4 * An implementation of the Nikoli game 'Loop the loop'.
5 * (c) Mike Pinna, 2005, 2006
6 * Substantially rewritten to allowing for more general types of grid.
7 * (c) Lambros Lambrou 2008
9 * vim: set shiftwidth=4 :set textwidth=80:
13 * Possible future solver enhancements:
15 * - There's an interesting deductive technique which makes use
16 * of topology rather than just graph theory. Each _face_ in
17 * the grid is either inside or outside the loop; you can tell
18 * that two faces are on the same side of the loop if they're
19 * separated by a LINE_NO (or, more generally, by a path
20 * crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
21 * and on the opposite side of the loop if they're separated by
22 * a LINE_YES (or an odd number of LINE_YESes and no
23 * LINE_UNKNOWNs). Oh, and any face separated from the outside
24 * of the grid by a LINE_YES or a LINE_NO is on the inside or
25 * outside respectively. So if you can track this for all
26 * faces, you figure out the state of the line between a pair
27 * once their relative insideness is known.
28 * + The way I envisage this working is simply to keep an edsf
29 * of all _faces_, which indicates whether they're on
30 * opposite sides of the loop from one another. We also
31 * include a special entry in the edsf for the infinite
33 * + So, the simple way to do this is to just go through the
34 * edges: every time we see an edge in a state other than
35 * LINE_UNKNOWN which separates two faces that aren't in the
36 * same edsf class, we can rectify that by merging the
37 * classes. Then, conversely, an edge in LINE_UNKNOWN state
38 * which separates two faces that _are_ in the same edsf
39 * class can immediately have its state determined.
40 * + But you can go one better, if you're prepared to loop
41 * over all _pairs_ of edges. Suppose we have edges A and B,
42 * which respectively separate faces A1,A2 and B1,B2.
43 * Suppose that A,B are in the same edge-edsf class and that
44 * A1,B1 (wlog) are in the same face-edsf class; then we can
45 * immediately place A2,B2 into the same face-edsf class (as
46 * each other, not as A1 and A2) one way round or the other.
47 * And conversely again, if A1,B1 are in the same face-edsf
48 * class and so are A2,B2, then we can put A,B into the same
50 * * Of course, this deduction requires a quadratic-time
51 * loop over all pairs of edges in the grid, so it should
52 * be reserved until there's nothing easier left to be
55 * - The generalised grid support has made me (SGT) notice a
56 * possible extension to the loop-avoidance code. When you have
57 * a path of connected edges such that no other edges at all
58 * are incident on any vertex in the middle of the path - or,
59 * alternatively, such that any such edges are already known to
60 * be LINE_NO - then you know those edges are either all
61 * LINE_YES or all LINE_NO. Hence you can mentally merge the
62 * entire path into a single long curly edge for the purposes
63 * of loop avoidance, and look directly at whether or not the
64 * extreme endpoints of the path are connected by some other
65 * route. I find this coming up fairly often when I play on the
66 * octagonal grid setting, so it might be worth implementing in
69 * - (Just a speed optimisation.) Consider some todo list queue where every
70 * time we modify something we mark it for consideration by other bits of
71 * the solver, to save iteration over things that have already been done.
86 /* Debugging options */
94 /* ----------------------------------------------------------------------
95 * Struct, enum and function declarations
112 /* Put -1 in a face that doesn't get a clue */
115 /* Array of line states, to store whether each line is
116 * YES, NO or UNKNOWN */
119 unsigned char *line_errors;
124 /* Used in game_text_format(), so that it knows what type of
125 * grid it's trying to render as ASCII text. */
130 SOLVER_SOLVED, /* This is the only solution the solver could find */
131 SOLVER_MISTAKE, /* This is definitely not a solution */
132 SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
133 SOLVER_INCOMPLETE /* This may be a partial solution */
136 /* ------ Solver state ------ */
137 typedef struct solver_state {
139 enum solver_status solver_status;
140 /* NB looplen is the number of dots that are joined together at a point, ie a
141 * looplen of 1 means there are no lines to a particular dot */
144 /* Difficulty level of solver. Used by solver functions that want to
145 * vary their behaviour depending on the requested difficulty level. */
151 char *face_yes_count;
153 char *dot_solved, *face_solved;
156 /* Information for Normal level deductions:
157 * For each dline, store a bitmask for whether we know:
158 * (bit 0) at least one is YES
159 * (bit 1) at most one is YES */
162 /* Hard level information */
167 * Difficulty levels. I do some macro ickery here to ensure that my
168 * enum and the various forms of my name list always match up.
171 #define DIFFLIST(A) \
176 #define ENUM(upper,title,lower) DIFF_ ## upper,
177 #define TITLE(upper,title,lower) #title,
178 #define ENCODE(upper,title,lower) #lower
179 #define CONFIG(upper,title,lower) ":" #title
180 enum { DIFFLIST(ENUM) DIFF_MAX };
181 static char const *const diffnames[] = { DIFFLIST(TITLE) };
182 static char const diffchars[] = DIFFLIST(ENCODE);
183 #define DIFFCONFIG DIFFLIST(CONFIG)
186 * Solver routines, sorted roughly in order of computational cost.
187 * The solver will run the faster deductions first, and slower deductions are
188 * only invoked when the faster deductions are unable to make progress.
189 * Each function is associated with a difficulty level, so that the generated
190 * puzzles are solvable by applying only the functions with the chosen
191 * difficulty level or lower.
193 #define SOLVERLIST(A) \
194 A(trivial_deductions, DIFF_EASY) \
195 A(dline_deductions, DIFF_NORMAL) \
196 A(linedsf_deductions, DIFF_HARD) \
197 A(loop_deductions, DIFF_EASY)
198 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
199 #define SOLVER_FN(fn,diff) &fn,
200 #define SOLVER_DIFF(fn,diff) diff,
201 SOLVERLIST(SOLVER_FN_DECL)
202 static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
203 static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
204 const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
211 /* Grid generation is expensive, so keep a (ref-counted) reference to the
212 * grid for these parameters, and only generate when required. */
216 /* line_drawstate is the same as line_state, but with the extra ERROR
217 * possibility. The drawing code copies line_state to line_drawstate,
218 * except in the case that the line is an error. */
219 enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
220 enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
221 DS_LINE_NO, DS_LINE_ERROR };
223 #define OPP(line_state) \
227 struct game_drawstate {
234 char *clue_satisfied;
237 static char *validate_desc(game_params *params, char *desc);
238 static int dot_order(const game_state* state, int i, char line_type);
239 static int face_order(const game_state* state, int i, char line_type);
240 static solver_state *solve_game_rec(const solver_state *sstate);
243 static void check_caches(const solver_state* sstate);
245 #define check_caches(s)
248 /* ------- List of grid generators ------- */
249 #define GRIDLIST(A) \
250 A(Squares,grid_new_square,3,3) \
251 A(Triangular,grid_new_triangular,3,3) \
252 A(Honeycomb,grid_new_honeycomb,3,3) \
253 A(Snub-Square,grid_new_snubsquare,3,3) \
254 A(Cairo,grid_new_cairo,3,4) \
255 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
256 A(Octagonal,grid_new_octagonal,3,3) \
257 A(Kites,grid_new_kites,3,3) \
258 A(Floret,grid_new_floret,1,2) \
259 A(Dodecagonal,grid_new_dodecagonal,2,2) \
260 A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2)
262 #define GRID_NAME(title,fn,amin,omin) #title,
263 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
264 #define GRID_FN(title,fn,amin,omin) &fn,
265 #define GRID_SIZES(title,fn,amin,omin) \
267 "Width and height for this grid type must both be at least " #amin, \
268 "At least one of width and height for this grid type must be at least " #omin,},
269 static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
270 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
271 static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) };
272 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
273 static const struct {
276 } grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
278 /* Generates a (dynamically allocated) new grid, according to the
279 * type and size requested in params. Does nothing if the grid is already
280 * generated. The allocated grid is owned by the params object, and will be
281 * freed in free_params(). */
282 static void params_generate_grid(game_params *params)
284 if (!params->game_grid) {
285 params->game_grid = grid_fns[params->type](params->w, params->h);
289 /* ----------------------------------------------------------------------
293 /* General constants */
294 #define PREFERRED_TILE_SIZE 32
295 #define BORDER(tilesize) ((tilesize) / 2)
296 #define FLASH_TIME 0.5F
298 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
300 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
301 ((field) |= (1<<(bit)), TRUE))
303 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
304 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
306 #define CLUE2CHAR(c) \
307 ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
309 /* ----------------------------------------------------------------------
310 * General struct manipulation and other straightforward code
313 static game_state *dup_game(game_state *state)
315 game_state *ret = snew(game_state);
317 ret->game_grid = state->game_grid;
318 ret->game_grid->refcount++;
320 ret->solved = state->solved;
321 ret->cheated = state->cheated;
323 ret->clues = snewn(state->game_grid->num_faces, signed char);
324 memcpy(ret->clues, state->clues, state->game_grid->num_faces);
326 ret->lines = snewn(state->game_grid->num_edges, char);
327 memcpy(ret->lines, state->lines, state->game_grid->num_edges);
329 ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
330 memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
332 ret->grid_type = state->grid_type;
336 static void free_game(game_state *state)
339 grid_free(state->game_grid);
342 sfree(state->line_errors);
347 static solver_state *new_solver_state(game_state *state, int diff) {
349 int num_dots = state->game_grid->num_dots;
350 int num_faces = state->game_grid->num_faces;
351 int num_edges = state->game_grid->num_edges;
352 solver_state *ret = snew(solver_state);
354 ret->state = dup_game(state);
356 ret->solver_status = SOLVER_INCOMPLETE;
359 ret->dotdsf = snew_dsf(num_dots);
360 ret->looplen = snewn(num_dots, int);
362 for (i = 0; i < num_dots; i++) {
366 ret->dot_solved = snewn(num_dots, char);
367 ret->face_solved = snewn(num_faces, char);
368 memset(ret->dot_solved, FALSE, num_dots);
369 memset(ret->face_solved, FALSE, num_faces);
371 ret->dot_yes_count = snewn(num_dots, char);
372 memset(ret->dot_yes_count, 0, num_dots);
373 ret->dot_no_count = snewn(num_dots, char);
374 memset(ret->dot_no_count, 0, num_dots);
375 ret->face_yes_count = snewn(num_faces, char);
376 memset(ret->face_yes_count, 0, num_faces);
377 ret->face_no_count = snewn(num_faces, char);
378 memset(ret->face_no_count, 0, num_faces);
380 if (diff < DIFF_NORMAL) {
383 ret->dlines = snewn(2*num_edges, char);
384 memset(ret->dlines, 0, 2*num_edges);
387 if (diff < DIFF_HARD) {
390 ret->linedsf = snew_dsf(state->game_grid->num_edges);
396 static void free_solver_state(solver_state *sstate) {
398 free_game(sstate->state);
399 sfree(sstate->dotdsf);
400 sfree(sstate->looplen);
401 sfree(sstate->dot_solved);
402 sfree(sstate->face_solved);
403 sfree(sstate->dot_yes_count);
404 sfree(sstate->dot_no_count);
405 sfree(sstate->face_yes_count);
406 sfree(sstate->face_no_count);
408 /* OK, because sfree(NULL) is a no-op */
409 sfree(sstate->dlines);
410 sfree(sstate->linedsf);
416 static solver_state *dup_solver_state(const solver_state *sstate) {
417 game_state *state = sstate->state;
418 int num_dots = state->game_grid->num_dots;
419 int num_faces = state->game_grid->num_faces;
420 int num_edges = state->game_grid->num_edges;
421 solver_state *ret = snew(solver_state);
423 ret->state = state = dup_game(sstate->state);
425 ret->solver_status = sstate->solver_status;
426 ret->diff = sstate->diff;
428 ret->dotdsf = snewn(num_dots, int);
429 ret->looplen = snewn(num_dots, int);
430 memcpy(ret->dotdsf, sstate->dotdsf,
431 num_dots * sizeof(int));
432 memcpy(ret->looplen, sstate->looplen,
433 num_dots * sizeof(int));
435 ret->dot_solved = snewn(num_dots, char);
436 ret->face_solved = snewn(num_faces, char);
437 memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
438 memcpy(ret->face_solved, sstate->face_solved, num_faces);
440 ret->dot_yes_count = snewn(num_dots, char);
441 memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
442 ret->dot_no_count = snewn(num_dots, char);
443 memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
445 ret->face_yes_count = snewn(num_faces, char);
446 memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
447 ret->face_no_count = snewn(num_faces, char);
448 memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
450 if (sstate->dlines) {
451 ret->dlines = snewn(2*num_edges, char);
452 memcpy(ret->dlines, sstate->dlines,
458 if (sstate->linedsf) {
459 ret->linedsf = snewn(num_edges, int);
460 memcpy(ret->linedsf, sstate->linedsf,
461 num_edges * sizeof(int));
469 static game_params *default_params(void)
471 game_params *ret = snew(game_params);
480 ret->diff = DIFF_EASY;
483 ret->game_grid = NULL;
488 static game_params *dup_params(game_params *params)
490 game_params *ret = snew(game_params);
492 *ret = *params; /* structure copy */
493 if (ret->game_grid) {
494 ret->game_grid->refcount++;
499 static const game_params presets[] = {
501 { 7, 7, DIFF_EASY, 0, NULL },
502 { 7, 7, DIFF_NORMAL, 0, NULL },
503 { 7, 7, DIFF_HARD, 0, NULL },
504 { 7, 7, DIFF_HARD, 1, NULL },
505 { 7, 7, DIFF_HARD, 2, NULL },
506 { 5, 5, DIFF_HARD, 3, NULL },
507 { 7, 7, DIFF_HARD, 4, NULL },
508 { 5, 4, DIFF_HARD, 5, NULL },
509 { 5, 5, DIFF_HARD, 6, NULL },
510 { 5, 5, DIFF_HARD, 7, NULL },
511 { 3, 3, DIFF_HARD, 8, NULL },
512 { 3, 3, DIFF_HARD, 9, NULL },
513 { 3, 3, DIFF_HARD, 10, NULL },
515 { 7, 7, DIFF_EASY, 0, NULL },
516 { 10, 10, DIFF_EASY, 0, NULL },
517 { 7, 7, DIFF_NORMAL, 0, NULL },
518 { 10, 10, DIFF_NORMAL, 0, NULL },
519 { 7, 7, DIFF_HARD, 0, NULL },
520 { 10, 10, DIFF_HARD, 0, NULL },
521 { 10, 10, DIFF_HARD, 1, NULL },
522 { 12, 10, DIFF_HARD, 2, NULL },
523 { 7, 7, DIFF_HARD, 3, NULL },
524 { 9, 9, DIFF_HARD, 4, NULL },
525 { 5, 4, DIFF_HARD, 5, NULL },
526 { 7, 7, DIFF_HARD, 6, NULL },
527 { 5, 5, DIFF_HARD, 7, NULL },
528 { 5, 5, DIFF_HARD, 8, NULL },
529 { 5, 4, DIFF_HARD, 9, NULL },
530 { 5, 4, DIFF_HARD, 10, NULL },
534 static int game_fetch_preset(int i, char **name, game_params **params)
539 if (i < 0 || i >= lenof(presets))
542 tmppar = snew(game_params);
543 *tmppar = presets[i];
545 sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
546 gridnames[tmppar->type], diffnames[tmppar->diff]);
552 static void free_params(game_params *params)
554 if (params->game_grid) {
555 grid_free(params->game_grid);
560 static void decode_params(game_params *params, char const *string)
562 if (params->game_grid) {
563 grid_free(params->game_grid);
564 params->game_grid = NULL;
566 params->h = params->w = atoi(string);
567 params->diff = DIFF_EASY;
568 while (*string && isdigit((unsigned char)*string)) string++;
569 if (*string == 'x') {
571 params->h = atoi(string);
572 while (*string && isdigit((unsigned char)*string)) string++;
574 if (*string == 't') {
576 params->type = atoi(string);
577 while (*string && isdigit((unsigned char)*string)) string++;
579 if (*string == 'd') {
582 for (i = 0; i < DIFF_MAX; i++)
583 if (*string == diffchars[i])
585 if (*string) string++;
589 static char *encode_params(game_params *params, int full)
592 sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
594 sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
598 static config_item *game_configure(game_params *params)
603 ret = snewn(5, config_item);
605 ret[0].name = "Width";
606 ret[0].type = C_STRING;
607 sprintf(buf, "%d", params->w);
608 ret[0].sval = dupstr(buf);
611 ret[1].name = "Height";
612 ret[1].type = C_STRING;
613 sprintf(buf, "%d", params->h);
614 ret[1].sval = dupstr(buf);
617 ret[2].name = "Grid type";
618 ret[2].type = C_CHOICES;
619 ret[2].sval = GRID_CONFIGS;
620 ret[2].ival = params->type;
622 ret[3].name = "Difficulty";
623 ret[3].type = C_CHOICES;
624 ret[3].sval = DIFFCONFIG;
625 ret[3].ival = params->diff;
635 static game_params *custom_params(config_item *cfg)
637 game_params *ret = snew(game_params);
639 ret->w = atoi(cfg[0].sval);
640 ret->h = atoi(cfg[1].sval);
641 ret->type = cfg[2].ival;
642 ret->diff = cfg[3].ival;
644 ret->game_grid = NULL;
648 static char *validate_params(game_params *params, int full)
650 if (params->type < 0 || params->type >= NUM_GRID_TYPES)
651 return "Illegal grid type";
652 if (params->w < grid_size_limits[params->type].amin ||
653 params->h < grid_size_limits[params->type].amin)
654 return grid_size_limits[params->type].aerr;
655 if (params->w < grid_size_limits[params->type].omin &&
656 params->h < grid_size_limits[params->type].omin)
657 return grid_size_limits[params->type].oerr;
660 * This shouldn't be able to happen at all, since decode_params
661 * and custom_params will never generate anything that isn't
664 assert(params->diff < DIFF_MAX);
669 /* Returns a newly allocated string describing the current puzzle */
670 static char *state_to_text(const game_state *state)
672 grid *g = state->game_grid;
674 int num_faces = g->num_faces;
675 char *description = snewn(num_faces + 1, char);
676 char *dp = description;
680 for (i = 0; i < num_faces; i++) {
681 if (state->clues[i] < 0) {
682 if (empty_count > 25) {
683 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
689 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
692 dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
697 dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
699 retval = dupstr(description);
705 /* We require that the params pass the test in validate_params and that the
706 * description fills the entire game area */
707 static char *validate_desc(game_params *params, char *desc)
711 params_generate_grid(params);
712 g = params->game_grid;
714 for (; *desc; ++desc) {
715 if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
720 count += *desc - 'a' + 1;
723 return "Unknown character in description";
726 if (count < g->num_faces)
727 return "Description too short for board size";
728 if (count > g->num_faces)
729 return "Description too long for board size";
734 /* Sums the lengths of the numbers in range [0,n) */
735 /* See equivalent function in solo.c for justification of this. */
736 static int len_0_to_n(int n)
738 int len = 1; /* Counting 0 as a bit of a special case */
741 for (i = 1; i < n; i *= 10) {
742 len += max(n - i, 0);
748 static char *encode_solve_move(const game_state *state)
753 int num_edges = state->game_grid->num_edges;
755 /* This is going to return a string representing the moves needed to set
756 * every line in a grid to be the same as the ones in 'state'. The exact
757 * length of this string is predictable. */
759 len = 1; /* Count the 'S' prefix */
760 /* Numbers in all lines */
761 len += len_0_to_n(num_edges);
762 /* For each line we also have a letter */
765 ret = snewn(len + 1, char);
768 p += sprintf(p, "S");
770 for (i = 0; i < num_edges; i++) {
771 switch (state->lines[i]) {
773 p += sprintf(p, "%dy", i);
776 p += sprintf(p, "%dn", i);
781 /* No point in doing sums like that if they're going to be wrong */
782 assert(strlen(ret) <= (size_t)len);
786 static game_ui *new_ui(game_state *state)
791 static void free_ui(game_ui *ui)
795 static char *encode_ui(game_ui *ui)
800 static void decode_ui(game_ui *ui, char *encoding)
804 static void game_changed_state(game_ui *ui, game_state *oldstate,
805 game_state *newstate)
809 static void game_compute_size(game_params *params, int tilesize,
813 int grid_width, grid_height, rendered_width, rendered_height;
815 params_generate_grid(params);
816 g = params->game_grid;
817 grid_width = g->highest_x - g->lowest_x;
818 grid_height = g->highest_y - g->lowest_y;
819 /* multiply first to minimise rounding error on integer division */
820 rendered_width = grid_width * tilesize / g->tilesize;
821 rendered_height = grid_height * tilesize / g->tilesize;
822 *x = rendered_width + 2 * BORDER(tilesize) + 1;
823 *y = rendered_height + 2 * BORDER(tilesize) + 1;
826 static void game_set_size(drawing *dr, game_drawstate *ds,
827 game_params *params, int tilesize)
829 ds->tilesize = tilesize;
832 static float *game_colours(frontend *fe, int *ncolours)
834 float *ret = snewn(4 * NCOLOURS, float);
836 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
838 ret[COL_FOREGROUND * 3 + 0] = 0.0F;
839 ret[COL_FOREGROUND * 3 + 1] = 0.0F;
840 ret[COL_FOREGROUND * 3 + 2] = 0.0F;
842 ret[COL_LINEUNKNOWN * 3 + 0] = 0.8F;
843 ret[COL_LINEUNKNOWN * 3 + 1] = 0.8F;
844 ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
846 ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
847 ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
848 ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;
850 ret[COL_MISTAKE * 3 + 0] = 1.0F;
851 ret[COL_MISTAKE * 3 + 1] = 0.0F;
852 ret[COL_MISTAKE * 3 + 2] = 0.0F;
854 ret[COL_SATISFIED * 3 + 0] = 0.0F;
855 ret[COL_SATISFIED * 3 + 1] = 0.0F;
856 ret[COL_SATISFIED * 3 + 2] = 0.0F;
858 /* We want the faint lines to be a bit darker than the background.
859 * Except if the background is pretty dark already; then it ought to be a
860 * bit lighter. Oy vey.
862 ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
863 ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
864 ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F;
866 *ncolours = NCOLOURS;
870 static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
872 struct game_drawstate *ds = snew(struct game_drawstate);
873 int num_faces = state->game_grid->num_faces;
874 int num_edges = state->game_grid->num_edges;
879 ds->lines = snewn(num_edges, char);
880 ds->clue_error = snewn(num_faces, char);
881 ds->clue_satisfied = snewn(num_faces, char);
882 ds->textx = snewn(num_faces, int);
883 ds->texty = snewn(num_faces, int);
886 memset(ds->lines, LINE_UNKNOWN, num_edges);
887 memset(ds->clue_error, 0, num_faces);
888 memset(ds->clue_satisfied, 0, num_faces);
889 for (i = 0; i < num_faces; i++)
890 ds->textx[i] = ds->texty[i] = -1;
895 static void game_free_drawstate(drawing *dr, game_drawstate *ds)
897 sfree(ds->clue_error);
898 sfree(ds->clue_satisfied);
903 static int game_timing_state(game_state *state, game_ui *ui)
908 static float game_anim_length(game_state *oldstate, game_state *newstate,
909 int dir, game_ui *ui)
914 static int game_can_format_as_text_now(game_params *params)
916 if (params->type != 0)
921 static char *game_text_format(game_state *state)
927 grid *g = state->game_grid;
930 assert(state->grid_type == 0);
932 /* Work out the basic size unit */
933 f = g->faces; /* first face */
934 assert(f->order == 4);
935 /* The dots are ordered clockwise, so the two opposite
936 * corners are guaranteed to span the square */
937 cell_size = abs(f->dots[0]->x - f->dots[2]->x);
939 w = (g->highest_x - g->lowest_x) / cell_size;
940 h = (g->highest_y - g->lowest_y) / cell_size;
942 /* Create a blank "canvas" to "draw" on */
945 ret = snewn(W * H + 1, char);
946 for (y = 0; y < H; y++) {
947 for (x = 0; x < W-1; x++) {
950 ret[y*W + W-1] = '\n';
954 /* Fill in edge info */
955 for (i = 0; i < g->num_edges; i++) {
956 grid_edge *e = g->edges + i;
957 /* Cell coordinates, from (0,0) to (w-1,h-1) */
958 int x1 = (e->dot1->x - g->lowest_x) / cell_size;
959 int x2 = (e->dot2->x - g->lowest_x) / cell_size;
960 int y1 = (e->dot1->y - g->lowest_y) / cell_size;
961 int y2 = (e->dot2->y - g->lowest_y) / cell_size;
962 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
963 * cell coordinates) */
966 switch (state->lines[i]) {
968 ret[y*W + x] = (y1 == y2) ? '-' : '|';
974 break; /* already a space */
976 assert(!"Illegal line state");
981 for (i = 0; i < g->num_faces; i++) {
985 assert(f->order == 4);
986 /* Cell coordinates, from (0,0) to (w-1,h-1) */
987 x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
988 x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
989 y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
990 y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
991 /* Midpoint, in canvas coordinates */
994 ret[y*W + x] = CLUE2CHAR(state->clues[i]);
999 /* ----------------------------------------------------------------------
1004 static void check_caches(const solver_state* sstate)
1007 const game_state *state = sstate->state;
1008 const grid *g = state->game_grid;
1010 for (i = 0; i < g->num_dots; i++) {
1011 assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
1012 assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
1015 for (i = 0; i < g->num_faces; i++) {
1016 assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
1017 assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
1022 #define check_caches(s) \
1024 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1028 #endif /* DEBUG_CACHES */
1030 /* ----------------------------------------------------------------------
1031 * Solver utility functions
1034 /* Sets the line (with index i) to the new state 'line_new', and updates
1035 * the cached counts of any affected faces and dots.
1036 * Returns TRUE if this actually changed the line's state. */
1037 static int solver_set_line(solver_state *sstate, int i,
1038 enum line_state line_new
1040 , const char *reason
1044 game_state *state = sstate->state;
1048 assert(line_new != LINE_UNKNOWN);
1050 check_caches(sstate);
1052 if (state->lines[i] == line_new) {
1053 return FALSE; /* nothing changed */
1055 state->lines[i] = line_new;
1058 fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
1059 i, line_new == LINE_YES ? "YES" : "NO",
1063 g = state->game_grid;
1066 /* Update the cache for both dots and both faces affected by this. */
1067 if (line_new == LINE_YES) {
1068 sstate->dot_yes_count[e->dot1 - g->dots]++;
1069 sstate->dot_yes_count[e->dot2 - g->dots]++;
1071 sstate->face_yes_count[e->face1 - g->faces]++;
1074 sstate->face_yes_count[e->face2 - g->faces]++;
1077 sstate->dot_no_count[e->dot1 - g->dots]++;
1078 sstate->dot_no_count[e->dot2 - g->dots]++;
1080 sstate->face_no_count[e->face1 - g->faces]++;
1083 sstate->face_no_count[e->face2 - g->faces]++;
1087 check_caches(sstate);
1092 #define solver_set_line(a, b, c) \
1093 solver_set_line(a, b, c, __FUNCTION__)
1097 * Merge two dots due to the existence of an edge between them.
1098 * Updates the dsf tracking equivalence classes, and keeps track of
1099 * the length of path each dot is currently a part of.
1100 * Returns TRUE if the dots were already linked, ie if they are part of a
1101 * closed loop, and false otherwise.
1103 static int merge_dots(solver_state *sstate, int edge_index)
1106 grid *g = sstate->state->game_grid;
1107 grid_edge *e = g->edges + edge_index;
1109 i = e->dot1 - g->dots;
1110 j = e->dot2 - g->dots;
1112 i = dsf_canonify(sstate->dotdsf, i);
1113 j = dsf_canonify(sstate->dotdsf, j);
1118 len = sstate->looplen[i] + sstate->looplen[j];
1119 dsf_merge(sstate->dotdsf, i, j);
1120 i = dsf_canonify(sstate->dotdsf, i);
1121 sstate->looplen[i] = len;
1126 /* Merge two lines because the solver has deduced that they must be either
1127 * identical or opposite. Returns TRUE if this is new information, otherwise
1129 static int merge_lines(solver_state *sstate, int i, int j, int inverse
1131 , const char *reason
1137 assert(i < sstate->state->game_grid->num_edges);
1138 assert(j < sstate->state->game_grid->num_edges);
1140 i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
1142 j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
1145 edsf_merge(sstate->linedsf, i, j, inverse);
1149 fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
1151 inverse ? "inverse " : "", reason);
1158 #define merge_lines(a, b, c, d) \
1159 merge_lines(a, b, c, d, __FUNCTION__)
1162 /* Count the number of lines of a particular type currently going into the
1164 static int dot_order(const game_state* state, int dot, char line_type)
1167 grid *g = state->game_grid;
1168 grid_dot *d = g->dots + dot;
1171 for (i = 0; i < d->order; i++) {
1172 grid_edge *e = d->edges[i];
1173 if (state->lines[e - g->edges] == line_type)
1179 /* Count the number of lines of a particular type currently surrounding the
1181 static int face_order(const game_state* state, int face, char line_type)
1184 grid *g = state->game_grid;
1185 grid_face *f = g->faces + face;
1188 for (i = 0; i < f->order; i++) {
1189 grid_edge *e = f->edges[i];
1190 if (state->lines[e - g->edges] == line_type)
1196 /* Set all lines bordering a dot of type old_type to type new_type
1197 * Return value tells caller whether this function actually did anything */
1198 static int dot_setall(solver_state *sstate, int dot,
1199 char old_type, char new_type)
1201 int retval = FALSE, r;
1202 game_state *state = sstate->state;
1207 if (old_type == new_type)
1210 g = state->game_grid;
1213 for (i = 0; i < d->order; i++) {
1214 int line_index = d->edges[i] - g->edges;
1215 if (state->lines[line_index] == old_type) {
1216 r = solver_set_line(sstate, line_index, new_type);
1224 /* Set all lines bordering a face of type old_type to type new_type */
1225 static int face_setall(solver_state *sstate, int face,
1226 char old_type, char new_type)
1228 int retval = FALSE, r;
1229 game_state *state = sstate->state;
1234 if (old_type == new_type)
1237 g = state->game_grid;
1238 f = g->faces + face;
1240 for (i = 0; i < f->order; i++) {
1241 int line_index = f->edges[i] - g->edges;
1242 if (state->lines[line_index] == old_type) {
1243 r = solver_set_line(sstate, line_index, new_type);
1251 /* ----------------------------------------------------------------------
1252 * Loop generation and clue removal
1255 /* We're going to store lists of current candidate faces for colouring black
1257 * Each face gets a 'score', which tells us how adding that face right
1258 * now would affect the curliness of the solution loop. We're trying to
1259 * maximise that quantity so will bias our random selection of faces to
1260 * colour those with high scores */
1264 unsigned long random;
1265 /* No need to store a grid_face* here. The 'face_scores' array will
1266 * be a list of 'face_score' objects, one for each face of the grid, so
1267 * the position (index) within the 'face_scores' array will determine
1268 * which face corresponds to a particular face_score.
1269 * Having a single 'face_scores' array for all faces simplifies memory
1270 * management, and probably improves performance, because we don't have to
1271 * malloc/free each individual face_score, and we don't have to maintain
1272 * a mapping from grid_face* pointers to face_score* pointers.
1276 static int generic_sort_cmpfn(void *v1, void *v2, size_t offset)
1278 struct face_score *f1 = v1;
1279 struct face_score *f2 = v2;
1282 r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset);
1287 if (f1->random < f2->random)
1289 else if (f1->random > f2->random)
1293 * It's _just_ possible that two faces might have been given
1294 * the same random value. In that situation, fall back to
1295 * comparing based on the positions within the face_scores list.
1296 * This introduces a tiny directional bias, but not a significant one.
1301 static int white_sort_cmpfn(void *v1, void *v2)
1303 return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score));
1306 static int black_sort_cmpfn(void *v1, void *v2)
1308 return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score));
1311 enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK };
1313 /* face should be of type grid_face* here. */
1314 #define FACE_COLOUR(face) \
1315 ( (face) == NULL ? FACE_BLACK : \
1316 board[(face) - g->faces] )
1318 /* 'board' is an array of these enums, indicating which faces are
1319 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1320 * Returns whether it's legal to colour the given face with this colour. */
1321 static int can_colour_face(grid *g, char* board, int face_index,
1322 enum face_colour colour)
1325 grid_face *test_face = g->faces + face_index;
1326 grid_face *starting_face, *current_face;
1327 grid_dot *starting_dot;
1329 int current_state, s; /* booleans: equal or not-equal to 'colour' */
1330 int found_same_coloured_neighbour = FALSE;
1331 assert(board[face_index] != colour);
1333 /* Can only consider a face for colouring if it's adjacent to a face
1334 * with the same colour. */
1335 for (i = 0; i < test_face->order; i++) {
1336 grid_edge *e = test_face->edges[i];
1337 grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1;
1338 if (FACE_COLOUR(f) == colour) {
1339 found_same_coloured_neighbour = TRUE;
1343 if (!found_same_coloured_neighbour)
1346 /* Need to avoid creating a loop of faces of this colour around some
1347 * differently-coloured faces.
1348 * Also need to avoid meeting a same-coloured face at a corner, with
1349 * other-coloured faces in between. Here's a simple test that (I believe)
1350 * takes care of both these conditions:
1352 * Take the circular path formed by this face's edges, and inflate it
1353 * slightly outwards. Imagine walking around this path and consider
1354 * the faces that you visit in sequence. This will include all faces
1355 * touching the given face, either along an edge or just at a corner.
1356 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1357 * you walk along the complete loop. This will obviously turn out to be
1359 * If 0, we're either in the middle of an "island" of this colour (should
1360 * be impossible as we're not supposed to create black or white loops),
1361 * or we're about to start a new island - also not allowed.
1362 * If 4 or greater, there are too many separate coloured regions touching
1363 * this face, and colouring it would create a loop or a corner-violation.
1364 * The only allowed case is when the count is exactly 2. */
1366 /* i points to a dot around the test face.
1367 * j points to a face around the i^th dot.
1368 * The current face will always be:
1369 * test_face->dots[i]->faces[j]
1370 * We assume dots go clockwise around the test face,
1371 * and faces go clockwise around dots. */
1374 * The end condition is slightly fiddly. In sufficiently strange
1375 * degenerate grids, our test face may be adjacent to the same
1376 * other face multiple times (typically if it's the exterior
1377 * face). Consider this, in particular:
1385 * The bottom left face there is adjacent to the exterior face
1386 * twice, so we can't just terminate our iteration when we reach
1387 * the same _face_ we started at. Furthermore, we can't
1388 * condition on having the same (i,j) pair either, because
1389 * several (i,j) pairs identify the bottom left contiguity with
1390 * the exterior face! We canonicalise the (i,j) pair by taking
1391 * one step around before we set the termination tracking.
1395 current_face = test_face->dots[0]->faces[0];
1396 if (current_face == test_face) {
1398 current_face = test_face->dots[0]->faces[1];
1401 current_state = (FACE_COLOUR(current_face) == colour);
1402 starting_dot = NULL;
1403 starting_face = NULL;
1405 /* Advance to next face.
1406 * Need to loop here because it might take several goes to
1410 if (j == test_face->dots[i]->order)
1413 if (test_face->dots[i]->faces[j] == test_face) {
1414 /* Advance to next dot round test_face, then
1415 * find current_face around new dot
1416 * and advance to the next face clockwise */
1418 if (i == test_face->order)
1420 for (j = 0; j < test_face->dots[i]->order; j++) {
1421 if (test_face->dots[i]->faces[j] == current_face)
1424 /* Must actually find current_face around new dot,
1425 * or else something's wrong with the grid. */
1426 assert(j != test_face->dots[i]->order);
1427 /* Found, so advance to next face and try again */
1432 /* (i,j) are now advanced to next face */
1433 current_face = test_face->dots[i]->faces[j];
1434 s = (FACE_COLOUR(current_face) == colour);
1435 if (!starting_dot) {
1436 starting_dot = test_face->dots[i];
1437 starting_face = current_face;
1440 if (s != current_state) {
1443 if (transitions > 2)
1446 if (test_face->dots[i] == starting_dot &&
1447 current_face == starting_face)
1452 return (transitions == 2) ? TRUE : FALSE;
1455 /* Count the number of neighbours of 'face', having colour 'colour' */
1456 static int face_num_neighbours(grid *g, char *board, grid_face *face,
1457 enum face_colour colour)
1459 int colour_count = 0;
1463 for (i = 0; i < face->order; i++) {
1465 f = (e->face1 == face) ? e->face2 : e->face1;
1466 if (FACE_COLOUR(f) == colour)
1469 return colour_count;
1472 /* The 'score' of a face reflects its current desirability for selection
1473 * as the next face to colour white or black. We want to encourage moving
1474 * into grey areas and increasing loopiness, so we give scores according to
1475 * how many of the face's neighbours are currently coloured the same as the
1476 * proposed colour. */
1477 static int face_score(grid *g, char *board, grid_face *face,
1478 enum face_colour colour)
1480 /* Simple formula: score = 0 - num. same-coloured neighbours,
1481 * so a higher score means fewer same-coloured neighbours. */
1482 return -face_num_neighbours(g, board, face, colour);
1485 /* Generate a new complete set of clues for the given game_state.
1486 * The method is to generate a WHITE/BLACK colouring of all the faces,
1487 * such that the WHITE faces will define the inside of the path, and the
1488 * BLACK faces define the outside.
1489 * To do this, we initially colour all faces GREY. The infinite space outside
1490 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1491 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1492 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1493 * we avoid creating loops of a single colour, to preserve the topological
1494 * shape of the WHITE and BLACK regions.
1495 * We also try to make the boundary as loopy and twisty as possible, to avoid
1496 * generating paths that are uninteresting.
1497 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1498 * face that can be coloured with that colour (without violating the
1499 * topological shape of that region). It's not obvious, but I think this
1500 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1501 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1502 * regions can be grown.
1503 * This is checked using assert()ions, and I haven't seen any failures yet.
1505 * Hand-wavy proof: imagine what can go wrong...
1507 * Could the white faces get completely cut off by the black faces, and still
1508 * leave some grey faces remaining?
1509 * No, because then the black faces would form a loop around both the white
1510 * faces and the grey faces, which is disallowed because we continually
1511 * maintain the correct topological shape of the black region.
1512 * Similarly, the black faces can never get cut off by the white faces. That
1513 * means both the WHITE and BLACK regions always have some room to grow into
1515 * Could it be that we can't colour some GREY face, because there are too many
1516 * WHITE/BLACK transitions as we walk round the face? (see the
1517 * can_colour_face() function for details)
1518 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1519 * around the face. The two WHITE faces would be connected by a WHITE path,
1520 * and the BLACK faces would be connected by a BLACK path. These paths would
1521 * have to cross, which is impossible.
1522 * Another thing that could go wrong: perhaps we can't find any GREY face to
1523 * colour WHITE, because it would create a loop-violation or a corner-violation
1524 * with the other WHITE faces?
1525 * This is a little bit tricky to prove impossible. Imagine you have such a
1526 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1527 * or corner violation).
1528 * That would cut all the non-white area into two blobs. One of those blobs
1529 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1530 * So we have a connected GREY area, completely surrounded by WHITE
1531 * (including the GREY face we've tentatively coloured WHITE).
1532 * A well-known result in graph theory says that you can always find a GREY
1533 * face whose removal leaves the remaining GREY area connected. And it says
1534 * there are at least two such faces, so we can always choose the one that
1535 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1536 * everything nice and connected, including that "tentative" GREY face which
1537 * acts as a gateway to the rest of the non-WHITE grid.
1539 static void add_full_clues(game_state *state, random_state *rs)
1541 signed char *clues = state->clues;
1543 grid *g = state->game_grid;
1545 int num_faces = g->num_faces;
1546 struct face_score *face_scores; /* Array of face_score objects */
1547 struct face_score *fs; /* Points somewhere in the above list */
1548 struct grid_face *cur_face;
1549 tree234 *lightable_faces_sorted;
1550 tree234 *darkable_faces_sorted;
1554 board = snewn(num_faces, char);
1557 memset(board, FACE_GREY, num_faces);
1559 /* Create and initialise the list of face_scores */
1560 face_scores = snewn(num_faces, struct face_score);
1561 for (i = 0; i < num_faces; i++) {
1562 face_scores[i].random = random_bits(rs, 31);
1563 face_scores[i].black_score = face_scores[i].white_score = 0;
1566 /* Colour a random, finite face white. The infinite face is implicitly
1567 * coloured black. Together, they will seed the random growth process
1568 * for the black and white areas. */
1569 i = random_upto(rs, num_faces);
1570 board[i] = FACE_WHITE;
1572 /* We need a way of favouring faces that will increase our loopiness.
1573 * We do this by maintaining a list of all candidate faces sorted by
1574 * their score and choose randomly from that with appropriate skew.
1575 * In order to avoid consistently biasing towards particular faces, we
1576 * need the sort order _within_ each group of scores to be completely
1577 * random. But it would be abusing the hospitality of the tree234 data
1578 * structure if our comparison function were nondeterministic :-). So with
1579 * each face we associate a random number that does not change during a
1580 * particular run of the generator, and use that as a secondary sort key.
1581 * Yes, this means we will be biased towards particular random faces in
1582 * any one run but that doesn't actually matter. */
1584 lightable_faces_sorted = newtree234(white_sort_cmpfn);
1585 darkable_faces_sorted = newtree234(black_sort_cmpfn);
1587 /* Initialise the lists of lightable and darkable faces. This is
1588 * slightly different from the code inside the while-loop, because we need
1589 * to check every face of the board (the grid structure does not keep a
1590 * list of the infinite face's neighbours). */
1591 for (i = 0; i < num_faces; i++) {
1592 grid_face *f = g->faces + i;
1593 struct face_score *fs = face_scores + i;
1594 if (board[i] != FACE_GREY) continue;
1595 /* We need the full colourability check here, it's not enough simply
1596 * to check neighbourhood. On some grids, a neighbour of the infinite
1597 * face is not necessarily darkable. */
1598 if (can_colour_face(g, board, i, FACE_BLACK)) {
1599 fs->black_score = face_score(g, board, f, FACE_BLACK);
1600 add234(darkable_faces_sorted, fs);
1602 if (can_colour_face(g, board, i, FACE_WHITE)) {
1603 fs->white_score = face_score(g, board, f, FACE_WHITE);
1604 add234(lightable_faces_sorted, fs);
1608 /* Colour faces one at a time until no more faces are colourable. */
1611 enum face_colour colour;
1612 struct face_score *fs_white, *fs_black;
1613 int c_lightable = count234(lightable_faces_sorted);
1614 int c_darkable = count234(darkable_faces_sorted);
1615 if (c_lightable == 0 && c_darkable == 0) {
1616 /* No more faces we can use at all. */
1619 assert(c_lightable != 0 && c_darkable != 0);
1621 fs_white = (struct face_score *)index234(lightable_faces_sorted, 0);
1622 fs_black = (struct face_score *)index234(darkable_faces_sorted, 0);
1624 /* Choose a colour, and colour the best available face
1625 * with that colour. */
1626 colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK;
1628 if (colour == FACE_WHITE)
1633 i = fs - face_scores;
1634 assert(board[i] == FACE_GREY);
1637 /* Remove this newly-coloured face from the lists. These lists should
1638 * only contain grey faces. */
1639 del234(lightable_faces_sorted, fs);
1640 del234(darkable_faces_sorted, fs);
1642 /* Remember which face we've just coloured */
1643 cur_face = g->faces + i;
1645 /* The face we've just coloured potentially affects the colourability
1646 * and the scores of any neighbouring faces (touching at a corner or
1647 * edge). So the search needs to be conducted around all faces
1648 * touching the one we've just lit. Iterate over its corners, then
1649 * over each corner's faces. For each such face, we remove it from
1650 * the lists, recalculate any scores, then add it back to the lists
1651 * (depending on whether it is lightable, darkable or both). */
1652 for (i = 0; i < cur_face->order; i++) {
1653 grid_dot *d = cur_face->dots[i];
1654 for (j = 0; j < d->order; j++) {
1655 grid_face *f = d->faces[j];
1656 int fi; /* face index of f */
1663 /* If the face is already coloured, it won't be on our
1664 * lightable/darkable lists anyway, so we can skip it without
1665 * bothering with the removal step. */
1666 if (FACE_COLOUR(f) != FACE_GREY) continue;
1668 /* Find the face index and face_score* corresponding to f */
1670 fs = face_scores + fi;
1672 /* Remove from lightable list if it's in there. We do this,
1673 * even if it is still lightable, because the score might
1674 * be different, and we need to remove-then-add to maintain
1675 * correct sort order. */
1676 del234(lightable_faces_sorted, fs);
1677 if (can_colour_face(g, board, fi, FACE_WHITE)) {
1678 fs->white_score = face_score(g, board, f, FACE_WHITE);
1679 add234(lightable_faces_sorted, fs);
1681 /* Do the same for darkable list. */
1682 del234(darkable_faces_sorted, fs);
1683 if (can_colour_face(g, board, fi, FACE_BLACK)) {
1684 fs->black_score = face_score(g, board, f, FACE_BLACK);
1685 add234(darkable_faces_sorted, fs);
1692 freetree234(lightable_faces_sorted);
1693 freetree234(darkable_faces_sorted);
1696 /* The next step requires a shuffled list of all faces */
1697 face_list = snewn(num_faces, int);
1698 for (i = 0; i < num_faces; ++i) {
1701 shuffle(face_list, num_faces, sizeof(int), rs);
1703 /* The above loop-generation algorithm can often leave large clumps
1704 * of faces of one colour. In extreme cases, the resulting path can be
1705 * degenerate and not very satisfying to solve.
1706 * This next step alleviates this problem:
1707 * Go through the shuffled list, and flip the colour of any face we can
1708 * legally flip, and which is adjacent to only one face of the opposite
1709 * colour - this tends to grow 'tendrils' into any clumps.
1710 * Repeat until we can find no more faces to flip. This will
1711 * eventually terminate, because each flip increases the loop's
1712 * perimeter, which cannot increase for ever.
1713 * The resulting path will have maximal loopiness (in the sense that it
1714 * cannot be improved "locally". Unfortunately, this allows a player to
1715 * make some illicit deductions. To combat this (and make the path more
1716 * interesting), we do one final pass making random flips. */
1718 /* Set to TRUE for final pass */
1719 do_random_pass = FALSE;
1722 /* Remember whether a flip occurred during this pass */
1723 int flipped = FALSE;
1725 for (i = 0; i < num_faces; ++i) {
1726 int j = face_list[i];
1727 enum face_colour opp =
1728 (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE;
1729 if (can_colour_face(g, board, j, opp)) {
1730 grid_face *face = g->faces +j;
1731 if (do_random_pass) {
1732 /* final random pass */
1733 if (!random_upto(rs, 10))
1736 /* normal pass - flip when neighbour count is 1 */
1737 if (face_num_neighbours(g, board, face, opp) == 1) {
1745 if (do_random_pass) break;
1746 if (!flipped) do_random_pass = TRUE;
1751 /* Fill out all the clues by initialising to 0, then iterating over
1752 * all edges and incrementing each clue as we find edges that border
1753 * between BLACK/WHITE faces. While we're at it, we verify that the
1754 * algorithm does work, and there aren't any GREY faces still there. */
1755 memset(clues, 0, num_faces);
1756 for (i = 0; i < g->num_edges; i++) {
1757 grid_edge *e = g->edges + i;
1758 grid_face *f1 = e->face1;
1759 grid_face *f2 = e->face2;
1760 enum face_colour c1 = FACE_COLOUR(f1);
1761 enum face_colour c2 = FACE_COLOUR(f2);
1762 assert(c1 != FACE_GREY);
1763 assert(c2 != FACE_GREY);
1765 if (f1) clues[f1 - g->faces]++;
1766 if (f2) clues[f2 - g->faces]++;
1774 static int game_has_unique_soln(const game_state *state, int diff)
1777 solver_state *sstate_new;
1778 solver_state *sstate = new_solver_state((game_state *)state, diff);
1780 sstate_new = solve_game_rec(sstate);
1782 assert(sstate_new->solver_status != SOLVER_MISTAKE);
1783 ret = (sstate_new->solver_status == SOLVER_SOLVED);
1785 free_solver_state(sstate_new);
1786 free_solver_state(sstate);
1792 /* Remove clues one at a time at random. */
1793 static game_state *remove_clues(game_state *state, random_state *rs,
1797 int num_faces = state->game_grid->num_faces;
1798 game_state *ret = dup_game(state), *saved_ret;
1801 /* We need to remove some clues. We'll do this by forming a list of all
1802 * available clues, shuffling it, then going along one at a
1803 * time clearing each clue in turn for which doing so doesn't render the
1804 * board unsolvable. */
1805 face_list = snewn(num_faces, int);
1806 for (n = 0; n < num_faces; ++n) {
1810 shuffle(face_list, num_faces, sizeof(int), rs);
1812 for (n = 0; n < num_faces; ++n) {
1813 saved_ret = dup_game(ret);
1814 ret->clues[face_list[n]] = -1;
1816 if (game_has_unique_soln(ret, diff)) {
1817 free_game(saved_ret);
1829 static char *new_game_desc(game_params *params, random_state *rs,
1830 char **aux, int interactive)
1832 /* solution and description both use run-length encoding in obvious ways */
1835 game_state *state = snew(game_state);
1836 game_state *state_new;
1837 params_generate_grid(params);
1838 state->game_grid = g = params->game_grid;
1840 state->clues = snewn(g->num_faces, signed char);
1841 state->lines = snewn(g->num_edges, char);
1842 state->line_errors = snewn(g->num_edges, unsigned char);
1844 state->grid_type = params->type;
1848 memset(state->lines, LINE_UNKNOWN, g->num_edges);
1849 memset(state->line_errors, 0, g->num_edges);
1851 state->solved = state->cheated = FALSE;
1853 /* Get a new random solvable board with all its clues filled in. Yes, this
1854 * can loop for ever if the params are suitably unfavourable, but
1855 * preventing games smaller than 4x4 seems to stop this happening */
1857 add_full_clues(state, rs);
1858 } while (!game_has_unique_soln(state, params->diff));
1860 state_new = remove_clues(state, rs, params->diff);
1865 if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
1867 fprintf(stderr, "Rejecting board, it is too easy\n");
1869 goto newboard_please;
1872 retval = state_to_text(state);
1876 assert(!validate_desc(params, retval));
1881 static game_state *new_game(midend *me, game_params *params, char *desc)
1884 game_state *state = snew(game_state);
1885 int empties_to_make = 0;
1887 const char *dp = desc;
1889 int num_faces, num_edges;
1891 params_generate_grid(params);
1892 state->game_grid = g = params->game_grid;
1894 num_faces = g->num_faces;
1895 num_edges = g->num_edges;
1897 state->clues = snewn(num_faces, signed char);
1898 state->lines = snewn(num_edges, char);
1899 state->line_errors = snewn(num_edges, unsigned char);
1901 state->solved = state->cheated = FALSE;
1903 state->grid_type = params->type;
1905 for (i = 0; i < num_faces; i++) {
1906 if (empties_to_make) {
1908 state->clues[i] = -1;
1914 n2 = *dp - 'A' + 10;
1915 if (n >= 0 && n < 10) {
1916 state->clues[i] = n;
1917 } else if (n2 >= 10 && n2 < 36) {
1918 state->clues[i] = n2;
1922 state->clues[i] = -1;
1923 empties_to_make = n - 1;
1928 memset(state->lines, LINE_UNKNOWN, num_edges);
1929 memset(state->line_errors, 0, num_edges);
1933 /* Calculates the line_errors data, and checks if the current state is a
1935 static int check_completion(game_state *state)
1937 grid *g = state->game_grid;
1939 int num_faces = g->num_faces;
1941 int infinite_area, finite_area;
1942 int loops_found = 0;
1943 int found_edge_not_in_loop = FALSE;
1945 memset(state->line_errors, 0, g->num_edges);
1947 /* LL implementation of SGT's idea:
1948 * A loop will partition the grid into an inside and an outside.
1949 * If there is more than one loop, the grid will be partitioned into
1950 * even more distinct regions. We can therefore track equivalence of
1951 * faces, by saying that two faces are equivalent when there is a non-YES
1952 * edge between them.
1953 * We could keep track of the number of connected components, by counting
1954 * the number of dsf-merges that aren't no-ops.
1955 * But we're only interested in 3 separate cases:
1956 * no loops, one loop, more than one loop.
1958 * No loops: all faces are equivalent to the infinite face.
1959 * One loop: only two equivalence classes - finite and infinite.
1960 * >= 2 loops: there are 2 distinct finite regions.
1962 * So we simply make two passes through all the edges.
1963 * In the first pass, we dsf-merge the two faces bordering each non-YES
1965 * In the second pass, we look for YES-edges bordering:
1966 * a) two non-equivalent faces.
1967 * b) two non-equivalent faces, and one of them is part of a different
1968 * finite area from the first finite area we've seen.
1970 * An occurrence of a) means there is at least one loop.
1971 * An occurrence of b) means there is more than one loop.
1972 * Edges satisfying a) are marked as errors.
1974 * While we're at it, we set a flag if we find a YES edge that is not
1976 * This information will help decide, if there's a single loop, whether it
1977 * is a candidate for being a solution (that is, all YES edges are part of
1980 * If there is a candidate loop, we then go through all clues and check
1981 * they are all satisfied. If so, we have found a solution and we can
1982 * unmark all line_errors.
1985 /* Infinite face is at the end - its index is num_faces.
1986 * This macro is just to make this obvious! */
1987 #define INF_FACE num_faces
1988 dsf = snewn(num_faces + 1, int);
1989 dsf_init(dsf, num_faces + 1);
1992 for (i = 0; i < g->num_edges; i++) {
1993 grid_edge *e = g->edges + i;
1994 int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
1995 int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
1996 if (state->lines[i] != LINE_YES)
1997 dsf_merge(dsf, f1, f2);
2001 infinite_area = dsf_canonify(dsf, INF_FACE);
2003 for (i = 0; i < g->num_edges; i++) {
2004 grid_edge *e = g->edges + i;
2005 int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
2006 int can1 = dsf_canonify(dsf, f1);
2007 int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
2008 int can2 = dsf_canonify(dsf, f2);
2009 if (state->lines[i] != LINE_YES) continue;
2012 /* Faces are equivalent, so this edge not part of a loop */
2013 found_edge_not_in_loop = TRUE;
2016 state->line_errors[i] = TRUE;
2017 if (loops_found == 0) loops_found = 1;
2019 /* Don't bother with further checks if we've already found 2 loops */
2020 if (loops_found == 2) continue;
2022 if (finite_area == -1) {
2023 /* Found our first finite area */
2024 if (can1 != infinite_area)
2030 /* Have we found a second area? */
2031 if (finite_area != -1) {
2032 if (can1 != infinite_area && can1 != finite_area) {
2036 if (can2 != infinite_area && can2 != finite_area) {
2043 printf("loops_found = %d\n", loops_found);
2044 printf("found_edge_not_in_loop = %s\n",
2045 found_edge_not_in_loop ? "TRUE" : "FALSE");
2048 sfree(dsf); /* No longer need the dsf */
2050 /* Have we found a candidate loop? */
2051 if (loops_found == 1 && !found_edge_not_in_loop) {
2052 /* Yes, so check all clues are satisfied */
2053 int found_clue_violation = FALSE;
2054 for (i = 0; i < num_faces; i++) {
2055 int c = state->clues[i];
2057 if (face_order(state, i, LINE_YES) != c) {
2058 found_clue_violation = TRUE;
2064 if (!found_clue_violation) {
2065 /* The loop is good */
2066 memset(state->line_errors, 0, g->num_edges);
2067 return TRUE; /* No need to bother checking for dot violations */
2071 /* Check for dot violations */
2072 for (i = 0; i < g->num_dots; i++) {
2073 int yes = dot_order(state, i, LINE_YES);
2074 int unknown = dot_order(state, i, LINE_UNKNOWN);
2075 if ((yes == 1 && unknown == 0) || (yes >= 3)) {
2076 /* violation, so mark all YES edges as errors */
2077 grid_dot *d = g->dots + i;
2079 for (j = 0; j < d->order; j++) {
2080 int e = d->edges[j] - g->edges;
2081 if (state->lines[e] == LINE_YES)
2082 state->line_errors[e] = TRUE;
2089 /* ----------------------------------------------------------------------
2092 * Our solver modes operate as follows. Each mode also uses the modes above it.
2095 * Just implement the rules of the game.
2097 * Normal and Tricky Modes
2098 * For each (adjacent) pair of lines through each dot we store a bit for
2099 * whether at least one of them is on and whether at most one is on. (If we
2100 * know both or neither is on that's already stored more directly.)
2103 * Use edsf data structure to make equivalence classes of lines that are
2104 * known identical to or opposite to one another.
2109 * For general grids, we consider "dlines" to be pairs of lines joined
2110 * at a dot. The lines must be adjacent around the dot, so we can think of
2111 * a dline as being a dot+face combination. Or, a dot+edge combination where
2112 * the second edge is taken to be the next clockwise edge from the dot.
2113 * Original loopy code didn't have this extra restriction of the lines being
2114 * adjacent. From my tests with square grids, this extra restriction seems to
2115 * take little, if anything, away from the quality of the puzzles.
2116 * A dline can be uniquely identified by an edge/dot combination, given that
2117 * a dline-pair always goes clockwise around its common dot. The edge/dot
2118 * combination can be represented by an edge/bool combination - if bool is
2119 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2120 * exactly twice the number of edges in the grid - although the dlines
2121 * spanning the infinite face are not all that useful to the solver.
2122 * Note that, by convention, a dline goes clockwise around its common dot,
2123 * which means the dline goes anti-clockwise around its common face.
2126 /* Helper functions for obtaining an index into an array of dlines, given
2127 * various information. We assume the grid layout conventions about how
2128 * the various lists are interleaved - see grid_make_consistent() for
2131 /* i points to the first edge of the dline pair, reading clockwise around
2133 static int dline_index_from_dot(grid *g, grid_dot *d, int i)
2135 grid_edge *e = d->edges[i];
2140 if (i2 == d->order) i2 = 0;
2143 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
2145 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2146 (int)(d - g->dots), i, (int)(e - g->edges),
2147 (int)(e2 - g->edges), ret);
2151 /* i points to the second edge of the dline pair, reading clockwise around
2152 * the face. That is, the edges of the dline, starting at edge{i}, read
2153 * anti-clockwise around the face. By layout conventions, the common dot
2154 * of the dline will be f->dots[i] */
2155 static int dline_index_from_face(grid *g, grid_face *f, int i)
2157 grid_edge *e = f->edges[i];
2158 grid_dot *d = f->dots[i];
2163 if (i2 < 0) i2 += f->order;
2166 ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
2168 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2169 (int)(f - g->faces), i, (int)(e - g->edges),
2170 (int)(e2 - g->edges), ret);
2174 static int is_atleastone(const char *dline_array, int index)
2176 return BIT_SET(dline_array[index], 0);
2178 static int set_atleastone(char *dline_array, int index)
2180 return SET_BIT(dline_array[index], 0);
2182 static int is_atmostone(const char *dline_array, int index)
2184 return BIT_SET(dline_array[index], 1);
2186 static int set_atmostone(char *dline_array, int index)
2188 return SET_BIT(dline_array[index], 1);
2191 static void array_setall(char *array, char from, char to, int len)
2193 char *p = array, *p_old = p;
2194 int len_remaining = len;
2196 while ((p = memchr(p, from, len_remaining))) {
2198 len_remaining -= p - p_old;
2203 /* Helper, called when doing dline dot deductions, in the case where we
2204 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2205 * them (because of dline atmostone/atleastone).
2206 * On entry, edge points to the first of these two UNKNOWNs. This function
2207 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2208 * and set their corresponding dline to atleastone. (Setting atmostone
2209 * already happens in earlier dline deductions) */
2210 static int dline_set_opp_atleastone(solver_state *sstate,
2211 grid_dot *d, int edge)
2213 game_state *state = sstate->state;
2214 grid *g = state->game_grid;
2217 for (opp = 0; opp < N; opp++) {
2218 int opp_dline_index;
2219 if (opp == edge || opp == edge+1 || opp == edge-1)
2221 if (opp == 0 && edge == N-1)
2223 if (opp == N-1 && edge == 0)
2226 if (opp2 == N) opp2 = 0;
2227 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2228 if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
2230 if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
2232 /* Found opposite UNKNOWNS and they're next to each other */
2233 opp_dline_index = dline_index_from_dot(g, d, opp);
2234 return set_atleastone(sstate->dlines, opp_dline_index);
2240 /* Set pairs of lines around this face which are known to be identical, to
2241 * the given line_state */
2242 static int face_setall_identical(solver_state *sstate, int face_index,
2243 enum line_state line_new)
2245 /* can[dir] contains the canonical line associated with the line in
2246 * direction dir from the square in question. Similarly inv[dir] is
2247 * whether or not the line in question is inverse to its canonical
2250 game_state *state = sstate->state;
2251 grid *g = state->game_grid;
2252 grid_face *f = g->faces + face_index;
2255 int can1, can2, inv1, inv2;
2257 for (i = 0; i < N; i++) {
2258 int line1_index = f->edges[i] - g->edges;
2259 if (state->lines[line1_index] != LINE_UNKNOWN)
2261 for (j = i + 1; j < N; j++) {
2262 int line2_index = f->edges[j] - g->edges;
2263 if (state->lines[line2_index] != LINE_UNKNOWN)
2266 /* Found two UNKNOWNS */
2267 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
2268 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
2269 if (can1 == can2 && inv1 == inv2) {
2270 solver_set_line(sstate, line1_index, line_new);
2271 solver_set_line(sstate, line2_index, line_new);
2278 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2279 * return the edge indices into e. */
2280 static void find_unknowns(game_state *state,
2281 grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
2282 int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
2283 int *e /* Returned edge indices */)
2286 grid *g = state->game_grid;
2287 while (c < expected_count) {
2288 int line_index = *edge_list - g->edges;
2289 if (state->lines[line_index] == LINE_UNKNOWN) {
2297 /* If we have a list of edges, and we know whether the number of YESs should
2298 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2299 * linedsf deductions. This can be used for both face and dot deductions.
2300 * Returns the difficulty level of the next solver that should be used,
2301 * or DIFF_MAX if no progress was made. */
2302 static int parity_deductions(solver_state *sstate,
2303 grid_edge **edge_list, /* Edge list (from a face or a dot) */
2304 int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
2307 game_state *state = sstate->state;
2308 int diff = DIFF_MAX;
2309 int *linedsf = sstate->linedsf;
2311 if (unknown_count == 2) {
2312 /* Lines are known alike/opposite, depending on inv. */
2314 find_unknowns(state, edge_list, 2, e);
2315 if (merge_lines(sstate, e[0], e[1], total_parity))
2316 diff = min(diff, DIFF_HARD);
2317 } else if (unknown_count == 3) {
2319 int can[3]; /* canonical edges */
2320 int inv[3]; /* whether can[x] is inverse to e[x] */
2321 find_unknowns(state, edge_list, 3, e);
2322 can[0] = edsf_canonify(linedsf, e[0], inv);
2323 can[1] = edsf_canonify(linedsf, e[1], inv+1);
2324 can[2] = edsf_canonify(linedsf, e[2], inv+2);
2325 if (can[0] == can[1]) {
2326 if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
2327 LINE_YES : LINE_NO))
2328 diff = min(diff, DIFF_EASY);
2330 if (can[0] == can[2]) {
2331 if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
2332 LINE_YES : LINE_NO))
2333 diff = min(diff, DIFF_EASY);
2335 if (can[1] == can[2]) {
2336 if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
2337 LINE_YES : LINE_NO))
2338 diff = min(diff, DIFF_EASY);
2340 } else if (unknown_count == 4) {
2342 int can[4]; /* canonical edges */
2343 int inv[4]; /* whether can[x] is inverse to e[x] */
2344 find_unknowns(state, edge_list, 4, e);
2345 can[0] = edsf_canonify(linedsf, e[0], inv);
2346 can[1] = edsf_canonify(linedsf, e[1], inv+1);
2347 can[2] = edsf_canonify(linedsf, e[2], inv+2);
2348 can[3] = edsf_canonify(linedsf, e[3], inv+3);
2349 if (can[0] == can[1]) {
2350 if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
2351 diff = min(diff, DIFF_HARD);
2352 } else if (can[0] == can[2]) {
2353 if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
2354 diff = min(diff, DIFF_HARD);
2355 } else if (can[0] == can[3]) {
2356 if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
2357 diff = min(diff, DIFF_HARD);
2358 } else if (can[1] == can[2]) {
2359 if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
2360 diff = min(diff, DIFF_HARD);
2361 } else if (can[1] == can[3]) {
2362 if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
2363 diff = min(diff, DIFF_HARD);
2364 } else if (can[2] == can[3]) {
2365 if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
2366 diff = min(diff, DIFF_HARD);
2374 * These are the main solver functions.
2376 * Their return values are diff values corresponding to the lowest mode solver
2377 * that would notice the work that they have done. For example if the normal
2378 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2379 * easy mode solver might be able to make progress using that. It doesn't make
2380 * sense for one of them to return a diff value higher than that of the
2383 * Each function returns the lowest value it can, as early as possible, in
2384 * order to try and pass as much work as possible back to the lower level
2385 * solvers which progress more quickly.
2388 /* PROPOSED NEW DESIGN:
2389 * We have a work queue consisting of 'events' notifying us that something has
2390 * happened that a particular solver mode might be interested in. For example
2391 * the hard mode solver might do something that helps the normal mode solver at
2392 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2393 * we pull events off the work queue, and hand each in turn to the solver that
2394 * is interested in them. If a solver reports that it failed we pass the same
2395 * event on to progressively more advanced solvers and the loop detector. Once
2396 * we've exhausted an event, or it has helped us progress, we drop it and
2397 * continue to the next one. The events are sorted first in order of solver
2398 * complexity (easy first) then order of insertion (oldest first).
2399 * Once we run out of events we loop over each permitted solver in turn
2400 * (easiest first) until either a deduction is made (and an event therefore
2401 * emerges) or no further deductions can be made (in which case we've failed).
2404 * * How do we 'loop over' a solver when both dots and squares are concerned.
2405 * Answer: first all squares then all dots.
2408 static int trivial_deductions(solver_state *sstate)
2410 int i, current_yes, current_no;
2411 game_state *state = sstate->state;
2412 grid *g = state->game_grid;
2413 int diff = DIFF_MAX;
2415 /* Per-face deductions */
2416 for (i = 0; i < g->num_faces; i++) {
2417 grid_face *f = g->faces + i;
2419 if (sstate->face_solved[i])
2422 current_yes = sstate->face_yes_count[i];
2423 current_no = sstate->face_no_count[i];
2425 if (current_yes + current_no == f->order) {
2426 sstate->face_solved[i] = TRUE;
2430 if (state->clues[i] < 0)
2434 * This code checks whether the numeric clue on a face is so
2435 * large as to permit all its remaining LINE_UNKNOWNs to be
2436 * filled in as LINE_YES, or alternatively so small as to
2437 * permit them all to be filled in as LINE_NO.
2440 if (state->clues[i] < current_yes) {
2441 sstate->solver_status = SOLVER_MISTAKE;
2444 if (state->clues[i] == current_yes) {
2445 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
2446 diff = min(diff, DIFF_EASY);
2447 sstate->face_solved[i] = TRUE;
2451 if (f->order - state->clues[i] < current_no) {
2452 sstate->solver_status = SOLVER_MISTAKE;
2455 if (f->order - state->clues[i] == current_no) {
2456 if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
2457 diff = min(diff, DIFF_EASY);
2458 sstate->face_solved[i] = TRUE;
2462 if (f->order - state->clues[i] == current_no + 1 &&
2463 f->order - current_yes - current_no > 2) {
2465 * One small refinement to the above: we also look for any
2466 * adjacent pair of LINE_UNKNOWNs around the face with
2467 * some LINE_YES incident on it from elsewhere. If we find
2468 * one, then we know that pair of LINE_UNKNOWNs can't
2469 * _both_ be LINE_YES, and hence that pushes us one line
2470 * closer to being able to determine all the rest.
2472 int j, k, e1, e2, e, d;
2474 for (j = 0; j < f->order; j++) {
2475 e1 = f->edges[j] - g->edges;
2476 e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges;
2478 if (g->edges[e1].dot1 == g->edges[e2].dot1 ||
2479 g->edges[e1].dot1 == g->edges[e2].dot2) {
2480 d = g->edges[e1].dot1 - g->dots;
2482 assert(g->edges[e1].dot2 == g->edges[e2].dot1 ||
2483 g->edges[e1].dot2 == g->edges[e2].dot2);
2484 d = g->edges[e1].dot2 - g->dots;
2487 if (state->lines[e1] == LINE_UNKNOWN &&
2488 state->lines[e2] == LINE_UNKNOWN) {
2489 for (k = 0; k < g->dots[d].order; k++) {
2490 int e = g->dots[d].edges[k] - g->edges;
2491 if (state->lines[e] == LINE_YES)
2492 goto found; /* multi-level break */
2500 * If we get here, we've found such a pair of edges, and
2501 * they're e1 and e2.
2503 for (j = 0; j < f->order; j++) {
2504 e = f->edges[j] - g->edges;
2505 if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) {
2506 int r = solver_set_line(sstate, e, LINE_YES);
2508 diff = min(diff, DIFF_EASY);
2514 check_caches(sstate);
2516 /* Per-dot deductions */
2517 for (i = 0; i < g->num_dots; i++) {
2518 grid_dot *d = g->dots + i;
2519 int yes, no, unknown;
2521 if (sstate->dot_solved[i])
2524 yes = sstate->dot_yes_count[i];
2525 no = sstate->dot_no_count[i];
2526 unknown = d->order - yes - no;
2530 sstate->dot_solved[i] = TRUE;
2531 } else if (unknown == 1) {
2532 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2533 diff = min(diff, DIFF_EASY);
2534 sstate->dot_solved[i] = TRUE;
2536 } else if (yes == 1) {
2538 sstate->solver_status = SOLVER_MISTAKE;
2540 } else if (unknown == 1) {
2541 dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
2542 diff = min(diff, DIFF_EASY);
2544 } else if (yes == 2) {
2546 dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
2547 diff = min(diff, DIFF_EASY);
2549 sstate->dot_solved[i] = TRUE;
2551 sstate->solver_status = SOLVER_MISTAKE;
2556 check_caches(sstate);
2561 static int dline_deductions(solver_state *sstate)
2563 game_state *state = sstate->state;
2564 grid *g = state->game_grid;
2565 char *dlines = sstate->dlines;
2567 int diff = DIFF_MAX;
2569 /* ------ Face deductions ------ */
2571 /* Given a set of dline atmostone/atleastone constraints, need to figure
2572 * out if we can deduce any further info. For more general faces than
2573 * squares, this turns out to be a tricky problem.
2574 * The approach taken here is to define (per face) NxN matrices:
2575 * "maxs" and "mins".
2576 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2577 * for the possible number of edges that are YES between positions j and k
2578 * going clockwise around the face. Can think of j and k as marking dots
2579 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2580 * edge1 joins dot1 to dot2 etc).
2581 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2582 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2583 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2584 * the dline atmostone/atleastone status for edges j and j+1.
2586 * Then we calculate the remaining entries recursively. We definitely
2588 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2589 * This is because any valid placement of YESs between j and k must give
2590 * a valid placement between j and u, and also between u and k.
2591 * I believe it's sufficient to use just the two values of u:
2592 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2593 * are rigorous, even if they might not be best-possible.
2595 * Once we have maxs and mins calculated, we can make inferences about
2596 * each dline{j,j+1} by looking at the possible complementary edge-counts
2597 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2598 * As well as dlines, we can make similar inferences about single edges.
2599 * For example, consider a pentagon with clue 3, and we know at most one
2600 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2601 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2602 * that final edge would have to be YES to make the count up to 3.
2605 /* Much quicker to allocate arrays on the stack than the heap, so
2606 * define the largest possible face size, and base our array allocations
2607 * on that. We check this with an assertion, in case someone decides to
2608 * make a grid which has larger faces than this. Note, this algorithm
2609 * could get quite expensive if there are many large faces. */
2610 #define MAX_FACE_SIZE 12
2612 for (i = 0; i < g->num_faces; i++) {
2613 int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
2614 int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
2615 grid_face *f = g->faces + i;
2618 int clue = state->clues[i];
2619 assert(N <= MAX_FACE_SIZE);
2620 if (sstate->face_solved[i])
2622 if (clue < 0) continue;
2624 /* Calculate the (j,j+1) entries */
2625 for (j = 0; j < N; j++) {
2626 int edge_index = f->edges[j] - g->edges;
2628 enum line_state line1 = state->lines[edge_index];
2629 enum line_state line2;
2633 maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
2634 mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
2635 /* Calculate the (j,j+2) entries */
2636 dline_index = dline_index_from_face(g, f, k);
2637 edge_index = f->edges[k] - g->edges;
2638 line2 = state->lines[edge_index];
2644 if (line1 == LINE_NO) tmp--;
2645 if (line2 == LINE_NO) tmp--;
2646 if (tmp == 2 && is_atmostone(dlines, dline_index))
2652 if (line1 == LINE_YES) tmp++;
2653 if (line2 == LINE_YES) tmp++;
2654 if (tmp == 0 && is_atleastone(dlines, dline_index))
2659 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2660 for (m = 3; m < N; m++) {
2661 for (j = 0; j < N; j++) {
2669 maxs[j][k] = maxs[j][u] + maxs[u][k];
2670 mins[j][k] = mins[j][u] + mins[u][k];
2671 tmp = maxs[j][v] + maxs[v][k];
2672 maxs[j][k] = min(maxs[j][k], tmp);
2673 tmp = mins[j][v] + mins[v][k];
2674 mins[j][k] = max(mins[j][k], tmp);
2678 /* See if we can make any deductions */
2679 for (j = 0; j < N; j++) {
2681 grid_edge *e = f->edges[j];
2682 int line_index = e - g->edges;
2685 if (state->lines[line_index] != LINE_UNKNOWN)
2690 /* minimum YESs in the complement of this edge */
2691 if (mins[k][j] > clue) {
2692 sstate->solver_status = SOLVER_MISTAKE;
2695 if (mins[k][j] == clue) {
2696 /* setting this edge to YES would make at least
2697 * (clue+1) edges - contradiction */
2698 solver_set_line(sstate, line_index, LINE_NO);
2699 diff = min(diff, DIFF_EASY);
2701 if (maxs[k][j] < clue - 1) {
2702 sstate->solver_status = SOLVER_MISTAKE;
2705 if (maxs[k][j] == clue - 1) {
2706 /* Only way to satisfy the clue is to set edge{j} as YES */
2707 solver_set_line(sstate, line_index, LINE_YES);
2708 diff = min(diff, DIFF_EASY);
2711 /* More advanced deduction that allows propagation along diagonal
2712 * chains of faces connected by dots, for example, 3-2-...-2-3
2713 * in square grids. */
2714 if (sstate->diff >= DIFF_TRICKY) {
2715 /* Now see if we can make dline deduction for edges{j,j+1} */
2717 if (state->lines[e - g->edges] != LINE_UNKNOWN)
2718 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2719 * Dlines where one of the edges is known, are handled in the
2723 dline_index = dline_index_from_face(g, f, k);
2727 /* minimum YESs in the complement of this dline */
2728 if (mins[k][j] > clue - 2) {
2729 /* Adding 2 YESs would break the clue */
2730 if (set_atmostone(dlines, dline_index))
2731 diff = min(diff, DIFF_NORMAL);
2733 /* maximum YESs in the complement of this dline */
2734 if (maxs[k][j] < clue) {
2735 /* Adding 2 NOs would mean not enough YESs */
2736 if (set_atleastone(dlines, dline_index))
2737 diff = min(diff, DIFF_NORMAL);
2743 if (diff < DIFF_NORMAL)
2746 /* ------ Dot deductions ------ */
2748 for (i = 0; i < g->num_dots; i++) {
2749 grid_dot *d = g->dots + i;
2751 int yes, no, unknown;
2753 if (sstate->dot_solved[i])
2755 yes = sstate->dot_yes_count[i];
2756 no = sstate->dot_no_count[i];
2757 unknown = N - yes - no;
2759 for (j = 0; j < N; j++) {
2762 int line1_index, line2_index;
2763 enum line_state line1, line2;
2766 dline_index = dline_index_from_dot(g, d, j);
2767 line1_index = d->edges[j] - g->edges;
2768 line2_index = d->edges[k] - g->edges;
2769 line1 = state->lines[line1_index];
2770 line2 = state->lines[line2_index];
2772 /* Infer dline state from line state */
2773 if (line1 == LINE_NO || line2 == LINE_NO) {
2774 if (set_atmostone(dlines, dline_index))
2775 diff = min(diff, DIFF_NORMAL);
2777 if (line1 == LINE_YES || line2 == LINE_YES) {
2778 if (set_atleastone(dlines, dline_index))
2779 diff = min(diff, DIFF_NORMAL);
2781 /* Infer line state from dline state */
2782 if (is_atmostone(dlines, dline_index)) {
2783 if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
2784 solver_set_line(sstate, line2_index, LINE_NO);
2785 diff = min(diff, DIFF_EASY);
2787 if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
2788 solver_set_line(sstate, line1_index, LINE_NO);
2789 diff = min(diff, DIFF_EASY);
2792 if (is_atleastone(dlines, dline_index)) {
2793 if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
2794 solver_set_line(sstate, line2_index, LINE_YES);
2795 diff = min(diff, DIFF_EASY);
2797 if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
2798 solver_set_line(sstate, line1_index, LINE_YES);
2799 diff = min(diff, DIFF_EASY);
2802 /* Deductions that depend on the numbers of lines.
2803 * Only bother if both lines are UNKNOWN, otherwise the
2804 * easy-mode solver (or deductions above) would have taken
2806 if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
2809 if (yes == 0 && unknown == 2) {
2810 /* Both these unknowns must be identical. If we know
2811 * atmostone or atleastone, we can make progress. */
2812 if (is_atmostone(dlines, dline_index)) {
2813 solver_set_line(sstate, line1_index, LINE_NO);
2814 solver_set_line(sstate, line2_index, LINE_NO);
2815 diff = min(diff, DIFF_EASY);
2817 if (is_atleastone(dlines, dline_index)) {
2818 solver_set_line(sstate, line1_index, LINE_YES);
2819 solver_set_line(sstate, line2_index, LINE_YES);
2820 diff = min(diff, DIFF_EASY);
2824 if (set_atmostone(dlines, dline_index))
2825 diff = min(diff, DIFF_NORMAL);
2827 if (set_atleastone(dlines, dline_index))
2828 diff = min(diff, DIFF_NORMAL);
2832 /* More advanced deduction that allows propagation along diagonal
2833 * chains of faces connected by dots, for example: 3-2-...-2-3
2834 * in square grids. */
2835 if (sstate->diff >= DIFF_TRICKY) {
2836 /* If we have atleastone set for this dline, infer
2837 * atmostone for each "opposite" dline (that is, each
2838 * dline without edges in common with this one).
2839 * Again, this test is only worth doing if both these
2840 * lines are UNKNOWN. For if one of these lines were YES,
2841 * the (yes == 1) test above would kick in instead. */
2842 if (is_atleastone(dlines, dline_index)) {
2844 for (opp = 0; opp < N; opp++) {
2845 int opp_dline_index;
2846 if (opp == j || opp == j+1 || opp == j-1)
2848 if (j == 0 && opp == N-1)
2850 if (j == N-1 && opp == 0)
2852 opp_dline_index = dline_index_from_dot(g, d, opp);
2853 if (set_atmostone(dlines, opp_dline_index))
2854 diff = min(diff, DIFF_NORMAL);
2856 if (yes == 0 && is_atmostone(dlines, dline_index)) {
2857 /* This dline has *exactly* one YES and there are no
2858 * other YESs. This allows more deductions. */
2860 /* Third unknown must be YES */
2861 for (opp = 0; opp < N; opp++) {
2863 if (opp == j || opp == k)
2865 opp_index = d->edges[opp] - g->edges;
2866 if (state->lines[opp_index] == LINE_UNKNOWN) {
2867 solver_set_line(sstate, opp_index,
2869 diff = min(diff, DIFF_EASY);
2872 } else if (unknown == 4) {
2873 /* Exactly one of opposite UNKNOWNS is YES. We've
2874 * already set atmostone, so set atleastone as
2877 if (dline_set_opp_atleastone(sstate, d, j))
2878 diff = min(diff, DIFF_NORMAL);
2888 static int linedsf_deductions(solver_state *sstate)
2890 game_state *state = sstate->state;
2891 grid *g = state->game_grid;
2892 char *dlines = sstate->dlines;
2894 int diff = DIFF_MAX;
2897 /* ------ Face deductions ------ */
2899 /* A fully-general linedsf deduction seems overly complicated
2900 * (I suspect the problem is NP-complete, though in practice it might just
2901 * be doable because faces are limited in size).
2902 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2903 * known to be identical. If setting them both to YES (or NO) would break
2904 * the clue, set them to NO (or YES). */
2906 for (i = 0; i < g->num_faces; i++) {
2907 int N, yes, no, unknown;
2910 if (sstate->face_solved[i])
2912 clue = state->clues[i];
2916 N = g->faces[i].order;
2917 yes = sstate->face_yes_count[i];
2918 if (yes + 1 == clue) {
2919 if (face_setall_identical(sstate, i, LINE_NO))
2920 diff = min(diff, DIFF_EASY);
2922 no = sstate->face_no_count[i];
2923 if (no + 1 == N - clue) {
2924 if (face_setall_identical(sstate, i, LINE_YES))
2925 diff = min(diff, DIFF_EASY);
2928 /* Reload YES count, it might have changed */
2929 yes = sstate->face_yes_count[i];
2930 unknown = N - no - yes;
2932 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2933 * parity of lines. */
2934 diff_tmp = parity_deductions(sstate, g->faces[i].edges,
2935 (clue - yes) % 2, unknown);
2936 diff = min(diff, diff_tmp);
2939 /* ------ Dot deductions ------ */
2940 for (i = 0; i < g->num_dots; i++) {
2941 grid_dot *d = g->dots + i;
2944 int yes, no, unknown;
2945 /* Go through dlines, and do any dline<->linedsf deductions wherever
2946 * we find two UNKNOWNS. */
2947 for (j = 0; j < N; j++) {
2948 int dline_index = dline_index_from_dot(g, d, j);
2951 int can1, can2, inv1, inv2;
2953 line1_index = d->edges[j] - g->edges;
2954 if (state->lines[line1_index] != LINE_UNKNOWN)
2957 if (j2 == N) j2 = 0;
2958 line2_index = d->edges[j2] - g->edges;
2959 if (state->lines[line2_index] != LINE_UNKNOWN)
2961 /* Infer dline flags from linedsf */
2962 can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
2963 can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
2964 if (can1 == can2 && inv1 != inv2) {
2965 /* These are opposites, so set dline atmostone/atleastone */
2966 if (set_atmostone(dlines, dline_index))
2967 diff = min(diff, DIFF_NORMAL);
2968 if (set_atleastone(dlines, dline_index))
2969 diff = min(diff, DIFF_NORMAL);
2972 /* Infer linedsf from dline flags */
2973 if (is_atmostone(dlines, dline_index)
2974 && is_atleastone(dlines, dline_index)) {
2975 if (merge_lines(sstate, line1_index, line2_index, 1))
2976 diff = min(diff, DIFF_HARD);
2980 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2981 * parity of lines. */
2982 yes = sstate->dot_yes_count[i];
2983 no = sstate->dot_no_count[i];
2984 unknown = N - yes - no;
2985 diff_tmp = parity_deductions(sstate, d->edges,
2987 diff = min(diff, diff_tmp);
2990 /* ------ Edge dsf deductions ------ */
2992 /* If the state of a line is known, deduce the state of its canonical line
2993 * too, and vice versa. */
2994 for (i = 0; i < g->num_edges; i++) {
2997 can = edsf_canonify(sstate->linedsf, i, &inv);
3000 s = sstate->state->lines[can];
3001 if (s != LINE_UNKNOWN) {
3002 if (solver_set_line(sstate, i, inv ? OPP(s) : s))
3003 diff = min(diff, DIFF_EASY);
3005 s = sstate->state->lines[i];
3006 if (s != LINE_UNKNOWN) {
3007 if (solver_set_line(sstate, can, inv ? OPP(s) : s))
3008 diff = min(diff, DIFF_EASY);
3016 static int loop_deductions(solver_state *sstate)
3018 int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
3019 game_state *state = sstate->state;
3020 grid *g = state->game_grid;
3021 int shortest_chainlen = g->num_dots;
3022 int loop_found = FALSE;
3024 int progress = FALSE;
3028 * Go through the grid and update for all the new edges.
3029 * Since merge_dots() is idempotent, the simplest way to
3030 * do this is just to update for _all_ the edges.
3031 * Also, while we're here, we count the edges.
3033 for (i = 0; i < g->num_edges; i++) {
3034 if (state->lines[i] == LINE_YES) {
3035 loop_found |= merge_dots(sstate, i);
3041 * Count the clues, count the satisfied clues, and count the
3042 * satisfied-minus-one clues.
3044 for (i = 0; i < g->num_faces; i++) {
3045 int c = state->clues[i];
3047 int o = sstate->face_yes_count[i];
3056 for (i = 0; i < g->num_dots; ++i) {
3058 sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
3059 if (dots_connected > 1)
3060 shortest_chainlen = min(shortest_chainlen, dots_connected);
3063 assert(sstate->solver_status == SOLVER_INCOMPLETE);
3065 if (satclues == clues && shortest_chainlen == edgecount) {
3066 sstate->solver_status = SOLVER_SOLVED;
3067 /* This discovery clearly counts as progress, even if we haven't
3068 * just added any lines or anything */
3070 goto finished_loop_deductionsing;
3074 * Now go through looking for LINE_UNKNOWN edges which
3075 * connect two dots that are already in the same
3076 * equivalence class. If we find one, test to see if the
3077 * loop it would create is a solution.
3079 for (i = 0; i < g->num_edges; i++) {
3080 grid_edge *e = g->edges + i;
3081 int d1 = e->dot1 - g->dots;
3082 int d2 = e->dot2 - g->dots;
3084 if (state->lines[i] != LINE_UNKNOWN)
3087 eqclass = dsf_canonify(sstate->dotdsf, d1);
3088 if (eqclass != dsf_canonify(sstate->dotdsf, d2))
3091 val = LINE_NO; /* loop is bad until proven otherwise */
3094 * This edge would form a loop. Next
3095 * question: how long would the loop be?
3096 * Would it equal the total number of edges
3097 * (plus the one we'd be adding if we added
3100 if (sstate->looplen[eqclass] == edgecount + 1) {
3104 * This edge would form a loop which
3105 * took in all the edges in the entire
3106 * grid. So now we need to work out
3107 * whether it would be a valid solution
3108 * to the puzzle, which means we have to
3109 * check if it satisfies all the clues.
3110 * This means that every clue must be
3111 * either satisfied or satisfied-minus-
3112 * 1, and also that the number of
3113 * satisfied-minus-1 clues must be at
3114 * most two and they must lie on either
3115 * side of this edge.
3119 int f = e->face1 - g->faces;
3120 int c = state->clues[f];
3121 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
3125 int f = e->face2 - g->faces;
3126 int c = state->clues[f];
3127 if (c >= 0 && sstate->face_yes_count[f] == c - 1)
3130 if (sm1clues == sm1_nearby &&
3131 sm1clues + satclues == clues) {
3132 val = LINE_YES; /* loop is good! */
3137 * Right. Now we know that adding this edge
3138 * would form a loop, and we know whether
3139 * that loop would be a viable solution or
3142 * If adding this edge produces a solution,
3143 * then we know we've found _a_ solution but
3144 * we don't know that it's _the_ solution -
3145 * if it were provably the solution then
3146 * we'd have deduced this edge some time ago
3147 * without the need to do loop detection. So
3148 * in this state we return SOLVER_AMBIGUOUS,
3149 * which has the effect that hitting Solve
3150 * on a user-provided puzzle will fill in a
3151 * solution but using the solver to
3152 * construct new puzzles won't consider this
3153 * a reasonable deduction for the user to
3156 progress = solver_set_line(sstate, i, val);
3157 assert(progress == TRUE);
3158 if (val == LINE_YES) {
3159 sstate->solver_status = SOLVER_AMBIGUOUS;
3160 goto finished_loop_deductionsing;
3164 finished_loop_deductionsing:
3165 return progress ? DIFF_EASY : DIFF_MAX;
3168 /* This will return a dynamically allocated solver_state containing the (more)
3170 static solver_state *solve_game_rec(const solver_state *sstate_start)
3172 solver_state *sstate;
3174 /* Index of the solver we should call next. */
3177 /* As a speed-optimisation, we avoid re-running solvers that we know
3178 * won't make any progress. This happens when a high-difficulty
3179 * solver makes a deduction that can only help other high-difficulty
3181 * For example: if a new 'dline' flag is set by dline_deductions, the
3182 * trivial_deductions solver cannot do anything with this information.
3183 * If we've already run the trivial_deductions solver (because it's
3184 * earlier in the list), there's no point running it again.
3186 * Therefore: if a solver is earlier in the list than "threshold_index",
3187 * we don't bother running it if it's difficulty level is less than
3190 int threshold_diff = 0;
3191 int threshold_index = 0;
3193 sstate = dup_solver_state(sstate_start);
3195 check_caches(sstate);
3197 while (i < NUM_SOLVERS) {
3198 if (sstate->solver_status == SOLVER_MISTAKE)
3200 if (sstate->solver_status == SOLVER_SOLVED ||
3201 sstate->solver_status == SOLVER_AMBIGUOUS) {
3202 /* solver finished */
3206 if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
3207 && solver_diffs[i] <= sstate->diff) {
3208 /* current_solver is eligible, so use it */
3209 int next_diff = solver_fns[i](sstate);
3210 if (next_diff != DIFF_MAX) {
3211 /* solver made progress, so use new thresholds and
3212 * start again at top of list. */
3213 threshold_diff = next_diff;
3214 threshold_index = i;
3219 /* current_solver is ineligible, or failed to make progress, so
3220 * go to the next solver in the list */
3224 if (sstate->solver_status == SOLVER_SOLVED ||
3225 sstate->solver_status == SOLVER_AMBIGUOUS) {
3226 /* s/LINE_UNKNOWN/LINE_NO/g */
3227 array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
3228 sstate->state->game_grid->num_edges);
3235 static char *solve_game(game_state *state, game_state *currstate,
3236 char *aux, char **error)
3239 solver_state *sstate, *new_sstate;
3241 sstate = new_solver_state(state, DIFF_MAX);
3242 new_sstate = solve_game_rec(sstate);
3244 if (new_sstate->solver_status == SOLVER_SOLVED) {
3245 soln = encode_solve_move(new_sstate->state);
3246 } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
3247 soln = encode_solve_move(new_sstate->state);
3248 /**error = "Solver found ambiguous solutions"; */
3250 soln = encode_solve_move(new_sstate->state);
3251 /**error = "Solver failed"; */
3254 free_solver_state(new_sstate);
3255 free_solver_state(sstate);
3260 /* ----------------------------------------------------------------------
3261 * Drawing and mouse-handling
3264 static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
3265 int x, int y, int button)
3267 grid *g = state->game_grid;
3271 char button_char = ' ';
3272 enum line_state old_state;
3274 button &= ~MOD_MASK;
3276 /* Convert mouse-click (x,y) to grid coordinates */
3277 x -= BORDER(ds->tilesize);
3278 y -= BORDER(ds->tilesize);
3279 x = x * g->tilesize / ds->tilesize;
3280 y = y * g->tilesize / ds->tilesize;
3284 e = grid_nearest_edge(g, x, y);
3290 /* I think it's only possible to play this game with mouse clicks, sorry */
3291 /* Maybe will add mouse drag support some time */
3292 old_state = state->lines[i];
3296 switch (old_state) {
3314 switch (old_state) {
3333 sprintf(buf, "%d%c", i, (int)button_char);
3339 static game_state *execute_move(game_state *state, char *move)
3342 game_state *newstate = dup_game(state);
3344 if (move[0] == 'S') {
3346 newstate->cheated = TRUE;
3351 if (i < 0 || i >= newstate->game_grid->num_edges)
3353 move += strspn(move, "1234567890");
3354 switch (*(move++)) {
3356 newstate->lines[i] = LINE_YES;
3359 newstate->lines[i] = LINE_NO;
3362 newstate->lines[i] = LINE_UNKNOWN;
3370 * Check for completion.
3372 if (check_completion(newstate))
3373 newstate->solved = TRUE;
3378 free_game(newstate);
3382 /* ----------------------------------------------------------------------
3386 /* Convert from grid coordinates to screen coordinates */
3387 static void grid_to_screen(const game_drawstate *ds, const grid *g,
3388 int grid_x, int grid_y, int *x, int *y)
3390 *x = grid_x - g->lowest_x;
3391 *y = grid_y - g->lowest_y;
3392 *x = *x * ds->tilesize / g->tilesize;
3393 *y = *y * ds->tilesize / g->tilesize;
3394 *x += BORDER(ds->tilesize);
3395 *y += BORDER(ds->tilesize);
3398 /* Returns (into x,y) position of centre of face for rendering the text clue.
3400 static void face_text_pos(const game_drawstate *ds, const grid *g,
3401 const grid_face *f, int *xret, int *yret)
3403 int x, y, x0, y0, x1, y1, xbest, ybest, i, shift;
3405 int faceindex = f - g->faces;
3408 * Return the cached position for this face, if we've already
3411 if (ds->textx[faceindex] >= 0) {
3412 *xret = ds->textx[faceindex];
3413 *yret = ds->texty[faceindex];
3418 * Otherwise, try to find the point in the polygon with the
3419 * maximum distance to any edge or corner.
3421 * Start by working out the face's bounding box, in grid
3424 x0 = x1 = f->dots[0]->x;
3425 y0 = y1 = f->dots[0]->y;
3426 for (i = 1; i < f->order; i++) {
3427 if (x0 > f->dots[i]->x) x0 = f->dots[i]->x;
3428 if (x1 < f->dots[i]->x) x1 = f->dots[i]->x;
3429 if (y0 > f->dots[i]->y) y0 = f->dots[i]->y;
3430 if (y1 < f->dots[i]->y) y1 = f->dots[i]->y;
3434 * If the grid is at excessive resolution, decide on a scaling
3435 * factor to bring it within reasonable bounds so we don't have to
3436 * think too hard or suffer integer overflow.
3439 while (x1 - x0 > 128 || y1 - y0 > 128) {
3448 * Now iterate over every point in that bounding box.
3452 for (y = y0; y <= y1; y++) {
3453 for (x = x0; x <= x1; x++) {
3455 * First, disqualify the point if it's not inside the
3456 * polygon, which we work out by counting the edges to the
3457 * right of the point. (For tiebreaking purposes when
3458 * edges start or end on our y-coordinate or go right
3459 * through it, we consider our point to be offset by a
3460 * small _positive_ epsilon in both the x- and
3464 for (i = 0; i < f->order; i++) {
3465 int xs = f->edges[i]->dot1->x >> shift;
3466 int xe = f->edges[i]->dot2->x >> shift;
3467 int ys = f->edges[i]->dot1->y >> shift;
3468 int ye = f->edges[i]->dot2->y >> shift;
3469 if ((y >= ys && y < ye) || (y >= ye && y < ys)) {
3471 * The line goes past our y-position. Now we need
3472 * to know if its x-coordinate when it does so is
3475 * The x-coordinate in question is mathematically
3476 * (y - ys) * (xe - xs) / (ye - ys), and we want
3477 * to know whether (x - xs) >= that. Of course we
3478 * avoid the division, so we can work in integers;
3479 * to do this we must multiply both sides of the
3480 * inequality by ye - ys, which means we must
3481 * first check that's not negative.
3483 int num = xe - xs, denom = ye - ys;
3488 if ((x - xs) * denom >= (y - ys) * num)
3494 long mindist = LONG_MAX;
3497 * This point is inside the polygon, so now we check
3498 * its minimum distance to every edge and corner.
3499 * First the corners ...
3501 for (i = 0; i < f->order; i++) {
3502 int xp = f->dots[i]->x >> shift;
3503 int yp = f->dots[i]->y >> shift;
3504 int dx = x - xp, dy = y - yp;
3505 long dist = (long)dx*dx + (long)dy*dy;
3511 * ... and now also check the perpendicular distance
3512 * to every edge, if the perpendicular lies between
3513 * the edge's endpoints.
3515 for (i = 0; i < f->order; i++) {
3516 int xs = f->edges[i]->dot1->x >> shift;
3517 int xe = f->edges[i]->dot2->x >> shift;
3518 int ys = f->edges[i]->dot1->y >> shift;
3519 int ye = f->edges[i]->dot2->y >> shift;
3522 * If s and e are our endpoints, and p our
3523 * candidate circle centre, the foot of a
3524 * perpendicular from p to the line se lies
3525 * between s and e if and only if (p-s).(e-s) lies
3526 * strictly between 0 and (e-s).(e-s).
3528 int edx = xe - xs, edy = ye - ys;
3529 int pdx = x - xs, pdy = y - ys;
3530 long pde = (long)pdx * edx + (long)pdy * edy;
3531 long ede = (long)edx * edx + (long)edy * edy;
3532 if (0 < pde && pde < ede) {
3534 * Yes, the nearest point on this edge is
3535 * closer than either endpoint, so we must
3536 * take it into account by measuring the
3537 * perpendicular distance to the edge and
3538 * checking its square against mindist.
3541 long pdre = (long)pdx * edy - (long)pdy * edx;
3542 long sqlen = pdre * pdre / ede;
3544 if (mindist > sqlen)
3550 * Right. Now we know the biggest circle around this
3551 * point, so we can check it against bestdist.
3553 if (bestdist < mindist) {
3562 assert(bestdist >= 0);
3564 /* convert to screen coordinates */
3565 grid_to_screen(ds, g, xbest << shift, ybest << shift,
3566 &ds->textx[faceindex], &ds->texty[faceindex]);
3568 *xret = ds->textx[faceindex];
3569 *yret = ds->texty[faceindex];
3572 static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f,
3573 int *x, int *y, int *w, int *h)
3576 face_text_pos(ds, g, f, &xx, &yy);
3578 /* There seems to be a certain amount of trial-and-error involved
3579 * in working out the correct bounding-box for the text. */
3581 *x = xx - ds->tilesize/4 - 1;
3582 *y = yy - ds->tilesize/4 - 3;
3583 *w = ds->tilesize/2 + 2;
3584 *h = ds->tilesize/2 + 5;
3587 static void game_redraw_clue(drawing *dr, game_drawstate *ds,
3588 game_state *state, int i)
3590 grid *g = state->game_grid;
3591 grid_face *f = g->faces + i;
3595 if (state->clues[i] < 10) {
3596 c[0] = CLUE2CHAR(state->clues[i]);
3599 sprintf(c, "%d", state->clues[i]);
3602 face_text_pos(ds, g, f, &x, &y);
3604 FONT_VARIABLE, ds->tilesize/2,
3605 ALIGN_VCENTRE | ALIGN_HCENTRE,
3606 ds->clue_error[i] ? COL_MISTAKE :
3607 ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
3610 static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e,
3611 int *x, int *y, int *w, int *h)
3613 int x1 = e->dot1->x;
3614 int y1 = e->dot1->y;
3615 int x2 = e->dot2->x;
3616 int y2 = e->dot2->y;
3617 int xmin, xmax, ymin, ymax;
3619 grid_to_screen(ds, g, x1, y1, &x1, &y1);
3620 grid_to_screen(ds, g, x2, y2, &x2, &y2);
3621 /* Allow extra margin for dots, and thickness of lines */
3622 xmin = min(x1, x2) - 2;
3623 xmax = max(x1, x2) + 2;
3624 ymin = min(y1, y2) - 2;
3625 ymax = max(y1, y2) + 2;
3629 *w = xmax - xmin + 1;
3630 *h = ymax - ymin + 1;
3633 static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d,
3634 int *x, int *y, int *w, int *h)
3638 grid_to_screen(ds, g, d->x, d->y, &x1, &y1);
3646 static const int loopy_line_redraw_phases[] = {
3647 COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE
3649 #define NPHASES lenof(loopy_line_redraw_phases)
3651 static void game_redraw_line(drawing *dr, game_drawstate *ds,
3652 game_state *state, int i, int phase)
3654 grid *g = state->game_grid;
3655 grid_edge *e = g->edges + i;
3657 int xmin, ymin, xmax, ymax;
3660 if (state->line_errors[i])
3661 line_colour = COL_MISTAKE;
3662 else if (state->lines[i] == LINE_UNKNOWN)
3663 line_colour = COL_LINEUNKNOWN;
3664 else if (state->lines[i] == LINE_NO)
3665 line_colour = COL_FAINT;
3666 else if (ds->flashing)
3667 line_colour = COL_HIGHLIGHT;
3669 line_colour = COL_FOREGROUND;
3670 if (line_colour != loopy_line_redraw_phases[phase])
3673 /* Convert from grid to screen coordinates */
3674 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3675 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3682 if (line_colour == COL_FAINT) {
3683 static int draw_faint_lines = -1;
3684 if (draw_faint_lines < 0) {
3685 char *env = getenv("LOOPY_FAINT_LINES");
3686 draw_faint_lines = (!env || (env[0] == 'y' ||
3689 if (draw_faint_lines)
3690 draw_line(dr, x1, y1, x2, y2, line_colour);
3692 draw_thick_line(dr, 3.0,
3699 static void game_redraw_dot(drawing *dr, game_drawstate *ds,
3700 game_state *state, int i)
3702 grid *g = state->game_grid;
3703 grid_dot *d = g->dots + i;
3706 grid_to_screen(ds, g, d->x, d->y, &x, &y);
3707 draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
3710 static int boxes_intersect(int x0, int y0, int w0, int h0,
3711 int x1, int y1, int w1, int h1)
3714 * Two intervals intersect iff neither is wholly on one side of
3715 * the other. Two boxes intersect iff their horizontal and
3716 * vertical intervals both intersect.
3718 return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0);
3721 static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
3722 game_state *state, int x, int y, int w, int h)
3724 grid *g = state->game_grid;
3728 clip(dr, x, y, w, h);
3729 draw_rect(dr, x, y, w, h, COL_BACKGROUND);
3731 for (i = 0; i < g->num_faces; i++) {
3732 face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh);
3733 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3734 game_redraw_clue(dr, ds, state, i);
3736 for (phase = 0; phase < NPHASES; phase++) {
3737 for (i = 0; i < g->num_edges; i++) {
3738 edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh);
3739 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3740 game_redraw_line(dr, ds, state, i, phase);
3743 for (i = 0; i < g->num_dots; i++) {
3744 dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh);
3745 if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
3746 game_redraw_dot(dr, ds, state, i);
3750 draw_update(dr, x, y, w, h);
3753 static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
3754 game_state *state, int dir, game_ui *ui,
3755 float animtime, float flashtime)
3757 #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
3759 grid *g = state->game_grid;
3760 int border = BORDER(ds->tilesize);
3763 int redraw_everything = FALSE;
3765 int edges[REDRAW_OBJECTS_LIMIT], nedges = 0;
3766 int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0;
3768 /* Redrawing is somewhat involved.
3770 * An update can theoretically affect an arbitrary number of edges
3771 * (consider, for example, completing or breaking a cycle which doesn't
3772 * satisfy all the clues -- we'll switch many edges between error and
3773 * normal states). On the other hand, redrawing the whole grid takes a
3774 * while, making the game feel sluggish, and many updates are actually
3775 * quite well localized.
3777 * This redraw algorithm attempts to cope with both situations gracefully
3778 * and correctly. For localized changes, we set a clip rectangle, fill
3779 * it with background, and then redraw (a plausible but conservative
3780 * guess at) the objects which intersect the rectangle; if several
3781 * objects need redrawing, we'll do them individually. However, if lots
3782 * of objects are affected, we'll just redraw everything.
3784 * The reason for all of this is that it's just not safe to do the redraw
3785 * piecemeal. If you try to draw an antialiased diagonal line over
3786 * itself, you get a slightly thicker antialiased diagonal line, which
3787 * looks rather ugly after a while.
3789 * So, we take two passes over the grid. The first attempts to work out
3790 * what needs doing, and the second actually does it.
3794 redraw_everything = TRUE;
3797 /* First, trundle through the faces. */
3798 for (i = 0; i < g->num_faces; i++) {
3799 grid_face *f = g->faces + i;
3800 int sides = f->order;
3803 int n = state->clues[i];
3807 clue_mistake = (face_order(state, i, LINE_YES) > n ||
3808 face_order(state, i, LINE_NO ) > (sides-n));
3809 clue_satisfied = (face_order(state, i, LINE_YES) == n &&
3810 face_order(state, i, LINE_NO ) == (sides-n));
3812 if (clue_mistake != ds->clue_error[i] ||
3813 clue_satisfied != ds->clue_satisfied[i]) {
3814 ds->clue_error[i] = clue_mistake;
3815 ds->clue_satisfied[i] = clue_satisfied;
3816 if (nfaces == REDRAW_OBJECTS_LIMIT)
3817 redraw_everything = TRUE;
3819 faces[nfaces++] = i;
3823 /* Work out what the flash state needs to be. */
3824 if (flashtime > 0 &&
3825 (flashtime <= FLASH_TIME/3 ||
3826 flashtime >= FLASH_TIME*2/3)) {
3827 flash_changed = !ds->flashing;
3828 ds->flashing = TRUE;
3830 flash_changed = ds->flashing;
3831 ds->flashing = FALSE;
3834 /* Now, trundle through the edges. */
3835 for (i = 0; i < g->num_edges; i++) {
3837 state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
3838 if (new_ds != ds->lines[i] ||
3839 (flash_changed && state->lines[i] == LINE_YES)) {
3840 ds->lines[i] = new_ds;
3841 if (nedges == REDRAW_OBJECTS_LIMIT)
3842 redraw_everything = TRUE;
3844 edges[nedges++] = i;
3849 /* Pass one is now done. Now we do the actual drawing. */
3850 if (redraw_everything) {
3851 int grid_width = g->highest_x - g->lowest_x;
3852 int grid_height = g->highest_y - g->lowest_y;
3853 int w = grid_width * ds->tilesize / g->tilesize;
3854 int h = grid_height * ds->tilesize / g->tilesize;
3856 game_redraw_in_rect(dr, ds, state,
3857 0, 0, w + 2*border + 1, h + 2*border + 1);
3860 /* Right. Now we roll up our sleeves. */
3862 for (i = 0; i < nfaces; i++) {
3863 grid_face *f = g->faces + faces[i];
3866 face_text_bbox(ds, g, f, &x, &y, &w, &h);
3867 game_redraw_in_rect(dr, ds, state, x, y, w, h);
3870 for (i = 0; i < nedges; i++) {
3871 grid_edge *e = g->edges + edges[i];
3874 edge_bbox(ds, g, e, &x, &y, &w, &h);
3875 game_redraw_in_rect(dr, ds, state, x, y, w, h);
3882 static float game_flash_length(game_state *oldstate, game_state *newstate,
3883 int dir, game_ui *ui)
3885 if (!oldstate->solved && newstate->solved &&
3886 !oldstate->cheated && !newstate->cheated) {
3893 static int game_is_solved(game_state *state)
3895 return state->solved;
3898 static void game_print_size(game_params *params, float *x, float *y)
3903 * I'll use 7mm "squares" by default.
3905 game_compute_size(params, 700, &pw, &ph);
3910 static void game_print(drawing *dr, game_state *state, int tilesize)
3912 int ink = print_mono_colour(dr, 0);
3914 game_drawstate ads, *ds = &ads;
3915 grid *g = state->game_grid;
3917 ds->tilesize = tilesize;
3919 for (i = 0; i < g->num_dots; i++) {
3921 grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
3922 draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
3928 for (i = 0; i < g->num_faces; i++) {
3929 grid_face *f = g->faces + i;
3930 int clue = state->clues[i];
3934 c[0] = CLUE2CHAR(clue);
3936 face_text_pos(ds, g, f, &x, &y);
3938 FONT_VARIABLE, ds->tilesize / 2,
3939 ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
3946 for (i = 0; i < g->num_edges; i++) {
3947 int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
3948 grid_edge *e = g->edges + i;
3950 grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
3951 grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
3952 if (state->lines[i] == LINE_YES)
3954 /* (dx, dy) points from (x1, y1) to (x2, y2).
3955 * The line is then "fattened" in a perpendicular
3956 * direction to create a thin rectangle. */
3957 double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
3958 double dx = (x2 - x1) / d;
3959 double dy = (y2 - y1) / d;
3962 dx = (dx * ds->tilesize) / thickness;
3963 dy = (dy * ds->tilesize) / thickness;
3964 points[0] = x1 + (int)dy;
3965 points[1] = y1 - (int)dx;
3966 points[2] = x1 - (int)dy;
3967 points[3] = y1 + (int)dx;
3968 points[4] = x2 - (int)dy;
3969 points[5] = y2 + (int)dx;
3970 points[6] = x2 + (int)dy;
3971 points[7] = y2 - (int)dx;
3972 draw_polygon(dr, points, 4, ink, ink);
3976 /* Draw a dotted line */
3979 for (j = 1; j < divisions; j++) {
3980 /* Weighted average */
3981 int x = (x1 * (divisions -j) + x2 * j) / divisions;
3982 int y = (y1 * (divisions -j) + y2 * j) / divisions;
3983 draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
3990 #define thegame loopy
3993 const struct game thegame = {
3994 "Loopy", "games.loopy", "loopy",
4001 TRUE, game_configure, custom_params,
4009 TRUE, game_can_format_as_text_now, game_text_format,
4017 PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
4020 game_free_drawstate,
4025 TRUE, FALSE, game_print_size, game_print,
4026 FALSE /* wants_statusbar */,
4027 FALSE, game_timing_state,
4028 0, /* mouse_priorities */
4031 #ifdef STANDALONE_SOLVER
4034 * Half-hearted standalone solver. It can't output the solution to
4035 * anything but a square puzzle, and it can't log the deductions
4036 * it makes either. But it can solve square puzzles, and more
4037 * importantly it can use its solver to grade the difficulty of
4038 * any puzzle you give it.
4043 int main(int argc, char **argv)
4047 char *id = NULL, *desc, *err;
4050 #if 0 /* verbose solver not supported here (yet) */
4051 int really_verbose = FALSE;
4054 while (--argc > 0) {
4056 #if 0 /* verbose solver not supported here (yet) */
4057 if (!strcmp(p, "-v")) {
4058 really_verbose = TRUE;
4061 if (!strcmp(p, "-g")) {
4063 } else if (*p == '-') {
4064 fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
4072 fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
4076 desc = strchr(id, ':');
4078 fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
4083 p = default_params();
4084 decode_params(p, id);
4085 err = validate_desc(p, desc);
4087 fprintf(stderr, "%s: %s\n", argv[0], err);
4090 s = new_game(NULL, p, desc);
4093 * When solving an Easy puzzle, we don't want to bother the
4094 * user with Hard-level deductions. For this reason, we grade
4095 * the puzzle internally before doing anything else.
4097 ret = -1; /* placate optimiser */
4098 for (diff = 0; diff < DIFF_MAX; diff++) {
4099 solver_state *sstate_new;
4100 solver_state *sstate = new_solver_state((game_state *)s, diff);
4102 sstate_new = solve_game_rec(sstate);
4104 if (sstate_new->solver_status == SOLVER_MISTAKE)
4106 else if (sstate_new->solver_status == SOLVER_SOLVED)
4111 free_solver_state(sstate_new);
4112 free_solver_state(sstate);
4118 if (diff == DIFF_MAX) {
4120 printf("Difficulty rating: harder than Hard, or ambiguous\n");
4122 printf("Unable to find a unique solution\n");
4126 printf("Difficulty rating: impossible (no solution exists)\n");
4128 printf("Difficulty rating: %s\n", diffnames[diff]);
4130 solver_state *sstate_new;
4131 solver_state *sstate = new_solver_state((game_state *)s, diff);
4133 /* If we supported a verbose solver, we'd set verbosity here */
4135 sstate_new = solve_game_rec(sstate);
4137 if (sstate_new->solver_status == SOLVER_MISTAKE)
4138 printf("Puzzle is inconsistent\n");
4140 assert(sstate_new->solver_status == SOLVER_SOLVED);
4141 if (s->grid_type == 0) {
4142 fputs(game_text_format(sstate_new->state), stdout);
4144 printf("Unable to output non-square grids\n");
4148 free_solver_state(sstate_new);
4149 free_solver_state(sstate);