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.
87 /* Debugging options */
95 /* ----------------------------------------------------------------------
96 * Struct, enum and function declarations
111 grid *game_grid; /* ref-counted (internally) */
113 /* Put -1 in a face that doesn't get a clue */
116 /* Array of line states, to store whether each line is
117 * YES, NO or UNKNOWN */
120 unsigned char *line_errors;
121 int exactly_one_loop;
126 /* Used in game_text_format(), so that it knows what type of
127 * grid it's trying to render as ASCII text. */
132 SOLVER_SOLVED, /* This is the only solution the solver could find */
133 SOLVER_MISTAKE, /* This is definitely not a solution */
134 SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
135 SOLVER_INCOMPLETE /* This may be a partial solution */
138 /* ------ Solver state ------ */
139 typedef struct solver_state {
141 enum solver_status solver_status;
142 /* NB looplen is the number of dots that are joined together at a point, ie a
143 * looplen of 1 means there are no lines to a particular dot */
146 /* Difficulty level of solver. Used by solver functions that want to
147 * vary their behaviour depending on the requested difficulty level. */
153 char *face_yes_count;
155 char *dot_solved, *face_solved;
158 /* Information for Normal level deductions:
159 * For each dline, store a bitmask for whether we know:
160 * (bit 0) at least one is YES
161 * (bit 1) at most one is YES */
164 /* Hard level information */
169 * Difficulty levels. I do some macro ickery here to ensure that my
170 * enum and the various forms of my name list always match up.
173 #define DIFFLIST(A) \
178 #define ENUM(upper,title,lower) DIFF_ ## upper,
179 #define TITLE(upper,title,lower) #title,
180 #define ENCODE(upper,title,lower) #lower
181 #define CONFIG(upper,title,lower) ":" #title
182 enum { DIFFLIST(ENUM) DIFF_MAX };
183 static char const *const diffnames[] = { DIFFLIST(TITLE) };
184 static char const diffchars[] = DIFFLIST(ENCODE);
185 #define DIFFCONFIG DIFFLIST(CONFIG)
188 * Solver routines, sorted roughly in order of computational cost.
189 * The solver will run the faster deductions first, and slower deductions are
190 * only invoked when the faster deductions are unable to make progress.
191 * Each function is associated with a difficulty level, so that the generated
192 * puzzles are solvable by applying only the functions with the chosen
193 * difficulty level or lower.
195 #define SOLVERLIST(A) \
196 A(trivial_deductions, DIFF_EASY) \
197 A(dline_deductions, DIFF_NORMAL) \
198 A(linedsf_deductions, DIFF_HARD) \
199 A(loop_deductions, DIFF_EASY)
200 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
201 #define SOLVER_FN(fn,diff) &fn,
202 #define SOLVER_DIFF(fn,diff) diff,
203 SOLVERLIST(SOLVER_FN_DECL)
204 static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
205 static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
206 static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
214 /* line_drawstate is the same as line_state, but with the extra ERROR
215 * possibility. The drawing code copies line_state to line_drawstate,
216 * except in the case that the line is an error. */
217 enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
218 enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
219 DS_LINE_NO, DS_LINE_ERROR };
221 #define OPP(line_state) \
225 struct game_drawstate {
232 char *clue_satisfied;
235 static char *validate_desc(const game_params *params, const char *desc);
236 static int dot_order(const game_state* state, int i, char line_type);
237 static int face_order(const game_state* state, int i, char line_type);
238 static solver_state *solve_game_rec(const solver_state *sstate);
241 static void check_caches(const solver_state* sstate);
243 #define check_caches(s)
247 * Grid type config options available in Loopy.
249 * Annoyingly, we have to use an enum here which doesn't match up
250 * exactly to the grid-type enum in grid.h. Values in params->types
251 * are given by names such as LOOPY_GRID_SQUARE, which shouldn't be
252 * confused with GRID_SQUARE which is the value you pass to grid_new()
253 * and friends. So beware!
255 * (This is partly for historical reasons - Loopy's version of the
256 * enum is encoded in game parameter strings, so we keep it for
257 * backwards compatibility. But also, we need to store additional data
258 * here alongside each enum value, such as names for the presets menu,
259 * which isn't stored in grid.h; so we have to have our own list macro
260 * here anyway, and C doesn't make it easy to enforce that that lines
261 * up exactly with grid.h.)
263 * Do not add values to this list _except_ at the end, or old game ids
266 #define GRIDLIST(A) \
267 A("Squares",SQUARE,3,3) \
268 A("Triangular",TRIANGULAR,3,3) \
269 A("Honeycomb",HONEYCOMB,3,3) \
270 A("Snub-Square",SNUBSQUARE,3,3) \
271 A("Cairo",CAIRO,3,4) \
272 A("Great-Hexagonal",GREATHEXAGONAL,3,3) \
273 A("Octagonal",OCTAGONAL,3,3) \
274 A("Kites",KITE,3,3) \
275 A("Floret",FLORET,1,2) \
276 A("Dodecagonal",DODECAGONAL,2,2) \
277 A("Great-Dodecagonal",GREATDODECAGONAL,2,2) \
278 A("Penrose (kite/dart)",PENROSE_P2,3,3) \
279 A("Penrose (rhombs)",PENROSE_P3,3,3) \
280 A("Great-Great-Dodecagonal",GREATGREATDODECAGONAL,2,2) \
283 #define GRID_NAME(title,type,amin,omin) title,
284 #define GRID_CONFIG(title,type,amin,omin) ":" title
285 #define GRID_LOOPYTYPE(title,type,amin,omin) LOOPY_GRID_ ## type,
286 #define GRID_GRIDTYPE(title,type,amin,omin) GRID_ ## type,
287 #define GRID_SIZES(title,type,amin,omin) \
289 "Width and height for this grid type must both be at least " #amin, \
290 "At least one of width and height for this grid type must be at least " #omin,},
291 enum { GRIDLIST(GRID_LOOPYTYPE) };
292 static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
293 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
294 static grid_type grid_types[] = { GRIDLIST(GRID_GRIDTYPE) };
295 #define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0]))
296 static const struct {
299 } grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
301 /* Generates a (dynamically allocated) new grid, according to the
302 * type and size requested in params. Does nothing if the grid is already
304 static grid *loopy_generate_grid(const game_params *params,
305 const char *grid_desc)
307 return grid_new(grid_types[params->type], params->w, params->h, grid_desc);
310 /* ----------------------------------------------------------------------
314 /* General constants */
315 #define PREFERRED_TILE_SIZE 32
316 #define BORDER(tilesize) ((tilesize) / 2)
317 #define FLASH_TIME 0.5F
319 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
321 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
322 ((field) |= (1<<(bit)), TRUE))
324 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
325 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
327 #define CLUE2CHAR(c) \
328 ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
330 /* ----------------------------------------------------------------------
331 * General struct manipulation and other straightforward code
334 static game_state *dup_game(const game_state *state)
336 game_state *ret = snew(game_state);
338 ret->game_grid = state->game_grid;
339 ret->game_grid->refcount++;
341 ret->solved = state->solved;
342 ret->cheated = state->cheated;
344 ret->clues = snewn(state->game_grid->num_faces, signed char);
345 memcpy(ret->clues, state->clues, state->game_grid->num_faces);
347 ret->lines = snewn(state->game_grid->num_edges, char);
348 memcpy(ret->lines, state->lines, state->game_grid->num_edges);
350 ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
351 memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
352 ret->exactly_one_loop = state->exactly_one_loop;
354 ret->grid_type = state->grid_type;
358 static void free_game(game_state *state)
361 grid_free(state->game_grid);
364 sfree(state->line_errors);
369 static solver_state *new_solver_state(const game_state *state, int diff) {
371 int num_dots = state->game_grid->num_dots;
372 int num_faces = state->game_grid->num_faces;
373 int num_edges = state->game_grid->num_edges;
374 solver_state *ret = snew(solver_state);
376 ret->state = dup_game(state);
378 ret->solver_status = SOLVER_INCOMPLETE;
381 ret->dotdsf = snew_dsf(num_dots);
382 ret->looplen = snewn(num_dots, int);
384 for (i = 0; i < num_dots; i++) {
388 ret->dot_solved = snewn(num_dots, char);
389 ret->face_solved = snewn(num_faces, char);
390 memset(ret->dot_solved, FALSE, num_dots);
391 memset(ret->face_solved, FALSE, num_faces);
393 ret->dot_yes_count = snewn(num_dots, char);
394 memset(ret->dot_yes_count, 0, num_dots);
395 ret->dot_no_count = snewn(num_dots, char);
396 memset(ret->dot_no_count, 0, num_dots);
397 ret->face_yes_count = snewn(num_faces, char);
398 memset(ret->face_yes_count, 0, num_faces);
399 ret->face_no_count = snewn(num_faces, char);
400 memset(ret->face_no_count, 0, num_faces);
402 if (diff < DIFF_NORMAL) {
405 ret->dlines = snewn(2*num_edges, char);
406 memset(ret->dlines, 0, 2*num_edges);
409 if (diff < DIFF_HARD) {
412 ret->linedsf = snew_dsf(state->game_grid->num_edges);
418 static void free_solver_state(solver_state *sstate) {
420 free_game(sstate->state);
421 sfree(sstate->dotdsf);
422 sfree(sstate->looplen);
423 sfree(sstate->dot_solved);
424 sfree(sstate->face_solved);
425 sfree(sstate->dot_yes_count);
426 sfree(sstate->dot_no_count);
427 sfree(sstate->face_yes_count);
428 sfree(sstate->face_no_count);
430 /* OK, because sfree(NULL) is a no-op */
431 sfree(sstate->dlines);
432 sfree(sstate->linedsf);
438 static solver_state *dup_solver_state(const solver_state *sstate) {
439 game_state *state = sstate->state;
440 int num_dots = state->game_grid->num_dots;
441 int num_faces = state->game_grid->num_faces;
442 int num_edges = state->game_grid->num_edges;
443 solver_state *ret = snew(solver_state);
445 ret->state = state = dup_game(sstate->state);
447 ret->solver_status = sstate->solver_status;
448 ret->diff = sstate->diff;
450 ret->dotdsf = snewn(num_dots, int);
451 ret->looplen = snewn(num_dots, int);
452 memcpy(ret->dotdsf, sstate->dotdsf,
453 num_dots * sizeof(int));
454 memcpy(ret->looplen, sstate->looplen,
455 num_dots * sizeof(int));
457 ret->dot_solved = snewn(num_dots, char);
458 ret->face_solved = snewn(num_faces, char);
459 memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
460 memcpy(ret->face_solved, sstate->face_solved, num_faces);
462 ret->dot_yes_count = snewn(num_dots, char);
463 memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
464 ret->dot_no_count = snewn(num_dots, char);
465 memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
467 ret->face_yes_count = snewn(num_faces, char);
468 memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
469 ret->face_no_count = snewn(num_faces, char);
470 memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
472 if (sstate->dlines) {
473 ret->dlines = snewn(2*num_edges, char);
474 memcpy(ret->dlines, sstate->dlines,
480 if (sstate->linedsf) {
481 ret->linedsf = snewn(num_edges, int);
482 memcpy(ret->linedsf, sstate->linedsf,
483 num_edges * sizeof(int));
491 static game_params *default_params(void)
493 game_params *ret = snew(game_params);
502 ret->diff = DIFF_EASY;
508 static game_params *dup_params(const game_params *params)
510 game_params *ret = snew(game_params);
512 *ret = *params; /* structure copy */
516 static const game_params presets[] = {
518 { 7, 7, DIFF_EASY, LOOPY_GRID_SQUARE },
519 { 7, 7, DIFF_NORMAL, LOOPY_GRID_SQUARE },
520 { 7, 7, DIFF_HARD, LOOPY_GRID_SQUARE },
521 { 7, 7, DIFF_HARD, LOOPY_GRID_HONEYCOMB },
522 { 7, 7, DIFF_HARD, LOOPY_GRID_TRIANGULAR },
523 { 5, 5, DIFF_HARD, LOOPY_GRID_SNUBSQUARE },
524 { 7, 7, DIFF_HARD, LOOPY_GRID_CAIRO },
525 { 5, 4, DIFF_HARD, LOOPY_GRID_GREATHEXAGONAL },
526 { 5, 5, DIFF_HARD, LOOPY_GRID_OCTAGONAL },
527 { 5, 5, DIFF_HARD, LOOPY_GRID_KITE },
528 { 3, 3, DIFF_HARD, LOOPY_GRID_FLORET },
529 { 3, 3, DIFF_HARD, LOOPY_GRID_DODECAGONAL },
530 { 3, 3, DIFF_HARD, LOOPY_GRID_GREATDODECAGONAL },
531 { 3, 2, DIFF_HARD, LOOPY_GRID_GREATGREATDODECAGONAL },
532 { 6, 6, DIFF_HARD, LOOPY_GRID_PENROSE_P2 },
533 { 6, 6, DIFF_HARD, LOOPY_GRID_PENROSE_P3 },
535 { 7, 7, DIFF_EASY, LOOPY_GRID_SQUARE },
536 { 10, 10, DIFF_EASY, LOOPY_GRID_SQUARE },
537 { 7, 7, DIFF_NORMAL, LOOPY_GRID_SQUARE },
538 { 10, 10, DIFF_NORMAL, LOOPY_GRID_SQUARE },
539 { 7, 7, DIFF_HARD, LOOPY_GRID_SQUARE },
540 { 10, 10, DIFF_HARD, LOOPY_GRID_SQUARE },
541 { 10, 10, DIFF_HARD, LOOPY_GRID_HONEYCOMB },
542 { 12, 10, DIFF_HARD, LOOPY_GRID_TRIANGULAR },
543 { 7, 7, DIFF_HARD, LOOPY_GRID_SNUBSQUARE },
544 { 9, 9, DIFF_HARD, LOOPY_GRID_CAIRO },
545 { 5, 4, DIFF_HARD, LOOPY_GRID_GREATHEXAGONAL },
546 { 7, 7, DIFF_HARD, LOOPY_GRID_OCTAGONAL },
547 { 5, 5, DIFF_HARD, LOOPY_GRID_KITE },
548 { 5, 5, DIFF_HARD, LOOPY_GRID_FLORET },
549 { 5, 4, DIFF_HARD, LOOPY_GRID_DODECAGONAL },
550 { 5, 4, DIFF_HARD, LOOPY_GRID_GREATDODECAGONAL },
551 { 5, 3, DIFF_HARD, LOOPY_GRID_GREATGREATDODECAGONAL },
552 { 10, 10, DIFF_HARD, LOOPY_GRID_PENROSE_P2 },
553 { 10, 10, DIFF_HARD, LOOPY_GRID_PENROSE_P3 },
557 static int game_fetch_preset(int i, char **name, game_params **params)
562 if (i < 0 || i >= lenof(presets))
565 tmppar = snew(game_params);
566 *tmppar = presets[i];
568 sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
569 gridnames[tmppar->type], diffnames[tmppar->diff]);
575 static void free_params(game_params *params)
580 static void decode_params(game_params *params, char const *string)
582 params->h = params->w = atoi(string);
583 params->diff = DIFF_EASY;
584 while (*string && isdigit((unsigned char)*string)) string++;
585 if (*string == 'x') {
587 params->h = atoi(string);
588 while (*string && isdigit((unsigned char)*string)) string++;
590 if (*string == 't') {
592 params->type = atoi(string);
593 while (*string && isdigit((unsigned char)*string)) string++;
595 if (*string == 'd') {
598 for (i = 0; i < DIFF_MAX; i++)
599 if (*string == diffchars[i])
601 if (*string) string++;
605 static char *encode_params(const game_params *params, int full)
608 sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
610 sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
614 static config_item *game_configure(const game_params *params)
619 ret = snewn(5, config_item);
621 ret[0].name = "Width";
622 ret[0].type = C_STRING;
623 sprintf(buf, "%d", params->w);
624 ret[0].sval = dupstr(buf);
627 ret[1].name = "Height";
628 ret[1].type = C_STRING;
629 sprintf(buf, "%d", params->h);
630 ret[1].sval = dupstr(buf);
633 ret[2].name = "Grid type";
634 ret[2].type = C_CHOICES;
635 ret[2].sval = GRID_CONFIGS;
636 ret[2].ival = params->type;
638 ret[3].name = "Difficulty";
639 ret[3].type = C_CHOICES;
640 ret[3].sval = DIFFCONFIG;
641 ret[3].ival = params->diff;
651 static game_params *custom_params(const config_item *cfg)
653 game_params *ret = snew(game_params);
655 ret->w = atoi(cfg[0].sval);
656 ret->h = atoi(cfg[1].sval);
657 ret->type = cfg[2].ival;
658 ret->diff = cfg[3].ival;
663 static char *validate_params(const game_params *params, int full)
665 if (params->type < 0 || params->type >= NUM_GRID_TYPES)
666 return "Illegal grid type";
667 if (params->w < grid_size_limits[params->type].amin ||
668 params->h < grid_size_limits[params->type].amin)
669 return grid_size_limits[params->type].aerr;
670 if (params->w < grid_size_limits[params->type].omin &&
671 params->h < grid_size_limits[params->type].omin)
672 return grid_size_limits[params->type].oerr;
675 * This shouldn't be able to happen at all, since decode_params
676 * and custom_params will never generate anything that isn't
679 assert(params->diff < DIFF_MAX);
684 /* Returns a newly allocated string describing the current puzzle */
685 static char *state_to_text(const game_state *state)
687 grid *g = state->game_grid;
689 int num_faces = g->num_faces;
690 char *description = snewn(num_faces + 1, char);
691 char *dp = description;
695 for (i = 0; i < num_faces; i++) {
696 if (state->clues[i] < 0) {
697 if (empty_count > 25) {
698 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
704 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
707 dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
712 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
714 retval = dupstr(description);
720 #define GRID_DESC_SEP '_'
722 /* Splits up a (optional) grid_desc from the game desc. Returns the
723 * grid_desc (which needs freeing) and updates the desc pointer to
724 * start of real desc, or returns NULL if no desc. */
725 static char *extract_grid_desc(const char **desc)
727 char *sep = strchr(*desc, GRID_DESC_SEP), *gd;
730 if (!sep) return NULL;
732 gd_len = sep - (*desc);
733 gd = snewn(gd_len+1, char);
734 memcpy(gd, *desc, gd_len);
742 /* We require that the params pass the test in validate_params and that the
743 * description fills the entire game area */
744 static char *validate_desc(const game_params *params, const char *desc)
748 char *grid_desc, *ret;
750 /* It's pretty inefficient to do this just for validation. All we need to
751 * know is the precise number of faces. */
752 grid_desc = extract_grid_desc(&desc);
753 ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc);
756 g = loopy_generate_grid(params, grid_desc);
757 if (grid_desc) sfree(grid_desc);
759 for (; *desc; ++desc) {
760 if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
765 count += *desc - 'a' + 1;
768 return "Unknown character in description";
771 if (count < g->num_faces)
772 return "Description too short for board size";
773 if (count > g->num_faces)
774 return "Description too long for board size";
781 /* Sums the lengths of the numbers in range [0,n) */
782 /* See equivalent function in solo.c for justification of this. */
783 static int len_0_to_n(int n)
785 int len = 1; /* Counting 0 as a bit of a special case */
788 for (i = 1; i < n; i *= 10) {
789 len += max(n - i, 0);
795 static char *encode_solve_move(const game_state *state)
800 int num_edges = state->game_grid->num_edges;
802 /* This is going to return a string representing the moves needed to set
803 * every line in a grid to be the same as the ones in 'state'. The exact
804 * length of this string is predictable. */
806 len = 1; /* Count the 'S' prefix */
807 /* Numbers in all lines */
808 len += len_0_to_n(num_edges);
809 /* For each line we also have a letter */
812 ret = snewn(len + 1, char);
815 p += sprintf(p, "S");
817 for (i = 0; i < num_edges; i++) {
818 switch (state->lines[i]) {
820 p += sprintf(p, "%dy", i);
823 p += sprintf(p, "%dn", i);
828 /* No point in doing sums like that if they're going to be wrong */
829 assert(strlen(ret) <= (size_t)len);
833 static game_ui *new_ui(const game_state *state)
838 static void free_ui(game_ui *ui)
842 static char *encode_ui(const game_ui *ui)
847 static void decode_ui(game_ui *ui, const char *encoding)
851 static void game_changed_state(game_ui *ui, const game_state *oldstate,
852 const game_state *newstate)
856 static void game_compute_size(const game_params *params, int tilesize,
859 int grid_width, grid_height, rendered_width, rendered_height;
862 grid_compute_size(grid_types[params->type], params->w, params->h,
863 &g_tilesize, &grid_width, &grid_height);
865 /* multiply first to minimise rounding error on integer division */
866 rendered_width = grid_width * tilesize / g_tilesize;
867 rendered_height = grid_height * tilesize / g_tilesize;
868 *x = rendered_width + 2 * BORDER(tilesize) + 1;
869 *y = rendered_height + 2 * BORDER(tilesize) + 1;
872 static void game_set_size(drawing *dr, game_drawstate *ds,
873 const game_params *params, int tilesize)
875 ds->tilesize = tilesize;
878 static float *game_colours(frontend *fe, int *ncolours)
880 float *ret = snewn(3 * NCOLOURS, float);
882 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
884 ret[COL_FOREGROUND * 3 + 0] = 0.0F;
885 ret[COL_FOREGROUND * 3 + 1] = 0.0F;
886 ret[COL_FOREGROUND * 3 + 2] = 0.0F;
889 * We want COL_LINEUNKNOWN to be a yellow which is a bit darker
890 * than the background. (I previously set it to 0.8,0.8,0, but
891 * found that this went badly with the 0.8,0.8,0.8 favoured as a
892 * background by the Java frontend.)
894 ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
895 ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
896 ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
898 ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
899 ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
900 ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;
902 ret[COL_MISTAKE * 3 + 0] = 1.0F;
903 ret[COL_MISTAKE * 3 + 1] = 0.0F;
904 ret[COL_MISTAKE * 3 + 2] = 0.0F;
906 ret[COL_SATISFIED * 3 + 0] = 0.0F;
907 ret[COL_SATISFIED * 3 + 1] = 0.0F;
908 ret[COL_SATISFIED * 3 + 2] = 0.0F;
910 /* We want the faint lines to be a bit darker than the background.
911 * Except if the background is pretty dark already; then it ought to be a
912 * bit lighter. Oy vey.
914 ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
915 ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
916 ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F;
918 *ncolours = NCOLOURS;
922 static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
924 struct game_drawstate *ds = snew(struct game_drawstate);
925 int num_faces = state->game_grid->num_faces;
926 int num_edges = state->game_grid->num_edges;
931 ds->lines = snewn(num_edges, char);
932 ds->clue_error = snewn(num_faces, char);
933 ds->clue_satisfied = snewn(num_faces, char);
934 ds->textx = snewn(num_faces, int);
935 ds->texty = snewn(num_faces, int);
938 memset(ds->lines, LINE_UNKNOWN, num_edges);
939 memset(ds->clue_error, 0, num_faces);
940 memset(ds->clue_satisfied, 0, num_faces);
941 for (i = 0; i < num_faces; i++)
942 ds->textx[i] = ds->texty[i] = -1;
947 static void game_free_drawstate(drawing *dr, game_drawstate *ds)
951 sfree(ds->clue_error);
952 sfree(ds->clue_satisfied);
957 static int game_timing_state(const game_state *state, game_ui *ui)
962 static float game_anim_length(const game_state *oldstate,
963 const game_state *newstate, int dir, game_ui *ui)
968 static int game_can_format_as_text_now(const game_params *params)
970 if (params->type != 0)
975 static char *game_text_format(const game_state *state)
981 grid *g = state->game_grid;
984 assert(state->grid_type == 0);
986 /* Work out the basic size unit */
987 f = g->faces; /* first face */
988 assert(f->order == 4);
989 /* The dots are ordered clockwise, so the two opposite
990 * corners are guaranteed to span the square */
991 cell_size = abs(f->dots[0]->x - f->dots[2]->x);
993 w = (g->highest_x - g->lowest_x) / cell_size;
994 h = (g->highest_y - g->lowest_y) / cell_size;
996 /* Create a blank "canvas" to "draw" on */
999 ret = snewn(W * H + 1, char);
1000 for (y = 0; y < H; y++) {
1001 for (x = 0; x < W-1; x++) {
1004 ret[y*W + W-1] = '\n';
1008 /* Fill in edge info */
1009 for (i = 0; i < g->num_edges; i++) {
1010 grid_edge *e = g->edges + i;
1011 /* Cell coordinates, from (0,0) to (w-1,h-1) */
1012 int x1 = (e->dot1->x - g->lowest_x) / cell_size;
1013 int x2 = (e->dot2->x - g->lowest_x) / cell_size;
1014 int y1 = (e->dot1->y - g->lowest_y) / cell_size;
1015 int y2 = (e->dot2->y - g->lowest_y) / cell_size;
1016 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
1017 * cell coordinates) */
1020 switch (state->lines[i]) {
1022 ret[y*W + x] = (y1 == y2) ? '-' : '|';
1028 break; /* already a space */
1030 assert(!"Illegal line state");
1035 for (i = 0; i < g->num_faces; i++) {
1039 assert(f->order == 4);
1040 /* Cell coordinates, from (0,0) to (w-1,h-1) */
1041 x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
1042 x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
1043 y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
1044 y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
1045 /* Midpoint, in canvas coordinates */
1048 ret[y*W + x] = CLUE2CHAR(state->clues[i]);
1053 /* ----------------------------------------------------------------------
1058 static void check_caches(const solver_state* sstate)
1061 const game_state *state = sstate->state;
1062 const grid *g = state->game_grid;
1064 for (i = 0; i < g->num_dots; i++) {
1065 assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
1066 assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
1069 for (i = 0; i < g->num_faces; i++) {
1070 assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
1071 assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
1076 #define check_caches(s) \
1078 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1082 #endif /* DEBUG_CACHES */
1084 /* ----------------------------------------------------------------------
1085 * Solver utility functions
1088 /* Sets the line (with index i) to the new state 'line_new', and updates
1089 * the cached counts of any affected faces and dots.
1090 * Returns TRUE if this actually changed the line's state. */
1091 static int solver_set_line(solver_state *sstate, int i,
1092 enum line_state line_new
1094 , const char *reason
1098 game_state *state = sstate->state;
1102 assert(line_new != LINE_UNKNOWN);
1104 check_caches(sstate);
1106 if (state->lines[i] == line_new) {
1107 return FALSE; /* nothing changed */
1109 state->lines[i] = line_new;
1112 fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
1113 i, line_new == LINE_YES ? "YES" : "NO",
1117 g = state->game_grid;
1120 /* Update the cache for both dots and both faces affected by this. */
1121 if (line_new == LINE_YES) {
1122 sstate->dot_yes_count[e->dot1 - g->dots]++;
1123 sstate->dot_yes_count[e->dot2 - g->dots]++;
1125 sstate->face_yes_count[e->face1 - g->faces]++;
1128 sstate->face_yes_count[e->face2 - g->faces]++;
1131 sstate->dot_no_count[e->dot1 - g->dots]++;
1132 sstate->dot_no_count[e->dot2 - g->dots]++;
1134 sstate->face_no_count[e->face1 - g->faces]++;
1137 sstate->face_no_count[e->face2 - g->faces]++;
1141 check_caches(sstate);
1146 #define solver_set_line(a, b, c) \
1147 solver_set_line(a, b, c, __FUNCTION__)
1151 * Merge two dots due to the existence of an edge between them.
1152 * Updates the dsf tracking equivalence classes, and keeps track of
1153 * the length of path each dot is currently a part of.
1154 * Returns TRUE if the dots were already linked, ie if they are part of a
1155 * closed loop, and false otherwise.
1157 static int merge_dots(solver_state *sstate, int edge_index)
1160 grid *g = sstate->state->game_grid;
1161 grid_edge *e = g->edges + edge_index;
1163 i = e->dot1 - g->dots;
1164 j = e->dot2 - g->dots;
1166 i = dsf_canonify(sstate->dotdsf, i);
1167 j = dsf_canonify(sstate->dotdsf, j);
1172 len = sstate->looplen[i] + sstate->looplen[j];
1173 dsf_merge(sstate->dotdsf, i, j);
1174 i = dsf_canonify(sstate->dotdsf, i);
1175 sstate->looplen[i] = len;
1180 /* Merge two lines because the solver has deduced that they must be either
1181 * identical or opposite. Returns TRUE if this is new information, otherwise
1183 static int merge_lines(solver_state *sstate, int i, int j, int inverse
1185 , const char *reason
1191 assert(i < sstate->state->game_grid->num_edges);
1192 assert(j < sstate->state->game_grid->num_edges);
1194 i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
1196 j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
1199 edsf_merge(sstate->linedsf, i, j, inverse);
1203 fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
1205 inverse ? "inverse " : "", reason);
1212 #define merge_lines(a, b, c, d) \
1213 merge_lines(a, b, c, d, __FUNCTION__)
1216 /* Count the number of lines of a particular type currently going into the
1218 static int dot_order(const game_state* state, int dot, char line_type)
1221 grid *g = state->game_grid;
1222 grid_dot *d = g->dots + dot;
1225 for (i = 0; i < d->order; i++) {
1226 grid_edge *e = d->edges[i];
1227 if (state->lines[e - g->edges] == line_type)
1233 /* Count the number of lines of a particular type currently surrounding the
1235 static int face_order(const game_state* state, int face, char line_type)
1238 grid *g = state->game_grid;
1239 grid_face *f = g->faces + face;
1242 for (i = 0; i < f->order; i++) {
1243 grid_edge *e = f->edges[i];
1244 if (state->lines[e - g->edges] == line_type)
1250 /* Set all lines bordering a dot of type old_type to type new_type
1251 * Return value tells caller whether this function actually did anything */
1252 static int dot_setall(solver_state *sstate, int dot,
1253 char old_type, char new_type)
1255 int retval = FALSE, r;
1256 game_state *state = sstate->state;
1261 if (old_type == new_type)
1264 g = state->game_grid;
1267 for (i = 0; i < d->order; i++) {
1268 int line_index = d->edges[i] - g->edges;
1269 if (state->lines[line_index] == old_type) {
1270 r = solver_set_line(sstate, line_index, new_type);
1278 /* Set all lines bordering a face of type old_type to type new_type */
1279 static int face_setall(solver_state *sstate, int face,
1280 char old_type, char new_type)
1282 int retval = FALSE, r;
1283 game_state *state = sstate->state;
1288 if (old_type == new_type)
1291 g = state->game_grid;
1292 f = g->faces + face;
1294 for (i = 0; i < f->order; i++) {
1295 int line_index = f->edges[i] - g->edges;
1296 if (state->lines[line_index] == old_type) {
1297 r = solver_set_line(sstate, line_index, new_type);
1305 /* ----------------------------------------------------------------------
1306 * Loop generation and clue removal
1309 static void add_full_clues(game_state *state, random_state *rs)
1311 signed char *clues = state->clues;
1312 grid *g = state->game_grid;
1313 char *board = snewn(g->num_faces, char);
1316 generate_loop(g, board, rs, NULL, NULL);
1318 /* Fill out all the clues by initialising to 0, then iterating over
1319 * all edges and incrementing each clue as we find edges that border
1320 * between BLACK/WHITE faces. While we're at it, we verify that the
1321 * algorithm does work, and there aren't any GREY faces still there. */
1322 memset(clues, 0, g->num_faces);
1323 for (i = 0; i < g->num_edges; i++) {
1324 grid_edge *e = g->edges + i;
1325 grid_face *f1 = e->face1;
1326 grid_face *f2 = e->face2;
1327 enum face_colour c1 = FACE_COLOUR(f1);
1328 enum face_colour c2 = FACE_COLOUR(f2);
1329 assert(c1 != FACE_GREY);
1330 assert(c2 != FACE_GREY);
1332 if (f1) clues[f1 - g->faces]++;
1333 if (f2) clues[f2 - g->faces]++;
1340 static int game_has_unique_soln(const game_state *state, int diff)
1343 solver_state *sstate_new;
1344 solver_state *sstate = new_solver_state((game_state *)state, diff);
1346 sstate_new = solve_game_rec(sstate);
1348 assert(sstate_new->solver_status != SOLVER_MISTAKE);
1349 ret = (sstate_new->solver_status == SOLVER_SOLVED);
1351 free_solver_state(sstate_new);
1352 free_solver_state(sstate);
1358 /* Remove clues one at a time at random. */
1359 static game_state *remove_clues(game_state *state, random_state *rs,
1363 int num_faces = state->game_grid->num_faces;
1364 game_state *ret = dup_game(state), *saved_ret;
1367 /* We need to remove some clues. We'll do this by forming a list of all
1368 * available clues, shuffling it, then going along one at a
1369 * time clearing each clue in turn for which doing so doesn't render the
1370 * board unsolvable. */
1371 face_list = snewn(num_faces, int);
1372 for (n = 0; n < num_faces; ++n) {
1376 shuffle(face_list, num_faces, sizeof(int), rs);
1378 for (n = 0; n < num_faces; ++n) {
1379 saved_ret = dup_game(ret);
1380 ret->clues[face_list[n]] = -1;
1382 if (game_has_unique_soln(ret, diff)) {
1383 free_game(saved_ret);
1395 static char *new_game_desc(const game_params *params, random_state *rs,
1396 char **aux, int interactive)
1398 /* solution and description both use run-length encoding in obvious ways */
1399 char *retval, *game_desc, *grid_desc;
1401 game_state *state = snew(game_state);
1402 game_state *state_new;
1404 grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs);
1405 state->game_grid = g = loopy_generate_grid(params, grid_desc);
1407 state->clues = snewn(g->num_faces, signed char);
1408 state->lines = snewn(g->num_edges, char);
1409 state->line_errors = snewn(g->num_edges, unsigned char);
1410 state->exactly_one_loop = FALSE;
1412 state->grid_type = params->type;
1416 memset(state->lines, LINE_UNKNOWN, g->num_edges);
1417 memset(state->line_errors, 0, g->num_edges);
1419 state->solved = state->cheated = FALSE;
1421 /* Get a new random solvable board with all its clues filled in. Yes, this
1422 * can loop for ever if the params are suitably unfavourable, but
1423 * preventing games smaller than 4x4 seems to stop this happening */
1425 add_full_clues(state, rs);
1426 } while (!game_has_unique_soln(state, params->diff));
1428 state_new = remove_clues(state, rs, params->diff);
1433 if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
1435 fprintf(stderr, "Rejecting board, it is too easy\n");
1437 goto newboard_please;
1440 game_desc = state_to_text(state);
1445 retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char);
1446 sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc);
1453 assert(!validate_desc(params, retval));
1458 static game_state *new_game(midend *me, const game_params *params,
1462 game_state *state = snew(game_state);
1463 int empties_to_make = 0;
1468 int num_faces, num_edges;
1470 grid_desc = extract_grid_desc(&desc);
1471 state->game_grid = g = loopy_generate_grid(params, grid_desc);
1472 if (grid_desc) sfree(grid_desc);
1476 num_faces = g->num_faces;
1477 num_edges = g->num_edges;
1479 state->clues = snewn(num_faces, signed char);
1480 state->lines = snewn(num_edges, char);
1481 state->line_errors = snewn(num_edges, unsigned char);
1482 state->exactly_one_loop = FALSE;
1484 state->solved = state->cheated = FALSE;
1486 state->grid_type = params->type;
1488 for (i = 0; i < num_faces; i++) {
1489 if (empties_to_make) {
1491 state->clues[i] = -1;
1497 n2 = *dp - 'A' + 10;
1498 if (n >= 0 && n < 10) {
1499 state->clues[i] = n;
1500 } else if (n2 >= 10 && n2 < 36) {
1501 state->clues[i] = n2;
1505 state->clues[i] = -1;
1506 empties_to_make = n - 1;
1511 memset(state->lines, LINE_UNKNOWN, num_edges);
1512 memset(state->line_errors, 0, num_edges);
1516 /* Calculates the line_errors data, and checks if the current state is a
1518 static int check_completion(game_state *state)
1520 grid *g = state->game_grid;
1522 int *dsf, *component_state;
1523 int nsilly, nloop, npath, largest_comp, largest_size, total_pathsize;
1524 enum { COMP_NONE, COMP_LOOP, COMP_PATH, COMP_SILLY, COMP_EMPTY };
1526 memset(state->line_errors, 0, g->num_edges);
1529 * Find loops in the grid, and determine whether the puzzle is
1532 * Loopy is a bit more complicated than most puzzles that care
1533 * about loop detection. In most of them, loops are simply
1534 * _forbidden_; so the obviously right way to do
1535 * error-highlighting during play is to light up a graph edge red
1536 * iff it is part of a loop, which is exactly what the centralised
1537 * findloop.c makes easy.
1539 * But Loopy is unusual in that you're _supposed_ to be making a
1540 * loop - and yet _some_ loops are not the right loop. So we need
1541 * to be more discriminating, by identifying loops one by one and
1542 * then thinking about which ones to highlight, and so findloop.c
1543 * isn't quite the right tool for the job in this case.
1545 * Worse still, consider situations in which the grid contains a
1546 * loop and also some non-loop edges: there are some cases like
1547 * this in which the user's intuitive expectation would be to
1548 * highlight the loop (if you're only about half way through the
1549 * puzzle and have accidentally made a little loop in some corner
1550 * of the grid), and others in which they'd be more likely to
1551 * expect you to highlight the non-loop edges (if you've just
1552 * closed off a whole loop that you thought was the entire
1553 * solution, but forgot some disconnected edges in a corner
1554 * somewhere). So while it's easy enough to check whether the
1555 * solution is _right_, highlighting the wrong parts is a tricky
1556 * problem for this puzzle!
1558 * I'd quite like, in some situations, to identify the largest
1559 * loop among the player's YES edges, and then light up everything
1560 * other than that. But finding the longest cycle in a graph is an
1561 * NP-complete problem (because, in particular, it must return a
1562 * Hamilton cycle if one exists).
1564 * However, I think we can make the problem tractable by
1565 * exercising the Puzzles principle that it isn't absolutely
1566 * necessary to highlight _all_ errors: the key point is that by
1567 * the time the user has filled in the whole grid, they should
1568 * either have seen a completion flash, or have _some_ error
1569 * highlight showing them why the solution isn't right. So in
1570 * principle it would be *just about* good enough to highlight
1571 * just one error in the whole grid, if there was really no better
1572 * way. But we'd like to highlight as many errors as possible.
1574 * In this case, I think the simple approach is to make use of the
1575 * fact that no vertex may have degree > 2, and that's really
1576 * simple to detect. So the plan goes like this:
1578 * - Form the dsf of connected components of the graph vertices.
1580 * - Highlight an error at any vertex with degree > 2. (It so
1581 * happens that we do this by lighting up all the edges
1582 * incident to that vertex, but that's an output detail.)
1584 * - Any component that contains such a vertex is now excluded
1585 * from further consideration, because it already has a
1588 * - The remaining components have no vertex with degree > 2, and
1589 * hence they all consist of either a simple loop, or a simple
1590 * path with two endpoints.
1592 * - For these purposes, group together all the paths and imagine
1593 * them to be a single component (because in most normal
1594 * situations the player will gradually build up the solution
1595 * _not_ all in one connected segment, but as lots of separate
1596 * little path pieces that gradually connect to each other).
1598 * - After doing that, if there is exactly one (sensible)
1599 * component - be it a collection of paths or a loop - then
1600 * highlight no further edge errors. (The former case is normal
1601 * during play, and the latter is a potentially solved puzzle.)
1603 * - Otherwise, find the largest of the sensible components,
1604 * leave that one unhighlighted, and light the rest up in red.
1607 dsf = snew_dsf(g->num_dots);
1609 /* Build the dsf. */
1610 for (i = 0; i < g->num_edges; i++) {
1611 if (state->lines[i] == LINE_YES) {
1612 grid_edge *e = g->edges + i;
1613 int d1 = e->dot1 - g->dots, d2 = e->dot2 - g->dots;
1614 dsf_merge(dsf, d1, d2);
1618 /* Initialise a state variable for each connected component. */
1619 component_state = snewn(g->num_dots, int);
1620 for (i = 0; i < g->num_dots; i++) {
1621 if (dsf_canonify(dsf, i) == i)
1622 component_state[i] = COMP_LOOP;
1624 component_state[i] = COMP_NONE;
1627 /* Check for dots with degree > 3. Here we also spot dots of
1628 * degree 1 in which the user has marked all the non-edges as
1629 * LINE_NO, because those are also clear vertex-level errors, so
1630 * we give them the same treatment of excluding their connected
1631 * component from the subsequent loop analysis. */
1632 for (i = 0; i < g->num_dots; i++) {
1633 int comp = dsf_canonify(dsf, i);
1634 int yes = dot_order(state, i, LINE_YES);
1635 int unknown = dot_order(state, i, LINE_UNKNOWN);
1636 if ((yes == 1 && unknown == 0) || (yes >= 3)) {
1637 /* violation, so mark all YES edges as errors */
1638 grid_dot *d = g->dots + i;
1640 for (j = 0; j < d->order; j++) {
1641 int e = d->edges[j] - g->edges;
1642 if (state->lines[e] == LINE_YES)
1643 state->line_errors[e] = TRUE;
1645 /* And mark this component as not worthy of further
1647 component_state[comp] = COMP_SILLY;
1649 } else if (yes == 0) {
1650 /* A completely isolated dot must also be excluded it from
1651 * the subsequent loop highlighting pass, but we tag it
1652 * with a different enum value to avoid it counting
1653 * towards the components that inhibit returning a win
1655 component_state[comp] = COMP_EMPTY;
1656 } else if (yes == 1) {
1657 /* A dot with degree 1 that didn't fall into the 'clearly
1658 * erroneous' case above indicates that this connected
1659 * component will be a path rather than a loop - unless
1660 * something worse elsewhere in the component has
1661 * classified it as silly. */
1662 if (component_state[comp] != COMP_SILLY)
1663 component_state[comp] = COMP_PATH;
1667 /* Count up the components. Also, find the largest sensible
1668 * component. (Tie-breaking condition is derived from the order of
1669 * vertices in the grid data structure, which is fairly arbitrary
1670 * but at least stays stable throughout the game.) */
1671 nsilly = nloop = npath = 0;
1673 largest_comp = largest_size = -1;
1674 for (i = 0; i < g->num_dots; i++) {
1675 if (component_state[i] == COMP_SILLY) {
1677 } else if (component_state[i] == COMP_PATH) {
1678 total_pathsize += dsf_size(dsf, i);
1680 } else if (component_state[i] == COMP_LOOP) {
1685 if ((this_size = dsf_size(dsf, i)) > largest_size) {
1687 largest_size = this_size;
1691 if (largest_size < total_pathsize) {
1692 largest_comp = -1; /* means the paths */
1693 largest_size = total_pathsize;
1696 if (nloop > 0 && nloop + npath > 1) {
1698 * If there are at least two sensible components including at
1699 * least one loop, highlight all edges in every sensible
1700 * component that is not the largest one.
1702 for (i = 0; i < g->num_edges; i++) {
1703 if (state->lines[i] == LINE_YES) {
1704 grid_edge *e = g->edges + i;
1705 int d1 = e->dot1 - g->dots; /* either endpoint is good enough */
1706 int comp = dsf_canonify(dsf, d1);
1707 if ((component_state[comp] == COMP_PATH &&
1708 -1 != largest_comp) ||
1709 (component_state[comp] == COMP_LOOP &&
1710 comp != largest_comp))
1711 state->line_errors[i] = TRUE;
1716 if (nloop == 1 && npath == 0 && nsilly == 0) {
1718 * If there is exactly one component and it is a loop, then
1719 * the puzzle is potentially complete, so check the clues.
1723 for (i = 0; i < g->num_faces; i++) {
1724 int c = state->clues[i];
1725 if (c >= 0 && face_order(state, i, LINE_YES) != c) {
1732 * Also, whether or not the puzzle is actually complete, set
1733 * the flag that says this game_state has exactly one loop and
1734 * nothing else, which will be used to vary the semantics of
1735 * clue highlighting at display time.
1737 state->exactly_one_loop = TRUE;
1740 state->exactly_one_loop = FALSE;
1743 sfree(component_state);
1749 /* ----------------------------------------------------------------------
1752 * Our solver modes operate as follows. Each mode also uses the modes above it.
1755 * Just implement the rules of the game.
1757 * Normal and Tricky Modes
1758 * For each (adjacent) pair of lines through each dot we store a bit for
1759 * whether at least one of them is on and whether at most one is on. (If we
1760 * know both or neither is on that's already stored more directly.)
1763 * Use edsf data structure to make equivalence classes of lines that are
1764 * known identical to or opposite to one another.
1769 * For general grids, we consider "dlines" to be pairs of lines joined
1770 * at a dot. The lines must be adjacent around the dot, so we can think of
1771 * a dline as being a dot+face combination. Or, a dot+edge combination where
1772 * the second edge is taken to be the next clockwise edge from the dot.
1773 * Original loopy code didn't have this extra restriction of the lines being
1774 * adjacent. From my tests with square grids, this extra restriction seems to
1775 * take little, if anything, away from the quality of the puzzles.
1776 * A dline can be uniquely identified by an edge/dot combination, given that
1777 * a dline-pair always goes clockwise around its common dot. The edge/dot
1778 * combination can be represented by an edge/bool combination - if bool is
1779 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
1780 * exactly twice the number of edges in the grid - although the dlines
1781 * spanning the infinite face are not all that useful to the solver.
1782 * Note that, by convention, a dline goes clockwise around its common dot,
1783 * which means the dline goes anti-clockwise around its common face.
1786 /* Helper functions for obtaining an index into an array of dlines, given
1787 * various information. We assume the grid layout conventions about how
1788 * the various lists are interleaved - see grid_make_consistent() for
1791 /* i points to the first edge of the dline pair, reading clockwise around
1793 static int dline_index_from_dot(grid *g, grid_dot *d, int i)
1795 grid_edge *e = d->edges[i];
1800 if (i2 == d->order) i2 = 0;
1803 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
1805 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
1806 (int)(d - g->dots), i, (int)(e - g->edges),
1807 (int)(e2 - g->edges), ret);
1811 /* i points to the second edge of the dline pair, reading clockwise around
1812 * the face. That is, the edges of the dline, starting at edge{i}, read
1813 * anti-clockwise around the face. By layout conventions, the common dot
1814 * of the dline will be f->dots[i] */
1815 static int dline_index_from_face(grid *g, grid_face *f, int i)
1817 grid_edge *e = f->edges[i];
1818 grid_dot *d = f->dots[i];
1823 if (i2 < 0) i2 += f->order;
1826 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
1828 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
1829 (int)(f - g->faces), i, (int)(e - g->edges),
1830 (int)(e2 - g->edges), ret);
1834 static int is_atleastone(const char *dline_array, int index)
1836 return BIT_SET(dline_array[index], 0);
1838 static int set_atleastone(char *dline_array, int index)
1840 return SET_BIT(dline_array[index], 0);
1842 static int is_atmostone(const char *dline_array, int index)
1844 return BIT_SET(dline_array[index], 1);
1846 static int set_atmostone(char *dline_array, int index)
1848 return SET_BIT(dline_array[index], 1);
1851 static void array_setall(char *array, char from, char to, int len)
1853 char *p = array, *p_old = p;
1854 int len_remaining = len;
1856 while ((p = memchr(p, from, len_remaining))) {
1858 len_remaining -= p - p_old;
1863 /* Helper, called when doing dline dot deductions, in the case where we
1864 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
1865 * them (because of dline atmostone/atleastone).
1866 * On entry, edge points to the first of these two UNKNOWNs. This function
1867 * will find the opposite UNKNOWNS (if they are adjacent to one another)
1868 * and set their corresponding dline to atleastone. (Setting atmostone
1869 * already happens in earlier dline deductions) */
1870 static int dline_set_opp_atleastone(solver_state *sstate,
1871 grid_dot *d, int edge)
1873 game_state *state = sstate->state;
1874 grid *g = state->game_grid;
1877 for (opp = 0; opp < N; opp++) {
1878 int opp_dline_index;
1879 if (opp == edge || opp == edge+1 || opp == edge-1)
1881 if (opp == 0 && edge == N-1)
1883 if (opp == N-1 && edge == 0)
1886 if (opp2 == N) opp2 = 0;
1887 /* Check if opp, opp2 point to LINE_UNKNOWNs */
1888 if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
1890 if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
1892 /* Found opposite UNKNOWNS and they're next to each other */
1893 opp_dline_index = dline_index_from_dot(g, d, opp);
1894 return set_atleastone(sstate->dlines, opp_dline_index);
1900 /* Set pairs of lines around this face which are known to be identical, to
1901 * the given line_state */
1902 static int face_setall_identical(solver_state *sstate, int face_index,
1903 enum line_state line_new)
1905 /* can[dir] contains the canonical line associated with the line in
1906 * direction dir from the square in question. Similarly inv[dir] is
1907 * whether or not the line in question is inverse to its canonical
1910 game_state *state = sstate->state;
1911 grid *g = state->game_grid;
1912 grid_face *f = g->faces + face_index;
1915 int can1, can2, inv1, inv2;
1917 for (i = 0; i < N; i++) {
1918 int line1_index = f->edges[i] - g->edges;
1919 if (state->lines[line1_index] != LINE_UNKNOWN)
1921 for (j = i + 1; j < N; j++) {
1922 int line2_index = f->edges[j] - g->edges;
1923 if (state->lines[line2_index] != LINE_UNKNOWN)
1926 /* Found two UNKNOWNS */
1927 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
1928 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
1929 if (can1 == can2 && inv1 == inv2) {
1930 solver_set_line(sstate, line1_index, line_new);
1931 solver_set_line(sstate, line2_index, line_new);
1938 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
1939 * return the edge indices into e. */
1940 static void find_unknowns(game_state *state,
1941 grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
1942 int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
1943 int *e /* Returned edge indices */)
1946 grid *g = state->game_grid;
1947 while (c < expected_count) {
1948 int line_index = *edge_list - g->edges;
1949 if (state->lines[line_index] == LINE_UNKNOWN) {
1957 /* If we have a list of edges, and we know whether the number of YESs should
1958 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
1959 * linedsf deductions. This can be used for both face and dot deductions.
1960 * Returns the difficulty level of the next solver that should be used,
1961 * or DIFF_MAX if no progress was made. */
1962 static int parity_deductions(solver_state *sstate,
1963 grid_edge **edge_list, /* Edge list (from a face or a dot) */
1964 int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
1967 game_state *state = sstate->state;
1968 int diff = DIFF_MAX;
1969 int *linedsf = sstate->linedsf;
1971 if (unknown_count == 2) {
1972 /* Lines are known alike/opposite, depending on inv. */
1974 find_unknowns(state, edge_list, 2, e);
1975 if (merge_lines(sstate, e[0], e[1], total_parity))
1976 diff = min(diff, DIFF_HARD);
1977 } else if (unknown_count == 3) {
1979 int can[3]; /* canonical edges */
1980 int inv[3]; /* whether can[x] is inverse to e[x] */
1981 find_unknowns(state, edge_list, 3, e);
1982 can[0] = edsf_canonify(linedsf, e[0], inv);
1983 can[1] = edsf_canonify(linedsf, e[1], inv+1);
1984 can[2] = edsf_canonify(linedsf, e[2], inv+2);
1985 if (can[0] == can[1]) {
1986 if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
1987 LINE_YES : LINE_NO))
1988 diff = min(diff, DIFF_EASY);
1990 if (can[0] == can[2]) {
1991 if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
1992 LINE_YES : LINE_NO))
1993 diff = min(diff, DIFF_EASY);
1995 if (can[1] == can[2]) {
1996 if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
1997 LINE_YES : LINE_NO))
1998 diff = min(diff, DIFF_EASY);
2000 } else if (unknown_count == 4) {
2002 int can[4]; /* canonical edges */
2003 int inv[4]; /* whether can[x] is inverse to e[x] */
2004 find_unknowns(state, edge_list, 4, e);
2005 can[0] = edsf_canonify(linedsf, e[0], inv);
2006 can[1] = edsf_canonify(linedsf, e[1], inv+1);
2007 can[2] = edsf_canonify(linedsf, e[2], inv+2);
2008 can[3] = edsf_canonify(linedsf, e[3], inv+3);
2009 if (can[0] == can[1]) {
2010 if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
2011 diff = min(diff, DIFF_HARD);
2012 } else if (can[0] == can[2]) {
2013 if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
2014 diff = min(diff, DIFF_HARD);
2015 } else if (can[0] == can[3]) {
2016 if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
2017 diff = min(diff, DIFF_HARD);
2018 } else if (can[1] == can[2]) {
2019 if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
2020 diff = min(diff, DIFF_HARD);
2021 } else if (can[1] == can[3]) {
2022 if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
2023 diff = min(diff, DIFF_HARD);
2024 } else if (can[2] == can[3]) {
2025 if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
2026 diff = min(diff, DIFF_HARD);
2034 * These are the main solver functions.
2036 * Their return values are diff values corresponding to the lowest mode solver
2037 * that would notice the work that they have done. For example if the normal
2038 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2039 * easy mode solver might be able to make progress using that. It doesn't make
2040 * sense for one of them to return a diff value higher than that of the
2043 * Each function returns the lowest value it can, as early as possible, in
2044 * order to try and pass as much work as possible back to the lower level
2045 * solvers which progress more quickly.
2048 /* PROPOSED NEW DESIGN:
2049 * We have a work queue consisting of 'events' notifying us that something has
2050 * happened that a particular solver mode might be interested in. For example
2051 * the hard mode solver might do something that helps the normal mode solver at
2052 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2053 * we pull events off the work queue, and hand each in turn to the solver that
2054 * is interested in them. If a solver reports that it failed we pass the same
2055 * event on to progressively more advanced solvers and the loop detector. Once
2056 * we've exhausted an event, or it has helped us progress, we drop it and
2057 * continue to the next one. The events are sorted first in order of solver
2058 * complexity (easy first) then order of insertion (oldest first).
2059 * Once we run out of events we loop over each permitted solver in turn
2060 * (easiest first) until either a deduction is made (and an event therefore
2061 * emerges) or no further deductions can be made (in which case we've failed).
2064 * * How do we 'loop over' a solver when both dots and squares are concerned.
2065 * Answer: first all squares then all dots.
2068 static int trivial_deductions(solver_state *sstate)
2070 int i, current_yes, current_no;
2071 game_state *state = sstate->state;
2072 grid *g = state->game_grid;
2073 int diff = DIFF_MAX;
2075 /* Per-face deductions */
2076 for (i = 0; i < g->num_faces; i++) {
2077 grid_face *f = g->faces + i;
2079 if (sstate->face_solved[i])
2082 current_yes = sstate->face_yes_count[i];
2083 current_no = sstate->face_no_count[i];
2085 if (current_yes + current_no == f->order) {
2086 sstate->face_solved[i] = TRUE;
2090 if (state->clues[i] < 0)
2094 * This code checks whether the numeric clue on a face is so
2095 * large as to permit all its remaining LINE_UNKNOWNs to be
2096 * filled in as LINE_YES, or alternatively so small as to
2097 * permit them all to be filled in as LINE_NO.
2100 if (state->clues[i] < current_yes) {
2101 sstate->solver_status = SOLVER_MISTAKE;
2104 if (state->clues[i] == current_yes) {
2105 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
2106 diff = min(diff, DIFF_EASY);
2107 sstate->face_solved[i] = TRUE;
2111 if (f->order - state->clues[i] < current_no) {
2112 sstate->solver_status = SOLVER_MISTAKE;
2115 if (f->order - state->clues[i] == current_no) {
2116 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
2117 diff = min(diff, DIFF_EASY);
2118 sstate->face_solved[i] = TRUE;
2122 if (f->order - state->clues[i] == current_no + 1 &&
2123 f->order - current_yes - current_no > 2) {
2125 * One small refinement to the above: we also look for any
2126 * adjacent pair of LINE_UNKNOWNs around the face with
2127 * some LINE_YES incident on it from elsewhere. If we find
2128 * one, then we know that pair of LINE_UNKNOWNs can't
2129 * _both_ be LINE_YES, and hence that pushes us one line
2130 * closer to being able to determine all the rest.
2132 int j, k, e1, e2, e, d;
2134 for (j = 0; j < f->order; j++) {
2135 e1 = f->edges[j] - g->edges;
2136 e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges;
2138 if (g->edges[e1].dot1 == g->edges[e2].dot1 ||
2139 g->edges[e1].dot1 == g->edges[e2].dot2) {
2140 d = g->edges[e1].dot1 - g->dots;
2142 assert(g->edges[e1].dot2 == g->edges[e2].dot1 ||
2143 g->edges[e1].dot2 == g->edges[e2].dot2);
2144 d = g->edges[e1].dot2 - g->dots;
2147 if (state->lines[e1] == LINE_UNKNOWN &&
2148 state->lines[e2] == LINE_UNKNOWN) {
2149 for (k = 0; k < g->dots[d].order; k++) {
2150 int e = g->dots[d].edges[k] - g->edges;
2151 if (state->lines[e] == LINE_YES)
2152 goto found; /* multi-level break */
2160 * If we get here, we've found such a pair of edges, and
2161 * they're e1 and e2.
2163 for (j = 0; j < f->order; j++) {
2164 e = f->edges[j] - g->edges;
2165 if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) {
2166 int r = solver_set_line(sstate, e, LINE_YES);
2168 diff = min(diff, DIFF_EASY);
2174 check_caches(sstate);
2176 /* Per-dot deductions */
2177 for (i = 0; i < g->num_dots; i++) {
2178 grid_dot *d = g->dots + i;
2179 int yes, no, unknown;
2181 if (sstate->dot_solved[i])
2184 yes = sstate->dot_yes_count[i];
2185 no = sstate->dot_no_count[i];
2186 unknown = d->order - yes - no;
2190 sstate->dot_solved[i] = TRUE;
2191 } else if (unknown == 1) {
2192 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2193 diff = min(diff, DIFF_EASY);
2194 sstate->dot_solved[i] = TRUE;
2196 } else if (yes == 1) {
2198 sstate->solver_status = SOLVER_MISTAKE;
2200 } else if (unknown == 1) {
2201 dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
2202 diff = min(diff, DIFF_EASY);
2204 } else if (yes == 2) {
2206 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2207 diff = min(diff, DIFF_EASY);
2209 sstate->dot_solved[i] = TRUE;
2211 sstate->solver_status = SOLVER_MISTAKE;
2216 check_caches(sstate);
2221 static int dline_deductions(solver_state *sstate)
2223 game_state *state = sstate->state;
2224 grid *g = state->game_grid;
2225 char *dlines = sstate->dlines;
2227 int diff = DIFF_MAX;
2229 /* ------ Face deductions ------ */
2231 /* Given a set of dline atmostone/atleastone constraints, need to figure
2232 * out if we can deduce any further info. For more general faces than
2233 * squares, this turns out to be a tricky problem.
2234 * The approach taken here is to define (per face) NxN matrices:
2235 * "maxs" and "mins".
2236 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2237 * for the possible number of edges that are YES between positions j and k
2238 * going clockwise around the face. Can think of j and k as marking dots
2239 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2240 * edge1 joins dot1 to dot2 etc).
2241 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2242 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2243 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2244 * the dline atmostone/atleastone status for edges j and j+1.
2246 * Then we calculate the remaining entries recursively. We definitely
2248 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2249 * This is because any valid placement of YESs between j and k must give
2250 * a valid placement between j and u, and also between u and k.
2251 * I believe it's sufficient to use just the two values of u:
2252 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2253 * are rigorous, even if they might not be best-possible.
2255 * Once we have maxs and mins calculated, we can make inferences about
2256 * each dline{j,j+1} by looking at the possible complementary edge-counts
2257 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2258 * As well as dlines, we can make similar inferences about single edges.
2259 * For example, consider a pentagon with clue 3, and we know at most one
2260 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2261 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2262 * that final edge would have to be YES to make the count up to 3.
2265 /* Much quicker to allocate arrays on the stack than the heap, so
2266 * define the largest possible face size, and base our array allocations
2267 * on that. We check this with an assertion, in case someone decides to
2268 * make a grid which has larger faces than this. Note, this algorithm
2269 * could get quite expensive if there are many large faces. */
2270 #define MAX_FACE_SIZE 12
2272 for (i = 0; i < g->num_faces; i++) {
2273 int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
2274 int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
2275 grid_face *f = g->faces + i;
2278 int clue = state->clues[i];
2279 assert(N <= MAX_FACE_SIZE);
2280 if (sstate->face_solved[i])
2282 if (clue < 0) continue;
2284 /* Calculate the (j,j+1) entries */
2285 for (j = 0; j < N; j++) {
2286 int edge_index = f->edges[j] - g->edges;
2288 enum line_state line1 = state->lines[edge_index];
2289 enum line_state line2;
2293 maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
2294 mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
2295 /* Calculate the (j,j+2) entries */
2296 dline_index = dline_index_from_face(g, f, k);
2297 edge_index = f->edges[k] - g->edges;
2298 line2 = state->lines[edge_index];
2304 if (line1 == LINE_NO) tmp--;
2305 if (line2 == LINE_NO) tmp--;
2306 if (tmp == 2 && is_atmostone(dlines, dline_index))
2312 if (line1 == LINE_YES) tmp++;
2313 if (line2 == LINE_YES) tmp++;
2314 if (tmp == 0 && is_atleastone(dlines, dline_index))
2319 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2320 for (m = 3; m < N; m++) {
2321 for (j = 0; j < N; j++) {
2329 maxs[j][k] = maxs[j][u] + maxs[u][k];
2330 mins[j][k] = mins[j][u] + mins[u][k];
2331 tmp = maxs[j][v] + maxs[v][k];
2332 maxs[j][k] = min(maxs[j][k], tmp);
2333 tmp = mins[j][v] + mins[v][k];
2334 mins[j][k] = max(mins[j][k], tmp);
2338 /* See if we can make any deductions */
2339 for (j = 0; j < N; j++) {
2341 grid_edge *e = f->edges[j];
2342 int line_index = e - g->edges;
2345 if (state->lines[line_index] != LINE_UNKNOWN)
2350 /* minimum YESs in the complement of this edge */
2351 if (mins[k][j] > clue) {
2352 sstate->solver_status = SOLVER_MISTAKE;
2355 if (mins[k][j] == clue) {
2356 /* setting this edge to YES would make at least
2357 * (clue+1) edges - contradiction */
2358 solver_set_line(sstate, line_index, LINE_NO);
2359 diff = min(diff, DIFF_EASY);
2361 if (maxs[k][j] < clue - 1) {
2362 sstate->solver_status = SOLVER_MISTAKE;
2365 if (maxs[k][j] == clue - 1) {
2366 /* Only way to satisfy the clue is to set edge{j} as YES */
2367 solver_set_line(sstate, line_index, LINE_YES);
2368 diff = min(diff, DIFF_EASY);
2371 /* More advanced deduction that allows propagation along diagonal
2372 * chains of faces connected by dots, for example, 3-2-...-2-3
2373 * in square grids. */
2374 if (sstate->diff >= DIFF_TRICKY) {
2375 /* Now see if we can make dline deduction for edges{j,j+1} */
2377 if (state->lines[e - g->edges] != LINE_UNKNOWN)
2378 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2379 * Dlines where one of the edges is known, are handled in the
2383 dline_index = dline_index_from_face(g, f, k);
2387 /* minimum YESs in the complement of this dline */
2388 if (mins[k][j] > clue - 2) {
2389 /* Adding 2 YESs would break the clue */
2390 if (set_atmostone(dlines, dline_index))
2391 diff = min(diff, DIFF_NORMAL);
2393 /* maximum YESs in the complement of this dline */
2394 if (maxs[k][j] < clue) {
2395 /* Adding 2 NOs would mean not enough YESs */
2396 if (set_atleastone(dlines, dline_index))
2397 diff = min(diff, DIFF_NORMAL);
2403 if (diff < DIFF_NORMAL)
2406 /* ------ Dot deductions ------ */
2408 for (i = 0; i < g->num_dots; i++) {
2409 grid_dot *d = g->dots + i;
2411 int yes, no, unknown;
2413 if (sstate->dot_solved[i])
2415 yes = sstate->dot_yes_count[i];
2416 no = sstate->dot_no_count[i];
2417 unknown = N - yes - no;
2419 for (j = 0; j < N; j++) {
2422 int line1_index, line2_index;
2423 enum line_state line1, line2;
2426 dline_index = dline_index_from_dot(g, d, j);
2427 line1_index = d->edges[j] - g->edges;
2428 line2_index = d->edges[k] - g->edges;
2429 line1 = state->lines[line1_index];
2430 line2 = state->lines[line2_index];
2432 /* Infer dline state from line state */
2433 if (line1 == LINE_NO || line2 == LINE_NO) {
2434 if (set_atmostone(dlines, dline_index))
2435 diff = min(diff, DIFF_NORMAL);
2437 if (line1 == LINE_YES || line2 == LINE_YES) {
2438 if (set_atleastone(dlines, dline_index))
2439 diff = min(diff, DIFF_NORMAL);
2441 /* Infer line state from dline state */
2442 if (is_atmostone(dlines, dline_index)) {
2443 if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
2444 solver_set_line(sstate, line2_index, LINE_NO);
2445 diff = min(diff, DIFF_EASY);
2447 if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
2448 solver_set_line(sstate, line1_index, LINE_NO);
2449 diff = min(diff, DIFF_EASY);
2452 if (is_atleastone(dlines, dline_index)) {
2453 if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
2454 solver_set_line(sstate, line2_index, LINE_YES);
2455 diff = min(diff, DIFF_EASY);
2457 if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
2458 solver_set_line(sstate, line1_index, LINE_YES);
2459 diff = min(diff, DIFF_EASY);
2462 /* Deductions that depend on the numbers of lines.
2463 * Only bother if both lines are UNKNOWN, otherwise the
2464 * easy-mode solver (or deductions above) would have taken
2466 if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
2469 if (yes == 0 && unknown == 2) {
2470 /* Both these unknowns must be identical. If we know
2471 * atmostone or atleastone, we can make progress. */
2472 if (is_atmostone(dlines, dline_index)) {
2473 solver_set_line(sstate, line1_index, LINE_NO);
2474 solver_set_line(sstate, line2_index, LINE_NO);
2475 diff = min(diff, DIFF_EASY);
2477 if (is_atleastone(dlines, dline_index)) {
2478 solver_set_line(sstate, line1_index, LINE_YES);
2479 solver_set_line(sstate, line2_index, LINE_YES);
2480 diff = min(diff, DIFF_EASY);
2484 if (set_atmostone(dlines, dline_index))
2485 diff = min(diff, DIFF_NORMAL);
2487 if (set_atleastone(dlines, dline_index))
2488 diff = min(diff, DIFF_NORMAL);
2492 /* More advanced deduction that allows propagation along diagonal
2493 * chains of faces connected by dots, for example: 3-2-...-2-3
2494 * in square grids. */
2495 if (sstate->diff >= DIFF_TRICKY) {
2496 /* If we have atleastone set for this dline, infer
2497 * atmostone for each "opposite" dline (that is, each
2498 * dline without edges in common with this one).
2499 * Again, this test is only worth doing if both these
2500 * lines are UNKNOWN. For if one of these lines were YES,
2501 * the (yes == 1) test above would kick in instead. */
2502 if (is_atleastone(dlines, dline_index)) {
2504 for (opp = 0; opp < N; opp++) {
2505 int opp_dline_index;
2506 if (opp == j || opp == j+1 || opp == j-1)
2508 if (j == 0 && opp == N-1)
2510 if (j == N-1 && opp == 0)
2512 opp_dline_index = dline_index_from_dot(g, d, opp);
2513 if (set_atmostone(dlines, opp_dline_index))
2514 diff = min(diff, DIFF_NORMAL);
2516 if (yes == 0 && is_atmostone(dlines, dline_index)) {
2517 /* This dline has *exactly* one YES and there are no
2518 * other YESs. This allows more deductions. */
2520 /* Third unknown must be YES */
2521 for (opp = 0; opp < N; opp++) {
2523 if (opp == j || opp == k)
2525 opp_index = d->edges[opp] - g->edges;
2526 if (state->lines[opp_index] == LINE_UNKNOWN) {
2527 solver_set_line(sstate, opp_index,
2529 diff = min(diff, DIFF_EASY);
2532 } else if (unknown == 4) {
2533 /* Exactly one of opposite UNKNOWNS is YES. We've
2534 * already set atmostone, so set atleastone as
2537 if (dline_set_opp_atleastone(sstate, d, j))
2538 diff = min(diff, DIFF_NORMAL);
2548 static int linedsf_deductions(solver_state *sstate)
2550 game_state *state = sstate->state;
2551 grid *g = state->game_grid;
2552 char *dlines = sstate->dlines;
2554 int diff = DIFF_MAX;
2557 /* ------ Face deductions ------ */
2559 /* A fully-general linedsf deduction seems overly complicated
2560 * (I suspect the problem is NP-complete, though in practice it might just
2561 * be doable because faces are limited in size).
2562 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2563 * known to be identical. If setting them both to YES (or NO) would break
2564 * the clue, set them to NO (or YES). */
2566 for (i = 0; i < g->num_faces; i++) {
2567 int N, yes, no, unknown;
2570 if (sstate->face_solved[i])
2572 clue = state->clues[i];
2576 N = g->faces[i].order;
2577 yes = sstate->face_yes_count[i];
2578 if (yes + 1 == clue) {
2579 if (face_setall_identical(sstate, i, LINE_NO))
2580 diff = min(diff, DIFF_EASY);
2582 no = sstate->face_no_count[i];
2583 if (no + 1 == N - clue) {
2584 if (face_setall_identical(sstate, i, LINE_YES))
2585 diff = min(diff, DIFF_EASY);
2588 /* Reload YES count, it might have changed */
2589 yes = sstate->face_yes_count[i];
2590 unknown = N - no - yes;
2592 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2593 * parity of lines. */
2594 diff_tmp = parity_deductions(sstate, g->faces[i].edges,
2595 (clue - yes) % 2, unknown);
2596 diff = min(diff, diff_tmp);
2599 /* ------ Dot deductions ------ */
2600 for (i = 0; i < g->num_dots; i++) {
2601 grid_dot *d = g->dots + i;
2604 int yes, no, unknown;
2605 /* Go through dlines, and do any dline<->linedsf deductions wherever
2606 * we find two UNKNOWNS. */
2607 for (j = 0; j < N; j++) {
2608 int dline_index = dline_index_from_dot(g, d, j);
2611 int can1, can2, inv1, inv2;
2613 line1_index = d->edges[j] - g->edges;
2614 if (state->lines[line1_index] != LINE_UNKNOWN)
2617 if (j2 == N) j2 = 0;
2618 line2_index = d->edges[j2] - g->edges;
2619 if (state->lines[line2_index] != LINE_UNKNOWN)
2621 /* Infer dline flags from linedsf */
2622 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
2623 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
2624 if (can1 == can2 && inv1 != inv2) {
2625 /* These are opposites, so set dline atmostone/atleastone */
2626 if (set_atmostone(dlines, dline_index))
2627 diff = min(diff, DIFF_NORMAL);
2628 if (set_atleastone(dlines, dline_index))
2629 diff = min(diff, DIFF_NORMAL);
2632 /* Infer linedsf from dline flags */
2633 if (is_atmostone(dlines, dline_index)
2634 && is_atleastone(dlines, dline_index)) {
2635 if (merge_lines(sstate, line1_index, line2_index, 1))
2636 diff = min(diff, DIFF_HARD);
2640 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2641 * parity of lines. */
2642 yes = sstate->dot_yes_count[i];
2643 no = sstate->dot_no_count[i];
2644 unknown = N - yes - no;
2645 diff_tmp = parity_deductions(sstate, d->edges,
2647 diff = min(diff, diff_tmp);
2650 /* ------ Edge dsf deductions ------ */
2652 /* If the state of a line is known, deduce the state of its canonical line
2653 * too, and vice versa. */
2654 for (i = 0; i < g->num_edges; i++) {
2657 can = edsf_canonify(sstate->linedsf, i, &inv);
2660 s = sstate->state->lines[can];
2661 if (s != LINE_UNKNOWN) {
2662 if (solver_set_line(sstate, i, inv ? OPP(s) : s))
2663 diff = min(diff, DIFF_EASY);
2665 s = sstate->state->lines[i];
2666 if (s != LINE_UNKNOWN) {
2667 if (solver_set_line(sstate, can, inv ? OPP(s) : s))
2668 diff = min(diff, DIFF_EASY);
2676 static int loop_deductions(solver_state *sstate)
2678 int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
2679 game_state *state = sstate->state;
2680 grid *g = state->game_grid;
2681 int shortest_chainlen = g->num_dots;
2682 int loop_found = FALSE;
2684 int progress = FALSE;
2688 * Go through the grid and update for all the new edges.
2689 * Since merge_dots() is idempotent, the simplest way to
2690 * do this is just to update for _all_ the edges.
2691 * Also, while we're here, we count the edges.
2693 for (i = 0; i < g->num_edges; i++) {
2694 if (state->lines[i] == LINE_YES) {
2695 loop_found |= merge_dots(sstate, i);
2701 * Count the clues, count the satisfied clues, and count the
2702 * satisfied-minus-one clues.
2704 for (i = 0; i < g->num_faces; i++) {
2705 int c = state->clues[i];
2707 int o = sstate->face_yes_count[i];
2716 for (i = 0; i < g->num_dots; ++i) {
2718 sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
2719 if (dots_connected > 1)
2720 shortest_chainlen = min(shortest_chainlen, dots_connected);
2723 assert(sstate->solver_status == SOLVER_INCOMPLETE);
2725 if (satclues == clues && shortest_chainlen == edgecount) {
2726 sstate->solver_status = SOLVER_SOLVED;
2727 /* This discovery clearly counts as progress, even if we haven't
2728 * just added any lines or anything */
2730 goto finished_loop_deductionsing;
2734 * Now go through looking for LINE_UNKNOWN edges which
2735 * connect two dots that are already in the same
2736 * equivalence class. If we find one, test to see if the
2737 * loop it would create is a solution.
2739 for (i = 0; i < g->num_edges; i++) {
2740 grid_edge *e = g->edges + i;
2741 int d1 = e->dot1 - g->dots;
2742 int d2 = e->dot2 - g->dots;
2744 if (state->lines[i] != LINE_UNKNOWN)
2747 eqclass = dsf_canonify(sstate->dotdsf, d1);
2748 if (eqclass != dsf_canonify(sstate->dotdsf, d2))
2751 val = LINE_NO; /* loop is bad until proven otherwise */
2754 * This edge would form a loop. Next
2755 * question: how long would the loop be?
2756 * Would it equal the total number of edges
2757 * (plus the one we'd be adding if we added
2760 if (sstate->looplen[eqclass] == edgecount + 1) {
2764 * This edge would form a loop which
2765 * took in all the edges in the entire
2766 * grid. So now we need to work out
2767 * whether it would be a valid solution
2768 * to the puzzle, which means we have to
2769 * check if it satisfies all the clues.
2770 * This means that every clue must be
2771 * either satisfied or satisfied-minus-
2772 * 1, and also that the number of
2773 * satisfied-minus-1 clues must be at
2774 * most two and they must lie on either
2775 * side of this edge.
2779 int f = e->face1 - g->faces;
2780 int c = state->clues[f];
2781 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
2785 int f = e->face2 - g->faces;
2786 int c = state->clues[f];
2787 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
2790 if (sm1clues == sm1_nearby &&
2791 sm1clues + satclues == clues) {
2792 val = LINE_YES; /* loop is good! */
2797 * Right. Now we know that adding this edge
2798 * would form a loop, and we know whether
2799 * that loop would be a viable solution or
2802 * If adding this edge produces a solution,
2803 * then we know we've found _a_ solution but
2804 * we don't know that it's _the_ solution -
2805 * if it were provably the solution then
2806 * we'd have deduced this edge some time ago
2807 * without the need to do loop detection. So
2808 * in this state we return SOLVER_AMBIGUOUS,
2809 * which has the effect that hitting Solve
2810 * on a user-provided puzzle will fill in a
2811 * solution but using the solver to
2812 * construct new puzzles won't consider this
2813 * a reasonable deduction for the user to
2816 progress = solver_set_line(sstate, i, val);
2817 assert(progress == TRUE);
2818 if (val == LINE_YES) {
2819 sstate->solver_status = SOLVER_AMBIGUOUS;
2820 goto finished_loop_deductionsing;
2824 finished_loop_deductionsing:
2825 return progress ? DIFF_EASY : DIFF_MAX;
2828 /* This will return a dynamically allocated solver_state containing the (more)
2830 static solver_state *solve_game_rec(const solver_state *sstate_start)
2832 solver_state *sstate;
2834 /* Index of the solver we should call next. */
2837 /* As a speed-optimisation, we avoid re-running solvers that we know
2838 * won't make any progress. This happens when a high-difficulty
2839 * solver makes a deduction that can only help other high-difficulty
2841 * For example: if a new 'dline' flag is set by dline_deductions, the
2842 * trivial_deductions solver cannot do anything with this information.
2843 * If we've already run the trivial_deductions solver (because it's
2844 * earlier in the list), there's no point running it again.
2846 * Therefore: if a solver is earlier in the list than "threshold_index",
2847 * we don't bother running it if it's difficulty level is less than
2850 int threshold_diff = 0;
2851 int threshold_index = 0;
2853 sstate = dup_solver_state(sstate_start);
2855 check_caches(sstate);
2857 while (i < NUM_SOLVERS) {
2858 if (sstate->solver_status == SOLVER_MISTAKE)
2860 if (sstate->solver_status == SOLVER_SOLVED ||
2861 sstate->solver_status == SOLVER_AMBIGUOUS) {
2862 /* solver finished */
2866 if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
2867 && solver_diffs[i] <= sstate->diff) {
2868 /* current_solver is eligible, so use it */
2869 int next_diff = solver_fns[i](sstate);
2870 if (next_diff != DIFF_MAX) {
2871 /* solver made progress, so use new thresholds and
2872 * start again at top of list. */
2873 threshold_diff = next_diff;
2874 threshold_index = i;
2879 /* current_solver is ineligible, or failed to make progress, so
2880 * go to the next solver in the list */
2884 if (sstate->solver_status == SOLVER_SOLVED ||
2885 sstate->solver_status == SOLVER_AMBIGUOUS) {
2886 /* s/LINE_UNKNOWN/LINE_NO/g */
2887 array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
2888 sstate->state->game_grid->num_edges);
2895 static char *solve_game(const game_state *state, const game_state *currstate,
2896 const char *aux, char **error)
2899 solver_state *sstate, *new_sstate;
2901 sstate = new_solver_state(state, DIFF_MAX);
2902 new_sstate = solve_game_rec(sstate);
2904 if (new_sstate->solver_status == SOLVER_SOLVED) {
2905 soln = encode_solve_move(new_sstate->state);
2906 } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
2907 soln = encode_solve_move(new_sstate->state);
2908 /**error = "Solver found ambiguous solutions"; */
2910 soln = encode_solve_move(new_sstate->state);
2911 /**error = "Solver failed"; */
2914 free_solver_state(new_sstate);
2915 free_solver_state(sstate);
2920 /* ----------------------------------------------------------------------
2921 * Drawing and mouse-handling
2924 static char *interpret_move(const game_state *state, game_ui *ui,
2925 const game_drawstate *ds,
2926 int x, int y, int button)
2928 grid *g = state->game_grid;
2932 char button_char = ' ';
2933 enum line_state old_state;
2935 button &= ~MOD_MASK;
2937 /* Convert mouse-click (x,y) to grid coordinates */
2938 x -= BORDER(ds->tilesize);
2939 y -= BORDER(ds->tilesize);
2940 x = x * g->tilesize / ds->tilesize;
2941 y = y * g->tilesize / ds->tilesize;
2945 e = grid_nearest_edge(g, x, y);
2951 /* I think it's only possible to play this game with mouse clicks, sorry */
2952 /* Maybe will add mouse drag support some time */
2953 old_state = state->lines[i];
2957 switch (old_state) {
2975 switch (old_state) {
2994 sprintf(buf, "%d%c", i, (int)button_char);
3000 static game_state *execute_move(const game_state *state, const char *move)
3003 game_state *newstate = dup_game(state);
3005 if (move[0] == 'S') {
3007 newstate->cheated = TRUE;
3012 if (i < 0 || i >= newstate->game_grid->num_edges)
3014 move += strspn(move, "1234567890");
3015 switch (*(move++)) {
3017 newstate->lines[i] = LINE_YES;
3020 newstate->lines[i] = LINE_NO;
3023 newstate->lines[i] = LINE_UNKNOWN;
3031 * Check for completion.
3033 if (check_completion(newstate))
3034 newstate->solved = TRUE;
3039 free_game(newstate);
3043 /* ----------------------------------------------------------------------
3047 /* Convert from grid coordinates to screen coordinates */
3048 static void grid_to_screen(const game_drawstate *ds, const grid *g,
3049 int grid_x, int grid_y, int *x, int *y)
3051 *x = grid_x - g->lowest_x;
3052 *y = grid_y - g->lowest_y;
3053 *x = *x * ds->tilesize / g->tilesize;
3054 *y = *y * ds->tilesize / g->tilesize;
3055 *x += BORDER(ds->tilesize);
3056 *y += BORDER(ds->tilesize);
3059 /* Returns (into x,y) position of centre of face for rendering the text clue.
3061 static void face_text_pos(const game_drawstate *ds, const grid *g,
3062 grid_face *f, int *xret, int *yret)
3064 int faceindex = f - g->faces;
3067 * Return the cached position for this face, if we've already
3070 if (ds->textx[faceindex] >= 0) {
3071 *xret = ds->textx[faceindex];
3072 *yret = ds->texty[faceindex];
3077 * Otherwise, use the incentre computed by grid.c and convert it
3078 * to screen coordinates.
3080 grid_find_incentre(f);
3081 grid_to_screen(ds, g, f->ix, f->iy,
3082 &ds->textx[faceindex], &ds->texty[faceindex]);
3084 *xret = ds->textx[faceindex];
3085 *yret = ds->texty[faceindex];
3088 static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f,
3089 int *x, int *y, int *w, int *h)
3092 face_text_pos(ds, g, f, &xx, &yy);
3094 /* There seems to be a certain amount of trial-and-error involved
3095 * in working out the correct bounding-box for the text. */
3097 *x = xx - ds->tilesize/4 - 1;
3098 *y = yy - ds->tilesize/4 - 3;
3099 *w = ds->tilesize/2 + 2;
3100 *h = ds->tilesize/2 + 5;
3103 static void game_redraw_clue(drawing *dr, game_drawstate *ds,
3104 const game_state *state, int i)
3106 grid *g = state->game_grid;
3107 grid_face *f = g->faces + i;
3111 sprintf(c, "%d", state->clues[i]);
3113 face_text_pos(ds, g, f, &x, &y);
3115 FONT_VARIABLE, ds->tilesize/2,
3116 ALIGN_VCENTRE | ALIGN_HCENTRE,
3117 ds->clue_error[i] ? COL_MISTAKE :
3118 ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
3121 static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e,
3122 int *x, int *y, int *w, int *h)
3124 int x1 = e->dot1->x;
3125 int y1 = e->dot1->y;
3126 int x2 = e->dot2->x;
3127 int y2 = e->dot2->y;
3128 int xmin, xmax, ymin, ymax;
3130 grid_to_screen(ds, g, x1, y1, &x1, &y1);
3131 grid_to_screen(ds, g, x2, y2, &x2, &y2);
3132 /* Allow extra margin for dots, and thickness of lines */
3133 xmin = min(x1, x2) - 2;
3134 xmax = max(x1, x2) + 2;
3135 ymin = min(y1, y2) - 2;
3136 ymax = max(y1, y2) + 2;
3140 *w = xmax - xmin + 1;
3141 *h = ymax - ymin + 1;
3144 static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d,
3145 int *x, int *y, int *w, int *h)
3149 grid_to_screen(ds, g, d->x, d->y, &x1, &y1);
3157 static const int loopy_line_redraw_phases[] = {
3158 COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE
3160 #define NPHASES lenof(loopy_line_redraw_phases)
3162 static void game_redraw_line(drawing *dr, game_drawstate *ds,
3163 const game_state *state, int i, int phase)
3165 grid *g = state->game_grid;
3166 grid_edge *e = g->edges + i;
3170 if (state->line_errors[i])
3171 line_colour = COL_MISTAKE;
3172 else if (state->lines[i] == LINE_UNKNOWN)
3173 line_colour = COL_LINEUNKNOWN;
3174 else if (state->lines[i] == LINE_NO)
3175 line_colour = COL_FAINT;
3176 else if (ds->flashing)
3177 line_colour = COL_HIGHLIGHT;
3179 line_colour = COL_FOREGROUND;
3180 if (line_colour != loopy_line_redraw_phases[phase])
3183 /* Convert from grid to screen coordinates */
3184 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3185 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3187 if (line_colour == COL_FAINT) {
3188 static int draw_faint_lines = -1;
3189 if (draw_faint_lines < 0) {
3190 char *env = getenv("LOOPY_FAINT_LINES");
3191 draw_faint_lines = (!env || (env[0] == 'y' ||
3194 if (draw_faint_lines)
3195 draw_line(dr, x1, y1, x2, y2, line_colour);
3197 draw_thick_line(dr, 3.0,
3204 static void game_redraw_dot(drawing *dr, game_drawstate *ds,
3205 const game_state *state, int i)
3207 grid *g = state->game_grid;
3208 grid_dot *d = g->dots + i;
3211 grid_to_screen(ds, g, d->x, d->y, &x, &y);
3212 draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
3215 static int boxes_intersect(int x0, int y0, int w0, int h0,
3216 int x1, int y1, int w1, int h1)
3219 * Two intervals intersect iff neither is wholly on one side of
3220 * the other. Two boxes intersect iff their horizontal and
3221 * vertical intervals both intersect.
3223 return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0);
3226 static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
3227 const game_state *state,
3228 int x, int y, int w, int h)
3230 grid *g = state->game_grid;
3234 clip(dr, x, y, w, h);
3235 draw_rect(dr, x, y, w, h, COL_BACKGROUND);
3237 for (i = 0; i < g->num_faces; i++) {
3238 if (state->clues[i] >= 0) {
3239 face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh);
3240 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3241 game_redraw_clue(dr, ds, state, i);
3244 for (phase = 0; phase < NPHASES; phase++) {
3245 for (i = 0; i < g->num_edges; i++) {
3246 edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh);
3247 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3248 game_redraw_line(dr, ds, state, i, phase);
3251 for (i = 0; i < g->num_dots; i++) {
3252 dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh);
3253 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3254 game_redraw_dot(dr, ds, state, i);
3258 draw_update(dr, x, y, w, h);
3261 static void game_redraw(drawing *dr, game_drawstate *ds,
3262 const game_state *oldstate, const game_state *state,
3263 int dir, const game_ui *ui,
3264 float animtime, float flashtime)
3266 #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
3268 grid *g = state->game_grid;
3269 int border = BORDER(ds->tilesize);
3272 int redraw_everything = FALSE;
3274 int edges[REDRAW_OBJECTS_LIMIT], nedges = 0;
3275 int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0;
3277 /* Redrawing is somewhat involved.
3279 * An update can theoretically affect an arbitrary number of edges
3280 * (consider, for example, completing or breaking a cycle which doesn't
3281 * satisfy all the clues -- we'll switch many edges between error and
3282 * normal states). On the other hand, redrawing the whole grid takes a
3283 * while, making the game feel sluggish, and many updates are actually
3284 * quite well localized.
3286 * This redraw algorithm attempts to cope with both situations gracefully
3287 * and correctly. For localized changes, we set a clip rectangle, fill
3288 * it with background, and then redraw (a plausible but conservative
3289 * guess at) the objects which intersect the rectangle; if several
3290 * objects need redrawing, we'll do them individually. However, if lots
3291 * of objects are affected, we'll just redraw everything.
3293 * The reason for all of this is that it's just not safe to do the redraw
3294 * piecemeal. If you try to draw an antialiased diagonal line over
3295 * itself, you get a slightly thicker antialiased diagonal line, which
3296 * looks rather ugly after a while.
3298 * So, we take two passes over the grid. The first attempts to work out
3299 * what needs doing, and the second actually does it.
3303 redraw_everything = TRUE;
3305 * But we must still go through the upcoming loops, so that we
3306 * set up stuff in ds correctly for the initial redraw.
3310 /* First, trundle through the faces. */
3311 for (i = 0; i < g->num_faces; i++) {
3312 grid_face *f = g->faces + i;
3313 int sides = f->order;
3314 int yes_order, no_order;
3317 int n = state->clues[i];
3321 yes_order = face_order(state, i, LINE_YES);
3322 if (state->exactly_one_loop) {
3324 * Special case: if the set of LINE_YES edges in the grid
3325 * consists of exactly one loop and nothing else, then we
3326 * switch to treating LINE_UNKNOWN the same as LINE_NO for
3327 * purposes of clue checking.
3329 * This is because some people like to play Loopy without
3330 * using the right-click, i.e. never setting anything to
3331 * LINE_NO. Without this special case, if a person playing
3332 * in that style fills in what they think is a correct
3333 * solution loop but in fact it has an underfilled clue,
3334 * then we will display no victory flash and also no error
3335 * highlight explaining why not. With this special case,
3336 * we light up underfilled clues at the instant the loop
3337 * is closed. (Of course, *overfilled* clues are fine
3340 * (It might still be considered unfortunate that we can't
3341 * warn this style of player any earlier, if they make a
3342 * mistake very near the beginning which doesn't show up
3343 * until they close the last edge of the loop. One other
3344 * thing we _could_ do here is to treat any LINE_UNKNOWN
3345 * as LINE_NO if either of its endpoints has yes-degree 2,
3346 * reflecting the fact that setting that line to YES would
3347 * be an obvious error. But I don't think even that could
3348 * catch _all_ clue errors in a timely manner; I think
3349 * there are some that won't be displayed until the loop
3350 * is filled in, even so, and there's no way to avoid that
3351 * with complete reliability except to switch to being a
3352 * player who sets things to LINE_NO.)
3354 no_order = sides - yes_order;
3356 no_order = face_order(state, i, LINE_NO);
3359 clue_mistake = (yes_order > n || no_order > (sides-n));
3360 clue_satisfied = (yes_order == n && no_order == (sides-n));
3362 if (clue_mistake != ds->clue_error[i] ||
3363 clue_satisfied != ds->clue_satisfied[i]) {
3364 ds->clue_error[i] = clue_mistake;
3365 ds->clue_satisfied[i] = clue_satisfied;
3366 if (nfaces == REDRAW_OBJECTS_LIMIT)
3367 redraw_everything = TRUE;
3369 faces[nfaces++] = i;
3373 /* Work out what the flash state needs to be. */
3374 if (flashtime > 0 &&
3375 (flashtime <= FLASH_TIME/3 ||
3376 flashtime >= FLASH_TIME*2/3)) {
3377 flash_changed = !ds->flashing;
3378 ds->flashing = TRUE;
3380 flash_changed = ds->flashing;
3381 ds->flashing = FALSE;
3384 /* Now, trundle through the edges. */
3385 for (i = 0; i < g->num_edges; i++) {
3387 state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
3388 if (new_ds != ds->lines[i] ||
3389 (flash_changed && state->lines[i] == LINE_YES)) {
3390 ds->lines[i] = new_ds;
3391 if (nedges == REDRAW_OBJECTS_LIMIT)
3392 redraw_everything = TRUE;
3394 edges[nedges++] = i;
3398 /* Pass one is now done. Now we do the actual drawing. */
3399 if (redraw_everything) {
3400 int grid_width = g->highest_x - g->lowest_x;
3401 int grid_height = g->highest_y - g->lowest_y;
3402 int w = grid_width * ds->tilesize / g->tilesize;
3403 int h = grid_height * ds->tilesize / g->tilesize;
3405 game_redraw_in_rect(dr, ds, state,
3406 0, 0, w + 2*border + 1, h + 2*border + 1);
3409 /* Right. Now we roll up our sleeves. */
3411 for (i = 0; i < nfaces; i++) {
3412 grid_face *f = g->faces + faces[i];
3415 face_text_bbox(ds, g, f, &x, &y, &w, &h);
3416 game_redraw_in_rect(dr, ds, state, x, y, w, h);
3419 for (i = 0; i < nedges; i++) {
3420 grid_edge *e = g->edges + edges[i];
3423 edge_bbox(ds, g, e, &x, &y, &w, &h);
3424 game_redraw_in_rect(dr, ds, state, x, y, w, h);
3431 static float game_flash_length(const game_state *oldstate,
3432 const game_state *newstate, int dir, game_ui *ui)
3434 if (!oldstate->solved && newstate->solved &&
3435 !oldstate->cheated && !newstate->cheated) {
3442 static int game_status(const game_state *state)
3444 return state->solved ? +1 : 0;
3447 static void game_print_size(const game_params *params, float *x, float *y)
3452 * I'll use 7mm "squares" by default.
3454 game_compute_size(params, 700, &pw, &ph);
3459 static void game_print(drawing *dr, const game_state *state, int tilesize)
3461 int ink = print_mono_colour(dr, 0);
3463 game_drawstate ads, *ds = &ads;
3464 grid *g = state->game_grid;
3466 ds->tilesize = tilesize;
3467 ds->textx = snewn(g->num_faces, int);
3468 ds->texty = snewn(g->num_faces, int);
3469 for (i = 0; i < g->num_faces; i++)
3470 ds->textx[i] = ds->texty[i] = -1;
3472 for (i = 0; i < g->num_dots; i++) {
3474 grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
3475 draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
3481 for (i = 0; i < g->num_faces; i++) {
3482 grid_face *f = g->faces + i;
3483 int clue = state->clues[i];
3487 sprintf(c, "%d", state->clues[i]);
3488 face_text_pos(ds, g, f, &x, &y);
3490 FONT_VARIABLE, ds->tilesize / 2,
3491 ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
3498 for (i = 0; i < g->num_edges; i++) {
3499 int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
3500 grid_edge *e = g->edges + i;
3502 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3503 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3504 if (state->lines[i] == LINE_YES)
3506 /* (dx, dy) points from (x1, y1) to (x2, y2).
3507 * The line is then "fattened" in a perpendicular
3508 * direction to create a thin rectangle. */
3509 double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
3510 double dx = (x2 - x1) / d;
3511 double dy = (y2 - y1) / d;
3514 dx = (dx * ds->tilesize) / thickness;
3515 dy = (dy * ds->tilesize) / thickness;
3516 points[0] = x1 + (int)dy;
3517 points[1] = y1 - (int)dx;
3518 points[2] = x1 - (int)dy;
3519 points[3] = y1 + (int)dx;
3520 points[4] = x2 - (int)dy;
3521 points[5] = y2 + (int)dx;
3522 points[6] = x2 + (int)dy;
3523 points[7] = y2 - (int)dx;
3524 draw_polygon(dr, points, 4, ink, ink);
3528 /* Draw a dotted line */
3531 for (j = 1; j < divisions; j++) {
3532 /* Weighted average */
3533 int x = (x1 * (divisions -j) + x2 * j) / divisions;
3534 int y = (y1 * (divisions -j) + y2 * j) / divisions;
3535 draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
3545 #define thegame loopy
3548 const struct game thegame = {
3549 "Loopy", "games.loopy", "loopy",
3551 game_fetch_preset, NULL,
3556 TRUE, game_configure, custom_params,
3564 TRUE, game_can_format_as_text_now, game_text_format,
3572 PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
3575 game_free_drawstate,
3580 TRUE, FALSE, game_print_size, game_print,
3581 FALSE /* wants_statusbar */,
3582 FALSE, game_timing_state,
3583 0, /* mouse_priorities */
3586 #ifdef STANDALONE_SOLVER
3589 * Half-hearted standalone solver. It can't output the solution to
3590 * anything but a square puzzle, and it can't log the deductions
3591 * it makes either. But it can solve square puzzles, and more
3592 * importantly it can use its solver to grade the difficulty of
3593 * any puzzle you give it.
3598 int main(int argc, char **argv)
3602 char *id = NULL, *desc, *err;
3605 #if 0 /* verbose solver not supported here (yet) */
3606 int really_verbose = FALSE;
3609 while (--argc > 0) {
3611 #if 0 /* verbose solver not supported here (yet) */
3612 if (!strcmp(p, "-v")) {
3613 really_verbose = TRUE;
3616 if (!strcmp(p, "-g")) {
3618 } else if (*p == '-') {
3619 fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
3627 fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
3631 desc = strchr(id, ':');
3633 fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
3638 p = default_params();
3639 decode_params(p, id);
3640 err = validate_desc(p, desc);
3642 fprintf(stderr, "%s: %s\n", argv[0], err);
3645 s = new_game(NULL, p, desc);
3648 * When solving an Easy puzzle, we don't want to bother the
3649 * user with Hard-level deductions. For this reason, we grade
3650 * the puzzle internally before doing anything else.
3652 ret = -1; /* placate optimiser */
3653 for (diff = 0; diff < DIFF_MAX; diff++) {
3654 solver_state *sstate_new;
3655 solver_state *sstate = new_solver_state((game_state *)s, diff);
3657 sstate_new = solve_game_rec(sstate);
3659 if (sstate_new->solver_status == SOLVER_MISTAKE)
3661 else if (sstate_new->solver_status == SOLVER_SOLVED)
3666 free_solver_state(sstate_new);
3667 free_solver_state(sstate);
3673 if (diff == DIFF_MAX) {
3675 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3677 printf("Unable to find a unique solution\n");
3681 printf("Difficulty rating: impossible (no solution exists)\n");
3683 printf("Difficulty rating: %s\n", diffnames[diff]);
3685 solver_state *sstate_new;
3686 solver_state *sstate = new_solver_state((game_state *)s, diff);
3688 /* If we supported a verbose solver, we'd set verbosity here */
3690 sstate_new = solve_game_rec(sstate);
3692 if (sstate_new->solver_status == SOLVER_MISTAKE)
3693 printf("Puzzle is inconsistent\n");
3695 assert(sstate_new->solver_status == SOLVER_SOLVED);
3696 if (s->grid_type == 0) {
3697 fputs(game_text_format(sstate_new->state), stdout);
3699 printf("Unable to output non-square grids\n");
3703 free_solver_state(sstate_new);
3704 free_solver_state(sstate);
3713 /* vim: set shiftwidth=4 tabstop=8: */