86e60e3d |
1 | /* |
2 | * tents.c: Puzzle involving placing tents next to trees subject to |
3 | * some confusing conditions. |
4 | * |
5 | * TODO: |
6 | * |
7 | * - error highlighting? |
8 | * * highlighting adjacent tents is easy |
9 | * * highlighting violated numeric clues is almost as easy |
10 | * (might want to pay attention to NONTENTs here) |
11 | * * but how in hell do we highlight a failure of maxflow |
12 | * during completion checking? |
13 | * + well, the _obvious_ approach is to use maxflow's own |
14 | * error report: it will provide, via the `cut' parameter, |
15 | * a set of trees which have too few tents between them. |
16 | * It's unclear that this will be particularly obvious to |
17 | * a user, however. Is there any other way? |
18 | * |
19 | * - it might be nice to make setter-provided tent/nontent clues |
20 | * inviolable? |
21 | * * on the other hand, this would introduce considerable extra |
22 | * complexity and size into the game state; also inviolable |
23 | * clues would have to be marked as such somehow, in an |
24 | * intrusive and annoying manner. Since they're never |
25 | * generated by _my_ generator, I'm currently more inclined |
26 | * not to bother. |
27 | * |
28 | * - more difficult levels at the top end? |
29 | * * for example, sometimes we can deduce that two BLANKs in |
30 | * the same row are each adjacent to the same unattached tree |
31 | * and to nothing else, implying that they can't both be |
32 | * tents; this enables us to rule out some extra combinations |
33 | * in the row-based deduction loop, and hence deduce more |
34 | * from the number in that row than we could otherwise do. |
35 | * * that by itself doesn't seem worth implementing a new |
36 | * difficulty level for, but if I can find a few more things |
37 | * like that then it might become worthwhile. |
38 | * * I wonder if there's a sensible heuristic for where to |
39 | * guess which would make a recursive solver viable? |
40 | */ |
41 | |
42 | #include <stdio.h> |
43 | #include <stdlib.h> |
44 | #include <string.h> |
45 | #include <assert.h> |
46 | #include <ctype.h> |
47 | #include <math.h> |
48 | |
49 | #include "puzzles.h" |
50 | #include "maxflow.h" |
51 | |
52 | /* |
53 | * Design discussion |
54 | * ----------------- |
55 | * |
56 | * The rules of this puzzle as available on the WWW are poorly |
57 | * specified. The bits about tents having to be orthogonally |
58 | * adjacent to trees, tents not being even diagonally adjacent to |
59 | * one another, and the number of tents in each row and column |
60 | * being given are simple enough; the difficult bit is the |
61 | * tent-to-tree matching. |
62 | * |
63 | * Some sources use simplistic wordings such as `each tree is |
64 | * exactly connected to only one tent', which is extremely unclear: |
65 | * it's easy to read erroneously as `each tree is _orthogonally |
66 | * adjacent_ to exactly one tent', which is definitely incorrect. |
67 | * Even the most coherent sources I've found don't do a much better |
68 | * job of stating the rule. |
69 | * |
70 | * A more precise statement of the rule is that it must be possible |
71 | * to find a bijection f between tents and trees such that each |
72 | * tree T is orthogonally adjacent to the tent f(T), but that a |
73 | * tent is permitted to be adjacent to other trees in addition to |
74 | * its own. This slightly non-obvious criterion is what gives this |
75 | * puzzle most of its subtlety. |
76 | * |
77 | * However, there's a particularly subtle ambiguity left over. Is |
78 | * the bijection between tents and trees required to be _unique_? |
79 | * In other words, is that bijection conceptually something the |
80 | * player should be able to exhibit as part of the solution (even |
81 | * if they aren't actually required to do so)? Or is it sufficient |
82 | * to have a unique _placement_ of the tents which gives rise to at |
83 | * least one suitable bijection? |
84 | * |
85 | * The puzzle shown to the right of this .T. 2 *T* 2 |
86 | * paragraph illustrates the problem. There T.T 0 -> T-T 0 |
87 | * are two distinct bijections available. .T. 2 *T* 2 |
88 | * The answer to the above question will |
89 | * determine whether it's a valid puzzle. 202 202 |
90 | * |
91 | * This is an important question, because it affects both the |
92 | * player and the generator. Eventually I found all the instances |
93 | * of this puzzle I could Google up, solved them all by hand, and |
94 | * verified that in all cases the tree/tent matching was uniquely |
95 | * determined given the tree and tent positions. Therefore, the |
96 | * puzzle as implemented in this source file takes the following |
97 | * policy: |
98 | * |
99 | * - When checking a user-supplied solution for correctness, only |
100 | * verify that there exists _at least_ one matching. |
101 | * - When generating a puzzle, enforce that there must be |
102 | * _exactly_ one. |
103 | * |
104 | * Algorithmic implications |
105 | * ------------------------ |
106 | * |
107 | * Another way of phrasing the tree/tent matching criterion is to |
108 | * say that the bipartite adjacency graph between trees and tents |
109 | * has a perfect matching. That is, if you construct a graph which |
110 | * has a vertex per tree and a vertex per tent, and an edge between |
111 | * any tree and tent which are orthogonally adjacent, it is |
112 | * possible to find a set of N edges of that graph (where N is the |
113 | * number of trees and also the number of tents) which between them |
114 | * connect every tree to every tent. |
115 | * |
116 | * The most efficient known algorithms for finding such a matching |
117 | * given a graph, as far as I'm aware, are the Munkres assignment |
118 | * algorithm (also known as the Hungarian algorithm) and the |
119 | * Ford-Fulkerson algorithm (for finding optimal flows in |
120 | * networks). Each of these takes O(N^3) running time; so we're |
121 | * talking O(N^3) time to verify any candidate solution to this |
122 | * puzzle. That's just about OK if you're doing it once per mouse |
123 | * click (and in fact not even that, since the sensible thing to do |
124 | * is check all the _other_ puzzle criteria and only wade into this |
125 | * quagmire if none are violated); but if the solver had to keep |
126 | * doing N^3 work internally, then it would probably end up with |
127 | * more like N^5 or N^6 running time, and grid generation would |
128 | * become very clunky. |
129 | * |
130 | * Fortunately, I've been able to prove a very useful property of |
131 | * _unique_ perfect matchings, by adapting the proof of Hall's |
132 | * Marriage Theorem. For those unaware of Hall's Theorem, I'll |
133 | * recap it and its proof: it states that a bipartite graph |
134 | * contains a perfect matching iff every set of vertices on the |
135 | * left side of the graph have a neighbourhood _at least_ as big on |
136 | * the right. |
137 | * |
138 | * This condition is obviously satisfied if a perfect matching does |
139 | * exist; each left-side node has a distinct right-side node which |
140 | * is the one assigned to it by the matching, and thus any set of n |
141 | * left vertices must have a combined neighbourhood containing at |
142 | * least the n corresponding right vertices, and possibly others |
143 | * too. Alternatively, imagine if you had (say) three left-side |
144 | * nodes all of which were connected to only two right-side nodes |
145 | * between them: any perfect matching would have to assign one of |
146 | * those two right nodes to each of the three left nodes, and still |
147 | * give the three left nodes a different right node each. This is |
148 | * of course impossible. |
149 | * |
150 | * To prove the converse (that if every subset of left vertices |
151 | * satisfies the Hall condition then a perfect matching exists), |
152 | * consider trying to find a proper subset of the left vertices |
153 | * which _exactly_ satisfies the Hall condition: that is, its right |
154 | * neighbourhood is precisely the same size as it. If we can find |
155 | * such a subset, then we can split the bipartite graph into two |
156 | * smaller ones: one consisting of the left subset and its right |
157 | * neighbourhood, the other consisting of everything else. Edges |
158 | * from the left side of the former graph to the right side of the |
159 | * latter do not exist, by construction; edges from the right side |
160 | * of the former to the left of the latter cannot be part of any |
161 | * perfect matching because otherwise the left subset would not be |
162 | * left with enough distinct right vertices to connect to (this is |
163 | * exactly the same deduction used in Solo's set analysis). You can |
164 | * then prove (left as an exercise) that both these smaller graphs |
165 | * still satisfy the Hall condition, and therefore the proof will |
166 | * follow by induction. |
167 | * |
168 | * There's one other possibility, which is the case where _no_ |
169 | * proper subset of the left vertices has a right neighbourhood of |
170 | * exactly the same size. That is, every left subset has a strictly |
171 | * _larger_ right neighbourhood. In this situation, we can simply |
172 | * remove an _arbitrary_ edge from the graph. This cannot reduce |
173 | * the size of any left subset's right neighbourhood by more than |
174 | * one, so if all neighbourhoods were strictly bigger than they |
175 | * needed to be initially, they must now still be _at least as big_ |
176 | * as they need to be. So we can keep throwing out arbitrary edges |
177 | * until we find a set which exactly satisfies the Hall condition, |
178 | * and then proceed as above. [] |
179 | * |
180 | * That's Hall's theorem. I now build on this by examining the |
181 | * circumstances in which a bipartite graph can have a _unique_ |
182 | * perfect matching. It is clear that in the second case, where no |
183 | * left subset exactly satisfies the Hall condition and so we can |
184 | * remove an arbitrary edge, there cannot be a unique perfect |
185 | * matching: given one perfect matching, we choose our arbitrary |
186 | * removed edge to be one of those contained in it, and then we can |
187 | * still find a perfect matching in the remaining graph, which will |
188 | * be a distinct perfect matching in the original. |
189 | * |
190 | * So it is a necessary condition for a unique perfect matching |
191 | * that there must be at least one proper left subset which |
192 | * _exactly_ satisfies the Hall condition. But now consider the |
193 | * smaller graph constructed by taking that left subset and its |
194 | * neighbourhood: if the graph as a whole had a unique perfect |
195 | * matching, then so must this smaller one, which means we can find |
196 | * a proper left subset _again_, and so on. Repeating this process |
197 | * must eventually reduce us to a graph with only one left-side |
198 | * vertex (so there are no proper subsets at all); this vertex must |
199 | * be connected to only one right-side vertex, and hence must be so |
200 | * in the original graph as well (by construction). So we can |
201 | * discard this vertex pair from the graph, and any other edges |
202 | * that involved it (which will by construction be from other left |
203 | * vertices only), and the resulting smaller graph still has a |
204 | * unique perfect matching which means we can do the same thing |
205 | * again. |
206 | * |
207 | * In other words, given any bipartite graph with a unique perfect |
208 | * matching, we can find that matching by the following extremely |
209 | * simple algorithm: |
210 | * |
211 | * - Find a left-side vertex which is only connected to one |
212 | * right-side vertex. |
213 | * - Assign those vertices to one another, and therefore discard |
214 | * any other edges connecting to that right vertex. |
215 | * - Repeat until all vertices have been matched. |
216 | * |
217 | * This algorithm can be run in O(V+E) time (where V is the number |
218 | * of vertices and E is the number of edges in the graph), and the |
219 | * only way it can fail is if there is not a unique perfect |
220 | * matching (either because there is no matching at all, or because |
221 | * it isn't unique; but it can't distinguish those cases). |
222 | * |
223 | * Thus, the internal solver in this source file can be confident |
224 | * that if the tree/tent matching is uniquely determined by the |
225 | * tree and tent positions, it can find it using only this kind of |
226 | * obvious and simple operation: assign a tree to a tent if it |
227 | * cannot possibly belong to any other tent, and vice versa. If the |
228 | * solver were _only_ trying to determine the matching, even that |
229 | * `vice versa' wouldn't be required; but it can come in handy when |
230 | * not all the tents have been placed yet. I can therefore be |
231 | * reasonably confident that as long as my solver doesn't need to |
232 | * cope with grids that have a non-unique matching, it will also |
233 | * not need to do anything complicated like set analysis between |
234 | * trees and tents. |
235 | */ |
236 | |
237 | /* |
238 | * In standalone solver mode, `verbose' is a variable which can be |
239 | * set by command-line option; in debugging mode it's simply always |
240 | * true. |
241 | */ |
242 | #if defined STANDALONE_SOLVER |
243 | #define SOLVER_DIAGNOSTICS |
244 | int verbose = FALSE; |
245 | #elif defined SOLVER_DIAGNOSTICS |
246 | #define verbose TRUE |
247 | #endif |
248 | |
249 | /* |
250 | * Difficulty levels. I do some macro ickery here to ensure that my |
251 | * enum and the various forms of my name list always match up. |
252 | */ |
253 | #define DIFFLIST(A) \ |
254 | A(EASY,Easy,e) \ |
255 | A(TRICKY,Tricky,t) |
256 | #define ENUM(upper,title,lower) DIFF_ ## upper, |
257 | #define TITLE(upper,title,lower) #title, |
258 | #define ENCODE(upper,title,lower) #lower |
259 | #define CONFIG(upper,title,lower) ":" #title |
260 | enum { DIFFLIST(ENUM) DIFFCOUNT }; |
261 | static char const *const tents_diffnames[] = { DIFFLIST(TITLE) }; |
262 | static char const tents_diffchars[] = DIFFLIST(ENCODE); |
263 | #define DIFFCONFIG DIFFLIST(CONFIG) |
264 | |
265 | enum { |
266 | COL_BACKGROUND, |
267 | COL_GRID, |
268 | COL_GRASS, |
269 | COL_TREETRUNK, |
270 | COL_TREELEAF, |
271 | COL_TENT, |
272 | NCOLOURS |
273 | }; |
274 | |
275 | enum { BLANK, TREE, TENT, NONTENT, MAGIC }; |
276 | |
277 | struct game_params { |
278 | int w, h; |
279 | int diff; |
280 | }; |
281 | |
282 | struct numbers { |
283 | int refcount; |
284 | int *numbers; |
285 | }; |
286 | |
287 | struct game_state { |
288 | game_params p; |
289 | char *grid; |
290 | struct numbers *numbers; |
291 | int completed, used_solve; |
292 | }; |
293 | |
294 | static game_params *default_params(void) |
295 | { |
296 | game_params *ret = snew(game_params); |
297 | |
298 | ret->w = ret->h = 8; |
299 | ret->diff = DIFF_EASY; |
300 | |
301 | return ret; |
302 | } |
303 | |
304 | static const struct game_params tents_presets[] = { |
305 | {8, 8, DIFF_EASY}, |
306 | {8, 8, DIFF_TRICKY}, |
307 | {10, 10, DIFF_EASY}, |
308 | {10, 10, DIFF_TRICKY}, |
309 | {15, 15, DIFF_EASY}, |
310 | {15, 15, DIFF_TRICKY}, |
311 | }; |
312 | |
313 | static int game_fetch_preset(int i, char **name, game_params **params) |
314 | { |
315 | game_params *ret; |
316 | char str[80]; |
317 | |
318 | if (i < 0 || i >= lenof(tents_presets)) |
319 | return FALSE; |
320 | |
321 | ret = snew(game_params); |
322 | *ret = tents_presets[i]; |
323 | |
324 | sprintf(str, "%dx%d %s", ret->w, ret->h, tents_diffnames[ret->diff]); |
325 | |
326 | *name = dupstr(str); |
327 | *params = ret; |
328 | return TRUE; |
329 | } |
330 | |
331 | static void free_params(game_params *params) |
332 | { |
333 | sfree(params); |
334 | } |
335 | |
336 | static game_params *dup_params(game_params *params) |
337 | { |
338 | game_params *ret = snew(game_params); |
339 | *ret = *params; /* structure copy */ |
340 | return ret; |
341 | } |
342 | |
343 | static void decode_params(game_params *params, char const *string) |
344 | { |
345 | params->w = params->h = atoi(string); |
346 | while (*string && isdigit((unsigned char)*string)) string++; |
347 | if (*string == 'x') { |
348 | string++; |
349 | params->h = atoi(string); |
350 | while (*string && isdigit((unsigned char)*string)) string++; |
351 | } |
352 | if (*string == 'd') { |
353 | int i; |
354 | string++; |
355 | for (i = 0; i < DIFFCOUNT; i++) |
356 | if (*string == tents_diffchars[i]) |
357 | params->diff = i; |
358 | if (*string) string++; |
359 | } |
360 | } |
361 | |
362 | static char *encode_params(game_params *params, int full) |
363 | { |
364 | char buf[120]; |
365 | |
366 | sprintf(buf, "%dx%d", params->w, params->h); |
367 | if (full) |
368 | sprintf(buf + strlen(buf), "d%c", |
369 | tents_diffchars[params->diff]); |
370 | return dupstr(buf); |
371 | } |
372 | |
373 | static config_item *game_configure(game_params *params) |
374 | { |
375 | config_item *ret; |
376 | char buf[80]; |
377 | |
378 | ret = snewn(4, config_item); |
379 | |
380 | ret[0].name = "Width"; |
381 | ret[0].type = C_STRING; |
382 | sprintf(buf, "%d", params->w); |
383 | ret[0].sval = dupstr(buf); |
384 | ret[0].ival = 0; |
385 | |
386 | ret[1].name = "Height"; |
387 | ret[1].type = C_STRING; |
388 | sprintf(buf, "%d", params->h); |
389 | ret[1].sval = dupstr(buf); |
390 | ret[1].ival = 0; |
391 | |
392 | ret[2].name = "Difficulty"; |
393 | ret[2].type = C_CHOICES; |
394 | ret[2].sval = DIFFCONFIG; |
395 | ret[2].ival = params->diff; |
396 | |
397 | ret[3].name = NULL; |
398 | ret[3].type = C_END; |
399 | ret[3].sval = NULL; |
400 | ret[3].ival = 0; |
401 | |
402 | return ret; |
403 | } |
404 | |
405 | static game_params *custom_params(config_item *cfg) |
406 | { |
407 | game_params *ret = snew(game_params); |
408 | |
409 | ret->w = atoi(cfg[0].sval); |
410 | ret->h = atoi(cfg[1].sval); |
411 | ret->diff = cfg[2].ival; |
412 | |
413 | return ret; |
414 | } |
415 | |
416 | static char *validate_params(game_params *params, int full) |
417 | { |
418 | if (params->w < 2 || params->h < 2) |
419 | return "Width and height must both be at least two"; |
420 | return NULL; |
421 | } |
422 | |
423 | /* |
424 | * Scratch space for solver. |
425 | */ |
426 | enum { N, U, L, R, D, MAXDIR }; /* link directions */ |
427 | #define dx(d) ( ((d)==R) - ((d)==L) ) |
428 | #define dy(d) ( ((d)==D) - ((d)==U) ) |
429 | #define F(d) ( U + D - (d) ) |
430 | struct solver_scratch { |
431 | char *links; /* mapping between trees and tents */ |
432 | int *locs; |
433 | char *place, *mrows, *trows; |
434 | }; |
435 | |
436 | static struct solver_scratch *new_scratch(int w, int h) |
437 | { |
438 | struct solver_scratch *ret = snew(struct solver_scratch); |
439 | |
440 | ret->links = snewn(w*h, char); |
441 | ret->locs = snewn(max(w, h), int); |
442 | ret->place = snewn(max(w, h), char); |
443 | ret->mrows = snewn(3 * max(w, h), char); |
444 | ret->trows = snewn(3 * max(w, h), char); |
445 | |
446 | return ret; |
447 | } |
448 | |
449 | static void free_scratch(struct solver_scratch *sc) |
450 | { |
451 | sfree(sc->trows); |
452 | sfree(sc->mrows); |
453 | sfree(sc->place); |
454 | sfree(sc->locs); |
455 | sfree(sc->links); |
456 | sfree(sc); |
457 | } |
458 | |
459 | /* |
460 | * Solver. Returns 0 for impossibility, 1 for success, 2 for |
461 | * ambiguity or failure to converge. |
462 | */ |
463 | static int tents_solve(int w, int h, const char *grid, int *numbers, |
464 | char *soln, struct solver_scratch *sc, int diff) |
465 | { |
466 | int x, y, d, i, j; |
467 | char *mrow, *mrow1, *mrow2, *trow, *trow1, *trow2; |
468 | |
469 | /* |
470 | * Set up solver data. |
471 | */ |
472 | memset(sc->links, N, w*h); |
473 | |
474 | /* |
475 | * Set up solution array. |
476 | */ |
477 | memcpy(soln, grid, w*h); |
478 | |
479 | /* |
480 | * Main solver loop. |
481 | */ |
482 | while (1) { |
483 | int done_something = FALSE; |
484 | |
485 | /* |
486 | * Any tent which has only one unattached tree adjacent to |
487 | * it can be tied to that tree. |
488 | */ |
489 | for (y = 0; y < h; y++) |
490 | for (x = 0; x < w; x++) |
491 | if (soln[y*w+x] == TENT && !sc->links[y*w+x]) { |
492 | int linkd = 0; |
493 | |
494 | for (d = 1; d < MAXDIR; d++) { |
495 | int x2 = x + dx(d), y2 = y + dy(d); |
496 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
497 | soln[y2*w+x2] == TREE && |
498 | !sc->links[y2*w+x2]) { |
499 | if (linkd) |
500 | break; /* found more than one */ |
501 | else |
502 | linkd = d; |
503 | } |
504 | } |
505 | |
506 | if (d == MAXDIR && linkd == 0) { |
507 | #ifdef SOLVER_DIAGNOSTICS |
508 | if (verbose) |
509 | printf("tent at %d,%d cannot link to anything\n", |
510 | x, y); |
511 | #endif |
512 | return 0; /* no solution exists */ |
513 | } else if (d == MAXDIR) { |
514 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
515 | |
516 | #ifdef SOLVER_DIAGNOSTICS |
517 | if (verbose) |
518 | printf("tent at %d,%d can only link to tree at" |
519 | " %d,%d\n", x, y, x2, y2); |
520 | #endif |
521 | |
522 | sc->links[y*w+x] = linkd; |
523 | sc->links[y2*w+x2] = F(linkd); |
524 | done_something = TRUE; |
525 | } |
526 | } |
527 | |
528 | if (done_something) |
529 | continue; |
530 | if (diff < 0) |
531 | break; /* don't do anything else! */ |
532 | |
533 | /* |
534 | * Mark a blank square as NONTENT if it is not orthogonally |
535 | * adjacent to any unmatched tree. |
536 | */ |
537 | for (y = 0; y < h; y++) |
538 | for (x = 0; x < w; x++) |
539 | if (soln[y*w+x] == BLANK) { |
540 | int can_be_tent = FALSE; |
541 | |
542 | for (d = 1; d < MAXDIR; d++) { |
543 | int x2 = x + dx(d), y2 = y + dy(d); |
544 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
545 | soln[y2*w+x2] == TREE && |
546 | !sc->links[y2*w+x2]) |
547 | can_be_tent = TRUE; |
548 | } |
549 | |
550 | if (!can_be_tent) { |
551 | #ifdef SOLVER_DIAGNOSTICS |
552 | if (verbose) |
553 | printf("%d,%d cannot be a tent (no adjacent" |
554 | " unmatched tree)\n", x, y); |
555 | #endif |
556 | soln[y*w+x] = NONTENT; |
557 | done_something = TRUE; |
558 | } |
559 | } |
560 | |
561 | if (done_something) |
562 | continue; |
563 | |
564 | /* |
565 | * Mark a blank square as NONTENT if it is (perhaps |
566 | * diagonally) adjacent to any other tent. |
567 | */ |
568 | for (y = 0; y < h; y++) |
569 | for (x = 0; x < w; x++) |
570 | if (soln[y*w+x] == BLANK) { |
571 | int dx, dy, imposs = FALSE; |
572 | |
573 | for (dy = -1; dy <= +1; dy++) |
574 | for (dx = -1; dx <= +1; dx++) |
575 | if (dy || dx) { |
576 | int x2 = x + dx, y2 = y + dy; |
577 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
578 | soln[y2*w+x2] == TENT) |
579 | imposs = TRUE; |
580 | } |
581 | |
582 | if (imposs) { |
583 | #ifdef SOLVER_DIAGNOSTICS |
584 | if (verbose) |
585 | printf("%d,%d cannot be a tent (adjacent tent)\n", |
586 | x, y); |
587 | #endif |
588 | soln[y*w+x] = NONTENT; |
589 | done_something = TRUE; |
590 | } |
591 | } |
592 | |
593 | if (done_something) |
594 | continue; |
595 | |
596 | /* |
597 | * Any tree which has exactly one {unattached tent, BLANK} |
598 | * adjacent to it must have its tent in that square. |
599 | */ |
600 | for (y = 0; y < h; y++) |
601 | for (x = 0; x < w; x++) |
602 | if (soln[y*w+x] == TREE && !sc->links[y*w+x]) { |
603 | int linkd = 0, linkd2 = 0, nd = 0; |
604 | |
605 | for (d = 1; d < MAXDIR; d++) { |
606 | int x2 = x + dx(d), y2 = y + dy(d); |
607 | if (!(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h)) |
608 | continue; |
609 | if (soln[y2*w+x2] == BLANK || |
610 | (soln[y2*w+x2] == TENT && !sc->links[y2*w+x2])) { |
611 | if (linkd) |
612 | linkd2 = d; |
613 | else |
614 | linkd = d; |
615 | nd++; |
616 | } |
617 | } |
618 | |
619 | if (nd == 0) { |
620 | #ifdef SOLVER_DIAGNOSTICS |
621 | if (verbose) |
622 | printf("tree at %d,%d cannot link to anything\n", |
623 | x, y); |
624 | #endif |
625 | return 0; /* no solution exists */ |
626 | } else if (nd == 1) { |
627 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
628 | |
629 | #ifdef SOLVER_DIAGNOSTICS |
630 | if (verbose) |
631 | printf("tree at %d,%d can only link to tent at" |
632 | " %d,%d\n", x, y, x2, y2); |
633 | #endif |
634 | soln[y2*w+x2] = TENT; |
635 | sc->links[y*w+x] = linkd; |
636 | sc->links[y2*w+x2] = F(linkd); |
637 | done_something = TRUE; |
638 | } else if (nd == 2 && (!dx(linkd) != !dx(linkd2)) && |
639 | diff >= DIFF_TRICKY) { |
640 | /* |
641 | * If there are two possible places where |
642 | * this tree's tent can go, and they are |
643 | * diagonally separated rather than being |
644 | * on opposite sides of the tree, then the |
645 | * square (other than the tree square) |
646 | * which is adjacent to both of them must |
647 | * be a non-tent. |
648 | */ |
649 | int x2 = x + dx(linkd) + dx(linkd2); |
650 | int y2 = y + dy(linkd) + dy(linkd2); |
651 | assert(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h); |
652 | if (soln[y2*w+x2] == BLANK) { |
653 | #ifdef SOLVER_DIAGNOSTICS |
654 | if (verbose) |
655 | printf("possible tent locations for tree at" |
656 | " %d,%d rule out tent at %d,%d\n", |
657 | x, y, x2, y2); |
658 | #endif |
659 | soln[y2*w+x2] = NONTENT; |
660 | done_something = TRUE; |
661 | } |
662 | } |
663 | } |
664 | |
665 | if (done_something) |
666 | continue; |
667 | |
668 | /* |
669 | * If localised deductions about the trees and tents |
670 | * themselves haven't helped us, it's time to resort to the |
671 | * numbers round the grid edge. For each row and column, we |
672 | * go through all possible combinations of locations for |
673 | * the unplaced tents, rule out any which have adjacent |
674 | * tents, and spot any square which is given the same state |
675 | * by all remaining combinations. |
676 | */ |
677 | for (i = 0; i < w+h; i++) { |
678 | int start, step, len, start1, start2, n, k; |
679 | |
680 | if (i < w) { |
681 | /* |
682 | * This is the number for a column. |
683 | */ |
684 | start = i; |
685 | step = w; |
686 | len = h; |
687 | if (i > 0) |
688 | start1 = start - 1; |
689 | else |
690 | start1 = -1; |
691 | if (i+1 < w) |
692 | start2 = start + 1; |
693 | else |
694 | start2 = -1; |
695 | } else { |
696 | /* |
697 | * This is the number for a row. |
698 | */ |
699 | start = (i-w)*w; |
700 | step = 1; |
701 | len = w; |
702 | if (i > w) |
703 | start1 = start - w; |
704 | else |
705 | start1 = -1; |
706 | if (i+1 < w+h) |
707 | start2 = start + w; |
708 | else |
709 | start2 = -1; |
710 | } |
711 | |
712 | if (diff < DIFF_TRICKY) { |
713 | /* |
714 | * In Easy mode, we don't look at the effect of one |
715 | * row on the next (i.e. ruling out a square if all |
716 | * possibilities for an adjacent row place a tent |
717 | * next to it). |
718 | */ |
719 | start1 = start2 = -1; |
720 | } |
721 | |
722 | k = numbers[i]; |
723 | |
724 | /* |
725 | * Count and store the locations of the free squares, |
726 | * and also count the number of tents already placed. |
727 | */ |
728 | n = 0; |
729 | for (j = 0; j < len; j++) { |
730 | if (soln[start+j*step] == TENT) |
731 | k--; /* one fewer tent to place */ |
732 | else if (soln[start+j*step] == BLANK) |
733 | sc->locs[n++] = j; |
734 | } |
735 | |
736 | if (n == 0) |
737 | continue; /* nothing left to do here */ |
738 | |
739 | /* |
740 | * Now we know we're placing k tents in n squares. Set |
741 | * up the first possibility. |
742 | */ |
743 | for (j = 0; j < n; j++) |
744 | sc->place[j] = (j < k ? TENT : NONTENT); |
745 | |
746 | /* |
747 | * We're aiming to find squares in this row which are |
748 | * invariant over all valid possibilities. Thus, we |
749 | * maintain the current state of that invariance. We |
750 | * start everything off at MAGIC to indicate that it |
751 | * hasn't been set up yet. |
752 | */ |
753 | mrow = sc->mrows; |
754 | mrow1 = sc->mrows + len; |
755 | mrow2 = sc->mrows + 2*len; |
756 | trow = sc->trows; |
757 | trow1 = sc->trows + len; |
758 | trow2 = sc->trows + 2*len; |
759 | memset(mrow, MAGIC, 3*len); |
760 | |
761 | /* |
762 | * And iterate over all possibilities. |
763 | */ |
764 | while (1) { |
765 | int p, valid; |
766 | |
767 | /* |
768 | * See if this possibility is valid. The only way |
769 | * it can fail to be valid is if it contains two |
770 | * adjacent tents. (Other forms of invalidity, such |
771 | * as containing a tent adjacent to one already |
772 | * placed, will have been dealt with already by |
773 | * other parts of the solver.) |
774 | */ |
775 | valid = TRUE; |
776 | for (j = 0; j+1 < n; j++) |
777 | if (sc->place[j] == TENT && |
778 | sc->place[j+1] == TENT && |
779 | sc->locs[j+1] == sc->locs[j]+1) { |
780 | valid = FALSE; |
781 | break; |
782 | } |
783 | |
784 | if (valid) { |
785 | /* |
786 | * Merge this valid combination into mrow. |
787 | */ |
788 | memset(trow, MAGIC, len); |
789 | memset(trow+len, BLANK, 2*len); |
790 | for (j = 0; j < n; j++) { |
791 | trow[sc->locs[j]] = sc->place[j]; |
792 | if (sc->place[j] == TENT) { |
793 | int jj; |
794 | for (jj = sc->locs[j]-1; jj <= sc->locs[j]+1; jj++) |
795 | if (jj >= 0 && jj < len) |
796 | trow1[jj] = trow2[jj] = NONTENT; |
797 | } |
798 | } |
799 | |
800 | for (j = 0; j < 3*len; j++) { |
801 | if (trow[j] == MAGIC) |
802 | continue; |
803 | if (mrow[j] == MAGIC || mrow[j] == trow[j]) { |
804 | /* |
805 | * Either this is the first valid |
806 | * placement we've found at all, or |
807 | * this square's contents are |
808 | * consistent with every previous valid |
809 | * combination. |
810 | */ |
811 | mrow[j] = trow[j]; |
812 | } else { |
813 | /* |
814 | * This square's contents fail to match |
815 | * what they were in a different |
816 | * combination, so we cannot deduce |
817 | * anything about this square. |
818 | */ |
819 | mrow[j] = BLANK; |
820 | } |
821 | } |
822 | } |
823 | |
824 | /* |
825 | * Find the next combination of k choices from n. |
826 | * We do this by finding the rightmost tent which |
827 | * can be moved one place right, doing so, and |
828 | * shunting all tents to the right of that as far |
829 | * left as they can go. |
830 | */ |
831 | p = 0; |
832 | for (j = n-1; j > 0; j--) { |
833 | if (sc->place[j] == TENT) |
834 | p++; |
835 | if (sc->place[j] == NONTENT && sc->place[j-1] == TENT) { |
836 | sc->place[j-1] = NONTENT; |
837 | sc->place[j] = TENT; |
838 | while (p--) |
839 | sc->place[++j] = TENT; |
840 | while (++j < n) |
841 | sc->place[j] = NONTENT; |
842 | break; |
843 | } |
844 | } |
845 | if (j <= 0) |
846 | break; /* we've finished */ |
847 | } |
848 | |
849 | /* |
850 | * It's just possible that _no_ placement was valid, in |
851 | * which case we have an internally inconsistent |
852 | * puzzle. |
853 | */ |
854 | if (mrow[sc->locs[0]] == MAGIC) |
855 | return 0; /* inconsistent */ |
856 | |
857 | /* |
858 | * Now go through mrow and see if there's anything |
859 | * we've deduced which wasn't already mentioned in soln. |
860 | */ |
861 | for (j = 0; j < len; j++) { |
862 | int whichrow; |
863 | |
864 | for (whichrow = 0; whichrow < 3; whichrow++) { |
865 | char *mthis = mrow + whichrow * len; |
866 | int tstart = (whichrow == 0 ? start : |
867 | whichrow == 1 ? start1 : start2); |
868 | if (tstart >= 0 && |
869 | mthis[j] != MAGIC && mthis[j] != BLANK && |
870 | soln[tstart+j*step] == BLANK) { |
871 | int pos = tstart+j*step; |
872 | |
873 | #ifdef SOLVER_DIAGNOSTICS |
874 | if (verbose) |
875 | printf("%s %d forces %s at %d,%d\n", |
876 | step==1 ? "row" : "column", |
877 | step==1 ? start/w : start, |
878 | mrow[j] == TENT ? "tent" : "non-tent", |
879 | pos % w, pos / w); |
880 | #endif |
881 | soln[pos] = mthis[j]; |
882 | done_something = TRUE; |
883 | } |
884 | } |
885 | } |
886 | } |
887 | |
888 | if (done_something) |
889 | continue; |
890 | |
891 | if (!done_something) |
892 | break; |
893 | } |
894 | |
895 | /* |
896 | * The solver has nothing further it can do. Return 1 if both |
897 | * soln and sc->links are completely filled in, or 2 otherwise. |
898 | */ |
899 | for (y = 0; y < h; y++) |
900 | for (x = 0; x < w; x++) { |
901 | if (soln[y*w+x] == BLANK) |
902 | return 2; |
903 | if (soln[y*w+x] != NONTENT && sc->links[y*w+x] == 0) |
904 | return 2; |
905 | } |
906 | |
907 | return 1; |
908 | } |
909 | |
910 | static char *new_game_desc(game_params *params, random_state *rs, |
911 | char **aux, int interactive) |
912 | { |
913 | int w = params->w, h = params->h; |
914 | int ntrees = w * h / 5; |
915 | char *grid = snewn(w*h, char); |
916 | char *puzzle = snewn(w*h, char); |
917 | int *numbers = snewn(w+h, int); |
918 | char *soln = snewn(w*h, char); |
919 | int *temp = snewn(2*w*h, int), *itemp = temp + w*h; |
920 | int maxedges = ntrees*4 + w*h; |
921 | int *edges = snewn(2*maxedges, int); |
922 | int *capacity = snewn(maxedges, int); |
923 | int *flow = snewn(maxedges, int); |
924 | struct solver_scratch *sc = new_scratch(w, h); |
925 | char *ret, *p; |
926 | int i, j, nedges; |
927 | |
928 | /* |
929 | * Since this puzzle has many global deductions and doesn't |
930 | * permit limited clue sets, generating grids for this puzzle |
931 | * is hard enough that I see no better option than to simply |
932 | * generate a solution and see if it's unique and has the |
933 | * required difficulty. This turns out to be computationally |
934 | * plausible as well. |
935 | * |
936 | * We chose our tree count (hence also tent count) by dividing |
937 | * the total grid area by five above. Why five? Well, w*h/4 is |
938 | * the maximum number of tents you can _possibly_ fit into the |
939 | * grid without violating the separation criterion, and to |
940 | * achieve that you are constrained to a very small set of |
941 | * possible layouts (the obvious one with a tent at every |
942 | * (even,even) coordinate, and trivial variations thereon). So |
943 | * if we reduce the tent count a bit more, we enable more |
944 | * random-looking placement; 5 turns out to be a plausible |
945 | * figure which yields sensible puzzles. Increasing the tent |
946 | * count would give puzzles whose solutions were too regimented |
947 | * and could be solved by the use of that knowledge (and would |
948 | * also take longer to find a viable placement); decreasing it |
949 | * would make the grids emptier and more boring. |
950 | * |
951 | * Actually generating a grid is a matter of first placing the |
952 | * tents, and then placing the trees by the use of maxflow |
953 | * (finding a distinct square adjacent to every tent). We do it |
954 | * this way round because otherwise satisfying the tent |
955 | * separation condition would become onerous: most randomly |
956 | * chosen tent layouts do not satisfy this condition, so we'd |
957 | * have gone to a lot of work before finding that a candidate |
958 | * layout was unusable. Instead, we place the tents first and |
959 | * ensure they meet the separation criterion _before_ doing |
960 | * lots of computation; this works much better. |
961 | * |
962 | * The maxflow algorithm is not randomised, so employed naively |
963 | * it would give rise to grids with clear structure and |
964 | * directional bias. Hence, I assign the network nodes as seen |
965 | * by maxflow to be a _random_ permutation the squares of the |
966 | * grid, so that any bias shown by maxflow towards low-numbered |
967 | * nodes is turned into a random bias. |
968 | * |
969 | * This generation strategy can fail at many points, including |
970 | * as early as tent placement (if you get a bad random order in |
971 | * which to greedily try the grid squares, you won't even |
972 | * manage to find enough mutually non-adjacent squares to put |
973 | * the tents in). Then it can fail if maxflow doesn't manage to |
974 | * find a good enough matching (i.e. the tent placements don't |
975 | * admit any adequate tree placements); and finally it can fail |
976 | * if the solver finds that the problem has the wrong |
977 | * difficulty (including being actually non-unique). All of |
978 | * these, however, are insufficiently frequent to cause |
979 | * trouble. |
980 | */ |
981 | |
982 | while (1) { |
983 | /* |
984 | * Arrange the grid squares into a random order, and invert |
985 | * that order so we can find a square's index as well. |
986 | */ |
987 | for (i = 0; i < w*h; i++) |
988 | temp[i] = i; |
989 | shuffle(temp, w*h, sizeof(*temp), rs); |
990 | for (i = 0; i < w*h; i++) |
991 | itemp[temp[i]] = i; |
992 | |
993 | /* |
994 | * The first `ntrees' entries in temp which we can get |
995 | * without making two tents adjacent will be the tent |
996 | * locations. |
997 | */ |
998 | memset(grid, BLANK, w*h); |
999 | j = ntrees; |
1000 | for (i = 0; i < w*h && j > 0; i++) { |
1001 | int x = temp[i] % w, y = temp[i] / w; |
1002 | int dy, dx, ok = TRUE; |
1003 | |
1004 | for (dy = -1; dy <= +1; dy++) |
1005 | for (dx = -1; dx <= +1; dx++) |
1006 | if (x+dx >= 0 && x+dx < w && |
1007 | y+dy >= 0 && y+dy < h && |
1008 | grid[(y+dy)*w+(x+dx)] == TENT) |
1009 | ok = FALSE; |
1010 | |
1011 | if (ok) { |
1012 | grid[temp[i]] = TENT; |
1013 | j--; |
1014 | } |
1015 | } |
1016 | if (j > 0) |
1017 | continue; /* couldn't place all the tents */ |
1018 | |
1019 | /* |
1020 | * Now we build up the list of graph edges. |
1021 | */ |
1022 | nedges = 0; |
1023 | for (i = 0; i < w*h; i++) { |
1024 | if (grid[temp[i]] == TENT) { |
1025 | for (j = 0; j < w*h; j++) { |
1026 | if (grid[temp[j]] != TENT) { |
1027 | int xi = temp[i] % w, yi = temp[i] / w; |
1028 | int xj = temp[j] % w, yj = temp[j] / w; |
1029 | if (abs(xi-xj) + abs(yi-yj) == 1) { |
1030 | edges[nedges*2] = i; |
1031 | edges[nedges*2+1] = j; |
1032 | capacity[nedges] = 1; |
1033 | nedges++; |
1034 | } |
1035 | } |
1036 | } |
1037 | } else { |
1038 | /* |
1039 | * Special node w*h is the sink node; any non-tent node |
1040 | * has an edge going to it. |
1041 | */ |
1042 | edges[nedges*2] = i; |
1043 | edges[nedges*2+1] = w*h; |
1044 | capacity[nedges] = 1; |
1045 | nedges++; |
1046 | } |
1047 | } |
1048 | |
1049 | /* |
1050 | * Special node w*h+1 is the source node, with an edge going to |
1051 | * every tent. |
1052 | */ |
1053 | for (i = 0; i < w*h; i++) { |
1054 | if (grid[temp[i]] == TENT) { |
1055 | edges[nedges*2] = w*h+1; |
1056 | edges[nedges*2+1] = i; |
1057 | capacity[nedges] = 1; |
1058 | nedges++; |
1059 | } |
1060 | } |
1061 | |
1062 | assert(nedges <= maxedges); |
1063 | |
1064 | /* |
1065 | * Now we're ready to call the maxflow algorithm to place the |
1066 | * trees. |
1067 | */ |
1068 | j = maxflow(w*h+2, w*h+1, w*h, nedges, edges, capacity, flow, NULL); |
1069 | |
1070 | if (j < ntrees) |
1071 | continue; /* couldn't place all the tents */ |
1072 | |
1073 | /* |
1074 | * We've placed the trees. Now we need to work out _where_ |
1075 | * we've placed them, which is a matter of reading back out |
1076 | * from the `flow' array. |
1077 | */ |
1078 | for (i = 0; i < nedges; i++) { |
1079 | if (edges[2*i] < w*h && edges[2*i+1] < w*h && flow[i] > 0) |
1080 | grid[temp[edges[2*i+1]]] = TREE; |
1081 | } |
1082 | |
1083 | /* |
1084 | * I think it looks ugly if there isn't at least one of |
1085 | * _something_ (tent or tree) in each row and each column |
1086 | * of the grid. This doesn't give any information away |
1087 | * since a completely empty row/column is instantly obvious |
1088 | * from the clues (it has no trees and a zero). |
1089 | */ |
1090 | for (i = 0; i < w; i++) { |
1091 | for (j = 0; j < h; j++) { |
1092 | if (grid[j*w+i] != BLANK) |
1093 | break; /* found something in this column */ |
1094 | } |
1095 | if (j == h) |
1096 | break; /* found empty column */ |
1097 | } |
1098 | if (i < w) |
1099 | continue; /* a column was empty */ |
1100 | |
1101 | for (j = 0; j < h; j++) { |
1102 | for (i = 0; i < w; i++) { |
1103 | if (grid[j*w+i] != BLANK) |
1104 | break; /* found something in this row */ |
1105 | } |
1106 | if (i == w) |
1107 | break; /* found empty row */ |
1108 | } |
1109 | if (j < h) |
1110 | continue; /* a row was empty */ |
1111 | |
1112 | /* |
1113 | * Now set up the numbers round the edge. |
1114 | */ |
1115 | for (i = 0; i < w; i++) { |
1116 | int n = 0; |
1117 | for (j = 0; j < h; j++) |
1118 | if (grid[j*w+i] == TENT) |
1119 | n++; |
1120 | numbers[i] = n; |
1121 | } |
1122 | for (i = 0; i < h; i++) { |
1123 | int n = 0; |
1124 | for (j = 0; j < w; j++) |
1125 | if (grid[i*w+j] == TENT) |
1126 | n++; |
1127 | numbers[w+i] = n; |
1128 | } |
1129 | |
1130 | /* |
1131 | * And now actually solve the puzzle, to see whether it's |
1132 | * unique and has the required difficulty. |
1133 | */ |
1134 | for (i = 0; i < w*h; i++) |
1135 | puzzle[i] = grid[i] == TREE ? TREE : BLANK; |
1136 | i = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff-1); |
1137 | j = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff); |
1138 | |
1139 | /* |
1140 | * We expect solving with difficulty params->diff to have |
1141 | * succeeded (otherwise the problem is too hard), and |
1142 | * solving with diff-1 to have failed (otherwise it's too |
1143 | * easy). |
1144 | */ |
1145 | if (i == 2 && j == 1) |
1146 | break; |
1147 | } |
1148 | |
1149 | /* |
1150 | * That's it. Encode as a game ID. |
1151 | */ |
1152 | ret = snewn((w+h)*40 + ntrees + (w*h)/26 + 1, char); |
1153 | p = ret; |
1154 | j = 0; |
1155 | for (i = 0; i <= w*h; i++) { |
1156 | int c = (i < w*h ? grid[i] == TREE : 1); |
1157 | if (c) { |
1158 | *p++ = (j == 0 ? '_' : j-1 + 'a'); |
1159 | j = 0; |
1160 | } else { |
1161 | j++; |
1162 | while (j > 25) { |
1163 | *p++ = 'z'; |
1164 | j -= 25; |
1165 | } |
1166 | } |
1167 | } |
1168 | for (i = 0; i < w+h; i++) |
1169 | p += sprintf(p, ",%d", numbers[i]); |
1170 | *p++ = '\0'; |
1171 | ret = sresize(ret, p - ret, char); |
1172 | |
1173 | /* |
1174 | * And encode the solution as an aux_info. |
1175 | */ |
1176 | *aux = snewn(ntrees * 40, char); |
1177 | p = *aux; |
1178 | *p++ = 'S'; |
1179 | for (i = 0; i < w*h; i++) |
1180 | if (grid[i] == TENT) |
1181 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1182 | *p++ = '\0'; |
1183 | *aux = sresize(*aux, p - *aux, char); |
1184 | |
1185 | free_scratch(sc); |
1186 | sfree(flow); |
1187 | sfree(capacity); |
1188 | sfree(edges); |
1189 | sfree(temp); |
1190 | sfree(soln); |
1191 | sfree(numbers); |
1192 | sfree(puzzle); |
1193 | sfree(grid); |
1194 | |
1195 | return ret; |
1196 | } |
1197 | |
1198 | static char *validate_desc(game_params *params, char *desc) |
1199 | { |
1200 | int w = params->w, h = params->h; |
1201 | int area, i; |
1202 | |
1203 | area = 0; |
1204 | while (*desc && *desc != ',') { |
1205 | if (*desc == '_') |
1206 | area++; |
1207 | else if (*desc >= 'a' && *desc < 'z') |
1208 | area += *desc - 'a' + 2; |
1209 | else if (*desc == 'z') |
1210 | area += 25; |
1211 | else if (*desc == '!' || *desc == '-') |
1212 | /* do nothing */; |
1213 | else |
1214 | return "Invalid character in grid specification"; |
1215 | |
1216 | desc++; |
1217 | } |
1218 | |
1219 | for (i = 0; i < w+h; i++) { |
1220 | if (!*desc) |
1221 | return "Not enough numbers given after grid specification"; |
1222 | else if (*desc != ',') |
1223 | return "Invalid character in number list"; |
1224 | desc++; |
1225 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1226 | } |
1227 | |
1228 | if (*desc) |
1229 | return "Unexpected additional data at end of game description"; |
1230 | return NULL; |
1231 | } |
1232 | |
1233 | static game_state *new_game(midend *me, game_params *params, char *desc) |
1234 | { |
1235 | int w = params->w, h = params->h; |
1236 | game_state *state = snew(game_state); |
1237 | int i; |
1238 | |
1239 | state->p = *params; /* structure copy */ |
1240 | state->grid = snewn(w*h, char); |
1241 | state->numbers = snew(struct numbers); |
1242 | state->numbers->refcount = 1; |
1243 | state->numbers->numbers = snewn(w+h, int); |
1244 | state->completed = state->used_solve = FALSE; |
1245 | |
1246 | i = 0; |
1247 | memset(state->grid, BLANK, w*h); |
1248 | |
1249 | while (*desc) { |
1250 | int run, type; |
1251 | |
1252 | type = TREE; |
1253 | |
1254 | if (*desc == '_') |
1255 | run = 0; |
1256 | else if (*desc >= 'a' && *desc < 'z') |
1257 | run = *desc - ('a'-1); |
1258 | else if (*desc == 'z') { |
1259 | run = 25; |
1260 | type = BLANK; |
1261 | } else { |
1262 | assert(*desc == '!' || *desc == '-'); |
1263 | run = -1; |
1264 | type = (*desc == '!' ? TENT : NONTENT); |
1265 | } |
1266 | |
1267 | desc++; |
1268 | |
1269 | i += run; |
1270 | assert(i >= 0 && i <= w*h); |
1271 | if (i == w*h) { |
1272 | assert(type == TREE); |
1273 | break; |
1274 | } else { |
1275 | if (type != BLANK) |
1276 | state->grid[i++] = type; |
1277 | } |
1278 | } |
1279 | |
1280 | for (i = 0; i < w+h; i++) { |
1281 | assert(*desc == ','); |
1282 | desc++; |
1283 | state->numbers->numbers[i] = atoi(desc); |
1284 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1285 | } |
1286 | |
1287 | assert(!*desc); |
1288 | |
1289 | return state; |
1290 | } |
1291 | |
1292 | static game_state *dup_game(game_state *state) |
1293 | { |
1294 | int w = state->p.w, h = state->p.h; |
1295 | game_state *ret = snew(game_state); |
1296 | |
1297 | ret->p = state->p; /* structure copy */ |
1298 | ret->grid = snewn(w*h, char); |
1299 | memcpy(ret->grid, state->grid, w*h); |
1300 | ret->numbers = state->numbers; |
1301 | state->numbers->refcount++; |
1302 | ret->completed = state->completed; |
1303 | ret->used_solve = state->used_solve; |
1304 | |
1305 | return ret; |
1306 | } |
1307 | |
1308 | static void free_game(game_state *state) |
1309 | { |
1310 | if (--state->numbers->refcount <= 0) { |
1311 | sfree(state->numbers->numbers); |
1312 | sfree(state->numbers); |
1313 | } |
1314 | sfree(state->grid); |
1315 | sfree(state); |
1316 | } |
1317 | |
1318 | static char *solve_game(game_state *state, game_state *currstate, |
1319 | char *aux, char **error) |
1320 | { |
1321 | int w = state->p.w, h = state->p.h; |
1322 | |
1323 | if (aux) { |
1324 | /* |
1325 | * If we already have the solution, save ourselves some |
1326 | * time. |
1327 | */ |
1328 | return dupstr(aux); |
1329 | } else { |
1330 | struct solver_scratch *sc = new_scratch(w, h); |
1331 | char *soln; |
1332 | int ret; |
1333 | char *move, *p; |
1334 | int i; |
1335 | |
1336 | soln = snewn(w*h, char); |
1337 | ret = tents_solve(w, h, state->grid, state->numbers->numbers, |
1338 | soln, sc, DIFFCOUNT-1); |
1339 | free_scratch(sc); |
1340 | if (ret != 1) { |
1341 | sfree(soln); |
1342 | if (ret == 0) |
1343 | *error = "This puzzle is not self-consistent"; |
1344 | else |
1345 | *error = "Unable to find a unique solution for this puzzle"; |
1346 | return NULL; |
1347 | } |
1348 | |
1349 | /* |
1350 | * Construct a move string which turns the current state |
1351 | * into the solved state. |
1352 | */ |
1353 | move = snewn(w*h * 40, char); |
1354 | p = move; |
1355 | *p++ = 'S'; |
1356 | for (i = 0; i < w*h; i++) |
1357 | if (soln[i] == TENT) |
1358 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1359 | *p++ = '\0'; |
1360 | move = sresize(move, p - move, char); |
1361 | |
1362 | sfree(soln); |
1363 | |
1364 | return move; |
1365 | } |
1366 | } |
1367 | |
1368 | static char *game_text_format(game_state *state) |
1369 | { |
1370 | int w = state->p.w, h = state->p.h; |
1371 | char *ret, *p; |
1372 | int x, y; |
1373 | |
1374 | /* |
1375 | * FIXME: We currently do not print the numbers round the edges |
1376 | * of the grid. I need to work out a sensible way of doing this |
1377 | * even when the column numbers exceed 9. |
1378 | * |
1379 | * In the absence of those numbers, the result size is h lines |
1380 | * of w+1 characters each, plus a NUL. |
1381 | * |
1382 | * This function is currently only used by the standalone |
1383 | * solver; until I make it look more sensible, I won't enable |
1384 | * it in the main game structure. |
1385 | */ |
1386 | ret = snewn(h*(w+1) + 1, char); |
1387 | p = ret; |
1388 | for (y = 0; y < h; y++) { |
1389 | for (x = 0; x < w; x++) { |
1390 | *p = (state->grid[y*w+x] == BLANK ? '.' : |
1391 | state->grid[y*w+x] == TREE ? 'T' : |
1392 | state->grid[y*w+x] == TENT ? '*' : |
1393 | state->grid[y*w+x] == NONTENT ? '-' : '?'); |
1394 | p++; |
1395 | } |
1396 | *p++ = '\n'; |
1397 | } |
1398 | *p++ = '\0'; |
1399 | |
1400 | return ret; |
1401 | } |
1402 | |
1403 | static game_ui *new_ui(game_state *state) |
1404 | { |
1405 | return NULL; |
1406 | } |
1407 | |
1408 | static void free_ui(game_ui *ui) |
1409 | { |
1410 | } |
1411 | |
1412 | static char *encode_ui(game_ui *ui) |
1413 | { |
1414 | return NULL; |
1415 | } |
1416 | |
1417 | static void decode_ui(game_ui *ui, char *encoding) |
1418 | { |
1419 | } |
1420 | |
1421 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
1422 | game_state *newstate) |
1423 | { |
1424 | } |
1425 | |
1426 | struct game_drawstate { |
1427 | int tilesize; |
1428 | int started; |
1429 | game_params p; |
1430 | char *drawn; |
1431 | }; |
1432 | |
1433 | #define PREFERRED_TILESIZE 32 |
1434 | #define TILESIZE (ds->tilesize) |
1435 | #define TLBORDER (TILESIZE/2) |
1436 | #define BRBORDER (TILESIZE*3/2) |
1437 | #define COORD(x) ( (x) * TILESIZE + TLBORDER ) |
1438 | #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 ) |
1439 | |
1440 | #define FLASH_TIME 0.30F |
1441 | |
1442 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
1443 | int x, int y, int button) |
1444 | { |
1445 | int w = state->p.w, h = state->p.h; |
1446 | |
1447 | if (button == LEFT_BUTTON || button == RIGHT_BUTTON) { |
1448 | int v; |
1449 | char buf[80]; |
1450 | |
1451 | x = FROMCOORD(x); |
1452 | y = FROMCOORD(y); |
1453 | if (x < 0 || y < 0 || x >= w || y >= h) |
1454 | return NULL; |
1455 | |
1456 | if (state->grid[y*w+x] == TREE) |
1457 | return NULL; |
1458 | |
1459 | if (button == LEFT_BUTTON) { |
1460 | v = (state->grid[y*w+x] == BLANK ? TENT : BLANK); |
1461 | } else { |
1462 | v = (state->grid[y*w+x] == BLANK ? NONTENT : BLANK); |
1463 | } |
1464 | |
1465 | sprintf(buf, "%c%d,%d", (int)(v==BLANK ? 'B' : |
1466 | v==TENT ? 'T' : 'N'), x, y); |
1467 | return dupstr(buf); |
1468 | } |
1469 | |
1470 | return NULL; |
1471 | } |
1472 | |
1473 | static game_state *execute_move(game_state *state, char *move) |
1474 | { |
1475 | int w = state->p.w, h = state->p.h; |
1476 | char c; |
1477 | int x, y, m, n, i, j; |
1478 | game_state *ret = dup_game(state); |
1479 | |
1480 | while (*move) { |
1481 | c = *move; |
1482 | if (c == 'S') { |
1483 | int i; |
1484 | ret->used_solve = TRUE; |
1485 | /* |
1486 | * Set all non-tree squares to NONTENT. The rest of the |
1487 | * solve move will fill the tents in over the top. |
1488 | */ |
1489 | for (i = 0; i < w*h; i++) |
1490 | if (ret->grid[i] != TREE) |
1491 | ret->grid[i] = NONTENT; |
1492 | move++; |
1493 | } else if (c == 'B' || c == 'T' || c == 'N') { |
1494 | move++; |
1495 | if (sscanf(move, "%d,%d%n", &x, &y, &n) != 2 || |
1496 | x < 0 || y < 0 || x >= w || y >= h) { |
1497 | free_game(ret); |
1498 | return NULL; |
1499 | } |
1500 | if (ret->grid[y*w+x] == TREE) { |
1501 | free_game(ret); |
1502 | return NULL; |
1503 | } |
1504 | ret->grid[y*w+x] = (c == 'B' ? BLANK : c == 'T' ? TENT : NONTENT); |
1505 | move += n; |
1506 | } else { |
1507 | free_game(ret); |
1508 | return NULL; |
1509 | } |
1510 | if (*move == ';') |
1511 | move++; |
1512 | else if (*move) { |
1513 | free_game(ret); |
1514 | return NULL; |
1515 | } |
1516 | } |
1517 | |
1518 | /* |
1519 | * Check for completion. |
1520 | */ |
1521 | for (i = n = m = 0; i < w*h; i++) { |
1522 | if (ret->grid[i] == TENT) |
1523 | n++; |
1524 | else if (ret->grid[i] == TREE) |
1525 | m++; |
1526 | } |
1527 | if (n == m) { |
1528 | int nedges, maxedges, *edges, *capacity, *flow; |
1529 | |
1530 | /* |
1531 | * We have the right number of tents, which is a |
1532 | * precondition for the game being complete. Now check that |
1533 | * the numbers add up. |
1534 | */ |
1535 | for (i = 0; i < w; i++) { |
1536 | n = 0; |
1537 | for (j = 0; j < h; j++) |
1538 | if (ret->grid[j*w+i] == TENT) |
1539 | n++; |
1540 | if (ret->numbers->numbers[i] != n) |
1541 | goto completion_check_done; |
1542 | } |
1543 | for (i = 0; i < h; i++) { |
1544 | n = 0; |
1545 | for (j = 0; j < w; j++) |
1546 | if (ret->grid[i*w+j] == TENT) |
1547 | n++; |
1548 | if (ret->numbers->numbers[w+i] != n) |
1549 | goto completion_check_done; |
1550 | } |
1551 | /* |
1552 | * Also, check that no two tents are adjacent. |
1553 | */ |
1554 | for (y = 0; y < h; y++) |
1555 | for (x = 0; x < w; x++) { |
1556 | if (x+1 < w && |
1557 | ret->grid[y*w+x] == TENT && ret->grid[y*w+x+1] == TENT) |
1558 | goto completion_check_done; |
1559 | if (y+1 < h && |
1560 | ret->grid[y*w+x] == TENT && ret->grid[(y+1)*w+x] == TENT) |
1561 | goto completion_check_done; |
1562 | if (x+1 < w && y+1 < h) { |
1563 | if (ret->grid[y*w+x] == TENT && |
1564 | ret->grid[(y+1)*w+(x+1)] == TENT) |
1565 | goto completion_check_done; |
1566 | if (ret->grid[(y+1)*w+x] == TENT && |
1567 | ret->grid[y*w+(x+1)] == TENT) |
1568 | goto completion_check_done; |
1569 | } |
1570 | } |
1571 | |
1572 | /* |
1573 | * OK; we have the right number of tents, they match the |
1574 | * numeric clues, and they satisfy the non-adjacency |
1575 | * criterion. Finally, we need to verify that they can be |
1576 | * placed in a one-to-one matching with the trees such that |
1577 | * every tent is orthogonally adjacent to its tree. |
1578 | * |
1579 | * This bit is where the hard work comes in: we have to do |
1580 | * it by finding such a matching using maxflow. |
1581 | * |
1582 | * So we construct a network with one special source node, |
1583 | * one special sink node, one node per tent, and one node |
1584 | * per tree. |
1585 | */ |
1586 | maxedges = 6 * m; |
1587 | edges = snewn(2 * maxedges, int); |
1588 | capacity = snewn(maxedges, int); |
1589 | flow = snewn(maxedges, int); |
1590 | nedges = 0; |
1591 | /* |
1592 | * Node numbering: |
1593 | * |
1594 | * 0..w*h trees/tents |
1595 | * w*h source |
1596 | * w*h+1 sink |
1597 | */ |
1598 | for (y = 0; y < h; y++) |
1599 | for (x = 0; x < w; x++) |
1600 | if (ret->grid[y*w+x] == TREE) { |
1601 | int d; |
1602 | |
1603 | /* |
1604 | * Here we use the direction enum declared for |
1605 | * the solver. We make use of the fact that the |
1606 | * directions are declared in the order |
1607 | * U,L,R,D, meaning that we go through the four |
1608 | * neighbours of any square in numerically |
1609 | * increasing order. |
1610 | */ |
1611 | for (d = 1; d < MAXDIR; d++) { |
1612 | int x2 = x + dx(d), y2 = y + dy(d); |
1613 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
1614 | ret->grid[y2*w+x2] == TENT) { |
1615 | assert(nedges < maxedges); |
1616 | edges[nedges*2] = y*w+x; |
1617 | edges[nedges*2+1] = y2*w+x2; |
1618 | capacity[nedges] = 1; |
1619 | nedges++; |
1620 | } |
1621 | } |
1622 | } else if (ret->grid[y*w+x] == TENT) { |
1623 | assert(nedges < maxedges); |
1624 | edges[nedges*2] = y*w+x; |
1625 | edges[nedges*2+1] = w*h+1; /* edge going to sink */ |
1626 | capacity[nedges] = 1; |
1627 | nedges++; |
1628 | } |
1629 | for (y = 0; y < h; y++) |
1630 | for (x = 0; x < w; x++) |
1631 | if (ret->grid[y*w+x] == TREE) { |
1632 | assert(nedges < maxedges); |
1633 | edges[nedges*2] = w*h; /* edge coming from source */ |
1634 | edges[nedges*2+1] = y*w+x; |
1635 | capacity[nedges] = 1; |
1636 | nedges++; |
1637 | } |
1638 | n = maxflow(w*h+2, w*h, w*h+1, nedges, edges, capacity, flow, NULL); |
1639 | |
1640 | sfree(flow); |
1641 | sfree(capacity); |
1642 | sfree(edges); |
1643 | |
1644 | if (n != m) |
1645 | goto completion_check_done; |
1646 | |
1647 | /* |
1648 | * We haven't managed to fault the grid on any count. Score! |
1649 | */ |
1650 | ret->completed = TRUE; |
1651 | } |
1652 | completion_check_done: |
1653 | |
1654 | return ret; |
1655 | } |
1656 | |
1657 | /* ---------------------------------------------------------------------- |
1658 | * Drawing routines. |
1659 | */ |
1660 | |
1661 | static void game_compute_size(game_params *params, int tilesize, |
1662 | int *x, int *y) |
1663 | { |
1664 | /* fool the macros */ |
1665 | struct dummy { int tilesize; } dummy = { tilesize }, *ds = &dummy; |
1666 | |
1667 | *x = TLBORDER + BRBORDER + TILESIZE * params->w; |
1668 | *y = TLBORDER + BRBORDER + TILESIZE * params->h; |
1669 | } |
1670 | |
1671 | static void game_set_size(drawing *dr, game_drawstate *ds, |
1672 | game_params *params, int tilesize) |
1673 | { |
1674 | ds->tilesize = tilesize; |
1675 | } |
1676 | |
1677 | static float *game_colours(frontend *fe, game_state *state, int *ncolours) |
1678 | { |
1679 | float *ret = snewn(3 * NCOLOURS, float); |
1680 | |
1681 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
1682 | |
1683 | ret[COL_GRID * 3 + 0] = 0.0F; |
1684 | ret[COL_GRID * 3 + 1] = 0.0F; |
1685 | ret[COL_GRID * 3 + 2] = 0.0F; |
1686 | |
1687 | ret[COL_GRASS * 3 + 0] = 0.7F; |
1688 | ret[COL_GRASS * 3 + 1] = 1.0F; |
1689 | ret[COL_GRASS * 3 + 2] = 0.5F; |
1690 | |
1691 | ret[COL_TREETRUNK * 3 + 0] = 0.6F; |
1692 | ret[COL_TREETRUNK * 3 + 1] = 0.4F; |
1693 | ret[COL_TREETRUNK * 3 + 2] = 0.0F; |
1694 | |
1695 | ret[COL_TREELEAF * 3 + 0] = 0.0F; |
1696 | ret[COL_TREELEAF * 3 + 1] = 0.7F; |
1697 | ret[COL_TREELEAF * 3 + 2] = 0.0F; |
1698 | |
1699 | ret[COL_TENT * 3 + 0] = 0.8F; |
1700 | ret[COL_TENT * 3 + 1] = 0.7F; |
1701 | ret[COL_TENT * 3 + 2] = 0.0F; |
1702 | |
1703 | *ncolours = NCOLOURS; |
1704 | return ret; |
1705 | } |
1706 | |
1707 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
1708 | { |
1709 | int w = state->p.w, h = state->p.h; |
1710 | struct game_drawstate *ds = snew(struct game_drawstate); |
1711 | |
1712 | ds->tilesize = 0; |
1713 | ds->started = FALSE; |
1714 | ds->p = state->p; /* structure copy */ |
1715 | ds->drawn = snewn(w*h, char); |
1716 | memset(ds->drawn, MAGIC, w*h); |
1717 | |
1718 | return ds; |
1719 | } |
1720 | |
1721 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
1722 | { |
1723 | sfree(ds->drawn); |
1724 | sfree(ds); |
1725 | } |
1726 | |
1727 | static void draw_tile(drawing *dr, game_drawstate *ds, |
1728 | int x, int y, int v, int printing) |
1729 | { |
1730 | int tx = COORD(x), ty = COORD(y); |
1731 | int cx = tx + TILESIZE/2, cy = ty + TILESIZE/2; |
1732 | |
1733 | clip(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1); |
1734 | |
1735 | if (!printing) |
1736 | draw_rect(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1, |
1737 | (v == BLANK ? COL_BACKGROUND : COL_GRASS)); |
1738 | |
1739 | if (v == TREE) { |
1740 | int i; |
1741 | |
1742 | (printing ? draw_rect_outline : draw_rect) |
1743 | (dr, cx-TILESIZE/15, ty+TILESIZE*3/10, |
1744 | 2*(TILESIZE/15)+1, (TILESIZE*9/10 - TILESIZE*3/10), |
1745 | COL_TREETRUNK); |
1746 | |
1747 | for (i = 0; i < (printing ? 2 : 1); i++) { |
1748 | int col = (i == 1 ? COL_BACKGROUND : COL_TREELEAF); |
1749 | int sub = i * (TILESIZE/32); |
1750 | draw_circle(dr, cx, ty+TILESIZE*4/10, TILESIZE/4 - sub, |
1751 | col, col); |
1752 | draw_circle(dr, cx+TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
1753 | col, col); |
1754 | draw_circle(dr, cx-TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
1755 | col, col); |
1756 | draw_circle(dr, cx+TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
1757 | col, col); |
1758 | draw_circle(dr, cx-TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
1759 | col, col); |
1760 | } |
1761 | } else if (v == TENT) { |
1762 | int coords[6]; |
1763 | coords[0] = cx - TILESIZE/3; |
1764 | coords[1] = cy + TILESIZE/3; |
1765 | coords[2] = cx + TILESIZE/3; |
1766 | coords[3] = cy + TILESIZE/3; |
1767 | coords[4] = cx; |
1768 | coords[5] = cy - TILESIZE/3; |
1769 | draw_polygon(dr, coords, 3, (printing ? -1 : COL_TENT), COL_TENT); |
1770 | } |
1771 | |
1772 | unclip(dr); |
1773 | draw_update(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1); |
1774 | } |
1775 | |
1776 | /* |
1777 | * Internal redraw function, used for printing as well as drawing. |
1778 | */ |
1779 | static void int_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
1780 | game_state *state, int dir, game_ui *ui, |
1781 | float animtime, float flashtime, int printing) |
1782 | { |
1783 | int w = state->p.w, h = state->p.h; |
1784 | int x, y, flashing; |
1785 | |
1786 | if (printing || !ds->started) { |
1787 | if (!printing) { |
1788 | int ww, wh; |
1789 | game_compute_size(&state->p, TILESIZE, &ww, &wh); |
1790 | draw_rect(dr, 0, 0, ww, wh, COL_BACKGROUND); |
1791 | draw_update(dr, 0, 0, ww, wh); |
1792 | ds->started = TRUE; |
1793 | } |
1794 | |
1795 | if (printing) |
1796 | print_line_width(dr, TILESIZE/64); |
1797 | |
1798 | /* |
1799 | * Draw the grid. |
1800 | */ |
1801 | for (y = 0; y <= h; y++) |
1802 | draw_line(dr, COORD(0), COORD(y), COORD(w), COORD(y), COL_GRID); |
1803 | for (x = 0; x <= w; x++) |
1804 | draw_line(dr, COORD(x), COORD(0), COORD(x), COORD(h), COL_GRID); |
1805 | |
1806 | /* |
1807 | * Draw the numbers. |
1808 | */ |
1809 | for (y = 0; y < h; y++) { |
1810 | char buf[80]; |
1811 | sprintf(buf, "%d", state->numbers->numbers[y+w]); |
1812 | draw_text(dr, COORD(w+1), COORD(y) + TILESIZE/2, |
1813 | FONT_VARIABLE, TILESIZE/2, ALIGN_HRIGHT|ALIGN_VCENTRE, |
1814 | COL_GRID, buf); |
1815 | } |
1816 | for (x = 0; x < w; x++) { |
1817 | char buf[80]; |
1818 | sprintf(buf, "%d", state->numbers->numbers[x]); |
1819 | draw_text(dr, COORD(x) + TILESIZE/2, COORD(h+1), |
1820 | FONT_VARIABLE, TILESIZE/2, ALIGN_HCENTRE|ALIGN_VNORMAL, |
1821 | COL_GRID, buf); |
1822 | } |
1823 | } |
1824 | |
1825 | if (flashtime > 0) |
1826 | flashing = (int)(flashtime * 3 / FLASH_TIME) != 1; |
1827 | else |
1828 | flashing = FALSE; |
1829 | |
1830 | /* |
1831 | * Draw the grid. |
1832 | */ |
1833 | for (y = 0; y < h; y++) |
1834 | for (x = 0; x < w; x++) { |
1835 | int v = state->grid[y*w+x]; |
1836 | |
1837 | if (flashing && (v == TREE || v == TENT)) |
1838 | v = NONTENT; |
1839 | |
1840 | if (printing || ds->drawn[y*w+x] != v) { |
1841 | draw_tile(dr, ds, x, y, v, printing); |
1842 | if (!printing) |
1843 | ds->drawn[y*w+x] = v; |
1844 | } |
1845 | } |
1846 | } |
1847 | |
1848 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
1849 | game_state *state, int dir, game_ui *ui, |
1850 | float animtime, float flashtime) |
1851 | { |
1852 | int_redraw(dr, ds, oldstate, state, dir, ui, animtime, flashtime, FALSE); |
1853 | } |
1854 | |
1855 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
1856 | int dir, game_ui *ui) |
1857 | { |
1858 | return 0.0F; |
1859 | } |
1860 | |
1861 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
1862 | int dir, game_ui *ui) |
1863 | { |
1864 | if (!oldstate->completed && newstate->completed && |
1865 | !oldstate->used_solve && !newstate->used_solve) |
1866 | return FLASH_TIME; |
1867 | |
1868 | return 0.0F; |
1869 | } |
1870 | |
1871 | static int game_wants_statusbar(void) |
1872 | { |
1873 | return FALSE; |
1874 | } |
1875 | |
1876 | static int game_timing_state(game_state *state, game_ui *ui) |
1877 | { |
1878 | return TRUE; |
1879 | } |
1880 | |
1881 | static void game_print_size(game_params *params, float *x, float *y) |
1882 | { |
1883 | int pw, ph; |
1884 | |
1885 | /* |
1886 | * I'll use 6mm squares by default. |
1887 | */ |
1888 | game_compute_size(params, 600, &pw, &ph); |
1889 | *x = pw / 100.0; |
1890 | *y = ph / 100.0; |
1891 | } |
1892 | |
1893 | static void game_print(drawing *dr, game_state *state, int tilesize) |
1894 | { |
1895 | int c; |
1896 | |
1897 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
1898 | game_drawstate ads, *ds = &ads; |
1899 | game_set_size(dr, ds, NULL, tilesize); |
1900 | |
1901 | c = print_mono_colour(dr, 1); assert(c == COL_BACKGROUND); |
1902 | c = print_mono_colour(dr, 0); assert(c == COL_GRID); |
1903 | c = print_mono_colour(dr, 1); assert(c == COL_GRASS); |
1904 | c = print_mono_colour(dr, 0); assert(c == COL_TREETRUNK); |
1905 | c = print_mono_colour(dr, 0); assert(c == COL_TREELEAF); |
1906 | c = print_mono_colour(dr, 0); assert(c == COL_TENT); |
1907 | |
1908 | int_redraw(dr, ds, NULL, state, +1, NULL, 0.0F, 0.0F, TRUE); |
1909 | } |
1910 | |
1911 | #ifdef COMBINED |
1912 | #define thegame tents |
1913 | #endif |
1914 | |
1915 | const struct game thegame = { |
1916 | "Tents", "games.tents", |
1917 | default_params, |
1918 | game_fetch_preset, |
1919 | decode_params, |
1920 | encode_params, |
1921 | free_params, |
1922 | dup_params, |
1923 | TRUE, game_configure, custom_params, |
1924 | validate_params, |
1925 | new_game_desc, |
1926 | validate_desc, |
1927 | new_game, |
1928 | dup_game, |
1929 | free_game, |
1930 | TRUE, solve_game, |
1931 | FALSE, game_text_format, |
1932 | new_ui, |
1933 | free_ui, |
1934 | encode_ui, |
1935 | decode_ui, |
1936 | game_changed_state, |
1937 | interpret_move, |
1938 | execute_move, |
1939 | PREFERRED_TILESIZE, game_compute_size, game_set_size, |
1940 | game_colours, |
1941 | game_new_drawstate, |
1942 | game_free_drawstate, |
1943 | game_redraw, |
1944 | game_anim_length, |
1945 | game_flash_length, |
1946 | TRUE, FALSE, game_print_size, game_print, |
1947 | game_wants_statusbar, |
1948 | FALSE, game_timing_state, |
1949 | 0, /* mouse_priorities */ |
1950 | }; |
1951 | |
1952 | #ifdef STANDALONE_SOLVER |
1953 | |
1954 | #include <stdarg.h> |
1955 | |
1956 | int main(int argc, char **argv) |
1957 | { |
1958 | game_params *p; |
1959 | game_state *s, *s2; |
1960 | char *id = NULL, *desc, *err; |
1961 | int grade = FALSE; |
1962 | int ret, diff, really_verbose = FALSE; |
1963 | struct solver_scratch *sc; |
1964 | |
1965 | while (--argc > 0) { |
1966 | char *p = *++argv; |
1967 | if (!strcmp(p, "-v")) { |
1968 | really_verbose = TRUE; |
1969 | } else if (!strcmp(p, "-g")) { |
1970 | grade = TRUE; |
1971 | } else if (*p == '-') { |
1972 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
1973 | return 1; |
1974 | } else { |
1975 | id = p; |
1976 | } |
1977 | } |
1978 | |
1979 | if (!id) { |
1980 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
1981 | return 1; |
1982 | } |
1983 | |
1984 | desc = strchr(id, ':'); |
1985 | if (!desc) { |
1986 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
1987 | return 1; |
1988 | } |
1989 | *desc++ = '\0'; |
1990 | |
1991 | p = default_params(); |
1992 | decode_params(p, id); |
1993 | err = validate_desc(p, desc); |
1994 | if (err) { |
1995 | fprintf(stderr, "%s: %s\n", argv[0], err); |
1996 | return 1; |
1997 | } |
1998 | s = new_game(NULL, p, desc); |
1999 | s2 = new_game(NULL, p, desc); |
2000 | |
2001 | sc = new_scratch(p->w, p->h); |
2002 | |
2003 | /* |
2004 | * When solving an Easy puzzle, we don't want to bother the |
2005 | * user with Hard-level deductions. For this reason, we grade |
2006 | * the puzzle internally before doing anything else. |
2007 | */ |
2008 | ret = -1; /* placate optimiser */ |
2009 | for (diff = 0; diff < DIFFCOUNT; diff++) { |
2010 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2011 | s2->grid, sc, diff); |
2012 | if (ret < 2) |
2013 | break; |
2014 | } |
2015 | |
2016 | if (diff == DIFFCOUNT) { |
2017 | if (grade) |
2018 | printf("Difficulty rating: too hard to solve internally\n"); |
2019 | else |
2020 | printf("Unable to find a unique solution\n"); |
2021 | } else { |
2022 | if (grade) { |
2023 | if (ret == 0) |
2024 | printf("Difficulty rating: impossible (no solution exists)\n"); |
2025 | else if (ret == 1) |
2026 | printf("Difficulty rating: %s\n", tents_diffnames[diff]); |
2027 | } else { |
2028 | verbose = really_verbose; |
2029 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2030 | s2->grid, sc, diff); |
2031 | if (ret == 0) |
2032 | printf("Puzzle is inconsistent\n"); |
2033 | else |
2034 | fputs(game_text_format(s2), stdout); |
2035 | } |
2036 | } |
2037 | |
2038 | return 0; |
2039 | } |
2040 | |
2041 | #endif |