86e60e3d |
1 | /* |
2 | * tents.c: Puzzle involving placing tents next to trees subject to |
3 | * some confusing conditions. |
4 | * |
5 | * TODO: |
e8df451f |
6 | * |
7 | * - my technique for highlighting errors in the tent/tree matching |
8 | * is not perfect. It currently works by finding the connected |
9 | * components of the bipartite adjacency graph between tents and |
10 | * trees, and highlighting red all the tents in such a component |
11 | * if they outnumber the trees (the red meaning "these tents have |
12 | * too few trees between them") and vice versa if the trees |
13 | * outnumber the tents (but this time considering BLANKs as |
14 | * potential tents as yet unplaced, to avoid highlighting |
15 | * 'errors' from the word go before the player has actually made |
16 | * any mistake). However, something more subtle can go wrong |
17 | * within a component: consider, for instance, the setup |
18 | * |
19 | * T |
20 | * tTtT |
21 | * t |
22 | * |
23 | * in which there is one connected component containing equal |
24 | * numbers of trees and tents, but nonetheless there is no |
25 | * perfect matching that can link the two sensibly. This will be |
26 | * rejected by the rigorous solution checker, but the error |
27 | * highlighter won't currently spot it. |
86e60e3d |
28 | * |
e8df451f |
29 | * Well, the _matching_ error highlighter won't spot it, anyway. |
30 | * In that diagram, there are two pairs of diagonally adjacent |
31 | * tents, which will be flagged as erroneous because that's much |
32 | * easier. So if I could prove that _all_ such setups require |
33 | * diagonally adjacent tents, I could safely ignore this problem. |
34 | * If not, however, then a proper treatment will require running |
35 | * the maxflow matcher over each component once I've identified |
36 | * them. |
37 | * |
86e60e3d |
38 | * - it might be nice to make setter-provided tent/nontent clues |
39 | * inviolable? |
40 | * * on the other hand, this would introduce considerable extra |
41 | * complexity and size into the game state; also inviolable |
42 | * clues would have to be marked as such somehow, in an |
43 | * intrusive and annoying manner. Since they're never |
44 | * generated by _my_ generator, I'm currently more inclined |
45 | * not to bother. |
46 | * |
47 | * - more difficult levels at the top end? |
48 | * * for example, sometimes we can deduce that two BLANKs in |
49 | * the same row are each adjacent to the same unattached tree |
50 | * and to nothing else, implying that they can't both be |
51 | * tents; this enables us to rule out some extra combinations |
52 | * in the row-based deduction loop, and hence deduce more |
53 | * from the number in that row than we could otherwise do. |
54 | * * that by itself doesn't seem worth implementing a new |
55 | * difficulty level for, but if I can find a few more things |
56 | * like that then it might become worthwhile. |
57 | * * I wonder if there's a sensible heuristic for where to |
58 | * guess which would make a recursive solver viable? |
59 | */ |
60 | |
61 | #include <stdio.h> |
62 | #include <stdlib.h> |
63 | #include <string.h> |
64 | #include <assert.h> |
65 | #include <ctype.h> |
66 | #include <math.h> |
67 | |
68 | #include "puzzles.h" |
69 | #include "maxflow.h" |
70 | |
71 | /* |
72 | * Design discussion |
73 | * ----------------- |
74 | * |
75 | * The rules of this puzzle as available on the WWW are poorly |
76 | * specified. The bits about tents having to be orthogonally |
77 | * adjacent to trees, tents not being even diagonally adjacent to |
78 | * one another, and the number of tents in each row and column |
79 | * being given are simple enough; the difficult bit is the |
80 | * tent-to-tree matching. |
81 | * |
82 | * Some sources use simplistic wordings such as `each tree is |
83 | * exactly connected to only one tent', which is extremely unclear: |
84 | * it's easy to read erroneously as `each tree is _orthogonally |
85 | * adjacent_ to exactly one tent', which is definitely incorrect. |
86 | * Even the most coherent sources I've found don't do a much better |
87 | * job of stating the rule. |
88 | * |
89 | * A more precise statement of the rule is that it must be possible |
90 | * to find a bijection f between tents and trees such that each |
91 | * tree T is orthogonally adjacent to the tent f(T), but that a |
92 | * tent is permitted to be adjacent to other trees in addition to |
93 | * its own. This slightly non-obvious criterion is what gives this |
94 | * puzzle most of its subtlety. |
95 | * |
96 | * However, there's a particularly subtle ambiguity left over. Is |
97 | * the bijection between tents and trees required to be _unique_? |
98 | * In other words, is that bijection conceptually something the |
99 | * player should be able to exhibit as part of the solution (even |
100 | * if they aren't actually required to do so)? Or is it sufficient |
101 | * to have a unique _placement_ of the tents which gives rise to at |
102 | * least one suitable bijection? |
103 | * |
104 | * The puzzle shown to the right of this .T. 2 *T* 2 |
105 | * paragraph illustrates the problem. There T.T 0 -> T-T 0 |
106 | * are two distinct bijections available. .T. 2 *T* 2 |
107 | * The answer to the above question will |
108 | * determine whether it's a valid puzzle. 202 202 |
109 | * |
110 | * This is an important question, because it affects both the |
111 | * player and the generator. Eventually I found all the instances |
112 | * of this puzzle I could Google up, solved them all by hand, and |
113 | * verified that in all cases the tree/tent matching was uniquely |
114 | * determined given the tree and tent positions. Therefore, the |
115 | * puzzle as implemented in this source file takes the following |
116 | * policy: |
117 | * |
118 | * - When checking a user-supplied solution for correctness, only |
119 | * verify that there exists _at least_ one matching. |
120 | * - When generating a puzzle, enforce that there must be |
121 | * _exactly_ one. |
122 | * |
123 | * Algorithmic implications |
124 | * ------------------------ |
125 | * |
126 | * Another way of phrasing the tree/tent matching criterion is to |
127 | * say that the bipartite adjacency graph between trees and tents |
128 | * has a perfect matching. That is, if you construct a graph which |
129 | * has a vertex per tree and a vertex per tent, and an edge between |
130 | * any tree and tent which are orthogonally adjacent, it is |
131 | * possible to find a set of N edges of that graph (where N is the |
132 | * number of trees and also the number of tents) which between them |
133 | * connect every tree to every tent. |
134 | * |
135 | * The most efficient known algorithms for finding such a matching |
136 | * given a graph, as far as I'm aware, are the Munkres assignment |
137 | * algorithm (also known as the Hungarian algorithm) and the |
138 | * Ford-Fulkerson algorithm (for finding optimal flows in |
139 | * networks). Each of these takes O(N^3) running time; so we're |
140 | * talking O(N^3) time to verify any candidate solution to this |
141 | * puzzle. That's just about OK if you're doing it once per mouse |
142 | * click (and in fact not even that, since the sensible thing to do |
143 | * is check all the _other_ puzzle criteria and only wade into this |
144 | * quagmire if none are violated); but if the solver had to keep |
145 | * doing N^3 work internally, then it would probably end up with |
146 | * more like N^5 or N^6 running time, and grid generation would |
147 | * become very clunky. |
148 | * |
149 | * Fortunately, I've been able to prove a very useful property of |
150 | * _unique_ perfect matchings, by adapting the proof of Hall's |
151 | * Marriage Theorem. For those unaware of Hall's Theorem, I'll |
152 | * recap it and its proof: it states that a bipartite graph |
153 | * contains a perfect matching iff every set of vertices on the |
154 | * left side of the graph have a neighbourhood _at least_ as big on |
155 | * the right. |
156 | * |
157 | * This condition is obviously satisfied if a perfect matching does |
158 | * exist; each left-side node has a distinct right-side node which |
159 | * is the one assigned to it by the matching, and thus any set of n |
160 | * left vertices must have a combined neighbourhood containing at |
161 | * least the n corresponding right vertices, and possibly others |
162 | * too. Alternatively, imagine if you had (say) three left-side |
163 | * nodes all of which were connected to only two right-side nodes |
164 | * between them: any perfect matching would have to assign one of |
165 | * those two right nodes to each of the three left nodes, and still |
166 | * give the three left nodes a different right node each. This is |
167 | * of course impossible. |
168 | * |
169 | * To prove the converse (that if every subset of left vertices |
170 | * satisfies the Hall condition then a perfect matching exists), |
171 | * consider trying to find a proper subset of the left vertices |
172 | * which _exactly_ satisfies the Hall condition: that is, its right |
173 | * neighbourhood is precisely the same size as it. If we can find |
174 | * such a subset, then we can split the bipartite graph into two |
175 | * smaller ones: one consisting of the left subset and its right |
176 | * neighbourhood, the other consisting of everything else. Edges |
177 | * from the left side of the former graph to the right side of the |
178 | * latter do not exist, by construction; edges from the right side |
179 | * of the former to the left of the latter cannot be part of any |
180 | * perfect matching because otherwise the left subset would not be |
181 | * left with enough distinct right vertices to connect to (this is |
182 | * exactly the same deduction used in Solo's set analysis). You can |
183 | * then prove (left as an exercise) that both these smaller graphs |
184 | * still satisfy the Hall condition, and therefore the proof will |
185 | * follow by induction. |
186 | * |
187 | * There's one other possibility, which is the case where _no_ |
188 | * proper subset of the left vertices has a right neighbourhood of |
189 | * exactly the same size. That is, every left subset has a strictly |
190 | * _larger_ right neighbourhood. In this situation, we can simply |
191 | * remove an _arbitrary_ edge from the graph. This cannot reduce |
192 | * the size of any left subset's right neighbourhood by more than |
193 | * one, so if all neighbourhoods were strictly bigger than they |
194 | * needed to be initially, they must now still be _at least as big_ |
195 | * as they need to be. So we can keep throwing out arbitrary edges |
196 | * until we find a set which exactly satisfies the Hall condition, |
197 | * and then proceed as above. [] |
198 | * |
199 | * That's Hall's theorem. I now build on this by examining the |
200 | * circumstances in which a bipartite graph can have a _unique_ |
201 | * perfect matching. It is clear that in the second case, where no |
202 | * left subset exactly satisfies the Hall condition and so we can |
203 | * remove an arbitrary edge, there cannot be a unique perfect |
204 | * matching: given one perfect matching, we choose our arbitrary |
205 | * removed edge to be one of those contained in it, and then we can |
206 | * still find a perfect matching in the remaining graph, which will |
207 | * be a distinct perfect matching in the original. |
208 | * |
209 | * So it is a necessary condition for a unique perfect matching |
210 | * that there must be at least one proper left subset which |
211 | * _exactly_ satisfies the Hall condition. But now consider the |
212 | * smaller graph constructed by taking that left subset and its |
213 | * neighbourhood: if the graph as a whole had a unique perfect |
214 | * matching, then so must this smaller one, which means we can find |
215 | * a proper left subset _again_, and so on. Repeating this process |
216 | * must eventually reduce us to a graph with only one left-side |
217 | * vertex (so there are no proper subsets at all); this vertex must |
218 | * be connected to only one right-side vertex, and hence must be so |
219 | * in the original graph as well (by construction). So we can |
220 | * discard this vertex pair from the graph, and any other edges |
221 | * that involved it (which will by construction be from other left |
222 | * vertices only), and the resulting smaller graph still has a |
223 | * unique perfect matching which means we can do the same thing |
224 | * again. |
225 | * |
226 | * In other words, given any bipartite graph with a unique perfect |
227 | * matching, we can find that matching by the following extremely |
228 | * simple algorithm: |
229 | * |
230 | * - Find a left-side vertex which is only connected to one |
231 | * right-side vertex. |
232 | * - Assign those vertices to one another, and therefore discard |
233 | * any other edges connecting to that right vertex. |
234 | * - Repeat until all vertices have been matched. |
235 | * |
236 | * This algorithm can be run in O(V+E) time (where V is the number |
237 | * of vertices and E is the number of edges in the graph), and the |
238 | * only way it can fail is if there is not a unique perfect |
239 | * matching (either because there is no matching at all, or because |
240 | * it isn't unique; but it can't distinguish those cases). |
241 | * |
242 | * Thus, the internal solver in this source file can be confident |
243 | * that if the tree/tent matching is uniquely determined by the |
244 | * tree and tent positions, it can find it using only this kind of |
245 | * obvious and simple operation: assign a tree to a tent if it |
246 | * cannot possibly belong to any other tent, and vice versa. If the |
247 | * solver were _only_ trying to determine the matching, even that |
248 | * `vice versa' wouldn't be required; but it can come in handy when |
249 | * not all the tents have been placed yet. I can therefore be |
250 | * reasonably confident that as long as my solver doesn't need to |
251 | * cope with grids that have a non-unique matching, it will also |
252 | * not need to do anything complicated like set analysis between |
253 | * trees and tents. |
254 | */ |
255 | |
256 | /* |
257 | * In standalone solver mode, `verbose' is a variable which can be |
258 | * set by command-line option; in debugging mode it's simply always |
259 | * true. |
260 | */ |
261 | #if defined STANDALONE_SOLVER |
262 | #define SOLVER_DIAGNOSTICS |
263 | int verbose = FALSE; |
264 | #elif defined SOLVER_DIAGNOSTICS |
265 | #define verbose TRUE |
266 | #endif |
267 | |
268 | /* |
269 | * Difficulty levels. I do some macro ickery here to ensure that my |
270 | * enum and the various forms of my name list always match up. |
271 | */ |
272 | #define DIFFLIST(A) \ |
273 | A(EASY,Easy,e) \ |
274 | A(TRICKY,Tricky,t) |
275 | #define ENUM(upper,title,lower) DIFF_ ## upper, |
276 | #define TITLE(upper,title,lower) #title, |
277 | #define ENCODE(upper,title,lower) #lower |
278 | #define CONFIG(upper,title,lower) ":" #title |
279 | enum { DIFFLIST(ENUM) DIFFCOUNT }; |
280 | static char const *const tents_diffnames[] = { DIFFLIST(TITLE) }; |
281 | static char const tents_diffchars[] = DIFFLIST(ENCODE); |
282 | #define DIFFCONFIG DIFFLIST(CONFIG) |
283 | |
284 | enum { |
285 | COL_BACKGROUND, |
286 | COL_GRID, |
287 | COL_GRASS, |
288 | COL_TREETRUNK, |
289 | COL_TREELEAF, |
290 | COL_TENT, |
2a27ffcf |
291 | COL_ERROR, |
292 | COL_ERRTEXT, |
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293 | COL_ERRTRUNK, |
86e60e3d |
294 | NCOLOURS |
295 | }; |
296 | |
297 | enum { BLANK, TREE, TENT, NONTENT, MAGIC }; |
298 | |
299 | struct game_params { |
300 | int w, h; |
301 | int diff; |
302 | }; |
303 | |
304 | struct numbers { |
305 | int refcount; |
306 | int *numbers; |
307 | }; |
308 | |
309 | struct game_state { |
310 | game_params p; |
311 | char *grid; |
312 | struct numbers *numbers; |
313 | int completed, used_solve; |
314 | }; |
315 | |
316 | static game_params *default_params(void) |
317 | { |
318 | game_params *ret = snew(game_params); |
319 | |
320 | ret->w = ret->h = 8; |
321 | ret->diff = DIFF_EASY; |
322 | |
323 | return ret; |
324 | } |
325 | |
326 | static const struct game_params tents_presets[] = { |
327 | {8, 8, DIFF_EASY}, |
328 | {8, 8, DIFF_TRICKY}, |
329 | {10, 10, DIFF_EASY}, |
330 | {10, 10, DIFF_TRICKY}, |
331 | {15, 15, DIFF_EASY}, |
332 | {15, 15, DIFF_TRICKY}, |
333 | }; |
334 | |
335 | static int game_fetch_preset(int i, char **name, game_params **params) |
336 | { |
337 | game_params *ret; |
338 | char str[80]; |
339 | |
340 | if (i < 0 || i >= lenof(tents_presets)) |
341 | return FALSE; |
342 | |
343 | ret = snew(game_params); |
344 | *ret = tents_presets[i]; |
345 | |
346 | sprintf(str, "%dx%d %s", ret->w, ret->h, tents_diffnames[ret->diff]); |
347 | |
348 | *name = dupstr(str); |
349 | *params = ret; |
350 | return TRUE; |
351 | } |
352 | |
353 | static void free_params(game_params *params) |
354 | { |
355 | sfree(params); |
356 | } |
357 | |
358 | static game_params *dup_params(game_params *params) |
359 | { |
360 | game_params *ret = snew(game_params); |
361 | *ret = *params; /* structure copy */ |
362 | return ret; |
363 | } |
364 | |
365 | static void decode_params(game_params *params, char const *string) |
366 | { |
367 | params->w = params->h = atoi(string); |
368 | while (*string && isdigit((unsigned char)*string)) string++; |
369 | if (*string == 'x') { |
370 | string++; |
371 | params->h = atoi(string); |
372 | while (*string && isdigit((unsigned char)*string)) string++; |
373 | } |
374 | if (*string == 'd') { |
375 | int i; |
376 | string++; |
377 | for (i = 0; i < DIFFCOUNT; i++) |
378 | if (*string == tents_diffchars[i]) |
379 | params->diff = i; |
380 | if (*string) string++; |
381 | } |
382 | } |
383 | |
384 | static char *encode_params(game_params *params, int full) |
385 | { |
386 | char buf[120]; |
387 | |
388 | sprintf(buf, "%dx%d", params->w, params->h); |
389 | if (full) |
390 | sprintf(buf + strlen(buf), "d%c", |
391 | tents_diffchars[params->diff]); |
392 | return dupstr(buf); |
393 | } |
394 | |
395 | static config_item *game_configure(game_params *params) |
396 | { |
397 | config_item *ret; |
398 | char buf[80]; |
399 | |
400 | ret = snewn(4, config_item); |
401 | |
402 | ret[0].name = "Width"; |
403 | ret[0].type = C_STRING; |
404 | sprintf(buf, "%d", params->w); |
405 | ret[0].sval = dupstr(buf); |
406 | ret[0].ival = 0; |
407 | |
408 | ret[1].name = "Height"; |
409 | ret[1].type = C_STRING; |
410 | sprintf(buf, "%d", params->h); |
411 | ret[1].sval = dupstr(buf); |
412 | ret[1].ival = 0; |
413 | |
414 | ret[2].name = "Difficulty"; |
415 | ret[2].type = C_CHOICES; |
416 | ret[2].sval = DIFFCONFIG; |
417 | ret[2].ival = params->diff; |
418 | |
419 | ret[3].name = NULL; |
420 | ret[3].type = C_END; |
421 | ret[3].sval = NULL; |
422 | ret[3].ival = 0; |
423 | |
424 | return ret; |
425 | } |
426 | |
427 | static game_params *custom_params(config_item *cfg) |
428 | { |
429 | game_params *ret = snew(game_params); |
430 | |
431 | ret->w = atoi(cfg[0].sval); |
432 | ret->h = atoi(cfg[1].sval); |
433 | ret->diff = cfg[2].ival; |
434 | |
435 | return ret; |
436 | } |
437 | |
438 | static char *validate_params(game_params *params, int full) |
439 | { |
e4b6a85b |
440 | /* |
441 | * Generating anything under 4x4 runs into trouble of one kind |
442 | * or another. |
443 | */ |
444 | if (params->w < 4 || params->h < 4) |
445 | return "Width and height must both be at least four"; |
86e60e3d |
446 | return NULL; |
447 | } |
448 | |
449 | /* |
450 | * Scratch space for solver. |
451 | */ |
452 | enum { N, U, L, R, D, MAXDIR }; /* link directions */ |
453 | #define dx(d) ( ((d)==R) - ((d)==L) ) |
454 | #define dy(d) ( ((d)==D) - ((d)==U) ) |
455 | #define F(d) ( U + D - (d) ) |
456 | struct solver_scratch { |
457 | char *links; /* mapping between trees and tents */ |
458 | int *locs; |
459 | char *place, *mrows, *trows; |
460 | }; |
461 | |
462 | static struct solver_scratch *new_scratch(int w, int h) |
463 | { |
464 | struct solver_scratch *ret = snew(struct solver_scratch); |
465 | |
466 | ret->links = snewn(w*h, char); |
467 | ret->locs = snewn(max(w, h), int); |
468 | ret->place = snewn(max(w, h), char); |
469 | ret->mrows = snewn(3 * max(w, h), char); |
470 | ret->trows = snewn(3 * max(w, h), char); |
471 | |
472 | return ret; |
473 | } |
474 | |
475 | static void free_scratch(struct solver_scratch *sc) |
476 | { |
477 | sfree(sc->trows); |
478 | sfree(sc->mrows); |
479 | sfree(sc->place); |
480 | sfree(sc->locs); |
481 | sfree(sc->links); |
482 | sfree(sc); |
483 | } |
484 | |
485 | /* |
486 | * Solver. Returns 0 for impossibility, 1 for success, 2 for |
487 | * ambiguity or failure to converge. |
488 | */ |
489 | static int tents_solve(int w, int h, const char *grid, int *numbers, |
490 | char *soln, struct solver_scratch *sc, int diff) |
491 | { |
492 | int x, y, d, i, j; |
493 | char *mrow, *mrow1, *mrow2, *trow, *trow1, *trow2; |
494 | |
495 | /* |
496 | * Set up solver data. |
497 | */ |
498 | memset(sc->links, N, w*h); |
499 | |
500 | /* |
501 | * Set up solution array. |
502 | */ |
503 | memcpy(soln, grid, w*h); |
504 | |
505 | /* |
506 | * Main solver loop. |
507 | */ |
508 | while (1) { |
509 | int done_something = FALSE; |
510 | |
511 | /* |
512 | * Any tent which has only one unattached tree adjacent to |
513 | * it can be tied to that tree. |
514 | */ |
515 | for (y = 0; y < h; y++) |
516 | for (x = 0; x < w; x++) |
517 | if (soln[y*w+x] == TENT && !sc->links[y*w+x]) { |
518 | int linkd = 0; |
519 | |
520 | for (d = 1; d < MAXDIR; d++) { |
521 | int x2 = x + dx(d), y2 = y + dy(d); |
522 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
523 | soln[y2*w+x2] == TREE && |
524 | !sc->links[y2*w+x2]) { |
525 | if (linkd) |
526 | break; /* found more than one */ |
527 | else |
528 | linkd = d; |
529 | } |
530 | } |
531 | |
532 | if (d == MAXDIR && linkd == 0) { |
533 | #ifdef SOLVER_DIAGNOSTICS |
534 | if (verbose) |
535 | printf("tent at %d,%d cannot link to anything\n", |
536 | x, y); |
537 | #endif |
538 | return 0; /* no solution exists */ |
539 | } else if (d == MAXDIR) { |
540 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
541 | |
542 | #ifdef SOLVER_DIAGNOSTICS |
543 | if (verbose) |
544 | printf("tent at %d,%d can only link to tree at" |
545 | " %d,%d\n", x, y, x2, y2); |
546 | #endif |
547 | |
548 | sc->links[y*w+x] = linkd; |
549 | sc->links[y2*w+x2] = F(linkd); |
550 | done_something = TRUE; |
551 | } |
552 | } |
553 | |
554 | if (done_something) |
555 | continue; |
556 | if (diff < 0) |
557 | break; /* don't do anything else! */ |
558 | |
559 | /* |
560 | * Mark a blank square as NONTENT if it is not orthogonally |
561 | * adjacent to any unmatched tree. |
562 | */ |
563 | for (y = 0; y < h; y++) |
564 | for (x = 0; x < w; x++) |
565 | if (soln[y*w+x] == BLANK) { |
566 | int can_be_tent = FALSE; |
567 | |
568 | for (d = 1; d < MAXDIR; d++) { |
569 | int x2 = x + dx(d), y2 = y + dy(d); |
570 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
571 | soln[y2*w+x2] == TREE && |
572 | !sc->links[y2*w+x2]) |
573 | can_be_tent = TRUE; |
574 | } |
575 | |
576 | if (!can_be_tent) { |
577 | #ifdef SOLVER_DIAGNOSTICS |
578 | if (verbose) |
579 | printf("%d,%d cannot be a tent (no adjacent" |
580 | " unmatched tree)\n", x, y); |
581 | #endif |
582 | soln[y*w+x] = NONTENT; |
583 | done_something = TRUE; |
584 | } |
585 | } |
586 | |
587 | if (done_something) |
588 | continue; |
589 | |
590 | /* |
591 | * Mark a blank square as NONTENT if it is (perhaps |
592 | * diagonally) adjacent to any other tent. |
593 | */ |
594 | for (y = 0; y < h; y++) |
595 | for (x = 0; x < w; x++) |
596 | if (soln[y*w+x] == BLANK) { |
597 | int dx, dy, imposs = FALSE; |
598 | |
599 | for (dy = -1; dy <= +1; dy++) |
600 | for (dx = -1; dx <= +1; dx++) |
601 | if (dy || dx) { |
602 | int x2 = x + dx, y2 = y + dy; |
603 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
604 | soln[y2*w+x2] == TENT) |
605 | imposs = TRUE; |
606 | } |
607 | |
608 | if (imposs) { |
609 | #ifdef SOLVER_DIAGNOSTICS |
610 | if (verbose) |
611 | printf("%d,%d cannot be a tent (adjacent tent)\n", |
612 | x, y); |
613 | #endif |
614 | soln[y*w+x] = NONTENT; |
615 | done_something = TRUE; |
616 | } |
617 | } |
618 | |
619 | if (done_something) |
620 | continue; |
621 | |
622 | /* |
623 | * Any tree which has exactly one {unattached tent, BLANK} |
624 | * adjacent to it must have its tent in that square. |
625 | */ |
626 | for (y = 0; y < h; y++) |
627 | for (x = 0; x < w; x++) |
628 | if (soln[y*w+x] == TREE && !sc->links[y*w+x]) { |
629 | int linkd = 0, linkd2 = 0, nd = 0; |
630 | |
631 | for (d = 1; d < MAXDIR; d++) { |
632 | int x2 = x + dx(d), y2 = y + dy(d); |
633 | if (!(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h)) |
634 | continue; |
635 | if (soln[y2*w+x2] == BLANK || |
636 | (soln[y2*w+x2] == TENT && !sc->links[y2*w+x2])) { |
637 | if (linkd) |
638 | linkd2 = d; |
639 | else |
640 | linkd = d; |
641 | nd++; |
642 | } |
643 | } |
644 | |
645 | if (nd == 0) { |
646 | #ifdef SOLVER_DIAGNOSTICS |
647 | if (verbose) |
648 | printf("tree at %d,%d cannot link to anything\n", |
649 | x, y); |
650 | #endif |
651 | return 0; /* no solution exists */ |
652 | } else if (nd == 1) { |
653 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
654 | |
655 | #ifdef SOLVER_DIAGNOSTICS |
656 | if (verbose) |
657 | printf("tree at %d,%d can only link to tent at" |
658 | " %d,%d\n", x, y, x2, y2); |
659 | #endif |
660 | soln[y2*w+x2] = TENT; |
661 | sc->links[y*w+x] = linkd; |
662 | sc->links[y2*w+x2] = F(linkd); |
663 | done_something = TRUE; |
664 | } else if (nd == 2 && (!dx(linkd) != !dx(linkd2)) && |
665 | diff >= DIFF_TRICKY) { |
666 | /* |
667 | * If there are two possible places where |
668 | * this tree's tent can go, and they are |
669 | * diagonally separated rather than being |
670 | * on opposite sides of the tree, then the |
671 | * square (other than the tree square) |
672 | * which is adjacent to both of them must |
673 | * be a non-tent. |
674 | */ |
675 | int x2 = x + dx(linkd) + dx(linkd2); |
676 | int y2 = y + dy(linkd) + dy(linkd2); |
677 | assert(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h); |
678 | if (soln[y2*w+x2] == BLANK) { |
679 | #ifdef SOLVER_DIAGNOSTICS |
680 | if (verbose) |
681 | printf("possible tent locations for tree at" |
682 | " %d,%d rule out tent at %d,%d\n", |
683 | x, y, x2, y2); |
684 | #endif |
685 | soln[y2*w+x2] = NONTENT; |
686 | done_something = TRUE; |
687 | } |
688 | } |
689 | } |
690 | |
691 | if (done_something) |
692 | continue; |
693 | |
694 | /* |
695 | * If localised deductions about the trees and tents |
696 | * themselves haven't helped us, it's time to resort to the |
697 | * numbers round the grid edge. For each row and column, we |
698 | * go through all possible combinations of locations for |
699 | * the unplaced tents, rule out any which have adjacent |
700 | * tents, and spot any square which is given the same state |
701 | * by all remaining combinations. |
702 | */ |
703 | for (i = 0; i < w+h; i++) { |
704 | int start, step, len, start1, start2, n, k; |
705 | |
706 | if (i < w) { |
707 | /* |
708 | * This is the number for a column. |
709 | */ |
710 | start = i; |
711 | step = w; |
712 | len = h; |
713 | if (i > 0) |
714 | start1 = start - 1; |
715 | else |
716 | start1 = -1; |
717 | if (i+1 < w) |
718 | start2 = start + 1; |
719 | else |
720 | start2 = -1; |
721 | } else { |
722 | /* |
723 | * This is the number for a row. |
724 | */ |
725 | start = (i-w)*w; |
726 | step = 1; |
727 | len = w; |
728 | if (i > w) |
729 | start1 = start - w; |
730 | else |
731 | start1 = -1; |
732 | if (i+1 < w+h) |
733 | start2 = start + w; |
734 | else |
735 | start2 = -1; |
736 | } |
737 | |
738 | if (diff < DIFF_TRICKY) { |
739 | /* |
740 | * In Easy mode, we don't look at the effect of one |
741 | * row on the next (i.e. ruling out a square if all |
742 | * possibilities for an adjacent row place a tent |
743 | * next to it). |
744 | */ |
745 | start1 = start2 = -1; |
746 | } |
747 | |
748 | k = numbers[i]; |
749 | |
750 | /* |
751 | * Count and store the locations of the free squares, |
752 | * and also count the number of tents already placed. |
753 | */ |
754 | n = 0; |
755 | for (j = 0; j < len; j++) { |
756 | if (soln[start+j*step] == TENT) |
757 | k--; /* one fewer tent to place */ |
758 | else if (soln[start+j*step] == BLANK) |
759 | sc->locs[n++] = j; |
760 | } |
761 | |
762 | if (n == 0) |
763 | continue; /* nothing left to do here */ |
764 | |
765 | /* |
766 | * Now we know we're placing k tents in n squares. Set |
767 | * up the first possibility. |
768 | */ |
769 | for (j = 0; j < n; j++) |
770 | sc->place[j] = (j < k ? TENT : NONTENT); |
771 | |
772 | /* |
773 | * We're aiming to find squares in this row which are |
774 | * invariant over all valid possibilities. Thus, we |
775 | * maintain the current state of that invariance. We |
776 | * start everything off at MAGIC to indicate that it |
777 | * hasn't been set up yet. |
778 | */ |
779 | mrow = sc->mrows; |
780 | mrow1 = sc->mrows + len; |
781 | mrow2 = sc->mrows + 2*len; |
782 | trow = sc->trows; |
783 | trow1 = sc->trows + len; |
784 | trow2 = sc->trows + 2*len; |
785 | memset(mrow, MAGIC, 3*len); |
786 | |
787 | /* |
788 | * And iterate over all possibilities. |
789 | */ |
790 | while (1) { |
791 | int p, valid; |
792 | |
793 | /* |
794 | * See if this possibility is valid. The only way |
795 | * it can fail to be valid is if it contains two |
796 | * adjacent tents. (Other forms of invalidity, such |
797 | * as containing a tent adjacent to one already |
798 | * placed, will have been dealt with already by |
799 | * other parts of the solver.) |
800 | */ |
801 | valid = TRUE; |
802 | for (j = 0; j+1 < n; j++) |
803 | if (sc->place[j] == TENT && |
804 | sc->place[j+1] == TENT && |
805 | sc->locs[j+1] == sc->locs[j]+1) { |
806 | valid = FALSE; |
807 | break; |
808 | } |
809 | |
810 | if (valid) { |
811 | /* |
812 | * Merge this valid combination into mrow. |
813 | */ |
814 | memset(trow, MAGIC, len); |
815 | memset(trow+len, BLANK, 2*len); |
816 | for (j = 0; j < n; j++) { |
817 | trow[sc->locs[j]] = sc->place[j]; |
818 | if (sc->place[j] == TENT) { |
819 | int jj; |
820 | for (jj = sc->locs[j]-1; jj <= sc->locs[j]+1; jj++) |
821 | if (jj >= 0 && jj < len) |
822 | trow1[jj] = trow2[jj] = NONTENT; |
823 | } |
824 | } |
825 | |
826 | for (j = 0; j < 3*len; j++) { |
827 | if (trow[j] == MAGIC) |
828 | continue; |
829 | if (mrow[j] == MAGIC || mrow[j] == trow[j]) { |
830 | /* |
831 | * Either this is the first valid |
832 | * placement we've found at all, or |
833 | * this square's contents are |
834 | * consistent with every previous valid |
835 | * combination. |
836 | */ |
837 | mrow[j] = trow[j]; |
838 | } else { |
839 | /* |
840 | * This square's contents fail to match |
841 | * what they were in a different |
842 | * combination, so we cannot deduce |
843 | * anything about this square. |
844 | */ |
845 | mrow[j] = BLANK; |
846 | } |
847 | } |
848 | } |
849 | |
850 | /* |
851 | * Find the next combination of k choices from n. |
852 | * We do this by finding the rightmost tent which |
853 | * can be moved one place right, doing so, and |
854 | * shunting all tents to the right of that as far |
855 | * left as they can go. |
856 | */ |
857 | p = 0; |
858 | for (j = n-1; j > 0; j--) { |
859 | if (sc->place[j] == TENT) |
860 | p++; |
861 | if (sc->place[j] == NONTENT && sc->place[j-1] == TENT) { |
862 | sc->place[j-1] = NONTENT; |
863 | sc->place[j] = TENT; |
864 | while (p--) |
865 | sc->place[++j] = TENT; |
866 | while (++j < n) |
867 | sc->place[j] = NONTENT; |
868 | break; |
869 | } |
870 | } |
871 | if (j <= 0) |
872 | break; /* we've finished */ |
873 | } |
874 | |
875 | /* |
876 | * It's just possible that _no_ placement was valid, in |
877 | * which case we have an internally inconsistent |
878 | * puzzle. |
879 | */ |
880 | if (mrow[sc->locs[0]] == MAGIC) |
881 | return 0; /* inconsistent */ |
882 | |
883 | /* |
884 | * Now go through mrow and see if there's anything |
885 | * we've deduced which wasn't already mentioned in soln. |
886 | */ |
887 | for (j = 0; j < len; j++) { |
888 | int whichrow; |
889 | |
890 | for (whichrow = 0; whichrow < 3; whichrow++) { |
891 | char *mthis = mrow + whichrow * len; |
892 | int tstart = (whichrow == 0 ? start : |
893 | whichrow == 1 ? start1 : start2); |
894 | if (tstart >= 0 && |
895 | mthis[j] != MAGIC && mthis[j] != BLANK && |
896 | soln[tstart+j*step] == BLANK) { |
897 | int pos = tstart+j*step; |
898 | |
899 | #ifdef SOLVER_DIAGNOSTICS |
900 | if (verbose) |
901 | printf("%s %d forces %s at %d,%d\n", |
902 | step==1 ? "row" : "column", |
903 | step==1 ? start/w : start, |
9d8d4f9f |
904 | mthis[j] == TENT ? "tent" : "non-tent", |
86e60e3d |
905 | pos % w, pos / w); |
906 | #endif |
907 | soln[pos] = mthis[j]; |
908 | done_something = TRUE; |
909 | } |
910 | } |
911 | } |
912 | } |
913 | |
914 | if (done_something) |
915 | continue; |
916 | |
917 | if (!done_something) |
918 | break; |
919 | } |
920 | |
921 | /* |
922 | * The solver has nothing further it can do. Return 1 if both |
923 | * soln and sc->links are completely filled in, or 2 otherwise. |
924 | */ |
925 | for (y = 0; y < h; y++) |
926 | for (x = 0; x < w; x++) { |
927 | if (soln[y*w+x] == BLANK) |
928 | return 2; |
929 | if (soln[y*w+x] != NONTENT && sc->links[y*w+x] == 0) |
930 | return 2; |
931 | } |
932 | |
933 | return 1; |
934 | } |
935 | |
936 | static char *new_game_desc(game_params *params, random_state *rs, |
937 | char **aux, int interactive) |
938 | { |
939 | int w = params->w, h = params->h; |
940 | int ntrees = w * h / 5; |
941 | char *grid = snewn(w*h, char); |
942 | char *puzzle = snewn(w*h, char); |
943 | int *numbers = snewn(w+h, int); |
944 | char *soln = snewn(w*h, char); |
e4b6a85b |
945 | int *temp = snewn(2*w*h, int); |
86e60e3d |
946 | int maxedges = ntrees*4 + w*h; |
947 | int *edges = snewn(2*maxedges, int); |
948 | int *capacity = snewn(maxedges, int); |
949 | int *flow = snewn(maxedges, int); |
950 | struct solver_scratch *sc = new_scratch(w, h); |
951 | char *ret, *p; |
952 | int i, j, nedges; |
953 | |
954 | /* |
955 | * Since this puzzle has many global deductions and doesn't |
956 | * permit limited clue sets, generating grids for this puzzle |
957 | * is hard enough that I see no better option than to simply |
958 | * generate a solution and see if it's unique and has the |
959 | * required difficulty. This turns out to be computationally |
960 | * plausible as well. |
961 | * |
962 | * We chose our tree count (hence also tent count) by dividing |
963 | * the total grid area by five above. Why five? Well, w*h/4 is |
964 | * the maximum number of tents you can _possibly_ fit into the |
965 | * grid without violating the separation criterion, and to |
966 | * achieve that you are constrained to a very small set of |
967 | * possible layouts (the obvious one with a tent at every |
968 | * (even,even) coordinate, and trivial variations thereon). So |
969 | * if we reduce the tent count a bit more, we enable more |
970 | * random-looking placement; 5 turns out to be a plausible |
971 | * figure which yields sensible puzzles. Increasing the tent |
972 | * count would give puzzles whose solutions were too regimented |
973 | * and could be solved by the use of that knowledge (and would |
974 | * also take longer to find a viable placement); decreasing it |
975 | * would make the grids emptier and more boring. |
976 | * |
977 | * Actually generating a grid is a matter of first placing the |
978 | * tents, and then placing the trees by the use of maxflow |
979 | * (finding a distinct square adjacent to every tent). We do it |
980 | * this way round because otherwise satisfying the tent |
981 | * separation condition would become onerous: most randomly |
982 | * chosen tent layouts do not satisfy this condition, so we'd |
983 | * have gone to a lot of work before finding that a candidate |
984 | * layout was unusable. Instead, we place the tents first and |
985 | * ensure they meet the separation criterion _before_ doing |
986 | * lots of computation; this works much better. |
987 | * |
988 | * The maxflow algorithm is not randomised, so employed naively |
989 | * it would give rise to grids with clear structure and |
990 | * directional bias. Hence, I assign the network nodes as seen |
e4b6a85b |
991 | * by maxflow to be a _random_ permutation of the squares of |
992 | * the grid, so that any bias shown by maxflow towards |
993 | * low-numbered nodes is turned into a random bias. |
86e60e3d |
994 | * |
995 | * This generation strategy can fail at many points, including |
996 | * as early as tent placement (if you get a bad random order in |
997 | * which to greedily try the grid squares, you won't even |
998 | * manage to find enough mutually non-adjacent squares to put |
999 | * the tents in). Then it can fail if maxflow doesn't manage to |
1000 | * find a good enough matching (i.e. the tent placements don't |
1001 | * admit any adequate tree placements); and finally it can fail |
1002 | * if the solver finds that the problem has the wrong |
1003 | * difficulty (including being actually non-unique). All of |
1004 | * these, however, are insufficiently frequent to cause |
1005 | * trouble. |
1006 | */ |
1007 | |
e4b6a85b |
1008 | if (params->diff > DIFF_EASY && params->w <= 4 && params->h <= 4) |
1009 | params->diff = DIFF_EASY; /* downgrade to prevent tight loop */ |
1010 | |
86e60e3d |
1011 | while (1) { |
1012 | /* |
e4b6a85b |
1013 | * Arrange the grid squares into a random order. |
86e60e3d |
1014 | */ |
1015 | for (i = 0; i < w*h; i++) |
1016 | temp[i] = i; |
1017 | shuffle(temp, w*h, sizeof(*temp), rs); |
86e60e3d |
1018 | |
1019 | /* |
1020 | * The first `ntrees' entries in temp which we can get |
1021 | * without making two tents adjacent will be the tent |
1022 | * locations. |
1023 | */ |
1024 | memset(grid, BLANK, w*h); |
1025 | j = ntrees; |
1026 | for (i = 0; i < w*h && j > 0; i++) { |
1027 | int x = temp[i] % w, y = temp[i] / w; |
1028 | int dy, dx, ok = TRUE; |
1029 | |
1030 | for (dy = -1; dy <= +1; dy++) |
1031 | for (dx = -1; dx <= +1; dx++) |
1032 | if (x+dx >= 0 && x+dx < w && |
1033 | y+dy >= 0 && y+dy < h && |
1034 | grid[(y+dy)*w+(x+dx)] == TENT) |
1035 | ok = FALSE; |
1036 | |
1037 | if (ok) { |
1038 | grid[temp[i]] = TENT; |
1039 | j--; |
1040 | } |
1041 | } |
1042 | if (j > 0) |
1043 | continue; /* couldn't place all the tents */ |
1044 | |
1045 | /* |
1046 | * Now we build up the list of graph edges. |
1047 | */ |
1048 | nedges = 0; |
1049 | for (i = 0; i < w*h; i++) { |
1050 | if (grid[temp[i]] == TENT) { |
1051 | for (j = 0; j < w*h; j++) { |
1052 | if (grid[temp[j]] != TENT) { |
1053 | int xi = temp[i] % w, yi = temp[i] / w; |
1054 | int xj = temp[j] % w, yj = temp[j] / w; |
1055 | if (abs(xi-xj) + abs(yi-yj) == 1) { |
1056 | edges[nedges*2] = i; |
1057 | edges[nedges*2+1] = j; |
1058 | capacity[nedges] = 1; |
1059 | nedges++; |
1060 | } |
1061 | } |
1062 | } |
1063 | } else { |
1064 | /* |
1065 | * Special node w*h is the sink node; any non-tent node |
1066 | * has an edge going to it. |
1067 | */ |
1068 | edges[nedges*2] = i; |
1069 | edges[nedges*2+1] = w*h; |
1070 | capacity[nedges] = 1; |
1071 | nedges++; |
1072 | } |
1073 | } |
1074 | |
1075 | /* |
1076 | * Special node w*h+1 is the source node, with an edge going to |
1077 | * every tent. |
1078 | */ |
1079 | for (i = 0; i < w*h; i++) { |
1080 | if (grid[temp[i]] == TENT) { |
1081 | edges[nedges*2] = w*h+1; |
1082 | edges[nedges*2+1] = i; |
1083 | capacity[nedges] = 1; |
1084 | nedges++; |
1085 | } |
1086 | } |
1087 | |
1088 | assert(nedges <= maxedges); |
1089 | |
1090 | /* |
1091 | * Now we're ready to call the maxflow algorithm to place the |
1092 | * trees. |
1093 | */ |
1094 | j = maxflow(w*h+2, w*h+1, w*h, nedges, edges, capacity, flow, NULL); |
1095 | |
1096 | if (j < ntrees) |
1097 | continue; /* couldn't place all the tents */ |
1098 | |
1099 | /* |
1100 | * We've placed the trees. Now we need to work out _where_ |
1101 | * we've placed them, which is a matter of reading back out |
1102 | * from the `flow' array. |
1103 | */ |
1104 | for (i = 0; i < nedges; i++) { |
1105 | if (edges[2*i] < w*h && edges[2*i+1] < w*h && flow[i] > 0) |
1106 | grid[temp[edges[2*i+1]]] = TREE; |
1107 | } |
1108 | |
1109 | /* |
1110 | * I think it looks ugly if there isn't at least one of |
1111 | * _something_ (tent or tree) in each row and each column |
1112 | * of the grid. This doesn't give any information away |
1113 | * since a completely empty row/column is instantly obvious |
1114 | * from the clues (it has no trees and a zero). |
1115 | */ |
1116 | for (i = 0; i < w; i++) { |
1117 | for (j = 0; j < h; j++) { |
1118 | if (grid[j*w+i] != BLANK) |
1119 | break; /* found something in this column */ |
1120 | } |
1121 | if (j == h) |
1122 | break; /* found empty column */ |
1123 | } |
1124 | if (i < w) |
1125 | continue; /* a column was empty */ |
1126 | |
1127 | for (j = 0; j < h; j++) { |
1128 | for (i = 0; i < w; i++) { |
1129 | if (grid[j*w+i] != BLANK) |
1130 | break; /* found something in this row */ |
1131 | } |
1132 | if (i == w) |
1133 | break; /* found empty row */ |
1134 | } |
1135 | if (j < h) |
1136 | continue; /* a row was empty */ |
1137 | |
1138 | /* |
1139 | * Now set up the numbers round the edge. |
1140 | */ |
1141 | for (i = 0; i < w; i++) { |
1142 | int n = 0; |
1143 | for (j = 0; j < h; j++) |
1144 | if (grid[j*w+i] == TENT) |
1145 | n++; |
1146 | numbers[i] = n; |
1147 | } |
1148 | for (i = 0; i < h; i++) { |
1149 | int n = 0; |
1150 | for (j = 0; j < w; j++) |
1151 | if (grid[i*w+j] == TENT) |
1152 | n++; |
1153 | numbers[w+i] = n; |
1154 | } |
1155 | |
1156 | /* |
1157 | * And now actually solve the puzzle, to see whether it's |
1158 | * unique and has the required difficulty. |
1159 | */ |
1160 | for (i = 0; i < w*h; i++) |
1161 | puzzle[i] = grid[i] == TREE ? TREE : BLANK; |
1162 | i = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff-1); |
1163 | j = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff); |
1164 | |
1165 | /* |
1166 | * We expect solving with difficulty params->diff to have |
1167 | * succeeded (otherwise the problem is too hard), and |
1168 | * solving with diff-1 to have failed (otherwise it's too |
1169 | * easy). |
1170 | */ |
1171 | if (i == 2 && j == 1) |
1172 | break; |
1173 | } |
1174 | |
1175 | /* |
1176 | * That's it. Encode as a game ID. |
1177 | */ |
1178 | ret = snewn((w+h)*40 + ntrees + (w*h)/26 + 1, char); |
1179 | p = ret; |
1180 | j = 0; |
1181 | for (i = 0; i <= w*h; i++) { |
1182 | int c = (i < w*h ? grid[i] == TREE : 1); |
1183 | if (c) { |
1184 | *p++ = (j == 0 ? '_' : j-1 + 'a'); |
1185 | j = 0; |
1186 | } else { |
1187 | j++; |
1188 | while (j > 25) { |
1189 | *p++ = 'z'; |
1190 | j -= 25; |
1191 | } |
1192 | } |
1193 | } |
1194 | for (i = 0; i < w+h; i++) |
1195 | p += sprintf(p, ",%d", numbers[i]); |
1196 | *p++ = '\0'; |
1197 | ret = sresize(ret, p - ret, char); |
1198 | |
1199 | /* |
1200 | * And encode the solution as an aux_info. |
1201 | */ |
1202 | *aux = snewn(ntrees * 40, char); |
1203 | p = *aux; |
1204 | *p++ = 'S'; |
1205 | for (i = 0; i < w*h; i++) |
1206 | if (grid[i] == TENT) |
1207 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1208 | *p++ = '\0'; |
1209 | *aux = sresize(*aux, p - *aux, char); |
1210 | |
1211 | free_scratch(sc); |
1212 | sfree(flow); |
1213 | sfree(capacity); |
1214 | sfree(edges); |
1215 | sfree(temp); |
1216 | sfree(soln); |
1217 | sfree(numbers); |
1218 | sfree(puzzle); |
1219 | sfree(grid); |
1220 | |
1221 | return ret; |
1222 | } |
1223 | |
1224 | static char *validate_desc(game_params *params, char *desc) |
1225 | { |
1226 | int w = params->w, h = params->h; |
1227 | int area, i; |
1228 | |
1229 | area = 0; |
1230 | while (*desc && *desc != ',') { |
1231 | if (*desc == '_') |
1232 | area++; |
1233 | else if (*desc >= 'a' && *desc < 'z') |
1234 | area += *desc - 'a' + 2; |
1235 | else if (*desc == 'z') |
1236 | area += 25; |
1237 | else if (*desc == '!' || *desc == '-') |
1238 | /* do nothing */; |
1239 | else |
1240 | return "Invalid character in grid specification"; |
1241 | |
1242 | desc++; |
1243 | } |
1244 | |
1245 | for (i = 0; i < w+h; i++) { |
1246 | if (!*desc) |
1247 | return "Not enough numbers given after grid specification"; |
1248 | else if (*desc != ',') |
1249 | return "Invalid character in number list"; |
1250 | desc++; |
1251 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1252 | } |
1253 | |
1254 | if (*desc) |
1255 | return "Unexpected additional data at end of game description"; |
1256 | return NULL; |
1257 | } |
1258 | |
1259 | static game_state *new_game(midend *me, game_params *params, char *desc) |
1260 | { |
1261 | int w = params->w, h = params->h; |
1262 | game_state *state = snew(game_state); |
1263 | int i; |
1264 | |
1265 | state->p = *params; /* structure copy */ |
1266 | state->grid = snewn(w*h, char); |
1267 | state->numbers = snew(struct numbers); |
1268 | state->numbers->refcount = 1; |
1269 | state->numbers->numbers = snewn(w+h, int); |
1270 | state->completed = state->used_solve = FALSE; |
1271 | |
1272 | i = 0; |
1273 | memset(state->grid, BLANK, w*h); |
1274 | |
1275 | while (*desc) { |
1276 | int run, type; |
1277 | |
1278 | type = TREE; |
1279 | |
1280 | if (*desc == '_') |
1281 | run = 0; |
1282 | else if (*desc >= 'a' && *desc < 'z') |
1283 | run = *desc - ('a'-1); |
1284 | else if (*desc == 'z') { |
1285 | run = 25; |
1286 | type = BLANK; |
1287 | } else { |
1288 | assert(*desc == '!' || *desc == '-'); |
1289 | run = -1; |
1290 | type = (*desc == '!' ? TENT : NONTENT); |
1291 | } |
1292 | |
1293 | desc++; |
1294 | |
1295 | i += run; |
1296 | assert(i >= 0 && i <= w*h); |
1297 | if (i == w*h) { |
1298 | assert(type == TREE); |
1299 | break; |
1300 | } else { |
1301 | if (type != BLANK) |
1302 | state->grid[i++] = type; |
1303 | } |
1304 | } |
1305 | |
1306 | for (i = 0; i < w+h; i++) { |
1307 | assert(*desc == ','); |
1308 | desc++; |
1309 | state->numbers->numbers[i] = atoi(desc); |
1310 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1311 | } |
1312 | |
1313 | assert(!*desc); |
1314 | |
1315 | return state; |
1316 | } |
1317 | |
1318 | static game_state *dup_game(game_state *state) |
1319 | { |
1320 | int w = state->p.w, h = state->p.h; |
1321 | game_state *ret = snew(game_state); |
1322 | |
1323 | ret->p = state->p; /* structure copy */ |
1324 | ret->grid = snewn(w*h, char); |
1325 | memcpy(ret->grid, state->grid, w*h); |
1326 | ret->numbers = state->numbers; |
1327 | state->numbers->refcount++; |
1328 | ret->completed = state->completed; |
1329 | ret->used_solve = state->used_solve; |
1330 | |
1331 | return ret; |
1332 | } |
1333 | |
1334 | static void free_game(game_state *state) |
1335 | { |
1336 | if (--state->numbers->refcount <= 0) { |
1337 | sfree(state->numbers->numbers); |
1338 | sfree(state->numbers); |
1339 | } |
1340 | sfree(state->grid); |
1341 | sfree(state); |
1342 | } |
1343 | |
1344 | static char *solve_game(game_state *state, game_state *currstate, |
1345 | char *aux, char **error) |
1346 | { |
1347 | int w = state->p.w, h = state->p.h; |
1348 | |
1349 | if (aux) { |
1350 | /* |
1351 | * If we already have the solution, save ourselves some |
1352 | * time. |
1353 | */ |
1354 | return dupstr(aux); |
1355 | } else { |
1356 | struct solver_scratch *sc = new_scratch(w, h); |
1357 | char *soln; |
1358 | int ret; |
1359 | char *move, *p; |
1360 | int i; |
1361 | |
1362 | soln = snewn(w*h, char); |
1363 | ret = tents_solve(w, h, state->grid, state->numbers->numbers, |
1364 | soln, sc, DIFFCOUNT-1); |
1365 | free_scratch(sc); |
1366 | if (ret != 1) { |
1367 | sfree(soln); |
1368 | if (ret == 0) |
1369 | *error = "This puzzle is not self-consistent"; |
1370 | else |
1371 | *error = "Unable to find a unique solution for this puzzle"; |
1372 | return NULL; |
1373 | } |
1374 | |
1375 | /* |
1376 | * Construct a move string which turns the current state |
1377 | * into the solved state. |
1378 | */ |
1379 | move = snewn(w*h * 40, char); |
1380 | p = move; |
1381 | *p++ = 'S'; |
1382 | for (i = 0; i < w*h; i++) |
1383 | if (soln[i] == TENT) |
1384 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1385 | *p++ = '\0'; |
1386 | move = sresize(move, p - move, char); |
1387 | |
1388 | sfree(soln); |
1389 | |
1390 | return move; |
1391 | } |
1392 | } |
1393 | |
fa3abef5 |
1394 | static int game_can_format_as_text_now(game_params *params) |
1395 | { |
1396 | return TRUE; |
1397 | } |
1398 | |
86e60e3d |
1399 | static char *game_text_format(game_state *state) |
1400 | { |
1401 | int w = state->p.w, h = state->p.h; |
1402 | char *ret, *p; |
1403 | int x, y; |
1404 | |
1405 | /* |
1406 | * FIXME: We currently do not print the numbers round the edges |
1407 | * of the grid. I need to work out a sensible way of doing this |
1408 | * even when the column numbers exceed 9. |
1409 | * |
1410 | * In the absence of those numbers, the result size is h lines |
1411 | * of w+1 characters each, plus a NUL. |
1412 | * |
1413 | * This function is currently only used by the standalone |
1414 | * solver; until I make it look more sensible, I won't enable |
1415 | * it in the main game structure. |
1416 | */ |
1417 | ret = snewn(h*(w+1) + 1, char); |
1418 | p = ret; |
1419 | for (y = 0; y < h; y++) { |
1420 | for (x = 0; x < w; x++) { |
1421 | *p = (state->grid[y*w+x] == BLANK ? '.' : |
1422 | state->grid[y*w+x] == TREE ? 'T' : |
1423 | state->grid[y*w+x] == TENT ? '*' : |
1424 | state->grid[y*w+x] == NONTENT ? '-' : '?'); |
1425 | p++; |
1426 | } |
1427 | *p++ = '\n'; |
1428 | } |
1429 | *p++ = '\0'; |
1430 | |
1431 | return ret; |
1432 | } |
1433 | |
565394e7 |
1434 | struct game_ui { |
1435 | int dsx, dsy; /* coords of drag start */ |
1436 | int dex, dey; /* coords of drag end */ |
1437 | int drag_button; /* -1 for none, or a button code */ |
1438 | int drag_ok; /* dragged off the window, to cancel */ |
505ea4e5 |
1439 | |
1440 | int cx, cy, cdisp; /* cursor position, and ?display. */ |
565394e7 |
1441 | }; |
1442 | |
86e60e3d |
1443 | static game_ui *new_ui(game_state *state) |
1444 | { |
565394e7 |
1445 | game_ui *ui = snew(game_ui); |
1446 | ui->dsx = ui->dsy = -1; |
1447 | ui->dex = ui->dey = -1; |
1448 | ui->drag_button = -1; |
1449 | ui->drag_ok = FALSE; |
505ea4e5 |
1450 | ui->cx = ui->cy = ui->cdisp = 0; |
565394e7 |
1451 | return ui; |
86e60e3d |
1452 | } |
1453 | |
1454 | static void free_ui(game_ui *ui) |
1455 | { |
565394e7 |
1456 | sfree(ui); |
86e60e3d |
1457 | } |
1458 | |
1459 | static char *encode_ui(game_ui *ui) |
1460 | { |
1461 | return NULL; |
1462 | } |
1463 | |
1464 | static void decode_ui(game_ui *ui, char *encoding) |
1465 | { |
1466 | } |
1467 | |
1468 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
1469 | game_state *newstate) |
1470 | { |
1471 | } |
1472 | |
1473 | struct game_drawstate { |
1474 | int tilesize; |
1475 | int started; |
1476 | game_params p; |
2a27ffcf |
1477 | int *drawn, *numbersdrawn; |
505ea4e5 |
1478 | int cx, cy; /* last-drawn cursor pos, or (-1,-1) if absent. */ |
86e60e3d |
1479 | }; |
1480 | |
1481 | #define PREFERRED_TILESIZE 32 |
1482 | #define TILESIZE (ds->tilesize) |
1483 | #define TLBORDER (TILESIZE/2) |
1484 | #define BRBORDER (TILESIZE*3/2) |
1485 | #define COORD(x) ( (x) * TILESIZE + TLBORDER ) |
1486 | #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 ) |
1487 | |
1488 | #define FLASH_TIME 0.30F |
1489 | |
565394e7 |
1490 | static int drag_xform(game_ui *ui, int x, int y, int v) |
1491 | { |
1492 | int xmin, ymin, xmax, ymax; |
1493 | |
1494 | xmin = min(ui->dsx, ui->dex); |
1495 | xmax = max(ui->dsx, ui->dex); |
1496 | ymin = min(ui->dsy, ui->dey); |
1497 | ymax = max(ui->dsy, ui->dey); |
1498 | |
1499 | /* |
1500 | * Left-dragging has no effect, so we treat a left-drag as a |
1501 | * single click on dsx,dsy. |
1502 | */ |
1503 | if (ui->drag_button == LEFT_BUTTON) { |
1504 | xmin = xmax = ui->dsx; |
1505 | ymin = ymax = ui->dsy; |
1506 | } |
1507 | |
1508 | if (x < xmin || x > xmax || y < ymin || y > ymax) |
1509 | return v; /* no change outside drag area */ |
1510 | |
1511 | if (v == TREE) |
1512 | return v; /* trees are inviolate always */ |
1513 | |
1514 | if (xmin == xmax && ymin == ymax) { |
1515 | /* |
1516 | * Results of a simple click. Left button sets blanks to |
1517 | * tents; right button sets blanks to non-tents; either |
1518 | * button clears a non-blank square. |
1519 | */ |
1520 | if (ui->drag_button == LEFT_BUTTON) |
1521 | v = (v == BLANK ? TENT : BLANK); |
1522 | else |
1523 | v = (v == BLANK ? NONTENT : BLANK); |
1524 | } else { |
1525 | /* |
1526 | * Results of a drag. Left-dragging has no effect. |
1527 | * Right-dragging sets all blank squares to non-tents and |
1528 | * has no effect on anything else. |
1529 | */ |
1530 | if (ui->drag_button == RIGHT_BUTTON) |
1531 | v = (v == BLANK ? NONTENT : v); |
1532 | else |
1533 | /* do nothing */; |
1534 | } |
1535 | |
1536 | return v; |
1537 | } |
1538 | |
86e60e3d |
1539 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
1540 | int x, int y, int button) |
1541 | { |
1542 | int w = state->p.w, h = state->p.h; |
505ea4e5 |
1543 | char tmpbuf[80]; |
86e60e3d |
1544 | |
1545 | if (button == LEFT_BUTTON || button == RIGHT_BUTTON) { |
86e60e3d |
1546 | x = FROMCOORD(x); |
1547 | y = FROMCOORD(y); |
1548 | if (x < 0 || y < 0 || x >= w || y >= h) |
1549 | return NULL; |
1550 | |
565394e7 |
1551 | ui->drag_button = button; |
1552 | ui->dsx = ui->dex = x; |
1553 | ui->dsy = ui->dey = y; |
1554 | ui->drag_ok = TRUE; |
505ea4e5 |
1555 | ui->cdisp = 0; |
565394e7 |
1556 | return ""; /* ui updated */ |
1557 | } |
86e60e3d |
1558 | |
565394e7 |
1559 | if ((IS_MOUSE_DRAG(button) || IS_MOUSE_RELEASE(button)) && |
1560 | ui->drag_button > 0) { |
1561 | int xmin, ymin, xmax, ymax; |
505ea4e5 |
1562 | char *buf, *sep; |
565394e7 |
1563 | int buflen, bufsize, tmplen; |
1564 | |
1565 | x = FROMCOORD(x); |
1566 | y = FROMCOORD(y); |
1567 | if (x < 0 || y < 0 || x >= w || y >= h) { |
1568 | ui->drag_ok = FALSE; |
86e60e3d |
1569 | } else { |
565394e7 |
1570 | /* |
1571 | * Drags are limited to one row or column. Hence, we |
1572 | * work out which coordinate is closer to the drag |
1573 | * start, and move it _to_ the drag start. |
1574 | */ |
1575 | if (abs(x - ui->dsx) < abs(y - ui->dsy)) |
1576 | x = ui->dsx; |
1577 | else |
1578 | y = ui->dsy; |
1579 | |
1580 | ui->dex = x; |
1581 | ui->dey = y; |
1582 | |
1583 | ui->drag_ok = TRUE; |
86e60e3d |
1584 | } |
1585 | |
565394e7 |
1586 | if (IS_MOUSE_DRAG(button)) |
1587 | return ""; /* ui updated */ |
1588 | |
1589 | /* |
1590 | * The drag has been released. Enact it. |
1591 | */ |
1592 | if (!ui->drag_ok) { |
1593 | ui->drag_button = -1; |
1594 | return ""; /* drag was just cancelled */ |
1595 | } |
1596 | |
1597 | xmin = min(ui->dsx, ui->dex); |
1598 | xmax = max(ui->dsx, ui->dex); |
1599 | ymin = min(ui->dsy, ui->dey); |
1600 | ymax = max(ui->dsy, ui->dey); |
1601 | assert(0 <= xmin && xmin <= xmax && xmax < w); |
e4b6a85b |
1602 | assert(0 <= ymin && ymin <= ymax && ymax < h); |
565394e7 |
1603 | |
1604 | buflen = 0; |
1605 | bufsize = 256; |
1606 | buf = snewn(bufsize, char); |
1607 | sep = ""; |
1608 | for (y = ymin; y <= ymax; y++) |
1609 | for (x = xmin; x <= xmax; x++) { |
1610 | int v = drag_xform(ui, x, y, state->grid[y*w+x]); |
1611 | if (state->grid[y*w+x] != v) { |
1612 | tmplen = sprintf(tmpbuf, "%s%c%d,%d", sep, |
626cd8ac |
1613 | (int)(v == BLANK ? 'B' : |
1614 | v == TENT ? 'T' : 'N'), |
565394e7 |
1615 | x, y); |
1616 | sep = ";"; |
1617 | |
1618 | if (buflen + tmplen >= bufsize) { |
1619 | bufsize = buflen + tmplen + 256; |
1620 | buf = sresize(buf, bufsize, char); |
1621 | } |
1622 | |
1623 | strcpy(buf+buflen, tmpbuf); |
1624 | buflen += tmplen; |
1625 | } |
1626 | } |
1627 | |
1628 | ui->drag_button = -1; /* drag is terminated */ |
1629 | |
1630 | if (buflen == 0) { |
1631 | sfree(buf); |
1632 | return ""; /* ui updated (drag was terminated) */ |
1633 | } else { |
1634 | buf[buflen] = '\0'; |
1635 | return buf; |
1636 | } |
86e60e3d |
1637 | } |
1638 | |
505ea4e5 |
1639 | if (IS_CURSOR_MOVE(button)) { |
1640 | move_cursor(button, &ui->cx, &ui->cy, w, h, 0); |
1641 | ui->cdisp = 1; |
1642 | return ""; |
1643 | } |
1644 | if (ui->cdisp) { |
1645 | char rep = 0; |
1646 | int v = state->grid[ui->cy*w+ui->cx]; |
1647 | |
1648 | if (v != TREE) { |
1649 | #ifdef SINGLE_CURSOR_SELECT |
1650 | if (button == CURSOR_SELECT) |
1651 | /* SELECT cycles T, N, B */ |
1652 | rep = v == BLANK ? 'T' : v == TENT ? 'N' : 'B'; |
1653 | #else |
1654 | if (button == CURSOR_SELECT) |
1655 | rep = v == BLANK ? 'T' : 'B'; |
1656 | else if (button == CURSOR_SELECT2) |
1657 | rep = v == BLANK ? 'N' : 'B'; |
1658 | else if (button == 'T' || button == 'N' || button == 'B') |
1659 | rep = (char)button; |
1660 | #endif |
1661 | } |
1662 | |
1663 | if (rep) { |
1664 | sprintf(tmpbuf, "%c%d,%d", (int)rep, ui->cx, ui->cy); |
1665 | return dupstr(tmpbuf); |
1666 | } |
1667 | } else if (IS_CURSOR_SELECT(button)) { |
1668 | ui->cdisp = 1; |
1669 | return ""; |
1670 | } |
1671 | |
86e60e3d |
1672 | return NULL; |
1673 | } |
1674 | |
1675 | static game_state *execute_move(game_state *state, char *move) |
1676 | { |
1677 | int w = state->p.w, h = state->p.h; |
1678 | char c; |
1679 | int x, y, m, n, i, j; |
1680 | game_state *ret = dup_game(state); |
1681 | |
1682 | while (*move) { |
1683 | c = *move; |
1684 | if (c == 'S') { |
1685 | int i; |
1686 | ret->used_solve = TRUE; |
1687 | /* |
1688 | * Set all non-tree squares to NONTENT. The rest of the |
1689 | * solve move will fill the tents in over the top. |
1690 | */ |
1691 | for (i = 0; i < w*h; i++) |
1692 | if (ret->grid[i] != TREE) |
1693 | ret->grid[i] = NONTENT; |
1694 | move++; |
1695 | } else if (c == 'B' || c == 'T' || c == 'N') { |
1696 | move++; |
1697 | if (sscanf(move, "%d,%d%n", &x, &y, &n) != 2 || |
1698 | x < 0 || y < 0 || x >= w || y >= h) { |
1699 | free_game(ret); |
1700 | return NULL; |
1701 | } |
1702 | if (ret->grid[y*w+x] == TREE) { |
1703 | free_game(ret); |
1704 | return NULL; |
1705 | } |
1706 | ret->grid[y*w+x] = (c == 'B' ? BLANK : c == 'T' ? TENT : NONTENT); |
1707 | move += n; |
1708 | } else { |
1709 | free_game(ret); |
1710 | return NULL; |
1711 | } |
1712 | if (*move == ';') |
1713 | move++; |
1714 | else if (*move) { |
1715 | free_game(ret); |
1716 | return NULL; |
1717 | } |
1718 | } |
1719 | |
1720 | /* |
1721 | * Check for completion. |
1722 | */ |
1723 | for (i = n = m = 0; i < w*h; i++) { |
1724 | if (ret->grid[i] == TENT) |
1725 | n++; |
1726 | else if (ret->grid[i] == TREE) |
1727 | m++; |
1728 | } |
1729 | if (n == m) { |
1730 | int nedges, maxedges, *edges, *capacity, *flow; |
1731 | |
1732 | /* |
1733 | * We have the right number of tents, which is a |
1734 | * precondition for the game being complete. Now check that |
1735 | * the numbers add up. |
1736 | */ |
1737 | for (i = 0; i < w; i++) { |
1738 | n = 0; |
1739 | for (j = 0; j < h; j++) |
1740 | if (ret->grid[j*w+i] == TENT) |
1741 | n++; |
1742 | if (ret->numbers->numbers[i] != n) |
1743 | goto completion_check_done; |
1744 | } |
1745 | for (i = 0; i < h; i++) { |
1746 | n = 0; |
1747 | for (j = 0; j < w; j++) |
1748 | if (ret->grid[i*w+j] == TENT) |
1749 | n++; |
1750 | if (ret->numbers->numbers[w+i] != n) |
1751 | goto completion_check_done; |
1752 | } |
1753 | /* |
1754 | * Also, check that no two tents are adjacent. |
1755 | */ |
1756 | for (y = 0; y < h; y++) |
1757 | for (x = 0; x < w; x++) { |
1758 | if (x+1 < w && |
1759 | ret->grid[y*w+x] == TENT && ret->grid[y*w+x+1] == TENT) |
1760 | goto completion_check_done; |
1761 | if (y+1 < h && |
1762 | ret->grid[y*w+x] == TENT && ret->grid[(y+1)*w+x] == TENT) |
1763 | goto completion_check_done; |
1764 | if (x+1 < w && y+1 < h) { |
1765 | if (ret->grid[y*w+x] == TENT && |
1766 | ret->grid[(y+1)*w+(x+1)] == TENT) |
1767 | goto completion_check_done; |
1768 | if (ret->grid[(y+1)*w+x] == TENT && |
1769 | ret->grid[y*w+(x+1)] == TENT) |
1770 | goto completion_check_done; |
1771 | } |
1772 | } |
1773 | |
1774 | /* |
1775 | * OK; we have the right number of tents, they match the |
1776 | * numeric clues, and they satisfy the non-adjacency |
1777 | * criterion. Finally, we need to verify that they can be |
1778 | * placed in a one-to-one matching with the trees such that |
1779 | * every tent is orthogonally adjacent to its tree. |
1780 | * |
1781 | * This bit is where the hard work comes in: we have to do |
1782 | * it by finding such a matching using maxflow. |
1783 | * |
1784 | * So we construct a network with one special source node, |
1785 | * one special sink node, one node per tent, and one node |
1786 | * per tree. |
1787 | */ |
1788 | maxedges = 6 * m; |
1789 | edges = snewn(2 * maxedges, int); |
1790 | capacity = snewn(maxedges, int); |
1791 | flow = snewn(maxedges, int); |
1792 | nedges = 0; |
1793 | /* |
1794 | * Node numbering: |
1795 | * |
1796 | * 0..w*h trees/tents |
1797 | * w*h source |
1798 | * w*h+1 sink |
1799 | */ |
1800 | for (y = 0; y < h; y++) |
1801 | for (x = 0; x < w; x++) |
1802 | if (ret->grid[y*w+x] == TREE) { |
1803 | int d; |
1804 | |
1805 | /* |
1806 | * Here we use the direction enum declared for |
1807 | * the solver. We make use of the fact that the |
1808 | * directions are declared in the order |
1809 | * U,L,R,D, meaning that we go through the four |
1810 | * neighbours of any square in numerically |
1811 | * increasing order. |
1812 | */ |
1813 | for (d = 1; d < MAXDIR; d++) { |
1814 | int x2 = x + dx(d), y2 = y + dy(d); |
1815 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
1816 | ret->grid[y2*w+x2] == TENT) { |
1817 | assert(nedges < maxedges); |
1818 | edges[nedges*2] = y*w+x; |
1819 | edges[nedges*2+1] = y2*w+x2; |
1820 | capacity[nedges] = 1; |
1821 | nedges++; |
1822 | } |
1823 | } |
1824 | } else if (ret->grid[y*w+x] == TENT) { |
1825 | assert(nedges < maxedges); |
1826 | edges[nedges*2] = y*w+x; |
1827 | edges[nedges*2+1] = w*h+1; /* edge going to sink */ |
1828 | capacity[nedges] = 1; |
1829 | nedges++; |
1830 | } |
1831 | for (y = 0; y < h; y++) |
1832 | for (x = 0; x < w; x++) |
1833 | if (ret->grid[y*w+x] == TREE) { |
1834 | assert(nedges < maxedges); |
1835 | edges[nedges*2] = w*h; /* edge coming from source */ |
1836 | edges[nedges*2+1] = y*w+x; |
1837 | capacity[nedges] = 1; |
1838 | nedges++; |
1839 | } |
1840 | n = maxflow(w*h+2, w*h, w*h+1, nedges, edges, capacity, flow, NULL); |
1841 | |
1842 | sfree(flow); |
1843 | sfree(capacity); |
1844 | sfree(edges); |
1845 | |
1846 | if (n != m) |
1847 | goto completion_check_done; |
1848 | |
1849 | /* |
1850 | * We haven't managed to fault the grid on any count. Score! |
1851 | */ |
1852 | ret->completed = TRUE; |
1853 | } |
1854 | completion_check_done: |
1855 | |
1856 | return ret; |
1857 | } |
1858 | |
1859 | /* ---------------------------------------------------------------------- |
1860 | * Drawing routines. |
1861 | */ |
1862 | |
1863 | static void game_compute_size(game_params *params, int tilesize, |
1864 | int *x, int *y) |
1865 | { |
1866 | /* fool the macros */ |
3466f373 |
1867 | struct dummy { int tilesize; } dummy, *ds = &dummy; |
1868 | dummy.tilesize = tilesize; |
86e60e3d |
1869 | |
1870 | *x = TLBORDER + BRBORDER + TILESIZE * params->w; |
1871 | *y = TLBORDER + BRBORDER + TILESIZE * params->h; |
1872 | } |
1873 | |
1874 | static void game_set_size(drawing *dr, game_drawstate *ds, |
1875 | game_params *params, int tilesize) |
1876 | { |
1877 | ds->tilesize = tilesize; |
1878 | } |
1879 | |
8266f3fc |
1880 | static float *game_colours(frontend *fe, int *ncolours) |
86e60e3d |
1881 | { |
1882 | float *ret = snewn(3 * NCOLOURS, float); |
1883 | |
1884 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
1885 | |
1886 | ret[COL_GRID * 3 + 0] = 0.0F; |
1887 | ret[COL_GRID * 3 + 1] = 0.0F; |
1888 | ret[COL_GRID * 3 + 2] = 0.0F; |
1889 | |
1890 | ret[COL_GRASS * 3 + 0] = 0.7F; |
1891 | ret[COL_GRASS * 3 + 1] = 1.0F; |
1892 | ret[COL_GRASS * 3 + 2] = 0.5F; |
1893 | |
1894 | ret[COL_TREETRUNK * 3 + 0] = 0.6F; |
1895 | ret[COL_TREETRUNK * 3 + 1] = 0.4F; |
1896 | ret[COL_TREETRUNK * 3 + 2] = 0.0F; |
1897 | |
1898 | ret[COL_TREELEAF * 3 + 0] = 0.0F; |
1899 | ret[COL_TREELEAF * 3 + 1] = 0.7F; |
1900 | ret[COL_TREELEAF * 3 + 2] = 0.0F; |
1901 | |
1902 | ret[COL_TENT * 3 + 0] = 0.8F; |
1903 | ret[COL_TENT * 3 + 1] = 0.7F; |
1904 | ret[COL_TENT * 3 + 2] = 0.0F; |
1905 | |
2a27ffcf |
1906 | ret[COL_ERROR * 3 + 0] = 1.0F; |
1907 | ret[COL_ERROR * 3 + 1] = 0.0F; |
1908 | ret[COL_ERROR * 3 + 2] = 0.0F; |
1909 | |
1910 | ret[COL_ERRTEXT * 3 + 0] = 1.0F; |
1911 | ret[COL_ERRTEXT * 3 + 1] = 1.0F; |
1912 | ret[COL_ERRTEXT * 3 + 2] = 1.0F; |
1913 | |
e8df451f |
1914 | ret[COL_ERRTRUNK * 3 + 0] = 0.6F; |
1915 | ret[COL_ERRTRUNK * 3 + 1] = 0.0F; |
1916 | ret[COL_ERRTRUNK * 3 + 2] = 0.0F; |
1917 | |
86e60e3d |
1918 | *ncolours = NCOLOURS; |
1919 | return ret; |
1920 | } |
1921 | |
1922 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
1923 | { |
1924 | int w = state->p.w, h = state->p.h; |
1925 | struct game_drawstate *ds = snew(struct game_drawstate); |
2a27ffcf |
1926 | int i; |
86e60e3d |
1927 | |
1928 | ds->tilesize = 0; |
1929 | ds->started = FALSE; |
1930 | ds->p = state->p; /* structure copy */ |
2a27ffcf |
1931 | ds->drawn = snewn(w*h, int); |
1932 | for (i = 0; i < w*h; i++) |
1933 | ds->drawn[i] = MAGIC; |
1934 | ds->numbersdrawn = snewn(w+h, int); |
1935 | for (i = 0; i < w+h; i++) |
1936 | ds->numbersdrawn[i] = 2; |
505ea4e5 |
1937 | ds->cx = ds->cy = -1; |
86e60e3d |
1938 | |
1939 | return ds; |
1940 | } |
1941 | |
1942 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
1943 | { |
1944 | sfree(ds->drawn); |
2a27ffcf |
1945 | sfree(ds->numbersdrawn); |
86e60e3d |
1946 | sfree(ds); |
1947 | } |
1948 | |
2a27ffcf |
1949 | enum { |
1950 | ERR_ADJ_TOPLEFT = 4, |
1951 | ERR_ADJ_TOP, |
1952 | ERR_ADJ_TOPRIGHT, |
1953 | ERR_ADJ_LEFT, |
1954 | ERR_ADJ_RIGHT, |
1955 | ERR_ADJ_BOTLEFT, |
1956 | ERR_ADJ_BOT, |
1957 | ERR_ADJ_BOTRIGHT, |
1958 | ERR_OVERCOMMITTED |
1959 | }; |
1960 | |
271cf5c1 |
1961 | static int *find_errors(game_state *state, char *grid) |
2a27ffcf |
1962 | { |
1963 | int w = state->p.w, h = state->p.h; |
1964 | int *ret = snewn(w*h + w + h, int); |
1965 | int *tmp = snewn(w*h*2, int), *dsf = tmp + w*h; |
1966 | int x, y; |
1967 | |
1968 | /* |
1969 | * ret[0] through to ret[w*h-1] give error markers for the grid |
1970 | * squares. After that, ret[w*h] to ret[w*h+w-1] give error |
1971 | * markers for the column numbers, and ret[w*h+w] to |
1972 | * ret[w*h+w+h-1] for the row numbers. |
1973 | */ |
1974 | |
1975 | /* |
1976 | * Spot tent-adjacency violations. |
1977 | */ |
1978 | for (x = 0; x < w*h; x++) |
1979 | ret[x] = 0; |
1980 | for (y = 0; y < h; y++) { |
1981 | for (x = 0; x < w; x++) { |
1982 | if (y+1 < h && x+1 < w && |
271cf5c1 |
1983 | ((grid[y*w+x] == TENT && |
1984 | grid[(y+1)*w+(x+1)] == TENT) || |
1985 | (grid[(y+1)*w+x] == TENT && |
1986 | grid[y*w+(x+1)] == TENT))) { |
2a27ffcf |
1987 | ret[y*w+x] |= 1 << ERR_ADJ_BOTRIGHT; |
1988 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOPRIGHT; |
1989 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_BOTLEFT; |
1990 | ret[(y+1)*w+(x+1)] |= 1 << ERR_ADJ_TOPLEFT; |
1991 | } |
1992 | if (y+1 < h && |
271cf5c1 |
1993 | grid[y*w+x] == TENT && |
1994 | grid[(y+1)*w+x] == TENT) { |
2a27ffcf |
1995 | ret[y*w+x] |= 1 << ERR_ADJ_BOT; |
1996 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOP; |
1997 | } |
1998 | if (x+1 < w && |
271cf5c1 |
1999 | grid[y*w+x] == TENT && |
2000 | grid[y*w+(x+1)] == TENT) { |
2a27ffcf |
2001 | ret[y*w+x] |= 1 << ERR_ADJ_RIGHT; |
2002 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_LEFT; |
2003 | } |
2004 | } |
2005 | } |
2006 | |
2007 | /* |
2008 | * Spot numeric clue violations. |
2009 | */ |
2010 | for (x = 0; x < w; x++) { |
2011 | int tents = 0, maybetents = 0; |
2012 | for (y = 0; y < h; y++) { |
271cf5c1 |
2013 | if (grid[y*w+x] == TENT) |
2a27ffcf |
2014 | tents++; |
271cf5c1 |
2015 | else if (grid[y*w+x] == BLANK) |
2a27ffcf |
2016 | maybetents++; |
2017 | } |
2018 | ret[w*h+x] = (tents > state->numbers->numbers[x] || |
2019 | tents + maybetents < state->numbers->numbers[x]); |
2020 | } |
2021 | for (y = 0; y < h; y++) { |
2022 | int tents = 0, maybetents = 0; |
2023 | for (x = 0; x < w; x++) { |
271cf5c1 |
2024 | if (grid[y*w+x] == TENT) |
2a27ffcf |
2025 | tents++; |
271cf5c1 |
2026 | else if (grid[y*w+x] == BLANK) |
2a27ffcf |
2027 | maybetents++; |
2028 | } |
2029 | ret[w*h+w+y] = (tents > state->numbers->numbers[w+y] || |
2030 | tents + maybetents < state->numbers->numbers[w+y]); |
2031 | } |
2032 | |
2033 | /* |
2034 | * Identify groups of tents with too few trees between them, |
2035 | * which we do by constructing the connected components of the |
2036 | * bipartite adjacency graph between tents and trees |
2037 | * ('bipartite' in the sense that we deliberately ignore |
2038 | * adjacency between tents or between trees), and highlighting |
2039 | * all the tents in any component which has a smaller tree |
2040 | * count. |
2041 | */ |
2042 | dsf_init(dsf, w*h); |
2043 | /* Construct the equivalence classes. */ |
2044 | for (y = 0; y < h; y++) { |
2045 | for (x = 0; x < w-1; x++) { |
271cf5c1 |
2046 | if ((grid[y*w+x] == TREE && grid[y*w+x+1] == TENT) || |
2047 | (grid[y*w+x] == TENT && grid[y*w+x+1] == TREE)) |
2a27ffcf |
2048 | dsf_merge(dsf, y*w+x, y*w+x+1); |
2049 | } |
2050 | } |
2051 | for (y = 0; y < h-1; y++) { |
2052 | for (x = 0; x < w; x++) { |
271cf5c1 |
2053 | if ((grid[y*w+x] == TREE && grid[(y+1)*w+x] == TENT) || |
2054 | (grid[y*w+x] == TENT && grid[(y+1)*w+x] == TREE)) |
2a27ffcf |
2055 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
2056 | } |
2057 | } |
2058 | /* Count up the tent/tree difference in each one. */ |
2059 | for (x = 0; x < w*h; x++) |
2060 | tmp[x] = 0; |
2061 | for (x = 0; x < w*h; x++) { |
2062 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2063 | if (grid[x] == TREE) |
2a27ffcf |
2064 | tmp[y]++; |
271cf5c1 |
2065 | else if (grid[x] == TENT) |
2a27ffcf |
2066 | tmp[y]--; |
2067 | } |
2068 | /* And highlight any tent belonging to an equivalence class with |
2069 | * a score less than zero. */ |
2070 | for (x = 0; x < w*h; x++) { |
2071 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2072 | if (grid[x] == TENT && tmp[y] < 0) |
2a27ffcf |
2073 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
2074 | } |
2075 | |
2076 | /* |
2077 | * Identify groups of trees with too few tents between them. |
2078 | * This is done similarly, except that we now count BLANK as |
2079 | * equivalent to TENT, i.e. we only highlight such trees when |
2080 | * the user hasn't even left _room_ to provide tents for them |
2081 | * all. (Otherwise, we'd highlight all trees red right at the |
2082 | * start of the game, before the user had done anything wrong!) |
2083 | */ |
2084 | #define TENT(x) ((x)==TENT || (x)==BLANK) |
2085 | dsf_init(dsf, w*h); |
2086 | /* Construct the equivalence classes. */ |
2087 | for (y = 0; y < h; y++) { |
2088 | for (x = 0; x < w-1; x++) { |
271cf5c1 |
2089 | if ((grid[y*w+x] == TREE && TENT(grid[y*w+x+1])) || |
2090 | (TENT(grid[y*w+x]) && grid[y*w+x+1] == TREE)) |
2a27ffcf |
2091 | dsf_merge(dsf, y*w+x, y*w+x+1); |
2092 | } |
2093 | } |
2094 | for (y = 0; y < h-1; y++) { |
2095 | for (x = 0; x < w; x++) { |
271cf5c1 |
2096 | if ((grid[y*w+x] == TREE && TENT(grid[(y+1)*w+x])) || |
2097 | (TENT(grid[y*w+x]) && grid[(y+1)*w+x] == TREE)) |
2a27ffcf |
2098 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
2099 | } |
2100 | } |
2101 | /* Count up the tent/tree difference in each one. */ |
2102 | for (x = 0; x < w*h; x++) |
2103 | tmp[x] = 0; |
2104 | for (x = 0; x < w*h; x++) { |
2105 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2106 | if (grid[x] == TREE) |
2a27ffcf |
2107 | tmp[y]++; |
271cf5c1 |
2108 | else if (TENT(grid[x])) |
2a27ffcf |
2109 | tmp[y]--; |
2110 | } |
2111 | /* And highlight any tree belonging to an equivalence class with |
2112 | * a score more than zero. */ |
2113 | for (x = 0; x < w*h; x++) { |
2114 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2115 | if (grid[x] == TREE && tmp[y] > 0) |
2a27ffcf |
2116 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
2117 | } |
2118 | #undef TENT |
2119 | |
2120 | sfree(tmp); |
2121 | return ret; |
2122 | } |
2123 | |
2124 | static void draw_err_adj(drawing *dr, game_drawstate *ds, int x, int y) |
2125 | { |
2126 | int coords[8]; |
2127 | int yext, xext; |
2128 | |
2129 | /* |
2130 | * Draw a diamond. |
2131 | */ |
2132 | coords[0] = x - TILESIZE*2/5; |
2133 | coords[1] = y; |
2134 | coords[2] = x; |
2135 | coords[3] = y - TILESIZE*2/5; |
2136 | coords[4] = x + TILESIZE*2/5; |
2137 | coords[5] = y; |
2138 | coords[6] = x; |
2139 | coords[7] = y + TILESIZE*2/5; |
2140 | draw_polygon(dr, coords, 4, COL_ERROR, COL_GRID); |
2141 | |
2142 | /* |
2143 | * Draw an exclamation mark in the diamond. This turns out to |
2144 | * look unpleasantly off-centre if done via draw_text, so I do |
2145 | * it by hand on the basis that exclamation marks aren't that |
2146 | * difficult to draw... |
2147 | */ |
2148 | xext = TILESIZE/16; |
2149 | yext = TILESIZE*2/5 - (xext*2+2); |
2150 | draw_rect(dr, x-xext, y-yext, xext*2+1, yext*2+1 - (xext*3), |
2151 | COL_ERRTEXT); |
2152 | draw_rect(dr, x-xext, y+yext-xext*2+1, xext*2+1, xext*2, COL_ERRTEXT); |
2153 | } |
2154 | |
86e60e3d |
2155 | static void draw_tile(drawing *dr, game_drawstate *ds, |
505ea4e5 |
2156 | int x, int y, int v, int cur, int printing) |
86e60e3d |
2157 | { |
2a27ffcf |
2158 | int err; |
86e60e3d |
2159 | int tx = COORD(x), ty = COORD(y); |
2160 | int cx = tx + TILESIZE/2, cy = ty + TILESIZE/2; |
2161 | |
2a27ffcf |
2162 | err = v & ~15; |
2163 | v &= 15; |
86e60e3d |
2164 | |
2a27ffcf |
2165 | clip(dr, tx, ty, TILESIZE, TILESIZE); |
2166 | |
2167 | if (!printing) { |
2168 | draw_rect(dr, tx, ty, TILESIZE, TILESIZE, COL_GRID); |
86e60e3d |
2169 | draw_rect(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1, |
2170 | (v == BLANK ? COL_BACKGROUND : COL_GRASS)); |
2a27ffcf |
2171 | } |
86e60e3d |
2172 | |
2173 | if (v == TREE) { |
2174 | int i; |
2175 | |
2176 | (printing ? draw_rect_outline : draw_rect) |
2177 | (dr, cx-TILESIZE/15, ty+TILESIZE*3/10, |
2178 | 2*(TILESIZE/15)+1, (TILESIZE*9/10 - TILESIZE*3/10), |
e8df451f |
2179 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERRTRUNK : COL_TREETRUNK)); |
86e60e3d |
2180 | |
2181 | for (i = 0; i < (printing ? 2 : 1); i++) { |
2a27ffcf |
2182 | int col = (i == 1 ? COL_BACKGROUND : |
2183 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : |
2184 | COL_TREELEAF)); |
86e60e3d |
2185 | int sub = i * (TILESIZE/32); |
2186 | draw_circle(dr, cx, ty+TILESIZE*4/10, TILESIZE/4 - sub, |
2187 | col, col); |
2188 | draw_circle(dr, cx+TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
2189 | col, col); |
2190 | draw_circle(dr, cx-TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
2191 | col, col); |
2192 | draw_circle(dr, cx+TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
2193 | col, col); |
2194 | draw_circle(dr, cx-TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
2195 | col, col); |
2196 | } |
2197 | } else if (v == TENT) { |
2198 | int coords[6]; |
2a27ffcf |
2199 | int col; |
86e60e3d |
2200 | coords[0] = cx - TILESIZE/3; |
2201 | coords[1] = cy + TILESIZE/3; |
2202 | coords[2] = cx + TILESIZE/3; |
2203 | coords[3] = cy + TILESIZE/3; |
2204 | coords[4] = cx; |
2205 | coords[5] = cy - TILESIZE/3; |
2a27ffcf |
2206 | col = (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : COL_TENT); |
2207 | draw_polygon(dr, coords, 3, (printing ? -1 : col), col); |
86e60e3d |
2208 | } |
2209 | |
2a27ffcf |
2210 | if (err & (1 << ERR_ADJ_TOPLEFT)) |
2211 | draw_err_adj(dr, ds, tx, ty); |
2212 | if (err & (1 << ERR_ADJ_TOP)) |
2213 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty); |
2214 | if (err & (1 << ERR_ADJ_TOPRIGHT)) |
2215 | draw_err_adj(dr, ds, tx+TILESIZE, ty); |
2216 | if (err & (1 << ERR_ADJ_LEFT)) |
2217 | draw_err_adj(dr, ds, tx, ty+TILESIZE/2); |
2218 | if (err & (1 << ERR_ADJ_RIGHT)) |
2219 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE/2); |
2220 | if (err & (1 << ERR_ADJ_BOTLEFT)) |
2221 | draw_err_adj(dr, ds, tx, ty+TILESIZE); |
2222 | if (err & (1 << ERR_ADJ_BOT)) |
2223 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty+TILESIZE); |
2224 | if (err & (1 << ERR_ADJ_BOTRIGHT)) |
2225 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE); |
2226 | |
505ea4e5 |
2227 | if (cur) { |
2228 | int coff = TILESIZE/8; |
2229 | draw_rect_outline(dr, tx + coff, ty + coff, |
fe5d2401 |
2230 | TILESIZE - coff*2 + 1, TILESIZE - coff*2 + 1, |
2231 | COL_GRID); |
505ea4e5 |
2232 | } |
2233 | |
86e60e3d |
2234 | unclip(dr); |
2235 | draw_update(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1); |
2236 | } |
2237 | |
2238 | /* |
2239 | * Internal redraw function, used for printing as well as drawing. |
2240 | */ |
2241 | static void int_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
2242 | game_state *state, int dir, game_ui *ui, |
2243 | float animtime, float flashtime, int printing) |
2244 | { |
2245 | int w = state->p.w, h = state->p.h; |
2246 | int x, y, flashing; |
505ea4e5 |
2247 | int cx = -1, cy = -1; |
2248 | int cmoved = 0; |
271cf5c1 |
2249 | char *tmpgrid; |
2a27ffcf |
2250 | int *errors; |
505ea4e5 |
2251 | |
2252 | if (ui) { |
2253 | if (ui->cdisp) { cx = ui->cx; cy = ui->cy; } |
2254 | if (cx != ds->cx || cy != ds->cy) cmoved = 1; |
2255 | } |
86e60e3d |
2256 | |
2257 | if (printing || !ds->started) { |
2258 | if (!printing) { |
2259 | int ww, wh; |
2260 | game_compute_size(&state->p, TILESIZE, &ww, &wh); |
2261 | draw_rect(dr, 0, 0, ww, wh, COL_BACKGROUND); |
2262 | draw_update(dr, 0, 0, ww, wh); |
2263 | ds->started = TRUE; |
2264 | } |
2265 | |
2266 | if (printing) |
2267 | print_line_width(dr, TILESIZE/64); |
2268 | |
2269 | /* |
2270 | * Draw the grid. |
2271 | */ |
2272 | for (y = 0; y <= h; y++) |
2273 | draw_line(dr, COORD(0), COORD(y), COORD(w), COORD(y), COL_GRID); |
2274 | for (x = 0; x <= w; x++) |
2275 | draw_line(dr, COORD(x), COORD(0), COORD(x), COORD(h), COL_GRID); |
86e60e3d |
2276 | } |
2277 | |
2278 | if (flashtime > 0) |
2279 | flashing = (int)(flashtime * 3 / FLASH_TIME) != 1; |
2280 | else |
2281 | flashing = FALSE; |
2282 | |
2283 | /* |
271cf5c1 |
2284 | * Find errors. For this we use _part_ of the information from a |
2285 | * currently active drag: we transform dsx,dsy but not anything |
2286 | * else. (This seems to strike a good compromise between having |
2287 | * the error highlights respond instantly to single clicks, but |
2288 | * not give constant feedback during a right-drag.) |
2a27ffcf |
2289 | */ |
271cf5c1 |
2290 | if (ui && ui->drag_button >= 0) { |
2291 | tmpgrid = snewn(w*h, char); |
2292 | memcpy(tmpgrid, state->grid, w*h); |
2293 | tmpgrid[ui->dsy * w + ui->dsx] = |
2294 | drag_xform(ui, ui->dsx, ui->dsy, tmpgrid[ui->dsy * w + ui->dsx]); |
2295 | errors = find_errors(state, tmpgrid); |
2296 | sfree(tmpgrid); |
2297 | } else { |
2298 | errors = find_errors(state, state->grid); |
2299 | } |
2a27ffcf |
2300 | |
2301 | /* |
86e60e3d |
2302 | * Draw the grid. |
2303 | */ |
2a27ffcf |
2304 | for (y = 0; y < h; y++) { |
86e60e3d |
2305 | for (x = 0; x < w; x++) { |
2306 | int v = state->grid[y*w+x]; |
505ea4e5 |
2307 | int credraw = 0; |
86e60e3d |
2308 | |
565394e7 |
2309 | /* |
2310 | * We deliberately do not take drag_ok into account |
2311 | * here, because user feedback suggests that it's |
2312 | * marginally nicer not to have the drag effects |
2313 | * flickering on and off disconcertingly. |
2314 | */ |
5bcb1aa3 |
2315 | if (ui && ui->drag_button >= 0) |
565394e7 |
2316 | v = drag_xform(ui, x, y, v); |
2317 | |
86e60e3d |
2318 | if (flashing && (v == TREE || v == TENT)) |
2319 | v = NONTENT; |
2320 | |
505ea4e5 |
2321 | if (cmoved) { |
2322 | if ((x == cx && y == cy) || |
2323 | (x == ds->cx && y == ds->cy)) credraw = 1; |
2324 | } |
2325 | |
2a27ffcf |
2326 | v |= errors[y*w+x]; |
2327 | |
505ea4e5 |
2328 | if (printing || ds->drawn[y*w+x] != v || credraw) { |
2329 | draw_tile(dr, ds, x, y, v, (x == cx && y == cy), printing); |
86e60e3d |
2330 | if (!printing) |
2331 | ds->drawn[y*w+x] = v; |
2332 | } |
2333 | } |
2a27ffcf |
2334 | } |
2335 | |
2336 | /* |
2337 | * Draw (or redraw, if their error-highlighted state has |
2338 | * changed) the numbers. |
2339 | */ |
2340 | for (x = 0; x < w; x++) { |
2341 | if (ds->numbersdrawn[x] != errors[w*h+x]) { |
2342 | char buf[80]; |
2343 | draw_rect(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1, |
2344 | COL_BACKGROUND); |
2345 | sprintf(buf, "%d", state->numbers->numbers[x]); |
2346 | draw_text(dr, COORD(x) + TILESIZE/2, COORD(h+1), |
2347 | FONT_VARIABLE, TILESIZE/2, ALIGN_HCENTRE|ALIGN_VNORMAL, |
2348 | (errors[w*h+x] ? COL_ERROR : COL_GRID), buf); |
2349 | draw_update(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1); |
2350 | ds->numbersdrawn[x] = errors[w*h+x]; |
2351 | } |
2352 | } |
2353 | for (y = 0; y < h; y++) { |
2354 | if (ds->numbersdrawn[w+y] != errors[w*h+w+y]) { |
2355 | char buf[80]; |
2356 | draw_rect(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE, |
2357 | COL_BACKGROUND); |
2358 | sprintf(buf, "%d", state->numbers->numbers[w+y]); |
2359 | draw_text(dr, COORD(w+1), COORD(y) + TILESIZE/2, |
2360 | FONT_VARIABLE, TILESIZE/2, ALIGN_HRIGHT|ALIGN_VCENTRE, |
2361 | (errors[w*h+w+y] ? COL_ERROR : COL_GRID), buf); |
2362 | draw_update(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE); |
2363 | ds->numbersdrawn[w+y] = errors[w*h+w+y]; |
2364 | } |
2365 | } |
2366 | |
2367 | if (cmoved) { |
2368 | ds->cx = cx; |
2369 | ds->cy = cy; |
2370 | } |
2371 | |
2372 | sfree(errors); |
86e60e3d |
2373 | } |
2374 | |
2375 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
2376 | game_state *state, int dir, game_ui *ui, |
2377 | float animtime, float flashtime) |
2378 | { |
2379 | int_redraw(dr, ds, oldstate, state, dir, ui, animtime, flashtime, FALSE); |
2380 | } |
2381 | |
2382 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
2383 | int dir, game_ui *ui) |
2384 | { |
2385 | return 0.0F; |
2386 | } |
2387 | |
2388 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
2389 | int dir, game_ui *ui) |
2390 | { |
2391 | if (!oldstate->completed && newstate->completed && |
2392 | !oldstate->used_solve && !newstate->used_solve) |
2393 | return FLASH_TIME; |
2394 | |
2395 | return 0.0F; |
2396 | } |
2397 | |
86e60e3d |
2398 | static int game_timing_state(game_state *state, game_ui *ui) |
2399 | { |
2400 | return TRUE; |
2401 | } |
2402 | |
2403 | static void game_print_size(game_params *params, float *x, float *y) |
2404 | { |
2405 | int pw, ph; |
2406 | |
2407 | /* |
2408 | * I'll use 6mm squares by default. |
2409 | */ |
2410 | game_compute_size(params, 600, &pw, &ph); |
505ea4e5 |
2411 | *x = pw / 100.0F; |
2412 | *y = ph / 100.0F; |
86e60e3d |
2413 | } |
2414 | |
2415 | static void game_print(drawing *dr, game_state *state, int tilesize) |
2416 | { |
2417 | int c; |
2418 | |
2419 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
2420 | game_drawstate ads, *ds = &ads; |
2421 | game_set_size(dr, ds, NULL, tilesize); |
2422 | |
2423 | c = print_mono_colour(dr, 1); assert(c == COL_BACKGROUND); |
2424 | c = print_mono_colour(dr, 0); assert(c == COL_GRID); |
2425 | c = print_mono_colour(dr, 1); assert(c == COL_GRASS); |
2426 | c = print_mono_colour(dr, 0); assert(c == COL_TREETRUNK); |
2427 | c = print_mono_colour(dr, 0); assert(c == COL_TREELEAF); |
2428 | c = print_mono_colour(dr, 0); assert(c == COL_TENT); |
2429 | |
2430 | int_redraw(dr, ds, NULL, state, +1, NULL, 0.0F, 0.0F, TRUE); |
2431 | } |
2432 | |
2433 | #ifdef COMBINED |
2434 | #define thegame tents |
2435 | #endif |
2436 | |
2437 | const struct game thegame = { |
750037d7 |
2438 | "Tents", "games.tents", "tents", |
86e60e3d |
2439 | default_params, |
2440 | game_fetch_preset, |
2441 | decode_params, |
2442 | encode_params, |
2443 | free_params, |
2444 | dup_params, |
2445 | TRUE, game_configure, custom_params, |
2446 | validate_params, |
2447 | new_game_desc, |
2448 | validate_desc, |
2449 | new_game, |
2450 | dup_game, |
2451 | free_game, |
2452 | TRUE, solve_game, |
fa3abef5 |
2453 | FALSE, game_can_format_as_text_now, game_text_format, |
86e60e3d |
2454 | new_ui, |
2455 | free_ui, |
2456 | encode_ui, |
2457 | decode_ui, |
2458 | game_changed_state, |
2459 | interpret_move, |
2460 | execute_move, |
2461 | PREFERRED_TILESIZE, game_compute_size, game_set_size, |
2462 | game_colours, |
2463 | game_new_drawstate, |
2464 | game_free_drawstate, |
2465 | game_redraw, |
2466 | game_anim_length, |
2467 | game_flash_length, |
2468 | TRUE, FALSE, game_print_size, game_print, |
ac9f41c4 |
2469 | FALSE, /* wants_statusbar */ |
86e60e3d |
2470 | FALSE, game_timing_state, |
cb0c7d4a |
2471 | REQUIRE_RBUTTON, /* flags */ |
86e60e3d |
2472 | }; |
2473 | |
2474 | #ifdef STANDALONE_SOLVER |
2475 | |
2476 | #include <stdarg.h> |
2477 | |
2478 | int main(int argc, char **argv) |
2479 | { |
2480 | game_params *p; |
2481 | game_state *s, *s2; |
2482 | char *id = NULL, *desc, *err; |
2483 | int grade = FALSE; |
2484 | int ret, diff, really_verbose = FALSE; |
2485 | struct solver_scratch *sc; |
2486 | |
2487 | while (--argc > 0) { |
2488 | char *p = *++argv; |
2489 | if (!strcmp(p, "-v")) { |
2490 | really_verbose = TRUE; |
2491 | } else if (!strcmp(p, "-g")) { |
2492 | grade = TRUE; |
2493 | } else if (*p == '-') { |
2494 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
2495 | return 1; |
2496 | } else { |
2497 | id = p; |
2498 | } |
2499 | } |
2500 | |
2501 | if (!id) { |
2502 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
2503 | return 1; |
2504 | } |
2505 | |
2506 | desc = strchr(id, ':'); |
2507 | if (!desc) { |
2508 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
2509 | return 1; |
2510 | } |
2511 | *desc++ = '\0'; |
2512 | |
2513 | p = default_params(); |
2514 | decode_params(p, id); |
2515 | err = validate_desc(p, desc); |
2516 | if (err) { |
2517 | fprintf(stderr, "%s: %s\n", argv[0], err); |
2518 | return 1; |
2519 | } |
2520 | s = new_game(NULL, p, desc); |
2521 | s2 = new_game(NULL, p, desc); |
2522 | |
2523 | sc = new_scratch(p->w, p->h); |
2524 | |
2525 | /* |
2526 | * When solving an Easy puzzle, we don't want to bother the |
2527 | * user with Hard-level deductions. For this reason, we grade |
2528 | * the puzzle internally before doing anything else. |
2529 | */ |
2530 | ret = -1; /* placate optimiser */ |
2531 | for (diff = 0; diff < DIFFCOUNT; diff++) { |
2532 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2533 | s2->grid, sc, diff); |
2534 | if (ret < 2) |
2535 | break; |
2536 | } |
2537 | |
2538 | if (diff == DIFFCOUNT) { |
2539 | if (grade) |
2540 | printf("Difficulty rating: too hard to solve internally\n"); |
2541 | else |
2542 | printf("Unable to find a unique solution\n"); |
2543 | } else { |
2544 | if (grade) { |
2545 | if (ret == 0) |
2546 | printf("Difficulty rating: impossible (no solution exists)\n"); |
2547 | else if (ret == 1) |
2548 | printf("Difficulty rating: %s\n", tents_diffnames[diff]); |
2549 | } else { |
2550 | verbose = really_verbose; |
2551 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2552 | s2->grid, sc, diff); |
2553 | if (ret == 0) |
2554 | printf("Puzzle is inconsistent\n"); |
2555 | else |
2556 | fputs(game_text_format(s2), stdout); |
2557 | } |
2558 | } |
2559 | |
2560 | return 0; |
2561 | } |
2562 | |
2563 | #endif |
505ea4e5 |
2564 | |
2565 | /* vim: set shiftwidth=4 tabstop=8: */ |