86e60e3d |
1 | /* |
2 | * tents.c: Puzzle involving placing tents next to trees subject to |
3 | * some confusing conditions. |
4 | * |
5 | * TODO: |
e8df451f |
6 | * |
86e60e3d |
7 | * - it might be nice to make setter-provided tent/nontent clues |
8 | * inviolable? |
9 | * * on the other hand, this would introduce considerable extra |
10 | * complexity and size into the game state; also inviolable |
11 | * clues would have to be marked as such somehow, in an |
12 | * intrusive and annoying manner. Since they're never |
13 | * generated by _my_ generator, I'm currently more inclined |
14 | * not to bother. |
15 | * |
16 | * - more difficult levels at the top end? |
17 | * * for example, sometimes we can deduce that two BLANKs in |
18 | * the same row are each adjacent to the same unattached tree |
19 | * and to nothing else, implying that they can't both be |
20 | * tents; this enables us to rule out some extra combinations |
21 | * in the row-based deduction loop, and hence deduce more |
22 | * from the number in that row than we could otherwise do. |
23 | * * that by itself doesn't seem worth implementing a new |
24 | * difficulty level for, but if I can find a few more things |
25 | * like that then it might become worthwhile. |
26 | * * I wonder if there's a sensible heuristic for where to |
27 | * guess which would make a recursive solver viable? |
28 | */ |
29 | |
30 | #include <stdio.h> |
31 | #include <stdlib.h> |
32 | #include <string.h> |
33 | #include <assert.h> |
34 | #include <ctype.h> |
35 | #include <math.h> |
36 | |
37 | #include "puzzles.h" |
38 | #include "maxflow.h" |
39 | |
40 | /* |
41 | * Design discussion |
42 | * ----------------- |
43 | * |
44 | * The rules of this puzzle as available on the WWW are poorly |
45 | * specified. The bits about tents having to be orthogonally |
46 | * adjacent to trees, tents not being even diagonally adjacent to |
47 | * one another, and the number of tents in each row and column |
48 | * being given are simple enough; the difficult bit is the |
49 | * tent-to-tree matching. |
50 | * |
51 | * Some sources use simplistic wordings such as `each tree is |
52 | * exactly connected to only one tent', which is extremely unclear: |
53 | * it's easy to read erroneously as `each tree is _orthogonally |
54 | * adjacent_ to exactly one tent', which is definitely incorrect. |
55 | * Even the most coherent sources I've found don't do a much better |
56 | * job of stating the rule. |
57 | * |
58 | * A more precise statement of the rule is that it must be possible |
59 | * to find a bijection f between tents and trees such that each |
60 | * tree T is orthogonally adjacent to the tent f(T), but that a |
61 | * tent is permitted to be adjacent to other trees in addition to |
62 | * its own. This slightly non-obvious criterion is what gives this |
63 | * puzzle most of its subtlety. |
64 | * |
65 | * However, there's a particularly subtle ambiguity left over. Is |
66 | * the bijection between tents and trees required to be _unique_? |
67 | * In other words, is that bijection conceptually something the |
68 | * player should be able to exhibit as part of the solution (even |
69 | * if they aren't actually required to do so)? Or is it sufficient |
70 | * to have a unique _placement_ of the tents which gives rise to at |
71 | * least one suitable bijection? |
72 | * |
73 | * The puzzle shown to the right of this .T. 2 *T* 2 |
74 | * paragraph illustrates the problem. There T.T 0 -> T-T 0 |
75 | * are two distinct bijections available. .T. 2 *T* 2 |
76 | * The answer to the above question will |
77 | * determine whether it's a valid puzzle. 202 202 |
78 | * |
79 | * This is an important question, because it affects both the |
80 | * player and the generator. Eventually I found all the instances |
81 | * of this puzzle I could Google up, solved them all by hand, and |
82 | * verified that in all cases the tree/tent matching was uniquely |
83 | * determined given the tree and tent positions. Therefore, the |
84 | * puzzle as implemented in this source file takes the following |
85 | * policy: |
86 | * |
87 | * - When checking a user-supplied solution for correctness, only |
88 | * verify that there exists _at least_ one matching. |
89 | * - When generating a puzzle, enforce that there must be |
90 | * _exactly_ one. |
91 | * |
92 | * Algorithmic implications |
93 | * ------------------------ |
94 | * |
95 | * Another way of phrasing the tree/tent matching criterion is to |
96 | * say that the bipartite adjacency graph between trees and tents |
97 | * has a perfect matching. That is, if you construct a graph which |
98 | * has a vertex per tree and a vertex per tent, and an edge between |
99 | * any tree and tent which are orthogonally adjacent, it is |
100 | * possible to find a set of N edges of that graph (where N is the |
101 | * number of trees and also the number of tents) which between them |
102 | * connect every tree to every tent. |
103 | * |
104 | * The most efficient known algorithms for finding such a matching |
105 | * given a graph, as far as I'm aware, are the Munkres assignment |
106 | * algorithm (also known as the Hungarian algorithm) and the |
107 | * Ford-Fulkerson algorithm (for finding optimal flows in |
108 | * networks). Each of these takes O(N^3) running time; so we're |
109 | * talking O(N^3) time to verify any candidate solution to this |
110 | * puzzle. That's just about OK if you're doing it once per mouse |
111 | * click (and in fact not even that, since the sensible thing to do |
112 | * is check all the _other_ puzzle criteria and only wade into this |
113 | * quagmire if none are violated); but if the solver had to keep |
114 | * doing N^3 work internally, then it would probably end up with |
115 | * more like N^5 or N^6 running time, and grid generation would |
116 | * become very clunky. |
117 | * |
118 | * Fortunately, I've been able to prove a very useful property of |
119 | * _unique_ perfect matchings, by adapting the proof of Hall's |
120 | * Marriage Theorem. For those unaware of Hall's Theorem, I'll |
121 | * recap it and its proof: it states that a bipartite graph |
122 | * contains a perfect matching iff every set of vertices on the |
123 | * left side of the graph have a neighbourhood _at least_ as big on |
124 | * the right. |
125 | * |
126 | * This condition is obviously satisfied if a perfect matching does |
127 | * exist; each left-side node has a distinct right-side node which |
128 | * is the one assigned to it by the matching, and thus any set of n |
129 | * left vertices must have a combined neighbourhood containing at |
130 | * least the n corresponding right vertices, and possibly others |
131 | * too. Alternatively, imagine if you had (say) three left-side |
132 | * nodes all of which were connected to only two right-side nodes |
133 | * between them: any perfect matching would have to assign one of |
134 | * those two right nodes to each of the three left nodes, and still |
135 | * give the three left nodes a different right node each. This is |
136 | * of course impossible. |
137 | * |
138 | * To prove the converse (that if every subset of left vertices |
139 | * satisfies the Hall condition then a perfect matching exists), |
140 | * consider trying to find a proper subset of the left vertices |
141 | * which _exactly_ satisfies the Hall condition: that is, its right |
142 | * neighbourhood is precisely the same size as it. If we can find |
143 | * such a subset, then we can split the bipartite graph into two |
144 | * smaller ones: one consisting of the left subset and its right |
145 | * neighbourhood, the other consisting of everything else. Edges |
146 | * from the left side of the former graph to the right side of the |
147 | * latter do not exist, by construction; edges from the right side |
148 | * of the former to the left of the latter cannot be part of any |
149 | * perfect matching because otherwise the left subset would not be |
150 | * left with enough distinct right vertices to connect to (this is |
151 | * exactly the same deduction used in Solo's set analysis). You can |
152 | * then prove (left as an exercise) that both these smaller graphs |
153 | * still satisfy the Hall condition, and therefore the proof will |
154 | * follow by induction. |
155 | * |
156 | * There's one other possibility, which is the case where _no_ |
157 | * proper subset of the left vertices has a right neighbourhood of |
158 | * exactly the same size. That is, every left subset has a strictly |
159 | * _larger_ right neighbourhood. In this situation, we can simply |
160 | * remove an _arbitrary_ edge from the graph. This cannot reduce |
161 | * the size of any left subset's right neighbourhood by more than |
162 | * one, so if all neighbourhoods were strictly bigger than they |
163 | * needed to be initially, they must now still be _at least as big_ |
164 | * as they need to be. So we can keep throwing out arbitrary edges |
165 | * until we find a set which exactly satisfies the Hall condition, |
166 | * and then proceed as above. [] |
167 | * |
168 | * That's Hall's theorem. I now build on this by examining the |
169 | * circumstances in which a bipartite graph can have a _unique_ |
170 | * perfect matching. It is clear that in the second case, where no |
171 | * left subset exactly satisfies the Hall condition and so we can |
172 | * remove an arbitrary edge, there cannot be a unique perfect |
173 | * matching: given one perfect matching, we choose our arbitrary |
174 | * removed edge to be one of those contained in it, and then we can |
175 | * still find a perfect matching in the remaining graph, which will |
176 | * be a distinct perfect matching in the original. |
177 | * |
178 | * So it is a necessary condition for a unique perfect matching |
179 | * that there must be at least one proper left subset which |
180 | * _exactly_ satisfies the Hall condition. But now consider the |
181 | * smaller graph constructed by taking that left subset and its |
182 | * neighbourhood: if the graph as a whole had a unique perfect |
183 | * matching, then so must this smaller one, which means we can find |
184 | * a proper left subset _again_, and so on. Repeating this process |
185 | * must eventually reduce us to a graph with only one left-side |
186 | * vertex (so there are no proper subsets at all); this vertex must |
187 | * be connected to only one right-side vertex, and hence must be so |
188 | * in the original graph as well (by construction). So we can |
189 | * discard this vertex pair from the graph, and any other edges |
190 | * that involved it (which will by construction be from other left |
191 | * vertices only), and the resulting smaller graph still has a |
192 | * unique perfect matching which means we can do the same thing |
193 | * again. |
194 | * |
195 | * In other words, given any bipartite graph with a unique perfect |
196 | * matching, we can find that matching by the following extremely |
197 | * simple algorithm: |
198 | * |
199 | * - Find a left-side vertex which is only connected to one |
200 | * right-side vertex. |
201 | * - Assign those vertices to one another, and therefore discard |
202 | * any other edges connecting to that right vertex. |
203 | * - Repeat until all vertices have been matched. |
204 | * |
205 | * This algorithm can be run in O(V+E) time (where V is the number |
206 | * of vertices and E is the number of edges in the graph), and the |
207 | * only way it can fail is if there is not a unique perfect |
208 | * matching (either because there is no matching at all, or because |
209 | * it isn't unique; but it can't distinguish those cases). |
210 | * |
211 | * Thus, the internal solver in this source file can be confident |
212 | * that if the tree/tent matching is uniquely determined by the |
213 | * tree and tent positions, it can find it using only this kind of |
214 | * obvious and simple operation: assign a tree to a tent if it |
215 | * cannot possibly belong to any other tent, and vice versa. If the |
216 | * solver were _only_ trying to determine the matching, even that |
217 | * `vice versa' wouldn't be required; but it can come in handy when |
218 | * not all the tents have been placed yet. I can therefore be |
219 | * reasonably confident that as long as my solver doesn't need to |
220 | * cope with grids that have a non-unique matching, it will also |
221 | * not need to do anything complicated like set analysis between |
222 | * trees and tents. |
223 | */ |
224 | |
225 | /* |
226 | * In standalone solver mode, `verbose' is a variable which can be |
227 | * set by command-line option; in debugging mode it's simply always |
228 | * true. |
229 | */ |
230 | #if defined STANDALONE_SOLVER |
231 | #define SOLVER_DIAGNOSTICS |
232 | int verbose = FALSE; |
233 | #elif defined SOLVER_DIAGNOSTICS |
234 | #define verbose TRUE |
235 | #endif |
236 | |
237 | /* |
238 | * Difficulty levels. I do some macro ickery here to ensure that my |
239 | * enum and the various forms of my name list always match up. |
240 | */ |
241 | #define DIFFLIST(A) \ |
242 | A(EASY,Easy,e) \ |
243 | A(TRICKY,Tricky,t) |
244 | #define ENUM(upper,title,lower) DIFF_ ## upper, |
245 | #define TITLE(upper,title,lower) #title, |
246 | #define ENCODE(upper,title,lower) #lower |
247 | #define CONFIG(upper,title,lower) ":" #title |
248 | enum { DIFFLIST(ENUM) DIFFCOUNT }; |
249 | static char const *const tents_diffnames[] = { DIFFLIST(TITLE) }; |
250 | static char const tents_diffchars[] = DIFFLIST(ENCODE); |
251 | #define DIFFCONFIG DIFFLIST(CONFIG) |
252 | |
253 | enum { |
254 | COL_BACKGROUND, |
255 | COL_GRID, |
256 | COL_GRASS, |
257 | COL_TREETRUNK, |
258 | COL_TREELEAF, |
259 | COL_TENT, |
2a27ffcf |
260 | COL_ERROR, |
261 | COL_ERRTEXT, |
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262 | COL_ERRTRUNK, |
86e60e3d |
263 | NCOLOURS |
264 | }; |
265 | |
266 | enum { BLANK, TREE, TENT, NONTENT, MAGIC }; |
267 | |
268 | struct game_params { |
269 | int w, h; |
270 | int diff; |
271 | }; |
272 | |
273 | struct numbers { |
274 | int refcount; |
275 | int *numbers; |
276 | }; |
277 | |
278 | struct game_state { |
279 | game_params p; |
280 | char *grid; |
281 | struct numbers *numbers; |
282 | int completed, used_solve; |
283 | }; |
284 | |
285 | static game_params *default_params(void) |
286 | { |
287 | game_params *ret = snew(game_params); |
288 | |
289 | ret->w = ret->h = 8; |
290 | ret->diff = DIFF_EASY; |
291 | |
292 | return ret; |
293 | } |
294 | |
295 | static const struct game_params tents_presets[] = { |
296 | {8, 8, DIFF_EASY}, |
297 | {8, 8, DIFF_TRICKY}, |
298 | {10, 10, DIFF_EASY}, |
299 | {10, 10, DIFF_TRICKY}, |
300 | {15, 15, DIFF_EASY}, |
301 | {15, 15, DIFF_TRICKY}, |
302 | }; |
303 | |
304 | static int game_fetch_preset(int i, char **name, game_params **params) |
305 | { |
306 | game_params *ret; |
307 | char str[80]; |
308 | |
309 | if (i < 0 || i >= lenof(tents_presets)) |
310 | return FALSE; |
311 | |
312 | ret = snew(game_params); |
313 | *ret = tents_presets[i]; |
314 | |
315 | sprintf(str, "%dx%d %s", ret->w, ret->h, tents_diffnames[ret->diff]); |
316 | |
317 | *name = dupstr(str); |
318 | *params = ret; |
319 | return TRUE; |
320 | } |
321 | |
322 | static void free_params(game_params *params) |
323 | { |
324 | sfree(params); |
325 | } |
326 | |
327 | static game_params *dup_params(game_params *params) |
328 | { |
329 | game_params *ret = snew(game_params); |
330 | *ret = *params; /* structure copy */ |
331 | return ret; |
332 | } |
333 | |
334 | static void decode_params(game_params *params, char const *string) |
335 | { |
336 | params->w = params->h = atoi(string); |
337 | while (*string && isdigit((unsigned char)*string)) string++; |
338 | if (*string == 'x') { |
339 | string++; |
340 | params->h = atoi(string); |
341 | while (*string && isdigit((unsigned char)*string)) string++; |
342 | } |
343 | if (*string == 'd') { |
344 | int i; |
345 | string++; |
346 | for (i = 0; i < DIFFCOUNT; i++) |
347 | if (*string == tents_diffchars[i]) |
348 | params->diff = i; |
349 | if (*string) string++; |
350 | } |
351 | } |
352 | |
353 | static char *encode_params(game_params *params, int full) |
354 | { |
355 | char buf[120]; |
356 | |
357 | sprintf(buf, "%dx%d", params->w, params->h); |
358 | if (full) |
359 | sprintf(buf + strlen(buf), "d%c", |
360 | tents_diffchars[params->diff]); |
361 | return dupstr(buf); |
362 | } |
363 | |
364 | static config_item *game_configure(game_params *params) |
365 | { |
366 | config_item *ret; |
367 | char buf[80]; |
368 | |
369 | ret = snewn(4, config_item); |
370 | |
371 | ret[0].name = "Width"; |
372 | ret[0].type = C_STRING; |
373 | sprintf(buf, "%d", params->w); |
374 | ret[0].sval = dupstr(buf); |
375 | ret[0].ival = 0; |
376 | |
377 | ret[1].name = "Height"; |
378 | ret[1].type = C_STRING; |
379 | sprintf(buf, "%d", params->h); |
380 | ret[1].sval = dupstr(buf); |
381 | ret[1].ival = 0; |
382 | |
383 | ret[2].name = "Difficulty"; |
384 | ret[2].type = C_CHOICES; |
385 | ret[2].sval = DIFFCONFIG; |
386 | ret[2].ival = params->diff; |
387 | |
388 | ret[3].name = NULL; |
389 | ret[3].type = C_END; |
390 | ret[3].sval = NULL; |
391 | ret[3].ival = 0; |
392 | |
393 | return ret; |
394 | } |
395 | |
396 | static game_params *custom_params(config_item *cfg) |
397 | { |
398 | game_params *ret = snew(game_params); |
399 | |
400 | ret->w = atoi(cfg[0].sval); |
401 | ret->h = atoi(cfg[1].sval); |
402 | ret->diff = cfg[2].ival; |
403 | |
404 | return ret; |
405 | } |
406 | |
407 | static char *validate_params(game_params *params, int full) |
408 | { |
e4b6a85b |
409 | /* |
410 | * Generating anything under 4x4 runs into trouble of one kind |
411 | * or another. |
412 | */ |
413 | if (params->w < 4 || params->h < 4) |
414 | return "Width and height must both be at least four"; |
86e60e3d |
415 | return NULL; |
416 | } |
417 | |
418 | /* |
419 | * Scratch space for solver. |
420 | */ |
421 | enum { N, U, L, R, D, MAXDIR }; /* link directions */ |
422 | #define dx(d) ( ((d)==R) - ((d)==L) ) |
423 | #define dy(d) ( ((d)==D) - ((d)==U) ) |
424 | #define F(d) ( U + D - (d) ) |
425 | struct solver_scratch { |
426 | char *links; /* mapping between trees and tents */ |
427 | int *locs; |
428 | char *place, *mrows, *trows; |
429 | }; |
430 | |
431 | static struct solver_scratch *new_scratch(int w, int h) |
432 | { |
433 | struct solver_scratch *ret = snew(struct solver_scratch); |
434 | |
435 | ret->links = snewn(w*h, char); |
436 | ret->locs = snewn(max(w, h), int); |
437 | ret->place = snewn(max(w, h), char); |
438 | ret->mrows = snewn(3 * max(w, h), char); |
439 | ret->trows = snewn(3 * max(w, h), char); |
440 | |
441 | return ret; |
442 | } |
443 | |
444 | static void free_scratch(struct solver_scratch *sc) |
445 | { |
446 | sfree(sc->trows); |
447 | sfree(sc->mrows); |
448 | sfree(sc->place); |
449 | sfree(sc->locs); |
450 | sfree(sc->links); |
451 | sfree(sc); |
452 | } |
453 | |
454 | /* |
455 | * Solver. Returns 0 for impossibility, 1 for success, 2 for |
456 | * ambiguity or failure to converge. |
457 | */ |
458 | static int tents_solve(int w, int h, const char *grid, int *numbers, |
459 | char *soln, struct solver_scratch *sc, int diff) |
460 | { |
461 | int x, y, d, i, j; |
462 | char *mrow, *mrow1, *mrow2, *trow, *trow1, *trow2; |
463 | |
464 | /* |
465 | * Set up solver data. |
466 | */ |
467 | memset(sc->links, N, w*h); |
468 | |
469 | /* |
470 | * Set up solution array. |
471 | */ |
472 | memcpy(soln, grid, w*h); |
473 | |
474 | /* |
475 | * Main solver loop. |
476 | */ |
477 | while (1) { |
478 | int done_something = FALSE; |
479 | |
480 | /* |
481 | * Any tent which has only one unattached tree adjacent to |
482 | * it can be tied to that tree. |
483 | */ |
484 | for (y = 0; y < h; y++) |
485 | for (x = 0; x < w; x++) |
486 | if (soln[y*w+x] == TENT && !sc->links[y*w+x]) { |
487 | int linkd = 0; |
488 | |
489 | for (d = 1; d < MAXDIR; d++) { |
490 | int x2 = x + dx(d), y2 = y + dy(d); |
491 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
492 | soln[y2*w+x2] == TREE && |
493 | !sc->links[y2*w+x2]) { |
494 | if (linkd) |
495 | break; /* found more than one */ |
496 | else |
497 | linkd = d; |
498 | } |
499 | } |
500 | |
501 | if (d == MAXDIR && linkd == 0) { |
502 | #ifdef SOLVER_DIAGNOSTICS |
503 | if (verbose) |
504 | printf("tent at %d,%d cannot link to anything\n", |
505 | x, y); |
506 | #endif |
507 | return 0; /* no solution exists */ |
508 | } else if (d == MAXDIR) { |
509 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
510 | |
511 | #ifdef SOLVER_DIAGNOSTICS |
512 | if (verbose) |
513 | printf("tent at %d,%d can only link to tree at" |
514 | " %d,%d\n", x, y, x2, y2); |
515 | #endif |
516 | |
517 | sc->links[y*w+x] = linkd; |
518 | sc->links[y2*w+x2] = F(linkd); |
519 | done_something = TRUE; |
520 | } |
521 | } |
522 | |
523 | if (done_something) |
524 | continue; |
525 | if (diff < 0) |
526 | break; /* don't do anything else! */ |
527 | |
528 | /* |
529 | * Mark a blank square as NONTENT if it is not orthogonally |
530 | * adjacent to any unmatched tree. |
531 | */ |
532 | for (y = 0; y < h; y++) |
533 | for (x = 0; x < w; x++) |
534 | if (soln[y*w+x] == BLANK) { |
535 | int can_be_tent = FALSE; |
536 | |
537 | for (d = 1; d < MAXDIR; d++) { |
538 | int x2 = x + dx(d), y2 = y + dy(d); |
539 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
540 | soln[y2*w+x2] == TREE && |
541 | !sc->links[y2*w+x2]) |
542 | can_be_tent = TRUE; |
543 | } |
544 | |
545 | if (!can_be_tent) { |
546 | #ifdef SOLVER_DIAGNOSTICS |
547 | if (verbose) |
548 | printf("%d,%d cannot be a tent (no adjacent" |
549 | " unmatched tree)\n", x, y); |
550 | #endif |
551 | soln[y*w+x] = NONTENT; |
552 | done_something = TRUE; |
553 | } |
554 | } |
555 | |
556 | if (done_something) |
557 | continue; |
558 | |
559 | /* |
560 | * Mark a blank square as NONTENT if it is (perhaps |
561 | * diagonally) adjacent to any other tent. |
562 | */ |
563 | for (y = 0; y < h; y++) |
564 | for (x = 0; x < w; x++) |
565 | if (soln[y*w+x] == BLANK) { |
566 | int dx, dy, imposs = FALSE; |
567 | |
568 | for (dy = -1; dy <= +1; dy++) |
569 | for (dx = -1; dx <= +1; dx++) |
570 | if (dy || dx) { |
571 | int x2 = x + dx, y2 = y + dy; |
572 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
573 | soln[y2*w+x2] == TENT) |
574 | imposs = TRUE; |
575 | } |
576 | |
577 | if (imposs) { |
578 | #ifdef SOLVER_DIAGNOSTICS |
579 | if (verbose) |
580 | printf("%d,%d cannot be a tent (adjacent tent)\n", |
581 | x, y); |
582 | #endif |
583 | soln[y*w+x] = NONTENT; |
584 | done_something = TRUE; |
585 | } |
586 | } |
587 | |
588 | if (done_something) |
589 | continue; |
590 | |
591 | /* |
592 | * Any tree which has exactly one {unattached tent, BLANK} |
593 | * adjacent to it must have its tent in that square. |
594 | */ |
595 | for (y = 0; y < h; y++) |
596 | for (x = 0; x < w; x++) |
597 | if (soln[y*w+x] == TREE && !sc->links[y*w+x]) { |
598 | int linkd = 0, linkd2 = 0, nd = 0; |
599 | |
600 | for (d = 1; d < MAXDIR; d++) { |
601 | int x2 = x + dx(d), y2 = y + dy(d); |
602 | if (!(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h)) |
603 | continue; |
604 | if (soln[y2*w+x2] == BLANK || |
605 | (soln[y2*w+x2] == TENT && !sc->links[y2*w+x2])) { |
606 | if (linkd) |
607 | linkd2 = d; |
608 | else |
609 | linkd = d; |
610 | nd++; |
611 | } |
612 | } |
613 | |
614 | if (nd == 0) { |
615 | #ifdef SOLVER_DIAGNOSTICS |
616 | if (verbose) |
617 | printf("tree at %d,%d cannot link to anything\n", |
618 | x, y); |
619 | #endif |
620 | return 0; /* no solution exists */ |
621 | } else if (nd == 1) { |
622 | int x2 = x + dx(linkd), y2 = y + dy(linkd); |
623 | |
624 | #ifdef SOLVER_DIAGNOSTICS |
625 | if (verbose) |
626 | printf("tree at %d,%d can only link to tent at" |
627 | " %d,%d\n", x, y, x2, y2); |
628 | #endif |
629 | soln[y2*w+x2] = TENT; |
630 | sc->links[y*w+x] = linkd; |
631 | sc->links[y2*w+x2] = F(linkd); |
632 | done_something = TRUE; |
633 | } else if (nd == 2 && (!dx(linkd) != !dx(linkd2)) && |
634 | diff >= DIFF_TRICKY) { |
635 | /* |
636 | * If there are two possible places where |
637 | * this tree's tent can go, and they are |
638 | * diagonally separated rather than being |
639 | * on opposite sides of the tree, then the |
640 | * square (other than the tree square) |
641 | * which is adjacent to both of them must |
642 | * be a non-tent. |
643 | */ |
644 | int x2 = x + dx(linkd) + dx(linkd2); |
645 | int y2 = y + dy(linkd) + dy(linkd2); |
646 | assert(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h); |
647 | if (soln[y2*w+x2] == BLANK) { |
648 | #ifdef SOLVER_DIAGNOSTICS |
649 | if (verbose) |
650 | printf("possible tent locations for tree at" |
651 | " %d,%d rule out tent at %d,%d\n", |
652 | x, y, x2, y2); |
653 | #endif |
654 | soln[y2*w+x2] = NONTENT; |
655 | done_something = TRUE; |
656 | } |
657 | } |
658 | } |
659 | |
660 | if (done_something) |
661 | continue; |
662 | |
663 | /* |
664 | * If localised deductions about the trees and tents |
665 | * themselves haven't helped us, it's time to resort to the |
666 | * numbers round the grid edge. For each row and column, we |
667 | * go through all possible combinations of locations for |
668 | * the unplaced tents, rule out any which have adjacent |
669 | * tents, and spot any square which is given the same state |
670 | * by all remaining combinations. |
671 | */ |
672 | for (i = 0; i < w+h; i++) { |
673 | int start, step, len, start1, start2, n, k; |
674 | |
675 | if (i < w) { |
676 | /* |
677 | * This is the number for a column. |
678 | */ |
679 | start = i; |
680 | step = w; |
681 | len = h; |
682 | if (i > 0) |
683 | start1 = start - 1; |
684 | else |
685 | start1 = -1; |
686 | if (i+1 < w) |
687 | start2 = start + 1; |
688 | else |
689 | start2 = -1; |
690 | } else { |
691 | /* |
692 | * This is the number for a row. |
693 | */ |
694 | start = (i-w)*w; |
695 | step = 1; |
696 | len = w; |
697 | if (i > w) |
698 | start1 = start - w; |
699 | else |
700 | start1 = -1; |
701 | if (i+1 < w+h) |
702 | start2 = start + w; |
703 | else |
704 | start2 = -1; |
705 | } |
706 | |
707 | if (diff < DIFF_TRICKY) { |
708 | /* |
709 | * In Easy mode, we don't look at the effect of one |
710 | * row on the next (i.e. ruling out a square if all |
711 | * possibilities for an adjacent row place a tent |
712 | * next to it). |
713 | */ |
714 | start1 = start2 = -1; |
715 | } |
716 | |
717 | k = numbers[i]; |
718 | |
719 | /* |
720 | * Count and store the locations of the free squares, |
721 | * and also count the number of tents already placed. |
722 | */ |
723 | n = 0; |
724 | for (j = 0; j < len; j++) { |
725 | if (soln[start+j*step] == TENT) |
726 | k--; /* one fewer tent to place */ |
727 | else if (soln[start+j*step] == BLANK) |
728 | sc->locs[n++] = j; |
729 | } |
730 | |
731 | if (n == 0) |
732 | continue; /* nothing left to do here */ |
733 | |
734 | /* |
735 | * Now we know we're placing k tents in n squares. Set |
736 | * up the first possibility. |
737 | */ |
738 | for (j = 0; j < n; j++) |
739 | sc->place[j] = (j < k ? TENT : NONTENT); |
740 | |
741 | /* |
742 | * We're aiming to find squares in this row which are |
743 | * invariant over all valid possibilities. Thus, we |
744 | * maintain the current state of that invariance. We |
745 | * start everything off at MAGIC to indicate that it |
746 | * hasn't been set up yet. |
747 | */ |
748 | mrow = sc->mrows; |
749 | mrow1 = sc->mrows + len; |
750 | mrow2 = sc->mrows + 2*len; |
751 | trow = sc->trows; |
752 | trow1 = sc->trows + len; |
753 | trow2 = sc->trows + 2*len; |
754 | memset(mrow, MAGIC, 3*len); |
755 | |
756 | /* |
757 | * And iterate over all possibilities. |
758 | */ |
759 | while (1) { |
760 | int p, valid; |
761 | |
762 | /* |
763 | * See if this possibility is valid. The only way |
764 | * it can fail to be valid is if it contains two |
765 | * adjacent tents. (Other forms of invalidity, such |
766 | * as containing a tent adjacent to one already |
767 | * placed, will have been dealt with already by |
768 | * other parts of the solver.) |
769 | */ |
770 | valid = TRUE; |
771 | for (j = 0; j+1 < n; j++) |
772 | if (sc->place[j] == TENT && |
773 | sc->place[j+1] == TENT && |
774 | sc->locs[j+1] == sc->locs[j]+1) { |
775 | valid = FALSE; |
776 | break; |
777 | } |
778 | |
779 | if (valid) { |
780 | /* |
781 | * Merge this valid combination into mrow. |
782 | */ |
783 | memset(trow, MAGIC, len); |
784 | memset(trow+len, BLANK, 2*len); |
785 | for (j = 0; j < n; j++) { |
786 | trow[sc->locs[j]] = sc->place[j]; |
787 | if (sc->place[j] == TENT) { |
788 | int jj; |
789 | for (jj = sc->locs[j]-1; jj <= sc->locs[j]+1; jj++) |
790 | if (jj >= 0 && jj < len) |
791 | trow1[jj] = trow2[jj] = NONTENT; |
792 | } |
793 | } |
794 | |
795 | for (j = 0; j < 3*len; j++) { |
796 | if (trow[j] == MAGIC) |
797 | continue; |
798 | if (mrow[j] == MAGIC || mrow[j] == trow[j]) { |
799 | /* |
800 | * Either this is the first valid |
801 | * placement we've found at all, or |
802 | * this square's contents are |
803 | * consistent with every previous valid |
804 | * combination. |
805 | */ |
806 | mrow[j] = trow[j]; |
807 | } else { |
808 | /* |
809 | * This square's contents fail to match |
810 | * what they were in a different |
811 | * combination, so we cannot deduce |
812 | * anything about this square. |
813 | */ |
814 | mrow[j] = BLANK; |
815 | } |
816 | } |
817 | } |
818 | |
819 | /* |
820 | * Find the next combination of k choices from n. |
821 | * We do this by finding the rightmost tent which |
822 | * can be moved one place right, doing so, and |
823 | * shunting all tents to the right of that as far |
824 | * left as they can go. |
825 | */ |
826 | p = 0; |
827 | for (j = n-1; j > 0; j--) { |
828 | if (sc->place[j] == TENT) |
829 | p++; |
830 | if (sc->place[j] == NONTENT && sc->place[j-1] == TENT) { |
831 | sc->place[j-1] = NONTENT; |
832 | sc->place[j] = TENT; |
833 | while (p--) |
834 | sc->place[++j] = TENT; |
835 | while (++j < n) |
836 | sc->place[j] = NONTENT; |
837 | break; |
838 | } |
839 | } |
840 | if (j <= 0) |
841 | break; /* we've finished */ |
842 | } |
843 | |
844 | /* |
845 | * It's just possible that _no_ placement was valid, in |
846 | * which case we have an internally inconsistent |
847 | * puzzle. |
848 | */ |
849 | if (mrow[sc->locs[0]] == MAGIC) |
850 | return 0; /* inconsistent */ |
851 | |
852 | /* |
853 | * Now go through mrow and see if there's anything |
854 | * we've deduced which wasn't already mentioned in soln. |
855 | */ |
856 | for (j = 0; j < len; j++) { |
857 | int whichrow; |
858 | |
859 | for (whichrow = 0; whichrow < 3; whichrow++) { |
860 | char *mthis = mrow + whichrow * len; |
861 | int tstart = (whichrow == 0 ? start : |
862 | whichrow == 1 ? start1 : start2); |
863 | if (tstart >= 0 && |
864 | mthis[j] != MAGIC && mthis[j] != BLANK && |
865 | soln[tstart+j*step] == BLANK) { |
866 | int pos = tstart+j*step; |
867 | |
868 | #ifdef SOLVER_DIAGNOSTICS |
869 | if (verbose) |
870 | printf("%s %d forces %s at %d,%d\n", |
871 | step==1 ? "row" : "column", |
872 | step==1 ? start/w : start, |
9d8d4f9f |
873 | mthis[j] == TENT ? "tent" : "non-tent", |
86e60e3d |
874 | pos % w, pos / w); |
875 | #endif |
876 | soln[pos] = mthis[j]; |
877 | done_something = TRUE; |
878 | } |
879 | } |
880 | } |
881 | } |
882 | |
883 | if (done_something) |
884 | continue; |
885 | |
886 | if (!done_something) |
887 | break; |
888 | } |
889 | |
890 | /* |
891 | * The solver has nothing further it can do. Return 1 if both |
892 | * soln and sc->links are completely filled in, or 2 otherwise. |
893 | */ |
894 | for (y = 0; y < h; y++) |
895 | for (x = 0; x < w; x++) { |
896 | if (soln[y*w+x] == BLANK) |
897 | return 2; |
898 | if (soln[y*w+x] != NONTENT && sc->links[y*w+x] == 0) |
899 | return 2; |
900 | } |
901 | |
902 | return 1; |
903 | } |
904 | |
905 | static char *new_game_desc(game_params *params, random_state *rs, |
906 | char **aux, int interactive) |
907 | { |
908 | int w = params->w, h = params->h; |
909 | int ntrees = w * h / 5; |
910 | char *grid = snewn(w*h, char); |
911 | char *puzzle = snewn(w*h, char); |
912 | int *numbers = snewn(w+h, int); |
913 | char *soln = snewn(w*h, char); |
e4b6a85b |
914 | int *temp = snewn(2*w*h, int); |
86e60e3d |
915 | int maxedges = ntrees*4 + w*h; |
916 | int *edges = snewn(2*maxedges, int); |
917 | int *capacity = snewn(maxedges, int); |
918 | int *flow = snewn(maxedges, int); |
919 | struct solver_scratch *sc = new_scratch(w, h); |
920 | char *ret, *p; |
921 | int i, j, nedges; |
922 | |
923 | /* |
924 | * Since this puzzle has many global deductions and doesn't |
925 | * permit limited clue sets, generating grids for this puzzle |
926 | * is hard enough that I see no better option than to simply |
927 | * generate a solution and see if it's unique and has the |
928 | * required difficulty. This turns out to be computationally |
929 | * plausible as well. |
930 | * |
931 | * We chose our tree count (hence also tent count) by dividing |
932 | * the total grid area by five above. Why five? Well, w*h/4 is |
933 | * the maximum number of tents you can _possibly_ fit into the |
934 | * grid without violating the separation criterion, and to |
935 | * achieve that you are constrained to a very small set of |
936 | * possible layouts (the obvious one with a tent at every |
937 | * (even,even) coordinate, and trivial variations thereon). So |
938 | * if we reduce the tent count a bit more, we enable more |
939 | * random-looking placement; 5 turns out to be a plausible |
940 | * figure which yields sensible puzzles. Increasing the tent |
941 | * count would give puzzles whose solutions were too regimented |
942 | * and could be solved by the use of that knowledge (and would |
943 | * also take longer to find a viable placement); decreasing it |
944 | * would make the grids emptier and more boring. |
945 | * |
946 | * Actually generating a grid is a matter of first placing the |
947 | * tents, and then placing the trees by the use of maxflow |
948 | * (finding a distinct square adjacent to every tent). We do it |
949 | * this way round because otherwise satisfying the tent |
950 | * separation condition would become onerous: most randomly |
951 | * chosen tent layouts do not satisfy this condition, so we'd |
952 | * have gone to a lot of work before finding that a candidate |
953 | * layout was unusable. Instead, we place the tents first and |
954 | * ensure they meet the separation criterion _before_ doing |
955 | * lots of computation; this works much better. |
956 | * |
957 | * The maxflow algorithm is not randomised, so employed naively |
958 | * it would give rise to grids with clear structure and |
959 | * directional bias. Hence, I assign the network nodes as seen |
e4b6a85b |
960 | * by maxflow to be a _random_ permutation of the squares of |
961 | * the grid, so that any bias shown by maxflow towards |
962 | * low-numbered nodes is turned into a random bias. |
86e60e3d |
963 | * |
964 | * This generation strategy can fail at many points, including |
965 | * as early as tent placement (if you get a bad random order in |
966 | * which to greedily try the grid squares, you won't even |
967 | * manage to find enough mutually non-adjacent squares to put |
968 | * the tents in). Then it can fail if maxflow doesn't manage to |
969 | * find a good enough matching (i.e. the tent placements don't |
970 | * admit any adequate tree placements); and finally it can fail |
971 | * if the solver finds that the problem has the wrong |
972 | * difficulty (including being actually non-unique). All of |
973 | * these, however, are insufficiently frequent to cause |
974 | * trouble. |
975 | */ |
976 | |
e4b6a85b |
977 | if (params->diff > DIFF_EASY && params->w <= 4 && params->h <= 4) |
978 | params->diff = DIFF_EASY; /* downgrade to prevent tight loop */ |
979 | |
86e60e3d |
980 | while (1) { |
981 | /* |
e4b6a85b |
982 | * Arrange the grid squares into a random order. |
86e60e3d |
983 | */ |
984 | for (i = 0; i < w*h; i++) |
985 | temp[i] = i; |
986 | shuffle(temp, w*h, sizeof(*temp), rs); |
86e60e3d |
987 | |
988 | /* |
989 | * The first `ntrees' entries in temp which we can get |
990 | * without making two tents adjacent will be the tent |
991 | * locations. |
992 | */ |
993 | memset(grid, BLANK, w*h); |
994 | j = ntrees; |
995 | for (i = 0; i < w*h && j > 0; i++) { |
996 | int x = temp[i] % w, y = temp[i] / w; |
997 | int dy, dx, ok = TRUE; |
998 | |
999 | for (dy = -1; dy <= +1; dy++) |
1000 | for (dx = -1; dx <= +1; dx++) |
1001 | if (x+dx >= 0 && x+dx < w && |
1002 | y+dy >= 0 && y+dy < h && |
1003 | grid[(y+dy)*w+(x+dx)] == TENT) |
1004 | ok = FALSE; |
1005 | |
1006 | if (ok) { |
1007 | grid[temp[i]] = TENT; |
1008 | j--; |
1009 | } |
1010 | } |
1011 | if (j > 0) |
1012 | continue; /* couldn't place all the tents */ |
1013 | |
1014 | /* |
1015 | * Now we build up the list of graph edges. |
1016 | */ |
1017 | nedges = 0; |
1018 | for (i = 0; i < w*h; i++) { |
1019 | if (grid[temp[i]] == TENT) { |
1020 | for (j = 0; j < w*h; j++) { |
1021 | if (grid[temp[j]] != TENT) { |
1022 | int xi = temp[i] % w, yi = temp[i] / w; |
1023 | int xj = temp[j] % w, yj = temp[j] / w; |
1024 | if (abs(xi-xj) + abs(yi-yj) == 1) { |
1025 | edges[nedges*2] = i; |
1026 | edges[nedges*2+1] = j; |
1027 | capacity[nedges] = 1; |
1028 | nedges++; |
1029 | } |
1030 | } |
1031 | } |
1032 | } else { |
1033 | /* |
1034 | * Special node w*h is the sink node; any non-tent node |
1035 | * has an edge going to it. |
1036 | */ |
1037 | edges[nedges*2] = i; |
1038 | edges[nedges*2+1] = w*h; |
1039 | capacity[nedges] = 1; |
1040 | nedges++; |
1041 | } |
1042 | } |
1043 | |
1044 | /* |
1045 | * Special node w*h+1 is the source node, with an edge going to |
1046 | * every tent. |
1047 | */ |
1048 | for (i = 0; i < w*h; i++) { |
1049 | if (grid[temp[i]] == TENT) { |
1050 | edges[nedges*2] = w*h+1; |
1051 | edges[nedges*2+1] = i; |
1052 | capacity[nedges] = 1; |
1053 | nedges++; |
1054 | } |
1055 | } |
1056 | |
1057 | assert(nedges <= maxedges); |
1058 | |
1059 | /* |
1060 | * Now we're ready to call the maxflow algorithm to place the |
1061 | * trees. |
1062 | */ |
1063 | j = maxflow(w*h+2, w*h+1, w*h, nedges, edges, capacity, flow, NULL); |
1064 | |
1065 | if (j < ntrees) |
1066 | continue; /* couldn't place all the tents */ |
1067 | |
1068 | /* |
1069 | * We've placed the trees. Now we need to work out _where_ |
1070 | * we've placed them, which is a matter of reading back out |
1071 | * from the `flow' array. |
1072 | */ |
1073 | for (i = 0; i < nedges; i++) { |
1074 | if (edges[2*i] < w*h && edges[2*i+1] < w*h && flow[i] > 0) |
1075 | grid[temp[edges[2*i+1]]] = TREE; |
1076 | } |
1077 | |
1078 | /* |
1079 | * I think it looks ugly if there isn't at least one of |
1080 | * _something_ (tent or tree) in each row and each column |
1081 | * of the grid. This doesn't give any information away |
1082 | * since a completely empty row/column is instantly obvious |
1083 | * from the clues (it has no trees and a zero). |
1084 | */ |
1085 | for (i = 0; i < w; i++) { |
1086 | for (j = 0; j < h; j++) { |
1087 | if (grid[j*w+i] != BLANK) |
1088 | break; /* found something in this column */ |
1089 | } |
1090 | if (j == h) |
1091 | break; /* found empty column */ |
1092 | } |
1093 | if (i < w) |
1094 | continue; /* a column was empty */ |
1095 | |
1096 | for (j = 0; j < h; j++) { |
1097 | for (i = 0; i < w; i++) { |
1098 | if (grid[j*w+i] != BLANK) |
1099 | break; /* found something in this row */ |
1100 | } |
1101 | if (i == w) |
1102 | break; /* found empty row */ |
1103 | } |
1104 | if (j < h) |
1105 | continue; /* a row was empty */ |
1106 | |
1107 | /* |
1108 | * Now set up the numbers round the edge. |
1109 | */ |
1110 | for (i = 0; i < w; i++) { |
1111 | int n = 0; |
1112 | for (j = 0; j < h; j++) |
1113 | if (grid[j*w+i] == TENT) |
1114 | n++; |
1115 | numbers[i] = n; |
1116 | } |
1117 | for (i = 0; i < h; i++) { |
1118 | int n = 0; |
1119 | for (j = 0; j < w; j++) |
1120 | if (grid[i*w+j] == TENT) |
1121 | n++; |
1122 | numbers[w+i] = n; |
1123 | } |
1124 | |
1125 | /* |
1126 | * And now actually solve the puzzle, to see whether it's |
1127 | * unique and has the required difficulty. |
1128 | */ |
1129 | for (i = 0; i < w*h; i++) |
1130 | puzzle[i] = grid[i] == TREE ? TREE : BLANK; |
1131 | i = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff-1); |
1132 | j = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff); |
1133 | |
1134 | /* |
1135 | * We expect solving with difficulty params->diff to have |
1136 | * succeeded (otherwise the problem is too hard), and |
1137 | * solving with diff-1 to have failed (otherwise it's too |
1138 | * easy). |
1139 | */ |
1140 | if (i == 2 && j == 1) |
1141 | break; |
1142 | } |
1143 | |
1144 | /* |
1145 | * That's it. Encode as a game ID. |
1146 | */ |
1147 | ret = snewn((w+h)*40 + ntrees + (w*h)/26 + 1, char); |
1148 | p = ret; |
1149 | j = 0; |
1150 | for (i = 0; i <= w*h; i++) { |
1151 | int c = (i < w*h ? grid[i] == TREE : 1); |
1152 | if (c) { |
1153 | *p++ = (j == 0 ? '_' : j-1 + 'a'); |
1154 | j = 0; |
1155 | } else { |
1156 | j++; |
1157 | while (j > 25) { |
1158 | *p++ = 'z'; |
1159 | j -= 25; |
1160 | } |
1161 | } |
1162 | } |
1163 | for (i = 0; i < w+h; i++) |
1164 | p += sprintf(p, ",%d", numbers[i]); |
1165 | *p++ = '\0'; |
1166 | ret = sresize(ret, p - ret, char); |
1167 | |
1168 | /* |
1169 | * And encode the solution as an aux_info. |
1170 | */ |
1171 | *aux = snewn(ntrees * 40, char); |
1172 | p = *aux; |
1173 | *p++ = 'S'; |
1174 | for (i = 0; i < w*h; i++) |
1175 | if (grid[i] == TENT) |
1176 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1177 | *p++ = '\0'; |
1178 | *aux = sresize(*aux, p - *aux, char); |
1179 | |
1180 | free_scratch(sc); |
1181 | sfree(flow); |
1182 | sfree(capacity); |
1183 | sfree(edges); |
1184 | sfree(temp); |
1185 | sfree(soln); |
1186 | sfree(numbers); |
1187 | sfree(puzzle); |
1188 | sfree(grid); |
1189 | |
1190 | return ret; |
1191 | } |
1192 | |
1193 | static char *validate_desc(game_params *params, char *desc) |
1194 | { |
1195 | int w = params->w, h = params->h; |
1196 | int area, i; |
1197 | |
1198 | area = 0; |
1199 | while (*desc && *desc != ',') { |
1200 | if (*desc == '_') |
1201 | area++; |
1202 | else if (*desc >= 'a' && *desc < 'z') |
1203 | area += *desc - 'a' + 2; |
1204 | else if (*desc == 'z') |
1205 | area += 25; |
1206 | else if (*desc == '!' || *desc == '-') |
1207 | /* do nothing */; |
1208 | else |
1209 | return "Invalid character in grid specification"; |
1210 | |
1211 | desc++; |
1212 | } |
724b4a49 |
1213 | if (area < w * h + 1) |
1214 | return "Not enough data to fill grid"; |
1215 | else if (area > w * h + 1) |
1216 | return "Too much data to fill grid"; |
86e60e3d |
1217 | |
1218 | for (i = 0; i < w+h; i++) { |
1219 | if (!*desc) |
1220 | return "Not enough numbers given after grid specification"; |
1221 | else if (*desc != ',') |
1222 | return "Invalid character in number list"; |
1223 | desc++; |
1224 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1225 | } |
1226 | |
1227 | if (*desc) |
1228 | return "Unexpected additional data at end of game description"; |
1229 | return NULL; |
1230 | } |
1231 | |
1232 | static game_state *new_game(midend *me, game_params *params, char *desc) |
1233 | { |
1234 | int w = params->w, h = params->h; |
1235 | game_state *state = snew(game_state); |
1236 | int i; |
1237 | |
1238 | state->p = *params; /* structure copy */ |
1239 | state->grid = snewn(w*h, char); |
1240 | state->numbers = snew(struct numbers); |
1241 | state->numbers->refcount = 1; |
1242 | state->numbers->numbers = snewn(w+h, int); |
1243 | state->completed = state->used_solve = FALSE; |
1244 | |
1245 | i = 0; |
1246 | memset(state->grid, BLANK, w*h); |
1247 | |
1248 | while (*desc) { |
1249 | int run, type; |
1250 | |
1251 | type = TREE; |
1252 | |
1253 | if (*desc == '_') |
1254 | run = 0; |
1255 | else if (*desc >= 'a' && *desc < 'z') |
1256 | run = *desc - ('a'-1); |
1257 | else if (*desc == 'z') { |
1258 | run = 25; |
1259 | type = BLANK; |
1260 | } else { |
1261 | assert(*desc == '!' || *desc == '-'); |
1262 | run = -1; |
1263 | type = (*desc == '!' ? TENT : NONTENT); |
1264 | } |
1265 | |
1266 | desc++; |
1267 | |
1268 | i += run; |
1269 | assert(i >= 0 && i <= w*h); |
1270 | if (i == w*h) { |
1271 | assert(type == TREE); |
1272 | break; |
1273 | } else { |
1274 | if (type != BLANK) |
1275 | state->grid[i++] = type; |
1276 | } |
1277 | } |
1278 | |
1279 | for (i = 0; i < w+h; i++) { |
1280 | assert(*desc == ','); |
1281 | desc++; |
1282 | state->numbers->numbers[i] = atoi(desc); |
1283 | while (*desc && isdigit((unsigned char)*desc)) desc++; |
1284 | } |
1285 | |
1286 | assert(!*desc); |
1287 | |
1288 | return state; |
1289 | } |
1290 | |
1291 | static game_state *dup_game(game_state *state) |
1292 | { |
1293 | int w = state->p.w, h = state->p.h; |
1294 | game_state *ret = snew(game_state); |
1295 | |
1296 | ret->p = state->p; /* structure copy */ |
1297 | ret->grid = snewn(w*h, char); |
1298 | memcpy(ret->grid, state->grid, w*h); |
1299 | ret->numbers = state->numbers; |
1300 | state->numbers->refcount++; |
1301 | ret->completed = state->completed; |
1302 | ret->used_solve = state->used_solve; |
1303 | |
1304 | return ret; |
1305 | } |
1306 | |
1307 | static void free_game(game_state *state) |
1308 | { |
1309 | if (--state->numbers->refcount <= 0) { |
1310 | sfree(state->numbers->numbers); |
1311 | sfree(state->numbers); |
1312 | } |
1313 | sfree(state->grid); |
1314 | sfree(state); |
1315 | } |
1316 | |
1317 | static char *solve_game(game_state *state, game_state *currstate, |
1318 | char *aux, char **error) |
1319 | { |
1320 | int w = state->p.w, h = state->p.h; |
1321 | |
1322 | if (aux) { |
1323 | /* |
1324 | * If we already have the solution, save ourselves some |
1325 | * time. |
1326 | */ |
1327 | return dupstr(aux); |
1328 | } else { |
1329 | struct solver_scratch *sc = new_scratch(w, h); |
1330 | char *soln; |
1331 | int ret; |
1332 | char *move, *p; |
1333 | int i; |
1334 | |
1335 | soln = snewn(w*h, char); |
1336 | ret = tents_solve(w, h, state->grid, state->numbers->numbers, |
1337 | soln, sc, DIFFCOUNT-1); |
1338 | free_scratch(sc); |
1339 | if (ret != 1) { |
1340 | sfree(soln); |
1341 | if (ret == 0) |
1342 | *error = "This puzzle is not self-consistent"; |
1343 | else |
1344 | *error = "Unable to find a unique solution for this puzzle"; |
1345 | return NULL; |
1346 | } |
1347 | |
1348 | /* |
1349 | * Construct a move string which turns the current state |
1350 | * into the solved state. |
1351 | */ |
1352 | move = snewn(w*h * 40, char); |
1353 | p = move; |
1354 | *p++ = 'S'; |
1355 | for (i = 0; i < w*h; i++) |
1356 | if (soln[i] == TENT) |
1357 | p += sprintf(p, ";T%d,%d", i%w, i/w); |
1358 | *p++ = '\0'; |
1359 | move = sresize(move, p - move, char); |
1360 | |
1361 | sfree(soln); |
1362 | |
1363 | return move; |
1364 | } |
1365 | } |
1366 | |
fa3abef5 |
1367 | static int game_can_format_as_text_now(game_params *params) |
1368 | { |
1369 | return TRUE; |
1370 | } |
1371 | |
86e60e3d |
1372 | static char *game_text_format(game_state *state) |
1373 | { |
1374 | int w = state->p.w, h = state->p.h; |
1375 | char *ret, *p; |
1376 | int x, y; |
1377 | |
1378 | /* |
1379 | * FIXME: We currently do not print the numbers round the edges |
1380 | * of the grid. I need to work out a sensible way of doing this |
1381 | * even when the column numbers exceed 9. |
1382 | * |
1383 | * In the absence of those numbers, the result size is h lines |
1384 | * of w+1 characters each, plus a NUL. |
1385 | * |
1386 | * This function is currently only used by the standalone |
1387 | * solver; until I make it look more sensible, I won't enable |
1388 | * it in the main game structure. |
1389 | */ |
1390 | ret = snewn(h*(w+1) + 1, char); |
1391 | p = ret; |
1392 | for (y = 0; y < h; y++) { |
1393 | for (x = 0; x < w; x++) { |
1394 | *p = (state->grid[y*w+x] == BLANK ? '.' : |
1395 | state->grid[y*w+x] == TREE ? 'T' : |
1396 | state->grid[y*w+x] == TENT ? '*' : |
1397 | state->grid[y*w+x] == NONTENT ? '-' : '?'); |
1398 | p++; |
1399 | } |
1400 | *p++ = '\n'; |
1401 | } |
1402 | *p++ = '\0'; |
1403 | |
1404 | return ret; |
1405 | } |
1406 | |
565394e7 |
1407 | struct game_ui { |
1408 | int dsx, dsy; /* coords of drag start */ |
1409 | int dex, dey; /* coords of drag end */ |
1410 | int drag_button; /* -1 for none, or a button code */ |
1411 | int drag_ok; /* dragged off the window, to cancel */ |
505ea4e5 |
1412 | |
1413 | int cx, cy, cdisp; /* cursor position, and ?display. */ |
565394e7 |
1414 | }; |
1415 | |
86e60e3d |
1416 | static game_ui *new_ui(game_state *state) |
1417 | { |
565394e7 |
1418 | game_ui *ui = snew(game_ui); |
1419 | ui->dsx = ui->dsy = -1; |
1420 | ui->dex = ui->dey = -1; |
1421 | ui->drag_button = -1; |
1422 | ui->drag_ok = FALSE; |
505ea4e5 |
1423 | ui->cx = ui->cy = ui->cdisp = 0; |
565394e7 |
1424 | return ui; |
86e60e3d |
1425 | } |
1426 | |
1427 | static void free_ui(game_ui *ui) |
1428 | { |
565394e7 |
1429 | sfree(ui); |
86e60e3d |
1430 | } |
1431 | |
1432 | static char *encode_ui(game_ui *ui) |
1433 | { |
1434 | return NULL; |
1435 | } |
1436 | |
1437 | static void decode_ui(game_ui *ui, char *encoding) |
1438 | { |
1439 | } |
1440 | |
1441 | static void game_changed_state(game_ui *ui, game_state *oldstate, |
1442 | game_state *newstate) |
1443 | { |
1444 | } |
1445 | |
1446 | struct game_drawstate { |
1447 | int tilesize; |
1448 | int started; |
1449 | game_params p; |
2a27ffcf |
1450 | int *drawn, *numbersdrawn; |
505ea4e5 |
1451 | int cx, cy; /* last-drawn cursor pos, or (-1,-1) if absent. */ |
86e60e3d |
1452 | }; |
1453 | |
1454 | #define PREFERRED_TILESIZE 32 |
1455 | #define TILESIZE (ds->tilesize) |
1456 | #define TLBORDER (TILESIZE/2) |
1457 | #define BRBORDER (TILESIZE*3/2) |
1458 | #define COORD(x) ( (x) * TILESIZE + TLBORDER ) |
1459 | #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 ) |
1460 | |
1461 | #define FLASH_TIME 0.30F |
1462 | |
565394e7 |
1463 | static int drag_xform(game_ui *ui, int x, int y, int v) |
1464 | { |
1465 | int xmin, ymin, xmax, ymax; |
1466 | |
1467 | xmin = min(ui->dsx, ui->dex); |
1468 | xmax = max(ui->dsx, ui->dex); |
1469 | ymin = min(ui->dsy, ui->dey); |
1470 | ymax = max(ui->dsy, ui->dey); |
1471 | |
1472 | /* |
1473 | * Left-dragging has no effect, so we treat a left-drag as a |
1474 | * single click on dsx,dsy. |
1475 | */ |
1476 | if (ui->drag_button == LEFT_BUTTON) { |
1477 | xmin = xmax = ui->dsx; |
1478 | ymin = ymax = ui->dsy; |
1479 | } |
1480 | |
1481 | if (x < xmin || x > xmax || y < ymin || y > ymax) |
1482 | return v; /* no change outside drag area */ |
1483 | |
1484 | if (v == TREE) |
1485 | return v; /* trees are inviolate always */ |
1486 | |
1487 | if (xmin == xmax && ymin == ymax) { |
1488 | /* |
1489 | * Results of a simple click. Left button sets blanks to |
1490 | * tents; right button sets blanks to non-tents; either |
1491 | * button clears a non-blank square. |
1492 | */ |
1493 | if (ui->drag_button == LEFT_BUTTON) |
1494 | v = (v == BLANK ? TENT : BLANK); |
1495 | else |
1496 | v = (v == BLANK ? NONTENT : BLANK); |
1497 | } else { |
1498 | /* |
1499 | * Results of a drag. Left-dragging has no effect. |
1500 | * Right-dragging sets all blank squares to non-tents and |
1501 | * has no effect on anything else. |
1502 | */ |
1503 | if (ui->drag_button == RIGHT_BUTTON) |
1504 | v = (v == BLANK ? NONTENT : v); |
1505 | else |
1506 | /* do nothing */; |
1507 | } |
1508 | |
1509 | return v; |
1510 | } |
1511 | |
86e60e3d |
1512 | static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, |
1513 | int x, int y, int button) |
1514 | { |
1515 | int w = state->p.w, h = state->p.h; |
505ea4e5 |
1516 | char tmpbuf[80]; |
86e60e3d |
1517 | |
1518 | if (button == LEFT_BUTTON || button == RIGHT_BUTTON) { |
86e60e3d |
1519 | x = FROMCOORD(x); |
1520 | y = FROMCOORD(y); |
1521 | if (x < 0 || y < 0 || x >= w || y >= h) |
1522 | return NULL; |
1523 | |
565394e7 |
1524 | ui->drag_button = button; |
1525 | ui->dsx = ui->dex = x; |
1526 | ui->dsy = ui->dey = y; |
1527 | ui->drag_ok = TRUE; |
505ea4e5 |
1528 | ui->cdisp = 0; |
565394e7 |
1529 | return ""; /* ui updated */ |
1530 | } |
86e60e3d |
1531 | |
565394e7 |
1532 | if ((IS_MOUSE_DRAG(button) || IS_MOUSE_RELEASE(button)) && |
1533 | ui->drag_button > 0) { |
1534 | int xmin, ymin, xmax, ymax; |
505ea4e5 |
1535 | char *buf, *sep; |
565394e7 |
1536 | int buflen, bufsize, tmplen; |
1537 | |
1538 | x = FROMCOORD(x); |
1539 | y = FROMCOORD(y); |
1540 | if (x < 0 || y < 0 || x >= w || y >= h) { |
1541 | ui->drag_ok = FALSE; |
86e60e3d |
1542 | } else { |
565394e7 |
1543 | /* |
1544 | * Drags are limited to one row or column. Hence, we |
1545 | * work out which coordinate is closer to the drag |
1546 | * start, and move it _to_ the drag start. |
1547 | */ |
1548 | if (abs(x - ui->dsx) < abs(y - ui->dsy)) |
1549 | x = ui->dsx; |
1550 | else |
1551 | y = ui->dsy; |
1552 | |
1553 | ui->dex = x; |
1554 | ui->dey = y; |
1555 | |
1556 | ui->drag_ok = TRUE; |
86e60e3d |
1557 | } |
1558 | |
565394e7 |
1559 | if (IS_MOUSE_DRAG(button)) |
1560 | return ""; /* ui updated */ |
1561 | |
1562 | /* |
1563 | * The drag has been released. Enact it. |
1564 | */ |
1565 | if (!ui->drag_ok) { |
1566 | ui->drag_button = -1; |
1567 | return ""; /* drag was just cancelled */ |
1568 | } |
1569 | |
1570 | xmin = min(ui->dsx, ui->dex); |
1571 | xmax = max(ui->dsx, ui->dex); |
1572 | ymin = min(ui->dsy, ui->dey); |
1573 | ymax = max(ui->dsy, ui->dey); |
1574 | assert(0 <= xmin && xmin <= xmax && xmax < w); |
e4b6a85b |
1575 | assert(0 <= ymin && ymin <= ymax && ymax < h); |
565394e7 |
1576 | |
1577 | buflen = 0; |
1578 | bufsize = 256; |
1579 | buf = snewn(bufsize, char); |
1580 | sep = ""; |
1581 | for (y = ymin; y <= ymax; y++) |
1582 | for (x = xmin; x <= xmax; x++) { |
1583 | int v = drag_xform(ui, x, y, state->grid[y*w+x]); |
1584 | if (state->grid[y*w+x] != v) { |
1585 | tmplen = sprintf(tmpbuf, "%s%c%d,%d", sep, |
626cd8ac |
1586 | (int)(v == BLANK ? 'B' : |
1587 | v == TENT ? 'T' : 'N'), |
565394e7 |
1588 | x, y); |
1589 | sep = ";"; |
1590 | |
1591 | if (buflen + tmplen >= bufsize) { |
1592 | bufsize = buflen + tmplen + 256; |
1593 | buf = sresize(buf, bufsize, char); |
1594 | } |
1595 | |
1596 | strcpy(buf+buflen, tmpbuf); |
1597 | buflen += tmplen; |
1598 | } |
1599 | } |
1600 | |
1601 | ui->drag_button = -1; /* drag is terminated */ |
1602 | |
1603 | if (buflen == 0) { |
1604 | sfree(buf); |
1605 | return ""; /* ui updated (drag was terminated) */ |
1606 | } else { |
1607 | buf[buflen] = '\0'; |
1608 | return buf; |
1609 | } |
86e60e3d |
1610 | } |
1611 | |
505ea4e5 |
1612 | if (IS_CURSOR_MOVE(button)) { |
1613 | move_cursor(button, &ui->cx, &ui->cy, w, h, 0); |
1614 | ui->cdisp = 1; |
1615 | return ""; |
1616 | } |
1617 | if (ui->cdisp) { |
1618 | char rep = 0; |
1619 | int v = state->grid[ui->cy*w+ui->cx]; |
1620 | |
1621 | if (v != TREE) { |
1622 | #ifdef SINGLE_CURSOR_SELECT |
1623 | if (button == CURSOR_SELECT) |
1624 | /* SELECT cycles T, N, B */ |
1625 | rep = v == BLANK ? 'T' : v == TENT ? 'N' : 'B'; |
1626 | #else |
1627 | if (button == CURSOR_SELECT) |
1628 | rep = v == BLANK ? 'T' : 'B'; |
1629 | else if (button == CURSOR_SELECT2) |
1630 | rep = v == BLANK ? 'N' : 'B'; |
1631 | else if (button == 'T' || button == 'N' || button == 'B') |
1632 | rep = (char)button; |
1633 | #endif |
1634 | } |
1635 | |
1636 | if (rep) { |
1637 | sprintf(tmpbuf, "%c%d,%d", (int)rep, ui->cx, ui->cy); |
1638 | return dupstr(tmpbuf); |
1639 | } |
1640 | } else if (IS_CURSOR_SELECT(button)) { |
1641 | ui->cdisp = 1; |
1642 | return ""; |
1643 | } |
1644 | |
86e60e3d |
1645 | return NULL; |
1646 | } |
1647 | |
1648 | static game_state *execute_move(game_state *state, char *move) |
1649 | { |
1650 | int w = state->p.w, h = state->p.h; |
1651 | char c; |
1652 | int x, y, m, n, i, j; |
1653 | game_state *ret = dup_game(state); |
1654 | |
1655 | while (*move) { |
1656 | c = *move; |
1657 | if (c == 'S') { |
1658 | int i; |
1659 | ret->used_solve = TRUE; |
1660 | /* |
1661 | * Set all non-tree squares to NONTENT. The rest of the |
1662 | * solve move will fill the tents in over the top. |
1663 | */ |
1664 | for (i = 0; i < w*h; i++) |
1665 | if (ret->grid[i] != TREE) |
1666 | ret->grid[i] = NONTENT; |
1667 | move++; |
1668 | } else if (c == 'B' || c == 'T' || c == 'N') { |
1669 | move++; |
1670 | if (sscanf(move, "%d,%d%n", &x, &y, &n) != 2 || |
1671 | x < 0 || y < 0 || x >= w || y >= h) { |
1672 | free_game(ret); |
1673 | return NULL; |
1674 | } |
1675 | if (ret->grid[y*w+x] == TREE) { |
1676 | free_game(ret); |
1677 | return NULL; |
1678 | } |
1679 | ret->grid[y*w+x] = (c == 'B' ? BLANK : c == 'T' ? TENT : NONTENT); |
1680 | move += n; |
1681 | } else { |
1682 | free_game(ret); |
1683 | return NULL; |
1684 | } |
1685 | if (*move == ';') |
1686 | move++; |
1687 | else if (*move) { |
1688 | free_game(ret); |
1689 | return NULL; |
1690 | } |
1691 | } |
1692 | |
1693 | /* |
1694 | * Check for completion. |
1695 | */ |
1696 | for (i = n = m = 0; i < w*h; i++) { |
1697 | if (ret->grid[i] == TENT) |
1698 | n++; |
1699 | else if (ret->grid[i] == TREE) |
1700 | m++; |
1701 | } |
1702 | if (n == m) { |
1703 | int nedges, maxedges, *edges, *capacity, *flow; |
1704 | |
1705 | /* |
1706 | * We have the right number of tents, which is a |
1707 | * precondition for the game being complete. Now check that |
1708 | * the numbers add up. |
1709 | */ |
1710 | for (i = 0; i < w; i++) { |
1711 | n = 0; |
1712 | for (j = 0; j < h; j++) |
1713 | if (ret->grid[j*w+i] == TENT) |
1714 | n++; |
1715 | if (ret->numbers->numbers[i] != n) |
1716 | goto completion_check_done; |
1717 | } |
1718 | for (i = 0; i < h; i++) { |
1719 | n = 0; |
1720 | for (j = 0; j < w; j++) |
1721 | if (ret->grid[i*w+j] == TENT) |
1722 | n++; |
1723 | if (ret->numbers->numbers[w+i] != n) |
1724 | goto completion_check_done; |
1725 | } |
1726 | /* |
1727 | * Also, check that no two tents are adjacent. |
1728 | */ |
1729 | for (y = 0; y < h; y++) |
1730 | for (x = 0; x < w; x++) { |
1731 | if (x+1 < w && |
1732 | ret->grid[y*w+x] == TENT && ret->grid[y*w+x+1] == TENT) |
1733 | goto completion_check_done; |
1734 | if (y+1 < h && |
1735 | ret->grid[y*w+x] == TENT && ret->grid[(y+1)*w+x] == TENT) |
1736 | goto completion_check_done; |
1737 | if (x+1 < w && y+1 < h) { |
1738 | if (ret->grid[y*w+x] == TENT && |
1739 | ret->grid[(y+1)*w+(x+1)] == TENT) |
1740 | goto completion_check_done; |
1741 | if (ret->grid[(y+1)*w+x] == TENT && |
1742 | ret->grid[y*w+(x+1)] == TENT) |
1743 | goto completion_check_done; |
1744 | } |
1745 | } |
1746 | |
1747 | /* |
1748 | * OK; we have the right number of tents, they match the |
1749 | * numeric clues, and they satisfy the non-adjacency |
1750 | * criterion. Finally, we need to verify that they can be |
1751 | * placed in a one-to-one matching with the trees such that |
1752 | * every tent is orthogonally adjacent to its tree. |
1753 | * |
1754 | * This bit is where the hard work comes in: we have to do |
1755 | * it by finding such a matching using maxflow. |
1756 | * |
1757 | * So we construct a network with one special source node, |
1758 | * one special sink node, one node per tent, and one node |
1759 | * per tree. |
1760 | */ |
1761 | maxedges = 6 * m; |
1762 | edges = snewn(2 * maxedges, int); |
1763 | capacity = snewn(maxedges, int); |
1764 | flow = snewn(maxedges, int); |
1765 | nedges = 0; |
1766 | /* |
1767 | * Node numbering: |
1768 | * |
1769 | * 0..w*h trees/tents |
1770 | * w*h source |
1771 | * w*h+1 sink |
1772 | */ |
1773 | for (y = 0; y < h; y++) |
1774 | for (x = 0; x < w; x++) |
1775 | if (ret->grid[y*w+x] == TREE) { |
1776 | int d; |
1777 | |
1778 | /* |
1779 | * Here we use the direction enum declared for |
1780 | * the solver. We make use of the fact that the |
1781 | * directions are declared in the order |
1782 | * U,L,R,D, meaning that we go through the four |
1783 | * neighbours of any square in numerically |
1784 | * increasing order. |
1785 | */ |
1786 | for (d = 1; d < MAXDIR; d++) { |
1787 | int x2 = x + dx(d), y2 = y + dy(d); |
1788 | if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h && |
1789 | ret->grid[y2*w+x2] == TENT) { |
1790 | assert(nedges < maxedges); |
1791 | edges[nedges*2] = y*w+x; |
1792 | edges[nedges*2+1] = y2*w+x2; |
1793 | capacity[nedges] = 1; |
1794 | nedges++; |
1795 | } |
1796 | } |
1797 | } else if (ret->grid[y*w+x] == TENT) { |
1798 | assert(nedges < maxedges); |
1799 | edges[nedges*2] = y*w+x; |
1800 | edges[nedges*2+1] = w*h+1; /* edge going to sink */ |
1801 | capacity[nedges] = 1; |
1802 | nedges++; |
1803 | } |
1804 | for (y = 0; y < h; y++) |
1805 | for (x = 0; x < w; x++) |
1806 | if (ret->grid[y*w+x] == TREE) { |
1807 | assert(nedges < maxedges); |
1808 | edges[nedges*2] = w*h; /* edge coming from source */ |
1809 | edges[nedges*2+1] = y*w+x; |
1810 | capacity[nedges] = 1; |
1811 | nedges++; |
1812 | } |
1813 | n = maxflow(w*h+2, w*h, w*h+1, nedges, edges, capacity, flow, NULL); |
1814 | |
1815 | sfree(flow); |
1816 | sfree(capacity); |
1817 | sfree(edges); |
1818 | |
1819 | if (n != m) |
1820 | goto completion_check_done; |
1821 | |
1822 | /* |
1823 | * We haven't managed to fault the grid on any count. Score! |
1824 | */ |
1825 | ret->completed = TRUE; |
1826 | } |
1827 | completion_check_done: |
1828 | |
1829 | return ret; |
1830 | } |
1831 | |
1832 | /* ---------------------------------------------------------------------- |
1833 | * Drawing routines. |
1834 | */ |
1835 | |
1836 | static void game_compute_size(game_params *params, int tilesize, |
1837 | int *x, int *y) |
1838 | { |
1839 | /* fool the macros */ |
3466f373 |
1840 | struct dummy { int tilesize; } dummy, *ds = &dummy; |
1841 | dummy.tilesize = tilesize; |
86e60e3d |
1842 | |
1843 | *x = TLBORDER + BRBORDER + TILESIZE * params->w; |
1844 | *y = TLBORDER + BRBORDER + TILESIZE * params->h; |
1845 | } |
1846 | |
1847 | static void game_set_size(drawing *dr, game_drawstate *ds, |
1848 | game_params *params, int tilesize) |
1849 | { |
1850 | ds->tilesize = tilesize; |
1851 | } |
1852 | |
8266f3fc |
1853 | static float *game_colours(frontend *fe, int *ncolours) |
86e60e3d |
1854 | { |
1855 | float *ret = snewn(3 * NCOLOURS, float); |
1856 | |
1857 | frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); |
1858 | |
1859 | ret[COL_GRID * 3 + 0] = 0.0F; |
1860 | ret[COL_GRID * 3 + 1] = 0.0F; |
1861 | ret[COL_GRID * 3 + 2] = 0.0F; |
1862 | |
1863 | ret[COL_GRASS * 3 + 0] = 0.7F; |
1864 | ret[COL_GRASS * 3 + 1] = 1.0F; |
1865 | ret[COL_GRASS * 3 + 2] = 0.5F; |
1866 | |
1867 | ret[COL_TREETRUNK * 3 + 0] = 0.6F; |
1868 | ret[COL_TREETRUNK * 3 + 1] = 0.4F; |
1869 | ret[COL_TREETRUNK * 3 + 2] = 0.0F; |
1870 | |
1871 | ret[COL_TREELEAF * 3 + 0] = 0.0F; |
1872 | ret[COL_TREELEAF * 3 + 1] = 0.7F; |
1873 | ret[COL_TREELEAF * 3 + 2] = 0.0F; |
1874 | |
1875 | ret[COL_TENT * 3 + 0] = 0.8F; |
1876 | ret[COL_TENT * 3 + 1] = 0.7F; |
1877 | ret[COL_TENT * 3 + 2] = 0.0F; |
1878 | |
2a27ffcf |
1879 | ret[COL_ERROR * 3 + 0] = 1.0F; |
1880 | ret[COL_ERROR * 3 + 1] = 0.0F; |
1881 | ret[COL_ERROR * 3 + 2] = 0.0F; |
1882 | |
1883 | ret[COL_ERRTEXT * 3 + 0] = 1.0F; |
1884 | ret[COL_ERRTEXT * 3 + 1] = 1.0F; |
1885 | ret[COL_ERRTEXT * 3 + 2] = 1.0F; |
1886 | |
e8df451f |
1887 | ret[COL_ERRTRUNK * 3 + 0] = 0.6F; |
1888 | ret[COL_ERRTRUNK * 3 + 1] = 0.0F; |
1889 | ret[COL_ERRTRUNK * 3 + 2] = 0.0F; |
1890 | |
86e60e3d |
1891 | *ncolours = NCOLOURS; |
1892 | return ret; |
1893 | } |
1894 | |
1895 | static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) |
1896 | { |
1897 | int w = state->p.w, h = state->p.h; |
1898 | struct game_drawstate *ds = snew(struct game_drawstate); |
2a27ffcf |
1899 | int i; |
86e60e3d |
1900 | |
1901 | ds->tilesize = 0; |
1902 | ds->started = FALSE; |
1903 | ds->p = state->p; /* structure copy */ |
2a27ffcf |
1904 | ds->drawn = snewn(w*h, int); |
1905 | for (i = 0; i < w*h; i++) |
1906 | ds->drawn[i] = MAGIC; |
1907 | ds->numbersdrawn = snewn(w+h, int); |
1908 | for (i = 0; i < w+h; i++) |
1909 | ds->numbersdrawn[i] = 2; |
505ea4e5 |
1910 | ds->cx = ds->cy = -1; |
86e60e3d |
1911 | |
1912 | return ds; |
1913 | } |
1914 | |
1915 | static void game_free_drawstate(drawing *dr, game_drawstate *ds) |
1916 | { |
1917 | sfree(ds->drawn); |
2a27ffcf |
1918 | sfree(ds->numbersdrawn); |
86e60e3d |
1919 | sfree(ds); |
1920 | } |
1921 | |
2a27ffcf |
1922 | enum { |
1923 | ERR_ADJ_TOPLEFT = 4, |
1924 | ERR_ADJ_TOP, |
1925 | ERR_ADJ_TOPRIGHT, |
1926 | ERR_ADJ_LEFT, |
1927 | ERR_ADJ_RIGHT, |
1928 | ERR_ADJ_BOTLEFT, |
1929 | ERR_ADJ_BOT, |
1930 | ERR_ADJ_BOTRIGHT, |
1931 | ERR_OVERCOMMITTED |
1932 | }; |
1933 | |
271cf5c1 |
1934 | static int *find_errors(game_state *state, char *grid) |
2a27ffcf |
1935 | { |
1936 | int w = state->p.w, h = state->p.h; |
1937 | int *ret = snewn(w*h + w + h, int); |
1938 | int *tmp = snewn(w*h*2, int), *dsf = tmp + w*h; |
1939 | int x, y; |
1940 | |
1941 | /* |
7463d8ef |
1942 | * This function goes through a grid and works out where to |
1943 | * highlight play errors in red. The aim is that it should |
1944 | * produce at least one error highlight for any complete grid |
1945 | * (or complete piece of grid) violating a puzzle constraint, so |
1946 | * that a grid containing no BLANK squares is either a win or is |
1947 | * marked up in some way that indicates why not. |
1948 | * |
1949 | * So it's easy enough to highlight errors in the numeric clues |
1950 | * - just light up any row or column number which is not |
1951 | * fulfilled - and it's just as easy to highlight adjacent |
1952 | * tents. The difficult bit is highlighting failures in the |
1953 | * tent/tree matching criterion. |
1954 | * |
1955 | * A natural approach would seem to be to apply the maxflow |
1956 | * algorithm to find the tent/tree matching; if this fails, it |
1957 | * must necessarily terminate with a min-cut which can be |
1958 | * reinterpreted as some set of trees which have too few tents |
1959 | * between them (or vice versa). However, it's bad for |
1960 | * localising errors, because it's not easy to make the |
1961 | * algorithm narrow down to the _smallest_ such set of trees: if |
1962 | * trees A and B have only one tent between them, for instance, |
1963 | * it might perfectly well highlight not only A and B but also |
1964 | * trees C and D which are correctly matched on the far side of |
1965 | * the grid, on the grounds that those four trees between them |
1966 | * have only three tents. |
1967 | * |
1968 | * Also, that approach fares badly when you introduce the |
1969 | * additional requirement that incomplete grids should have |
1970 | * errors highlighted only when they can be proved to be errors |
26dac649 |
1971 | * - so that trees should not be marked as having too few tents |
1972 | * if there are enough BLANK squares remaining around them that |
1973 | * could be turned into the missing tents (to do so would be |
1974 | * patronising, since the overwhelming likelihood is not that |
1975 | * the player has forgotten to put a tree there but that they |
1976 | * have merely not put one there _yet_). However, tents with too |
1977 | * few trees can be marked immediately, since those are |
1978 | * definitely player error. |
7463d8ef |
1979 | * |
1980 | * So I adopt an alternative approach, which is to consider the |
1981 | * bipartite adjacency graph between trees and tents |
1982 | * ('bipartite' in the sense that for these purposes I |
1983 | * deliberately ignore two adjacent trees or two adjacent |
1984 | * tents), divide that graph up into its connected components |
1985 | * using a dsf, and look for components which contain different |
1986 | * numbers of trees and tents. This allows me to highlight |
1987 | * groups of tents with too few trees between them immediately, |
1988 | * and then in order to find groups of trees with too few tents |
1989 | * I redo the same process but counting BLANKs as potential |
1990 | * tents (so that the only trees highlighted are those |
1991 | * surrounded by enough NONTENTs to make it impossible to give |
1992 | * them enough tents). |
1993 | * |
1994 | * However, this technique is incomplete: it is not a sufficient |
1995 | * condition for the existence of a perfect matching that every |
1996 | * connected component of the graph has the same number of tents |
1997 | * and trees. An example of a graph which satisfies the latter |
1998 | * condition but still has no perfect matching is |
1999 | * |
2000 | * A B C |
2001 | * | / ,/| |
2002 | * | / ,'/ | |
2003 | * | / ,' / | |
2004 | * |/,' / | |
2005 | * 1 2 3 |
2006 | * |
2007 | * which can be realised in Tents as |
2008 | * |
2009 | * B |
2010 | * A 1 C 2 |
2011 | * 3 |
2012 | * |
2013 | * The matching-error highlighter described above will not mark |
2014 | * this construction as erroneous. However, something else will: |
2015 | * the three tents in the above diagram (let us suppose A,B,C |
2016 | * are the tents, though it doesn't matter which) contain two |
2017 | * diagonally adjacent pairs. So there will be _an_ error |
2018 | * highlighted for the above layout, even though not all types |
2019 | * of error will be highlighted. |
2020 | * |
2021 | * And in fact we can prove that this will always be the case: |
2022 | * that the shortcomings of the matching-error highlighter will |
2023 | * always be made up for by the easy tent adjacency highlighter. |
2024 | * |
2025 | * Lemma: Let G be a bipartite graph between n trees and n |
2026 | * tents, which is connected, and in which no tree has degree |
2027 | * more than two (but a tent may). Then G has a perfect matching. |
2028 | * |
2029 | * (Note: in the statement and proof of the Lemma I will |
2030 | * consistently use 'tree' to indicate a type of graph vertex as |
2031 | * opposed to a tent, and not to indicate a tree in the graph- |
2032 | * theoretic sense.) |
2033 | * |
2034 | * Proof: |
2035 | * |
2036 | * If we can find a tent of degree 1 joined to a tree of degree |
2037 | * 2, then any perfect matching must pair that tent with that |
2038 | * tree. Hence, we can remove both, leaving a smaller graph G' |
2039 | * which still satisfies all the conditions of the Lemma, and |
2040 | * which has a perfect matching iff G does. |
2041 | * |
2042 | * So, wlog, we may assume G contains no tent of degree 1 joined |
2043 | * to a tree of degree 2; if it does, we can reduce it as above. |
2044 | * |
2045 | * If G has no tent of degree 1 at all, then every tent has |
2046 | * degree at least two, so there are at least 2n edges in the |
2047 | * graph. But every tree has degree at most two, so there are at |
2048 | * most 2n edges. Hence there must be exactly 2n edges, so every |
2049 | * tree and every tent must have degree exactly two, which means |
2050 | * that the whole graph consists of a single loop (by |
2051 | * connectedness), and therefore certainly has a perfect |
2052 | * matching. |
2053 | * |
2054 | * Alternatively, if G does have a tent of degree 1 but it is |
2055 | * not connected to a tree of degree 2, then the tree it is |
2056 | * connected to must have degree 1 - and, by connectedness, that |
2057 | * must mean that that tent and that tree between them form the |
2058 | * entire graph. This trivial graph has a trivial perfect |
2059 | * matching. [] |
2060 | * |
2061 | * That proves the lemma. Hence, in any case where the matching- |
2062 | * error highlighter fails to highlight an erroneous component |
2063 | * (because it has the same number of tents as trees, but they |
2064 | * cannot be matched up), the above lemma tells us that there |
2065 | * must be a tree with degree more than 2, i.e. a tree |
2066 | * orthogonally adjacent to at least three tents. But in that |
2067 | * case, there must be some pair of those three tents which are |
2068 | * diagonally adjacent to each other, so the tent-adjacency |
2069 | * highlighter will necessarily show an error. So any filled |
2070 | * layout in Tents which is not a correct solution to the puzzle |
2071 | * must have _some_ error highlighted by the subroutine below. |
2072 | * |
2073 | * (Of course it would be nicer if we could highlight all |
2074 | * errors: in the above example layout, we would like to |
2075 | * highlight tents A,B as having too few trees between them, and |
2076 | * trees 2,3 as having too few tents, in addition to marking the |
2077 | * adjacency problems. But I can't immediately think of any way |
2078 | * to find the smallest sets of such tents and trees without an |
2079 | * O(2^N) loop over all subsets of a given component.) |
2080 | */ |
2081 | |
2082 | /* |
2a27ffcf |
2083 | * ret[0] through to ret[w*h-1] give error markers for the grid |
2084 | * squares. After that, ret[w*h] to ret[w*h+w-1] give error |
2085 | * markers for the column numbers, and ret[w*h+w] to |
2086 | * ret[w*h+w+h-1] for the row numbers. |
2087 | */ |
2088 | |
2089 | /* |
2090 | * Spot tent-adjacency violations. |
2091 | */ |
2092 | for (x = 0; x < w*h; x++) |
2093 | ret[x] = 0; |
2094 | for (y = 0; y < h; y++) { |
2095 | for (x = 0; x < w; x++) { |
2096 | if (y+1 < h && x+1 < w && |
271cf5c1 |
2097 | ((grid[y*w+x] == TENT && |
2098 | grid[(y+1)*w+(x+1)] == TENT) || |
2099 | (grid[(y+1)*w+x] == TENT && |
2100 | grid[y*w+(x+1)] == TENT))) { |
2a27ffcf |
2101 | ret[y*w+x] |= 1 << ERR_ADJ_BOTRIGHT; |
2102 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOPRIGHT; |
2103 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_BOTLEFT; |
2104 | ret[(y+1)*w+(x+1)] |= 1 << ERR_ADJ_TOPLEFT; |
2105 | } |
2106 | if (y+1 < h && |
271cf5c1 |
2107 | grid[y*w+x] == TENT && |
2108 | grid[(y+1)*w+x] == TENT) { |
2a27ffcf |
2109 | ret[y*w+x] |= 1 << ERR_ADJ_BOT; |
2110 | ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOP; |
2111 | } |
2112 | if (x+1 < w && |
271cf5c1 |
2113 | grid[y*w+x] == TENT && |
2114 | grid[y*w+(x+1)] == TENT) { |
2a27ffcf |
2115 | ret[y*w+x] |= 1 << ERR_ADJ_RIGHT; |
2116 | ret[y*w+(x+1)] |= 1 << ERR_ADJ_LEFT; |
2117 | } |
2118 | } |
2119 | } |
2120 | |
2121 | /* |
2122 | * Spot numeric clue violations. |
2123 | */ |
2124 | for (x = 0; x < w; x++) { |
2125 | int tents = 0, maybetents = 0; |
2126 | for (y = 0; y < h; y++) { |
271cf5c1 |
2127 | if (grid[y*w+x] == TENT) |
2a27ffcf |
2128 | tents++; |
271cf5c1 |
2129 | else if (grid[y*w+x] == BLANK) |
2a27ffcf |
2130 | maybetents++; |
2131 | } |
2132 | ret[w*h+x] = (tents > state->numbers->numbers[x] || |
2133 | tents + maybetents < state->numbers->numbers[x]); |
2134 | } |
2135 | for (y = 0; y < h; y++) { |
2136 | int tents = 0, maybetents = 0; |
2137 | for (x = 0; x < w; x++) { |
271cf5c1 |
2138 | if (grid[y*w+x] == TENT) |
2a27ffcf |
2139 | tents++; |
271cf5c1 |
2140 | else if (grid[y*w+x] == BLANK) |
2a27ffcf |
2141 | maybetents++; |
2142 | } |
2143 | ret[w*h+w+y] = (tents > state->numbers->numbers[w+y] || |
2144 | tents + maybetents < state->numbers->numbers[w+y]); |
2145 | } |
2146 | |
2147 | /* |
2148 | * Identify groups of tents with too few trees between them, |
2149 | * which we do by constructing the connected components of the |
2150 | * bipartite adjacency graph between tents and trees |
2151 | * ('bipartite' in the sense that we deliberately ignore |
2152 | * adjacency between tents or between trees), and highlighting |
2153 | * all the tents in any component which has a smaller tree |
2154 | * count. |
2155 | */ |
2156 | dsf_init(dsf, w*h); |
2157 | /* Construct the equivalence classes. */ |
2158 | for (y = 0; y < h; y++) { |
2159 | for (x = 0; x < w-1; x++) { |
271cf5c1 |
2160 | if ((grid[y*w+x] == TREE && grid[y*w+x+1] == TENT) || |
2161 | (grid[y*w+x] == TENT && grid[y*w+x+1] == TREE)) |
2a27ffcf |
2162 | dsf_merge(dsf, y*w+x, y*w+x+1); |
2163 | } |
2164 | } |
2165 | for (y = 0; y < h-1; y++) { |
2166 | for (x = 0; x < w; x++) { |
271cf5c1 |
2167 | if ((grid[y*w+x] == TREE && grid[(y+1)*w+x] == TENT) || |
2168 | (grid[y*w+x] == TENT && grid[(y+1)*w+x] == TREE)) |
2a27ffcf |
2169 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
2170 | } |
2171 | } |
2172 | /* Count up the tent/tree difference in each one. */ |
2173 | for (x = 0; x < w*h; x++) |
2174 | tmp[x] = 0; |
2175 | for (x = 0; x < w*h; x++) { |
2176 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2177 | if (grid[x] == TREE) |
2a27ffcf |
2178 | tmp[y]++; |
271cf5c1 |
2179 | else if (grid[x] == TENT) |
2a27ffcf |
2180 | tmp[y]--; |
2181 | } |
2182 | /* And highlight any tent belonging to an equivalence class with |
2183 | * a score less than zero. */ |
2184 | for (x = 0; x < w*h; x++) { |
2185 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2186 | if (grid[x] == TENT && tmp[y] < 0) |
2a27ffcf |
2187 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
2188 | } |
2189 | |
2190 | /* |
2191 | * Identify groups of trees with too few tents between them. |
2192 | * This is done similarly, except that we now count BLANK as |
2193 | * equivalent to TENT, i.e. we only highlight such trees when |
2194 | * the user hasn't even left _room_ to provide tents for them |
2195 | * all. (Otherwise, we'd highlight all trees red right at the |
2196 | * start of the game, before the user had done anything wrong!) |
2197 | */ |
2198 | #define TENT(x) ((x)==TENT || (x)==BLANK) |
2199 | dsf_init(dsf, w*h); |
2200 | /* Construct the equivalence classes. */ |
2201 | for (y = 0; y < h; y++) { |
2202 | for (x = 0; x < w-1; x++) { |
271cf5c1 |
2203 | if ((grid[y*w+x] == TREE && TENT(grid[y*w+x+1])) || |
2204 | (TENT(grid[y*w+x]) && grid[y*w+x+1] == TREE)) |
2a27ffcf |
2205 | dsf_merge(dsf, y*w+x, y*w+x+1); |
2206 | } |
2207 | } |
2208 | for (y = 0; y < h-1; y++) { |
2209 | for (x = 0; x < w; x++) { |
271cf5c1 |
2210 | if ((grid[y*w+x] == TREE && TENT(grid[(y+1)*w+x])) || |
2211 | (TENT(grid[y*w+x]) && grid[(y+1)*w+x] == TREE)) |
2a27ffcf |
2212 | dsf_merge(dsf, y*w+x, (y+1)*w+x); |
2213 | } |
2214 | } |
2215 | /* Count up the tent/tree difference in each one. */ |
2216 | for (x = 0; x < w*h; x++) |
2217 | tmp[x] = 0; |
2218 | for (x = 0; x < w*h; x++) { |
2219 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2220 | if (grid[x] == TREE) |
2a27ffcf |
2221 | tmp[y]++; |
271cf5c1 |
2222 | else if (TENT(grid[x])) |
2a27ffcf |
2223 | tmp[y]--; |
2224 | } |
2225 | /* And highlight any tree belonging to an equivalence class with |
2226 | * a score more than zero. */ |
2227 | for (x = 0; x < w*h; x++) { |
2228 | y = dsf_canonify(dsf, x); |
271cf5c1 |
2229 | if (grid[x] == TREE && tmp[y] > 0) |
2a27ffcf |
2230 | ret[x] |= 1 << ERR_OVERCOMMITTED; |
2231 | } |
2232 | #undef TENT |
2233 | |
2234 | sfree(tmp); |
2235 | return ret; |
2236 | } |
2237 | |
2238 | static void draw_err_adj(drawing *dr, game_drawstate *ds, int x, int y) |
2239 | { |
2240 | int coords[8]; |
2241 | int yext, xext; |
2242 | |
2243 | /* |
2244 | * Draw a diamond. |
2245 | */ |
2246 | coords[0] = x - TILESIZE*2/5; |
2247 | coords[1] = y; |
2248 | coords[2] = x; |
2249 | coords[3] = y - TILESIZE*2/5; |
2250 | coords[4] = x + TILESIZE*2/5; |
2251 | coords[5] = y; |
2252 | coords[6] = x; |
2253 | coords[7] = y + TILESIZE*2/5; |
2254 | draw_polygon(dr, coords, 4, COL_ERROR, COL_GRID); |
2255 | |
2256 | /* |
2257 | * Draw an exclamation mark in the diamond. This turns out to |
2258 | * look unpleasantly off-centre if done via draw_text, so I do |
2259 | * it by hand on the basis that exclamation marks aren't that |
2260 | * difficult to draw... |
2261 | */ |
2262 | xext = TILESIZE/16; |
2263 | yext = TILESIZE*2/5 - (xext*2+2); |
2264 | draw_rect(dr, x-xext, y-yext, xext*2+1, yext*2+1 - (xext*3), |
2265 | COL_ERRTEXT); |
2266 | draw_rect(dr, x-xext, y+yext-xext*2+1, xext*2+1, xext*2, COL_ERRTEXT); |
2267 | } |
2268 | |
86e60e3d |
2269 | static void draw_tile(drawing *dr, game_drawstate *ds, |
505ea4e5 |
2270 | int x, int y, int v, int cur, int printing) |
86e60e3d |
2271 | { |
2a27ffcf |
2272 | int err; |
86e60e3d |
2273 | int tx = COORD(x), ty = COORD(y); |
2274 | int cx = tx + TILESIZE/2, cy = ty + TILESIZE/2; |
2275 | |
2a27ffcf |
2276 | err = v & ~15; |
2277 | v &= 15; |
86e60e3d |
2278 | |
2a27ffcf |
2279 | clip(dr, tx, ty, TILESIZE, TILESIZE); |
2280 | |
2281 | if (!printing) { |
2282 | draw_rect(dr, tx, ty, TILESIZE, TILESIZE, COL_GRID); |
86e60e3d |
2283 | draw_rect(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1, |
2284 | (v == BLANK ? COL_BACKGROUND : COL_GRASS)); |
2a27ffcf |
2285 | } |
86e60e3d |
2286 | |
2287 | if (v == TREE) { |
2288 | int i; |
2289 | |
2290 | (printing ? draw_rect_outline : draw_rect) |
2291 | (dr, cx-TILESIZE/15, ty+TILESIZE*3/10, |
2292 | 2*(TILESIZE/15)+1, (TILESIZE*9/10 - TILESIZE*3/10), |
e8df451f |
2293 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERRTRUNK : COL_TREETRUNK)); |
86e60e3d |
2294 | |
2295 | for (i = 0; i < (printing ? 2 : 1); i++) { |
2a27ffcf |
2296 | int col = (i == 1 ? COL_BACKGROUND : |
2297 | (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : |
2298 | COL_TREELEAF)); |
86e60e3d |
2299 | int sub = i * (TILESIZE/32); |
2300 | draw_circle(dr, cx, ty+TILESIZE*4/10, TILESIZE/4 - sub, |
2301 | col, col); |
2302 | draw_circle(dr, cx+TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
2303 | col, col); |
2304 | draw_circle(dr, cx-TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub, |
2305 | col, col); |
2306 | draw_circle(dr, cx+TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
2307 | col, col); |
2308 | draw_circle(dr, cx-TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub, |
2309 | col, col); |
2310 | } |
2311 | } else if (v == TENT) { |
2312 | int coords[6]; |
2a27ffcf |
2313 | int col; |
86e60e3d |
2314 | coords[0] = cx - TILESIZE/3; |
2315 | coords[1] = cy + TILESIZE/3; |
2316 | coords[2] = cx + TILESIZE/3; |
2317 | coords[3] = cy + TILESIZE/3; |
2318 | coords[4] = cx; |
2319 | coords[5] = cy - TILESIZE/3; |
2a27ffcf |
2320 | col = (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : COL_TENT); |
2321 | draw_polygon(dr, coords, 3, (printing ? -1 : col), col); |
86e60e3d |
2322 | } |
2323 | |
2a27ffcf |
2324 | if (err & (1 << ERR_ADJ_TOPLEFT)) |
2325 | draw_err_adj(dr, ds, tx, ty); |
2326 | if (err & (1 << ERR_ADJ_TOP)) |
2327 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty); |
2328 | if (err & (1 << ERR_ADJ_TOPRIGHT)) |
2329 | draw_err_adj(dr, ds, tx+TILESIZE, ty); |
2330 | if (err & (1 << ERR_ADJ_LEFT)) |
2331 | draw_err_adj(dr, ds, tx, ty+TILESIZE/2); |
2332 | if (err & (1 << ERR_ADJ_RIGHT)) |
2333 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE/2); |
2334 | if (err & (1 << ERR_ADJ_BOTLEFT)) |
2335 | draw_err_adj(dr, ds, tx, ty+TILESIZE); |
2336 | if (err & (1 << ERR_ADJ_BOT)) |
2337 | draw_err_adj(dr, ds, tx+TILESIZE/2, ty+TILESIZE); |
2338 | if (err & (1 << ERR_ADJ_BOTRIGHT)) |
2339 | draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE); |
2340 | |
505ea4e5 |
2341 | if (cur) { |
2342 | int coff = TILESIZE/8; |
2343 | draw_rect_outline(dr, tx + coff, ty + coff, |
fe5d2401 |
2344 | TILESIZE - coff*2 + 1, TILESIZE - coff*2 + 1, |
2345 | COL_GRID); |
505ea4e5 |
2346 | } |
2347 | |
86e60e3d |
2348 | unclip(dr); |
2349 | draw_update(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1); |
2350 | } |
2351 | |
2352 | /* |
2353 | * Internal redraw function, used for printing as well as drawing. |
2354 | */ |
2355 | static void int_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
2356 | game_state *state, int dir, game_ui *ui, |
2357 | float animtime, float flashtime, int printing) |
2358 | { |
2359 | int w = state->p.w, h = state->p.h; |
2360 | int x, y, flashing; |
505ea4e5 |
2361 | int cx = -1, cy = -1; |
2362 | int cmoved = 0; |
271cf5c1 |
2363 | char *tmpgrid; |
2a27ffcf |
2364 | int *errors; |
505ea4e5 |
2365 | |
2366 | if (ui) { |
2367 | if (ui->cdisp) { cx = ui->cx; cy = ui->cy; } |
2368 | if (cx != ds->cx || cy != ds->cy) cmoved = 1; |
2369 | } |
86e60e3d |
2370 | |
2371 | if (printing || !ds->started) { |
2372 | if (!printing) { |
2373 | int ww, wh; |
2374 | game_compute_size(&state->p, TILESIZE, &ww, &wh); |
2375 | draw_rect(dr, 0, 0, ww, wh, COL_BACKGROUND); |
2376 | draw_update(dr, 0, 0, ww, wh); |
2377 | ds->started = TRUE; |
2378 | } |
2379 | |
2380 | if (printing) |
2381 | print_line_width(dr, TILESIZE/64); |
2382 | |
2383 | /* |
2384 | * Draw the grid. |
2385 | */ |
2386 | for (y = 0; y <= h; y++) |
2387 | draw_line(dr, COORD(0), COORD(y), COORD(w), COORD(y), COL_GRID); |
2388 | for (x = 0; x <= w; x++) |
2389 | draw_line(dr, COORD(x), COORD(0), COORD(x), COORD(h), COL_GRID); |
86e60e3d |
2390 | } |
2391 | |
2392 | if (flashtime > 0) |
2393 | flashing = (int)(flashtime * 3 / FLASH_TIME) != 1; |
2394 | else |
2395 | flashing = FALSE; |
2396 | |
2397 | /* |
271cf5c1 |
2398 | * Find errors. For this we use _part_ of the information from a |
2399 | * currently active drag: we transform dsx,dsy but not anything |
2400 | * else. (This seems to strike a good compromise between having |
2401 | * the error highlights respond instantly to single clicks, but |
26dac649 |
2402 | * not giving constant feedback during a right-drag.) |
2a27ffcf |
2403 | */ |
271cf5c1 |
2404 | if (ui && ui->drag_button >= 0) { |
2405 | tmpgrid = snewn(w*h, char); |
2406 | memcpy(tmpgrid, state->grid, w*h); |
2407 | tmpgrid[ui->dsy * w + ui->dsx] = |
2408 | drag_xform(ui, ui->dsx, ui->dsy, tmpgrid[ui->dsy * w + ui->dsx]); |
2409 | errors = find_errors(state, tmpgrid); |
2410 | sfree(tmpgrid); |
2411 | } else { |
2412 | errors = find_errors(state, state->grid); |
2413 | } |
2a27ffcf |
2414 | |
2415 | /* |
86e60e3d |
2416 | * Draw the grid. |
2417 | */ |
2a27ffcf |
2418 | for (y = 0; y < h; y++) { |
86e60e3d |
2419 | for (x = 0; x < w; x++) { |
2420 | int v = state->grid[y*w+x]; |
505ea4e5 |
2421 | int credraw = 0; |
86e60e3d |
2422 | |
565394e7 |
2423 | /* |
2424 | * We deliberately do not take drag_ok into account |
2425 | * here, because user feedback suggests that it's |
2426 | * marginally nicer not to have the drag effects |
2427 | * flickering on and off disconcertingly. |
2428 | */ |
5bcb1aa3 |
2429 | if (ui && ui->drag_button >= 0) |
565394e7 |
2430 | v = drag_xform(ui, x, y, v); |
2431 | |
86e60e3d |
2432 | if (flashing && (v == TREE || v == TENT)) |
2433 | v = NONTENT; |
2434 | |
505ea4e5 |
2435 | if (cmoved) { |
2436 | if ((x == cx && y == cy) || |
2437 | (x == ds->cx && y == ds->cy)) credraw = 1; |
2438 | } |
2439 | |
2a27ffcf |
2440 | v |= errors[y*w+x]; |
2441 | |
505ea4e5 |
2442 | if (printing || ds->drawn[y*w+x] != v || credraw) { |
2443 | draw_tile(dr, ds, x, y, v, (x == cx && y == cy), printing); |
86e60e3d |
2444 | if (!printing) |
2445 | ds->drawn[y*w+x] = v; |
2446 | } |
2447 | } |
2a27ffcf |
2448 | } |
2449 | |
2450 | /* |
2451 | * Draw (or redraw, if their error-highlighted state has |
2452 | * changed) the numbers. |
2453 | */ |
2454 | for (x = 0; x < w; x++) { |
2455 | if (ds->numbersdrawn[x] != errors[w*h+x]) { |
2456 | char buf[80]; |
2457 | draw_rect(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1, |
2458 | COL_BACKGROUND); |
2459 | sprintf(buf, "%d", state->numbers->numbers[x]); |
2460 | draw_text(dr, COORD(x) + TILESIZE/2, COORD(h+1), |
2461 | FONT_VARIABLE, TILESIZE/2, ALIGN_HCENTRE|ALIGN_VNORMAL, |
2462 | (errors[w*h+x] ? COL_ERROR : COL_GRID), buf); |
2463 | draw_update(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1); |
2464 | ds->numbersdrawn[x] = errors[w*h+x]; |
2465 | } |
2466 | } |
2467 | for (y = 0; y < h; y++) { |
2468 | if (ds->numbersdrawn[w+y] != errors[w*h+w+y]) { |
2469 | char buf[80]; |
2470 | draw_rect(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE, |
2471 | COL_BACKGROUND); |
2472 | sprintf(buf, "%d", state->numbers->numbers[w+y]); |
2473 | draw_text(dr, COORD(w+1), COORD(y) + TILESIZE/2, |
2474 | FONT_VARIABLE, TILESIZE/2, ALIGN_HRIGHT|ALIGN_VCENTRE, |
2475 | (errors[w*h+w+y] ? COL_ERROR : COL_GRID), buf); |
2476 | draw_update(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE); |
2477 | ds->numbersdrawn[w+y] = errors[w*h+w+y]; |
2478 | } |
2479 | } |
2480 | |
2481 | if (cmoved) { |
2482 | ds->cx = cx; |
2483 | ds->cy = cy; |
2484 | } |
2485 | |
2486 | sfree(errors); |
86e60e3d |
2487 | } |
2488 | |
2489 | static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, |
2490 | game_state *state, int dir, game_ui *ui, |
2491 | float animtime, float flashtime) |
2492 | { |
2493 | int_redraw(dr, ds, oldstate, state, dir, ui, animtime, flashtime, FALSE); |
2494 | } |
2495 | |
2496 | static float game_anim_length(game_state *oldstate, game_state *newstate, |
2497 | int dir, game_ui *ui) |
2498 | { |
2499 | return 0.0F; |
2500 | } |
2501 | |
2502 | static float game_flash_length(game_state *oldstate, game_state *newstate, |
2503 | int dir, game_ui *ui) |
2504 | { |
2505 | if (!oldstate->completed && newstate->completed && |
2506 | !oldstate->used_solve && !newstate->used_solve) |
2507 | return FLASH_TIME; |
2508 | |
2509 | return 0.0F; |
2510 | } |
2511 | |
86e60e3d |
2512 | static int game_timing_state(game_state *state, game_ui *ui) |
2513 | { |
2514 | return TRUE; |
2515 | } |
2516 | |
2517 | static void game_print_size(game_params *params, float *x, float *y) |
2518 | { |
2519 | int pw, ph; |
2520 | |
2521 | /* |
2522 | * I'll use 6mm squares by default. |
2523 | */ |
2524 | game_compute_size(params, 600, &pw, &ph); |
505ea4e5 |
2525 | *x = pw / 100.0F; |
2526 | *y = ph / 100.0F; |
86e60e3d |
2527 | } |
2528 | |
2529 | static void game_print(drawing *dr, game_state *state, int tilesize) |
2530 | { |
2531 | int c; |
2532 | |
2533 | /* Ick: fake up `ds->tilesize' for macro expansion purposes */ |
2534 | game_drawstate ads, *ds = &ads; |
2535 | game_set_size(dr, ds, NULL, tilesize); |
2536 | |
2537 | c = print_mono_colour(dr, 1); assert(c == COL_BACKGROUND); |
2538 | c = print_mono_colour(dr, 0); assert(c == COL_GRID); |
2539 | c = print_mono_colour(dr, 1); assert(c == COL_GRASS); |
2540 | c = print_mono_colour(dr, 0); assert(c == COL_TREETRUNK); |
2541 | c = print_mono_colour(dr, 0); assert(c == COL_TREELEAF); |
2542 | c = print_mono_colour(dr, 0); assert(c == COL_TENT); |
2543 | |
2544 | int_redraw(dr, ds, NULL, state, +1, NULL, 0.0F, 0.0F, TRUE); |
2545 | } |
2546 | |
2547 | #ifdef COMBINED |
2548 | #define thegame tents |
2549 | #endif |
2550 | |
2551 | const struct game thegame = { |
750037d7 |
2552 | "Tents", "games.tents", "tents", |
86e60e3d |
2553 | default_params, |
2554 | game_fetch_preset, |
2555 | decode_params, |
2556 | encode_params, |
2557 | free_params, |
2558 | dup_params, |
2559 | TRUE, game_configure, custom_params, |
2560 | validate_params, |
2561 | new_game_desc, |
2562 | validate_desc, |
2563 | new_game, |
2564 | dup_game, |
2565 | free_game, |
2566 | TRUE, solve_game, |
fa3abef5 |
2567 | FALSE, game_can_format_as_text_now, game_text_format, |
86e60e3d |
2568 | new_ui, |
2569 | free_ui, |
2570 | encode_ui, |
2571 | decode_ui, |
2572 | game_changed_state, |
2573 | interpret_move, |
2574 | execute_move, |
2575 | PREFERRED_TILESIZE, game_compute_size, game_set_size, |
2576 | game_colours, |
2577 | game_new_drawstate, |
2578 | game_free_drawstate, |
2579 | game_redraw, |
2580 | game_anim_length, |
2581 | game_flash_length, |
2582 | TRUE, FALSE, game_print_size, game_print, |
ac9f41c4 |
2583 | FALSE, /* wants_statusbar */ |
86e60e3d |
2584 | FALSE, game_timing_state, |
cb0c7d4a |
2585 | REQUIRE_RBUTTON, /* flags */ |
86e60e3d |
2586 | }; |
2587 | |
2588 | #ifdef STANDALONE_SOLVER |
2589 | |
2590 | #include <stdarg.h> |
2591 | |
2592 | int main(int argc, char **argv) |
2593 | { |
2594 | game_params *p; |
2595 | game_state *s, *s2; |
2596 | char *id = NULL, *desc, *err; |
2597 | int grade = FALSE; |
2598 | int ret, diff, really_verbose = FALSE; |
2599 | struct solver_scratch *sc; |
2600 | |
2601 | while (--argc > 0) { |
2602 | char *p = *++argv; |
2603 | if (!strcmp(p, "-v")) { |
2604 | really_verbose = TRUE; |
2605 | } else if (!strcmp(p, "-g")) { |
2606 | grade = TRUE; |
2607 | } else if (*p == '-') { |
2608 | fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); |
2609 | return 1; |
2610 | } else { |
2611 | id = p; |
2612 | } |
2613 | } |
2614 | |
2615 | if (!id) { |
2616 | fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); |
2617 | return 1; |
2618 | } |
2619 | |
2620 | desc = strchr(id, ':'); |
2621 | if (!desc) { |
2622 | fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); |
2623 | return 1; |
2624 | } |
2625 | *desc++ = '\0'; |
2626 | |
2627 | p = default_params(); |
2628 | decode_params(p, id); |
2629 | err = validate_desc(p, desc); |
2630 | if (err) { |
2631 | fprintf(stderr, "%s: %s\n", argv[0], err); |
2632 | return 1; |
2633 | } |
2634 | s = new_game(NULL, p, desc); |
2635 | s2 = new_game(NULL, p, desc); |
2636 | |
2637 | sc = new_scratch(p->w, p->h); |
2638 | |
2639 | /* |
2640 | * When solving an Easy puzzle, we don't want to bother the |
2641 | * user with Hard-level deductions. For this reason, we grade |
2642 | * the puzzle internally before doing anything else. |
2643 | */ |
2644 | ret = -1; /* placate optimiser */ |
2645 | for (diff = 0; diff < DIFFCOUNT; diff++) { |
2646 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2647 | s2->grid, sc, diff); |
2648 | if (ret < 2) |
2649 | break; |
2650 | } |
2651 | |
2652 | if (diff == DIFFCOUNT) { |
2653 | if (grade) |
2654 | printf("Difficulty rating: too hard to solve internally\n"); |
2655 | else |
2656 | printf("Unable to find a unique solution\n"); |
2657 | } else { |
2658 | if (grade) { |
2659 | if (ret == 0) |
2660 | printf("Difficulty rating: impossible (no solution exists)\n"); |
2661 | else if (ret == 1) |
2662 | printf("Difficulty rating: %s\n", tents_diffnames[diff]); |
2663 | } else { |
2664 | verbose = really_verbose; |
2665 | ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers, |
2666 | s2->grid, sc, diff); |
2667 | if (ret == 0) |
2668 | printf("Puzzle is inconsistent\n"); |
2669 | else |
2670 | fputs(game_text_format(s2), stdout); |
2671 | } |
2672 | } |
2673 | |
2674 | return 0; |
2675 | } |
2676 | |
2677 | #endif |
505ea4e5 |
2678 | |
2679 | /* vim: set shiftwidth=4 tabstop=8: */ |