
| 932 | * the total grid area by five above. Why five? Well, w*h/4 is |
| 933 | * the maximum number of tents you can _possibly_ fit into the |
| 934 | * grid without violating the separation criterion, and to |
| 935 | * achieve that you are constrained to a very small set of |
| 936 | * possible layouts (the obvious one with a tent at every |
| 937 | * (even,even) coordinate, and trivial variations thereon). So |
| 938 | * if we reduce the tent count a bit more, we enable more |
| 939 | * random-looking placement; 5 turns out to be a plausible |
| 940 | * figure which yields sensible puzzles. Increasing the tent |
| 941 | * count would give puzzles whose solutions were too regimented |
| 942 | * and could be solved by the use of that knowledge (and would |
| 943 | * also take longer to find a viable placement); decreasing it |
| 944 | * would make the grids emptier and more boring. |
| 945 | * |
| 946 | * Actually generating a grid is a matter of first placing the |
| 947 | * tents, and then placing the trees by the use of maxflow |
| 948 | * (finding a distinct square adjacent to every tent). We do it |
| 949 | * this way round because otherwise satisfying the tent |
| 950 | * separation condition would become onerous: most randomly |
| 951 | * chosen tent layouts do not satisfy this condition, so we'd |
| 952 | * have gone to a lot of work before finding that a candidate |
| 953 | * layout was unusable. Instead, we place the tents first and |
| 954 | * ensure they meet the separation criterion _before_ doing |
| 955 | * lots of computation; this works much better. |
| 956 | * |
| 957 | * The maxflow algorithm is not randomised, so employed naively |
| 958 | * it would give rise to grids with clear structure and |
| 959 | * directional bias. Hence, I assign the network nodes as seen |
| 960 | * by maxflow to be a _random_ permutation of the squares of |
| 961 | * the grid, so that any bias shown by maxflow towards |
| 962 | * low-numbered nodes is turned into a random bias. |
| 963 | * |
| 964 | * This generation strategy can fail at many points, including |
| 965 | * as early as tent placement (if you get a bad random order in |
| 966 | * which to greedily try the grid squares, you won't even |
| 967 | * manage to find enough mutually non-adjacent squares to put |
| 968 | * the tents in). Then it can fail if maxflow doesn't manage to |
| 969 | * find a good enough matching (i.e. the tent placements don't |
| 970 | * admit any adequate tree placements); and finally it can fail |
| 971 | * if the solver finds that the problem has the wrong |
| 972 | * difficulty (including being actually non-unique). All of |
| 973 | * these, however, are insufficiently frequent to cause |
| 974 | * trouble. |
| 975 | */ |
| 976 | |
| 977 | if (params->diff > DIFF_EASY && params->w <= 4 && params->h <= 4) |
| 978 | params->diff = DIFF_EASY; /* downgrade to prevent tight loop */ |
| 979 | |
| 980 | while (1) { |
| 981 | /* |
| 982 | * Arrange the grid squares into a random order. |
| 983 | */ |
| 984 | for (i = 0; i < w*h; i++) |
| 985 | temp[i] = i; |
| 986 | shuffle(temp, w*h, sizeof(*temp), rs); |
| 987 | |
| 988 | /* |
| 989 | * The first `ntrees' entries in temp which we can get |
| 990 | * without making two tents adjacent will be the tent |
| 991 | * locations. |
| 992 | */ |
| 993 | memset(grid, BLANK, w*h); |
| 994 | j = ntrees; |
| 995 | for (i = 0; i < w*h && j > 0; i++) { |
| 996 | int x = temp[i] % w, y = temp[i] / w; |
| 997 | int dy, dx, ok = TRUE; |
| 998 | |
| 999 | for (dy = -1; dy <= +1; dy++) |
| 1000 | for (dx = -1; dx <= +1; dx++) |
| 1001 | if (x+dx >= 0 && x+dx < w && |
| 1002 | y+dy >= 0 && y+dy < h && |
| 1003 | grid[(y+dy)*w+(x+dx)] == TENT) |
| 1004 | ok = FALSE; |
| 1005 | |
| 1006 | if (ok) { |
| 1007 | grid[temp[i]] = TENT; |
| 1008 | j--; |
| 1009 | } |
| 1010 | } |
| 1011 | if (j > 0) |
| 1012 | continue; /* couldn't place all the tents */ |
| 1013 | |
| 1014 | /* |
| 1015 | * Now we build up the list of graph edges. |
| 1016 | */ |
| 1017 | nedges = 0; |
| 1018 | for (i = 0; i < w*h; i++) { |
| 1019 | if (grid[temp[i]] == TENT) { |
| 1020 | for (j = 0; j < w*h; j++) { |
| 1021 | if (grid[temp[j]] != TENT) { |
| 1022 | int xi = temp[i] % w, yi = temp[i] / w; |
| 1023 | int xj = temp[j] % w, yj = temp[j] / w; |
| 1024 | if (abs(xi-xj) + abs(yi-yj) == 1) { |
| 1025 | edges[nedges*2] = i; |
| 1026 | edges[nedges*2+1] = j; |
| 1027 | capacity[nedges] = 1; |
| 1028 | nedges++; |
| 1029 | } |
| 1030 | } |
| 1031 | } |
| 1032 | } else { |
| 1033 | /* |
| 1034 | * Special node w*h is the sink node; any non-tent node |
| 1035 | * has an edge going to it. |
| 1036 | */ |
| 1037 | edges[nedges*2] = i; |
| 1038 | edges[nedges*2+1] = w*h; |
| 1039 | capacity[nedges] = 1; |
| 1040 | nedges++; |
| 1041 | } |
| 1042 | } |
| 1043 | |
| 1044 | /* |
| 1045 | * Special node w*h+1 is the source node, with an edge going to |
| 1046 | * every tent. |
| 1047 | */ |
| 1048 | for (i = 0; i < w*h; i++) { |
| 1049 | if (grid[temp[i]] == TENT) { |
| 1050 | edges[nedges*2] = w*h+1; |
| 1051 | edges[nedges*2+1] = i; |
| 1052 | capacity[nedges] = 1; |
| 1053 | nedges++; |
| 1054 | } |
| 1055 | } |
| 1056 | |
| 1057 | assert(nedges <= maxedges); |
| 1058 | |
| 1059 | /* |
| 1060 | * Now we're ready to call the maxflow algorithm to place the |
| 1061 | * trees. |
| 1062 | */ |
| 1063 | j = maxflow(w*h+2, w*h+1, w*h, nedges, edges, capacity, flow, NULL); |
| 1064 | |
| 1065 | if (j < ntrees) |
| 1066 | continue; /* couldn't place all the tents */ |
| 1067 | |
| 1068 | /* |
| 1069 | * We've placed the trees. Now we need to work out _where_ |
| 1070 | * we've placed them, which is a matter of reading back out |
| 1071 | * from the `flow' array. |
| 1072 | */ |
| 1073 | for (i = 0; i < nedges; i++) { |
| 1074 | if (edges[2*i] < w*h && edges[2*i+1] < w*h && flow[i] > 0) |
| 1075 | grid[temp[edges[2*i+1]]] = TREE; |
| 1076 | } |
| 1077 | |
| 1078 | /* |
| 1079 | * I think it looks ugly if there isn't at least one of |
| 1080 | * _something_ (tent or tree) in each row and each column |
| 1081 | * of the grid. This doesn't give any information away |
| 1082 | * since a completely empty row/column is instantly obvious |
| 1083 | * from the clues (it has no trees and a zero). |
| 1084 | */ |
| 1085 | for (i = 0; i < w; i++) { |
| 1086 | for (j = 0; j < h; j++) { |
| 1087 | if (grid[j*w+i] != BLANK) |
| 1088 | break; /* found something in this column */ |
| 1089 | } |
| 1090 | if (j == h) |
| 1091 | break; /* found empty column */ |
| 1092 | } |
| 1093 | if (i < w) |
| 1094 | continue; /* a column was empty */ |
| 1095 | |
| 1096 | for (j = 0; j < h; j++) { |
| 1097 | for (i = 0; i < w; i++) { |
| 1098 | if (grid[j*w+i] != BLANK) |
| 1099 | break; /* found something in this row */ |
| 1100 | } |
| 1101 | if (i == w) |
| 1102 | break; /* found empty row */ |
| 1103 | } |
| 1104 | if (j < h) |
| 1105 | continue; /* a row was empty */ |
| 1106 | |
| 1107 | /* |
| 1108 | * Now set up the numbers round the edge. |
| 1109 | */ |
| 1110 | for (i = 0; i < w; i++) { |
| 1111 | int n = 0; |
| 1112 | for (j = 0; j < h; j++) |
| 1113 | if (grid[j*w+i] == TENT) |
| 1114 | n++; |
| 1115 | numbers[i] = n; |
| 1116 | } |
| 1117 | for (i = 0; i < h; i++) { |
| 1118 | int n = 0; |
| 1119 | for (j = 0; j < w; j++) |
| 1120 | if (grid[i*w+j] == TENT) |
| 1121 | n++; |
| 1122 | numbers[w+i] = n; |
| 1123 | } |
| 1124 | |
| 1125 | /* |
| 1126 | * And now actually solve the puzzle, to see whether it's |
| 1127 | * unique and has the required difficulty. |
| 1128 | */ |
| 1129 | for (i = 0; i < w*h; i++) |
| 1130 | puzzle[i] = grid[i] == TREE ? TREE : BLANK; |
| 1131 | i = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff-1); |
| 1132 | j = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff); |
| 1133 | |
| 1134 | /* |
| 1135 | * We expect solving with difficulty params->diff to have |
| 1136 | * succeeded (otherwise the problem is too hard), and |
| 1137 | * solving with diff-1 to have failed (otherwise it's too |
| 1138 | * easy). |
| 1139 | */ |
| 1140 | if (i == 2 && j == 1) |
| 1141 | break; |
| 1142 | } |
| 1143 | |
| 1144 | /* |
| 1145 | * That's it. Encode as a game ID. |
| 1146 | */ |
| 1147 | ret = snewn((w+h)*40 + ntrees + (w*h)/26 + 1, char); |
| 1148 | p = ret; |
| 1149 | j = 0; |
| 1150 | for (i = 0; i <= w*h; i++) { |
| 1151 | int c = (i < w*h ? grid[i] == TREE : 1); |
| 1152 | if (c) { |
| 1153 | *p++ = (j == 0 ? '_' : j-1 + 'a'); |
| 1154 | j = 0; |
| 1155 | } else { |
| 1156 | j++; |
| 1157 | while (j > 25) { |
| 1158 | *p++ = 'z'; |
| 1159 | j -= 25; |
| 1160 | } |
| 1161 | } |
| 1162 | } |
| 1163 | for (i = 0; i < w+h; i++) |
| 1164 | p += sprintf(p, ",%d", numbers[i]); |
| 1165 | *p++ = '\0'; |
| 1166 | ret = sresize(ret, p - ret, char); |
| 1167 | |
| 1168 | /* |
| 1169 | * And encode the solution as an aux_info. |
| 1170 | */ |
| 1171 | *aux = snewn(ntrees * 40, char); |
| 1172 | p = *aux; |
| 1173 | *p++ = 'S'; |
| 1174 | for (i = 0; i < w*h; i++) |
| 1175 | if (grid[i] == TENT) |
| 1176 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
| 1177 | *p++ = '\0'; |
| 1178 | *aux = sresize(*aux, p - *aux, char); |
| 1179 | |
| 1180 | free_scratch(sc); |
| 1181 | sfree(flow); |
| 1182 | sfree(capacity); |
| 1183 | sfree(edges); |
| 1184 | sfree(temp); |
| 1185 | sfree(soln); |
| 1186 | sfree(numbers); |
| 1187 | sfree(puzzle); |
| 1188 | sfree(grid); |
| 1189 | |
| 1190 | return ret; |
| 1191 | } |
| 1192 | |
| 1193 | static char *validate_desc(game_params *params, char *desc) |
| 1194 | { |
| 1195 | int w = params->w, h = params->h; |
| 1196 | int area, i; |
| 1197 | |
| 1198 | area = 0; |
| 1199 | while (*desc && *desc != ',') { |
| 1200 | if (*desc == '_') |
| 1201 | area++; |
| 1202 | else if (*desc >= 'a' && *desc < 'z') |
| 1203 | area += *desc - 'a' + 2; |
| 1204 | else if (*desc == 'z') |
| 1205 | area += 25; |
| 1206 | else if (*desc == '!' || *desc == '-') |
| 1207 | /* do nothing */; |
| 1208 | else |
| 1209 | return "Invalid character in grid specification"; |
| 1210 | |
| 1211 | desc++; |
| 1212 | } |
| 1213 | if (area < w * h + 1) |
| 1214 | return "Not enough data to fill grid"; |
| 1215 | else if (area > w * h + 1) |
| 1216 | return "Too much data to fill grid"; |
| 1217 | |
| 1218 | for (i = 0; i < w+h; i++) { |
| 1219 | if (!*desc) |
| 1220 | return "Not enough numbers given after grid specification"; |
| 1221 | else if (*desc != ',') |
| 1222 | return "Invalid character in number list"; |
| 1223 | desc++; |
| 1224 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
| 1225 | } |
| 1226 | |
| 1227 | if (*desc) |
| 1228 | return "Unexpected additional data at end of game description"; |
| 1229 | return NULL; |
| 1230 | } |
| 1231 | |
| 1232 | static game_state *new_game(midend *me, game_params *params, char *desc) |
| 1233 | { |
| 1234 | int w = params->w, h = params->h; |
| 1235 | game_state *state = snew(game_state); |
| 1236 | int i; |
| 1237 | |
| 1238 | state->p = *params; /* structure copy */ |
| 1239 | state->grid = snewn(w*h, char); |
| 1240 | state->numbers = snew(struct numbers); |
| 1241 | state->numbers->refcount = 1; |
| 1242 | state->numbers->numbers = snewn(w+h, int); |
| 1243 | state->completed = state->used_solve = FALSE; |
| 1244 | |
| 1245 | i = 0; |
| 1246 | memset(state->grid, BLANK, w*h); |
| 1247 | |
| 1248 | while (*desc) { |
| 1249 | int run, type; |
| 1250 | |
| 1251 | type = TREE; |
| 1252 | |
| 1253 | if (*desc == '_') |
| 1254 | run = 0; |
| 1255 | else if (*desc >= 'a' && *desc < 'z') |
| 1256 | run = *desc - ('a'-1); |
| 1257 | else if (*desc == 'z') { |
| 1258 | run = 25; |
| 1259 | type = BLANK; |
| 1260 | } else { |
| 1261 | assert(*desc == '!' || *desc == '-'); |
| 1262 | run = -1; |
| 1263 | type = (*desc == '!' ? TENT : NONTENT); |
| 1264 | } |
| 1265 | |
| 1266 | desc++; |
| 1267 | |
| 1268 | i += run; |
| 1269 | assert(i >= 0 && i <= w*h); |
| 1270 | if (i == w*h) { |
| 1271 | assert(type == TREE); |
| 1272 | break; |
| 1273 | } else { |
| 1274 | if (type != BLANK) |
| 1275 | state->grid[i++] = type; |
| 1276 | } |
| 1277 | } |
| 1278 | |
| 1279 | for (i = 0; i < w+h; i++) { |
| 1280 | assert(*desc == ','); |
| 1281 | desc++; |
| 1282 | state->numbers->numbers[i] = atoi(desc); |
| 1283 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
| 1284 | } |
| 1285 | |
| 1286 | assert(!*desc); |
| 1287 | |
| 1288 | return state; |
| 1289 | } |
| 1290 | |
| 1291 | static game_state *dup_game(game_state *state) |
| 1292 | { |
| 1293 | int w = state->p.w, h = state->p.h; |
| 1294 | game_state *ret = snew(game_state); |
| 1295 | |
| 1296 | ret->p = state->p; /* structure copy */ |
| 1297 | ret->grid = snewn(w*h, char); |
| 1298 | memcpy(ret->grid, state->grid, w*h); |
| 1299 | ret->numbers = state->numbers; |
| 1300 | state->numbers->refcount++; |
| 1301 | ret->completed = state->completed; |
| 1302 | ret->used_solve = state->used_solve; |
| 1303 | |
| 1304 | return ret; |
| 1305 | } |
| 1306 | |
| 1307 | static void free_game(game_state *state) |
| 1308 | { |
| 1309 | if (--state->numbers->refcount <= 0) { |
| 1310 | sfree(state->numbers->numbers); |
| 1311 | sfree(state->numbers); |
| 1312 | } |
| 1313 | sfree(state->grid); |
| 1314 | sfree(state); |
| 1315 | } |
| 1316 | |
| 1317 | static char *solve_game(game_state *state, game_state *currstate, |
| 1318 | char *aux, char **error) |
| 1319 | { |
| 1320 | int w = state->p.w, h = state->p.h; |
| 1321 | |
| 1322 | if (aux) { |
| 1323 | /* |
| 1324 | * If we already have the solution, save ourselves some |
| 1325 | * time. |
| 1326 | */ |
| 1327 | return dupstr(aux); |
| 1328 | } else { |
| 1329 | struct solver_scratch *sc = new_scratch(w, h); |
| 1330 | char *soln; |
| 1331 | int ret; |
| 1332 | char *move, *p; |
| 1333 | int i; |
| 1334 | |
| 1335 | soln = snewn(w*h, char); |
| 1336 | ret = tents_solve(w, h, state->grid, state->numbers->numbers, |
| 1337 | soln, sc, DIFFCOUNT-1); |
| 1338 | free_scratch(sc); |
| 1339 | if (ret != 1) { |
| 1340 | sfree(soln); |
| 1341 | if (ret == 0) |
| 1342 | *error = "This puzzle is not self-consistent"; |
| 1343 | else |
| 1344 | *error = "Unable to find a unique solution for this puzzle"; |
| 1345 | return NULL; |
| 1346 | } |
| 1347 | |
| 1348 | /* |
| 1349 | * Construct a move string which turns the current state |
| 1350 | * into the solved state. |
| 1351 | */ |
| 1352 | move = snewn(w*h * 40, char); |
| 1353 | p = move; |
| 1354 | *p++ = 'S'; |
| 1355 | for (i = 0; i < w*h; i++) |
| 1356 | if (soln[i] == TENT) |
| 1357 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
| 1358 | *p++ = '\0'; |
| 1359 | move = sresize(move, p - move, char); |
| 1360 | |
| 1361 | sfree(soln); |
| 1362 | |
| 1363 | return move; |
| 1364 | } |
| 1365 | } |
| 1366 | |
| 1367 | static int game_can_format_as_text_now(game_params *params) |
| 1368 | { |
| 1369 | return TRUE; |
| 1370 | } |
| 1371 | |
| 1372 | static char *game_text_format(game_state *state) |
| 1373 | { |
| 1374 | int w = state->p.w, h = state->p.h; |
| 1375 | char *ret, *p; |
| 1376 | int x, y; |
| 1377 | |
| 1378 | /* |
| 1379 | * FIXME: We currently do not print the numbers round the edges |
| 1380 | * of the grid. I need to work out a sensible way of doing this |
| 1381 | * even when the column numbers exceed 9. |
| 1382 | * |
| 1383 | * In the absence of those numbers, the result size is h lines |
| 1384 | * of w+1 characters each, plus a NUL. |
| 1385 | * |
| 1386 | * This function is currently only used by the standalone |
| 1387 | * solver; until I make it look more sensible, I won't enable |
| 1388 | * it in the main game structure. |
| 1389 | */ |
| 1390 | ret = snewn(h*(w+1) + 1, char); |
| 1391 | p = ret; |
| 1392 | for (y = 0; y < h; y++) { |
| 1393 | for (x = 0; x < w; x++) { |
| 1394 | *p = (state->grid[y*w+x] == BLANK ? '.' : |
| 1395 | state->grid[y*w+x] == TREE ? 'T' : |
| 1396 | state->grid[y*w+x] == TENT ? '*' : |
| 1397 | state->grid[y*w+x] == NONTENT ? '-' : '?'); |
| 1398 | p++; |
| 1399 | } |
| 1400 | *p++ = '\n'; |
| 1401 | } |
| 1402 | *p++ = '\0'; |
| 1403 | |
| 1404 | return ret; |
| 1405 | } |
| 1406 | |
| 1407 | struct game_ui { |
| 1408 | int dsx, dsy; /* coords of drag start */ |
| 1409 | int dex, dey; /* coords of drag end */ |
| 1410 | int drag_button; /* -1 for none, or a button code */ |
| 1411 | int drag_ok; /* dragged off the window, to cancel */ |
| 1412 | |
| 1413 | int cx, cy, cdisp; /* cursor position, and ?display. */ |
| 1414 | }; |
| 1415 | |
| 1416 | static game_ui *new_ui(game_state *state) |
| 1417 | { |
| 1418 | game_ui *ui = snew(game_ui); |
| 1419 | ui->dsx = ui->dsy = -1; |
| 1420 | ui->dex = ui->dey = -1; |
| 1421 | ui->drag_button = -1; |
| 1422 | ui->drag_ok = FALSE; |
| 1423 | ui->cx = ui->cy = ui->cdisp = 0; |
| 1424 | return ui; |
| 1425 | } |
| 1426 | |
| 1427 | static void free_ui(game_ui *ui) |
| 1428 | { |
| 1429 | sfree(ui); |
| 1430 | } |
| 1431 | |
| 1432 | static char *encode_ui(game_ui *ui) |
| 1433 | { |
| 1434 | return NULL; |
| 1435 | } |
| 1436 | |
| 1437 | static void decode_ui(game_ui *ui, char *encoding) |
| 1438 | { |
| 1439 | } |
| 1440 | |
| 1441 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
| 1442 | game_state *newstate) |
| 1443 | { |
| 1444 | } |
| 1445 | |
| 1446 | struct game_drawstate { |
| 1447 | int tilesize; |
| 1448 | int started; |
| 1449 | game_params p; |
| 1450 | int *drawn, *numbersdrawn; |
| 1451 | int cx, cy; /* last-drawn cursor pos, or (-1,-1) if absent. */ |
| 1452 | }; |
| 1453 | |
| 1454 | #define PREFERRED_TILESIZE 32 |
| 1455 | #define TILESIZE (ds->tilesize) |
| 1456 | #define TLBORDER (TILESIZE/2) |
| 1457 | #define BRBORDER (TILESIZE*3/2) |
| 1458 | #define COORD(x) ( (x) * TILESIZE + TLBORDER ) |
| 1459 | #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 ) |
| 1460 | |
| 1461 | #define FLASH_TIME 0.30F |
| 1462 | |
| 1463 | static int drag_xform(game_ui *ui, int x, int y, int v) |
| 1464 | { |
| 1465 | int xmin, ymin, xmax, ymax; |
| 1466 | |
| 1467 | xmin = min(ui->dsx, ui->dex); |
| 1468 | xmax = max(ui->dsx, ui->dex); |
| 1469 | ymin = min(ui->dsy, ui->dey); |
| 1470 | ymax = max(ui->dsy, ui->dey); |
| 1471 | |
| 1472 | /* |
| 1473 | * Left-dragging has no effect, so we treat a left-drag as a |
| 1474 | * single click on dsx,dsy. |
| 1475 | */ |
| 1476 | if (ui->drag_button == LEFT_BUTTON) { |
| 1477 | xmin = xmax = ui->dsx; |
| 1478 | ymin = ymax = ui->dsy; |
| 1479 | } |
| 1480 | |
| 1481 | if (x < xmin || x > xmax || y < ymin || y > ymax) |
| 1482 | return v; /* no change outside drag area */ |
| 1483 | |
| 1484 | if (v == TREE) |
| 1485 | return v; /* trees are inviolate always */ |
| 1486 | |
| 1487 | if (xmin == xmax && ymin == ymax) { |
| 1488 | /* |
| 1489 | * Results of a simple click. Left button sets blanks to |
| 1490 | * tents; right button sets blanks to non-tents; either |
| 1491 | * button clears a non-blank square. |
| 1492 | */ |
| 1493 | if (ui->drag_button == LEFT_BUTTON) |
| 1494 | v = (v == BLANK ? TENT : BLANK); |
| 1495 | else |
| 1496 | v = (v == BLANK ? NONTENT : BLANK); |
| 1497 | } else { |
| 1498 | /* |
| 1499 | * Results of a drag. Left-dragging has no effect. |
| 1500 | * Right-dragging sets all blank squares to non-tents and |
| 1501 | * has no effect on anything else. |
| 1502 | */ |
| 1503 | if (ui->drag_button == RIGHT_BUTTON) |
| 1504 | v = (v == BLANK ? NONTENT : v); |
| 1505 | else |
| 1506 | /* do nothing */; |
| 1507 | } |
| 1508 | |
| 1509 | return v; |
| 1510 | } |
| 1511 | |
| 1512 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
| 1513 | int x, int y, int button) |
| 1514 | { |
| 1515 | int w = state->p.w, h = state->p.h; |
| 1516 | char tmpbuf[80]; |
| 1517 | |
| 1518 | if (button == LEFT_BUTTON || button == RIGHT_BUTTON) { |
| 1519 | x = FROMCOORD(x); |
| 1520 | y = FROMCOORD(y); |
| 1521 | if (x < 0 || y < 0 || x >= w || y >= h) |
| 1522 | return NULL; |
| 1523 | |
| 1524 | ui->drag_button = button; |
| 1525 | ui->dsx = ui->dex = x; |
| 1526 | ui->dsy = ui->dey = y; |
| 1527 | ui->drag_ok = TRUE; |
| 1528 | ui->cdisp = 0; |
| 1529 | return ""; /* ui updated */ |
| 1530 | } |
| 1531 | |
| 1532 | if ((IS_MOUSE_DRAG(button) || IS_MOUSE_RELEASE(button)) && |
| 1533 | ui->drag_button > 0) { |
| 1534 | int xmin, ymin, xmax, ymax; |
| 1535 | char *buf, *sep; |
| 1536 | int buflen, bufsize, tmplen; |
| 1537 | |
| 1538 | x = FROMCOORD(x); |
| 1539 | y = FROMCOORD(y); |
| 1540 | if (x < 0 || y < 0 || x >= w || y >= h) { |
| 1541 | ui->drag_ok = FALSE; |
| 1542 | } else { |
| 1543 | /* |
| 1544 | * Drags are limited to one row or column. Hence, we |
| 1545 | * work out which coordinate is closer to the drag |
| 1546 | * start, and move it _to_ the drag start. |
| 1547 | */ |
| 1548 | if (abs(x - ui->dsx) < abs(y - ui->dsy)) |
| 1549 | x = ui->dsx; |
| 1550 | else |
| 1551 | y = ui->dsy; |
| 1552 | |
| 1553 | ui->dex = x; |
| 1554 | ui->dey = y; |
| 1555 | |
| 1556 | ui->drag_ok = TRUE; |
| 1557 | } |
| 1558 | |
| 1559 | if (IS_MOUSE_DRAG(button)) |
| 1560 | return ""; /* ui updated */ |
| 1561 | |
| 1562 | /* |
| 1563 | * The drag has been released. Enact it. |
| 1564 | */ |
| 1565 | if (!ui->drag_ok) { |
| 1566 | ui->drag_button = -1; |
| 1567 | return ""; /* drag was just cancelled */ |
| 1568 | } |
| 1569 | |
| 1570 | xmin = min(ui->dsx, ui->dex); |
| 1571 | xmax = max(ui->dsx, ui->dex); |
| 1572 | ymin = min(ui->dsy, ui->dey); |
| 1573 | ymax = max(ui->dsy, ui->dey); |
| 1574 | assert(0 <= xmin && xmin <= xmax && xmax < w); |
| 1575 | assert(0 <= ymin && ymin <= ymax && ymax < h); |
| 1576 | |
| 1577 | buflen = 0; |
| 1578 | bufsize = 256; |
| 1579 | buf = snewn(bufsize, char); |
| 1580 | sep = ""; |
| 1581 | for (y = ymin; y <= ymax; y++) |
| 1582 | for (x = xmin; x <= xmax; x++) { |
| 1583 | int v = drag_xform(ui, x, y, state->grid[y*w+x]); |
| 1584 | if (state->grid[y*w+x] != v) { |
| 1585 | tmplen = sprintf(tmpbuf, "%s%c%d,%d", sep, |
| 1586 | (int)(v == BLANK ? 'B' : |
| 1587 | v == TENT ? 'T' : 'N'), |
| 1588 | x, y); |
| 1589 | sep = ";"; |
| 1590 | |
| 1591 | if (buflen + tmplen >= bufsize) { |
| 1592 | bufsize = buflen + tmplen + 256; |
| 1593 | buf = sresize(buf, bufsize, char); |
| 1594 | } |
| 1595 | |
| 1596 | strcpy(buf+buflen, tmpbuf); |
| 1597 | buflen += tmplen; |
| 1598 | } |
| 1599 | } |
| 1600 | |
| 1601 | ui->drag_button = -1; /* drag is terminated */ |
| 1602 | |
| 1603 | if (buflen == 0) { |
| 1604 | sfree(buf); |
| 1605 | return ""; /* ui updated (drag was terminated) */ |
| 1606 | } else { |
| 1607 | buf[buflen] = '\0'; |
| 1608 | return buf; |
| 1609 | } |
| 1610 | } |
| 1611 | |
| 1612 | if (IS_CURSOR_MOVE(button)) { |
| 1613 | move_cursor(button, &ui->cx, &ui->cy, w, h, 0); |
| 1614 | ui->cdisp = 1; |
| 1615 | return ""; |
| 1616 | } |
| 1617 | if (ui->cdisp) { |
| 1618 | char rep = 0; |
| 1619 | int v = state->grid[ui->cy*w+ui->cx]; |
| 1620 | |
| 1621 | if (v != TREE) { |
| 1622 | #ifdef SINGLE_CURSOR_SELECT |
| 1623 | if (button == CURSOR_SELECT) |
| 1624 | /* SELECT cycles T, N, B */ |
| 1625 | rep = v == BLANK ? 'T' : v == TENT ? 'N' : 'B'; |
| 1626 | #else |
| 1627 | if (button == CURSOR_SELECT) |
| 1628 | rep = v == BLANK ? 'T' : 'B'; |
| 1629 | else if (button == CURSOR_SELECT2) |
| 1630 | rep = v == BLANK ? 'N' : 'B'; |
| 1631 | else if (button == 'T' || button == 'N' || button == 'B') |
| 1632 | rep = (char)button; |
| 1633 | #endif |
| 1634 | } |
| 1635 | |
| 1636 | if (rep) { |
| 1637 | sprintf(tmpbuf, "%c%d,%d", (int)rep, ui->cx, ui->cy); |
| 1638 | return dupstr(tmpbuf); |
| 1639 | } |
| 1640 | } else if (IS_CURSOR_SELECT(button)) { |
| 1641 | ui->cdisp = 1; |
| 1642 | return ""; |
| 1643 | } |
| 1644 | |
| 1645 | return NULL; |
| 1646 | } |
| 1647 | |
| 1648 | static game_state *execute_move(game_state *state, char *move) |
| 1649 | { |
| 1650 | int w = state->p.w, h = state->p.h; |
| 1651 | char c; |
| 1652 | int x, y, m, n, i, j; |
| 1653 | game_state *ret = dup_game(state); |
| 1654 | |
| 1655 | while (*move) { |
| 1656 | c = *move; |
| 1657 | if (c == 'S') { |
| 1658 | int i; |
| 1659 | ret->used_solve = TRUE; |
| 1660 | /* |
| 1661 | * Set all non-tree squares to NONTENT. The rest of the |
| 1662 | * solve move will fill the tents in over the top. |
| 1663 | */ |
| 1664 | for (i = 0; i < w*h; i++) |
| 1665 | if (ret->grid[i] != TREE) |
| 1666 | ret->grid[i] = NONTENT; |
| 1667 | move++; |
| 1668 | } else if (c == 'B' || c == 'T' || c == 'N') { |
| 1669 | move++; |
| 1670 | if (sscanf(move, "%d,%d%n", &x, &y, &n) != 2 || |
| 1671 | x < 0 || y < 0 || x >= w || y >= h) { |
| 1672 | free_game(ret); |
| 1673 | return NULL; |
| 1674 | } |
| 1675 | if (ret->grid[y*w+x] == TREE) { |
| 1676 | free_game(ret); |
| 1677 | return NULL; |
| 1678 | } |
| 1679 | ret->grid[y*w+x] = (c == 'B' ? BLANK : c == 'T' ? TENT : NONTENT); |
| 1680 | move += n; |
| 1681 | } else { |
| 1682 | free_game(ret); |
| 1683 | return NULL; |
| 1684 | } |
| 1685 | if (*move == ';') |
| 1686 | move++; |
| 1687 | else if (*move) { |
| 1688 | free_game(ret); |
| 1689 | return NULL; |
| 1690 | } |
| 1691 | } |
| 1692 | |
| 1693 | /* |
| 1694 | * Check for completion. |
| 1695 | */ |
| 1696 | for (i = n = m = 0; i < w*h; i++) { |
| 1697 | if (ret->grid[i] == TENT) |
| 1698 | n++; |
| 1699 | else if (ret->grid[i] == TREE) |
| 1700 | m++; |
| 1701 | } |
| 1702 | if (n == m) { |
| 1703 | int nedges, maxedges, *edges, *capacity, *flow; |
| 1704 | |
| 1705 | /* |
| 1706 | * We have the right number of tents, which is a |
| 1707 | * precondition for the game being complete. Now check that |
| 1708 | * the numbers add up. |
| 1709 | */ |
| 1710 | for (i = 0; i < w; i++) { |
| 1711 | n = 0; |
| 1712 | for (j = 0; j < h; j++) |
| 1713 | if (ret->grid[j*w+i] == TENT) |
| 1714 | n++; |
| 1715 | if (ret->numbers->numbers[i] != n) |
| 1716 | goto completion_check_done; |
| 1717 | } |
| 1718 | for (i = 0; i < h; i++) { |
| 1719 | n = 0; |
| 1720 | for (j = 0; j < w; j++) |
| 1721 | if (ret->grid[i*w+j] == TENT) |
| 1722 | n++; |
| 1723 | if (ret->numbers->numbers[w+i] != n) |
| 1724 | goto completion_check_done; |
| 1725 | } |
| 1726 | /* |
| 1727 | * Also, check that no two tents are adjacent. |
| 1728 | */ |
| 1729 | for (y = 0; y < h; y++) |
| 1730 | for (x = 0; x < w; x++) { |
| 1731 | if (x+1 < w && |
| 1732 | ret->grid[y*w+x] == TENT && ret->grid[y*w+x+1] == TENT) |
| 1733 | goto completion_check_done; |
| 1734 | if (y+1 < h && |
| 1735 | ret->grid[y*w+x] == TENT && ret->grid[(y+1)*w+x] == TENT) |
| 1736 | goto completion_check_done; |
| 1737 | if (x+1 < w && y+1 < h) { |
| 1738 | if (ret->grid[y*w+x] == TENT && |
| 1739 | ret->grid[(y+1)*w+(x+1)] == TENT) |
| 1740 | goto completion_check_done; |
| 1741 | if (ret->grid[(y+1)*w+x] == TENT && |
| 1742 | ret->grid[y*w+(x+1)] == TENT) |
| 1743 | goto completion_check_done; |
| 1744 | } |
| 1745 | } |
| 1746 | |
| 1747 | /* |
| 1748 | * OK; we have the right number of tents, they match the |
| 1749 | * numeric clues, and they satisfy the non-adjacency |
| 1750 | * criterion. Finally, we need to verify that they can be |
| 1751 | * placed in a one-to-one matching with the trees such that |
| 1752 | * every tent is orthogonally adjacent to its tree. |
| 1753 | * |
| 1754 | * This bit is where the hard work comes in: we have to do |
| 1755 | * it by finding such a matching using maxflow. |
| 1756 | * |
| 1757 | * So we construct a network with one special source node, |
| 1758 | * one special sink node, one node per tent, and one node |
| 1759 | * per tree. |
| 1760 | */ |
| 1761 | maxedges = 6 * m; |
| 1762 | edges = snewn(2 * maxedges, int); |
| 1763 | capacity = snewn(maxedges, int); |
| 1764 | flow = snewn(maxedges, int); |
| 1765 | nedges = 0; |
| 1766 | /* |
| 1767 | * Node numbering: |
| 1768 | * |
| 1769 | * 0..w*h trees/tents |
| 1770 | * w*h source |
| 1771 | * w*h+1 sink |
| 1772 | */ |
| 1773 | for (y = 0; y < h; y++) |
| 1774 | for (x = 0; x < w; x++) |
| 1775 | if (ret->grid[y*w+x] == TREE) { |
| 1776 | int d; |
| 1777 | |
| 1778 | /* |
| 1779 | * Here we use the direction enum declared for |
| 1780 | * the solver. We make use of the fact that the |
| 1781 | * directions are declared in the order |
| 1782 | * U,L,R,D, meaning that we go through the four |
| 1783 | * neighbours of any square in numerically |
| 1784 | * increasing order. |
| 1785 | */ |
| 1786 | for (d = 1; d < MAXDIR; d++) { |
| 1787 | int x2 = x + dx(d), y2 = y + dy(d); |
| 1788 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
| 1789 | ret->grid[y2*w+x2] == TENT) { |
| 1790 | assert(nedges < maxedges); |
| 1791 | edges[nedges*2] = y*w+x; |
| 1792 | edges[nedges*2+1] = y2*w+x2; |
| 1793 | capacity[nedges] = 1; |
| 1794 | nedges++; |
| 1795 | } |
| 1796 | } |
| 1797 | } else if (ret->grid[y*w+x] == TENT) { |
| 1798 | assert(nedges < maxedges); |
| 1799 | edges[nedges*2] = y*w+x; |
| 1800 | edges[nedges*2+1] = w*h+1; /* edge going to sink */ |
| 1801 | capacity[nedges] = 1; |
| 1802 | nedges++; |
| 1803 | } |
| 1804 | for (y = 0; y < h; y++) |
| 1805 | for (x = 0; x < w; x++) |
| 1806 | if (ret->grid[y*w+x] == TREE) { |
| 1807 | assert(nedges < maxedges); |
| 1808 | edges[nedges*2] = w*h; /* edge coming from source */ |
| 1809 | edges[nedges*2+1] = y*w+x; |
| 1810 | capacity[nedges] = 1; |
| 1811 | nedges++; |
| 1812 | } |
| 1813 | n = maxflow(w*h+2, w*h, w*h+1, nedges, edges, capacity, flow, NULL); |
| 1814 | |
| 1815 | sfree(flow); |
| 1816 | sfree(capacity); |
| 1817 | sfree(edges); |
| 1818 | |
| 1819 | if (n != m) |
| 1820 | goto completion_check_done; |
| 1821 | |
| 1822 | /* |
| 1823 | * We haven't managed to fault the grid on any count. Score! |
| 1824 | */ |
| 1825 | ret->completed = TRUE; |
| 1826 | } |
| 1827 | completion_check_done: |
| 1828 | |
| 1829 | return ret; |
| 1830 | } |
| 1831 | |
| 1832 | /* ---------------------------------------------------------------------- |
| 1833 | * Drawing routines. |
| 1834 | */ |
| 1835 | |
| 1836 | static void game_compute_size(game_params *params, int tilesize, |
| 1837 | int *x, int *y) |
| 1838 | { |
| 1839 | /* fool the macros */ |
| 1840 | struct dummy { int tilesize; } dummy, *ds = &dummy; |
| 1841 | dummy.tilesize = tilesize; |
| 1842 | |
| 1843 | *x = TLBORDER + BRBORDER + TILESIZE * params->w; |
| 1844 | *y = TLBORDER + BRBORDER + TILESIZE * params->h; |
| 1845 | } |
| 1846 | |
| 1847 | static void game_set_size(drawing *dr, game_drawstate *ds, |
| 1848 | game_params *params, int tilesize) |
| 1849 | { |
| 1850 | ds->tilesize = tilesize; |
| 1851 | } |
| 1852 | |
| 1853 | static float *game_colours(frontend *fe, int *ncolours) |
| 1854 | { |
| 1855 | float *ret = snewn(3 * NCOLOURS, float); |
| 1856 | |
| 1857 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
| 1858 | |
| 1859 | ret[COL_GRID * 3 + 0] = 0.0F; |
| 1860 | ret[COL_GRID * 3 + 1] = 0.0F; |
| 1861 | ret[COL_GRID * 3 + 2] = 0.0F; |
| 1862 | |
| 1863 | ret[COL_GRASS * 3 + 0] = 0.7F; |
| 1864 | ret[COL_GRASS * 3 + 1] = 1.0F; |
| 1865 | ret[COL_GRASS * 3 + 2] = 0.5F; |
| 1866 | |
| 1867 | ret[COL_TREETRUNK * 3 + 0] = 0.6F; |
| 1868 | ret[COL_TREETRUNK * 3 + 1] = 0.4F; |
| 1869 | ret[COL_TREETRUNK * 3 + 2] = 0.0F; |
| 1870 | |
| 1871 | ret[COL_TREELEAF * 3 + 0] = 0.0F; |
| 1872 | ret[COL_TREELEAF * 3 + 1] = 0.7F; |
| 1873 | ret[COL_TREELEAF * 3 + 2] = 0.0F; |
| 1874 | |
| 1875 | ret[COL_TENT * 3 + 0] = 0.8F; |
| 1876 | ret[COL_TENT * 3 + 1] = 0.7F; |
| 1877 | ret[COL_TENT * 3 + 2] = 0.0F; |
| 1878 | |
| 1879 | ret[COL_ERROR * 3 + 0] = 1.0F; |
| 1880 | ret[COL_ERROR * 3 + 1] = 0.0F; |
| 1881 | ret[COL_ERROR * 3 + 2] = 0.0F; |
| 1882 | |
| 1883 | ret[COL_ERRTEXT * 3 + 0] = 1.0F; |
| 1884 | ret[COL_ERRTEXT * 3 + 1] = 1.0F; |
| 1885 | ret[COL_ERRTEXT * 3 + 2] = 1.0F; |
| 1886 | |
| 1887 | ret[COL_ERRTRUNK * 3 + 0] = 0.6F; |
| 1888 | ret[COL_ERRTRUNK * 3 + 1] = 0.0F; |
| 1889 | ret[COL_ERRTRUNK * 3 + 2] = 0.0F; |
| 1890 | |
| 1891 | *ncolours = NCOLOURS; |
| 1892 | return ret; |
| 1893 | } |
| 1894 | |
| 1895 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
| 1896 | { |
| 1897 | int w = state->p.w, h = state->p.h; |
| 1898 | struct game_drawstate *ds = snew(struct game_drawstate); |
| 1899 | int i; |
| 1900 | |
| 1901 | ds->tilesize = 0; |
| 1902 | ds->started = FALSE; |
| 1903 | ds->p = state->p; /* structure copy */ |
| 1904 | ds->drawn = snewn(w*h, int); |
| 1905 | for (i = 0; i < w*h; i++) |
| 1906 | ds->drawn[i] = MAGIC; |
| 1907 | ds->numbersdrawn = snewn(w+h, int); |
| 1908 | for (i = 0; i < w+h; i++) |
| 1909 | ds->numbersdrawn[i] = 2; |
| 1910 | ds->cx = ds->cy = -1; |
| 1911 | |
| 1912 | return ds; |
| 1913 | } |
| 1914 | |
| 1915 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
| 1916 | { |
| 1917 | sfree(ds->drawn); |
| 1918 | sfree(ds->numbersdrawn); |
| 1919 | sfree(ds); |
| 1920 | } |
| 1921 | |
| 1922 | enum { |
| 1923 | ERR_ADJ_TOPLEFT = 4, |
| 1924 | ERR_ADJ_TOP, |
| 1925 | ERR_ADJ_TOPRIGHT, |
| 1926 | ERR_ADJ_LEFT, |
| 1927 | ERR_ADJ_RIGHT, |
| 1928 | ERR_ADJ_BOTLEFT, |
| 1929 | ERR_ADJ_BOT, |
| 1930 | ERR_ADJ_BOTRIGHT, |
| 1931 | ERR_OVERCOMMITTED |
| 1932 | }; |
| 1933 | |
| 1934 | static int *find_errors(game_state *state, char *grid) |
| 1935 | { |
| 1936 | int w = state->p.w, h = state->p.h; |
| 1937 | int *ret = snewn(w*h + w + h, int); |
| 1938 | int *tmp = snewn(w*h*2, int), *dsf = tmp + w*h; |
| 1939 | int x, y; |
| 1940 | |
| 1941 | /* |
| 1942 | * This function goes through a grid and works out where to |
| 1943 | * highlight play errors in red. The aim is that it should |
| 1944 | * produce at least one error highlight for any complete grid |
| 1945 | * (or complete piece of grid) violating a puzzle constraint, so |
| 1946 | * that a grid containing no BLANK squares is either a win or is |
| 1947 | * marked up in some way that indicates why not. |
| 1948 | * |
| 1949 | * So it's easy enough to highlight errors in the numeric clues |
| 1950 | * - just light up any row or column number which is not |
| 1951 | * fulfilled - and it's just as easy to highlight adjacent |
| 1952 | * tents. The difficult bit is highlighting failures in the |
| 1953 | * tent/tree matching criterion. |
| 1954 | * |
| 1955 | * A natural approach would seem to be to apply the maxflow |
| 1956 | * algorithm to find the tent/tree matching; if this fails, it |
| 1957 | * must necessarily terminate with a min-cut which can be |
| 1958 | * reinterpreted as some set of trees which have too few tents |
| 1959 | * between them (or vice versa). However, it's bad for |
| 1960 | * localising errors, because it's not easy to make the |
| 1961 | * algorithm narrow down to the _smallest_ such set of trees: if |
| 1962 | * trees A and B have only one tent between them, for instance, |
| 1963 | * it might perfectly well highlight not only A and B but also |
| 1964 | * trees C and D which are correctly matched on the far side of |
| 1965 | * the grid, on the grounds that those four trees between them |
| 1966 | * have only three tents. |
| 1967 | * |
| 1968 | * Also, that approach fares badly when you introduce the |
| 1969 | * additional requirement that incomplete grids should have |
| 1970 | * errors highlighted only when they can be proved to be errors |
| 1971 | * - so that trees should not be marked as having too few tents |
| 1972 | * if there are enough BLANK squares remaining around them that |
| 1973 | * could be turned into the missing tents (to do so would be |
| 1974 | * patronising, since the overwhelming likelihood is not that |
| 1975 | * the player has forgotten to put a tree there but that they |
| 1976 | * have merely not put one there _yet_). However, tents with too |
| 1977 | * few trees can be marked immediately, since those are |
| 1978 | * definitely player error. |
| 1979 | * |
| 1980 | * So I adopt an alternative approach, which is to consider the |
| 1981 | * bipartite adjacency graph between trees and tents |
| 1982 | * ('bipartite' in the sense that for these purposes I |
| 1983 | * deliberately ignore two adjacent trees or two adjacent |
| 1984 | * tents), divide that graph up into its connected components |
| 1985 | * using a dsf, and look for components which contain different |
| 1986 | * numbers of trees and tents. This allows me to highlight |
| 1987 | * groups of tents with too few trees between them immediately, |
| 1988 | * and then in order to find groups of trees with too few tents |
| 1989 | * I redo the same process but counting BLANKs as potential |
| 1990 | * tents (so that the only trees highlighted are those |
| 1991 | * surrounded by enough NONTENTs to make it impossible to give |
| 1992 | * them enough tents). |
| 1993 | * |
| 1994 | * However, this technique is incomplete: it is not a sufficient |
| 1995 | * condition for the existence of a perfect matching that every |
| 1996 | * connected component of the graph has the same number of tents |
| 1997 | * and trees. An example of a graph which satisfies the latter |
| 1998 | * condition but still has no perfect matching is |
| 1999 | * |
| 2000 | * A B C |
| 2001 | * | / ,/| |
| 2002 | * | / ,'/ | |
| 2003 | * | / ,' / | |
| 2004 | * |/,' / | |
| 2005 | * 1 2 3 |
| 2006 | * |
| 2007 | * which can be realised in Tents as |
| 2008 | * |
| 2009 | * B |
| 2010 | * A 1 C 2 |
| 2011 | * 3 |
| 2012 | * |
| 2013 | * The matching-error highlighter described above will not mark |
| 2014 | * this construction as erroneous. However, something else will: |
| 2015 | * the three tents in the above diagram (let us suppose A,B,C |
| 2016 | * are the tents, though it doesn't matter which) contain two |
| 2017 | * diagonally adjacent pairs. So there will be _an_ error |
| 2018 | * highlighted for the above layout, even though not all types |
| 2019 | * of error will be highlighted. |
| 2020 | * |
| 2021 | * And in fact we can prove that this will always be the case: |
| 2022 | * that the shortcomings of the matching-error highlighter will |
| 2023 | * always be made up for by the easy tent adjacency highlighter. |
| 2024 | * |
| 2025 | * Lemma: Let G be a bipartite graph between n trees and n |
| 2026 | * tents, which is connected, and in which no tree has degree |
| 2027 | * more than two (but a tent may). Then G has a perfect matching. |
| 2028 | * |
| 2029 | * (Note: in the statement and proof of the Lemma I will |
| 2030 | * consistently use 'tree' to indicate a type of graph vertex as |
| 2031 | * opposed to a tent, and not to indicate a tree in the graph- |
| 2032 | * theoretic sense.) |
| 2033 | * |
| 2034 | * Proof: |
| 2035 | * |
| 2036 | * If we can find a tent of degree 1 joined to a tree of degree |
| 2037 | * 2, then any perfect matching must pair that tent with that |
| 2038 | * tree. Hence, we can remove both, leaving a smaller graph G' |
| 2039 | * which still satisfies all the conditions of the Lemma, and |
| 2040 | * which has a perfect matching iff G does. |
| 2041 | * |
| 2042 | * So, wlog, we may assume G contains no tent of degree 1 joined |
| 2043 | * to a tree of degree 2; if it does, we can reduce it as above. |
| 2044 | * |
| 2045 | * If G has no tent of degree 1 at all, then every tent has |
| 2046 | * degree at least two, so there are at least 2n edges in the |
| 2047 | * graph. But every tree has degree at most two, so there are at |
| 2048 | * most 2n edges. Hence there must be exactly 2n edges, so every |
| 2049 | * tree and every tent must have degree exactly two, which means |
| 2050 | * that the whole graph consists of a single loop (by |
| 2051 | * connectedness), and therefore certainly has a perfect |
| 2052 | * matching. |
| 2053 | * |
| 2054 | * Alternatively, if G does have a tent of degree 1 but it is |
| 2055 | * not connected to a tree of degree 2, then the tree it is |
| 2056 | * connected to must have degree 1 - and, by connectedness, that |
| 2057 | * must mean that that tent and that tree between them form the |
| 2058 | * entire graph. This trivial graph has a trivial perfect |
| 2059 | * matching. [] |
| 2060 | * |
| 2061 | * That proves the lemma. Hence, in any case where the matching- |
| 2062 | * error highlighter fails to highlight an erroneous component |
| 2063 | * (because it has the same number of tents as trees, but they |
| 2064 | * cannot be matched up), the above lemma tells us that there |
| 2065 | * must be a tree with degree more than 2, i.e. a tree |
| 2066 | * orthogonally adjacent to at least three tents. But in that |
| 2067 | * case, there must be some pair of those three tents which are |
| 2068 | * diagonally adjacent to each other, so the tent-adjacency |
| 2069 | * highlighter will necessarily show an error. So any filled |
| 2070 | * layout in Tents which is not a correct solution to the puzzle |
| 2071 | * must have _some_ error highlighted by the subroutine below. |
| 2072 | * |
| 2073 | * (Of course it would be nicer if we could highlight all |
| 2074 | * errors: in the above example layout, we would like to |
| 2075 | * highlight tents A,B as having too few trees between them, and |
| 2076 | * trees 2,3 as having too few tents, in addition to marking the |
| 2077 | * adjacency problems. But I can't immediately think of any way |
| 2078 | * to find the smallest sets of such tents and trees without an |
| 2079 | * O(2^N) loop over all subsets of a given component.) |
| 2080 | */ |
| 2081 | |
| 2082 | /* |
| 2083 | * ret[0] through to ret[w*h-1] give error markers for the grid |
| 2084 | * squares. After that, ret[w*h] to ret[w*h+w-1] give error |
| 2085 | * markers for the column numbers, and ret[w*h+w] to |
| 2086 | * ret[w*h+w+h-1] for the row numbers. |
| 2087 | */ |
| 2088 | |
| 2089 | /* |
| 2090 | * Spot tent-adjacency violations. |
| 2091 | */ |
| 2092 | for (x = 0; x < w*h; x++) |
| 2093 | ret[x] = 0; |
| 2094 | for (y = 0; y < h; y++) { |
| 2095 | for (x = 0; x < w; x++) { |
| 2096 | if (y+1 < h && x+1 < w && |
| 2097 | ((grid[y*w+x] == TENT && |
| 2098 | grid[(y+1)*w+(x+1)] == TENT) || |
| 2099 | (grid[(y+1)*w+x] == TENT && |
| 2100 | grid[y*w+(x+1)] == TENT))) { |
| 2101 | ret[y*w+x] |= 1 << ERR_ADJ_BOTRIGHT; |
| 2102 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOPRIGHT; |
| 2103 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_BOTLEFT; |
| 2104 | ret[(y+1)*w+(x+1)] |= 1 << ERR_ADJ_TOPLEFT; |
| 2105 | } |
| 2106 | if (y+1 < h && |
| 2107 | grid[y*w+x] == TENT && |
| 2108 | grid[(y+1)*w+x] == TENT) { |
| 2109 | ret[y*w+x] |= 1 << ERR_ADJ_BOT; |
| 2110 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOP; |
| 2111 | } |
| 2112 | if (x+1 < w && |
| 2113 | grid[y*w+x] == TENT && |
| 2114 | grid[y*w+(x+1)] == TENT) { |
| 2115 | ret[y*w+x] |= 1 << ERR_ADJ_RIGHT; |
| 2116 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_LEFT; |
| 2117 | } |
| 2118 | } |
| 2119 | } |
| 2120 | |
| 2121 | /* |
| 2122 | * Spot numeric clue violations. |
| 2123 | */ |
| 2124 | for (x = 0; x < w; x++) { |
| 2125 | int tents = 0, maybetents = 0; |
| 2126 | for (y = 0; y < h; y++) { |
| 2127 | if (grid[y*w+x] == TENT) |
| 2128 | tents++; |
| 2129 | else if (grid[y*w+x] == BLANK) |
| 2130 | maybetents++; |
| 2131 | } |
| 2132 | ret[w*h+x] = (tents > state->numbers->numbers[x] || |
| 2133 | tents + maybetents < state->numbers->numbers[x]); |
| 2134 | } |
| 2135 | for (y = 0; y < h; y++) { |
| 2136 | int tents = 0, maybetents = 0; |
| 2137 | for (x = 0; x < w; x++) { |
| 2138 | if (grid[y*w+x] == TENT) |
| 2139 | tents++; |
| 2140 | else if (grid[y*w+x] == BLANK) |
| 2141 | maybetents++; |
| 2142 | } |
| 2143 | ret[w*h+w+y] = (tents > state->numbers->numbers[w+y] || |
| 2144 | tents + maybetents < state->numbers->numbers[w+y]); |
| 2145 | } |
| 2146 | |
| 2147 | /* |
| 2148 | * Identify groups of tents with too few trees between them, |
| 2149 | * which we do by constructing the connected components of the |
| 2150 | * bipartite adjacency graph between tents and trees |
| 2151 | * ('bipartite' in the sense that we deliberately ignore |
| 2152 | * adjacency between tents or between trees), and highlighting |
| 2153 | * all the tents in any component which has a smaller tree |
| 2154 | * count. |
| 2155 | */ |
| 2156 | dsf_init(dsf, w*h); |
| 2157 | /* Construct the equivalence classes. */ |
| 2158 | for (y = 0; y < h; y++) { |
| 2159 | for (x = 0; x < w-1; x++) { |
| 2160 | if ((grid[y*w+x] == TREE && grid[y*w+x+1] == TENT) || |
| 2161 | (grid[y*w+x] == TENT && grid[y*w+x+1] == TREE)) |
| 2162 | dsf_merge(dsf, y*w+x, y*w+x+1); |
| 2163 | } |
| 2164 | } |
| 2165 | for (y = 0; y < h-1; y++) { |
| 2166 | for (x = 0; x < w; x++) { |
| 2167 | if ((grid[y*w+x] == TREE && grid[(y+1)*w+x] == TENT) || |
| 2168 | (grid[y*w+x] == TENT && grid[(y+1)*w+x] == TREE)) |
| 2169 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
| 2170 | } |
| 2171 | } |
| 2172 | /* Count up the tent/tree difference in each one. */ |
| 2173 | for (x = 0; x < w*h; x++) |
| 2174 | tmp[x] = 0; |
| 2175 | for (x = 0; x < w*h; x++) { |
| 2176 | y = dsf_canonify(dsf, x); |
| 2177 | if (grid[x] == TREE) |
| 2178 | tmp[y]++; |
| 2179 | else if (grid[x] == TENT) |
| 2180 | tmp[y]--; |
| 2181 | } |
| 2182 | /* And highlight any tent belonging to an equivalence class with |
| 2183 | * a score less than zero. */ |
| 2184 | for (x = 0; x < w*h; x++) { |
| 2185 | y = dsf_canonify(dsf, x); |
| 2186 | if (grid[x] == TENT && tmp[y] < 0) |
| 2187 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
| 2188 | } |
| 2189 | |
| 2190 | /* |
| 2191 | * Identify groups of trees with too few tents between them. |
| 2192 | * This is done similarly, except that we now count BLANK as |
| 2193 | * equivalent to TENT, i.e. we only highlight such trees when |
| 2194 | * the user hasn't even left _room_ to provide tents for them |
| 2195 | * all. (Otherwise, we'd highlight all trees red right at the |
| 2196 | * start of the game, before the user had done anything wrong!) |
| 2197 | */ |
| 2198 | #define TENT(x) ((x)==TENT || (x)==BLANK) |
| 2199 | dsf_init(dsf, w*h); |
| 2200 | /* Construct the equivalence classes. */ |
| 2201 | for (y = 0; y < h; y++) { |
| 2202 | for (x = 0; x < w-1; x++) { |
| 2203 | if ((grid[y*w+x] == TREE && TENT(grid[y*w+x+1])) || |
| 2204 | (TENT(grid[y*w+x]) && grid[y*w+x+1] == TREE)) |
| 2205 | dsf_merge(dsf, y*w+x, y*w+x+1); |
| 2206 | } |
| 2207 | } |
| 2208 | for (y = 0; y < h-1; y++) { |
| 2209 | for (x = 0; x < w; x++) { |
| 2210 | if ((grid[y*w+x] == TREE && TENT(grid[(y+1)*w+x])) || |
| 2211 | (TENT(grid[y*w+x]) && grid[(y+1)*w+x] == TREE)) |
| 2212 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
| 2213 | } |
| 2214 | } |
| 2215 | /* Count up the tent/tree difference in each one. */ |
| 2216 | for (x = 0; x < w*h; x++) |
| 2217 | tmp[x] = 0; |
| 2218 | for (x = 0; x < w*h; x++) { |
| 2219 | y = dsf_canonify(dsf, x); |
| 2220 | if (grid[x] == TREE) |
| 2221 | tmp[y]++; |
| 2222 | else if (TENT(grid[x])) |
| 2223 | tmp[y]--; |
| 2224 | } |
| 2225 | /* And highlight any tree belonging to an equivalence class with |
| 2226 | * a score more than zero. */ |
| 2227 | for (x = 0; x < w*h; x++) { |
| 2228 | y = dsf_canonify(dsf, x); |
| 2229 | if (grid[x] == TREE && tmp[y] > 0) |
| 2230 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
| 2231 | } |
| 2232 | #undef TENT |
| 2233 | |
| 2234 | sfree(tmp); |
| 2235 | return ret; |
| 2236 | } |
| 2237 | |
| 2238 | static void draw_err_adj(drawing *dr, game_drawstate *ds, int x, int y) |
| 2239 | { |
| 2240 | int coords[8]; |
| 2241 | int yext, xext; |
| 2242 | |
| 2243 | /* |
| 2244 | * Draw a diamond. |
| 2245 | */ |
| 2246 | coords[0] = x - TILESIZE*2/5; |
| 2247 | coords[1] = y; |
| 2248 | coords[2] = x; |
| 2249 | coords[3] = y - TILESIZE*2/5; |
| 2250 | coords[4] = x + TILESIZE*2/5; |
| 2251 | coords[5] = y; |
| 2252 | coords[6] = x; |
| 2253 | coords[7] = y + TILESIZE*2/5; |
| 2254 | draw_polygon(dr, coords, 4, COL_ERROR, COL_GRID); |
| 2255 | |
| 2256 | /* |
| 2257 | * Draw an exclamation mark in the diamond. This turns out to |
| 2258 | * look unpleasantly off-centre if done via draw_text, so I do |
| 2259 | * it by hand on the basis that exclamation marks aren't that |
| 2260 | * difficult to draw... |
| 2261 | */ |
| 2262 | xext = TILESIZE/16; |
| 2263 | yext = TILESIZE*2/5 - (xext*2+2); |
| 2264 | draw_rect(dr, x-xext, y-yext, xext*2+1, yext*2+1 - (xext*3), |
| 2265 | COL_ERRTEXT); |
| 2266 | draw_rect(dr, x-xext, y+yext-xext*2+1, xext*2+1, xext*2, COL_ERRTEXT); |
| 2267 | } |
| 2268 | |
| 2269 | static void draw_tile(drawing *dr, game_drawstate *ds, |
| 2270 | int x, int y, int v, int cur, int printing) |
| 2271 | { |
| 2272 | int err; |
| 2273 | int tx = COORD(x), ty = COORD(y); |
| 2274 | int cx = tx + TILESIZE/2, cy = ty + TILESIZE/2; |
| 2275 | |
| 2276 | err = v & ~15; |
| 2277 | v &= 15; |
| 2278 | |
| 2279 | clip(dr, tx, ty, TILESIZE, TILESIZE); |
| 2280 | |
| 2281 | if (!printing) { |
| 2282 | draw_rect(dr, tx, ty, TILESIZE, TILESIZE, COL_GRID); |
| 2283 | draw_rect(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1, |
| 2284 | (v == BLANK ? COL_BACKGROUND : COL_GRASS)); |
| 2285 | } |
| 2286 | |
| 2287 | if (v == TREE) { |
| 2288 | int i; |
| 2289 | |
| 2290 | (printing ? draw_rect_outline : draw_rect) |
| 2291 | (dr, cx-TILESIZE/15, ty+TILESIZE*3/10, |
| 2292 | 2*(TILESIZE/15)+1, (TILESIZE*9/10 - TILESIZE*3/10), |
| 2293 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERRTRUNK : COL_TREETRUNK)); |
| 2294 | |
| 2295 | for (i = 0; i < (printing ? 2 : 1); i++) { |
| 2296 | int col = (i == 1 ? COL_BACKGROUND : |
| 2297 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : |
| 2298 | COL_TREELEAF)); |
| 2299 | int sub = i * (TILESIZE/32); |
| 2300 | draw_circle(dr, cx, ty+TILESIZE*4/10, TILESIZE/4 - sub, |
| 2301 | col, col); |
| 2302 | draw_circle(dr, cx+TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
| 2303 | col, col); |
| 2304 | draw_circle(dr, cx-TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
| 2305 | col, col); |
| 2306 | draw_circle(dr, cx+TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
| 2307 | col, col); |
| 2308 | draw_circle(dr, cx-TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
| 2309 | col, col); |
| 2310 | } |
| 2311 | } else if (v == TENT) { |
| 2312 | int coords[6]; |
| 2313 | int col; |
| 2314 | coords[0] = cx - TILESIZE/3; |
| 2315 | coords[1] = cy + TILESIZE/3; |
| 2316 | coords[2] = cx + TILESIZE/3; |
| 2317 | coords[3] = cy + TILESIZE/3; |
| 2318 | coords[4] = cx; |
| 2319 | coords[5] = cy - TILESIZE/3; |
| 2320 | col = (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : COL_TENT); |
| 2321 | draw_polygon(dr, coords, 3, (printing ? -1 : col), col); |
| 2322 | } |
| 2323 | |
| 2324 | if (err & (1 << ERR_ADJ_TOPLEFT)) |
| 2325 | draw_err_adj(dr, ds, tx, ty); |
| 2326 | if (err & (1 << ERR_ADJ_TOP)) |
| 2327 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty); |
| 2328 | if (err & (1 << ERR_ADJ_TOPRIGHT)) |
| 2329 | draw_err_adj(dr, ds, tx+TILESIZE, ty); |
| 2330 | if (err & (1 << ERR_ADJ_LEFT)) |
| 2331 | draw_err_adj(dr, ds, tx, ty+TILESIZE/2); |
| 2332 | if (err & (1 << ERR_ADJ_RIGHT)) |
| 2333 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE/2); |
| 2334 | if (err & (1 << ERR_ADJ_BOTLEFT)) |
| 2335 | draw_err_adj(dr, ds, tx, ty+TILESIZE); |
| 2336 | if (err & (1 << ERR_ADJ_BOT)) |
| 2337 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty+TILESIZE); |
| 2338 | if (err & (1 << ERR_ADJ_BOTRIGHT)) |
| 2339 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE); |
| 2340 | |
| 2341 | if (cur) { |
| 2342 | int coff = TILESIZE/8; |
| 2343 | draw_rect_outline(dr, tx + coff, ty + coff, |
| 2344 | TILESIZE - coff*2 + 1, TILESIZE - coff*2 + 1, |
| 2345 | COL_GRID); |
| 2346 | } |
| 2347 | |
| 2348 | unclip(dr); |
| 2349 | draw_update(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1); |
| 2350 | } |
| 2351 | |
| 2352 | /* |
| 2353 | * Internal redraw function, used for printing as well as drawing. |
| 2354 | */ |
| 2355 | static void int_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
| 2356 | game_state *state, int dir, game_ui *ui, |
| 2357 | float animtime, float flashtime, int printing) |
| 2358 | { |
| 2359 | int w = state->p.w, h = state->p.h; |
| 2360 | int x, y, flashing; |
| 2361 | int cx = -1, cy = -1; |
| 2362 | int cmoved = 0; |
| 2363 | char *tmpgrid; |
| 2364 | int *errors; |
| 2365 | |
| 2366 | if (ui) { |
| 2367 | if (ui->cdisp) { cx = ui->cx; cy = ui->cy; } |
| 2368 | if (cx != ds->cx || cy != ds->cy) cmoved = 1; |
| 2369 | } |
| 2370 | |
| 2371 | if (printing || !ds->started) { |
| 2372 | if (!printing) { |
| 2373 | int ww, wh; |
| 2374 | game_compute_size(&state->p, TILESIZE, &ww, &wh); |
| 2375 | draw_rect(dr, 0, 0, ww, wh, COL_BACKGROUND); |
| 2376 | draw_update(dr, 0, 0, ww, wh); |
| 2377 | ds->started = TRUE; |
| 2378 | } |
| 2379 | |
| 2380 | if (printing) |
| 2381 | print_line_width(dr, TILESIZE/64); |
| 2382 | |
| 2383 | /* |
| 2384 | * Draw the grid. |
| 2385 | */ |
| 2386 | for (y = 0; y <= h; y++) |
| 2387 | draw_line(dr, COORD(0), COORD(y), COORD(w), COORD(y), COL_GRID); |
| 2388 | for (x = 0; x <= w; x++) |
| 2389 | draw_line(dr, COORD(x), COORD(0), COORD(x), COORD(h), COL_GRID); |
| 2390 | } |
| 2391 | |
| 2392 | if (flashtime > 0) |
| 2393 | flashing = (int)(flashtime * 3 / FLASH_TIME) != 1; |
| 2394 | else |
| 2395 | flashing = FALSE; |
| 2396 | |
| 2397 | /* |
| 2398 | * Find errors. For this we use _part_ of the information from a |
| 2399 | * currently active drag: we transform dsx,dsy but not anything |
| 2400 | * else. (This seems to strike a good compromise between having |
| 2401 | * the error highlights respond instantly to single clicks, but |
| 2402 | * not giving constant feedback during a right-drag.) |
| 2403 | */ |
| 2404 | if (ui && ui->drag_button >= 0) { |
| 2405 | tmpgrid = snewn(w*h, char); |
| 2406 | memcpy(tmpgrid, state->grid, w*h); |
| 2407 | tmpgrid[ui->dsy * w + ui->dsx] = |
| 2408 | drag_xform(ui, ui->dsx, ui->dsy, tmpgrid[ui->dsy * w + ui->dsx]); |
| 2409 | errors = find_errors(state, tmpgrid); |
| 2410 | sfree(tmpgrid); |
| 2411 | } else { |
| 2412 | errors = find_errors(state, state->grid); |
| 2413 | } |
| 2414 | |
| 2415 | /* |
| 2416 | * Draw the grid. |
| 2417 | */ |
| 2418 | for (y = 0; y < h; y++) { |
| 2419 | for (x = 0; x < w; x++) { |
| 2420 | int v = state->grid[y*w+x]; |
| 2421 | int credraw = 0; |
| 2422 | |
| 2423 | /* |
| 2424 | * We deliberately do not take drag_ok into account |
| 2425 | * here, because user feedback suggests that it's |
| 2426 | * marginally nicer not to have the drag effects |
| 2427 | * flickering on and off disconcertingly. |
| 2428 | */ |
| 2429 | if (ui && ui->drag_button >= 0) |
| 2430 | v = drag_xform(ui, x, y, v); |
| 2431 | |
| 2432 | if (flashing && (v == TREE || v == TENT)) |
| 2433 | v = NONTENT; |
| 2434 | |
| 2435 | if (cmoved) { |
| 2436 | if ((x == cx && y == cy) || |
| 2437 | (x == ds->cx && y == ds->cy)) credraw = 1; |
| 2438 | } |
| 2439 | |
| 2440 | v |= errors[y*w+x]; |
| 2441 | |
| 2442 | if (printing || ds->drawn[y*w+x] != v || credraw) { |
| 2443 | draw_tile(dr, ds, x, y, v, (x == cx && y == cy), printing); |
| 2444 | if (!printing) |
| 2445 | ds->drawn[y*w+x] = v; |
| 2446 | } |
| 2447 | } |
| 2448 | } |
| 2449 | |
| 2450 | /* |
| 2451 | * Draw (or redraw, if their error-highlighted state has |
| 2452 | * changed) the numbers. |
| 2453 | */ |
| 2454 | for (x = 0; x < w; x++) { |
| 2455 | if (ds->numbersdrawn[x] != errors[w*h+x]) { |
| 2456 | char buf[80]; |
| 2457 | draw_rect(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1, |
| 2458 | COL_BACKGROUND); |
| 2459 | sprintf(buf, "%d", state->numbers->numbers[x]); |
| 2460 | draw_text(dr, COORD(x) + TILESIZE/2, COORD(h+1), |
| 2461 | FONT_VARIABLE, TILESIZE/2, ALIGN_HCENTRE|ALIGN_VNORMAL, |
| 2462 | (errors[w*h+x] ? COL_ERROR : COL_GRID), buf); |
| 2463 | draw_update(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1); |
| 2464 | ds->numbersdrawn[x] = errors[w*h+x]; |
| 2465 | } |
| 2466 | } |
| 2467 | for (y = 0; y < h; y++) { |
| 2468 | if (ds->numbersdrawn[w+y] != errors[w*h+w+y]) { |
| 2469 | char buf[80]; |
| 2470 | draw_rect(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE, |
| 2471 | COL_BACKGROUND); |
| 2472 | sprintf(buf, "%d", state->numbers->numbers[w+y]); |
| 2473 | draw_text(dr, COORD(w+1), COORD(y) + TILESIZE/2, |
| 2474 | FONT_VARIABLE, TILESIZE/2, ALIGN_HRIGHT|ALIGN_VCENTRE, |
| 2475 | (errors[w*h+w+y] ? COL_ERROR : COL_GRID), buf); |
| 2476 | draw_update(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE); |
| 2477 | ds->numbersdrawn[w+y] = errors[w*h+w+y]; |
| 2478 | } |
| 2479 | } |
| 2480 | |
| 2481 | if (cmoved) { |
| 2482 | ds->cx = cx; |
| 2483 | ds->cy = cy; |
| 2484 | } |
| 2485 | |
| 2486 | sfree(errors); |
| 2487 | } |
| 2488 | |
| 2489 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
| 2490 | game_state *state, int dir, game_ui *ui, |
| 2491 | float animtime, float flashtime) |
| 2492 | { |
| 2493 | int_redraw(dr, ds, oldstate, state, dir, ui, animtime, flashtime, FALSE); |
| 2494 | } |
| 2495 | |
| 2496 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
| 2497 | int dir, game_ui *ui) |
| 2498 | { |
| 2499 | return 0.0F; |
| 2500 | } |
| 2501 | |
| 2502 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
| 2503 | int dir, game_ui *ui) |
| 2504 | { |
| 2505 | if (!oldstate->completed && newstate->completed && |
| 2506 | !oldstate->used_solve && !newstate->used_solve) |
| 2507 | return FLASH_TIME; |
| 2508 | |
| 2509 | return 0.0F; |
| 2510 | } |
| 2511 | |
| 2512 | static int game_timing_state(game_state *state, game_ui *ui) |
| 2513 | { |
| 2514 | return TRUE; |
| 2515 | } |
| 2516 | |
| 2517 | static void game_print_size(game_params *params, float *x, float *y) |
| 2518 | { |
| 2519 | int pw, ph; |
| 2520 | |
| 2521 | /* |
| 2522 | * I'll use 6mm squares by default. |
| 2523 | */ |
| 2524 | game_compute_size(params, 600, &pw, &ph); |
| 2525 | *x = pw / 100.0F; |
| 2526 | *y = ph / 100.0F; |
| 2527 | } |
| 2528 | |
| 2529 | static void game_print(drawing *dr, game_state *state, int tilesize) |
| 2530 | { |
| 2531 | int c; |
| 2532 | |
| 2533 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
| 2534 | game_drawstate ads, *ds = &ads; |
| 2535 | game_set_size(dr, ds, NULL, tilesize); |
| 2536 | |
| 2537 | c = print_mono_colour(dr, 1); assert(c == COL_BACKGROUND); |
| 2538 | c = print_mono_colour(dr, 0); assert(c == COL_GRID); |
| 2539 | c = print_mono_colour(dr, 1); assert(c == COL_GRASS); |
| 2540 | c = print_mono_colour(dr, 0); assert(c == COL_TREETRUNK); |
| 2541 | c = print_mono_colour(dr, 0); assert(c == COL_TREELEAF); |
| 2542 | c = print_mono_colour(dr, 0); assert(c == COL_TENT); |
| 2543 | |
| 2544 | int_redraw(dr, ds, NULL, state, +1, NULL, 0.0F, 0.0F, TRUE); |
| 2545 | } |
| 2546 | |
| 2547 | #ifdef COMBINED |
| 2548 | #define thegame tents |
| 2549 | #endif |
| 2550 | |
| 2551 | const struct game thegame = { |
| 2552 | "Tents", "games.tents", "tents", |
| 2553 | default_params, |
| 2554 | game_fetch_preset, |
| 2555 | decode_params, |
| 2556 | encode_params, |
| 2557 | free_params, |
| 2558 | dup_params, |
| 2559 | TRUE, game_configure, custom_params, |
| 2560 | validate_params, |
| 2561 | new_game_desc, |
| 2562 | validate_desc, |
| 2563 | new_game, |
| 2564 | dup_game, |
| 2565 | free_game, |
| 2566 | TRUE, solve_game, |
| 2567 | FALSE, game_can_format_as_text_now, game_text_format, |
| 2568 | new_ui, |
| 2569 | free_ui, |
| 2570 | encode_ui, |
| 2571 | decode_ui, |
| 2572 | game_changed_state, |
| 2573 | interpret_move, |
| 2574 | execute_move, |
| 2575 | PREFERRED_TILESIZE, game_compute_size, game_set_size, |
| 2576 | game_colours, |
| 2577 | game_new_drawstate, |
| 2578 | game_free_drawstate, |
| 2579 | game_redraw, |
| 2580 | game_anim_length, |
| 2581 | game_flash_length, |
| 2582 | TRUE, FALSE, game_print_size, game_print, |
| 2583 | FALSE, /* wants_statusbar */ |
| 2584 | FALSE, game_timing_state, |
| 2585 | REQUIRE_RBUTTON, /* flags */ |
| 2586 | }; |
| 2587 | |
| 2588 | #ifdef STANDALONE_SOLVER |
| 2589 | |
| 2590 | #include <stdarg.h> |
| 2591 | |
| 2592 | int main(int argc, char **argv) |
| 2593 | { |
| 2594 | game_params *p; |
| 2595 | game_state *s, *s2; |
| 2596 | char *id = NULL, *desc, *err; |
| 2597 | int grade = FALSE; |
| 2598 | int ret, diff, really_verbose = FALSE; |
| 2599 | struct solver_scratch *sc; |
| 2600 | |
| 2601 | while (--argc > 0) { |
| 2602 | char *p = *++argv; |
| 2603 | if (!strcmp(p, "-v")) { |
| 2604 | really_verbose = TRUE; |
| 2605 | } else if (!strcmp(p, "-g")) { |
| 2606 | grade = TRUE; |
| 2607 | } else if (*p == '-') { |
| 2608 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
| 2609 | return 1; |
| 2610 | } else { |
| 2611 | id = p; |
| 2612 | } |
| 2613 | } |
| 2614 | |
| 2615 | if (!id) { |
| 2616 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
| 2617 | return 1; |
| 2618 | } |
| 2619 | |
| 2620 | desc = strchr(id, ':'); |
| 2621 | if (!desc) { |
| 2622 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
| 2623 | return 1; |
| 2624 | } |
| 2625 | *desc++ = '\0'; |
| 2626 | |
| 2627 | p = default_params(); |
| 2628 | decode_params(p, id); |
| 2629 | err = validate_desc(p, desc); |
| 2630 | if (err) { |
| 2631 | fprintf(stderr, "%s: %s\n", argv[0], err); |
| 2632 | return 1; |
| 2633 | } |
| 2634 | s = new_game(NULL, p, desc); |
| 2635 | s2 = new_game(NULL, p, desc); |
| 2636 | |
| 2637 | sc = new_scratch(p->w, p->h); |
| 2638 | |
| 2639 | /* |
| 2640 | * When solving an Easy puzzle, we don't want to bother the |
| 2641 | * user with Hard-level deductions. For this reason, we grade |
| 2642 | * the puzzle internally before doing anything else. |
| 2643 | */ |
| 2644 | ret = -1; /* placate optimiser */ |
| 2645 | for (diff = 0; diff < DIFFCOUNT; diff++) { |
| 2646 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
| 2647 | s2->grid, sc, diff); |
| 2648 | if (ret < 2) |
| 2649 | break; |
| 2650 | } |
| 2651 | |
| 2652 | if (diff == DIFFCOUNT) { |
| 2653 | if (grade) |
| 2654 | printf("Difficulty rating: too hard to solve internally\n"); |
| 2655 | else |
| 2656 | printf("Unable to find a unique solution\n"); |
| 2657 | } else { |
| 2658 | if (grade) { |
| 2659 | if (ret == 0) |
| 2660 | printf("Difficulty rating: impossible (no solution exists)\n"); |
| 2661 | else if (ret == 1) |
| 2662 | printf("Difficulty rating: %s\n", tents_diffnames[diff]); |
| 2663 | } else { |
| 2664 | verbose = really_verbose; |
| 2665 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
| 2666 | s2->grid, sc, diff); |
| 2667 | if (ret == 0) |
| 2668 | printf("Puzzle is inconsistent\n"); |
| 2669 | else |
| 2670 | fputs(game_text_format(s2), stdout); |
| 2671 | } |
| 2672 | } |
| 2673 | |
| 2674 | return 0; |
| 2675 | } |
| 2676 | |
| 2677 | #endif |
| 2678 | |
| 2679 | /* vim: set shiftwidth=4 tabstop=8: */ |