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
111 /* Put -1 in a face that doesn't get a clue */
114 /* Array of line states, to store whether each line is
115 * YES, NO or UNKNOWN */
118 unsigned char *line_errors
;
123 /* Used in game_text_format(), so that it knows what type of
124 * grid it's trying to render as ASCII text. */
129 SOLVER_SOLVED
, /* This is the only solution the solver could find */
130 SOLVER_MISTAKE
, /* This is definitely not a solution */
131 SOLVER_AMBIGUOUS
, /* This _might_ be an ambiguous solution */
132 SOLVER_INCOMPLETE
/* This may be a partial solution */
135 /* ------ Solver state ------ */
136 typedef struct solver_state
{
138 enum solver_status solver_status
;
139 /* NB looplen is the number of dots that are joined together at a point, ie a
140 * looplen of 1 means there are no lines to a particular dot */
143 /* Difficulty level of solver. Used by solver functions that want to
144 * vary their behaviour depending on the requested difficulty level. */
150 char *face_yes_count
;
152 char *dot_solved
, *face_solved
;
155 /* Information for Normal level deductions:
156 * For each dline, store a bitmask for whether we know:
157 * (bit 0) at least one is YES
158 * (bit 1) at most one is YES */
161 /* Hard level information */
166 * Difficulty levels. I do some macro ickery here to ensure that my
167 * enum and the various forms of my name list always match up.
170 #define DIFFLIST(A) \
175 #define ENUM(upper,title,lower) DIFF_ ## upper,
176 #define TITLE(upper,title,lower) #title,
177 #define ENCODE(upper,title,lower) #lower
178 #define CONFIG(upper,title,lower) ":" #title
179 enum { DIFFLIST(ENUM
) DIFF_MAX
};
180 static char const *const diffnames
[] = { DIFFLIST(TITLE
) };
181 static char const diffchars
[] = DIFFLIST(ENCODE
);
182 #define DIFFCONFIG DIFFLIST(CONFIG)
185 * Solver routines, sorted roughly in order of computational cost.
186 * The solver will run the faster deductions first, and slower deductions are
187 * only invoked when the faster deductions are unable to make progress.
188 * Each function is associated with a difficulty level, so that the generated
189 * puzzles are solvable by applying only the functions with the chosen
190 * difficulty level or lower.
192 #define SOLVERLIST(A) \
193 A(trivial_deductions, DIFF_EASY) \
194 A(dline_deductions, DIFF_NORMAL) \
195 A(linedsf_deductions, DIFF_HARD) \
196 A(loop_deductions, DIFF_EASY)
197 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
198 #define SOLVER_FN(fn,diff) &fn,
199 #define SOLVER_DIFF(fn,diff) diff,
200 SOLVERLIST(SOLVER_FN_DECL
)
201 static int (*(solver_fns
[]))(solver_state
*) = { SOLVERLIST(SOLVER_FN
) };
202 static int const solver_diffs
[] = { SOLVERLIST(SOLVER_DIFF
) };
203 const int NUM_SOLVERS
= sizeof(solver_diffs
)/sizeof(*solver_diffs
);
210 /* Grid generation is expensive, so keep a (ref-counted) reference to the
211 * grid for these parameters, and only generate when required. */
215 /* line_drawstate is the same as line_state, but with the extra ERROR
216 * possibility. The drawing code copies line_state to line_drawstate,
217 * except in the case that the line is an error. */
218 enum line_state
{ LINE_YES
, LINE_UNKNOWN
, LINE_NO
};
219 enum line_drawstate
{ DS_LINE_YES
, DS_LINE_UNKNOWN
,
220 DS_LINE_NO
, DS_LINE_ERROR
};
222 #define OPP(line_state) \
226 struct game_drawstate
{
232 char *clue_satisfied
;
235 static char *validate_desc(game_params
*params
, char *desc
);
236 static int dot_order(const game_state
* state
, int i
, char line_type
);
237 static int face_order(const game_state
* state
, int i
, char line_type
);
238 static solver_state
*solve_game_rec(const solver_state
*sstate
);
241 static void check_caches(const solver_state
* sstate
);
243 #define check_caches(s)
246 /* ------- List of grid generators ------- */
247 #define GRIDLIST(A) \
248 A(Squares,grid_new_square,3,3) \
249 A(Triangular,grid_new_triangular,3,3) \
250 A(Honeycomb,grid_new_honeycomb,3,3) \
251 A(Snub-Square,grid_new_snubsquare,3,3) \
252 A(Cairo,grid_new_cairo,3,4) \
253 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
254 A(Octagonal,grid_new_octagonal,3,3) \
255 A(Kites,grid_new_kites,3,3)
257 #define GRID_NAME(title,fn,amin,omin) #title,
258 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
259 #define GRID_FN(title,fn,amin,omin) &fn,
260 #define GRID_SIZES(title,fn,amin,omin) \
262 "Width and height for this grid type must both be at least " #amin, \
263 "At least one of width and height for this grid type must be at least " #omin,},
264 static char const *const gridnames
[] = { GRIDLIST(GRID_NAME
) };
265 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
266 static grid
* (*(grid_fns
[]))(int w
, int h
) = { GRIDLIST(GRID_FN
) };
267 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
268 static const struct {
271 } grid_size_limits
[] = { GRIDLIST(GRID_SIZES
) };
273 /* Generates a (dynamically allocated) new grid, according to the
274 * type and size requested in params. Does nothing if the grid is already
275 * generated. The allocated grid is owned by the params object, and will be
276 * freed in free_params(). */
277 static void params_generate_grid(game_params
*params
)
279 if (!params
->game_grid
) {
280 params
->game_grid
= grid_fns
[params
->type
](params
->w
, params
->h
);
284 /* ----------------------------------------------------------------------
288 /* General constants */
289 #define PREFERRED_TILE_SIZE 32
290 #define BORDER(tilesize) ((tilesize) / 2)
291 #define FLASH_TIME 0.5F
293 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
295 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
296 ((field) |= (1<<(bit)), TRUE))
298 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
299 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
301 #define CLUE2CHAR(c) \
302 ((c < 0) ? ' ' : c + '0')
304 /* ----------------------------------------------------------------------
305 * General struct manipulation and other straightforward code
308 static game_state
*dup_game(game_state
*state
)
310 game_state
*ret
= snew(game_state
);
312 ret
->game_grid
= state
->game_grid
;
313 ret
->game_grid
->refcount
++;
315 ret
->solved
= state
->solved
;
316 ret
->cheated
= state
->cheated
;
318 ret
->clues
= snewn(state
->game_grid
->num_faces
, signed char);
319 memcpy(ret
->clues
, state
->clues
, state
->game_grid
->num_faces
);
321 ret
->lines
= snewn(state
->game_grid
->num_edges
, char);
322 memcpy(ret
->lines
, state
->lines
, state
->game_grid
->num_edges
);
324 ret
->line_errors
= snewn(state
->game_grid
->num_edges
, unsigned char);
325 memcpy(ret
->line_errors
, state
->line_errors
, state
->game_grid
->num_edges
);
327 ret
->grid_type
= state
->grid_type
;
331 static void free_game(game_state
*state
)
334 grid_free(state
->game_grid
);
337 sfree(state
->line_errors
);
342 static solver_state
*new_solver_state(game_state
*state
, int diff
) {
344 int num_dots
= state
->game_grid
->num_dots
;
345 int num_faces
= state
->game_grid
->num_faces
;
346 int num_edges
= state
->game_grid
->num_edges
;
347 solver_state
*ret
= snew(solver_state
);
349 ret
->state
= dup_game(state
);
351 ret
->solver_status
= SOLVER_INCOMPLETE
;
354 ret
->dotdsf
= snew_dsf(num_dots
);
355 ret
->looplen
= snewn(num_dots
, int);
357 for (i
= 0; i
< num_dots
; i
++) {
361 ret
->dot_solved
= snewn(num_dots
, char);
362 ret
->face_solved
= snewn(num_faces
, char);
363 memset(ret
->dot_solved
, FALSE
, num_dots
);
364 memset(ret
->face_solved
, FALSE
, num_faces
);
366 ret
->dot_yes_count
= snewn(num_dots
, char);
367 memset(ret
->dot_yes_count
, 0, num_dots
);
368 ret
->dot_no_count
= snewn(num_dots
, char);
369 memset(ret
->dot_no_count
, 0, num_dots
);
370 ret
->face_yes_count
= snewn(num_faces
, char);
371 memset(ret
->face_yes_count
, 0, num_faces
);
372 ret
->face_no_count
= snewn(num_faces
, char);
373 memset(ret
->face_no_count
, 0, num_faces
);
375 if (diff
< DIFF_NORMAL
) {
378 ret
->dlines
= snewn(2*num_edges
, char);
379 memset(ret
->dlines
, 0, 2*num_edges
);
382 if (diff
< DIFF_HARD
) {
385 ret
->linedsf
= snew_dsf(state
->game_grid
->num_edges
);
391 static void free_solver_state(solver_state
*sstate
) {
393 free_game(sstate
->state
);
394 sfree(sstate
->dotdsf
);
395 sfree(sstate
->looplen
);
396 sfree(sstate
->dot_solved
);
397 sfree(sstate
->face_solved
);
398 sfree(sstate
->dot_yes_count
);
399 sfree(sstate
->dot_no_count
);
400 sfree(sstate
->face_yes_count
);
401 sfree(sstate
->face_no_count
);
403 /* OK, because sfree(NULL) is a no-op */
404 sfree(sstate
->dlines
);
405 sfree(sstate
->linedsf
);
411 static solver_state
*dup_solver_state(const solver_state
*sstate
) {
412 game_state
*state
= sstate
->state
;
413 int num_dots
= state
->game_grid
->num_dots
;
414 int num_faces
= state
->game_grid
->num_faces
;
415 int num_edges
= state
->game_grid
->num_edges
;
416 solver_state
*ret
= snew(solver_state
);
418 ret
->state
= state
= dup_game(sstate
->state
);
420 ret
->solver_status
= sstate
->solver_status
;
421 ret
->diff
= sstate
->diff
;
423 ret
->dotdsf
= snewn(num_dots
, int);
424 ret
->looplen
= snewn(num_dots
, int);
425 memcpy(ret
->dotdsf
, sstate
->dotdsf
,
426 num_dots
* sizeof(int));
427 memcpy(ret
->looplen
, sstate
->looplen
,
428 num_dots
* sizeof(int));
430 ret
->dot_solved
= snewn(num_dots
, char);
431 ret
->face_solved
= snewn(num_faces
, char);
432 memcpy(ret
->dot_solved
, sstate
->dot_solved
, num_dots
);
433 memcpy(ret
->face_solved
, sstate
->face_solved
, num_faces
);
435 ret
->dot_yes_count
= snewn(num_dots
, char);
436 memcpy(ret
->dot_yes_count
, sstate
->dot_yes_count
, num_dots
);
437 ret
->dot_no_count
= snewn(num_dots
, char);
438 memcpy(ret
->dot_no_count
, sstate
->dot_no_count
, num_dots
);
440 ret
->face_yes_count
= snewn(num_faces
, char);
441 memcpy(ret
->face_yes_count
, sstate
->face_yes_count
, num_faces
);
442 ret
->face_no_count
= snewn(num_faces
, char);
443 memcpy(ret
->face_no_count
, sstate
->face_no_count
, num_faces
);
445 if (sstate
->dlines
) {
446 ret
->dlines
= snewn(2*num_edges
, char);
447 memcpy(ret
->dlines
, sstate
->dlines
,
453 if (sstate
->linedsf
) {
454 ret
->linedsf
= snewn(num_edges
, int);
455 memcpy(ret
->linedsf
, sstate
->linedsf
,
456 num_edges
* sizeof(int));
464 static game_params
*default_params(void)
466 game_params
*ret
= snew(game_params
);
475 ret
->diff
= DIFF_EASY
;
478 ret
->game_grid
= NULL
;
483 static game_params
*dup_params(game_params
*params
)
485 game_params
*ret
= snew(game_params
);
487 *ret
= *params
; /* structure copy */
488 if (ret
->game_grid
) {
489 ret
->game_grid
->refcount
++;
494 static const game_params presets
[] = {
496 { 7, 7, DIFF_EASY
, 0, NULL
},
497 { 7, 7, DIFF_NORMAL
, 0, NULL
},
498 { 7, 7, DIFF_HARD
, 0, NULL
},
499 { 7, 7, DIFF_HARD
, 1, NULL
},
500 { 7, 7, DIFF_HARD
, 2, NULL
},
501 { 5, 5, DIFF_HARD
, 3, NULL
},
502 { 7, 7, DIFF_HARD
, 4, NULL
},
503 { 5, 4, DIFF_HARD
, 5, NULL
},
504 { 5, 5, DIFF_HARD
, 6, NULL
},
505 { 5, 5, DIFF_HARD
, 7, NULL
},
507 { 7, 7, DIFF_EASY
, 0, NULL
},
508 { 10, 10, DIFF_EASY
, 0, NULL
},
509 { 7, 7, DIFF_NORMAL
, 0, NULL
},
510 { 10, 10, DIFF_NORMAL
, 0, NULL
},
511 { 7, 7, DIFF_HARD
, 0, NULL
},
512 { 10, 10, DIFF_HARD
, 0, NULL
},
513 { 10, 10, DIFF_HARD
, 1, NULL
},
514 { 12, 10, DIFF_HARD
, 2, NULL
},
515 { 7, 7, DIFF_HARD
, 3, NULL
},
516 { 9, 9, DIFF_HARD
, 4, NULL
},
517 { 5, 4, DIFF_HARD
, 5, NULL
},
518 { 7, 7, DIFF_HARD
, 6, NULL
},
519 { 5, 5, DIFF_HARD
, 7, NULL
},
523 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
528 if (i
< 0 || i
>= lenof(presets
))
531 tmppar
= snew(game_params
);
532 *tmppar
= presets
[i
];
534 sprintf(buf
, "%dx%d %s - %s", tmppar
->h
, tmppar
->w
,
535 gridnames
[tmppar
->type
], diffnames
[tmppar
->diff
]);
541 static void free_params(game_params
*params
)
543 if (params
->game_grid
) {
544 grid_free(params
->game_grid
);
549 static void decode_params(game_params
*params
, char const *string
)
551 if (params
->game_grid
) {
552 grid_free(params
->game_grid
);
553 params
->game_grid
= NULL
;
555 params
->h
= params
->w
= atoi(string
);
556 params
->diff
= DIFF_EASY
;
557 while (*string
&& isdigit((unsigned char)*string
)) string
++;
558 if (*string
== 'x') {
560 params
->h
= atoi(string
);
561 while (*string
&& isdigit((unsigned char)*string
)) string
++;
563 if (*string
== 't') {
565 params
->type
= atoi(string
);
566 while (*string
&& isdigit((unsigned char)*string
)) string
++;
568 if (*string
== 'd') {
571 for (i
= 0; i
< DIFF_MAX
; i
++)
572 if (*string
== diffchars
[i
])
574 if (*string
) string
++;
578 static char *encode_params(game_params
*params
, int full
)
581 sprintf(str
, "%dx%dt%d", params
->w
, params
->h
, params
->type
);
583 sprintf(str
+ strlen(str
), "d%c", diffchars
[params
->diff
]);
587 static config_item
*game_configure(game_params
*params
)
592 ret
= snewn(5, config_item
);
594 ret
[0].name
= "Width";
595 ret
[0].type
= C_STRING
;
596 sprintf(buf
, "%d", params
->w
);
597 ret
[0].sval
= dupstr(buf
);
600 ret
[1].name
= "Height";
601 ret
[1].type
= C_STRING
;
602 sprintf(buf
, "%d", params
->h
);
603 ret
[1].sval
= dupstr(buf
);
606 ret
[2].name
= "Grid type";
607 ret
[2].type
= C_CHOICES
;
608 ret
[2].sval
= GRID_CONFIGS
;
609 ret
[2].ival
= params
->type
;
611 ret
[3].name
= "Difficulty";
612 ret
[3].type
= C_CHOICES
;
613 ret
[3].sval
= DIFFCONFIG
;
614 ret
[3].ival
= params
->diff
;
624 static game_params
*custom_params(config_item
*cfg
)
626 game_params
*ret
= snew(game_params
);
628 ret
->w
= atoi(cfg
[0].sval
);
629 ret
->h
= atoi(cfg
[1].sval
);
630 ret
->type
= cfg
[2].ival
;
631 ret
->diff
= cfg
[3].ival
;
633 ret
->game_grid
= NULL
;
637 static char *validate_params(game_params
*params
, int full
)
639 if (params
->type
< 0 || params
->type
>= NUM_GRID_TYPES
)
640 return "Illegal grid type";
641 if (params
->w
< grid_size_limits
[params
->type
].amin
||
642 params
->h
< grid_size_limits
[params
->type
].amin
)
643 return grid_size_limits
[params
->type
].aerr
;
644 if (params
->w
< grid_size_limits
[params
->type
].omin
&&
645 params
->h
< grid_size_limits
[params
->type
].omin
)
646 return grid_size_limits
[params
->type
].oerr
;
649 * This shouldn't be able to happen at all, since decode_params
650 * and custom_params will never generate anything that isn't
653 assert(params
->diff
< DIFF_MAX
);
658 /* Returns a newly allocated string describing the current puzzle */
659 static char *state_to_text(const game_state
*state
)
661 grid
*g
= state
->game_grid
;
663 int num_faces
= g
->num_faces
;
664 char *description
= snewn(num_faces
+ 1, char);
665 char *dp
= description
;
669 for (i
= 0; i
< num_faces
; i
++) {
670 if (state
->clues
[i
] < 0) {
671 if (empty_count
> 25) {
672 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
678 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
681 dp
+= sprintf(dp
, "%c", (int)CLUE2CHAR(state
->clues
[i
]));
686 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
688 retval
= dupstr(description
);
694 /* We require that the params pass the test in validate_params and that the
695 * description fills the entire game area */
696 static char *validate_desc(game_params
*params
, char *desc
)
700 params_generate_grid(params
);
701 g
= params
->game_grid
;
703 for (; *desc
; ++desc
) {
704 if (*desc
>= '0' && *desc
<= '9') {
709 count
+= *desc
- 'a' + 1;
712 return "Unknown character in description";
715 if (count
< g
->num_faces
)
716 return "Description too short for board size";
717 if (count
> g
->num_faces
)
718 return "Description too long for board size";
723 /* Sums the lengths of the numbers in range [0,n) */
724 /* See equivalent function in solo.c for justification of this. */
725 static int len_0_to_n(int n
)
727 int len
= 1; /* Counting 0 as a bit of a special case */
730 for (i
= 1; i
< n
; i
*= 10) {
731 len
+= max(n
- i
, 0);
737 static char *encode_solve_move(const game_state
*state
)
742 int num_edges
= state
->game_grid
->num_edges
;
744 /* This is going to return a string representing the moves needed to set
745 * every line in a grid to be the same as the ones in 'state'. The exact
746 * length of this string is predictable. */
748 len
= 1; /* Count the 'S' prefix */
749 /* Numbers in all lines */
750 len
+= len_0_to_n(num_edges
);
751 /* For each line we also have a letter */
754 ret
= snewn(len
+ 1, char);
757 p
+= sprintf(p
, "S");
759 for (i
= 0; i
< num_edges
; i
++) {
760 switch (state
->lines
[i
]) {
762 p
+= sprintf(p
, "%dy", i
);
765 p
+= sprintf(p
, "%dn", i
);
770 /* No point in doing sums like that if they're going to be wrong */
771 assert(strlen(ret
) <= (size_t)len
);
775 static game_ui
*new_ui(game_state
*state
)
780 static void free_ui(game_ui
*ui
)
784 static char *encode_ui(game_ui
*ui
)
789 static void decode_ui(game_ui
*ui
, char *encoding
)
793 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
794 game_state
*newstate
)
798 static void game_compute_size(game_params
*params
, int tilesize
,
802 int grid_width
, grid_height
, rendered_width
, rendered_height
;
804 params_generate_grid(params
);
805 g
= params
->game_grid
;
806 grid_width
= g
->highest_x
- g
->lowest_x
;
807 grid_height
= g
->highest_y
- g
->lowest_y
;
808 /* multiply first to minimise rounding error on integer division */
809 rendered_width
= grid_width
* tilesize
/ g
->tilesize
;
810 rendered_height
= grid_height
* tilesize
/ g
->tilesize
;
811 *x
= rendered_width
+ 2 * BORDER(tilesize
) + 1;
812 *y
= rendered_height
+ 2 * BORDER(tilesize
) + 1;
815 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
816 game_params
*params
, int tilesize
)
818 ds
->tilesize
= tilesize
;
821 static float *game_colours(frontend
*fe
, int *ncolours
)
823 float *ret
= snewn(4 * NCOLOURS
, float);
825 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
827 ret
[COL_FOREGROUND
* 3 + 0] = 0.0F
;
828 ret
[COL_FOREGROUND
* 3 + 1] = 0.0F
;
829 ret
[COL_FOREGROUND
* 3 + 2] = 0.0F
;
831 ret
[COL_LINEUNKNOWN
* 3 + 0] = 0.8F
;
832 ret
[COL_LINEUNKNOWN
* 3 + 1] = 0.8F
;
833 ret
[COL_LINEUNKNOWN
* 3 + 2] = 0.0F
;
835 ret
[COL_HIGHLIGHT
* 3 + 0] = 1.0F
;
836 ret
[COL_HIGHLIGHT
* 3 + 1] = 1.0F
;
837 ret
[COL_HIGHLIGHT
* 3 + 2] = 1.0F
;
839 ret
[COL_MISTAKE
* 3 + 0] = 1.0F
;
840 ret
[COL_MISTAKE
* 3 + 1] = 0.0F
;
841 ret
[COL_MISTAKE
* 3 + 2] = 0.0F
;
843 ret
[COL_SATISFIED
* 3 + 0] = 0.0F
;
844 ret
[COL_SATISFIED
* 3 + 1] = 0.0F
;
845 ret
[COL_SATISFIED
* 3 + 2] = 0.0F
;
847 *ncolours
= NCOLOURS
;
851 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
853 struct game_drawstate
*ds
= snew(struct game_drawstate
);
854 int num_faces
= state
->game_grid
->num_faces
;
855 int num_edges
= state
->game_grid
->num_edges
;
859 ds
->lines
= snewn(num_edges
, char);
860 ds
->clue_error
= snewn(num_faces
, char);
861 ds
->clue_satisfied
= snewn(num_faces
, char);
864 memset(ds
->lines
, LINE_UNKNOWN
, num_edges
);
865 memset(ds
->clue_error
, 0, num_faces
);
866 memset(ds
->clue_satisfied
, 0, num_faces
);
871 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
873 sfree(ds
->clue_error
);
874 sfree(ds
->clue_satisfied
);
879 static int game_timing_state(game_state
*state
, game_ui
*ui
)
884 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
885 int dir
, game_ui
*ui
)
890 static int game_can_format_as_text_now(game_params
*params
)
892 if (params
->type
!= 0)
897 static char *game_text_format(game_state
*state
)
903 grid
*g
= state
->game_grid
;
906 assert(state
->grid_type
== 0);
908 /* Work out the basic size unit */
909 f
= g
->faces
; /* first face */
910 assert(f
->order
== 4);
911 /* The dots are ordered clockwise, so the two opposite
912 * corners are guaranteed to span the square */
913 cell_size
= abs(f
->dots
[0]->x
- f
->dots
[2]->x
);
915 w
= (g
->highest_x
- g
->lowest_x
) / cell_size
;
916 h
= (g
->highest_y
- g
->lowest_y
) / cell_size
;
918 /* Create a blank "canvas" to "draw" on */
921 ret
= snewn(W
* H
+ 1, char);
922 for (y
= 0; y
< H
; y
++) {
923 for (x
= 0; x
< W
-1; x
++) {
926 ret
[y
*W
+ W
-1] = '\n';
930 /* Fill in edge info */
931 for (i
= 0; i
< g
->num_edges
; i
++) {
932 grid_edge
*e
= g
->edges
+ i
;
933 /* Cell coordinates, from (0,0) to (w-1,h-1) */
934 int x1
= (e
->dot1
->x
- g
->lowest_x
) / cell_size
;
935 int x2
= (e
->dot2
->x
- g
->lowest_x
) / cell_size
;
936 int y1
= (e
->dot1
->y
- g
->lowest_y
) / cell_size
;
937 int y2
= (e
->dot2
->y
- g
->lowest_y
) / cell_size
;
938 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
939 * cell coordinates) */
942 switch (state
->lines
[i
]) {
944 ret
[y
*W
+ x
] = (y1
== y2
) ?
'-' : '|';
950 break; /* already a space */
952 assert(!"Illegal line state");
957 for (i
= 0; i
< g
->num_faces
; i
++) {
961 assert(f
->order
== 4);
962 /* Cell coordinates, from (0,0) to (w-1,h-1) */
963 x1
= (f
->dots
[0]->x
- g
->lowest_x
) / cell_size
;
964 x2
= (f
->dots
[2]->x
- g
->lowest_x
) / cell_size
;
965 y1
= (f
->dots
[0]->y
- g
->lowest_y
) / cell_size
;
966 y2
= (f
->dots
[2]->y
- g
->lowest_y
) / cell_size
;
967 /* Midpoint, in canvas coordinates */
970 ret
[y
*W
+ x
] = CLUE2CHAR(state
->clues
[i
]);
975 /* ----------------------------------------------------------------------
980 static void check_caches(const solver_state
* sstate
)
983 const game_state
*state
= sstate
->state
;
984 const grid
*g
= state
->game_grid
;
986 for (i
= 0; i
< g
->num_dots
; i
++) {
987 assert(dot_order(state
, i
, LINE_YES
) == sstate
->dot_yes_count
[i
]);
988 assert(dot_order(state
, i
, LINE_NO
) == sstate
->dot_no_count
[i
]);
991 for (i
= 0; i
< g
->num_faces
; i
++) {
992 assert(face_order(state
, i
, LINE_YES
) == sstate
->face_yes_count
[i
]);
993 assert(face_order(state
, i
, LINE_NO
) == sstate
->face_no_count
[i
]);
998 #define check_caches(s) \
1000 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1004 #endif /* DEBUG_CACHES */
1006 /* ----------------------------------------------------------------------
1007 * Solver utility functions
1010 /* Sets the line (with index i) to the new state 'line_new', and updates
1011 * the cached counts of any affected faces and dots.
1012 * Returns TRUE if this actually changed the line's state. */
1013 static int solver_set_line(solver_state
*sstate
, int i
,
1014 enum line_state line_new
1016 , const char *reason
1020 game_state
*state
= sstate
->state
;
1024 assert(line_new
!= LINE_UNKNOWN
);
1026 check_caches(sstate
);
1028 if (state
->lines
[i
] == line_new
) {
1029 return FALSE
; /* nothing changed */
1031 state
->lines
[i
] = line_new
;
1034 fprintf(stderr
, "solver: set line [%d] to %s (%s)\n",
1035 i
, line_new
== LINE_YES ?
"YES" : "NO",
1039 g
= state
->game_grid
;
1042 /* Update the cache for both dots and both faces affected by this. */
1043 if (line_new
== LINE_YES
) {
1044 sstate
->dot_yes_count
[e
->dot1
- g
->dots
]++;
1045 sstate
->dot_yes_count
[e
->dot2
- g
->dots
]++;
1047 sstate
->face_yes_count
[e
->face1
- g
->faces
]++;
1050 sstate
->face_yes_count
[e
->face2
- g
->faces
]++;
1053 sstate
->dot_no_count
[e
->dot1
- g
->dots
]++;
1054 sstate
->dot_no_count
[e
->dot2
- g
->dots
]++;
1056 sstate
->face_no_count
[e
->face1
- g
->faces
]++;
1059 sstate
->face_no_count
[e
->face2
- g
->faces
]++;
1063 check_caches(sstate
);
1068 #define solver_set_line(a, b, c) \
1069 solver_set_line(a, b, c, __FUNCTION__)
1073 * Merge two dots due to the existence of an edge between them.
1074 * Updates the dsf tracking equivalence classes, and keeps track of
1075 * the length of path each dot is currently a part of.
1076 * Returns TRUE if the dots were already linked, ie if they are part of a
1077 * closed loop, and false otherwise.
1079 static int merge_dots(solver_state
*sstate
, int edge_index
)
1082 grid
*g
= sstate
->state
->game_grid
;
1083 grid_edge
*e
= g
->edges
+ edge_index
;
1085 i
= e
->dot1
- g
->dots
;
1086 j
= e
->dot2
- g
->dots
;
1088 i
= dsf_canonify(sstate
->dotdsf
, i
);
1089 j
= dsf_canonify(sstate
->dotdsf
, j
);
1094 len
= sstate
->looplen
[i
] + sstate
->looplen
[j
];
1095 dsf_merge(sstate
->dotdsf
, i
, j
);
1096 i
= dsf_canonify(sstate
->dotdsf
, i
);
1097 sstate
->looplen
[i
] = len
;
1102 /* Merge two lines because the solver has deduced that they must be either
1103 * identical or opposite. Returns TRUE if this is new information, otherwise
1105 static int merge_lines(solver_state
*sstate
, int i
, int j
, int inverse
1107 , const char *reason
1113 assert(i
< sstate
->state
->game_grid
->num_edges
);
1114 assert(j
< sstate
->state
->game_grid
->num_edges
);
1116 i
= edsf_canonify(sstate
->linedsf
, i
, &inv_tmp
);
1118 j
= edsf_canonify(sstate
->linedsf
, j
, &inv_tmp
);
1121 edsf_merge(sstate
->linedsf
, i
, j
, inverse
);
1125 fprintf(stderr
, "%s [%d] [%d] %s(%s)\n",
1127 inverse ?
"inverse " : "", reason
);
1134 #define merge_lines(a, b, c, d) \
1135 merge_lines(a, b, c, d, __FUNCTION__)
1138 /* Count the number of lines of a particular type currently going into the
1140 static int dot_order(const game_state
* state
, int dot
, char line_type
)
1143 grid
*g
= state
->game_grid
;
1144 grid_dot
*d
= g
->dots
+ dot
;
1147 for (i
= 0; i
< d
->order
; i
++) {
1148 grid_edge
*e
= d
->edges
[i
];
1149 if (state
->lines
[e
- g
->edges
] == line_type
)
1155 /* Count the number of lines of a particular type currently surrounding the
1157 static int face_order(const game_state
* state
, int face
, char line_type
)
1160 grid
*g
= state
->game_grid
;
1161 grid_face
*f
= g
->faces
+ face
;
1164 for (i
= 0; i
< f
->order
; i
++) {
1165 grid_edge
*e
= f
->edges
[i
];
1166 if (state
->lines
[e
- g
->edges
] == line_type
)
1172 /* Set all lines bordering a dot of type old_type to type new_type
1173 * Return value tells caller whether this function actually did anything */
1174 static int dot_setall(solver_state
*sstate
, int dot
,
1175 char old_type
, char new_type
)
1177 int retval
= FALSE
, r
;
1178 game_state
*state
= sstate
->state
;
1183 if (old_type
== new_type
)
1186 g
= state
->game_grid
;
1189 for (i
= 0; i
< d
->order
; i
++) {
1190 int line_index
= d
->edges
[i
] - g
->edges
;
1191 if (state
->lines
[line_index
] == old_type
) {
1192 r
= solver_set_line(sstate
, line_index
, new_type
);
1200 /* Set all lines bordering a face of type old_type to type new_type */
1201 static int face_setall(solver_state
*sstate
, int face
,
1202 char old_type
, char new_type
)
1204 int retval
= FALSE
, r
;
1205 game_state
*state
= sstate
->state
;
1210 if (old_type
== new_type
)
1213 g
= state
->game_grid
;
1214 f
= g
->faces
+ face
;
1216 for (i
= 0; i
< f
->order
; i
++) {
1217 int line_index
= f
->edges
[i
] - g
->edges
;
1218 if (state
->lines
[line_index
] == old_type
) {
1219 r
= solver_set_line(sstate
, line_index
, new_type
);
1227 /* ----------------------------------------------------------------------
1228 * Loop generation and clue removal
1231 /* We're going to store lists of current candidate faces for colouring black
1233 * Each face gets a 'score', which tells us how adding that face right
1234 * now would affect the curliness of the solution loop. We're trying to
1235 * maximise that quantity so will bias our random selection of faces to
1236 * colour those with high scores */
1240 unsigned long random
;
1241 /* No need to store a grid_face* here. The 'face_scores' array will
1242 * be a list of 'face_score' objects, one for each face of the grid, so
1243 * the position (index) within the 'face_scores' array will determine
1244 * which face corresponds to a particular face_score.
1245 * Having a single 'face_scores' array for all faces simplifies memory
1246 * management, and probably improves performance, because we don't have to
1247 * malloc/free each individual face_score, and we don't have to maintain
1248 * a mapping from grid_face* pointers to face_score* pointers.
1252 static int generic_sort_cmpfn(void *v1
, void *v2
, size_t offset
)
1254 struct face_score
*f1
= v1
;
1255 struct face_score
*f2
= v2
;
1258 r
= *(int *)((char *)f2
+ offset
) - *(int *)((char *)f1
+ offset
);
1263 if (f1
->random
< f2
->random
)
1265 else if (f1
->random
> f2
->random
)
1269 * It's _just_ possible that two faces might have been given
1270 * the same random value. In that situation, fall back to
1271 * comparing based on the positions within the face_scores list.
1272 * This introduces a tiny directional bias, but not a significant one.
1277 static int white_sort_cmpfn(void *v1
, void *v2
)
1279 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,white_score
));
1282 static int black_sort_cmpfn(void *v1
, void *v2
)
1284 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,black_score
));
1287 enum face_colour
{ FACE_WHITE
, FACE_GREY
, FACE_BLACK
};
1289 /* face should be of type grid_face* here. */
1290 #define FACE_COLOUR(face) \
1291 ( (face) == NULL ? FACE_BLACK : \
1292 board[(face) - g->faces] )
1294 /* 'board' is an array of these enums, indicating which faces are
1295 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1296 * Returns whether it's legal to colour the given face with this colour. */
1297 static int can_colour_face(grid
*g
, char* board
, int face_index
,
1298 enum face_colour colour
)
1301 grid_face
*test_face
= g
->faces
+ face_index
;
1302 grid_face
*starting_face
, *current_face
;
1304 int current_state
, s
; /* booleans: equal or not-equal to 'colour' */
1305 int found_same_coloured_neighbour
= FALSE
;
1306 assert(board
[face_index
] != colour
);
1308 /* Can only consider a face for colouring if it's adjacent to a face
1309 * with the same colour. */
1310 for (i
= 0; i
< test_face
->order
; i
++) {
1311 grid_edge
*e
= test_face
->edges
[i
];
1312 grid_face
*f
= (e
->face1
== test_face
) ? e
->face2
: e
->face1
;
1313 if (FACE_COLOUR(f
) == colour
) {
1314 found_same_coloured_neighbour
= TRUE
;
1318 if (!found_same_coloured_neighbour
)
1321 /* Need to avoid creating a loop of faces of this colour around some
1322 * differently-coloured faces.
1323 * Also need to avoid meeting a same-coloured face at a corner, with
1324 * other-coloured faces in between. Here's a simple test that (I believe)
1325 * takes care of both these conditions:
1327 * Take the circular path formed by this face's edges, and inflate it
1328 * slightly outwards. Imagine walking around this path and consider
1329 * the faces that you visit in sequence. This will include all faces
1330 * touching the given face, either along an edge or just at a corner.
1331 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1332 * you walk along the complete loop. This will obviously turn out to be
1334 * If 0, we're either in the middle of an "island" of this colour (should
1335 * be impossible as we're not supposed to create black or white loops),
1336 * or we're about to start a new island - also not allowed.
1337 * If 4 or greater, there are too many separate coloured regions touching
1338 * this face, and colouring it would create a loop or a corner-violation.
1339 * The only allowed case is when the count is exactly 2. */
1341 /* i points to a dot around the test face.
1342 * j points to a face around the i^th dot.
1343 * The current face will always be:
1344 * test_face->dots[i]->faces[j]
1345 * We assume dots go clockwise around the test face,
1346 * and faces go clockwise around dots. */
1348 starting_face
= test_face
->dots
[0]->faces
[0];
1349 if (starting_face
== test_face
) {
1351 starting_face
= test_face
->dots
[0]->faces
[1];
1353 current_face
= starting_face
;
1355 current_state
= (FACE_COLOUR(current_face
) == colour
);
1358 /* Advance to next face.
1359 * Need to loop here because it might take several goes to
1363 if (j
== test_face
->dots
[i
]->order
)
1366 if (test_face
->dots
[i
]->faces
[j
] == test_face
) {
1367 /* Advance to next dot round test_face, then
1368 * find current_face around new dot
1369 * and advance to the next face clockwise */
1371 if (i
== test_face
->order
)
1373 for (j
= 0; j
< test_face
->dots
[i
]->order
; j
++) {
1374 if (test_face
->dots
[i
]->faces
[j
] == current_face
)
1377 /* Must actually find current_face around new dot,
1378 * or else something's wrong with the grid. */
1379 assert(j
!= test_face
->dots
[i
]->order
);
1380 /* Found, so advance to next face and try again */
1385 /* (i,j) are now advanced to next face */
1386 current_face
= test_face
->dots
[i
]->faces
[j
];
1387 s
= (FACE_COLOUR(current_face
) == colour
);
1388 if (s
!= current_state
) {
1391 if (transitions
> 2)
1392 return FALSE
; /* no point in continuing */
1394 } while (current_face
!= starting_face
);
1396 return (transitions
== 2) ? TRUE
: FALSE
;
1399 /* Count the number of neighbours of 'face', having colour 'colour' */
1400 static int face_num_neighbours(grid
*g
, char *board
, grid_face
*face
,
1401 enum face_colour colour
)
1403 int colour_count
= 0;
1407 for (i
= 0; i
< face
->order
; i
++) {
1409 f
= (e
->face1
== face
) ? e
->face2
: e
->face1
;
1410 if (FACE_COLOUR(f
) == colour
)
1413 return colour_count
;
1416 /* The 'score' of a face reflects its current desirability for selection
1417 * as the next face to colour white or black. We want to encourage moving
1418 * into grey areas and increasing loopiness, so we give scores according to
1419 * how many of the face's neighbours are currently coloured the same as the
1420 * proposed colour. */
1421 static int face_score(grid
*g
, char *board
, grid_face
*face
,
1422 enum face_colour colour
)
1424 /* Simple formula: score = 0 - num. same-coloured neighbours,
1425 * so a higher score means fewer same-coloured neighbours. */
1426 return -face_num_neighbours(g
, board
, face
, colour
);
1429 /* Generate a new complete set of clues for the given game_state.
1430 * The method is to generate a WHITE/BLACK colouring of all the faces,
1431 * such that the WHITE faces will define the inside of the path, and the
1432 * BLACK faces define the outside.
1433 * To do this, we initially colour all faces GREY. The infinite space outside
1434 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1435 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1436 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1437 * we avoid creating loops of a single colour, to preserve the topological
1438 * shape of the WHITE and BLACK regions.
1439 * We also try to make the boundary as loopy and twisty as possible, to avoid
1440 * generating paths that are uninteresting.
1441 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1442 * face that can be coloured with that colour (without violating the
1443 * topological shape of that region). It's not obvious, but I think this
1444 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1445 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1446 * regions can be grown.
1447 * This is checked using assert()ions, and I haven't seen any failures yet.
1449 * Hand-wavy proof: imagine what can go wrong...
1451 * Could the white faces get completely cut off by the black faces, and still
1452 * leave some grey faces remaining?
1453 * No, because then the black faces would form a loop around both the white
1454 * faces and the grey faces, which is disallowed because we continually
1455 * maintain the correct topological shape of the black region.
1456 * Similarly, the black faces can never get cut off by the white faces. That
1457 * means both the WHITE and BLACK regions always have some room to grow into
1459 * Could it be that we can't colour some GREY face, because there are too many
1460 * WHITE/BLACK transitions as we walk round the face? (see the
1461 * can_colour_face() function for details)
1462 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1463 * around the face. The two WHITE faces would be connected by a WHITE path,
1464 * and the BLACK faces would be connected by a BLACK path. These paths would
1465 * have to cross, which is impossible.
1466 * Another thing that could go wrong: perhaps we can't find any GREY face to
1467 * colour WHITE, because it would create a loop-violation or a corner-violation
1468 * with the other WHITE faces?
1469 * This is a little bit tricky to prove impossible. Imagine you have such a
1470 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1471 * or corner violation).
1472 * That would cut all the non-white area into two blobs. One of those blobs
1473 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1474 * So we have a connected GREY area, completely surrounded by WHITE
1475 * (including the GREY face we've tentatively coloured WHITE).
1476 * A well-known result in graph theory says that you can always find a GREY
1477 * face whose removal leaves the remaining GREY area connected. And it says
1478 * there are at least two such faces, so we can always choose the one that
1479 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1480 * everything nice and connected, including that "tentative" GREY face which
1481 * acts as a gateway to the rest of the non-WHITE grid.
1483 static void add_full_clues(game_state
*state
, random_state
*rs
)
1485 signed char *clues
= state
->clues
;
1487 grid
*g
= state
->game_grid
;
1489 int num_faces
= g
->num_faces
;
1490 struct face_score
*face_scores
; /* Array of face_score objects */
1491 struct face_score
*fs
; /* Points somewhere in the above list */
1492 struct grid_face
*cur_face
;
1493 tree234
*lightable_faces_sorted
;
1494 tree234
*darkable_faces_sorted
;
1498 board
= snewn(num_faces
, char);
1501 memset(board
, FACE_GREY
, num_faces
);
1503 /* Create and initialise the list of face_scores */
1504 face_scores
= snewn(num_faces
, struct face_score
);
1505 for (i
= 0; i
< num_faces
; i
++) {
1506 face_scores
[i
].random
= random_bits(rs
, 31);
1509 /* Colour a random, finite face white. The infinite face is implicitly
1510 * coloured black. Together, they will seed the random growth process
1511 * for the black and white areas. */
1512 i
= random_upto(rs
, num_faces
);
1513 board
[i
] = FACE_WHITE
;
1515 /* We need a way of favouring faces that will increase our loopiness.
1516 * We do this by maintaining a list of all candidate faces sorted by
1517 * their score and choose randomly from that with appropriate skew.
1518 * In order to avoid consistently biasing towards particular faces, we
1519 * need the sort order _within_ each group of scores to be completely
1520 * random. But it would be abusing the hospitality of the tree234 data
1521 * structure if our comparison function were nondeterministic :-). So with
1522 * each face we associate a random number that does not change during a
1523 * particular run of the generator, and use that as a secondary sort key.
1524 * Yes, this means we will be biased towards particular random faces in
1525 * any one run but that doesn't actually matter. */
1527 lightable_faces_sorted
= newtree234(white_sort_cmpfn
);
1528 darkable_faces_sorted
= newtree234(black_sort_cmpfn
);
1530 /* Initialise the lists of lightable and darkable faces. This is
1531 * slightly different from the code inside the while-loop, because we need
1532 * to check every face of the board (the grid structure does not keep a
1533 * list of the infinite face's neighbours). */
1534 for (i
= 0; i
< num_faces
; i
++) {
1535 grid_face
*f
= g
->faces
+ i
;
1536 struct face_score
*fs
= face_scores
+ i
;
1537 if (board
[i
] != FACE_GREY
) continue;
1538 /* We need the full colourability check here, it's not enough simply
1539 * to check neighbourhood. On some grids, a neighbour of the infinite
1540 * face is not necessarily darkable. */
1541 if (can_colour_face(g
, board
, i
, FACE_BLACK
)) {
1542 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1543 add234(darkable_faces_sorted
, fs
);
1545 if (can_colour_face(g
, board
, i
, FACE_WHITE
)) {
1546 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1547 add234(lightable_faces_sorted
, fs
);
1551 /* Colour faces one at a time until no more faces are colourable. */
1554 enum face_colour colour
;
1555 struct face_score
*fs_white
, *fs_black
;
1556 int c_lightable
= count234(lightable_faces_sorted
);
1557 int c_darkable
= count234(darkable_faces_sorted
);
1558 if (c_lightable
== 0) {
1559 /* No more lightable faces. Because of how the algorithm
1560 * works, there should be no more darkable faces either. */
1561 assert(c_darkable
== 0);
1565 fs_white
= (struct face_score
*)index234(lightable_faces_sorted
, 0);
1566 fs_black
= (struct face_score
*)index234(darkable_faces_sorted
, 0);
1568 /* Choose a colour, and colour the best available face
1569 * with that colour. */
1570 colour
= random_upto(rs
, 2) ? FACE_WHITE
: FACE_BLACK
;
1572 if (colour
== FACE_WHITE
)
1577 i
= fs
- face_scores
;
1578 assert(board
[i
] == FACE_GREY
);
1581 /* Remove this newly-coloured face from the lists. These lists should
1582 * only contain grey faces. */
1583 del234(lightable_faces_sorted
, fs
);
1584 del234(darkable_faces_sorted
, fs
);
1586 /* Remember which face we've just coloured */
1587 cur_face
= g
->faces
+ i
;
1589 /* The face we've just coloured potentially affects the colourability
1590 * and the scores of any neighbouring faces (touching at a corner or
1591 * edge). So the search needs to be conducted around all faces
1592 * touching the one we've just lit. Iterate over its corners, then
1593 * over each corner's faces. For each such face, we remove it from
1594 * the lists, recalculate any scores, then add it back to the lists
1595 * (depending on whether it is lightable, darkable or both). */
1596 for (i
= 0; i
< cur_face
->order
; i
++) {
1597 grid_dot
*d
= cur_face
->dots
[i
];
1598 for (j
= 0; j
< d
->order
; j
++) {
1599 grid_face
*f
= d
->faces
[j
];
1600 int fi
; /* face index of f */
1607 /* If the face is already coloured, it won't be on our
1608 * lightable/darkable lists anyway, so we can skip it without
1609 * bothering with the removal step. */
1610 if (FACE_COLOUR(f
) != FACE_GREY
) continue;
1612 /* Find the face index and face_score* corresponding to f */
1614 fs
= face_scores
+ fi
;
1616 /* Remove from lightable list if it's in there. We do this,
1617 * even if it is still lightable, because the score might
1618 * be different, and we need to remove-then-add to maintain
1619 * correct sort order. */
1620 del234(lightable_faces_sorted
, fs
);
1621 if (can_colour_face(g
, board
, fi
, FACE_WHITE
)) {
1622 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1623 add234(lightable_faces_sorted
, fs
);
1625 /* Do the same for darkable list. */
1626 del234(darkable_faces_sorted
, fs
);
1627 if (can_colour_face(g
, board
, fi
, FACE_BLACK
)) {
1628 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1629 add234(darkable_faces_sorted
, fs
);
1636 freetree234(lightable_faces_sorted
);
1637 freetree234(darkable_faces_sorted
);
1640 /* The next step requires a shuffled list of all faces */
1641 face_list
= snewn(num_faces
, int);
1642 for (i
= 0; i
< num_faces
; ++i
) {
1645 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1647 /* The above loop-generation algorithm can often leave large clumps
1648 * of faces of one colour. In extreme cases, the resulting path can be
1649 * degenerate and not very satisfying to solve.
1650 * This next step alleviates this problem:
1651 * Go through the shuffled list, and flip the colour of any face we can
1652 * legally flip, and which is adjacent to only one face of the opposite
1653 * colour - this tends to grow 'tendrils' into any clumps.
1654 * Repeat until we can find no more faces to flip. This will
1655 * eventually terminate, because each flip increases the loop's
1656 * perimeter, which cannot increase for ever.
1657 * The resulting path will have maximal loopiness (in the sense that it
1658 * cannot be improved "locally". Unfortunately, this allows a player to
1659 * make some illicit deductions. To combat this (and make the path more
1660 * interesting), we do one final pass making random flips. */
1662 /* Set to TRUE for final pass */
1663 do_random_pass
= FALSE
;
1666 /* Remember whether a flip occurred during this pass */
1667 int flipped
= FALSE
;
1669 for (i
= 0; i
< num_faces
; ++i
) {
1670 int j
= face_list
[i
];
1671 enum face_colour opp
=
1672 (board
[j
] == FACE_WHITE
) ? FACE_BLACK
: FACE_WHITE
;
1673 if (can_colour_face(g
, board
, j
, opp
)) {
1674 grid_face
*face
= g
->faces
+j
;
1675 if (do_random_pass
) {
1676 /* final random pass */
1677 if (!random_upto(rs
, 10))
1680 /* normal pass - flip when neighbour count is 1 */
1681 if (face_num_neighbours(g
, board
, face
, opp
) == 1) {
1689 if (do_random_pass
) break;
1690 if (!flipped
) do_random_pass
= TRUE
;
1695 /* Fill out all the clues by initialising to 0, then iterating over
1696 * all edges and incrementing each clue as we find edges that border
1697 * between BLACK/WHITE faces. While we're at it, we verify that the
1698 * algorithm does work, and there aren't any GREY faces still there. */
1699 memset(clues
, 0, num_faces
);
1700 for (i
= 0; i
< g
->num_edges
; i
++) {
1701 grid_edge
*e
= g
->edges
+ i
;
1702 grid_face
*f1
= e
->face1
;
1703 grid_face
*f2
= e
->face2
;
1704 enum face_colour c1
= FACE_COLOUR(f1
);
1705 enum face_colour c2
= FACE_COLOUR(f2
);
1706 assert(c1
!= FACE_GREY
);
1707 assert(c2
!= FACE_GREY
);
1709 if (f1
) clues
[f1
- g
->faces
]++;
1710 if (f2
) clues
[f2
- g
->faces
]++;
1718 static int game_has_unique_soln(const game_state
*state
, int diff
)
1721 solver_state
*sstate_new
;
1722 solver_state
*sstate
= new_solver_state((game_state
*)state
, diff
);
1724 sstate_new
= solve_game_rec(sstate
);
1726 assert(sstate_new
->solver_status
!= SOLVER_MISTAKE
);
1727 ret
= (sstate_new
->solver_status
== SOLVER_SOLVED
);
1729 free_solver_state(sstate_new
);
1730 free_solver_state(sstate
);
1736 /* Remove clues one at a time at random. */
1737 static game_state
*remove_clues(game_state
*state
, random_state
*rs
,
1741 int num_faces
= state
->game_grid
->num_faces
;
1742 game_state
*ret
= dup_game(state
), *saved_ret
;
1745 /* We need to remove some clues. We'll do this by forming a list of all
1746 * available clues, shuffling it, then going along one at a
1747 * time clearing each clue in turn for which doing so doesn't render the
1748 * board unsolvable. */
1749 face_list
= snewn(num_faces
, int);
1750 for (n
= 0; n
< num_faces
; ++n
) {
1754 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1756 for (n
= 0; n
< num_faces
; ++n
) {
1757 saved_ret
= dup_game(ret
);
1758 ret
->clues
[face_list
[n
]] = -1;
1760 if (game_has_unique_soln(ret
, diff
)) {
1761 free_game(saved_ret
);
1773 static char *new_game_desc(game_params
*params
, random_state
*rs
,
1774 char **aux
, int interactive
)
1776 /* solution and description both use run-length encoding in obvious ways */
1779 game_state
*state
= snew(game_state
);
1780 game_state
*state_new
;
1781 params_generate_grid(params
);
1782 state
->game_grid
= g
= params
->game_grid
;
1784 state
->clues
= snewn(g
->num_faces
, signed char);
1785 state
->lines
= snewn(g
->num_edges
, char);
1786 state
->line_errors
= snewn(g
->num_edges
, unsigned char);
1788 state
->grid_type
= params
->type
;
1792 memset(state
->lines
, LINE_UNKNOWN
, g
->num_edges
);
1793 memset(state
->line_errors
, 0, g
->num_edges
);
1795 state
->solved
= state
->cheated
= FALSE
;
1797 /* Get a new random solvable board with all its clues filled in. Yes, this
1798 * can loop for ever if the params are suitably unfavourable, but
1799 * preventing games smaller than 4x4 seems to stop this happening */
1801 add_full_clues(state
, rs
);
1802 } while (!game_has_unique_soln(state
, params
->diff
));
1804 state_new
= remove_clues(state
, rs
, params
->diff
);
1809 if (params
->diff
> 0 && game_has_unique_soln(state
, params
->diff
-1)) {
1811 fprintf(stderr
, "Rejecting board, it is too easy\n");
1813 goto newboard_please
;
1816 retval
= state_to_text(state
);
1820 assert(!validate_desc(params
, retval
));
1825 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1828 game_state
*state
= snew(game_state
);
1829 int empties_to_make
= 0;
1831 const char *dp
= desc
;
1833 int num_faces
, num_edges
;
1835 params_generate_grid(params
);
1836 state
->game_grid
= g
= params
->game_grid
;
1838 num_faces
= g
->num_faces
;
1839 num_edges
= g
->num_edges
;
1841 state
->clues
= snewn(num_faces
, signed char);
1842 state
->lines
= snewn(num_edges
, char);
1843 state
->line_errors
= snewn(num_edges
, unsigned char);
1845 state
->solved
= state
->cheated
= FALSE
;
1847 state
->grid_type
= params
->type
;
1849 for (i
= 0; i
< num_faces
; i
++) {
1850 if (empties_to_make
) {
1852 state
->clues
[i
] = -1;
1858 if (n
>= 0 && n
< 10) {
1859 state
->clues
[i
] = n
;
1863 state
->clues
[i
] = -1;
1864 empties_to_make
= n
- 1;
1869 memset(state
->lines
, LINE_UNKNOWN
, num_edges
);
1870 memset(state
->line_errors
, 0, num_edges
);
1874 /* Calculates the line_errors data, and checks if the current state is a
1876 static int check_completion(game_state
*state
)
1878 grid
*g
= state
->game_grid
;
1880 int num_faces
= g
->num_faces
;
1882 int infinite_area
, finite_area
;
1883 int loops_found
= 0;
1884 int found_edge_not_in_loop
= FALSE
;
1886 memset(state
->line_errors
, 0, g
->num_edges
);
1888 /* LL implementation of SGT's idea:
1889 * A loop will partition the grid into an inside and an outside.
1890 * If there is more than one loop, the grid will be partitioned into
1891 * even more distinct regions. We can therefore track equivalence of
1892 * faces, by saying that two faces are equivalent when there is a non-YES
1893 * edge between them.
1894 * We could keep track of the number of connected components, by counting
1895 * the number of dsf-merges that aren't no-ops.
1896 * But we're only interested in 3 separate cases:
1897 * no loops, one loop, more than one loop.
1899 * No loops: all faces are equivalent to the infinite face.
1900 * One loop: only two equivalence classes - finite and infinite.
1901 * >= 2 loops: there are 2 distinct finite regions.
1903 * So we simply make two passes through all the edges.
1904 * In the first pass, we dsf-merge the two faces bordering each non-YES
1906 * In the second pass, we look for YES-edges bordering:
1907 * a) two non-equivalent faces.
1908 * b) two non-equivalent faces, and one of them is part of a different
1909 * finite area from the first finite area we've seen.
1911 * An occurrence of a) means there is at least one loop.
1912 * An occurrence of b) means there is more than one loop.
1913 * Edges satisfying a) are marked as errors.
1915 * While we're at it, we set a flag if we find a YES edge that is not
1917 * This information will help decide, if there's a single loop, whether it
1918 * is a candidate for being a solution (that is, all YES edges are part of
1921 * If there is a candidate loop, we then go through all clues and check
1922 * they are all satisfied. If so, we have found a solution and we can
1923 * unmark all line_errors.
1926 /* Infinite face is at the end - its index is num_faces.
1927 * This macro is just to make this obvious! */
1928 #define INF_FACE num_faces
1929 dsf
= snewn(num_faces
+ 1, int);
1930 dsf_init(dsf
, num_faces
+ 1);
1933 for (i
= 0; i
< g
->num_edges
; i
++) {
1934 grid_edge
*e
= g
->edges
+ i
;
1935 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1936 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1937 if (state
->lines
[i
] != LINE_YES
)
1938 dsf_merge(dsf
, f1
, f2
);
1942 infinite_area
= dsf_canonify(dsf
, INF_FACE
);
1944 for (i
= 0; i
< g
->num_edges
; i
++) {
1945 grid_edge
*e
= g
->edges
+ i
;
1946 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1947 int can1
= dsf_canonify(dsf
, f1
);
1948 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1949 int can2
= dsf_canonify(dsf
, f2
);
1950 if (state
->lines
[i
] != LINE_YES
) continue;
1953 /* Faces are equivalent, so this edge not part of a loop */
1954 found_edge_not_in_loop
= TRUE
;
1957 state
->line_errors
[i
] = TRUE
;
1958 if (loops_found
== 0) loops_found
= 1;
1960 /* Don't bother with further checks if we've already found 2 loops */
1961 if (loops_found
== 2) continue;
1963 if (finite_area
== -1) {
1964 /* Found our first finite area */
1965 if (can1
!= infinite_area
)
1971 /* Have we found a second area? */
1972 if (finite_area
!= -1) {
1973 if (can1
!= infinite_area
&& can1
!= finite_area
) {
1977 if (can2
!= infinite_area
&& can2
!= finite_area
) {
1984 printf("loops_found = %d\n", loops_found);
1985 printf("found_edge_not_in_loop = %s\n",
1986 found_edge_not_in_loop ? "TRUE" : "FALSE");
1989 sfree(dsf
); /* No longer need the dsf */
1991 /* Have we found a candidate loop? */
1992 if (loops_found
== 1 && !found_edge_not_in_loop
) {
1993 /* Yes, so check all clues are satisfied */
1994 int found_clue_violation
= FALSE
;
1995 for (i
= 0; i
< num_faces
; i
++) {
1996 int c
= state
->clues
[i
];
1998 if (face_order(state
, i
, LINE_YES
) != c
) {
1999 found_clue_violation
= TRUE
;
2005 if (!found_clue_violation
) {
2006 /* The loop is good */
2007 memset(state
->line_errors
, 0, g
->num_edges
);
2008 return TRUE
; /* No need to bother checking for dot violations */
2012 /* Check for dot violations */
2013 for (i
= 0; i
< g
->num_dots
; i
++) {
2014 int yes
= dot_order(state
, i
, LINE_YES
);
2015 int unknown
= dot_order(state
, i
, LINE_UNKNOWN
);
2016 if ((yes
== 1 && unknown
== 0) || (yes
>= 3)) {
2017 /* violation, so mark all YES edges as errors */
2018 grid_dot
*d
= g
->dots
+ i
;
2020 for (j
= 0; j
< d
->order
; j
++) {
2021 int e
= d
->edges
[j
] - g
->edges
;
2022 if (state
->lines
[e
] == LINE_YES
)
2023 state
->line_errors
[e
] = TRUE
;
2030 /* ----------------------------------------------------------------------
2033 * Our solver modes operate as follows. Each mode also uses the modes above it.
2036 * Just implement the rules of the game.
2038 * Normal and Tricky Modes
2039 * For each (adjacent) pair of lines through each dot we store a bit for
2040 * whether at least one of them is on and whether at most one is on. (If we
2041 * know both or neither is on that's already stored more directly.)
2044 * Use edsf data structure to make equivalence classes of lines that are
2045 * known identical to or opposite to one another.
2050 * For general grids, we consider "dlines" to be pairs of lines joined
2051 * at a dot. The lines must be adjacent around the dot, so we can think of
2052 * a dline as being a dot+face combination. Or, a dot+edge combination where
2053 * the second edge is taken to be the next clockwise edge from the dot.
2054 * Original loopy code didn't have this extra restriction of the lines being
2055 * adjacent. From my tests with square grids, this extra restriction seems to
2056 * take little, if anything, away from the quality of the puzzles.
2057 * A dline can be uniquely identified by an edge/dot combination, given that
2058 * a dline-pair always goes clockwise around its common dot. The edge/dot
2059 * combination can be represented by an edge/bool combination - if bool is
2060 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2061 * exactly twice the number of edges in the grid - although the dlines
2062 * spanning the infinite face are not all that useful to the solver.
2063 * Note that, by convention, a dline goes clockwise around its common dot,
2064 * which means the dline goes anti-clockwise around its common face.
2067 /* Helper functions for obtaining an index into an array of dlines, given
2068 * various information. We assume the grid layout conventions about how
2069 * the various lists are interleaved - see grid_make_consistent() for
2072 /* i points to the first edge of the dline pair, reading clockwise around
2074 static int dline_index_from_dot(grid
*g
, grid_dot
*d
, int i
)
2076 grid_edge
*e
= d
->edges
[i
];
2081 if (i2
== d
->order
) i2
= 0;
2084 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2086 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2087 (int)(d
- g
->dots
), i
, (int)(e
- g
->edges
),
2088 (int)(e2
- g
->edges
), ret
);
2092 /* i points to the second edge of the dline pair, reading clockwise around
2093 * the face. That is, the edges of the dline, starting at edge{i}, read
2094 * anti-clockwise around the face. By layout conventions, the common dot
2095 * of the dline will be f->dots[i] */
2096 static int dline_index_from_face(grid
*g
, grid_face
*f
, int i
)
2098 grid_edge
*e
= f
->edges
[i
];
2099 grid_dot
*d
= f
->dots
[i
];
2104 if (i2
< 0) i2
+= f
->order
;
2107 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2109 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2110 (int)(f
- g
->faces
), i
, (int)(e
- g
->edges
),
2111 (int)(e2
- g
->edges
), ret
);
2115 static int is_atleastone(const char *dline_array
, int index
)
2117 return BIT_SET(dline_array
[index
], 0);
2119 static int set_atleastone(char *dline_array
, int index
)
2121 return SET_BIT(dline_array
[index
], 0);
2123 static int is_atmostone(const char *dline_array
, int index
)
2125 return BIT_SET(dline_array
[index
], 1);
2127 static int set_atmostone(char *dline_array
, int index
)
2129 return SET_BIT(dline_array
[index
], 1);
2132 static void array_setall(char *array
, char from
, char to
, int len
)
2134 char *p
= array
, *p_old
= p
;
2135 int len_remaining
= len
;
2137 while ((p
= memchr(p
, from
, len_remaining
))) {
2139 len_remaining
-= p
- p_old
;
2144 /* Helper, called when doing dline dot deductions, in the case where we
2145 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2146 * them (because of dline atmostone/atleastone).
2147 * On entry, edge points to the first of these two UNKNOWNs. This function
2148 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2149 * and set their corresponding dline to atleastone. (Setting atmostone
2150 * already happens in earlier dline deductions) */
2151 static int dline_set_opp_atleastone(solver_state
*sstate
,
2152 grid_dot
*d
, int edge
)
2154 game_state
*state
= sstate
->state
;
2155 grid
*g
= state
->game_grid
;
2158 for (opp
= 0; opp
< N
; opp
++) {
2159 int opp_dline_index
;
2160 if (opp
== edge
|| opp
== edge
+1 || opp
== edge
-1)
2162 if (opp
== 0 && edge
== N
-1)
2164 if (opp
== N
-1 && edge
== 0)
2167 if (opp2
== N
) opp2
= 0;
2168 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2169 if (state
->lines
[d
->edges
[opp
] - g
->edges
] != LINE_UNKNOWN
)
2171 if (state
->lines
[d
->edges
[opp2
] - g
->edges
] != LINE_UNKNOWN
)
2173 /* Found opposite UNKNOWNS and they're next to each other */
2174 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2175 return set_atleastone(sstate
->dlines
, opp_dline_index
);
2181 /* Set pairs of lines around this face which are known to be identical, to
2182 * the given line_state */
2183 static int face_setall_identical(solver_state
*sstate
, int face_index
,
2184 enum line_state line_new
)
2186 /* can[dir] contains the canonical line associated with the line in
2187 * direction dir from the square in question. Similarly inv[dir] is
2188 * whether or not the line in question is inverse to its canonical
2191 game_state
*state
= sstate
->state
;
2192 grid
*g
= state
->game_grid
;
2193 grid_face
*f
= g
->faces
+ face_index
;
2196 int can1
, can2
, inv1
, inv2
;
2198 for (i
= 0; i
< N
; i
++) {
2199 int line1_index
= f
->edges
[i
] - g
->edges
;
2200 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2202 for (j
= i
+ 1; j
< N
; j
++) {
2203 int line2_index
= f
->edges
[j
] - g
->edges
;
2204 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2207 /* Found two UNKNOWNS */
2208 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2209 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2210 if (can1
== can2
&& inv1
== inv2
) {
2211 solver_set_line(sstate
, line1_index
, line_new
);
2212 solver_set_line(sstate
, line2_index
, line_new
);
2219 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2220 * return the edge indices into e. */
2221 static void find_unknowns(game_state
*state
,
2222 grid_edge
**edge_list
, /* Edge list to search (from a face or a dot) */
2223 int expected_count
, /* Number of UNKNOWNs (comes from solver's cache) */
2224 int *e
/* Returned edge indices */)
2227 grid
*g
= state
->game_grid
;
2228 while (c
< expected_count
) {
2229 int line_index
= *edge_list
- g
->edges
;
2230 if (state
->lines
[line_index
] == LINE_UNKNOWN
) {
2238 /* If we have a list of edges, and we know whether the number of YESs should
2239 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2240 * linedsf deductions. This can be used for both face and dot deductions.
2241 * Returns the difficulty level of the next solver that should be used,
2242 * or DIFF_MAX if no progress was made. */
2243 static int parity_deductions(solver_state
*sstate
,
2244 grid_edge
**edge_list
, /* Edge list (from a face or a dot) */
2245 int total_parity
, /* Expected number of YESs modulo 2 (either 0 or 1) */
2248 game_state
*state
= sstate
->state
;
2249 int diff
= DIFF_MAX
;
2250 int *linedsf
= sstate
->linedsf
;
2252 if (unknown_count
== 2) {
2253 /* Lines are known alike/opposite, depending on inv. */
2255 find_unknowns(state
, edge_list
, 2, e
);
2256 if (merge_lines(sstate
, e
[0], e
[1], total_parity
))
2257 diff
= min(diff
, DIFF_HARD
);
2258 } else if (unknown_count
== 3) {
2260 int can
[3]; /* canonical edges */
2261 int inv
[3]; /* whether can[x] is inverse to e[x] */
2262 find_unknowns(state
, edge_list
, 3, e
);
2263 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2264 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2265 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2266 if (can
[0] == can
[1]) {
2267 if (solver_set_line(sstate
, e
[2], (total_parity
^inv
[0]^inv
[1]) ?
2268 LINE_YES
: LINE_NO
))
2269 diff
= min(diff
, DIFF_EASY
);
2271 if (can
[0] == can
[2]) {
2272 if (solver_set_line(sstate
, e
[1], (total_parity
^inv
[0]^inv
[2]) ?
2273 LINE_YES
: LINE_NO
))
2274 diff
= min(diff
, DIFF_EASY
);
2276 if (can
[1] == can
[2]) {
2277 if (solver_set_line(sstate
, e
[0], (total_parity
^inv
[1]^inv
[2]) ?
2278 LINE_YES
: LINE_NO
))
2279 diff
= min(diff
, DIFF_EASY
);
2281 } else if (unknown_count
== 4) {
2283 int can
[4]; /* canonical edges */
2284 int inv
[4]; /* whether can[x] is inverse to e[x] */
2285 find_unknowns(state
, edge_list
, 4, e
);
2286 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2287 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2288 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2289 can
[3] = edsf_canonify(linedsf
, e
[3], inv
+3);
2290 if (can
[0] == can
[1]) {
2291 if (merge_lines(sstate
, e
[2], e
[3], total_parity
^inv
[0]^inv
[1]))
2292 diff
= min(diff
, DIFF_HARD
);
2293 } else if (can
[0] == can
[2]) {
2294 if (merge_lines(sstate
, e
[1], e
[3], total_parity
^inv
[0]^inv
[2]))
2295 diff
= min(diff
, DIFF_HARD
);
2296 } else if (can
[0] == can
[3]) {
2297 if (merge_lines(sstate
, e
[1], e
[2], total_parity
^inv
[0]^inv
[3]))
2298 diff
= min(diff
, DIFF_HARD
);
2299 } else if (can
[1] == can
[2]) {
2300 if (merge_lines(sstate
, e
[0], e
[3], total_parity
^inv
[1]^inv
[2]))
2301 diff
= min(diff
, DIFF_HARD
);
2302 } else if (can
[1] == can
[3]) {
2303 if (merge_lines(sstate
, e
[0], e
[2], total_parity
^inv
[1]^inv
[3]))
2304 diff
= min(diff
, DIFF_HARD
);
2305 } else if (can
[2] == can
[3]) {
2306 if (merge_lines(sstate
, e
[0], e
[1], total_parity
^inv
[2]^inv
[3]))
2307 diff
= min(diff
, DIFF_HARD
);
2315 * These are the main solver functions.
2317 * Their return values are diff values corresponding to the lowest mode solver
2318 * that would notice the work that they have done. For example if the normal
2319 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2320 * easy mode solver might be able to make progress using that. It doesn't make
2321 * sense for one of them to return a diff value higher than that of the
2324 * Each function returns the lowest value it can, as early as possible, in
2325 * order to try and pass as much work as possible back to the lower level
2326 * solvers which progress more quickly.
2329 /* PROPOSED NEW DESIGN:
2330 * We have a work queue consisting of 'events' notifying us that something has
2331 * happened that a particular solver mode might be interested in. For example
2332 * the hard mode solver might do something that helps the normal mode solver at
2333 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2334 * we pull events off the work queue, and hand each in turn to the solver that
2335 * is interested in them. If a solver reports that it failed we pass the same
2336 * event on to progressively more advanced solvers and the loop detector. Once
2337 * we've exhausted an event, or it has helped us progress, we drop it and
2338 * continue to the next one. The events are sorted first in order of solver
2339 * complexity (easy first) then order of insertion (oldest first).
2340 * Once we run out of events we loop over each permitted solver in turn
2341 * (easiest first) until either a deduction is made (and an event therefore
2342 * emerges) or no further deductions can be made (in which case we've failed).
2345 * * How do we 'loop over' a solver when both dots and squares are concerned.
2346 * Answer: first all squares then all dots.
2349 static int trivial_deductions(solver_state
*sstate
)
2351 int i
, current_yes
, current_no
;
2352 game_state
*state
= sstate
->state
;
2353 grid
*g
= state
->game_grid
;
2354 int diff
= DIFF_MAX
;
2356 /* Per-face deductions */
2357 for (i
= 0; i
< g
->num_faces
; i
++) {
2358 grid_face
*f
= g
->faces
+ i
;
2360 if (sstate
->face_solved
[i
])
2363 current_yes
= sstate
->face_yes_count
[i
];
2364 current_no
= sstate
->face_no_count
[i
];
2366 if (current_yes
+ current_no
== f
->order
) {
2367 sstate
->face_solved
[i
] = TRUE
;
2371 if (state
->clues
[i
] < 0)
2374 if (state
->clues
[i
] < current_yes
) {
2375 sstate
->solver_status
= SOLVER_MISTAKE
;
2378 if (state
->clues
[i
] == current_yes
) {
2379 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
))
2380 diff
= min(diff
, DIFF_EASY
);
2381 sstate
->face_solved
[i
] = TRUE
;
2385 if (f
->order
- state
->clues
[i
] < current_no
) {
2386 sstate
->solver_status
= SOLVER_MISTAKE
;
2389 if (f
->order
- state
->clues
[i
] == current_no
) {
2390 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
))
2391 diff
= min(diff
, DIFF_EASY
);
2392 sstate
->face_solved
[i
] = TRUE
;
2397 check_caches(sstate
);
2399 /* Per-dot deductions */
2400 for (i
= 0; i
< g
->num_dots
; i
++) {
2401 grid_dot
*d
= g
->dots
+ i
;
2402 int yes
, no
, unknown
;
2404 if (sstate
->dot_solved
[i
])
2407 yes
= sstate
->dot_yes_count
[i
];
2408 no
= sstate
->dot_no_count
[i
];
2409 unknown
= d
->order
- yes
- no
;
2413 sstate
->dot_solved
[i
] = TRUE
;
2414 } else if (unknown
== 1) {
2415 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2416 diff
= min(diff
, DIFF_EASY
);
2417 sstate
->dot_solved
[i
] = TRUE
;
2419 } else if (yes
== 1) {
2421 sstate
->solver_status
= SOLVER_MISTAKE
;
2423 } else if (unknown
== 1) {
2424 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
);
2425 diff
= min(diff
, DIFF_EASY
);
2427 } else if (yes
== 2) {
2429 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2430 diff
= min(diff
, DIFF_EASY
);
2432 sstate
->dot_solved
[i
] = TRUE
;
2434 sstate
->solver_status
= SOLVER_MISTAKE
;
2439 check_caches(sstate
);
2444 static int dline_deductions(solver_state
*sstate
)
2446 game_state
*state
= sstate
->state
;
2447 grid
*g
= state
->game_grid
;
2448 char *dlines
= sstate
->dlines
;
2450 int diff
= DIFF_MAX
;
2452 /* ------ Face deductions ------ */
2454 /* Given a set of dline atmostone/atleastone constraints, need to figure
2455 * out if we can deduce any further info. For more general faces than
2456 * squares, this turns out to be a tricky problem.
2457 * The approach taken here is to define (per face) NxN matrices:
2458 * "maxs" and "mins".
2459 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2460 * for the possible number of edges that are YES between positions j and k
2461 * going clockwise around the face. Can think of j and k as marking dots
2462 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2463 * edge1 joins dot1 to dot2 etc).
2464 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2465 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2466 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2467 * the dline atmostone/atleastone status for edges j and j+1.
2469 * Then we calculate the remaining entries recursively. We definitely
2471 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2472 * This is because any valid placement of YESs between j and k must give
2473 * a valid placement between j and u, and also between u and k.
2474 * I believe it's sufficient to use just the two values of u:
2475 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2476 * are rigorous, even if they might not be best-possible.
2478 * Once we have maxs and mins calculated, we can make inferences about
2479 * each dline{j,j+1} by looking at the possible complementary edge-counts
2480 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2481 * As well as dlines, we can make similar inferences about single edges.
2482 * For example, consider a pentagon with clue 3, and we know at most one
2483 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2484 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2485 * that final edge would have to be YES to make the count up to 3.
2488 /* Much quicker to allocate arrays on the stack than the heap, so
2489 * define the largest possible face size, and base our array allocations
2490 * on that. We check this with an assertion, in case someone decides to
2491 * make a grid which has larger faces than this. Note, this algorithm
2492 * could get quite expensive if there are many large faces. */
2493 #define MAX_FACE_SIZE 8
2495 for (i
= 0; i
< g
->num_faces
; i
++) {
2496 int maxs
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2497 int mins
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2498 grid_face
*f
= g
->faces
+ i
;
2501 int clue
= state
->clues
[i
];
2502 assert(N
<= MAX_FACE_SIZE
);
2503 if (sstate
->face_solved
[i
])
2505 if (clue
< 0) continue;
2507 /* Calculate the (j,j+1) entries */
2508 for (j
= 0; j
< N
; j
++) {
2509 int edge_index
= f
->edges
[j
] - g
->edges
;
2511 enum line_state line1
= state
->lines
[edge_index
];
2512 enum line_state line2
;
2516 maxs
[j
][k
] = (line1
== LINE_NO
) ?
0 : 1;
2517 mins
[j
][k
] = (line1
== LINE_YES
) ?
1 : 0;
2518 /* Calculate the (j,j+2) entries */
2519 dline_index
= dline_index_from_face(g
, f
, k
);
2520 edge_index
= f
->edges
[k
] - g
->edges
;
2521 line2
= state
->lines
[edge_index
];
2527 if (line1
== LINE_NO
) tmp
--;
2528 if (line2
== LINE_NO
) tmp
--;
2529 if (tmp
== 2 && is_atmostone(dlines
, dline_index
))
2535 if (line1
== LINE_YES
) tmp
++;
2536 if (line2
== LINE_YES
) tmp
++;
2537 if (tmp
== 0 && is_atleastone(dlines
, dline_index
))
2542 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2543 for (m
= 3; m
< N
; m
++) {
2544 for (j
= 0; j
< N
; j
++) {
2552 maxs
[j
][k
] = maxs
[j
][u
] + maxs
[u
][k
];
2553 mins
[j
][k
] = mins
[j
][u
] + mins
[u
][k
];
2554 tmp
= maxs
[j
][v
] + maxs
[v
][k
];
2555 maxs
[j
][k
] = min(maxs
[j
][k
], tmp
);
2556 tmp
= mins
[j
][v
] + mins
[v
][k
];
2557 mins
[j
][k
] = max(mins
[j
][k
], tmp
);
2561 /* See if we can make any deductions */
2562 for (j
= 0; j
< N
; j
++) {
2564 grid_edge
*e
= f
->edges
[j
];
2565 int line_index
= e
- g
->edges
;
2568 if (state
->lines
[line_index
] != LINE_UNKNOWN
)
2573 /* minimum YESs in the complement of this edge */
2574 if (mins
[k
][j
] > clue
) {
2575 sstate
->solver_status
= SOLVER_MISTAKE
;
2578 if (mins
[k
][j
] == clue
) {
2579 /* setting this edge to YES would make at least
2580 * (clue+1) edges - contradiction */
2581 solver_set_line(sstate
, line_index
, LINE_NO
);
2582 diff
= min(diff
, DIFF_EASY
);
2584 if (maxs
[k
][j
] < clue
- 1) {
2585 sstate
->solver_status
= SOLVER_MISTAKE
;
2588 if (maxs
[k
][j
] == clue
- 1) {
2589 /* Only way to satisfy the clue is to set edge{j} as YES */
2590 solver_set_line(sstate
, line_index
, LINE_YES
);
2591 diff
= min(diff
, DIFF_EASY
);
2594 /* More advanced deduction that allows propagation along diagonal
2595 * chains of faces connected by dots, for example, 3-2-...-2-3
2596 * in square grids. */
2597 if (sstate
->diff
>= DIFF_TRICKY
) {
2598 /* Now see if we can make dline deduction for edges{j,j+1} */
2600 if (state
->lines
[e
- g
->edges
] != LINE_UNKNOWN
)
2601 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2602 * Dlines where one of the edges is known, are handled in the
2606 dline_index
= dline_index_from_face(g
, f
, k
);
2610 /* minimum YESs in the complement of this dline */
2611 if (mins
[k
][j
] > clue
- 2) {
2612 /* Adding 2 YESs would break the clue */
2613 if (set_atmostone(dlines
, dline_index
))
2614 diff
= min(diff
, DIFF_NORMAL
);
2616 /* maximum YESs in the complement of this dline */
2617 if (maxs
[k
][j
] < clue
) {
2618 /* Adding 2 NOs would mean not enough YESs */
2619 if (set_atleastone(dlines
, dline_index
))
2620 diff
= min(diff
, DIFF_NORMAL
);
2626 if (diff
< DIFF_NORMAL
)
2629 /* ------ Dot deductions ------ */
2631 for (i
= 0; i
< g
->num_dots
; i
++) {
2632 grid_dot
*d
= g
->dots
+ i
;
2634 int yes
, no
, unknown
;
2636 if (sstate
->dot_solved
[i
])
2638 yes
= sstate
->dot_yes_count
[i
];
2639 no
= sstate
->dot_no_count
[i
];
2640 unknown
= N
- yes
- no
;
2642 for (j
= 0; j
< N
; j
++) {
2645 int line1_index
, line2_index
;
2646 enum line_state line1
, line2
;
2649 dline_index
= dline_index_from_dot(g
, d
, j
);
2650 line1_index
= d
->edges
[j
] - g
->edges
;
2651 line2_index
= d
->edges
[k
] - g
->edges
;
2652 line1
= state
->lines
[line1_index
];
2653 line2
= state
->lines
[line2_index
];
2655 /* Infer dline state from line state */
2656 if (line1
== LINE_NO
|| line2
== LINE_NO
) {
2657 if (set_atmostone(dlines
, dline_index
))
2658 diff
= min(diff
, DIFF_NORMAL
);
2660 if (line1
== LINE_YES
|| line2
== LINE_YES
) {
2661 if (set_atleastone(dlines
, dline_index
))
2662 diff
= min(diff
, DIFF_NORMAL
);
2664 /* Infer line state from dline state */
2665 if (is_atmostone(dlines
, dline_index
)) {
2666 if (line1
== LINE_YES
&& line2
== LINE_UNKNOWN
) {
2667 solver_set_line(sstate
, line2_index
, LINE_NO
);
2668 diff
= min(diff
, DIFF_EASY
);
2670 if (line2
== LINE_YES
&& line1
== LINE_UNKNOWN
) {
2671 solver_set_line(sstate
, line1_index
, LINE_NO
);
2672 diff
= min(diff
, DIFF_EASY
);
2675 if (is_atleastone(dlines
, dline_index
)) {
2676 if (line1
== LINE_NO
&& line2
== LINE_UNKNOWN
) {
2677 solver_set_line(sstate
, line2_index
, LINE_YES
);
2678 diff
= min(diff
, DIFF_EASY
);
2680 if (line2
== LINE_NO
&& line1
== LINE_UNKNOWN
) {
2681 solver_set_line(sstate
, line1_index
, LINE_YES
);
2682 diff
= min(diff
, DIFF_EASY
);
2685 /* Deductions that depend on the numbers of lines.
2686 * Only bother if both lines are UNKNOWN, otherwise the
2687 * easy-mode solver (or deductions above) would have taken
2689 if (line1
!= LINE_UNKNOWN
|| line2
!= LINE_UNKNOWN
)
2692 if (yes
== 0 && unknown
== 2) {
2693 /* Both these unknowns must be identical. If we know
2694 * atmostone or atleastone, we can make progress. */
2695 if (is_atmostone(dlines
, dline_index
)) {
2696 solver_set_line(sstate
, line1_index
, LINE_NO
);
2697 solver_set_line(sstate
, line2_index
, LINE_NO
);
2698 diff
= min(diff
, DIFF_EASY
);
2700 if (is_atleastone(dlines
, dline_index
)) {
2701 solver_set_line(sstate
, line1_index
, LINE_YES
);
2702 solver_set_line(sstate
, line2_index
, LINE_YES
);
2703 diff
= min(diff
, DIFF_EASY
);
2707 if (set_atmostone(dlines
, dline_index
))
2708 diff
= min(diff
, DIFF_NORMAL
);
2710 if (set_atleastone(dlines
, dline_index
))
2711 diff
= min(diff
, DIFF_NORMAL
);
2715 /* More advanced deduction that allows propagation along diagonal
2716 * chains of faces connected by dots, for example: 3-2-...-2-3
2717 * in square grids. */
2718 if (sstate
->diff
>= DIFF_TRICKY
) {
2719 /* If we have atleastone set for this dline, infer
2720 * atmostone for each "opposite" dline (that is, each
2721 * dline without edges in common with this one).
2722 * Again, this test is only worth doing if both these
2723 * lines are UNKNOWN. For if one of these lines were YES,
2724 * the (yes == 1) test above would kick in instead. */
2725 if (is_atleastone(dlines
, dline_index
)) {
2727 for (opp
= 0; opp
< N
; opp
++) {
2728 int opp_dline_index
;
2729 if (opp
== j
|| opp
== j
+1 || opp
== j
-1)
2731 if (j
== 0 && opp
== N
-1)
2733 if (j
== N
-1 && opp
== 0)
2735 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2736 if (set_atmostone(dlines
, opp_dline_index
))
2737 diff
= min(diff
, DIFF_NORMAL
);
2739 if (yes
== 0 && is_atmostone(dlines
, dline_index
)) {
2740 /* This dline has *exactly* one YES and there are no
2741 * other YESs. This allows more deductions. */
2743 /* Third unknown must be YES */
2744 for (opp
= 0; opp
< N
; opp
++) {
2746 if (opp
== j
|| opp
== k
)
2748 opp_index
= d
->edges
[opp
] - g
->edges
;
2749 if (state
->lines
[opp_index
] == LINE_UNKNOWN
) {
2750 solver_set_line(sstate
, opp_index
,
2752 diff
= min(diff
, DIFF_EASY
);
2755 } else if (unknown
== 4) {
2756 /* Exactly one of opposite UNKNOWNS is YES. We've
2757 * already set atmostone, so set atleastone as
2760 if (dline_set_opp_atleastone(sstate
, d
, j
))
2761 diff
= min(diff
, DIFF_NORMAL
);
2771 static int linedsf_deductions(solver_state
*sstate
)
2773 game_state
*state
= sstate
->state
;
2774 grid
*g
= state
->game_grid
;
2775 char *dlines
= sstate
->dlines
;
2777 int diff
= DIFF_MAX
;
2780 /* ------ Face deductions ------ */
2782 /* A fully-general linedsf deduction seems overly complicated
2783 * (I suspect the problem is NP-complete, though in practice it might just
2784 * be doable because faces are limited in size).
2785 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2786 * known to be identical. If setting them both to YES (or NO) would break
2787 * the clue, set them to NO (or YES). */
2789 for (i
= 0; i
< g
->num_faces
; i
++) {
2790 int N
, yes
, no
, unknown
;
2793 if (sstate
->face_solved
[i
])
2795 clue
= state
->clues
[i
];
2799 N
= g
->faces
[i
].order
;
2800 yes
= sstate
->face_yes_count
[i
];
2801 if (yes
+ 1 == clue
) {
2802 if (face_setall_identical(sstate
, i
, LINE_NO
))
2803 diff
= min(diff
, DIFF_EASY
);
2805 no
= sstate
->face_no_count
[i
];
2806 if (no
+ 1 == N
- clue
) {
2807 if (face_setall_identical(sstate
, i
, LINE_YES
))
2808 diff
= min(diff
, DIFF_EASY
);
2811 /* Reload YES count, it might have changed */
2812 yes
= sstate
->face_yes_count
[i
];
2813 unknown
= N
- no
- yes
;
2815 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2816 * parity of lines. */
2817 diff_tmp
= parity_deductions(sstate
, g
->faces
[i
].edges
,
2818 (clue
- yes
) % 2, unknown
);
2819 diff
= min(diff
, diff_tmp
);
2822 /* ------ Dot deductions ------ */
2823 for (i
= 0; i
< g
->num_dots
; i
++) {
2824 grid_dot
*d
= g
->dots
+ i
;
2827 int yes
, no
, unknown
;
2828 /* Go through dlines, and do any dline<->linedsf deductions wherever
2829 * we find two UNKNOWNS. */
2830 for (j
= 0; j
< N
; j
++) {
2831 int dline_index
= dline_index_from_dot(g
, d
, j
);
2834 int can1
, can2
, inv1
, inv2
;
2836 line1_index
= d
->edges
[j
] - g
->edges
;
2837 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2840 if (j2
== N
) j2
= 0;
2841 line2_index
= d
->edges
[j2
] - g
->edges
;
2842 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2844 /* Infer dline flags from linedsf */
2845 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2846 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2847 if (can1
== can2
&& inv1
!= inv2
) {
2848 /* These are opposites, so set dline atmostone/atleastone */
2849 if (set_atmostone(dlines
, dline_index
))
2850 diff
= min(diff
, DIFF_NORMAL
);
2851 if (set_atleastone(dlines
, dline_index
))
2852 diff
= min(diff
, DIFF_NORMAL
);
2855 /* Infer linedsf from dline flags */
2856 if (is_atmostone(dlines
, dline_index
)
2857 && is_atleastone(dlines
, dline_index
)) {
2858 if (merge_lines(sstate
, line1_index
, line2_index
, 1))
2859 diff
= min(diff
, DIFF_HARD
);
2863 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2864 * parity of lines. */
2865 yes
= sstate
->dot_yes_count
[i
];
2866 no
= sstate
->dot_no_count
[i
];
2867 unknown
= N
- yes
- no
;
2868 diff_tmp
= parity_deductions(sstate
, d
->edges
,
2870 diff
= min(diff
, diff_tmp
);
2873 /* ------ Edge dsf deductions ------ */
2875 /* If the state of a line is known, deduce the state of its canonical line
2876 * too, and vice versa. */
2877 for (i
= 0; i
< g
->num_edges
; i
++) {
2880 can
= edsf_canonify(sstate
->linedsf
, i
, &inv
);
2883 s
= sstate
->state
->lines
[can
];
2884 if (s
!= LINE_UNKNOWN
) {
2885 if (solver_set_line(sstate
, i
, inv ?
OPP(s
) : s
))
2886 diff
= min(diff
, DIFF_EASY
);
2888 s
= sstate
->state
->lines
[i
];
2889 if (s
!= LINE_UNKNOWN
) {
2890 if (solver_set_line(sstate
, can
, inv ?
OPP(s
) : s
))
2891 diff
= min(diff
, DIFF_EASY
);
2899 static int loop_deductions(solver_state
*sstate
)
2901 int edgecount
= 0, clues
= 0, satclues
= 0, sm1clues
= 0;
2902 game_state
*state
= sstate
->state
;
2903 grid
*g
= state
->game_grid
;
2904 int shortest_chainlen
= g
->num_dots
;
2905 int loop_found
= FALSE
;
2907 int progress
= FALSE
;
2911 * Go through the grid and update for all the new edges.
2912 * Since merge_dots() is idempotent, the simplest way to
2913 * do this is just to update for _all_ the edges.
2914 * Also, while we're here, we count the edges.
2916 for (i
= 0; i
< g
->num_edges
; i
++) {
2917 if (state
->lines
[i
] == LINE_YES
) {
2918 loop_found
|= merge_dots(sstate
, i
);
2924 * Count the clues, count the satisfied clues, and count the
2925 * satisfied-minus-one clues.
2927 for (i
= 0; i
< g
->num_faces
; i
++) {
2928 int c
= state
->clues
[i
];
2930 int o
= sstate
->face_yes_count
[i
];
2939 for (i
= 0; i
< g
->num_dots
; ++i
) {
2941 sstate
->looplen
[dsf_canonify(sstate
->dotdsf
, i
)];
2942 if (dots_connected
> 1)
2943 shortest_chainlen
= min(shortest_chainlen
, dots_connected
);
2946 assert(sstate
->solver_status
== SOLVER_INCOMPLETE
);
2948 if (satclues
== clues
&& shortest_chainlen
== edgecount
) {
2949 sstate
->solver_status
= SOLVER_SOLVED
;
2950 /* This discovery clearly counts as progress, even if we haven't
2951 * just added any lines or anything */
2953 goto finished_loop_deductionsing
;
2957 * Now go through looking for LINE_UNKNOWN edges which
2958 * connect two dots that are already in the same
2959 * equivalence class. If we find one, test to see if the
2960 * loop it would create is a solution.
2962 for (i
= 0; i
< g
->num_edges
; i
++) {
2963 grid_edge
*e
= g
->edges
+ i
;
2964 int d1
= e
->dot1
- g
->dots
;
2965 int d2
= e
->dot2
- g
->dots
;
2967 if (state
->lines
[i
] != LINE_UNKNOWN
)
2970 eqclass
= dsf_canonify(sstate
->dotdsf
, d1
);
2971 if (eqclass
!= dsf_canonify(sstate
->dotdsf
, d2
))
2974 val
= LINE_NO
; /* loop is bad until proven otherwise */
2977 * This edge would form a loop. Next
2978 * question: how long would the loop be?
2979 * Would it equal the total number of edges
2980 * (plus the one we'd be adding if we added
2983 if (sstate
->looplen
[eqclass
] == edgecount
+ 1) {
2987 * This edge would form a loop which
2988 * took in all the edges in the entire
2989 * grid. So now we need to work out
2990 * whether it would be a valid solution
2991 * to the puzzle, which means we have to
2992 * check if it satisfies all the clues.
2993 * This means that every clue must be
2994 * either satisfied or satisfied-minus-
2995 * 1, and also that the number of
2996 * satisfied-minus-1 clues must be at
2997 * most two and they must lie on either
2998 * side of this edge.
3002 int f
= e
->face1
- g
->faces
;
3003 int c
= state
->clues
[f
];
3004 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3008 int f
= e
->face2
- g
->faces
;
3009 int c
= state
->clues
[f
];
3010 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3013 if (sm1clues
== sm1_nearby
&&
3014 sm1clues
+ satclues
== clues
) {
3015 val
= LINE_YES
; /* loop is good! */
3020 * Right. Now we know that adding this edge
3021 * would form a loop, and we know whether
3022 * that loop would be a viable solution or
3025 * If adding this edge produces a solution,
3026 * then we know we've found _a_ solution but
3027 * we don't know that it's _the_ solution -
3028 * if it were provably the solution then
3029 * we'd have deduced this edge some time ago
3030 * without the need to do loop detection. So
3031 * in this state we return SOLVER_AMBIGUOUS,
3032 * which has the effect that hitting Solve
3033 * on a user-provided puzzle will fill in a
3034 * solution but using the solver to
3035 * construct new puzzles won't consider this
3036 * a reasonable deduction for the user to
3039 progress
= solver_set_line(sstate
, i
, val
);
3040 assert(progress
== TRUE
);
3041 if (val
== LINE_YES
) {
3042 sstate
->solver_status
= SOLVER_AMBIGUOUS
;
3043 goto finished_loop_deductionsing
;
3047 finished_loop_deductionsing
:
3048 return progress ? DIFF_EASY
: DIFF_MAX
;
3051 /* This will return a dynamically allocated solver_state containing the (more)
3053 static solver_state
*solve_game_rec(const solver_state
*sstate_start
)
3055 solver_state
*sstate
;
3057 /* Index of the solver we should call next. */
3060 /* As a speed-optimisation, we avoid re-running solvers that we know
3061 * won't make any progress. This happens when a high-difficulty
3062 * solver makes a deduction that can only help other high-difficulty
3064 * For example: if a new 'dline' flag is set by dline_deductions, the
3065 * trivial_deductions solver cannot do anything with this information.
3066 * If we've already run the trivial_deductions solver (because it's
3067 * earlier in the list), there's no point running it again.
3069 * Therefore: if a solver is earlier in the list than "threshold_index",
3070 * we don't bother running it if it's difficulty level is less than
3073 int threshold_diff
= 0;
3074 int threshold_index
= 0;
3076 sstate
= dup_solver_state(sstate_start
);
3078 check_caches(sstate
);
3080 while (i
< NUM_SOLVERS
) {
3081 if (sstate
->solver_status
== SOLVER_MISTAKE
)
3083 if (sstate
->solver_status
== SOLVER_SOLVED
||
3084 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3085 /* solver finished */
3089 if ((solver_diffs
[i
] >= threshold_diff
|| i
>= threshold_index
)
3090 && solver_diffs
[i
] <= sstate
->diff
) {
3091 /* current_solver is eligible, so use it */
3092 int next_diff
= solver_fns
[i
](sstate
);
3093 if (next_diff
!= DIFF_MAX
) {
3094 /* solver made progress, so use new thresholds and
3095 * start again at top of list. */
3096 threshold_diff
= next_diff
;
3097 threshold_index
= i
;
3102 /* current_solver is ineligible, or failed to make progress, so
3103 * go to the next solver in the list */
3107 if (sstate
->solver_status
== SOLVER_SOLVED
||
3108 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3109 /* s/LINE_UNKNOWN/LINE_NO/g */
3110 array_setall(sstate
->state
->lines
, LINE_UNKNOWN
, LINE_NO
,
3111 sstate
->state
->game_grid
->num_edges
);
3118 static char *solve_game(game_state
*state
, game_state
*currstate
,
3119 char *aux
, char **error
)
3122 solver_state
*sstate
, *new_sstate
;
3124 sstate
= new_solver_state(state
, DIFF_MAX
);
3125 new_sstate
= solve_game_rec(sstate
);
3127 if (new_sstate
->solver_status
== SOLVER_SOLVED
) {
3128 soln
= encode_solve_move(new_sstate
->state
);
3129 } else if (new_sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3130 soln
= encode_solve_move(new_sstate
->state
);
3131 /**error = "Solver found ambiguous solutions"; */
3133 soln
= encode_solve_move(new_sstate
->state
);
3134 /**error = "Solver failed"; */
3137 free_solver_state(new_sstate
);
3138 free_solver_state(sstate
);
3143 /* ----------------------------------------------------------------------
3144 * Drawing and mouse-handling
3147 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
3148 int x
, int y
, int button
)
3150 grid
*g
= state
->game_grid
;
3154 char button_char
= ' ';
3155 enum line_state old_state
;
3157 button
&= ~MOD_MASK
;
3159 /* Convert mouse-click (x,y) to grid coordinates */
3160 x
-= BORDER(ds
->tilesize
);
3161 y
-= BORDER(ds
->tilesize
);
3162 x
= x
* g
->tilesize
/ ds
->tilesize
;
3163 y
= y
* g
->tilesize
/ ds
->tilesize
;
3167 e
= grid_nearest_edge(g
, x
, y
);
3173 /* I think it's only possible to play this game with mouse clicks, sorry */
3174 /* Maybe will add mouse drag support some time */
3175 old_state
= state
->lines
[i
];
3179 switch (old_state
) {
3193 switch (old_state
) {
3208 sprintf(buf
, "%d%c", i
, (int)button_char
);
3214 static game_state
*execute_move(game_state
*state
, char *move
)
3217 game_state
*newstate
= dup_game(state
);
3219 if (move
[0] == 'S') {
3221 newstate
->cheated
= TRUE
;
3226 move
+= strspn(move
, "1234567890");
3227 switch (*(move
++)) {
3229 newstate
->lines
[i
] = LINE_YES
;
3232 newstate
->lines
[i
] = LINE_NO
;
3235 newstate
->lines
[i
] = LINE_UNKNOWN
;
3243 * Check for completion.
3245 if (check_completion(newstate
))
3246 newstate
->solved
= TRUE
;
3251 free_game(newstate
);
3255 /* ----------------------------------------------------------------------
3259 /* Convert from grid coordinates to screen coordinates */
3260 static void grid_to_screen(const game_drawstate
*ds
, const grid
*g
,
3261 int grid_x
, int grid_y
, int *x
, int *y
)
3263 *x
= grid_x
- g
->lowest_x
;
3264 *y
= grid_y
- g
->lowest_y
;
3265 *x
= *x
* ds
->tilesize
/ g
->tilesize
;
3266 *y
= *y
* ds
->tilesize
/ g
->tilesize
;
3267 *x
+= BORDER(ds
->tilesize
);
3268 *y
+= BORDER(ds
->tilesize
);
3271 /* Returns (into x,y) position of centre of face for rendering the text clue.
3273 static void face_text_pos(const game_drawstate
*ds
, const grid
*g
,
3274 const grid_face
*f
, int *x
, int *y
)
3278 /* Simplest solution is the centroid. Might not work in some cases. */
3280 /* Another algorithm to look into:
3281 * Find the midpoints of the sides, find the bounding-box,
3282 * then take the centre of that. */
3284 /* Best solution probably involves incentres (inscribed circles) */
3286 int sx
= 0, sy
= 0; /* sums */
3287 for (i
= 0; i
< f
->order
; i
++) {
3288 grid_dot
*d
= f
->dots
[i
];
3295 /* convert to screen coordinates */
3296 grid_to_screen(ds
, g
, sx
, sy
, x
, y
);
3299 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
3300 game_state
*state
, int dir
, game_ui
*ui
,
3301 float animtime
, float flashtime
)
3303 grid
*g
= state
->game_grid
;
3304 int border
= BORDER(ds
->tilesize
);
3307 int line_colour
, flash_changed
;
3313 * The initial contents of the window are not guaranteed and
3314 * can vary with front ends. To be on the safe side, all games
3315 * should start by drawing a big background-colour rectangle
3316 * covering the whole window.
3318 int grid_width
= g
->highest_x
- g
->lowest_x
;
3319 int grid_height
= g
->highest_y
- g
->lowest_y
;
3320 int w
= grid_width
* ds
->tilesize
/ g
->tilesize
;
3321 int h
= grid_height
* ds
->tilesize
/ g
->tilesize
;
3322 draw_rect(dr
, 0, 0, w
+ 2 * border
+ 1, h
+ 2 * border
+ 1,
3326 for (i
= 0; i
< g
->num_faces
; i
++) {
3330 c
[0] = CLUE2CHAR(state
->clues
[i
]);
3333 face_text_pos(ds
, g
, f
, &x
, &y
);
3334 draw_text(dr
, x
, y
, FONT_VARIABLE
, ds
->tilesize
/2,
3335 ALIGN_VCENTRE
| ALIGN_HCENTRE
, COL_FOREGROUND
, c
);
3337 draw_update(dr
, 0, 0, w
+ 2 * border
, h
+ 2 * border
);
3340 if (flashtime
> 0 &&
3341 (flashtime
<= FLASH_TIME
/3 ||
3342 flashtime
>= FLASH_TIME
*2/3)) {
3343 flash_changed
= !ds
->flashing
;
3344 ds
->flashing
= TRUE
;
3346 flash_changed
= ds
->flashing
;
3347 ds
->flashing
= FALSE
;
3350 /* Some platforms may perform anti-aliasing, which may prevent clean
3351 * repainting of lines when the colour is changed.
3352 * If a line needs to be over-drawn in a different colour, erase a
3353 * bounding-box around the line, then flag all nearby objects for redraw.
3356 const char redraw_flag
= (char)(1<<7);
3357 for (i
= 0; i
< g
->num_edges
; i
++) {
3358 char prev_ds
= (ds
->lines
[i
] & ~redraw_flag
);
3359 char new_ds
= state
->lines
[i
];
3360 if (state
->line_errors
[i
])
3361 new_ds
= DS_LINE_ERROR
;
3363 /* If we're changing state, AND
3364 * the previous state was a coloured line */
3365 if ((prev_ds
!= new_ds
) && (prev_ds
!= LINE_NO
)) {
3366 grid_edge
*e
= g
->edges
+ i
;
3367 int x1
= e
->dot1
->x
;
3368 int y1
= e
->dot1
->y
;
3369 int x2
= e
->dot2
->x
;
3370 int y2
= e
->dot2
->y
;
3371 int xmin
, xmax
, ymin
, ymax
;
3373 grid_to_screen(ds
, g
, x1
, y1
, &x1
, &y1
);
3374 grid_to_screen(ds
, g
, x2
, y2
, &x2
, &y2
);
3375 /* Allow extra margin for dots, and thickness of lines */
3376 xmin
= min(x1
, x2
) - 2;
3377 xmax
= max(x1
, x2
) + 2;
3378 ymin
= min(y1
, y2
) - 2;
3379 ymax
= max(y1
, y2
) + 2;
3380 /* For testing, I find it helpful to change COL_BACKGROUND
3381 * to COL_SATISFIED here. */
3382 draw_rect(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1,
3384 draw_update(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1);
3386 /* Mark nearby lines for redraw */
3387 for (j
= 0; j
< e
->dot1
->order
; j
++)
3388 ds
->lines
[e
->dot1
->edges
[j
] - g
->edges
] |= redraw_flag
;
3389 for (j
= 0; j
< e
->dot2
->order
; j
++)
3390 ds
->lines
[e
->dot2
->edges
[j
] - g
->edges
] |= redraw_flag
;
3391 /* Mark nearby clues for redraw. Use a value that is
3392 * neither TRUE nor FALSE for this. */
3394 ds
->clue_error
[e
->face1
- g
->faces
] = 2;
3396 ds
->clue_error
[e
->face2
- g
->faces
] = 2;
3401 /* Redraw clue colours if necessary */
3402 for (i
= 0; i
< g
->num_faces
; i
++) {
3403 grid_face
*f
= g
->faces
+ i
;
3404 int sides
= f
->order
;
3406 n
= state
->clues
[i
];
3410 c
[0] = CLUE2CHAR(n
);
3413 clue_mistake
= (face_order(state
, i
, LINE_YES
) > n
||
3414 face_order(state
, i
, LINE_NO
) > (sides
-n
));
3416 clue_satisfied
= (face_order(state
, i
, LINE_YES
) == n
&&
3417 face_order(state
, i
, LINE_NO
) == (sides
-n
));
3419 if (clue_mistake
!= ds
->clue_error
[i
]
3420 || clue_satisfied
!= ds
->clue_satisfied
[i
]) {
3422 face_text_pos(ds
, g
, f
, &x
, &y
);
3423 /* There seems to be a certain amount of trial-and-error
3424 * involved in working out the correct bounding-box for
3426 draw_rect(dr
, x
- ds
->tilesize
/4 - 1, y
- ds
->tilesize
/4 - 3,
3427 ds
->tilesize
/2 + 2, ds
->tilesize
/2 + 5,
3430 FONT_VARIABLE
, ds
->tilesize
/2,
3431 ALIGN_VCENTRE
| ALIGN_HCENTRE
,
3432 clue_mistake ? COL_MISTAKE
:
3433 clue_satisfied ? COL_SATISFIED
: COL_FOREGROUND
, c
);
3434 draw_update(dr
, x
- ds
->tilesize
/4 - 1, y
- ds
->tilesize
/4 - 3,
3435 ds
->tilesize
/2 + 2, ds
->tilesize
/2 + 5);
3437 ds
->clue_error
[i
] = clue_mistake
;
3438 ds
->clue_satisfied
[i
] = clue_satisfied
;
3440 /* Sometimes, the bounding-box encroaches into the surrounding
3441 * lines (particularly if the window is resized fairly small).
3442 * So redraw them. */
3443 for (j
= 0; j
< f
->order
; j
++)
3444 ds
->lines
[f
->edges
[j
] - g
->edges
] = -1;
3449 for (i
= 0; i
< g
->num_edges
; i
++) {
3450 grid_edge
*e
= g
->edges
+ i
;
3452 int xmin
, ymin
, xmax
, ymax
;
3453 char new_ds
, need_draw
;
3454 new_ds
= state
->lines
[i
];
3455 if (state
->line_errors
[i
])
3456 new_ds
= DS_LINE_ERROR
;
3457 need_draw
= (new_ds
!= ds
->lines
[i
]) ? TRUE
: FALSE
;
3458 if (flash_changed
&& (state
->lines
[i
] == LINE_YES
))
3461 need_draw
= TRUE
; /* draw everything at the start */
3462 ds
->lines
[i
] = new_ds
;
3465 if (state
->line_errors
[i
])
3466 line_colour
= COL_MISTAKE
;
3467 else if (state
->lines
[i
] == LINE_UNKNOWN
)
3468 line_colour
= COL_LINEUNKNOWN
;
3469 else if (state
->lines
[i
] == LINE_NO
)
3470 line_colour
= COL_BACKGROUND
;
3471 else if (ds
->flashing
)
3472 line_colour
= COL_HIGHLIGHT
;
3474 line_colour
= COL_FOREGROUND
;
3476 /* Convert from grid to screen coordinates */
3477 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3478 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3485 if (line_colour
!= COL_BACKGROUND
) {
3486 /* (dx, dy) points roughly from (x1, y1) to (x2, y2).
3487 * The line is then "fattened" in a (roughly) perpendicular
3488 * direction to create a thin rectangle. */
3489 int dx
= (x1
> x2
) ?
-1 : ((x1
< x2
) ?
1 : 0);
3490 int dy
= (y1
> y2
) ?
-1 : ((y1
< y2
) ?
1 : 0);
3492 points
[0] = x1
+ dy
;
3493 points
[1] = y1
- dx
;
3494 points
[2] = x1
- dy
;
3495 points
[3] = y1
+ dx
;
3496 points
[4] = x2
- dy
;
3497 points
[5] = y2
+ dx
;
3498 points
[6] = x2
+ dy
;
3499 points
[7] = y2
- dx
;
3500 draw_polygon(dr
, points
, 4, line_colour
, line_colour
);
3503 /* Draw dots at ends of the line */
3504 draw_circle(dr
, x1
, y1
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3505 draw_circle(dr
, x2
, y2
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3507 draw_update(dr
, xmin
-2, ymin
-2, xmax
- xmin
+ 4, ymax
- ymin
+ 4);
3512 for (i
= 0; i
< g
->num_dots
; i
++) {
3513 grid_dot
*d
= g
->dots
+ i
;
3515 grid_to_screen(ds
, g
, d
->x
, d
->y
, &x
, &y
);
3516 draw_circle(dr
, x
, y
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3522 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
3523 int dir
, game_ui
*ui
)
3525 if (!oldstate
->solved
&& newstate
->solved
&&
3526 !oldstate
->cheated
&& !newstate
->cheated
) {
3533 static void game_print_size(game_params
*params
, float *x
, float *y
)
3538 * I'll use 7mm "squares" by default.
3540 game_compute_size(params
, 700, &pw
, &ph
);
3545 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
3547 int ink
= print_mono_colour(dr
, 0);
3549 game_drawstate ads
, *ds
= &ads
;
3550 grid
*g
= state
->game_grid
;
3552 game_set_size(dr
, ds
, NULL
, tilesize
);
3554 for (i
= 0; i
< g
->num_dots
; i
++) {
3556 grid_to_screen(ds
, g
, g
->dots
[i
].x
, g
->dots
[i
].y
, &x
, &y
);
3557 draw_circle(dr
, x
, y
, ds
->tilesize
/ 15, ink
, ink
);
3563 for (i
= 0; i
< g
->num_faces
; i
++) {
3564 grid_face
*f
= g
->faces
+ i
;
3565 int clue
= state
->clues
[i
];
3569 c
[0] = CLUE2CHAR(clue
);
3571 face_text_pos(ds
, g
, f
, &x
, &y
);
3573 FONT_VARIABLE
, ds
->tilesize
/ 2,
3574 ALIGN_VCENTRE
| ALIGN_HCENTRE
, ink
, c
);
3581 for (i
= 0; i
< g
->num_edges
; i
++) {
3582 int thickness
= (state
->lines
[i
] == LINE_YES
) ?
30 : 150;
3583 grid_edge
*e
= g
->edges
+ i
;
3585 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3586 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3587 if (state
->lines
[i
] == LINE_YES
)
3589 /* (dx, dy) points from (x1, y1) to (x2, y2).
3590 * The line is then "fattened" in a perpendicular
3591 * direction to create a thin rectangle. */
3592 double d
= sqrt(SQ((double)x1
- x2
) + SQ((double)y1
- y2
));
3593 double dx
= (x2
- x1
) / d
;
3594 double dy
= (y2
- y1
) / d
;
3597 dx
= (dx
* ds
->tilesize
) / thickness
;
3598 dy
= (dy
* ds
->tilesize
) / thickness
;
3599 points
[0] = x1
+ (int)dy
;
3600 points
[1] = y1
- (int)dx
;
3601 points
[2] = x1
- (int)dy
;
3602 points
[3] = y1
+ (int)dx
;
3603 points
[4] = x2
- (int)dy
;
3604 points
[5] = y2
+ (int)dx
;
3605 points
[6] = x2
+ (int)dy
;
3606 points
[7] = y2
- (int)dx
;
3607 draw_polygon(dr
, points
, 4, ink
, ink
);
3611 /* Draw a dotted line */
3614 for (j
= 1; j
< divisions
; j
++) {
3615 /* Weighted average */
3616 int x
= (x1
* (divisions
-j
) + x2
* j
) / divisions
;
3617 int y
= (y1
* (divisions
-j
) + y2
* j
) / divisions
;
3618 draw_circle(dr
, x
, y
, ds
->tilesize
/ thickness
, ink
, ink
);
3625 #define thegame loopy
3628 const struct game thegame
= {
3629 "Loopy", "games.loopy", "loopy",
3636 TRUE
, game_configure
, custom_params
,
3644 TRUE
, game_can_format_as_text_now
, game_text_format
,
3652 PREFERRED_TILE_SIZE
, game_compute_size
, game_set_size
,
3655 game_free_drawstate
,
3659 TRUE
, FALSE
, game_print_size
, game_print
,
3660 FALSE
/* wants_statusbar */,
3661 FALSE
, game_timing_state
,
3662 0, /* mouse_priorities */
3665 #ifdef STANDALONE_SOLVER
3668 * Half-hearted standalone solver. It can't output the solution to
3669 * anything but a square puzzle, and it can't log the deductions
3670 * it makes either. But it can solve square puzzles, and more
3671 * importantly it can use its solver to grade the difficulty of
3672 * any puzzle you give it.
3677 int main(int argc
, char **argv
)
3681 char *id
= NULL
, *desc
, *err
;
3684 #if 0 /* verbose solver not supported here (yet) */
3685 int really_verbose
= FALSE
;
3688 while (--argc
> 0) {
3690 #if 0 /* verbose solver not supported here (yet) */
3691 if (!strcmp(p
, "-v")) {
3692 really_verbose
= TRUE
;
3695 if (!strcmp(p
, "-g")) {
3697 } else if (*p
== '-') {
3698 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
3706 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
3710 desc
= strchr(id
, ':');
3712 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
3717 p
= default_params();
3718 decode_params(p
, id
);
3719 err
= validate_desc(p
, desc
);
3721 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
3724 s
= new_game(NULL
, p
, desc
);
3727 * When solving an Easy puzzle, we don't want to bother the
3728 * user with Hard-level deductions. For this reason, we grade
3729 * the puzzle internally before doing anything else.
3731 ret
= -1; /* placate optimiser */
3732 for (diff
= 0; diff
< DIFF_MAX
; diff
++) {
3733 solver_state
*sstate_new
;
3734 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3736 sstate_new
= solve_game_rec(sstate
);
3738 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3740 else if (sstate_new
->solver_status
== SOLVER_SOLVED
)
3745 free_solver_state(sstate_new
);
3746 free_solver_state(sstate
);
3752 if (diff
== DIFF_MAX
) {
3754 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3756 printf("Unable to find a unique solution\n");
3760 printf("Difficulty rating: impossible (no solution exists)\n");
3762 printf("Difficulty rating: %s\n", diffnames
[diff
]);
3764 solver_state
*sstate_new
;
3765 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3767 /* If we supported a verbose solver, we'd set verbosity here */
3769 sstate_new
= solve_game_rec(sstate
);
3771 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3772 printf("Puzzle is inconsistent\n");
3774 assert(sstate_new
->solver_status
== SOLVER_SOLVED
);
3775 if (s
->grid_type
== 0) {
3776 fputs(game_text_format(sstate_new
->state
), stdout
);
3778 printf("Unable to output non-square grids\n");
3782 free_solver_state(sstate_new
);
3783 free_solver_state(sstate
);