4 * An implementation of the Nikoli game 'Loop the loop'.
5 * (c) Mike Pinna, 2005, 2006
6 * Substantially rewritten to allowing for more general types of grid.
7 * (c) Lambros Lambrou 2008
9 * vim: set shiftwidth=4 :set textwidth=80:
13 * Possible future solver enhancements:
15 * - There's an interesting deductive technique which makes use
16 * of topology rather than just graph theory. Each _face_ in
17 * the grid is either inside or outside the loop; you can tell
18 * that two faces are on the same side of the loop if they're
19 * separated by a LINE_NO (or, more generally, by a path
20 * crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
21 * and on the opposite side of the loop if they're separated by
22 * a LINE_YES (or an odd number of LINE_YESes and no
23 * LINE_UNKNOWNs). Oh, and any face separated from the outside
24 * of the grid by a LINE_YES or a LINE_NO is on the inside or
25 * outside respectively. So if you can track this for all
26 * faces, you figure out the state of the line between a pair
27 * once their relative insideness is known.
28 * + The way I envisage this working is simply to keep an edsf
29 * of all _faces_, which indicates whether they're on
30 * opposite sides of the loop from one another. We also
31 * include a special entry in the edsf for the infinite
33 * + So, the simple way to do this is to just go through the
34 * edges: every time we see an edge in a state other than
35 * LINE_UNKNOWN which separates two faces that aren't in the
36 * same edsf class, we can rectify that by merging the
37 * classes. Then, conversely, an edge in LINE_UNKNOWN state
38 * which separates two faces that _are_ in the same edsf
39 * class can immediately have its state determined.
40 * + But you can go one better, if you're prepared to loop
41 * over all _pairs_ of edges. Suppose we have edges A and B,
42 * which respectively separate faces A1,A2 and B1,B2.
43 * Suppose that A,B are in the same edge-edsf class and that
44 * A1,B1 (wlog) are in the same face-edsf class; then we can
45 * immediately place A2,B2 into the same face-edsf class (as
46 * each other, not as A1 and A2) one way round or the other.
47 * And conversely again, if A1,B1 are in the same face-edsf
48 * class and so are A2,B2, then we can put A,B into the same
50 * * Of course, this deduction requires a quadratic-time
51 * loop over all pairs of edges in the grid, so it should
52 * be reserved until there's nothing easier left to be
55 * - The generalised grid support has made me (SGT) notice a
56 * possible extension to the loop-avoidance code. When you have
57 * a path of connected edges such that no other edges at all
58 * are incident on any vertex in the middle of the path - or,
59 * alternatively, such that any such edges are already known to
60 * be LINE_NO - then you know those edges are either all
61 * LINE_YES or all LINE_NO. Hence you can mentally merge the
62 * entire path into a single long curly edge for the purposes
63 * of loop avoidance, and look directly at whether or not the
64 * extreme endpoints of the path are connected by some other
65 * route. I find this coming up fairly often when I play on the
66 * octagonal grid setting, so it might be worth implementing in
69 * - (Just a speed optimisation.) Consider some todo list queue where every
70 * time we modify something we mark it for consideration by other bits of
71 * the solver, to save iteration over things that have already been done.
86 /* Debugging options */
94 /* ----------------------------------------------------------------------
95 * Struct, enum and function declarations
112 /* Put -1 in a face that doesn't get a clue */
115 /* Array of line states, to store whether each line is
116 * YES, NO or UNKNOWN */
119 unsigned char *line_errors;
124 /* Used in game_text_format(), so that it knows what type of
125 * grid it's trying to render as ASCII text. */
130 SOLVER_SOLVED, /* This is the only solution the solver could find */
131 SOLVER_MISTAKE, /* This is definitely not a solution */
132 SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
133 SOLVER_INCOMPLETE /* This may be a partial solution */
136 /* ------ Solver state ------ */
137 typedef struct solver_state {
139 enum solver_status solver_status;
140 /* NB looplen is the number of dots that are joined together at a point, ie a
141 * looplen of 1 means there are no lines to a particular dot */
144 /* Difficulty level of solver. Used by solver functions that want to
145 * vary their behaviour depending on the requested difficulty level. */
151 char *face_yes_count;
153 char *dot_solved, *face_solved;
156 /* Information for Normal level deductions:
157 * For each dline, store a bitmask for whether we know:
158 * (bit 0) at least one is YES
159 * (bit 1) at most one is YES */
162 /* Hard level information */
167 * Difficulty levels. I do some macro ickery here to ensure that my
168 * enum and the various forms of my name list always match up.
171 #define DIFFLIST(A) \
176 #define ENUM(upper,title,lower) DIFF_ ## upper,
177 #define TITLE(upper,title,lower) #title,
178 #define ENCODE(upper,title,lower) #lower
179 #define CONFIG(upper,title,lower) ":" #title
180 enum { DIFFLIST(ENUM) DIFF_MAX };
181 static char const *const diffnames[] = { DIFFLIST(TITLE) };
182 static char const diffchars[] = DIFFLIST(ENCODE);
183 #define DIFFCONFIG DIFFLIST(CONFIG)
186 * Solver routines, sorted roughly in order of computational cost.
187 * The solver will run the faster deductions first, and slower deductions are
188 * only invoked when the faster deductions are unable to make progress.
189 * Each function is associated with a difficulty level, so that the generated
190 * puzzles are solvable by applying only the functions with the chosen
191 * difficulty level or lower.
193 #define SOLVERLIST(A) \
194 A(trivial_deductions, DIFF_EASY) \
195 A(dline_deductions, DIFF_NORMAL) \
196 A(linedsf_deductions, DIFF_HARD) \
197 A(loop_deductions, DIFF_EASY)
198 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
199 #define SOLVER_FN(fn,diff) &fn,
200 #define SOLVER_DIFF(fn,diff) diff,
201 SOLVERLIST(SOLVER_FN_DECL)
202 static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
203 static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
204 const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
211 /* Grid generation is expensive, so keep a (ref-counted) reference to the
212 * grid for these parameters, and only generate when required. */
216 /* line_drawstate is the same as line_state, but with the extra ERROR
217 * possibility. The drawing code copies line_state to line_drawstate,
218 * except in the case that the line is an error. */
219 enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
220 enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
221 DS_LINE_NO, DS_LINE_ERROR };
223 #define OPP(line_state) \
227 struct game_drawstate {
233 char *clue_satisfied;
236 static char *validate_desc(game_params *params, char *desc);
237 static int dot_order(const game_state* state, int i, char line_type);
238 static int face_order(const game_state* state, int i, char line_type);
239 static solver_state *solve_game_rec(const solver_state *sstate);
242 static void check_caches(const solver_state* sstate);
244 #define check_caches(s)
247 /* ------- List of grid generators ------- */
248 #define GRIDLIST(A) \
249 A(Squares,grid_new_square,3,3) \
250 A(Triangular,grid_new_triangular,3,3) \
251 A(Honeycomb,grid_new_honeycomb,3,3) \
252 A(Snub-Square,grid_new_snubsquare,3,3) \
253 A(Cairo,grid_new_cairo,3,4) \
254 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
255 A(Octagonal,grid_new_octagonal,3,3) \
256 A(Kites,grid_new_kites,3,3) \
257 A(Floret,grid_new_floret,1,2) \
258 A(Dodecagonal,grid_new_dodecagonal,2,2) \
259 A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2)
261 #define GRID_NAME(title,fn,amin,omin) #title,
262 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
263 #define GRID_FN(title,fn,amin,omin) &fn,
264 #define GRID_SIZES(title,fn,amin,omin) \
266 "Width and height for this grid type must both be at least " #amin, \
267 "At least one of width and height for this grid type must be at least " #omin,},
268 static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
269 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
270 static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) };
271 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
272 static const struct {
275 } grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
277 /* Generates a (dynamically allocated) new grid, according to the
278 * type and size requested in params. Does nothing if the grid is already
279 * generated. The allocated grid is owned by the params object, and will be
280 * freed in free_params(). */
281 static void params_generate_grid(game_params *params)
283 if (!params->game_grid) {
284 params->game_grid = grid_fns[params->type](params->w, params->h);
288 /* ----------------------------------------------------------------------
292 /* General constants */
293 #define PREFERRED_TILE_SIZE 32
294 #define BORDER(tilesize) ((tilesize) / 2)
295 #define FLASH_TIME 0.5F
297 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
299 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
300 ((field) |= (1<<(bit)), TRUE))
302 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
303 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
305 #define CLUE2CHAR(c) \
306 ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
308 /* ----------------------------------------------------------------------
309 * General struct manipulation and other straightforward code
312 static game_state *dup_game(game_state *state)
314 game_state *ret = snew(game_state);
316 ret->game_grid = state->game_grid;
317 ret->game_grid->refcount++;
319 ret->solved = state->solved;
320 ret->cheated = state->cheated;
322 ret->clues = snewn(state->game_grid->num_faces, signed char);
323 memcpy(ret->clues, state->clues, state->game_grid->num_faces);
325 ret->lines = snewn(state->game_grid->num_edges, char);
326 memcpy(ret->lines, state->lines, state->game_grid->num_edges);
328 ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
329 memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
331 ret->grid_type = state->grid_type;
335 static void free_game(game_state *state)
338 grid_free(state->game_grid);
341 sfree(state->line_errors);
346 static solver_state *new_solver_state(game_state *state, int diff) {
348 int num_dots = state->game_grid->num_dots;
349 int num_faces = state->game_grid->num_faces;
350 int num_edges = state->game_grid->num_edges;
351 solver_state *ret = snew(solver_state);
353 ret->state = dup_game(state);
355 ret->solver_status = SOLVER_INCOMPLETE;
358 ret->dotdsf = snew_dsf(num_dots);
359 ret->looplen = snewn(num_dots, int);
361 for (i = 0; i < num_dots; i++) {
365 ret->dot_solved = snewn(num_dots, char);
366 ret->face_solved = snewn(num_faces, char);
367 memset(ret->dot_solved, FALSE, num_dots);
368 memset(ret->face_solved, FALSE, num_faces);
370 ret->dot_yes_count = snewn(num_dots, char);
371 memset(ret->dot_yes_count, 0, num_dots);
372 ret->dot_no_count = snewn(num_dots, char);
373 memset(ret->dot_no_count, 0, num_dots);
374 ret->face_yes_count = snewn(num_faces, char);
375 memset(ret->face_yes_count, 0, num_faces);
376 ret->face_no_count = snewn(num_faces, char);
377 memset(ret->face_no_count, 0, num_faces);
379 if (diff < DIFF_NORMAL) {
382 ret->dlines = snewn(2*num_edges, char);
383 memset(ret->dlines, 0, 2*num_edges);
386 if (diff < DIFF_HARD) {
389 ret->linedsf = snew_dsf(state->game_grid->num_edges);
395 static void free_solver_state(solver_state *sstate) {
397 free_game(sstate->state);
398 sfree(sstate->dotdsf);
399 sfree(sstate->looplen);
400 sfree(sstate->dot_solved);
401 sfree(sstate->face_solved);
402 sfree(sstate->dot_yes_count);
403 sfree(sstate->dot_no_count);
404 sfree(sstate->face_yes_count);
405 sfree(sstate->face_no_count);
407 /* OK, because sfree(NULL) is a no-op */
408 sfree(sstate->dlines);
409 sfree(sstate->linedsf);
415 static solver_state *dup_solver_state(const solver_state *sstate) {
416 game_state *state = sstate->state;
417 int num_dots = state->game_grid->num_dots;
418 int num_faces = state->game_grid->num_faces;
419 int num_edges = state->game_grid->num_edges;
420 solver_state *ret = snew(solver_state);
422 ret->state = state = dup_game(sstate->state);
424 ret->solver_status = sstate->solver_status;
425 ret->diff = sstate->diff;
427 ret->dotdsf = snewn(num_dots, int);
428 ret->looplen = snewn(num_dots, int);
429 memcpy(ret->dotdsf, sstate->dotdsf,
430 num_dots * sizeof(int));
431 memcpy(ret->looplen, sstate->looplen,
432 num_dots * sizeof(int));
434 ret->dot_solved = snewn(num_dots, char);
435 ret->face_solved = snewn(num_faces, char);
436 memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
437 memcpy(ret->face_solved, sstate->face_solved, num_faces);
439 ret->dot_yes_count = snewn(num_dots, char);
440 memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
441 ret->dot_no_count = snewn(num_dots, char);
442 memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
444 ret->face_yes_count = snewn(num_faces, char);
445 memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
446 ret->face_no_count = snewn(num_faces, char);
447 memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
449 if (sstate->dlines) {
450 ret->dlines = snewn(2*num_edges, char);
451 memcpy(ret->dlines, sstate->dlines,
457 if (sstate->linedsf) {
458 ret->linedsf = snewn(num_edges, int);
459 memcpy(ret->linedsf, sstate->linedsf,
460 num_edges * sizeof(int));
468 static game_params *default_params(void)
470 game_params *ret = snew(game_params);
479 ret->diff = DIFF_EASY;
482 ret->game_grid = NULL;
487 static game_params *dup_params(game_params *params)
489 game_params *ret = snew(game_params);
491 *ret = *params; /* structure copy */
492 if (ret->game_grid) {
493 ret->game_grid->refcount++;
498 static const game_params presets[] = {
500 { 7, 7, DIFF_EASY, 0, NULL },
501 { 7, 7, DIFF_NORMAL, 0, NULL },
502 { 7, 7, DIFF_HARD, 0, NULL },
503 { 7, 7, DIFF_HARD, 1, NULL },
504 { 7, 7, DIFF_HARD, 2, NULL },
505 { 5, 5, DIFF_HARD, 3, NULL },
506 { 7, 7, DIFF_HARD, 4, NULL },
507 { 5, 4, DIFF_HARD, 5, NULL },
508 { 5, 5, DIFF_HARD, 6, NULL },
509 { 5, 5, DIFF_HARD, 7, NULL },
510 { 3, 3, DIFF_HARD, 8, NULL },
511 { 3, 3, DIFF_HARD, 9, NULL },
512 { 3, 3, DIFF_HARD, 10, NULL },
514 { 7, 7, DIFF_EASY, 0, NULL },
515 { 10, 10, DIFF_EASY, 0, NULL },
516 { 7, 7, DIFF_NORMAL, 0, NULL },
517 { 10, 10, DIFF_NORMAL, 0, NULL },
518 { 7, 7, DIFF_HARD, 0, NULL },
519 { 10, 10, DIFF_HARD, 0, NULL },
520 { 10, 10, DIFF_HARD, 1, NULL },
521 { 12, 10, DIFF_HARD, 2, NULL },
522 { 7, 7, DIFF_HARD, 3, NULL },
523 { 9, 9, DIFF_HARD, 4, NULL },
524 { 5, 4, DIFF_HARD, 5, NULL },
525 { 7, 7, DIFF_HARD, 6, NULL },
526 { 5, 5, DIFF_HARD, 7, NULL },
527 { 5, 5, DIFF_HARD, 8, NULL },
528 { 5, 4, DIFF_HARD, 9, NULL },
529 { 5, 4, DIFF_HARD, 10, NULL },
533 static int game_fetch_preset(int i, char **name, game_params **params)
538 if (i < 0 || i >= lenof(presets))
541 tmppar = snew(game_params);
542 *tmppar = presets[i];
544 sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
545 gridnames[tmppar->type], diffnames[tmppar->diff]);
551 static void free_params(game_params *params)
553 if (params->game_grid) {
554 grid_free(params->game_grid);
559 static void decode_params(game_params *params, char const *string)
561 if (params->game_grid) {
562 grid_free(params->game_grid);
563 params->game_grid = NULL;
565 params->h = params->w = atoi(string);
566 params->diff = DIFF_EASY;
567 while (*string && isdigit((unsigned char)*string)) string++;
568 if (*string == 'x') {
570 params->h = atoi(string);
571 while (*string && isdigit((unsigned char)*string)) string++;
573 if (*string == 't') {
575 params->type = atoi(string);
576 while (*string && isdigit((unsigned char)*string)) string++;
578 if (*string == 'd') {
581 for (i = 0; i < DIFF_MAX; i++)
582 if (*string == diffchars[i])
584 if (*string) string++;
588 static char *encode_params(game_params *params, int full)
591 sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
593 sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
597 static config_item *game_configure(game_params *params)
602 ret = snewn(5, config_item);
604 ret[0].name = "Width";
605 ret[0].type = C_STRING;
606 sprintf(buf, "%d", params->w);
607 ret[0].sval = dupstr(buf);
610 ret[1].name = "Height";
611 ret[1].type = C_STRING;
612 sprintf(buf, "%d", params->h);
613 ret[1].sval = dupstr(buf);
616 ret[2].name = "Grid type";
617 ret[2].type = C_CHOICES;
618 ret[2].sval = GRID_CONFIGS;
619 ret[2].ival = params->type;
621 ret[3].name = "Difficulty";
622 ret[3].type = C_CHOICES;
623 ret[3].sval = DIFFCONFIG;
624 ret[3].ival = params->diff;
634 static game_params *custom_params(config_item *cfg)
636 game_params *ret = snew(game_params);
638 ret->w = atoi(cfg[0].sval);
639 ret->h = atoi(cfg[1].sval);
640 ret->type = cfg[2].ival;
641 ret->diff = cfg[3].ival;
643 ret->game_grid = NULL;
647 static char *validate_params(game_params *params, int full)
649 if (params->type < 0 || params->type >= NUM_GRID_TYPES)
650 return "Illegal grid type";
651 if (params->w < grid_size_limits[params->type].amin ||
652 params->h < grid_size_limits[params->type].amin)
653 return grid_size_limits[params->type].aerr;
654 if (params->w < grid_size_limits[params->type].omin &&
655 params->h < grid_size_limits[params->type].omin)
656 return grid_size_limits[params->type].oerr;
659 * This shouldn't be able to happen at all, since decode_params
660 * and custom_params will never generate anything that isn't
663 assert(params->diff < DIFF_MAX);
668 /* Returns a newly allocated string describing the current puzzle */
669 static char *state_to_text(const game_state *state)
671 grid *g = state->game_grid;
673 int num_faces = g->num_faces;
674 char *description = snewn(num_faces + 1, char);
675 char *dp = description;
679 for (i = 0; i < num_faces; i++) {
680 if (state->clues[i] < 0) {
681 if (empty_count > 25) {
682 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
688 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
691 dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
696 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
698 retval = dupstr(description);
704 /* We require that the params pass the test in validate_params and that the
705 * description fills the entire game area */
706 static char *validate_desc(game_params *params, char *desc)
710 params_generate_grid(params);
711 g = params->game_grid;
713 for (; *desc; ++desc) {
714 if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
719 count += *desc - 'a' + 1;
722 return "Unknown character in description";
725 if (count < g->num_faces)
726 return "Description too short for board size";
727 if (count > g->num_faces)
728 return "Description too long for board size";
733 /* Sums the lengths of the numbers in range [0,n) */
734 /* See equivalent function in solo.c for justification of this. */
735 static int len_0_to_n(int n)
737 int len = 1; /* Counting 0 as a bit of a special case */
740 for (i = 1; i < n; i *= 10) {
741 len += max(n - i, 0);
747 static char *encode_solve_move(const game_state *state)
752 int num_edges = state->game_grid->num_edges;
754 /* This is going to return a string representing the moves needed to set
755 * every line in a grid to be the same as the ones in 'state'. The exact
756 * length of this string is predictable. */
758 len = 1; /* Count the 'S' prefix */
759 /* Numbers in all lines */
760 len += len_0_to_n(num_edges);
761 /* For each line we also have a letter */
764 ret = snewn(len + 1, char);
767 p += sprintf(p, "S");
769 for (i = 0; i < num_edges; i++) {
770 switch (state->lines[i]) {
772 p += sprintf(p, "%dy", i);
775 p += sprintf(p, "%dn", i);
780 /* No point in doing sums like that if they're going to be wrong */
781 assert(strlen(ret) <= (size_t)len);
785 static game_ui *new_ui(game_state *state)
790 static void free_ui(game_ui *ui)
794 static char *encode_ui(game_ui *ui)
799 static void decode_ui(game_ui *ui, char *encoding)
803 static void game_changed_state(game_ui *ui, game_state *oldstate,
804 game_state *newstate)
808 static void game_compute_size(game_params *params, int tilesize,
812 int grid_width, grid_height, rendered_width, rendered_height;
814 params_generate_grid(params);
815 g = params->game_grid;
816 grid_width = g->highest_x - g->lowest_x;
817 grid_height = g->highest_y - g->lowest_y;
818 /* multiply first to minimise rounding error on integer division */
819 rendered_width = grid_width * tilesize / g->tilesize;
820 rendered_height = grid_height * tilesize / g->tilesize;
821 *x = rendered_width + 2 * BORDER(tilesize) + 1;
822 *y = rendered_height + 2 * BORDER(tilesize) + 1;
825 static void game_set_size(drawing *dr, game_drawstate *ds,
826 game_params *params, int tilesize)
828 ds->tilesize = tilesize;
831 static float *game_colours(frontend *fe, int *ncolours)
833 float *ret = snewn(4 * NCOLOURS, float);
835 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
837 ret[COL_FOREGROUND * 3 + 0] = 0.0F;
838 ret[COL_FOREGROUND * 3 + 1] = 0.0F;
839 ret[COL_FOREGROUND * 3 + 2] = 0.0F;
841 ret[COL_LINEUNKNOWN * 3 + 0] = 0.8F;
842 ret[COL_LINEUNKNOWN * 3 + 1] = 0.8F;
843 ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
845 ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
846 ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
847 ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;
849 ret[COL_MISTAKE * 3 + 0] = 1.0F;
850 ret[COL_MISTAKE * 3 + 1] = 0.0F;
851 ret[COL_MISTAKE * 3 + 2] = 0.0F;
853 ret[COL_SATISFIED * 3 + 0] = 0.0F;
854 ret[COL_SATISFIED * 3 + 1] = 0.0F;
855 ret[COL_SATISFIED * 3 + 2] = 0.0F;
857 /* We want the faint lines to be a bit darker than the background.
858 * Except if the background is pretty dark already; then it ought to be a
859 * bit lighter. Oy vey.
861 ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
862 ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
863 ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F;
865 *ncolours = NCOLOURS;
869 static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
871 struct game_drawstate *ds = snew(struct game_drawstate);
872 int num_faces = state->game_grid->num_faces;
873 int num_edges = state->game_grid->num_edges;
877 ds->lines = snewn(num_edges, char);
878 ds->clue_error = snewn(num_faces, char);
879 ds->clue_satisfied = snewn(num_faces, char);
882 memset(ds->lines, LINE_UNKNOWN, num_edges);
883 memset(ds->clue_error, 0, num_faces);
884 memset(ds->clue_satisfied, 0, num_faces);
889 static void game_free_drawstate(drawing *dr, game_drawstate *ds)
891 sfree(ds->clue_error);
892 sfree(ds->clue_satisfied);
897 static int game_timing_state(game_state *state, game_ui *ui)
902 static float game_anim_length(game_state *oldstate, game_state *newstate,
903 int dir, game_ui *ui)
908 static int game_can_format_as_text_now(game_params *params)
910 if (params->type != 0)
915 static char *game_text_format(game_state *state)
921 grid *g = state->game_grid;
924 assert(state->grid_type == 0);
926 /* Work out the basic size unit */
927 f = g->faces; /* first face */
928 assert(f->order == 4);
929 /* The dots are ordered clockwise, so the two opposite
930 * corners are guaranteed to span the square */
931 cell_size = abs(f->dots[0]->x - f->dots[2]->x);
933 w = (g->highest_x - g->lowest_x) / cell_size;
934 h = (g->highest_y - g->lowest_y) / cell_size;
936 /* Create a blank "canvas" to "draw" on */
939 ret = snewn(W * H + 1, char);
940 for (y = 0; y < H; y++) {
941 for (x = 0; x < W-1; x++) {
944 ret[y*W + W-1] = '\n';
948 /* Fill in edge info */
949 for (i = 0; i < g->num_edges; i++) {
950 grid_edge *e = g->edges + i;
951 /* Cell coordinates, from (0,0) to (w-1,h-1) */
952 int x1 = (e->dot1->x - g->lowest_x) / cell_size;
953 int x2 = (e->dot2->x - g->lowest_x) / cell_size;
954 int y1 = (e->dot1->y - g->lowest_y) / cell_size;
955 int y2 = (e->dot2->y - g->lowest_y) / cell_size;
956 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
957 * cell coordinates) */
960 switch (state->lines[i]) {
962 ret[y*W + x] = (y1 == y2) ? '-' : '|';
968 break; /* already a space */
970 assert(!"Illegal line state");
975 for (i = 0; i < g->num_faces; i++) {
979 assert(f->order == 4);
980 /* Cell coordinates, from (0,0) to (w-1,h-1) */
981 x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
982 x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
983 y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
984 y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
985 /* Midpoint, in canvas coordinates */
988 ret[y*W + x] = CLUE2CHAR(state->clues[i]);
993 /* ----------------------------------------------------------------------
998 static void check_caches(const solver_state* sstate)
1001 const game_state *state = sstate->state;
1002 const grid *g = state->game_grid;
1004 for (i = 0; i < g->num_dots; i++) {
1005 assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
1006 assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
1009 for (i = 0; i < g->num_faces; i++) {
1010 assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
1011 assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
1016 #define check_caches(s) \
1018 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1022 #endif /* DEBUG_CACHES */
1024 /* ----------------------------------------------------------------------
1025 * Solver utility functions
1028 /* Sets the line (with index i) to the new state 'line_new', and updates
1029 * the cached counts of any affected faces and dots.
1030 * Returns TRUE if this actually changed the line's state. */
1031 static int solver_set_line(solver_state *sstate, int i,
1032 enum line_state line_new
1034 , const char *reason
1038 game_state *state = sstate->state;
1042 assert(line_new != LINE_UNKNOWN);
1044 check_caches(sstate);
1046 if (state->lines[i] == line_new) {
1047 return FALSE; /* nothing changed */
1049 state->lines[i] = line_new;
1052 fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
1053 i, line_new == LINE_YES ? "YES" : "NO",
1057 g = state->game_grid;
1060 /* Update the cache for both dots and both faces affected by this. */
1061 if (line_new == LINE_YES) {
1062 sstate->dot_yes_count[e->dot1 - g->dots]++;
1063 sstate->dot_yes_count[e->dot2 - g->dots]++;
1065 sstate->face_yes_count[e->face1 - g->faces]++;
1068 sstate->face_yes_count[e->face2 - g->faces]++;
1071 sstate->dot_no_count[e->dot1 - g->dots]++;
1072 sstate->dot_no_count[e->dot2 - g->dots]++;
1074 sstate->face_no_count[e->face1 - g->faces]++;
1077 sstate->face_no_count[e->face2 - g->faces]++;
1081 check_caches(sstate);
1086 #define solver_set_line(a, b, c) \
1087 solver_set_line(a, b, c, __FUNCTION__)
1091 * Merge two dots due to the existence of an edge between them.
1092 * Updates the dsf tracking equivalence classes, and keeps track of
1093 * the length of path each dot is currently a part of.
1094 * Returns TRUE if the dots were already linked, ie if they are part of a
1095 * closed loop, and false otherwise.
1097 static int merge_dots(solver_state *sstate, int edge_index)
1100 grid *g = sstate->state->game_grid;
1101 grid_edge *e = g->edges + edge_index;
1103 i = e->dot1 - g->dots;
1104 j = e->dot2 - g->dots;
1106 i = dsf_canonify(sstate->dotdsf, i);
1107 j = dsf_canonify(sstate->dotdsf, j);
1112 len = sstate->looplen[i] + sstate->looplen[j];
1113 dsf_merge(sstate->dotdsf, i, j);
1114 i = dsf_canonify(sstate->dotdsf, i);
1115 sstate->looplen[i] = len;
1120 /* Merge two lines because the solver has deduced that they must be either
1121 * identical or opposite. Returns TRUE if this is new information, otherwise
1123 static int merge_lines(solver_state *sstate, int i, int j, int inverse
1125 , const char *reason
1131 assert(i < sstate->state->game_grid->num_edges);
1132 assert(j < sstate->state->game_grid->num_edges);
1134 i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
1136 j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
1139 edsf_merge(sstate->linedsf, i, j, inverse);
1143 fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
1145 inverse ? "inverse " : "", reason);
1152 #define merge_lines(a, b, c, d) \
1153 merge_lines(a, b, c, d, __FUNCTION__)
1156 /* Count the number of lines of a particular type currently going into the
1158 static int dot_order(const game_state* state, int dot, char line_type)
1161 grid *g = state->game_grid;
1162 grid_dot *d = g->dots + dot;
1165 for (i = 0; i < d->order; i++) {
1166 grid_edge *e = d->edges[i];
1167 if (state->lines[e - g->edges] == line_type)
1173 /* Count the number of lines of a particular type currently surrounding the
1175 static int face_order(const game_state* state, int face, char line_type)
1178 grid *g = state->game_grid;
1179 grid_face *f = g->faces + face;
1182 for (i = 0; i < f->order; i++) {
1183 grid_edge *e = f->edges[i];
1184 if (state->lines[e - g->edges] == line_type)
1190 /* Set all lines bordering a dot of type old_type to type new_type
1191 * Return value tells caller whether this function actually did anything */
1192 static int dot_setall(solver_state *sstate, int dot,
1193 char old_type, char new_type)
1195 int retval = FALSE, r;
1196 game_state *state = sstate->state;
1201 if (old_type == new_type)
1204 g = state->game_grid;
1207 for (i = 0; i < d->order; i++) {
1208 int line_index = d->edges[i] - g->edges;
1209 if (state->lines[line_index] == old_type) {
1210 r = solver_set_line(sstate, line_index, new_type);
1218 /* Set all lines bordering a face of type old_type to type new_type */
1219 static int face_setall(solver_state *sstate, int face,
1220 char old_type, char new_type)
1222 int retval = FALSE, r;
1223 game_state *state = sstate->state;
1228 if (old_type == new_type)
1231 g = state->game_grid;
1232 f = g->faces + face;
1234 for (i = 0; i < f->order; i++) {
1235 int line_index = f->edges[i] - g->edges;
1236 if (state->lines[line_index] == old_type) {
1237 r = solver_set_line(sstate, line_index, new_type);
1245 /* ----------------------------------------------------------------------
1246 * Loop generation and clue removal
1249 /* We're going to store lists of current candidate faces for colouring black
1251 * Each face gets a 'score', which tells us how adding that face right
1252 * now would affect the curliness of the solution loop. We're trying to
1253 * maximise that quantity so will bias our random selection of faces to
1254 * colour those with high scores */
1258 unsigned long random;
1259 /* No need to store a grid_face* here. The 'face_scores' array will
1260 * be a list of 'face_score' objects, one for each face of the grid, so
1261 * the position (index) within the 'face_scores' array will determine
1262 * which face corresponds to a particular face_score.
1263 * Having a single 'face_scores' array for all faces simplifies memory
1264 * management, and probably improves performance, because we don't have to
1265 * malloc/free each individual face_score, and we don't have to maintain
1266 * a mapping from grid_face* pointers to face_score* pointers.
1270 static int generic_sort_cmpfn(void *v1, void *v2, size_t offset)
1272 struct face_score *f1 = v1;
1273 struct face_score *f2 = v2;
1276 r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset);
1281 if (f1->random < f2->random)
1283 else if (f1->random > f2->random)
1287 * It's _just_ possible that two faces might have been given
1288 * the same random value. In that situation, fall back to
1289 * comparing based on the positions within the face_scores list.
1290 * This introduces a tiny directional bias, but not a significant one.
1295 static int white_sort_cmpfn(void *v1, void *v2)
1297 return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score));
1300 static int black_sort_cmpfn(void *v1, void *v2)
1302 return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score));
1305 enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK };
1307 /* face should be of type grid_face* here. */
1308 #define FACE_COLOUR(face) \
1309 ( (face) == NULL ? FACE_BLACK : \
1310 board[(face) - g->faces] )
1312 /* 'board' is an array of these enums, indicating which faces are
1313 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1314 * Returns whether it's legal to colour the given face with this colour. */
1315 static int can_colour_face(grid *g, char* board, int face_index,
1316 enum face_colour colour)
1319 grid_face *test_face = g->faces + face_index;
1320 grid_face *starting_face, *current_face;
1321 grid_dot *starting_dot;
1323 int current_state, s; /* booleans: equal or not-equal to 'colour' */
1324 int found_same_coloured_neighbour = FALSE;
1325 assert(board[face_index] != colour);
1327 /* Can only consider a face for colouring if it's adjacent to a face
1328 * with the same colour. */
1329 for (i = 0; i < test_face->order; i++) {
1330 grid_edge *e = test_face->edges[i];
1331 grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1;
1332 if (FACE_COLOUR(f) == colour) {
1333 found_same_coloured_neighbour = TRUE;
1337 if (!found_same_coloured_neighbour)
1340 /* Need to avoid creating a loop of faces of this colour around some
1341 * differently-coloured faces.
1342 * Also need to avoid meeting a same-coloured face at a corner, with
1343 * other-coloured faces in between. Here's a simple test that (I believe)
1344 * takes care of both these conditions:
1346 * Take the circular path formed by this face's edges, and inflate it
1347 * slightly outwards. Imagine walking around this path and consider
1348 * the faces that you visit in sequence. This will include all faces
1349 * touching the given face, either along an edge or just at a corner.
1350 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1351 * you walk along the complete loop. This will obviously turn out to be
1353 * If 0, we're either in the middle of an "island" of this colour (should
1354 * be impossible as we're not supposed to create black or white loops),
1355 * or we're about to start a new island - also not allowed.
1356 * If 4 or greater, there are too many separate coloured regions touching
1357 * this face, and colouring it would create a loop or a corner-violation.
1358 * The only allowed case is when the count is exactly 2. */
1360 /* i points to a dot around the test face.
1361 * j points to a face around the i^th dot.
1362 * The current face will always be:
1363 * test_face->dots[i]->faces[j]
1364 * We assume dots go clockwise around the test face,
1365 * and faces go clockwise around dots. */
1368 * The end condition is slightly fiddly. In sufficiently strange
1369 * degenerate grids, our test face may be adjacent to the same
1370 * other face multiple times (typically if it's the exterior
1371 * face). Consider this, in particular:
1379 * The bottom left face there is adjacent to the exterior face
1380 * twice, so we can't just terminate our iteration when we reach
1381 * the same _face_ we started at. Furthermore, we can't
1382 * condition on having the same (i,j) pair either, because
1383 * several (i,j) pairs identify the bottom left contiguity with
1384 * the exterior face! We canonicalise the (i,j) pair by taking
1385 * one step around before we set the termination tracking.
1389 current_face = test_face->dots[0]->faces[0];
1390 if (current_face == test_face) {
1392 current_face = test_face->dots[0]->faces[1];
1395 current_state = (FACE_COLOUR(current_face) == colour);
1396 starting_dot = NULL;
1397 starting_face = NULL;
1399 /* Advance to next face.
1400 * Need to loop here because it might take several goes to
1404 if (j == test_face->dots[i]->order)
1407 if (test_face->dots[i]->faces[j] == test_face) {
1408 /* Advance to next dot round test_face, then
1409 * find current_face around new dot
1410 * and advance to the next face clockwise */
1412 if (i == test_face->order)
1414 for (j = 0; j < test_face->dots[i]->order; j++) {
1415 if (test_face->dots[i]->faces[j] == current_face)
1418 /* Must actually find current_face around new dot,
1419 * or else something's wrong with the grid. */
1420 assert(j != test_face->dots[i]->order);
1421 /* Found, so advance to next face and try again */
1426 /* (i,j) are now advanced to next face */
1427 current_face = test_face->dots[i]->faces[j];
1428 s = (FACE_COLOUR(current_face) == colour);
1429 if (!starting_dot) {
1430 starting_dot = test_face->dots[i];
1431 starting_face = current_face;
1434 if (s != current_state) {
1437 if (transitions > 2)
1440 if (test_face->dots[i] == starting_dot &&
1441 current_face == starting_face)
1446 return (transitions == 2) ? TRUE : FALSE;
1449 /* Count the number of neighbours of 'face', having colour 'colour' */
1450 static int face_num_neighbours(grid *g, char *board, grid_face *face,
1451 enum face_colour colour)
1453 int colour_count = 0;
1457 for (i = 0; i < face->order; i++) {
1459 f = (e->face1 == face) ? e->face2 : e->face1;
1460 if (FACE_COLOUR(f) == colour)
1463 return colour_count;
1466 /* The 'score' of a face reflects its current desirability for selection
1467 * as the next face to colour white or black. We want to encourage moving
1468 * into grey areas and increasing loopiness, so we give scores according to
1469 * how many of the face's neighbours are currently coloured the same as the
1470 * proposed colour. */
1471 static int face_score(grid *g, char *board, grid_face *face,
1472 enum face_colour colour)
1474 /* Simple formula: score = 0 - num. same-coloured neighbours,
1475 * so a higher score means fewer same-coloured neighbours. */
1476 return -face_num_neighbours(g, board, face, colour);
1479 /* Generate a new complete set of clues for the given game_state.
1480 * The method is to generate a WHITE/BLACK colouring of all the faces,
1481 * such that the WHITE faces will define the inside of the path, and the
1482 * BLACK faces define the outside.
1483 * To do this, we initially colour all faces GREY. The infinite space outside
1484 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1485 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1486 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1487 * we avoid creating loops of a single colour, to preserve the topological
1488 * shape of the WHITE and BLACK regions.
1489 * We also try to make the boundary as loopy and twisty as possible, to avoid
1490 * generating paths that are uninteresting.
1491 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1492 * face that can be coloured with that colour (without violating the
1493 * topological shape of that region). It's not obvious, but I think this
1494 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1495 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1496 * regions can be grown.
1497 * This is checked using assert()ions, and I haven't seen any failures yet.
1499 * Hand-wavy proof: imagine what can go wrong...
1501 * Could the white faces get completely cut off by the black faces, and still
1502 * leave some grey faces remaining?
1503 * No, because then the black faces would form a loop around both the white
1504 * faces and the grey faces, which is disallowed because we continually
1505 * maintain the correct topological shape of the black region.
1506 * Similarly, the black faces can never get cut off by the white faces. That
1507 * means both the WHITE and BLACK regions always have some room to grow into
1509 * Could it be that we can't colour some GREY face, because there are too many
1510 * WHITE/BLACK transitions as we walk round the face? (see the
1511 * can_colour_face() function for details)
1512 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1513 * around the face. The two WHITE faces would be connected by a WHITE path,
1514 * and the BLACK faces would be connected by a BLACK path. These paths would
1515 * have to cross, which is impossible.
1516 * Another thing that could go wrong: perhaps we can't find any GREY face to
1517 * colour WHITE, because it would create a loop-violation or a corner-violation
1518 * with the other WHITE faces?
1519 * This is a little bit tricky to prove impossible. Imagine you have such a
1520 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1521 * or corner violation).
1522 * That would cut all the non-white area into two blobs. One of those blobs
1523 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1524 * So we have a connected GREY area, completely surrounded by WHITE
1525 * (including the GREY face we've tentatively coloured WHITE).
1526 * A well-known result in graph theory says that you can always find a GREY
1527 * face whose removal leaves the remaining GREY area connected. And it says
1528 * there are at least two such faces, so we can always choose the one that
1529 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1530 * everything nice and connected, including that "tentative" GREY face which
1531 * acts as a gateway to the rest of the non-WHITE grid.
1533 static void add_full_clues(game_state *state, random_state *rs)
1535 signed char *clues = state->clues;
1537 grid *g = state->game_grid;
1539 int num_faces = g->num_faces;
1540 struct face_score *face_scores; /* Array of face_score objects */
1541 struct face_score *fs; /* Points somewhere in the above list */
1542 struct grid_face *cur_face;
1543 tree234 *lightable_faces_sorted;
1544 tree234 *darkable_faces_sorted;
1548 board = snewn(num_faces, char);
1551 memset(board, FACE_GREY, num_faces);
1553 /* Create and initialise the list of face_scores */
1554 face_scores = snewn(num_faces, struct face_score);
1555 for (i = 0; i < num_faces; i++) {
1556 face_scores[i].random = random_bits(rs, 31);
1557 face_scores[i].black_score = face_scores[i].white_score = 0;
1560 /* Colour a random, finite face white. The infinite face is implicitly
1561 * coloured black. Together, they will seed the random growth process
1562 * for the black and white areas. */
1563 i = random_upto(rs, num_faces);
1564 board[i] = FACE_WHITE;
1566 /* We need a way of favouring faces that will increase our loopiness.
1567 * We do this by maintaining a list of all candidate faces sorted by
1568 * their score and choose randomly from that with appropriate skew.
1569 * In order to avoid consistently biasing towards particular faces, we
1570 * need the sort order _within_ each group of scores to be completely
1571 * random. But it would be abusing the hospitality of the tree234 data
1572 * structure if our comparison function were nondeterministic :-). So with
1573 * each face we associate a random number that does not change during a
1574 * particular run of the generator, and use that as a secondary sort key.
1575 * Yes, this means we will be biased towards particular random faces in
1576 * any one run but that doesn't actually matter. */
1578 lightable_faces_sorted = newtree234(white_sort_cmpfn);
1579 darkable_faces_sorted = newtree234(black_sort_cmpfn);
1581 /* Initialise the lists of lightable and darkable faces. This is
1582 * slightly different from the code inside the while-loop, because we need
1583 * to check every face of the board (the grid structure does not keep a
1584 * list of the infinite face's neighbours). */
1585 for (i = 0; i < num_faces; i++) {
1586 grid_face *f = g->faces + i;
1587 struct face_score *fs = face_scores + i;
1588 if (board[i] != FACE_GREY) continue;
1589 /* We need the full colourability check here, it's not enough simply
1590 * to check neighbourhood. On some grids, a neighbour of the infinite
1591 * face is not necessarily darkable. */
1592 if (can_colour_face(g, board, i, FACE_BLACK)) {
1593 fs->black_score = face_score(g, board, f, FACE_BLACK);
1594 add234(darkable_faces_sorted, fs);
1596 if (can_colour_face(g, board, i, FACE_WHITE)) {
1597 fs->white_score = face_score(g, board, f, FACE_WHITE);
1598 add234(lightable_faces_sorted, fs);
1602 /* Colour faces one at a time until no more faces are colourable. */
1605 enum face_colour colour;
1606 struct face_score *fs_white, *fs_black;
1607 int c_lightable = count234(lightable_faces_sorted);
1608 int c_darkable = count234(darkable_faces_sorted);
1609 if (c_lightable == 0 && c_darkable == 0) {
1610 /* No more faces we can use at all. */
1613 assert(c_lightable != 0 && c_darkable != 0);
1615 fs_white = (struct face_score *)index234(lightable_faces_sorted, 0);
1616 fs_black = (struct face_score *)index234(darkable_faces_sorted, 0);
1618 /* Choose a colour, and colour the best available face
1619 * with that colour. */
1620 colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK;
1622 if (colour == FACE_WHITE)
1627 i = fs - face_scores;
1628 assert(board[i] == FACE_GREY);
1631 /* Remove this newly-coloured face from the lists. These lists should
1632 * only contain grey faces. */
1633 del234(lightable_faces_sorted, fs);
1634 del234(darkable_faces_sorted, fs);
1636 /* Remember which face we've just coloured */
1637 cur_face = g->faces + i;
1639 /* The face we've just coloured potentially affects the colourability
1640 * and the scores of any neighbouring faces (touching at a corner or
1641 * edge). So the search needs to be conducted around all faces
1642 * touching the one we've just lit. Iterate over its corners, then
1643 * over each corner's faces. For each such face, we remove it from
1644 * the lists, recalculate any scores, then add it back to the lists
1645 * (depending on whether it is lightable, darkable or both). */
1646 for (i = 0; i < cur_face->order; i++) {
1647 grid_dot *d = cur_face->dots[i];
1648 for (j = 0; j < d->order; j++) {
1649 grid_face *f = d->faces[j];
1650 int fi; /* face index of f */
1657 /* If the face is already coloured, it won't be on our
1658 * lightable/darkable lists anyway, so we can skip it without
1659 * bothering with the removal step. */
1660 if (FACE_COLOUR(f) != FACE_GREY) continue;
1662 /* Find the face index and face_score* corresponding to f */
1664 fs = face_scores + fi;
1666 /* Remove from lightable list if it's in there. We do this,
1667 * even if it is still lightable, because the score might
1668 * be different, and we need to remove-then-add to maintain
1669 * correct sort order. */
1670 del234(lightable_faces_sorted, fs);
1671 if (can_colour_face(g, board, fi, FACE_WHITE)) {
1672 fs->white_score = face_score(g, board, f, FACE_WHITE);
1673 add234(lightable_faces_sorted, fs);
1675 /* Do the same for darkable list. */
1676 del234(darkable_faces_sorted, fs);
1677 if (can_colour_face(g, board, fi, FACE_BLACK)) {
1678 fs->black_score = face_score(g, board, f, FACE_BLACK);
1679 add234(darkable_faces_sorted, fs);
1686 freetree234(lightable_faces_sorted);
1687 freetree234(darkable_faces_sorted);
1690 /* The next step requires a shuffled list of all faces */
1691 face_list = snewn(num_faces, int);
1692 for (i = 0; i < num_faces; ++i) {
1695 shuffle(face_list, num_faces, sizeof(int), rs);
1697 /* The above loop-generation algorithm can often leave large clumps
1698 * of faces of one colour. In extreme cases, the resulting path can be
1699 * degenerate and not very satisfying to solve.
1700 * This next step alleviates this problem:
1701 * Go through the shuffled list, and flip the colour of any face we can
1702 * legally flip, and which is adjacent to only one face of the opposite
1703 * colour - this tends to grow 'tendrils' into any clumps.
1704 * Repeat until we can find no more faces to flip. This will
1705 * eventually terminate, because each flip increases the loop's
1706 * perimeter, which cannot increase for ever.
1707 * The resulting path will have maximal loopiness (in the sense that it
1708 * cannot be improved "locally". Unfortunately, this allows a player to
1709 * make some illicit deductions. To combat this (and make the path more
1710 * interesting), we do one final pass making random flips. */
1712 /* Set to TRUE for final pass */
1713 do_random_pass = FALSE;
1716 /* Remember whether a flip occurred during this pass */
1717 int flipped = FALSE;
1719 for (i = 0; i < num_faces; ++i) {
1720 int j = face_list[i];
1721 enum face_colour opp =
1722 (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE;
1723 if (can_colour_face(g, board, j, opp)) {
1724 grid_face *face = g->faces +j;
1725 if (do_random_pass) {
1726 /* final random pass */
1727 if (!random_upto(rs, 10))
1730 /* normal pass - flip when neighbour count is 1 */
1731 if (face_num_neighbours(g, board, face, opp) == 1) {
1739 if (do_random_pass) break;
1740 if (!flipped) do_random_pass = TRUE;
1745 /* Fill out all the clues by initialising to 0, then iterating over
1746 * all edges and incrementing each clue as we find edges that border
1747 * between BLACK/WHITE faces. While we're at it, we verify that the
1748 * algorithm does work, and there aren't any GREY faces still there. */
1749 memset(clues, 0, num_faces);
1750 for (i = 0; i < g->num_edges; i++) {
1751 grid_edge *e = g->edges + i;
1752 grid_face *f1 = e->face1;
1753 grid_face *f2 = e->face2;
1754 enum face_colour c1 = FACE_COLOUR(f1);
1755 enum face_colour c2 = FACE_COLOUR(f2);
1756 assert(c1 != FACE_GREY);
1757 assert(c2 != FACE_GREY);
1759 if (f1) clues[f1 - g->faces]++;
1760 if (f2) clues[f2 - g->faces]++;
1768 static int game_has_unique_soln(const game_state *state, int diff)
1771 solver_state *sstate_new;
1772 solver_state *sstate = new_solver_state((game_state *)state, diff);
1774 sstate_new = solve_game_rec(sstate);
1776 assert(sstate_new->solver_status != SOLVER_MISTAKE);
1777 ret = (sstate_new->solver_status == SOLVER_SOLVED);
1779 free_solver_state(sstate_new);
1780 free_solver_state(sstate);
1786 /* Remove clues one at a time at random. */
1787 static game_state *remove_clues(game_state *state, random_state *rs,
1791 int num_faces = state->game_grid->num_faces;
1792 game_state *ret = dup_game(state), *saved_ret;
1795 /* We need to remove some clues. We'll do this by forming a list of all
1796 * available clues, shuffling it, then going along one at a
1797 * time clearing each clue in turn for which doing so doesn't render the
1798 * board unsolvable. */
1799 face_list = snewn(num_faces, int);
1800 for (n = 0; n < num_faces; ++n) {
1804 shuffle(face_list, num_faces, sizeof(int), rs);
1806 for (n = 0; n < num_faces; ++n) {
1807 saved_ret = dup_game(ret);
1808 ret->clues[face_list[n]] = -1;
1810 if (game_has_unique_soln(ret, diff)) {
1811 free_game(saved_ret);
1823 static char *new_game_desc(game_params *params, random_state *rs,
1824 char **aux, int interactive)
1826 /* solution and description both use run-length encoding in obvious ways */
1829 game_state *state = snew(game_state);
1830 game_state *state_new;
1832 params_generate_grid(params);
1833 state->game_grid = g = params->game_grid;
1835 state->clues = snewn(g->num_faces, signed char);
1836 state->lines = snewn(g->num_edges, char);
1837 state->line_errors = snewn(g->num_edges, unsigned char);
1839 state->grid_type = params->type;
1843 memset(state->lines, LINE_UNKNOWN, g->num_edges);
1844 memset(state->line_errors, 0, g->num_edges);
1846 state->solved = state->cheated = FALSE;
1848 /* Get a new random solvable board with all its clues filled in. Yes, this
1849 * can loop for ever if the params are suitably unfavourable, but
1850 * preventing games smaller than 4x4 seems to stop this happening */
1852 add_full_clues(state, rs);
1853 if (++count%100 == 0) printf("tried %d times to make a unique board\n", count);
1854 } while (!game_has_unique_soln(state, params->diff));
1856 state_new = remove_clues(state, rs, params->diff);
1861 if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
1863 fprintf(stderr, "Rejecting board, it is too easy\n");
1865 goto newboard_please;
1868 retval = state_to_text(state);
1872 assert(!validate_desc(params, retval));
1877 static game_state *new_game(midend *me, game_params *params, char *desc)
1880 game_state *state = snew(game_state);
1881 int empties_to_make = 0;
1883 const char *dp = desc;
1885 int num_faces, num_edges;
1887 params_generate_grid(params);
1888 state->game_grid = g = params->game_grid;
1890 num_faces = g->num_faces;
1891 num_edges = g->num_edges;
1893 state->clues = snewn(num_faces, signed char);
1894 state->lines = snewn(num_edges, char);
1895 state->line_errors = snewn(num_edges, unsigned char);
1897 state->solved = state->cheated = FALSE;
1899 state->grid_type = params->type;
1901 for (i = 0; i < num_faces; i++) {
1902 if (empties_to_make) {
1904 state->clues[i] = -1;
1910 n2 = *dp - 'A' + 10;
1911 if (n >= 0 && n < 10) {
1912 state->clues[i] = n;
1913 } else if (n2 >= 10 && n2 < 36) {
1914 state->clues[i] = n2;
1918 state->clues[i] = -1;
1919 empties_to_make = n - 1;
1924 memset(state->lines, LINE_UNKNOWN, num_edges);
1925 memset(state->line_errors, 0, num_edges);
1929 /* Calculates the line_errors data, and checks if the current state is a
1931 static int check_completion(game_state *state)
1933 grid *g = state->game_grid;
1935 int num_faces = g->num_faces;
1937 int infinite_area, finite_area;
1938 int loops_found = 0;
1939 int found_edge_not_in_loop = FALSE;
1941 memset(state->line_errors, 0, g->num_edges);
1943 /* LL implementation of SGT's idea:
1944 * A loop will partition the grid into an inside and an outside.
1945 * If there is more than one loop, the grid will be partitioned into
1946 * even more distinct regions. We can therefore track equivalence of
1947 * faces, by saying that two faces are equivalent when there is a non-YES
1948 * edge between them.
1949 * We could keep track of the number of connected components, by counting
1950 * the number of dsf-merges that aren't no-ops.
1951 * But we're only interested in 3 separate cases:
1952 * no loops, one loop, more than one loop.
1954 * No loops: all faces are equivalent to the infinite face.
1955 * One loop: only two equivalence classes - finite and infinite.
1956 * >= 2 loops: there are 2 distinct finite regions.
1958 * So we simply make two passes through all the edges.
1959 * In the first pass, we dsf-merge the two faces bordering each non-YES
1961 * In the second pass, we look for YES-edges bordering:
1962 * a) two non-equivalent faces.
1963 * b) two non-equivalent faces, and one of them is part of a different
1964 * finite area from the first finite area we've seen.
1966 * An occurrence of a) means there is at least one loop.
1967 * An occurrence of b) means there is more than one loop.
1968 * Edges satisfying a) are marked as errors.
1970 * While we're at it, we set a flag if we find a YES edge that is not
1972 * This information will help decide, if there's a single loop, whether it
1973 * is a candidate for being a solution (that is, all YES edges are part of
1976 * If there is a candidate loop, we then go through all clues and check
1977 * they are all satisfied. If so, we have found a solution and we can
1978 * unmark all line_errors.
1981 /* Infinite face is at the end - its index is num_faces.
1982 * This macro is just to make this obvious! */
1983 #define INF_FACE num_faces
1984 dsf = snewn(num_faces + 1, int);
1985 dsf_init(dsf, num_faces + 1);
1988 for (i = 0; i < g->num_edges; i++) {
1989 grid_edge *e = g->edges + i;
1990 int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
1991 int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
1992 if (state->lines[i] != LINE_YES)
1993 dsf_merge(dsf, f1, f2);
1997 infinite_area = dsf_canonify(dsf, INF_FACE);
1999 for (i = 0; i < g->num_edges; i++) {
2000 grid_edge *e = g->edges + i;
2001 int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
2002 int can1 = dsf_canonify(dsf, f1);
2003 int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
2004 int can2 = dsf_canonify(dsf, f2);
2005 if (state->lines[i] != LINE_YES) continue;
2008 /* Faces are equivalent, so this edge not part of a loop */
2009 found_edge_not_in_loop = TRUE;
2012 state->line_errors[i] = TRUE;
2013 if (loops_found == 0) loops_found = 1;
2015 /* Don't bother with further checks if we've already found 2 loops */
2016 if (loops_found == 2) continue;
2018 if (finite_area == -1) {
2019 /* Found our first finite area */
2020 if (can1 != infinite_area)
2026 /* Have we found a second area? */
2027 if (finite_area != -1) {
2028 if (can1 != infinite_area && can1 != finite_area) {
2032 if (can2 != infinite_area && can2 != finite_area) {
2039 printf("loops_found = %d\n", loops_found);
2040 printf("found_edge_not_in_loop = %s\n",
2041 found_edge_not_in_loop ? "TRUE" : "FALSE");
2044 sfree(dsf); /* No longer need the dsf */
2046 /* Have we found a candidate loop? */
2047 if (loops_found == 1 && !found_edge_not_in_loop) {
2048 /* Yes, so check all clues are satisfied */
2049 int found_clue_violation = FALSE;
2050 for (i = 0; i < num_faces; i++) {
2051 int c = state->clues[i];
2053 if (face_order(state, i, LINE_YES) != c) {
2054 found_clue_violation = TRUE;
2060 if (!found_clue_violation) {
2061 /* The loop is good */
2062 memset(state->line_errors, 0, g->num_edges);
2063 return TRUE; /* No need to bother checking for dot violations */
2067 /* Check for dot violations */
2068 for (i = 0; i < g->num_dots; i++) {
2069 int yes = dot_order(state, i, LINE_YES);
2070 int unknown = dot_order(state, i, LINE_UNKNOWN);
2071 if ((yes == 1 && unknown == 0) || (yes >= 3)) {
2072 /* violation, so mark all YES edges as errors */
2073 grid_dot *d = g->dots + i;
2075 for (j = 0; j < d->order; j++) {
2076 int e = d->edges[j] - g->edges;
2077 if (state->lines[e] == LINE_YES)
2078 state->line_errors[e] = TRUE;
2085 /* ----------------------------------------------------------------------
2088 * Our solver modes operate as follows. Each mode also uses the modes above it.
2091 * Just implement the rules of the game.
2093 * Normal and Tricky Modes
2094 * For each (adjacent) pair of lines through each dot we store a bit for
2095 * whether at least one of them is on and whether at most one is on. (If we
2096 * know both or neither is on that's already stored more directly.)
2099 * Use edsf data structure to make equivalence classes of lines that are
2100 * known identical to or opposite to one another.
2105 * For general grids, we consider "dlines" to be pairs of lines joined
2106 * at a dot. The lines must be adjacent around the dot, so we can think of
2107 * a dline as being a dot+face combination. Or, a dot+edge combination where
2108 * the second edge is taken to be the next clockwise edge from the dot.
2109 * Original loopy code didn't have this extra restriction of the lines being
2110 * adjacent. From my tests with square grids, this extra restriction seems to
2111 * take little, if anything, away from the quality of the puzzles.
2112 * A dline can be uniquely identified by an edge/dot combination, given that
2113 * a dline-pair always goes clockwise around its common dot. The edge/dot
2114 * combination can be represented by an edge/bool combination - if bool is
2115 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2116 * exactly twice the number of edges in the grid - although the dlines
2117 * spanning the infinite face are not all that useful to the solver.
2118 * Note that, by convention, a dline goes clockwise around its common dot,
2119 * which means the dline goes anti-clockwise around its common face.
2122 /* Helper functions for obtaining an index into an array of dlines, given
2123 * various information. We assume the grid layout conventions about how
2124 * the various lists are interleaved - see grid_make_consistent() for
2127 /* i points to the first edge of the dline pair, reading clockwise around
2129 static int dline_index_from_dot(grid *g, grid_dot *d, int i)
2131 grid_edge *e = d->edges[i];
2136 if (i2 == d->order) i2 = 0;
2139 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
2141 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2142 (int)(d - g->dots), i, (int)(e - g->edges),
2143 (int)(e2 - g->edges), ret);
2147 /* i points to the second edge of the dline pair, reading clockwise around
2148 * the face. That is, the edges of the dline, starting at edge{i}, read
2149 * anti-clockwise around the face. By layout conventions, the common dot
2150 * of the dline will be f->dots[i] */
2151 static int dline_index_from_face(grid *g, grid_face *f, int i)
2153 grid_edge *e = f->edges[i];
2154 grid_dot *d = f->dots[i];
2159 if (i2 < 0) i2 += f->order;
2162 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
2164 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2165 (int)(f - g->faces), i, (int)(e - g->edges),
2166 (int)(e2 - g->edges), ret);
2170 static int is_atleastone(const char *dline_array, int index)
2172 return BIT_SET(dline_array[index], 0);
2174 static int set_atleastone(char *dline_array, int index)
2176 return SET_BIT(dline_array[index], 0);
2178 static int is_atmostone(const char *dline_array, int index)
2180 return BIT_SET(dline_array[index], 1);
2182 static int set_atmostone(char *dline_array, int index)
2184 return SET_BIT(dline_array[index], 1);
2187 static void array_setall(char *array, char from, char to, int len)
2189 char *p = array, *p_old = p;
2190 int len_remaining = len;
2192 while ((p = memchr(p, from, len_remaining))) {
2194 len_remaining -= p - p_old;
2199 /* Helper, called when doing dline dot deductions, in the case where we
2200 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2201 * them (because of dline atmostone/atleastone).
2202 * On entry, edge points to the first of these two UNKNOWNs. This function
2203 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2204 * and set their corresponding dline to atleastone. (Setting atmostone
2205 * already happens in earlier dline deductions) */
2206 static int dline_set_opp_atleastone(solver_state *sstate,
2207 grid_dot *d, int edge)
2209 game_state *state = sstate->state;
2210 grid *g = state->game_grid;
2213 for (opp = 0; opp < N; opp++) {
2214 int opp_dline_index;
2215 if (opp == edge || opp == edge+1 || opp == edge-1)
2217 if (opp == 0 && edge == N-1)
2219 if (opp == N-1 && edge == 0)
2222 if (opp2 == N) opp2 = 0;
2223 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2224 if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
2226 if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
2228 /* Found opposite UNKNOWNS and they're next to each other */
2229 opp_dline_index = dline_index_from_dot(g, d, opp);
2230 return set_atleastone(sstate->dlines, opp_dline_index);
2236 /* Set pairs of lines around this face which are known to be identical, to
2237 * the given line_state */
2238 static int face_setall_identical(solver_state *sstate, int face_index,
2239 enum line_state line_new)
2241 /* can[dir] contains the canonical line associated with the line in
2242 * direction dir from the square in question. Similarly inv[dir] is
2243 * whether or not the line in question is inverse to its canonical
2246 game_state *state = sstate->state;
2247 grid *g = state->game_grid;
2248 grid_face *f = g->faces + face_index;
2251 int can1, can2, inv1, inv2;
2253 for (i = 0; i < N; i++) {
2254 int line1_index = f->edges[i] - g->edges;
2255 if (state->lines[line1_index] != LINE_UNKNOWN)
2257 for (j = i + 1; j < N; j++) {
2258 int line2_index = f->edges[j] - g->edges;
2259 if (state->lines[line2_index] != LINE_UNKNOWN)
2262 /* Found two UNKNOWNS */
2263 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
2264 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
2265 if (can1 == can2 && inv1 == inv2) {
2266 solver_set_line(sstate, line1_index, line_new);
2267 solver_set_line(sstate, line2_index, line_new);
2274 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2275 * return the edge indices into e. */
2276 static void find_unknowns(game_state *state,
2277 grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
2278 int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
2279 int *e /* Returned edge indices */)
2282 grid *g = state->game_grid;
2283 while (c < expected_count) {
2284 int line_index = *edge_list - g->edges;
2285 if (state->lines[line_index] == LINE_UNKNOWN) {
2293 /* If we have a list of edges, and we know whether the number of YESs should
2294 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2295 * linedsf deductions. This can be used for both face and dot deductions.
2296 * Returns the difficulty level of the next solver that should be used,
2297 * or DIFF_MAX if no progress was made. */
2298 static int parity_deductions(solver_state *sstate,
2299 grid_edge **edge_list, /* Edge list (from a face or a dot) */
2300 int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
2303 game_state *state = sstate->state;
2304 int diff = DIFF_MAX;
2305 int *linedsf = sstate->linedsf;
2307 if (unknown_count == 2) {
2308 /* Lines are known alike/opposite, depending on inv. */
2310 find_unknowns(state, edge_list, 2, e);
2311 if (merge_lines(sstate, e[0], e[1], total_parity))
2312 diff = min(diff, DIFF_HARD);
2313 } else if (unknown_count == 3) {
2315 int can[3]; /* canonical edges */
2316 int inv[3]; /* whether can[x] is inverse to e[x] */
2317 find_unknowns(state, edge_list, 3, e);
2318 can[0] = edsf_canonify(linedsf, e[0], inv);
2319 can[1] = edsf_canonify(linedsf, e[1], inv+1);
2320 can[2] = edsf_canonify(linedsf, e[2], inv+2);
2321 if (can[0] == can[1]) {
2322 if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
2323 LINE_YES : LINE_NO))
2324 diff = min(diff, DIFF_EASY);
2326 if (can[0] == can[2]) {
2327 if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
2328 LINE_YES : LINE_NO))
2329 diff = min(diff, DIFF_EASY);
2331 if (can[1] == can[2]) {
2332 if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
2333 LINE_YES : LINE_NO))
2334 diff = min(diff, DIFF_EASY);
2336 } else if (unknown_count == 4) {
2338 int can[4]; /* canonical edges */
2339 int inv[4]; /* whether can[x] is inverse to e[x] */
2340 find_unknowns(state, edge_list, 4, e);
2341 can[0] = edsf_canonify(linedsf, e[0], inv);
2342 can[1] = edsf_canonify(linedsf, e[1], inv+1);
2343 can[2] = edsf_canonify(linedsf, e[2], inv+2);
2344 can[3] = edsf_canonify(linedsf, e[3], inv+3);
2345 if (can[0] == can[1]) {
2346 if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
2347 diff = min(diff, DIFF_HARD);
2348 } else if (can[0] == can[2]) {
2349 if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
2350 diff = min(diff, DIFF_HARD);
2351 } else if (can[0] == can[3]) {
2352 if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
2353 diff = min(diff, DIFF_HARD);
2354 } else if (can[1] == can[2]) {
2355 if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
2356 diff = min(diff, DIFF_HARD);
2357 } else if (can[1] == can[3]) {
2358 if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
2359 diff = min(diff, DIFF_HARD);
2360 } else if (can[2] == can[3]) {
2361 if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
2362 diff = min(diff, DIFF_HARD);
2370 * These are the main solver functions.
2372 * Their return values are diff values corresponding to the lowest mode solver
2373 * that would notice the work that they have done. For example if the normal
2374 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2375 * easy mode solver might be able to make progress using that. It doesn't make
2376 * sense for one of them to return a diff value higher than that of the
2379 * Each function returns the lowest value it can, as early as possible, in
2380 * order to try and pass as much work as possible back to the lower level
2381 * solvers which progress more quickly.
2384 /* PROPOSED NEW DESIGN:
2385 * We have a work queue consisting of 'events' notifying us that something has
2386 * happened that a particular solver mode might be interested in. For example
2387 * the hard mode solver might do something that helps the normal mode solver at
2388 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2389 * we pull events off the work queue, and hand each in turn to the solver that
2390 * is interested in them. If a solver reports that it failed we pass the same
2391 * event on to progressively more advanced solvers and the loop detector. Once
2392 * we've exhausted an event, or it has helped us progress, we drop it and
2393 * continue to the next one. The events are sorted first in order of solver
2394 * complexity (easy first) then order of insertion (oldest first).
2395 * Once we run out of events we loop over each permitted solver in turn
2396 * (easiest first) until either a deduction is made (and an event therefore
2397 * emerges) or no further deductions can be made (in which case we've failed).
2400 * * How do we 'loop over' a solver when both dots and squares are concerned.
2401 * Answer: first all squares then all dots.
2404 static int trivial_deductions(solver_state *sstate)
2406 int i, current_yes, current_no;
2407 game_state *state = sstate->state;
2408 grid *g = state->game_grid;
2409 int diff = DIFF_MAX;
2411 /* Per-face deductions */
2412 for (i = 0; i < g->num_faces; i++) {
2413 grid_face *f = g->faces + i;
2415 if (sstate->face_solved[i])
2418 current_yes = sstate->face_yes_count[i];
2419 current_no = sstate->face_no_count[i];
2421 if (current_yes + current_no == f->order) {
2422 sstate->face_solved[i] = TRUE;
2426 if (state->clues[i] < 0)
2429 if (state->clues[i] < current_yes) {
2430 sstate->solver_status = SOLVER_MISTAKE;
2433 if (state->clues[i] == current_yes) {
2434 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
2435 diff = min(diff, DIFF_EASY);
2436 sstate->face_solved[i] = TRUE;
2440 if (f->order - state->clues[i] < current_no) {
2441 sstate->solver_status = SOLVER_MISTAKE;
2444 if (f->order - state->clues[i] == current_no) {
2445 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
2446 diff = min(diff, DIFF_EASY);
2447 sstate->face_solved[i] = TRUE;
2452 check_caches(sstate);
2454 /* Per-dot deductions */
2455 for (i = 0; i < g->num_dots; i++) {
2456 grid_dot *d = g->dots + i;
2457 int yes, no, unknown;
2459 if (sstate->dot_solved[i])
2462 yes = sstate->dot_yes_count[i];
2463 no = sstate->dot_no_count[i];
2464 unknown = d->order - yes - no;
2468 sstate->dot_solved[i] = TRUE;
2469 } else if (unknown == 1) {
2470 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2471 diff = min(diff, DIFF_EASY);
2472 sstate->dot_solved[i] = TRUE;
2474 } else if (yes == 1) {
2476 sstate->solver_status = SOLVER_MISTAKE;
2478 } else if (unknown == 1) {
2479 dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
2480 diff = min(diff, DIFF_EASY);
2482 } else if (yes == 2) {
2484 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2485 diff = min(diff, DIFF_EASY);
2487 sstate->dot_solved[i] = TRUE;
2489 sstate->solver_status = SOLVER_MISTAKE;
2494 check_caches(sstate);
2499 static int dline_deductions(solver_state *sstate)
2501 game_state *state = sstate->state;
2502 grid *g = state->game_grid;
2503 char *dlines = sstate->dlines;
2505 int diff = DIFF_MAX;
2507 /* ------ Face deductions ------ */
2509 /* Given a set of dline atmostone/atleastone constraints, need to figure
2510 * out if we can deduce any further info. For more general faces than
2511 * squares, this turns out to be a tricky problem.
2512 * The approach taken here is to define (per face) NxN matrices:
2513 * "maxs" and "mins".
2514 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2515 * for the possible number of edges that are YES between positions j and k
2516 * going clockwise around the face. Can think of j and k as marking dots
2517 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2518 * edge1 joins dot1 to dot2 etc).
2519 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2520 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2521 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2522 * the dline atmostone/atleastone status for edges j and j+1.
2524 * Then we calculate the remaining entries recursively. We definitely
2526 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2527 * This is because any valid placement of YESs between j and k must give
2528 * a valid placement between j and u, and also between u and k.
2529 * I believe it's sufficient to use just the two values of u:
2530 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2531 * are rigorous, even if they might not be best-possible.
2533 * Once we have maxs and mins calculated, we can make inferences about
2534 * each dline{j,j+1} by looking at the possible complementary edge-counts
2535 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2536 * As well as dlines, we can make similar inferences about single edges.
2537 * For example, consider a pentagon with clue 3, and we know at most one
2538 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2539 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2540 * that final edge would have to be YES to make the count up to 3.
2543 /* Much quicker to allocate arrays on the stack than the heap, so
2544 * define the largest possible face size, and base our array allocations
2545 * on that. We check this with an assertion, in case someone decides to
2546 * make a grid which has larger faces than this. Note, this algorithm
2547 * could get quite expensive if there are many large faces. */
2548 #define MAX_FACE_SIZE 12
2550 for (i = 0; i < g->num_faces; i++) {
2551 int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
2552 int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
2553 grid_face *f = g->faces + i;
2556 int clue = state->clues[i];
2557 assert(N <= MAX_FACE_SIZE);
2558 if (sstate->face_solved[i])
2560 if (clue < 0) continue;
2562 /* Calculate the (j,j+1) entries */
2563 for (j = 0; j < N; j++) {
2564 int edge_index = f->edges[j] - g->edges;
2566 enum line_state line1 = state->lines[edge_index];
2567 enum line_state line2;
2571 maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
2572 mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
2573 /* Calculate the (j,j+2) entries */
2574 dline_index = dline_index_from_face(g, f, k);
2575 edge_index = f->edges[k] - g->edges;
2576 line2 = state->lines[edge_index];
2582 if (line1 == LINE_NO) tmp--;
2583 if (line2 == LINE_NO) tmp--;
2584 if (tmp == 2 && is_atmostone(dlines, dline_index))
2590 if (line1 == LINE_YES) tmp++;
2591 if (line2 == LINE_YES) tmp++;
2592 if (tmp == 0 && is_atleastone(dlines, dline_index))
2597 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2598 for (m = 3; m < N; m++) {
2599 for (j = 0; j < N; j++) {
2607 maxs[j][k] = maxs[j][u] + maxs[u][k];
2608 mins[j][k] = mins[j][u] + mins[u][k];
2609 tmp = maxs[j][v] + maxs[v][k];
2610 maxs[j][k] = min(maxs[j][k], tmp);
2611 tmp = mins[j][v] + mins[v][k];
2612 mins[j][k] = max(mins[j][k], tmp);
2616 /* See if we can make any deductions */
2617 for (j = 0; j < N; j++) {
2619 grid_edge *e = f->edges[j];
2620 int line_index = e - g->edges;
2623 if (state->lines[line_index] != LINE_UNKNOWN)
2628 /* minimum YESs in the complement of this edge */
2629 if (mins[k][j] > clue) {
2630 sstate->solver_status = SOLVER_MISTAKE;
2633 if (mins[k][j] == clue) {
2634 /* setting this edge to YES would make at least
2635 * (clue+1) edges - contradiction */
2636 solver_set_line(sstate, line_index, LINE_NO);
2637 diff = min(diff, DIFF_EASY);
2639 if (maxs[k][j] < clue - 1) {
2640 sstate->solver_status = SOLVER_MISTAKE;
2643 if (maxs[k][j] == clue - 1) {
2644 /* Only way to satisfy the clue is to set edge{j} as YES */
2645 solver_set_line(sstate, line_index, LINE_YES);
2646 diff = min(diff, DIFF_EASY);
2649 /* More advanced deduction that allows propagation along diagonal
2650 * chains of faces connected by dots, for example, 3-2-...-2-3
2651 * in square grids. */
2652 if (sstate->diff >= DIFF_TRICKY) {
2653 /* Now see if we can make dline deduction for edges{j,j+1} */
2655 if (state->lines[e - g->edges] != LINE_UNKNOWN)
2656 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2657 * Dlines where one of the edges is known, are handled in the
2661 dline_index = dline_index_from_face(g, f, k);
2665 /* minimum YESs in the complement of this dline */
2666 if (mins[k][j] > clue - 2) {
2667 /* Adding 2 YESs would break the clue */
2668 if (set_atmostone(dlines, dline_index))
2669 diff = min(diff, DIFF_NORMAL);
2671 /* maximum YESs in the complement of this dline */
2672 if (maxs[k][j] < clue) {
2673 /* Adding 2 NOs would mean not enough YESs */
2674 if (set_atleastone(dlines, dline_index))
2675 diff = min(diff, DIFF_NORMAL);
2681 if (diff < DIFF_NORMAL)
2684 /* ------ Dot deductions ------ */
2686 for (i = 0; i < g->num_dots; i++) {
2687 grid_dot *d = g->dots + i;
2689 int yes, no, unknown;
2691 if (sstate->dot_solved[i])
2693 yes = sstate->dot_yes_count[i];
2694 no = sstate->dot_no_count[i];
2695 unknown = N - yes - no;
2697 for (j = 0; j < N; j++) {
2700 int line1_index, line2_index;
2701 enum line_state line1, line2;
2704 dline_index = dline_index_from_dot(g, d, j);
2705 line1_index = d->edges[j] - g->edges;
2706 line2_index = d->edges[k] - g->edges;
2707 line1 = state->lines[line1_index];
2708 line2 = state->lines[line2_index];
2710 /* Infer dline state from line state */
2711 if (line1 == LINE_NO || line2 == LINE_NO) {
2712 if (set_atmostone(dlines, dline_index))
2713 diff = min(diff, DIFF_NORMAL);
2715 if (line1 == LINE_YES || line2 == LINE_YES) {
2716 if (set_atleastone(dlines, dline_index))
2717 diff = min(diff, DIFF_NORMAL);
2719 /* Infer line state from dline state */
2720 if (is_atmostone(dlines, dline_index)) {
2721 if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
2722 solver_set_line(sstate, line2_index, LINE_NO);
2723 diff = min(diff, DIFF_EASY);
2725 if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
2726 solver_set_line(sstate, line1_index, LINE_NO);
2727 diff = min(diff, DIFF_EASY);
2730 if (is_atleastone(dlines, dline_index)) {
2731 if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
2732 solver_set_line(sstate, line2_index, LINE_YES);
2733 diff = min(diff, DIFF_EASY);
2735 if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
2736 solver_set_line(sstate, line1_index, LINE_YES);
2737 diff = min(diff, DIFF_EASY);
2740 /* Deductions that depend on the numbers of lines.
2741 * Only bother if both lines are UNKNOWN, otherwise the
2742 * easy-mode solver (or deductions above) would have taken
2744 if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
2747 if (yes == 0 && unknown == 2) {
2748 /* Both these unknowns must be identical. If we know
2749 * atmostone or atleastone, we can make progress. */
2750 if (is_atmostone(dlines, dline_index)) {
2751 solver_set_line(sstate, line1_index, LINE_NO);
2752 solver_set_line(sstate, line2_index, LINE_NO);
2753 diff = min(diff, DIFF_EASY);
2755 if (is_atleastone(dlines, dline_index)) {
2756 solver_set_line(sstate, line1_index, LINE_YES);
2757 solver_set_line(sstate, line2_index, LINE_YES);
2758 diff = min(diff, DIFF_EASY);
2762 if (set_atmostone(dlines, dline_index))
2763 diff = min(diff, DIFF_NORMAL);
2765 if (set_atleastone(dlines, dline_index))
2766 diff = min(diff, DIFF_NORMAL);
2770 /* More advanced deduction that allows propagation along diagonal
2771 * chains of faces connected by dots, for example: 3-2-...-2-3
2772 * in square grids. */
2773 if (sstate->diff >= DIFF_TRICKY) {
2774 /* If we have atleastone set for this dline, infer
2775 * atmostone for each "opposite" dline (that is, each
2776 * dline without edges in common with this one).
2777 * Again, this test is only worth doing if both these
2778 * lines are UNKNOWN. For if one of these lines were YES,
2779 * the (yes == 1) test above would kick in instead. */
2780 if (is_atleastone(dlines, dline_index)) {
2782 for (opp = 0; opp < N; opp++) {
2783 int opp_dline_index;
2784 if (opp == j || opp == j+1 || opp == j-1)
2786 if (j == 0 && opp == N-1)
2788 if (j == N-1 && opp == 0)
2790 opp_dline_index = dline_index_from_dot(g, d, opp);
2791 if (set_atmostone(dlines, opp_dline_index))
2792 diff = min(diff, DIFF_NORMAL);
2794 if (yes == 0 && is_atmostone(dlines, dline_index)) {
2795 /* This dline has *exactly* one YES and there are no
2796 * other YESs. This allows more deductions. */
2798 /* Third unknown must be YES */
2799 for (opp = 0; opp < N; opp++) {
2801 if (opp == j || opp == k)
2803 opp_index = d->edges[opp] - g->edges;
2804 if (state->lines[opp_index] == LINE_UNKNOWN) {
2805 solver_set_line(sstate, opp_index,
2807 diff = min(diff, DIFF_EASY);
2810 } else if (unknown == 4) {
2811 /* Exactly one of opposite UNKNOWNS is YES. We've
2812 * already set atmostone, so set atleastone as
2815 if (dline_set_opp_atleastone(sstate, d, j))
2816 diff = min(diff, DIFF_NORMAL);
2826 static int linedsf_deductions(solver_state *sstate)
2828 game_state *state = sstate->state;
2829 grid *g = state->game_grid;
2830 char *dlines = sstate->dlines;
2832 int diff = DIFF_MAX;
2835 /* ------ Face deductions ------ */
2837 /* A fully-general linedsf deduction seems overly complicated
2838 * (I suspect the problem is NP-complete, though in practice it might just
2839 * be doable because faces are limited in size).
2840 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2841 * known to be identical. If setting them both to YES (or NO) would break
2842 * the clue, set them to NO (or YES). */
2844 for (i = 0; i < g->num_faces; i++) {
2845 int N, yes, no, unknown;
2848 if (sstate->face_solved[i])
2850 clue = state->clues[i];
2854 N = g->faces[i].order;
2855 yes = sstate->face_yes_count[i];
2856 if (yes + 1 == clue) {
2857 if (face_setall_identical(sstate, i, LINE_NO))
2858 diff = min(diff, DIFF_EASY);
2860 no = sstate->face_no_count[i];
2861 if (no + 1 == N - clue) {
2862 if (face_setall_identical(sstate, i, LINE_YES))
2863 diff = min(diff, DIFF_EASY);
2866 /* Reload YES count, it might have changed */
2867 yes = sstate->face_yes_count[i];
2868 unknown = N - no - yes;
2870 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2871 * parity of lines. */
2872 diff_tmp = parity_deductions(sstate, g->faces[i].edges,
2873 (clue - yes) % 2, unknown);
2874 diff = min(diff, diff_tmp);
2877 /* ------ Dot deductions ------ */
2878 for (i = 0; i < g->num_dots; i++) {
2879 grid_dot *d = g->dots + i;
2882 int yes, no, unknown;
2883 /* Go through dlines, and do any dline<->linedsf deductions wherever
2884 * we find two UNKNOWNS. */
2885 for (j = 0; j < N; j++) {
2886 int dline_index = dline_index_from_dot(g, d, j);
2889 int can1, can2, inv1, inv2;
2891 line1_index = d->edges[j] - g->edges;
2892 if (state->lines[line1_index] != LINE_UNKNOWN)
2895 if (j2 == N) j2 = 0;
2896 line2_index = d->edges[j2] - g->edges;
2897 if (state->lines[line2_index] != LINE_UNKNOWN)
2899 /* Infer dline flags from linedsf */
2900 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
2901 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
2902 if (can1 == can2 && inv1 != inv2) {
2903 /* These are opposites, so set dline atmostone/atleastone */
2904 if (set_atmostone(dlines, dline_index))
2905 diff = min(diff, DIFF_NORMAL);
2906 if (set_atleastone(dlines, dline_index))
2907 diff = min(diff, DIFF_NORMAL);
2910 /* Infer linedsf from dline flags */
2911 if (is_atmostone(dlines, dline_index)
2912 && is_atleastone(dlines, dline_index)) {
2913 if (merge_lines(sstate, line1_index, line2_index, 1))
2914 diff = min(diff, DIFF_HARD);
2918 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2919 * parity of lines. */
2920 yes = sstate->dot_yes_count[i];
2921 no = sstate->dot_no_count[i];
2922 unknown = N - yes - no;
2923 diff_tmp = parity_deductions(sstate, d->edges,
2925 diff = min(diff, diff_tmp);
2928 /* ------ Edge dsf deductions ------ */
2930 /* If the state of a line is known, deduce the state of its canonical line
2931 * too, and vice versa. */
2932 for (i = 0; i < g->num_edges; i++) {
2935 can = edsf_canonify(sstate->linedsf, i, &inv);
2938 s = sstate->state->lines[can];
2939 if (s != LINE_UNKNOWN) {
2940 if (solver_set_line(sstate, i, inv ? OPP(s) : s))
2941 diff = min(diff, DIFF_EASY);
2943 s = sstate->state->lines[i];
2944 if (s != LINE_UNKNOWN) {
2945 if (solver_set_line(sstate, can, inv ? OPP(s) : s))
2946 diff = min(diff, DIFF_EASY);
2954 static int loop_deductions(solver_state *sstate)
2956 int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
2957 game_state *state = sstate->state;
2958 grid *g = state->game_grid;
2959 int shortest_chainlen = g->num_dots;
2960 int loop_found = FALSE;
2962 int progress = FALSE;
2966 * Go through the grid and update for all the new edges.
2967 * Since merge_dots() is idempotent, the simplest way to
2968 * do this is just to update for _all_ the edges.
2969 * Also, while we're here, we count the edges.
2971 for (i = 0; i < g->num_edges; i++) {
2972 if (state->lines[i] == LINE_YES) {
2973 loop_found |= merge_dots(sstate, i);
2979 * Count the clues, count the satisfied clues, and count the
2980 * satisfied-minus-one clues.
2982 for (i = 0; i < g->num_faces; i++) {
2983 int c = state->clues[i];
2985 int o = sstate->face_yes_count[i];
2994 for (i = 0; i < g->num_dots; ++i) {
2996 sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
2997 if (dots_connected > 1)
2998 shortest_chainlen = min(shortest_chainlen, dots_connected);
3001 assert(sstate->solver_status == SOLVER_INCOMPLETE);
3003 if (satclues == clues && shortest_chainlen == edgecount) {
3004 sstate->solver_status = SOLVER_SOLVED;
3005 /* This discovery clearly counts as progress, even if we haven't
3006 * just added any lines or anything */
3008 goto finished_loop_deductionsing;
3012 * Now go through looking for LINE_UNKNOWN edges which
3013 * connect two dots that are already in the same
3014 * equivalence class. If we find one, test to see if the
3015 * loop it would create is a solution.
3017 for (i = 0; i < g->num_edges; i++) {
3018 grid_edge *e = g->edges + i;
3019 int d1 = e->dot1 - g->dots;
3020 int d2 = e->dot2 - g->dots;
3022 if (state->lines[i] != LINE_UNKNOWN)
3025 eqclass = dsf_canonify(sstate->dotdsf, d1);
3026 if (eqclass != dsf_canonify(sstate->dotdsf, d2))
3029 val = LINE_NO; /* loop is bad until proven otherwise */
3032 * This edge would form a loop. Next
3033 * question: how long would the loop be?
3034 * Would it equal the total number of edges
3035 * (plus the one we'd be adding if we added
3038 if (sstate->looplen[eqclass] == edgecount + 1) {
3042 * This edge would form a loop which
3043 * took in all the edges in the entire
3044 * grid. So now we need to work out
3045 * whether it would be a valid solution
3046 * to the puzzle, which means we have to
3047 * check if it satisfies all the clues.
3048 * This means that every clue must be
3049 * either satisfied or satisfied-minus-
3050 * 1, and also that the number of
3051 * satisfied-minus-1 clues must be at
3052 * most two and they must lie on either
3053 * side of this edge.
3057 int f = e->face1 - g->faces;
3058 int c = state->clues[f];
3059 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
3063 int f = e->face2 - g->faces;
3064 int c = state->clues[f];
3065 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
3068 if (sm1clues == sm1_nearby &&
3069 sm1clues + satclues == clues) {
3070 val = LINE_YES; /* loop is good! */
3075 * Right. Now we know that adding this edge
3076 * would form a loop, and we know whether
3077 * that loop would be a viable solution or
3080 * If adding this edge produces a solution,
3081 * then we know we've found _a_ solution but
3082 * we don't know that it's _the_ solution -
3083 * if it were provably the solution then
3084 * we'd have deduced this edge some time ago
3085 * without the need to do loop detection. So
3086 * in this state we return SOLVER_AMBIGUOUS,
3087 * which has the effect that hitting Solve
3088 * on a user-provided puzzle will fill in a
3089 * solution but using the solver to
3090 * construct new puzzles won't consider this
3091 * a reasonable deduction for the user to
3094 progress = solver_set_line(sstate, i, val);
3095 assert(progress == TRUE);
3096 if (val == LINE_YES) {
3097 sstate->solver_status = SOLVER_AMBIGUOUS;
3098 goto finished_loop_deductionsing;
3102 finished_loop_deductionsing:
3103 return progress ? DIFF_EASY : DIFF_MAX;
3106 /* This will return a dynamically allocated solver_state containing the (more)
3108 static solver_state *solve_game_rec(const solver_state *sstate_start)
3110 solver_state *sstate;
3112 /* Index of the solver we should call next. */
3115 /* As a speed-optimisation, we avoid re-running solvers that we know
3116 * won't make any progress. This happens when a high-difficulty
3117 * solver makes a deduction that can only help other high-difficulty
3119 * For example: if a new 'dline' flag is set by dline_deductions, the
3120 * trivial_deductions solver cannot do anything with this information.
3121 * If we've already run the trivial_deductions solver (because it's
3122 * earlier in the list), there's no point running it again.
3124 * Therefore: if a solver is earlier in the list than "threshold_index",
3125 * we don't bother running it if it's difficulty level is less than
3128 int threshold_diff = 0;
3129 int threshold_index = 0;
3131 sstate = dup_solver_state(sstate_start);
3133 check_caches(sstate);
3135 while (i < NUM_SOLVERS) {
3136 if (sstate->solver_status == SOLVER_MISTAKE)
3138 if (sstate->solver_status == SOLVER_SOLVED ||
3139 sstate->solver_status == SOLVER_AMBIGUOUS) {
3140 /* solver finished */
3144 if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
3145 && solver_diffs[i] <= sstate->diff) {
3146 /* current_solver is eligible, so use it */
3147 int next_diff = solver_fns[i](sstate);
3148 if (next_diff != DIFF_MAX) {
3149 /* solver made progress, so use new thresholds and
3150 * start again at top of list. */
3151 threshold_diff = next_diff;
3152 threshold_index = i;
3157 /* current_solver is ineligible, or failed to make progress, so
3158 * go to the next solver in the list */
3162 if (sstate->solver_status == SOLVER_SOLVED ||
3163 sstate->solver_status == SOLVER_AMBIGUOUS) {
3164 /* s/LINE_UNKNOWN/LINE_NO/g */
3165 array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
3166 sstate->state->game_grid->num_edges);
3173 static char *solve_game(game_state *state, game_state *currstate,
3174 char *aux, char **error)
3177 solver_state *sstate, *new_sstate;
3179 sstate = new_solver_state(state, DIFF_MAX);
3180 new_sstate = solve_game_rec(sstate);
3182 if (new_sstate->solver_status == SOLVER_SOLVED) {
3183 soln = encode_solve_move(new_sstate->state);
3184 } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
3185 soln = encode_solve_move(new_sstate->state);
3186 /**error = "Solver found ambiguous solutions"; */
3188 soln = encode_solve_move(new_sstate->state);
3189 /**error = "Solver failed"; */
3192 free_solver_state(new_sstate);
3193 free_solver_state(sstate);
3198 /* ----------------------------------------------------------------------
3199 * Drawing and mouse-handling
3202 static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
3203 int x, int y, int button)
3205 grid *g = state->game_grid;
3209 char button_char = ' ';
3210 enum line_state old_state;
3212 button &= ~MOD_MASK;
3214 /* Convert mouse-click (x,y) to grid coordinates */
3215 x -= BORDER(ds->tilesize);
3216 y -= BORDER(ds->tilesize);
3217 x = x * g->tilesize / ds->tilesize;
3218 y = y * g->tilesize / ds->tilesize;
3222 e = grid_nearest_edge(g, x, y);
3228 /* I think it's only possible to play this game with mouse clicks, sorry */
3229 /* Maybe will add mouse drag support some time */
3230 old_state = state->lines[i];
3234 switch (old_state) {
3252 switch (old_state) {
3271 sprintf(buf, "%d%c", i, (int)button_char);
3277 static game_state *execute_move(game_state *state, char *move)
3280 game_state *newstate = dup_game(state);
3282 if (move[0] == 'S') {
3284 newstate->cheated = TRUE;
3289 if (i < 0 || i >= newstate->game_grid->num_edges)
3291 move += strspn(move, "1234567890");
3292 switch (*(move++)) {
3294 newstate->lines[i] = LINE_YES;
3297 newstate->lines[i] = LINE_NO;
3300 newstate->lines[i] = LINE_UNKNOWN;
3308 * Check for completion.
3310 if (check_completion(newstate))
3311 newstate->solved = TRUE;
3316 free_game(newstate);
3320 /* ----------------------------------------------------------------------
3324 /* Convert from grid coordinates to screen coordinates */
3325 static void grid_to_screen(const game_drawstate *ds, const grid *g,
3326 int grid_x, int grid_y, int *x, int *y)
3328 *x = grid_x - g->lowest_x;
3329 *y = grid_y - g->lowest_y;
3330 *x = *x * ds->tilesize / g->tilesize;
3331 *y = *y * ds->tilesize / g->tilesize;
3332 *x += BORDER(ds->tilesize);
3333 *y += BORDER(ds->tilesize);
3336 /* Returns (into x,y) position of centre of face for rendering the text clue.
3338 static void face_text_pos(const game_drawstate *ds, const grid *g,
3339 const grid_face *f, int *x, int *y)
3343 /* Simplest solution is the centroid. Might not work in some cases. */
3345 /* Another algorithm to look into:
3346 * Find the midpoints of the sides, find the bounding-box,
3347 * then take the centre of that. */
3349 /* Best solution probably involves incentres (inscribed circles) */
3351 int sx = 0, sy = 0; /* sums */
3352 for (i = 0; i < f->order; i++) {
3353 grid_dot *d = f->dots[i];
3360 /* convert to screen coordinates */
3361 grid_to_screen(ds, g, sx, sy, x, y);
3364 static void game_redraw_clue(drawing *dr, game_drawstate *ds,
3365 game_state *state, int i)
3367 grid *g = state->game_grid;
3368 grid_face *f = g->faces + i;
3372 if (state->clues[i] < 10) {
3373 c[0] = CLUE2CHAR(state->clues[i]);
3376 sprintf(c, "%d", state->clues[i]);
3379 face_text_pos(ds, g, f, &x, &y);
3381 FONT_VARIABLE, ds->tilesize/2,
3382 ALIGN_VCENTRE | ALIGN_HCENTRE,
3383 ds->clue_error[i] ? COL_MISTAKE :
3384 ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
3387 static void game_redraw_line(drawing *dr, game_drawstate *ds,
3388 game_state *state, int i)
3390 grid *g = state->game_grid;
3391 grid_edge *e = g->edges + i;
3393 int xmin, ymin, xmax, ymax;
3396 if (state->line_errors[i])
3397 line_colour = COL_MISTAKE;
3398 else if (state->lines[i] == LINE_UNKNOWN)
3399 line_colour = COL_LINEUNKNOWN;
3400 else if (state->lines[i] == LINE_NO)
3401 line_colour = COL_FAINT;
3402 else if (ds->flashing)
3403 line_colour = COL_HIGHLIGHT;
3405 line_colour = COL_FOREGROUND;
3407 /* Convert from grid to screen coordinates */
3408 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3409 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3416 if (line_colour == COL_FAINT) {
3417 static int draw_faint_lines = -1;
3418 if (draw_faint_lines < 0) {
3419 char *env = getenv("LOOPY_FAINT_LINES");
3420 draw_faint_lines = (!env || (env[0] == 'y' ||
3423 if (draw_faint_lines)
3424 draw_line(dr, x1, y1, x2, y2, line_colour);
3426 draw_thick_line(dr, 3.0,
3433 static void game_redraw_dot(drawing *dr, game_drawstate *ds,
3434 game_state *state, int i)
3436 grid *g = state->game_grid;
3437 grid_dot *d = g->dots + i;
3440 grid_to_screen(ds, g, d->x, d->y, &x, &y);
3441 draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
3444 static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
3445 game_state *state, int dir, game_ui *ui,
3446 float animtime, float flashtime)
3448 #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
3450 grid *g = state->game_grid;
3451 int border = BORDER(ds->tilesize);
3454 int redraw_everything = FALSE;
3456 int edges[REDRAW_OBJECTS_LIMIT], nedges = 0;
3457 int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0;
3459 /* Redrawing is somewhat involved.
3461 * An update can theoretically affect an arbitrary number of edges
3462 * (consider, for example, completing or breaking a cycle which doesn't
3463 * satisfy all the clues -- we'll switch many edges between error and
3464 * normal states). On the other hand, redrawing the whole grid takes a
3465 * while, making the game feel sluggish, and many updates are actually
3466 * quite well localized.
3468 * This redraw algorithm attempts to cope with both situations gracefully
3469 * and correctly. For localized changes, we set a clip rectangle, fill
3470 * it with background, and then redraw (a plausible but conservative
3471 * guess at) the objects which intersect the rectangle; if several
3472 * objects need redrawing, we'll do them individually. However, if lots
3473 * of objects are affected, we'll just redraw everything.
3475 * The reason for all of this is that it's just not safe to do the redraw
3476 * piecemeal. If you try to draw an antialiased diagonal line over
3477 * itself, you get a slightly thicker antialiased diagonal line, which
3478 * looks rather ugly after a while.
3480 * So, we take two passes over the grid. The first attempts to work out
3481 * what needs doing, and the second actually does it.
3485 redraw_everything = TRUE;
3488 /* First, trundle through the faces. */
3489 for (i = 0; i < g->num_faces; i++) {
3490 grid_face *f = g->faces + i;
3491 int sides = f->order;
3494 int n = state->clues[i];
3498 clue_mistake = (face_order(state, i, LINE_YES) > n ||
3499 face_order(state, i, LINE_NO ) > (sides-n));
3500 clue_satisfied = (face_order(state, i, LINE_YES) == n &&
3501 face_order(state, i, LINE_NO ) == (sides-n));
3503 if (clue_mistake != ds->clue_error[i] ||
3504 clue_satisfied != ds->clue_satisfied[i]) {
3505 ds->clue_error[i] = clue_mistake;
3506 ds->clue_satisfied[i] = clue_satisfied;
3507 if (nfaces == REDRAW_OBJECTS_LIMIT)
3508 redraw_everything = TRUE;
3510 faces[nfaces++] = i;
3514 /* Work out what the flash state needs to be. */
3515 if (flashtime > 0 &&
3516 (flashtime <= FLASH_TIME/3 ||
3517 flashtime >= FLASH_TIME*2/3)) {
3518 flash_changed = !ds->flashing;
3519 ds->flashing = TRUE;
3521 flash_changed = ds->flashing;
3522 ds->flashing = FALSE;
3525 /* Now, trundle through the edges. */
3526 for (i = 0; i < g->num_edges; i++) {
3528 state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
3529 if (new_ds != ds->lines[i] ||
3530 (flash_changed && state->lines[i] == LINE_YES)) {
3531 ds->lines[i] = new_ds;
3532 if (nedges == REDRAW_OBJECTS_LIMIT)
3533 redraw_everything = TRUE;
3535 edges[nedges++] = i;
3540 /* Pass one is now done. Now we do the actual drawing. */
3541 if (redraw_everything) {
3543 /* This is the unsubtle version. */
3545 int grid_width = g->highest_x - g->lowest_x;
3546 int grid_height = g->highest_y - g->lowest_y;
3547 int w = grid_width * ds->tilesize / g->tilesize;
3548 int h = grid_height * ds->tilesize / g->tilesize;
3550 draw_rect(dr, 0, 0, w + 2*border + 1, h + 2*border + 1,
3553 for (i = 0; i < g->num_faces; i++)
3554 game_redraw_clue(dr, ds, state, i);
3555 for (i = 0; i < g->num_edges; i++)
3556 game_redraw_line(dr, ds, state, i);
3557 for (i = 0; i < g->num_dots; i++)
3558 game_redraw_dot(dr, ds, state, i);
3560 draw_update(dr, 0, 0, w + 2*border + 1, h + 2*border + 1);
3563 /* Right. Now we roll up our sleeves. */
3565 for (i = 0; i < nfaces; i++) {
3566 grid_face *f = g->faces + faces[i];
3571 /* There seems to be a certain amount of trial-and-error
3572 * involved in working out the correct bounding-box for
3574 face_text_pos(ds, g, f, &xx, &yy);
3576 x = xx - ds->tilesize/4 - 1; w = ds->tilesize/2 + 2;
3577 y = yy - ds->tilesize/4 - 3; h = ds->tilesize/2 + 5;
3578 clip(dr, x, y, w, h);
3579 draw_rect(dr, x, y, w, h, COL_BACKGROUND);
3581 game_redraw_clue(dr, ds, state, faces[i]);
3582 for (j = 0; j < f->order; j++)
3583 game_redraw_line(dr, ds, state, f->edges[j] - g->edges);
3584 for (j = 0; j < f->order; j++)
3585 game_redraw_dot(dr, ds, state, f->dots[j] - g->dots);
3587 draw_update(dr, x, y, w, h);
3590 for (i = 0; i < nedges; i++) {
3591 grid_edge *e = g->edges + edges[i], *ee;
3592 int x1 = e->dot1->x;
3593 int y1 = e->dot1->y;
3594 int x2 = e->dot2->x;
3595 int y2 = e->dot2->y;
3596 int xmin, xmax, ymin, ymax;
3599 grid_to_screen(ds, g, x1, y1, &x1, &y1);
3600 grid_to_screen(ds, g, x2, y2, &x2, &y2);
3601 /* Allow extra margin for dots, and thickness of lines */
3602 xmin = min(x1, x2) - 2;
3603 xmax = max(x1, x2) + 2;
3604 ymin = min(y1, y2) - 2;
3605 ymax = max(y1, y2) + 2;
3606 /* For testing, I find it helpful to change COL_BACKGROUND
3607 * to COL_SATISFIED here. */
3608 clip(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
3609 draw_rect(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1,
3613 game_redraw_clue(dr, ds, state, e->face1 - g->faces);
3615 game_redraw_clue(dr, ds, state, e->face2 - g->faces);
3617 game_redraw_line(dr, ds, state, edges[i]);
3618 for (j = 0; j < e->dot1->order; j++) {
3619 ee = e->dot1->edges[j];
3621 game_redraw_line(dr, ds, state, ee - g->edges);
3623 for (j = 0; j < e->dot2->order; j++) {
3624 ee = e->dot2->edges[j];
3626 game_redraw_line(dr, ds, state, ee - g->edges);
3628 game_redraw_dot(dr, ds, state, e->dot1 - g->dots);
3629 game_redraw_dot(dr, ds, state, e->dot2 - g->dots);
3632 draw_update(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
3639 static float game_flash_length(game_state *oldstate, game_state *newstate,
3640 int dir, game_ui *ui)
3642 if (!oldstate->solved && newstate->solved &&
3643 !oldstate->cheated && !newstate->cheated) {
3650 static void game_print_size(game_params *params, float *x, float *y)
3655 * I'll use 7mm "squares" by default.
3657 game_compute_size(params, 700, &pw, &ph);
3662 static void game_print(drawing *dr, game_state *state, int tilesize)
3664 int ink = print_mono_colour(dr, 0);
3666 game_drawstate ads, *ds = &ads;
3667 grid *g = state->game_grid;
3669 ds->tilesize = tilesize;
3671 for (i = 0; i < g->num_dots; i++) {
3673 grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
3674 draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
3680 for (i = 0; i < g->num_faces; i++) {
3681 grid_face *f = g->faces + i;
3682 int clue = state->clues[i];
3686 c[0] = CLUE2CHAR(clue);
3688 face_text_pos(ds, g, f, &x, &y);
3690 FONT_VARIABLE, ds->tilesize / 2,
3691 ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
3698 for (i = 0; i < g->num_edges; i++) {
3699 int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
3700 grid_edge *e = g->edges + i;
3702 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3703 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3704 if (state->lines[i] == LINE_YES)
3706 /* (dx, dy) points from (x1, y1) to (x2, y2).
3707 * The line is then "fattened" in a perpendicular
3708 * direction to create a thin rectangle. */
3709 double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
3710 double dx = (x2 - x1) / d;
3711 double dy = (y2 - y1) / d;
3714 dx = (dx * ds->tilesize) / thickness;
3715 dy = (dy * ds->tilesize) / thickness;
3716 points[0] = x1 + (int)dy;
3717 points[1] = y1 - (int)dx;
3718 points[2] = x1 - (int)dy;
3719 points[3] = y1 + (int)dx;
3720 points[4] = x2 - (int)dy;
3721 points[5] = y2 + (int)dx;
3722 points[6] = x2 + (int)dy;
3723 points[7] = y2 - (int)dx;
3724 draw_polygon(dr, points, 4, ink, ink);
3728 /* Draw a dotted line */
3731 for (j = 1; j < divisions; j++) {
3732 /* Weighted average */
3733 int x = (x1 * (divisions -j) + x2 * j) / divisions;
3734 int y = (y1 * (divisions -j) + y2 * j) / divisions;
3735 draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
3742 #define thegame loopy
3745 const struct game thegame = {
3746 "Loopy", "games.loopy", "loopy",
3753 TRUE, game_configure, custom_params,
3761 TRUE, game_can_format_as_text_now, game_text_format,
3769 PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
3772 game_free_drawstate,
3776 TRUE, FALSE, game_print_size, game_print,
3777 FALSE /* wants_statusbar */,
3778 FALSE, game_timing_state,
3779 0, /* mouse_priorities */
3782 #ifdef STANDALONE_SOLVER
3785 * Half-hearted standalone solver. It can't output the solution to
3786 * anything but a square puzzle, and it can't log the deductions
3787 * it makes either. But it can solve square puzzles, and more
3788 * importantly it can use its solver to grade the difficulty of
3789 * any puzzle you give it.
3794 int main(int argc, char **argv)
3798 char *id = NULL, *desc, *err;
3801 #if 0 /* verbose solver not supported here (yet) */
3802 int really_verbose = FALSE;
3805 while (--argc > 0) {
3807 #if 0 /* verbose solver not supported here (yet) */
3808 if (!strcmp(p, "-v")) {
3809 really_verbose = TRUE;
3812 if (!strcmp(p, "-g")) {
3814 } else if (*p == '-') {
3815 fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
3823 fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
3827 desc = strchr(id, ':');
3829 fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
3834 p = default_params();
3835 decode_params(p, id);
3836 err = validate_desc(p, desc);
3838 fprintf(stderr, "%s: %s\n", argv[0], err);
3841 s = new_game(NULL, p, desc);
3844 * When solving an Easy puzzle, we don't want to bother the
3845 * user with Hard-level deductions. For this reason, we grade
3846 * the puzzle internally before doing anything else.
3848 ret = -1; /* placate optimiser */
3849 for (diff = 0; diff < DIFF_MAX; diff++) {
3850 solver_state *sstate_new;
3851 solver_state *sstate = new_solver_state((game_state *)s, diff);
3853 sstate_new = solve_game_rec(sstate);
3855 if (sstate_new->solver_status == SOLVER_MISTAKE)
3857 else if (sstate_new->solver_status == SOLVER_SOLVED)
3862 free_solver_state(sstate_new);
3863 free_solver_state(sstate);
3869 if (diff == DIFF_MAX) {
3871 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3873 printf("Unable to find a unique solution\n");
3877 printf("Difficulty rating: impossible (no solution exists)\n");
3879 printf("Difficulty rating: %s\n", diffnames[diff]);
3881 solver_state *sstate_new;
3882 solver_state *sstate = new_solver_state((game_state *)s, diff);
3884 /* If we supported a verbose solver, we'd set verbosity here */
3886 sstate_new = solve_game_rec(sstate);
3888 if (sstate_new->solver_status == SOLVER_MISTAKE)
3889 printf("Puzzle is inconsistent\n");
3891 assert(sstate_new->solver_status == SOLVER_SOLVED);
3892 if (s->grid_type == 0) {
3893 fputs(game_text_format(sstate_new->state), stdout);
3895 printf("Unable to output non-square grids\n");
3899 free_solver_state(sstate_new);
3900 free_solver_state(sstate);