b760b8bd |
1 | /* |
2 | * loopgen.c: loop generation functions for grid.[ch]. |
3 | */ |
4 | |
5 | #include <stdio.h> |
6 | #include <stdlib.h> |
7 | #include <stddef.h> |
8 | #include <string.h> |
9 | #include <assert.h> |
10 | #include <ctype.h> |
11 | #include <math.h> |
12 | |
13 | #include "puzzles.h" |
14 | #include "tree234.h" |
15 | #include "grid.h" |
16 | #include "loopgen.h" |
17 | |
18 | |
19 | /* We're going to store lists of current candidate faces for colouring black |
20 | * or white. |
21 | * Each face gets a 'score', which tells us how adding that face right |
22 | * now would affect the curliness of the solution loop. We're trying to |
23 | * maximise that quantity so will bias our random selection of faces to |
24 | * colour those with high scores */ |
25 | struct face_score { |
26 | int white_score; |
27 | int black_score; |
28 | unsigned long random; |
29 | /* No need to store a grid_face* here. The 'face_scores' array will |
30 | * be a list of 'face_score' objects, one for each face of the grid, so |
31 | * the position (index) within the 'face_scores' array will determine |
32 | * which face corresponds to a particular face_score. |
33 | * Having a single 'face_scores' array for all faces simplifies memory |
34 | * management, and probably improves performance, because we don't have to |
35 | * malloc/free each individual face_score, and we don't have to maintain |
36 | * a mapping from grid_face* pointers to face_score* pointers. |
37 | */ |
38 | }; |
39 | |
40 | static int generic_sort_cmpfn(void *v1, void *v2, size_t offset) |
41 | { |
42 | struct face_score *f1 = v1; |
43 | struct face_score *f2 = v2; |
44 | int r; |
45 | |
46 | r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset); |
47 | if (r) { |
48 | return r; |
49 | } |
50 | |
51 | if (f1->random < f2->random) |
52 | return -1; |
53 | else if (f1->random > f2->random) |
54 | return 1; |
55 | |
56 | /* |
57 | * It's _just_ possible that two faces might have been given |
58 | * the same random value. In that situation, fall back to |
59 | * comparing based on the positions within the face_scores list. |
60 | * This introduces a tiny directional bias, but not a significant one. |
61 | */ |
62 | return f1 - f2; |
63 | } |
64 | |
65 | static int white_sort_cmpfn(void *v1, void *v2) |
66 | { |
67 | return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score)); |
68 | } |
69 | |
70 | static int black_sort_cmpfn(void *v1, void *v2) |
71 | { |
72 | return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score)); |
73 | } |
74 | |
75 | /* 'board' is an array of enum face_colour, indicating which faces are |
76 | * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK. |
77 | * Returns whether it's legal to colour the given face with this colour. */ |
78 | static int can_colour_face(grid *g, char* board, int face_index, |
79 | enum face_colour colour) |
80 | { |
81 | int i, j; |
82 | grid_face *test_face = g->faces + face_index; |
83 | grid_face *starting_face, *current_face; |
84 | grid_dot *starting_dot; |
85 | int transitions; |
86 | int current_state, s; /* booleans: equal or not-equal to 'colour' */ |
87 | int found_same_coloured_neighbour = FALSE; |
88 | assert(board[face_index] != colour); |
89 | |
90 | /* Can only consider a face for colouring if it's adjacent to a face |
91 | * with the same colour. */ |
92 | for (i = 0; i < test_face->order; i++) { |
93 | grid_edge *e = test_face->edges[i]; |
94 | grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1; |
95 | if (FACE_COLOUR(f) == colour) { |
96 | found_same_coloured_neighbour = TRUE; |
97 | break; |
98 | } |
99 | } |
100 | if (!found_same_coloured_neighbour) |
101 | return FALSE; |
102 | |
103 | /* Need to avoid creating a loop of faces of this colour around some |
104 | * differently-coloured faces. |
105 | * Also need to avoid meeting a same-coloured face at a corner, with |
106 | * other-coloured faces in between. Here's a simple test that (I believe) |
107 | * takes care of both these conditions: |
108 | * |
109 | * Take the circular path formed by this face's edges, and inflate it |
110 | * slightly outwards. Imagine walking around this path and consider |
111 | * the faces that you visit in sequence. This will include all faces |
112 | * touching the given face, either along an edge or just at a corner. |
113 | * Count the number of 'colour'/not-'colour' transitions you encounter, as |
114 | * you walk along the complete loop. This will obviously turn out to be |
115 | * an even number. |
116 | * If 0, we're either in the middle of an "island" of this colour (should |
117 | * be impossible as we're not supposed to create black or white loops), |
118 | * or we're about to start a new island - also not allowed. |
119 | * If 4 or greater, there are too many separate coloured regions touching |
120 | * this face, and colouring it would create a loop or a corner-violation. |
121 | * The only allowed case is when the count is exactly 2. */ |
122 | |
123 | /* i points to a dot around the test face. |
124 | * j points to a face around the i^th dot. |
125 | * The current face will always be: |
126 | * test_face->dots[i]->faces[j] |
127 | * We assume dots go clockwise around the test face, |
128 | * and faces go clockwise around dots. */ |
129 | |
130 | /* |
131 | * The end condition is slightly fiddly. In sufficiently strange |
132 | * degenerate grids, our test face may be adjacent to the same |
133 | * other face multiple times (typically if it's the exterior |
134 | * face). Consider this, in particular: |
135 | * |
136 | * +--+ |
137 | * | | |
138 | * +--+--+ |
139 | * | | | |
140 | * +--+--+ |
141 | * |
142 | * The bottom left face there is adjacent to the exterior face |
143 | * twice, so we can't just terminate our iteration when we reach |
144 | * the same _face_ we started at. Furthermore, we can't |
145 | * condition on having the same (i,j) pair either, because |
146 | * several (i,j) pairs identify the bottom left contiguity with |
147 | * the exterior face! We canonicalise the (i,j) pair by taking |
148 | * one step around before we set the termination tracking. |
149 | */ |
150 | |
151 | i = j = 0; |
152 | current_face = test_face->dots[0]->faces[0]; |
153 | if (current_face == test_face) { |
154 | j = 1; |
155 | current_face = test_face->dots[0]->faces[1]; |
156 | } |
157 | transitions = 0; |
158 | current_state = (FACE_COLOUR(current_face) == colour); |
159 | starting_dot = NULL; |
160 | starting_face = NULL; |
161 | while (TRUE) { |
162 | /* Advance to next face. |
163 | * Need to loop here because it might take several goes to |
164 | * find it. */ |
165 | while (TRUE) { |
166 | j++; |
167 | if (j == test_face->dots[i]->order) |
168 | j = 0; |
169 | |
170 | if (test_face->dots[i]->faces[j] == test_face) { |
171 | /* Advance to next dot round test_face, then |
172 | * find current_face around new dot |
173 | * and advance to the next face clockwise */ |
174 | i++; |
175 | if (i == test_face->order) |
176 | i = 0; |
177 | for (j = 0; j < test_face->dots[i]->order; j++) { |
178 | if (test_face->dots[i]->faces[j] == current_face) |
179 | break; |
180 | } |
181 | /* Must actually find current_face around new dot, |
182 | * or else something's wrong with the grid. */ |
183 | assert(j != test_face->dots[i]->order); |
184 | /* Found, so advance to next face and try again */ |
185 | } else { |
186 | break; |
187 | } |
188 | } |
189 | /* (i,j) are now advanced to next face */ |
190 | current_face = test_face->dots[i]->faces[j]; |
191 | s = (FACE_COLOUR(current_face) == colour); |
192 | if (!starting_dot) { |
193 | starting_dot = test_face->dots[i]; |
194 | starting_face = current_face; |
195 | current_state = s; |
196 | } else { |
197 | if (s != current_state) { |
198 | ++transitions; |
199 | current_state = s; |
200 | if (transitions > 2) |
201 | break; |
202 | } |
203 | if (test_face->dots[i] == starting_dot && |
204 | current_face == starting_face) |
205 | break; |
206 | } |
207 | } |
208 | |
209 | return (transitions == 2) ? TRUE : FALSE; |
210 | } |
211 | |
212 | /* Count the number of neighbours of 'face', having colour 'colour' */ |
213 | static int face_num_neighbours(grid *g, char *board, grid_face *face, |
214 | enum face_colour colour) |
215 | { |
216 | int colour_count = 0; |
217 | int i; |
218 | grid_face *f; |
219 | grid_edge *e; |
220 | for (i = 0; i < face->order; i++) { |
221 | e = face->edges[i]; |
222 | f = (e->face1 == face) ? e->face2 : e->face1; |
223 | if (FACE_COLOUR(f) == colour) |
224 | ++colour_count; |
225 | } |
226 | return colour_count; |
227 | } |
228 | |
229 | /* The 'score' of a face reflects its current desirability for selection |
230 | * as the next face to colour white or black. We want to encourage moving |
231 | * into grey areas and increasing loopiness, so we give scores according to |
232 | * how many of the face's neighbours are currently coloured the same as the |
233 | * proposed colour. */ |
234 | static int face_score(grid *g, char *board, grid_face *face, |
235 | enum face_colour colour) |
236 | { |
237 | /* Simple formula: score = 0 - num. same-coloured neighbours, |
238 | * so a higher score means fewer same-coloured neighbours. */ |
239 | return -face_num_neighbours(g, board, face, colour); |
240 | } |
241 | |
242 | /* |
243 | * Generate a new complete random closed loop for the given grid. |
244 | * |
245 | * The method is to generate a WHITE/BLACK colouring of all the faces, |
246 | * such that the WHITE faces will define the inside of the path, and the |
247 | * BLACK faces define the outside. |
248 | * To do this, we initially colour all faces GREY. The infinite space outside |
249 | * the grid is coloured BLACK, and we choose a random face to colour WHITE. |
250 | * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY |
251 | * faces, until the grid is filled with BLACK/WHITE. As we grow the regions, |
252 | * we avoid creating loops of a single colour, to preserve the topological |
253 | * shape of the WHITE and BLACK regions. |
254 | * We also try to make the boundary as loopy and twisty as possible, to avoid |
255 | * generating paths that are uninteresting. |
256 | * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY |
257 | * face that can be coloured with that colour (without violating the |
258 | * topological shape of that region). It's not obvious, but I think this |
259 | * algorithm is guaranteed to terminate without leaving any GREY faces behind. |
260 | * Indeed, if there are any GREY faces at all, both the WHITE and BLACK |
261 | * regions can be grown. |
262 | * This is checked using assert()ions, and I haven't seen any failures yet. |
263 | * |
264 | * Hand-wavy proof: imagine what can go wrong... |
265 | * |
266 | * Could the white faces get completely cut off by the black faces, and still |
267 | * leave some grey faces remaining? |
268 | * No, because then the black faces would form a loop around both the white |
269 | * faces and the grey faces, which is disallowed because we continually |
270 | * maintain the correct topological shape of the black region. |
271 | * Similarly, the black faces can never get cut off by the white faces. That |
272 | * means both the WHITE and BLACK regions always have some room to grow into |
273 | * the GREY regions. |
274 | * Could it be that we can't colour some GREY face, because there are too many |
275 | * WHITE/BLACK transitions as we walk round the face? (see the |
276 | * can_colour_face() function for details) |
277 | * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk |
278 | * around the face. The two WHITE faces would be connected by a WHITE path, |
279 | * and the BLACK faces would be connected by a BLACK path. These paths would |
280 | * have to cross, which is impossible. |
281 | * Another thing that could go wrong: perhaps we can't find any GREY face to |
282 | * colour WHITE, because it would create a loop-violation or a corner-violation |
283 | * with the other WHITE faces? |
284 | * This is a little bit tricky to prove impossible. Imagine you have such a |
285 | * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop |
286 | * or corner violation). |
287 | * That would cut all the non-white area into two blobs. One of those blobs |
288 | * must be free of BLACK faces (because the BLACK stuff is a connected blob). |
289 | * So we have a connected GREY area, completely surrounded by WHITE |
290 | * (including the GREY face we've tentatively coloured WHITE). |
291 | * A well-known result in graph theory says that you can always find a GREY |
292 | * face whose removal leaves the remaining GREY area connected. And it says |
293 | * there are at least two such faces, so we can always choose the one that |
294 | * isn't the "tentative" GREY face. Colouring that face WHITE leaves |
295 | * everything nice and connected, including that "tentative" GREY face which |
296 | * acts as a gateway to the rest of the non-WHITE grid. |
297 | */ |
298 | void generate_loop(grid *g, char *board, random_state *rs, |
299 | loopgen_bias_fn_t bias, void *biasctx) |
300 | { |
301 | int i, j; |
302 | int num_faces = g->num_faces; |
303 | struct face_score *face_scores; /* Array of face_score objects */ |
304 | struct face_score *fs; /* Points somewhere in the above list */ |
305 | struct grid_face *cur_face; |
306 | tree234 *lightable_faces_sorted; |
307 | tree234 *darkable_faces_sorted; |
308 | int *face_list; |
309 | int do_random_pass; |
310 | |
311 | /* Make a board */ |
312 | memset(board, FACE_GREY, num_faces); |
313 | |
314 | /* Create and initialise the list of face_scores */ |
315 | face_scores = snewn(num_faces, struct face_score); |
316 | for (i = 0; i < num_faces; i++) { |
317 | face_scores[i].random = random_bits(rs, 31); |
318 | face_scores[i].black_score = face_scores[i].white_score = 0; |
319 | } |
320 | |
321 | /* Colour a random, finite face white. The infinite face is implicitly |
322 | * coloured black. Together, they will seed the random growth process |
323 | * for the black and white areas. */ |
324 | i = random_upto(rs, num_faces); |
325 | board[i] = FACE_WHITE; |
326 | |
327 | /* We need a way of favouring faces that will increase our loopiness. |
328 | * We do this by maintaining a list of all candidate faces sorted by |
329 | * their score and choose randomly from that with appropriate skew. |
330 | * In order to avoid consistently biasing towards particular faces, we |
331 | * need the sort order _within_ each group of scores to be completely |
332 | * random. But it would be abusing the hospitality of the tree234 data |
333 | * structure if our comparison function were nondeterministic :-). So with |
334 | * each face we associate a random number that does not change during a |
335 | * particular run of the generator, and use that as a secondary sort key. |
336 | * Yes, this means we will be biased towards particular random faces in |
337 | * any one run but that doesn't actually matter. */ |
338 | |
339 | lightable_faces_sorted = newtree234(white_sort_cmpfn); |
340 | darkable_faces_sorted = newtree234(black_sort_cmpfn); |
341 | |
342 | /* Initialise the lists of lightable and darkable faces. This is |
343 | * slightly different from the code inside the while-loop, because we need |
344 | * to check every face of the board (the grid structure does not keep a |
345 | * list of the infinite face's neighbours). */ |
346 | for (i = 0; i < num_faces; i++) { |
347 | grid_face *f = g->faces + i; |
348 | struct face_score *fs = face_scores + i; |
349 | if (board[i] != FACE_GREY) continue; |
350 | /* We need the full colourability check here, it's not enough simply |
351 | * to check neighbourhood. On some grids, a neighbour of the infinite |
352 | * face is not necessarily darkable. */ |
353 | if (can_colour_face(g, board, i, FACE_BLACK)) { |
354 | fs->black_score = face_score(g, board, f, FACE_BLACK); |
355 | add234(darkable_faces_sorted, fs); |
356 | } |
357 | if (can_colour_face(g, board, i, FACE_WHITE)) { |
358 | fs->white_score = face_score(g, board, f, FACE_WHITE); |
359 | add234(lightable_faces_sorted, fs); |
360 | } |
361 | } |
362 | |
363 | /* Colour faces one at a time until no more faces are colourable. */ |
364 | while (TRUE) |
365 | { |
366 | enum face_colour colour; |
367 | tree234 *faces_to_pick; |
368 | int c_lightable = count234(lightable_faces_sorted); |
369 | int c_darkable = count234(darkable_faces_sorted); |
370 | if (c_lightable == 0 && c_darkable == 0) { |
371 | /* No more faces we can use at all. */ |
372 | break; |
373 | } |
374 | assert(c_lightable != 0 && c_darkable != 0); |
375 | |
376 | /* Choose a colour, and colour the best available face |
377 | * with that colour. */ |
378 | colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK; |
379 | |
380 | if (colour == FACE_WHITE) |
381 | faces_to_pick = lightable_faces_sorted; |
382 | else |
383 | faces_to_pick = darkable_faces_sorted; |
384 | if (bias) { |
385 | /* |
386 | * Go through all the candidate faces and pick the one the |
387 | * bias function likes best, breaking ties using the |
388 | * ordering in our tree234 (which is why we replace only |
389 | * if score > bestscore, not >=). |
390 | */ |
391 | int j, k; |
392 | struct face_score *best = NULL; |
393 | int score, bestscore = 0; |
394 | |
395 | for (j = 0; |
396 | (fs = (struct face_score *)index234(faces_to_pick, j))!=NULL; |
397 | j++) { |
398 | |
399 | assert(fs); |
400 | k = fs - face_scores; |
401 | assert(board[k] == FACE_GREY); |
402 | board[k] = colour; |
403 | score = bias(biasctx, board, k); |
404 | board[k] = FACE_GREY; |
405 | bias(biasctx, board, k); /* let bias know we put it back */ |
406 | |
407 | if (!best || score > bestscore) { |
408 | bestscore = score; |
409 | best = fs; |
410 | } |
411 | } |
412 | fs = best; |
413 | } else { |
414 | fs = (struct face_score *)index234(faces_to_pick, 0); |
415 | } |
416 | assert(fs); |
417 | i = fs - face_scores; |
418 | assert(board[i] == FACE_GREY); |
419 | board[i] = colour; |
420 | if (bias) |
421 | bias(biasctx, board, i); /* notify bias function of the change */ |
422 | |
423 | /* Remove this newly-coloured face from the lists. These lists should |
424 | * only contain grey faces. */ |
425 | del234(lightable_faces_sorted, fs); |
426 | del234(darkable_faces_sorted, fs); |
427 | |
428 | /* Remember which face we've just coloured */ |
429 | cur_face = g->faces + i; |
430 | |
431 | /* The face we've just coloured potentially affects the colourability |
432 | * and the scores of any neighbouring faces (touching at a corner or |
433 | * edge). So the search needs to be conducted around all faces |
434 | * touching the one we've just lit. Iterate over its corners, then |
435 | * over each corner's faces. For each such face, we remove it from |
436 | * the lists, recalculate any scores, then add it back to the lists |
437 | * (depending on whether it is lightable, darkable or both). */ |
438 | for (i = 0; i < cur_face->order; i++) { |
439 | grid_dot *d = cur_face->dots[i]; |
440 | for (j = 0; j < d->order; j++) { |
441 | grid_face *f = d->faces[j]; |
442 | int fi; /* face index of f */ |
443 | |
444 | if (f == NULL) |
445 | continue; |
446 | if (f == cur_face) |
447 | continue; |
448 | |
449 | /* If the face is already coloured, it won't be on our |
450 | * lightable/darkable lists anyway, so we can skip it without |
451 | * bothering with the removal step. */ |
452 | if (FACE_COLOUR(f) != FACE_GREY) continue; |
453 | |
454 | /* Find the face index and face_score* corresponding to f */ |
455 | fi = f - g->faces; |
456 | fs = face_scores + fi; |
457 | |
458 | /* Remove from lightable list if it's in there. We do this, |
459 | * even if it is still lightable, because the score might |
460 | * be different, and we need to remove-then-add to maintain |
461 | * correct sort order. */ |
462 | del234(lightable_faces_sorted, fs); |
463 | if (can_colour_face(g, board, fi, FACE_WHITE)) { |
464 | fs->white_score = face_score(g, board, f, FACE_WHITE); |
465 | add234(lightable_faces_sorted, fs); |
466 | } |
467 | /* Do the same for darkable list. */ |
468 | del234(darkable_faces_sorted, fs); |
469 | if (can_colour_face(g, board, fi, FACE_BLACK)) { |
470 | fs->black_score = face_score(g, board, f, FACE_BLACK); |
471 | add234(darkable_faces_sorted, fs); |
472 | } |
473 | } |
474 | } |
475 | } |
476 | |
477 | /* Clean up */ |
478 | freetree234(lightable_faces_sorted); |
479 | freetree234(darkable_faces_sorted); |
480 | sfree(face_scores); |
481 | |
482 | /* The next step requires a shuffled list of all faces */ |
483 | face_list = snewn(num_faces, int); |
484 | for (i = 0; i < num_faces; ++i) { |
485 | face_list[i] = i; |
486 | } |
487 | shuffle(face_list, num_faces, sizeof(int), rs); |
488 | |
489 | /* The above loop-generation algorithm can often leave large clumps |
490 | * of faces of one colour. In extreme cases, the resulting path can be |
491 | * degenerate and not very satisfying to solve. |
492 | * This next step alleviates this problem: |
493 | * Go through the shuffled list, and flip the colour of any face we can |
494 | * legally flip, and which is adjacent to only one face of the opposite |
495 | * colour - this tends to grow 'tendrils' into any clumps. |
496 | * Repeat until we can find no more faces to flip. This will |
497 | * eventually terminate, because each flip increases the loop's |
498 | * perimeter, which cannot increase for ever. |
499 | * The resulting path will have maximal loopiness (in the sense that it |
500 | * cannot be improved "locally". Unfortunately, this allows a player to |
501 | * make some illicit deductions. To combat this (and make the path more |
502 | * interesting), we do one final pass making random flips. */ |
503 | |
504 | /* Set to TRUE for final pass */ |
505 | do_random_pass = FALSE; |
506 | |
507 | while (TRUE) { |
508 | /* Remember whether a flip occurred during this pass */ |
509 | int flipped = FALSE; |
510 | |
511 | for (i = 0; i < num_faces; ++i) { |
512 | int j = face_list[i]; |
513 | enum face_colour opp = |
514 | (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE; |
515 | if (can_colour_face(g, board, j, opp)) { |
516 | grid_face *face = g->faces +j; |
517 | if (do_random_pass) { |
518 | /* final random pass */ |
519 | if (!random_upto(rs, 10)) |
520 | board[j] = opp; |
521 | } else { |
522 | /* normal pass - flip when neighbour count is 1 */ |
523 | if (face_num_neighbours(g, board, face, opp) == 1) { |
524 | board[j] = opp; |
525 | flipped = TRUE; |
526 | } |
527 | } |
528 | } |
529 | } |
530 | |
531 | if (do_random_pass) break; |
532 | if (!flipped) do_random_pass = TRUE; |
533 | } |
534 | |
535 | sfree(face_list); |
536 | } |