4 * An implementation of the Nikoli game 'Loop the loop'.
5 * (c) Mike Pinna, 2005, 2006
6 * Substantially rewritten to allowing for more general types of grid.
7 * (c) Lambros Lambrou 2008
9 * vim: set shiftwidth=4 :set textwidth=80:
13 * Possible future solver enhancements:
15 * - There's an interesting deductive technique which makes use
16 * of topology rather than just graph theory. Each _face_ in
17 * the grid is either inside or outside the loop; you can tell
18 * that two faces are on the same side of the loop if they're
19 * separated by a LINE_NO (or, more generally, by a path
20 * crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
21 * and on the opposite side of the loop if they're separated by
22 * a LINE_YES (or an odd number of LINE_YESes and no
23 * LINE_UNKNOWNs). Oh, and any face separated from the outside
24 * of the grid by a LINE_YES or a LINE_NO is on the inside or
25 * outside respectively. So if you can track this for all
26 * faces, you figure out the state of the line between a pair
27 * once their relative insideness is known.
28 * + The way I envisage this working is simply to keep an edsf
29 * of all _faces_, which indicates whether they're on
30 * opposite sides of the loop from one another. We also
31 * include a special entry in the edsf for the infinite
33 * + So, the simple way to do this is to just go through the
34 * edges: every time we see an edge in a state other than
35 * LINE_UNKNOWN which separates two faces that aren't in the
36 * same edsf class, we can rectify that by merging the
37 * classes. Then, conversely, an edge in LINE_UNKNOWN state
38 * which separates two faces that _are_ in the same edsf
39 * class can immediately have its state determined.
40 * + But you can go one better, if you're prepared to loop
41 * over all _pairs_ of edges. Suppose we have edges A and B,
42 * which respectively separate faces A1,A2 and B1,B2.
43 * Suppose that A,B are in the same edge-edsf class and that
44 * A1,B1 (wlog) are in the same face-edsf class; then we can
45 * immediately place A2,B2 into the same face-edsf class (as
46 * each other, not as A1 and A2) one way round or the other.
47 * And conversely again, if A1,B1 are in the same face-edsf
48 * class and so are A2,B2, then we can put A,B into the same
50 * * Of course, this deduction requires a quadratic-time
51 * loop over all pairs of edges in the grid, so it should
52 * be reserved until there's nothing easier left to be
55 * - The generalised grid support has made me (SGT) notice a
56 * possible extension to the loop-avoidance code. When you have
57 * a path of connected edges such that no other edges at all
58 * are incident on any vertex in the middle of the path - or,
59 * alternatively, such that any such edges are already known to
60 * be LINE_NO - then you know those edges are either all
61 * LINE_YES or all LINE_NO. Hence you can mentally merge the
62 * entire path into a single long curly edge for the purposes
63 * of loop avoidance, and look directly at whether or not the
64 * extreme endpoints of the path are connected by some other
65 * route. I find this coming up fairly often when I play on the
66 * octagonal grid setting, so it might be worth implementing in
69 * - (Just a speed optimisation.) Consider some todo list queue where every
70 * time we modify something we mark it for consideration by other bits of
71 * the solver, to save iteration over things that have already been done.
86 /* Debugging options */
94 /* ----------------------------------------------------------------------
95 * Struct, enum and function declarations
112 /* Put -1 in a face that doesn't get a clue */
115 /* Array of line states, to store whether each line is
116 * YES, NO or UNKNOWN */
119 unsigned char *line_errors
;
124 /* Used in game_text_format(), so that it knows what type of
125 * grid it's trying to render as ASCII text. */
130 SOLVER_SOLVED
, /* This is the only solution the solver could find */
131 SOLVER_MISTAKE
, /* This is definitely not a solution */
132 SOLVER_AMBIGUOUS
, /* This _might_ be an ambiguous solution */
133 SOLVER_INCOMPLETE
/* This may be a partial solution */
136 /* ------ Solver state ------ */
137 typedef struct solver_state
{
139 enum solver_status solver_status
;
140 /* NB looplen is the number of dots that are joined together at a point, ie a
141 * looplen of 1 means there are no lines to a particular dot */
144 /* Difficulty level of solver. Used by solver functions that want to
145 * vary their behaviour depending on the requested difficulty level. */
151 char *face_yes_count
;
153 char *dot_solved
, *face_solved
;
156 /* Information for Normal level deductions:
157 * For each dline, store a bitmask for whether we know:
158 * (bit 0) at least one is YES
159 * (bit 1) at most one is YES */
162 /* Hard level information */
167 * Difficulty levels. I do some macro ickery here to ensure that my
168 * enum and the various forms of my name list always match up.
171 #define DIFFLIST(A) \
176 #define ENUM(upper,title,lower) DIFF_ ## upper,
177 #define TITLE(upper,title,lower) #title,
178 #define ENCODE(upper,title,lower) #lower
179 #define CONFIG(upper,title,lower) ":" #title
180 enum { DIFFLIST(ENUM
) DIFF_MAX
};
181 static char const *const diffnames
[] = { DIFFLIST(TITLE
) };
182 static char const diffchars
[] = DIFFLIST(ENCODE
);
183 #define DIFFCONFIG DIFFLIST(CONFIG)
186 * Solver routines, sorted roughly in order of computational cost.
187 * The solver will run the faster deductions first, and slower deductions are
188 * only invoked when the faster deductions are unable to make progress.
189 * Each function is associated with a difficulty level, so that the generated
190 * puzzles are solvable by applying only the functions with the chosen
191 * difficulty level or lower.
193 #define SOLVERLIST(A) \
194 A(trivial_deductions, DIFF_EASY) \
195 A(dline_deductions, DIFF_NORMAL) \
196 A(linedsf_deductions, DIFF_HARD) \
197 A(loop_deductions, DIFF_EASY)
198 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
199 #define SOLVER_FN(fn,diff) &fn,
200 #define SOLVER_DIFF(fn,diff) diff,
201 SOLVERLIST(SOLVER_FN_DECL
)
202 static int (*(solver_fns
[]))(solver_state
*) = { SOLVERLIST(SOLVER_FN
) };
203 static int const solver_diffs
[] = { SOLVERLIST(SOLVER_DIFF
) };
204 const int NUM_SOLVERS
= sizeof(solver_diffs
)/sizeof(*solver_diffs
);
211 /* Grid generation is expensive, so keep a (ref-counted) reference to the
212 * grid for these parameters, and only generate when required. */
216 /* line_drawstate is the same as line_state, but with the extra ERROR
217 * possibility. The drawing code copies line_state to line_drawstate,
218 * except in the case that the line is an error. */
219 enum line_state
{ LINE_YES
, LINE_UNKNOWN
, LINE_NO
};
220 enum line_drawstate
{ DS_LINE_YES
, DS_LINE_UNKNOWN
,
221 DS_LINE_NO
, DS_LINE_ERROR
};
223 #define OPP(line_state) \
227 struct game_drawstate
{
233 char *clue_satisfied
;
236 static char *validate_desc(game_params
*params
, char *desc
);
237 static int dot_order(const game_state
* state
, int i
, char line_type
);
238 static int face_order(const game_state
* state
, int i
, char line_type
);
239 static solver_state
*solve_game_rec(const solver_state
*sstate
);
242 static void check_caches(const solver_state
* sstate
);
244 #define check_caches(s)
247 /* ------- List of grid generators ------- */
248 #define GRIDLIST(A) \
249 A(Squares,grid_new_square,3,3) \
250 A(Triangular,grid_new_triangular,3,3) \
251 A(Honeycomb,grid_new_honeycomb,3,3) \
252 A(Snub-Square,grid_new_snubsquare,3,3) \
253 A(Cairo,grid_new_cairo,3,4) \
254 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
255 A(Octagonal,grid_new_octagonal,3,3) \
256 A(Kites,grid_new_kites,3,3)
258 #define GRID_NAME(title,fn,amin,omin) #title,
259 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
260 #define GRID_FN(title,fn,amin,omin) &fn,
261 #define GRID_SIZES(title,fn,amin,omin) \
263 "Width and height for this grid type must both be at least " #amin, \
264 "At least one of width and height for this grid type must be at least " #omin,},
265 static char const *const gridnames
[] = { GRIDLIST(GRID_NAME
) };
266 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
267 static grid
* (*(grid_fns
[]))(int w
, int h
) = { GRIDLIST(GRID_FN
) };
268 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
269 static const struct {
272 } grid_size_limits
[] = { GRIDLIST(GRID_SIZES
) };
274 /* Generates a (dynamically allocated) new grid, according to the
275 * type and size requested in params. Does nothing if the grid is already
276 * generated. The allocated grid is owned by the params object, and will be
277 * freed in free_params(). */
278 static void params_generate_grid(game_params
*params
)
280 if (!params
->game_grid
) {
281 params
->game_grid
= grid_fns
[params
->type
](params
->w
, params
->h
);
285 /* ----------------------------------------------------------------------
289 /* General constants */
290 #define PREFERRED_TILE_SIZE 32
291 #define BORDER(tilesize) ((tilesize) / 2)
292 #define FLASH_TIME 0.5F
294 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
296 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
297 ((field) |= (1<<(bit)), TRUE))
299 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
300 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
302 #define CLUE2CHAR(c) \
303 ((c < 0) ? ' ' : c + '0')
305 /* ----------------------------------------------------------------------
306 * General struct manipulation and other straightforward code
309 static game_state
*dup_game(game_state
*state
)
311 game_state
*ret
= snew(game_state
);
313 ret
->game_grid
= state
->game_grid
;
314 ret
->game_grid
->refcount
++;
316 ret
->solved
= state
->solved
;
317 ret
->cheated
= state
->cheated
;
319 ret
->clues
= snewn(state
->game_grid
->num_faces
, signed char);
320 memcpy(ret
->clues
, state
->clues
, state
->game_grid
->num_faces
);
322 ret
->lines
= snewn(state
->game_grid
->num_edges
, char);
323 memcpy(ret
->lines
, state
->lines
, state
->game_grid
->num_edges
);
325 ret
->line_errors
= snewn(state
->game_grid
->num_edges
, unsigned char);
326 memcpy(ret
->line_errors
, state
->line_errors
, state
->game_grid
->num_edges
);
328 ret
->grid_type
= state
->grid_type
;
332 static void free_game(game_state
*state
)
335 grid_free(state
->game_grid
);
338 sfree(state
->line_errors
);
343 static solver_state
*new_solver_state(game_state
*state
, int diff
) {
345 int num_dots
= state
->game_grid
->num_dots
;
346 int num_faces
= state
->game_grid
->num_faces
;
347 int num_edges
= state
->game_grid
->num_edges
;
348 solver_state
*ret
= snew(solver_state
);
350 ret
->state
= dup_game(state
);
352 ret
->solver_status
= SOLVER_INCOMPLETE
;
355 ret
->dotdsf
= snew_dsf(num_dots
);
356 ret
->looplen
= snewn(num_dots
, int);
358 for (i
= 0; i
< num_dots
; i
++) {
362 ret
->dot_solved
= snewn(num_dots
, char);
363 ret
->face_solved
= snewn(num_faces
, char);
364 memset(ret
->dot_solved
, FALSE
, num_dots
);
365 memset(ret
->face_solved
, FALSE
, num_faces
);
367 ret
->dot_yes_count
= snewn(num_dots
, char);
368 memset(ret
->dot_yes_count
, 0, num_dots
);
369 ret
->dot_no_count
= snewn(num_dots
, char);
370 memset(ret
->dot_no_count
, 0, num_dots
);
371 ret
->face_yes_count
= snewn(num_faces
, char);
372 memset(ret
->face_yes_count
, 0, num_faces
);
373 ret
->face_no_count
= snewn(num_faces
, char);
374 memset(ret
->face_no_count
, 0, num_faces
);
376 if (diff
< DIFF_NORMAL
) {
379 ret
->dlines
= snewn(2*num_edges
, char);
380 memset(ret
->dlines
, 0, 2*num_edges
);
383 if (diff
< DIFF_HARD
) {
386 ret
->linedsf
= snew_dsf(state
->game_grid
->num_edges
);
392 static void free_solver_state(solver_state
*sstate
) {
394 free_game(sstate
->state
);
395 sfree(sstate
->dotdsf
);
396 sfree(sstate
->looplen
);
397 sfree(sstate
->dot_solved
);
398 sfree(sstate
->face_solved
);
399 sfree(sstate
->dot_yes_count
);
400 sfree(sstate
->dot_no_count
);
401 sfree(sstate
->face_yes_count
);
402 sfree(sstate
->face_no_count
);
404 /* OK, because sfree(NULL) is a no-op */
405 sfree(sstate
->dlines
);
406 sfree(sstate
->linedsf
);
412 static solver_state
*dup_solver_state(const solver_state
*sstate
) {
413 game_state
*state
= sstate
->state
;
414 int num_dots
= state
->game_grid
->num_dots
;
415 int num_faces
= state
->game_grid
->num_faces
;
416 int num_edges
= state
->game_grid
->num_edges
;
417 solver_state
*ret
= snew(solver_state
);
419 ret
->state
= state
= dup_game(sstate
->state
);
421 ret
->solver_status
= sstate
->solver_status
;
422 ret
->diff
= sstate
->diff
;
424 ret
->dotdsf
= snewn(num_dots
, int);
425 ret
->looplen
= snewn(num_dots
, int);
426 memcpy(ret
->dotdsf
, sstate
->dotdsf
,
427 num_dots
* sizeof(int));
428 memcpy(ret
->looplen
, sstate
->looplen
,
429 num_dots
* sizeof(int));
431 ret
->dot_solved
= snewn(num_dots
, char);
432 ret
->face_solved
= snewn(num_faces
, char);
433 memcpy(ret
->dot_solved
, sstate
->dot_solved
, num_dots
);
434 memcpy(ret
->face_solved
, sstate
->face_solved
, num_faces
);
436 ret
->dot_yes_count
= snewn(num_dots
, char);
437 memcpy(ret
->dot_yes_count
, sstate
->dot_yes_count
, num_dots
);
438 ret
->dot_no_count
= snewn(num_dots
, char);
439 memcpy(ret
->dot_no_count
, sstate
->dot_no_count
, num_dots
);
441 ret
->face_yes_count
= snewn(num_faces
, char);
442 memcpy(ret
->face_yes_count
, sstate
->face_yes_count
, num_faces
);
443 ret
->face_no_count
= snewn(num_faces
, char);
444 memcpy(ret
->face_no_count
, sstate
->face_no_count
, num_faces
);
446 if (sstate
->dlines
) {
447 ret
->dlines
= snewn(2*num_edges
, char);
448 memcpy(ret
->dlines
, sstate
->dlines
,
454 if (sstate
->linedsf
) {
455 ret
->linedsf
= snewn(num_edges
, int);
456 memcpy(ret
->linedsf
, sstate
->linedsf
,
457 num_edges
* sizeof(int));
465 static game_params
*default_params(void)
467 game_params
*ret
= snew(game_params
);
476 ret
->diff
= DIFF_EASY
;
479 ret
->game_grid
= NULL
;
484 static game_params
*dup_params(game_params
*params
)
486 game_params
*ret
= snew(game_params
);
488 *ret
= *params
; /* structure copy */
489 if (ret
->game_grid
) {
490 ret
->game_grid
->refcount
++;
495 static const game_params presets
[] = {
497 { 7, 7, DIFF_EASY
, 0, NULL
},
498 { 7, 7, DIFF_NORMAL
, 0, NULL
},
499 { 7, 7, DIFF_HARD
, 0, NULL
},
500 { 7, 7, DIFF_HARD
, 1, NULL
},
501 { 7, 7, DIFF_HARD
, 2, NULL
},
502 { 5, 5, DIFF_HARD
, 3, NULL
},
503 { 7, 7, DIFF_HARD
, 4, NULL
},
504 { 5, 4, DIFF_HARD
, 5, NULL
},
505 { 5, 5, DIFF_HARD
, 6, NULL
},
506 { 5, 5, DIFF_HARD
, 7, NULL
},
508 { 7, 7, DIFF_EASY
, 0, NULL
},
509 { 10, 10, DIFF_EASY
, 0, NULL
},
510 { 7, 7, DIFF_NORMAL
, 0, NULL
},
511 { 10, 10, DIFF_NORMAL
, 0, NULL
},
512 { 7, 7, DIFF_HARD
, 0, NULL
},
513 { 10, 10, DIFF_HARD
, 0, NULL
},
514 { 10, 10, DIFF_HARD
, 1, NULL
},
515 { 12, 10, DIFF_HARD
, 2, NULL
},
516 { 7, 7, DIFF_HARD
, 3, NULL
},
517 { 9, 9, DIFF_HARD
, 4, NULL
},
518 { 5, 4, DIFF_HARD
, 5, NULL
},
519 { 7, 7, DIFF_HARD
, 6, NULL
},
520 { 5, 5, DIFF_HARD
, 7, NULL
},
524 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
529 if (i
< 0 || i
>= lenof(presets
))
532 tmppar
= snew(game_params
);
533 *tmppar
= presets
[i
];
535 sprintf(buf
, "%dx%d %s - %s", tmppar
->h
, tmppar
->w
,
536 gridnames
[tmppar
->type
], diffnames
[tmppar
->diff
]);
542 static void free_params(game_params
*params
)
544 if (params
->game_grid
) {
545 grid_free(params
->game_grid
);
550 static void decode_params(game_params
*params
, char const *string
)
552 if (params
->game_grid
) {
553 grid_free(params
->game_grid
);
554 params
->game_grid
= NULL
;
556 params
->h
= params
->w
= atoi(string
);
557 params
->diff
= DIFF_EASY
;
558 while (*string
&& isdigit((unsigned char)*string
)) string
++;
559 if (*string
== 'x') {
561 params
->h
= atoi(string
);
562 while (*string
&& isdigit((unsigned char)*string
)) string
++;
564 if (*string
== 't') {
566 params
->type
= atoi(string
);
567 while (*string
&& isdigit((unsigned char)*string
)) string
++;
569 if (*string
== 'd') {
572 for (i
= 0; i
< DIFF_MAX
; i
++)
573 if (*string
== diffchars
[i
])
575 if (*string
) string
++;
579 static char *encode_params(game_params
*params
, int full
)
582 sprintf(str
, "%dx%dt%d", params
->w
, params
->h
, params
->type
);
584 sprintf(str
+ strlen(str
), "d%c", diffchars
[params
->diff
]);
588 static config_item
*game_configure(game_params
*params
)
593 ret
= snewn(5, config_item
);
595 ret
[0].name
= "Width";
596 ret
[0].type
= C_STRING
;
597 sprintf(buf
, "%d", params
->w
);
598 ret
[0].sval
= dupstr(buf
);
601 ret
[1].name
= "Height";
602 ret
[1].type
= C_STRING
;
603 sprintf(buf
, "%d", params
->h
);
604 ret
[1].sval
= dupstr(buf
);
607 ret
[2].name
= "Grid type";
608 ret
[2].type
= C_CHOICES
;
609 ret
[2].sval
= GRID_CONFIGS
;
610 ret
[2].ival
= params
->type
;
612 ret
[3].name
= "Difficulty";
613 ret
[3].type
= C_CHOICES
;
614 ret
[3].sval
= DIFFCONFIG
;
615 ret
[3].ival
= params
->diff
;
625 static game_params
*custom_params(config_item
*cfg
)
627 game_params
*ret
= snew(game_params
);
629 ret
->w
= atoi(cfg
[0].sval
);
630 ret
->h
= atoi(cfg
[1].sval
);
631 ret
->type
= cfg
[2].ival
;
632 ret
->diff
= cfg
[3].ival
;
634 ret
->game_grid
= NULL
;
638 static char *validate_params(game_params
*params
, int full
)
640 if (params
->type
< 0 || params
->type
>= NUM_GRID_TYPES
)
641 return "Illegal grid type";
642 if (params
->w
< grid_size_limits
[params
->type
].amin
||
643 params
->h
< grid_size_limits
[params
->type
].amin
)
644 return grid_size_limits
[params
->type
].aerr
;
645 if (params
->w
< grid_size_limits
[params
->type
].omin
&&
646 params
->h
< grid_size_limits
[params
->type
].omin
)
647 return grid_size_limits
[params
->type
].oerr
;
650 * This shouldn't be able to happen at all, since decode_params
651 * and custom_params will never generate anything that isn't
654 assert(params
->diff
< DIFF_MAX
);
659 /* Returns a newly allocated string describing the current puzzle */
660 static char *state_to_text(const game_state
*state
)
662 grid
*g
= state
->game_grid
;
664 int num_faces
= g
->num_faces
;
665 char *description
= snewn(num_faces
+ 1, char);
666 char *dp
= description
;
670 for (i
= 0; i
< num_faces
; i
++) {
671 if (state
->clues
[i
] < 0) {
672 if (empty_count
> 25) {
673 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
679 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
682 dp
+= sprintf(dp
, "%c", (int)CLUE2CHAR(state
->clues
[i
]));
687 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
689 retval
= dupstr(description
);
695 /* We require that the params pass the test in validate_params and that the
696 * description fills the entire game area */
697 static char *validate_desc(game_params
*params
, char *desc
)
701 params_generate_grid(params
);
702 g
= params
->game_grid
;
704 for (; *desc
; ++desc
) {
705 if (*desc
>= '0' && *desc
<= '9') {
710 count
+= *desc
- 'a' + 1;
713 return "Unknown character in description";
716 if (count
< g
->num_faces
)
717 return "Description too short for board size";
718 if (count
> g
->num_faces
)
719 return "Description too long for board size";
724 /* Sums the lengths of the numbers in range [0,n) */
725 /* See equivalent function in solo.c for justification of this. */
726 static int len_0_to_n(int n
)
728 int len
= 1; /* Counting 0 as a bit of a special case */
731 for (i
= 1; i
< n
; i
*= 10) {
732 len
+= max(n
- i
, 0);
738 static char *encode_solve_move(const game_state
*state
)
743 int num_edges
= state
->game_grid
->num_edges
;
745 /* This is going to return a string representing the moves needed to set
746 * every line in a grid to be the same as the ones in 'state'. The exact
747 * length of this string is predictable. */
749 len
= 1; /* Count the 'S' prefix */
750 /* Numbers in all lines */
751 len
+= len_0_to_n(num_edges
);
752 /* For each line we also have a letter */
755 ret
= snewn(len
+ 1, char);
758 p
+= sprintf(p
, "S");
760 for (i
= 0; i
< num_edges
; i
++) {
761 switch (state
->lines
[i
]) {
763 p
+= sprintf(p
, "%dy", i
);
766 p
+= sprintf(p
, "%dn", i
);
771 /* No point in doing sums like that if they're going to be wrong */
772 assert(strlen(ret
) <= (size_t)len
);
776 static game_ui
*new_ui(game_state
*state
)
781 static void free_ui(game_ui
*ui
)
785 static char *encode_ui(game_ui
*ui
)
790 static void decode_ui(game_ui
*ui
, char *encoding
)
794 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
795 game_state
*newstate
)
799 static void game_compute_size(game_params
*params
, int tilesize
,
803 int grid_width
, grid_height
, rendered_width
, rendered_height
;
805 params_generate_grid(params
);
806 g
= params
->game_grid
;
807 grid_width
= g
->highest_x
- g
->lowest_x
;
808 grid_height
= g
->highest_y
- g
->lowest_y
;
809 /* multiply first to minimise rounding error on integer division */
810 rendered_width
= grid_width
* tilesize
/ g
->tilesize
;
811 rendered_height
= grid_height
* tilesize
/ g
->tilesize
;
812 *x
= rendered_width
+ 2 * BORDER(tilesize
) + 1;
813 *y
= rendered_height
+ 2 * BORDER(tilesize
) + 1;
816 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
817 game_params
*params
, int tilesize
)
819 ds
->tilesize
= tilesize
;
822 static float *game_colours(frontend
*fe
, int *ncolours
)
824 float *ret
= snewn(4 * NCOLOURS
, float);
826 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
828 ret
[COL_FOREGROUND
* 3 + 0] = 0.0F
;
829 ret
[COL_FOREGROUND
* 3 + 1] = 0.0F
;
830 ret
[COL_FOREGROUND
* 3 + 2] = 0.0F
;
832 ret
[COL_LINEUNKNOWN
* 3 + 0] = 0.8F
;
833 ret
[COL_LINEUNKNOWN
* 3 + 1] = 0.8F
;
834 ret
[COL_LINEUNKNOWN
* 3 + 2] = 0.0F
;
836 ret
[COL_HIGHLIGHT
* 3 + 0] = 1.0F
;
837 ret
[COL_HIGHLIGHT
* 3 + 1] = 1.0F
;
838 ret
[COL_HIGHLIGHT
* 3 + 2] = 1.0F
;
840 ret
[COL_MISTAKE
* 3 + 0] = 1.0F
;
841 ret
[COL_MISTAKE
* 3 + 1] = 0.0F
;
842 ret
[COL_MISTAKE
* 3 + 2] = 0.0F
;
844 ret
[COL_SATISFIED
* 3 + 0] = 0.0F
;
845 ret
[COL_SATISFIED
* 3 + 1] = 0.0F
;
846 ret
[COL_SATISFIED
* 3 + 2] = 0.0F
;
848 /* We want the faint lines to be a bit darker than the background.
849 * Except if the background is pretty dark already; then it ought to be a
850 * bit lighter. Oy vey.
852 ret
[COL_FAINT
* 3 + 0] = ret
[COL_BACKGROUND
* 3 + 0] * 0.9F
;
853 ret
[COL_FAINT
* 3 + 1] = ret
[COL_BACKGROUND
* 3 + 1] * 0.9F
;
854 ret
[COL_FAINT
* 3 + 2] = ret
[COL_BACKGROUND
* 3 + 2] * 0.9F
;
856 *ncolours
= NCOLOURS
;
860 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
862 struct game_drawstate
*ds
= snew(struct game_drawstate
);
863 int num_faces
= state
->game_grid
->num_faces
;
864 int num_edges
= state
->game_grid
->num_edges
;
868 ds
->lines
= snewn(num_edges
, char);
869 ds
->clue_error
= snewn(num_faces
, char);
870 ds
->clue_satisfied
= snewn(num_faces
, char);
873 memset(ds
->lines
, LINE_UNKNOWN
, num_edges
);
874 memset(ds
->clue_error
, 0, num_faces
);
875 memset(ds
->clue_satisfied
, 0, num_faces
);
880 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
882 sfree(ds
->clue_error
);
883 sfree(ds
->clue_satisfied
);
888 static int game_timing_state(game_state
*state
, game_ui
*ui
)
893 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
894 int dir
, game_ui
*ui
)
899 static int game_can_format_as_text_now(game_params
*params
)
901 if (params
->type
!= 0)
906 static char *game_text_format(game_state
*state
)
912 grid
*g
= state
->game_grid
;
915 assert(state
->grid_type
== 0);
917 /* Work out the basic size unit */
918 f
= g
->faces
; /* first face */
919 assert(f
->order
== 4);
920 /* The dots are ordered clockwise, so the two opposite
921 * corners are guaranteed to span the square */
922 cell_size
= abs(f
->dots
[0]->x
- f
->dots
[2]->x
);
924 w
= (g
->highest_x
- g
->lowest_x
) / cell_size
;
925 h
= (g
->highest_y
- g
->lowest_y
) / cell_size
;
927 /* Create a blank "canvas" to "draw" on */
930 ret
= snewn(W
* H
+ 1, char);
931 for (y
= 0; y
< H
; y
++) {
932 for (x
= 0; x
< W
-1; x
++) {
935 ret
[y
*W
+ W
-1] = '\n';
939 /* Fill in edge info */
940 for (i
= 0; i
< g
->num_edges
; i
++) {
941 grid_edge
*e
= g
->edges
+ i
;
942 /* Cell coordinates, from (0,0) to (w-1,h-1) */
943 int x1
= (e
->dot1
->x
- g
->lowest_x
) / cell_size
;
944 int x2
= (e
->dot2
->x
- g
->lowest_x
) / cell_size
;
945 int y1
= (e
->dot1
->y
- g
->lowest_y
) / cell_size
;
946 int y2
= (e
->dot2
->y
- g
->lowest_y
) / cell_size
;
947 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
948 * cell coordinates) */
951 switch (state
->lines
[i
]) {
953 ret
[y
*W
+ x
] = (y1
== y2
) ?
'-' : '|';
959 break; /* already a space */
961 assert(!"Illegal line state");
966 for (i
= 0; i
< g
->num_faces
; i
++) {
970 assert(f
->order
== 4);
971 /* Cell coordinates, from (0,0) to (w-1,h-1) */
972 x1
= (f
->dots
[0]->x
- g
->lowest_x
) / cell_size
;
973 x2
= (f
->dots
[2]->x
- g
->lowest_x
) / cell_size
;
974 y1
= (f
->dots
[0]->y
- g
->lowest_y
) / cell_size
;
975 y2
= (f
->dots
[2]->y
- g
->lowest_y
) / cell_size
;
976 /* Midpoint, in canvas coordinates */
979 ret
[y
*W
+ x
] = CLUE2CHAR(state
->clues
[i
]);
984 /* ----------------------------------------------------------------------
989 static void check_caches(const solver_state
* sstate
)
992 const game_state
*state
= sstate
->state
;
993 const grid
*g
= state
->game_grid
;
995 for (i
= 0; i
< g
->num_dots
; i
++) {
996 assert(dot_order(state
, i
, LINE_YES
) == sstate
->dot_yes_count
[i
]);
997 assert(dot_order(state
, i
, LINE_NO
) == sstate
->dot_no_count
[i
]);
1000 for (i
= 0; i
< g
->num_faces
; i
++) {
1001 assert(face_order(state
, i
, LINE_YES
) == sstate
->face_yes_count
[i
]);
1002 assert(face_order(state
, i
, LINE_NO
) == sstate
->face_no_count
[i
]);
1007 #define check_caches(s) \
1009 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1013 #endif /* DEBUG_CACHES */
1015 /* ----------------------------------------------------------------------
1016 * Solver utility functions
1019 /* Sets the line (with index i) to the new state 'line_new', and updates
1020 * the cached counts of any affected faces and dots.
1021 * Returns TRUE if this actually changed the line's state. */
1022 static int solver_set_line(solver_state
*sstate
, int i
,
1023 enum line_state line_new
1025 , const char *reason
1029 game_state
*state
= sstate
->state
;
1033 assert(line_new
!= LINE_UNKNOWN
);
1035 check_caches(sstate
);
1037 if (state
->lines
[i
] == line_new
) {
1038 return FALSE
; /* nothing changed */
1040 state
->lines
[i
] = line_new
;
1043 fprintf(stderr
, "solver: set line [%d] to %s (%s)\n",
1044 i
, line_new
== LINE_YES ?
"YES" : "NO",
1048 g
= state
->game_grid
;
1051 /* Update the cache for both dots and both faces affected by this. */
1052 if (line_new
== LINE_YES
) {
1053 sstate
->dot_yes_count
[e
->dot1
- g
->dots
]++;
1054 sstate
->dot_yes_count
[e
->dot2
- g
->dots
]++;
1056 sstate
->face_yes_count
[e
->face1
- g
->faces
]++;
1059 sstate
->face_yes_count
[e
->face2
- g
->faces
]++;
1062 sstate
->dot_no_count
[e
->dot1
- g
->dots
]++;
1063 sstate
->dot_no_count
[e
->dot2
- g
->dots
]++;
1065 sstate
->face_no_count
[e
->face1
- g
->faces
]++;
1068 sstate
->face_no_count
[e
->face2
- g
->faces
]++;
1072 check_caches(sstate
);
1077 #define solver_set_line(a, b, c) \
1078 solver_set_line(a, b, c, __FUNCTION__)
1082 * Merge two dots due to the existence of an edge between them.
1083 * Updates the dsf tracking equivalence classes, and keeps track of
1084 * the length of path each dot is currently a part of.
1085 * Returns TRUE if the dots were already linked, ie if they are part of a
1086 * closed loop, and false otherwise.
1088 static int merge_dots(solver_state
*sstate
, int edge_index
)
1091 grid
*g
= sstate
->state
->game_grid
;
1092 grid_edge
*e
= g
->edges
+ edge_index
;
1094 i
= e
->dot1
- g
->dots
;
1095 j
= e
->dot2
- g
->dots
;
1097 i
= dsf_canonify(sstate
->dotdsf
, i
);
1098 j
= dsf_canonify(sstate
->dotdsf
, j
);
1103 len
= sstate
->looplen
[i
] + sstate
->looplen
[j
];
1104 dsf_merge(sstate
->dotdsf
, i
, j
);
1105 i
= dsf_canonify(sstate
->dotdsf
, i
);
1106 sstate
->looplen
[i
] = len
;
1111 /* Merge two lines because the solver has deduced that they must be either
1112 * identical or opposite. Returns TRUE if this is new information, otherwise
1114 static int merge_lines(solver_state
*sstate
, int i
, int j
, int inverse
1116 , const char *reason
1122 assert(i
< sstate
->state
->game_grid
->num_edges
);
1123 assert(j
< sstate
->state
->game_grid
->num_edges
);
1125 i
= edsf_canonify(sstate
->linedsf
, i
, &inv_tmp
);
1127 j
= edsf_canonify(sstate
->linedsf
, j
, &inv_tmp
);
1130 edsf_merge(sstate
->linedsf
, i
, j
, inverse
);
1134 fprintf(stderr
, "%s [%d] [%d] %s(%s)\n",
1136 inverse ?
"inverse " : "", reason
);
1143 #define merge_lines(a, b, c, d) \
1144 merge_lines(a, b, c, d, __FUNCTION__)
1147 /* Count the number of lines of a particular type currently going into the
1149 static int dot_order(const game_state
* state
, int dot
, char line_type
)
1152 grid
*g
= state
->game_grid
;
1153 grid_dot
*d
= g
->dots
+ dot
;
1156 for (i
= 0; i
< d
->order
; i
++) {
1157 grid_edge
*e
= d
->edges
[i
];
1158 if (state
->lines
[e
- g
->edges
] == line_type
)
1164 /* Count the number of lines of a particular type currently surrounding the
1166 static int face_order(const game_state
* state
, int face
, char line_type
)
1169 grid
*g
= state
->game_grid
;
1170 grid_face
*f
= g
->faces
+ face
;
1173 for (i
= 0; i
< f
->order
; i
++) {
1174 grid_edge
*e
= f
->edges
[i
];
1175 if (state
->lines
[e
- g
->edges
] == line_type
)
1181 /* Set all lines bordering a dot of type old_type to type new_type
1182 * Return value tells caller whether this function actually did anything */
1183 static int dot_setall(solver_state
*sstate
, int dot
,
1184 char old_type
, char new_type
)
1186 int retval
= FALSE
, r
;
1187 game_state
*state
= sstate
->state
;
1192 if (old_type
== new_type
)
1195 g
= state
->game_grid
;
1198 for (i
= 0; i
< d
->order
; i
++) {
1199 int line_index
= d
->edges
[i
] - g
->edges
;
1200 if (state
->lines
[line_index
] == old_type
) {
1201 r
= solver_set_line(sstate
, line_index
, new_type
);
1209 /* Set all lines bordering a face of type old_type to type new_type */
1210 static int face_setall(solver_state
*sstate
, int face
,
1211 char old_type
, char new_type
)
1213 int retval
= FALSE
, r
;
1214 game_state
*state
= sstate
->state
;
1219 if (old_type
== new_type
)
1222 g
= state
->game_grid
;
1223 f
= g
->faces
+ face
;
1225 for (i
= 0; i
< f
->order
; i
++) {
1226 int line_index
= f
->edges
[i
] - g
->edges
;
1227 if (state
->lines
[line_index
] == old_type
) {
1228 r
= solver_set_line(sstate
, line_index
, new_type
);
1236 /* ----------------------------------------------------------------------
1237 * Loop generation and clue removal
1240 /* We're going to store lists of current candidate faces for colouring black
1242 * Each face gets a 'score', which tells us how adding that face right
1243 * now would affect the curliness of the solution loop. We're trying to
1244 * maximise that quantity so will bias our random selection of faces to
1245 * colour those with high scores */
1249 unsigned long random
;
1250 /* No need to store a grid_face* here. The 'face_scores' array will
1251 * be a list of 'face_score' objects, one for each face of the grid, so
1252 * the position (index) within the 'face_scores' array will determine
1253 * which face corresponds to a particular face_score.
1254 * Having a single 'face_scores' array for all faces simplifies memory
1255 * management, and probably improves performance, because we don't have to
1256 * malloc/free each individual face_score, and we don't have to maintain
1257 * a mapping from grid_face* pointers to face_score* pointers.
1261 static int generic_sort_cmpfn(void *v1
, void *v2
, size_t offset
)
1263 struct face_score
*f1
= v1
;
1264 struct face_score
*f2
= v2
;
1267 r
= *(int *)((char *)f2
+ offset
) - *(int *)((char *)f1
+ offset
);
1272 if (f1
->random
< f2
->random
)
1274 else if (f1
->random
> f2
->random
)
1278 * It's _just_ possible that two faces might have been given
1279 * the same random value. In that situation, fall back to
1280 * comparing based on the positions within the face_scores list.
1281 * This introduces a tiny directional bias, but not a significant one.
1286 static int white_sort_cmpfn(void *v1
, void *v2
)
1288 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,white_score
));
1291 static int black_sort_cmpfn(void *v1
, void *v2
)
1293 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,black_score
));
1296 enum face_colour
{ FACE_WHITE
, FACE_GREY
, FACE_BLACK
};
1298 /* face should be of type grid_face* here. */
1299 #define FACE_COLOUR(face) \
1300 ( (face) == NULL ? FACE_BLACK : \
1301 board[(face) - g->faces] )
1303 /* 'board' is an array of these enums, indicating which faces are
1304 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1305 * Returns whether it's legal to colour the given face with this colour. */
1306 static int can_colour_face(grid
*g
, char* board
, int face_index
,
1307 enum face_colour colour
)
1310 grid_face
*test_face
= g
->faces
+ face_index
;
1311 grid_face
*starting_face
, *current_face
;
1312 grid_dot
*starting_dot
;
1314 int current_state
, s
; /* booleans: equal or not-equal to 'colour' */
1315 int found_same_coloured_neighbour
= FALSE
;
1316 assert(board
[face_index
] != colour
);
1318 /* Can only consider a face for colouring if it's adjacent to a face
1319 * with the same colour. */
1320 for (i
= 0; i
< test_face
->order
; i
++) {
1321 grid_edge
*e
= test_face
->edges
[i
];
1322 grid_face
*f
= (e
->face1
== test_face
) ? e
->face2
: e
->face1
;
1323 if (FACE_COLOUR(f
) == colour
) {
1324 found_same_coloured_neighbour
= TRUE
;
1328 if (!found_same_coloured_neighbour
)
1331 /* Need to avoid creating a loop of faces of this colour around some
1332 * differently-coloured faces.
1333 * Also need to avoid meeting a same-coloured face at a corner, with
1334 * other-coloured faces in between. Here's a simple test that (I believe)
1335 * takes care of both these conditions:
1337 * Take the circular path formed by this face's edges, and inflate it
1338 * slightly outwards. Imagine walking around this path and consider
1339 * the faces that you visit in sequence. This will include all faces
1340 * touching the given face, either along an edge or just at a corner.
1341 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1342 * you walk along the complete loop. This will obviously turn out to be
1344 * If 0, we're either in the middle of an "island" of this colour (should
1345 * be impossible as we're not supposed to create black or white loops),
1346 * or we're about to start a new island - also not allowed.
1347 * If 4 or greater, there are too many separate coloured regions touching
1348 * this face, and colouring it would create a loop or a corner-violation.
1349 * The only allowed case is when the count is exactly 2. */
1351 /* i points to a dot around the test face.
1352 * j points to a face around the i^th dot.
1353 * The current face will always be:
1354 * test_face->dots[i]->faces[j]
1355 * We assume dots go clockwise around the test face,
1356 * and faces go clockwise around dots. */
1359 * The end condition is slightly fiddly. In sufficiently strange
1360 * degenerate grids, our test face may be adjacent to the same
1361 * other face multiple times (typically if it's the exterior
1362 * face). Consider this, in particular:
1370 * The bottom left face there is adjacent to the exterior face
1371 * twice, so we can't just terminate our iteration when we reach
1372 * the same _face_ we started at. Furthermore, we can't
1373 * condition on having the same (i,j) pair either, because
1374 * several (i,j) pairs identify the bottom left contiguity with
1375 * the exterior face! We canonicalise the (i,j) pair by taking
1376 * one step around before we set the termination tracking.
1380 current_face
= test_face
->dots
[0]->faces
[0];
1381 if (current_face
== test_face
) {
1383 current_face
= test_face
->dots
[0]->faces
[1];
1386 current_state
= (FACE_COLOUR(current_face
) == colour
);
1387 starting_dot
= NULL
;
1388 starting_face
= NULL
;
1390 /* Advance to next face.
1391 * Need to loop here because it might take several goes to
1395 if (j
== test_face
->dots
[i
]->order
)
1398 if (test_face
->dots
[i
]->faces
[j
] == test_face
) {
1399 /* Advance to next dot round test_face, then
1400 * find current_face around new dot
1401 * and advance to the next face clockwise */
1403 if (i
== test_face
->order
)
1405 for (j
= 0; j
< test_face
->dots
[i
]->order
; j
++) {
1406 if (test_face
->dots
[i
]->faces
[j
] == current_face
)
1409 /* Must actually find current_face around new dot,
1410 * or else something's wrong with the grid. */
1411 assert(j
!= test_face
->dots
[i
]->order
);
1412 /* Found, so advance to next face and try again */
1417 /* (i,j) are now advanced to next face */
1418 current_face
= test_face
->dots
[i
]->faces
[j
];
1419 s
= (FACE_COLOUR(current_face
) == colour
);
1420 if (!starting_dot
) {
1421 starting_dot
= test_face
->dots
[i
];
1422 starting_face
= current_face
;
1425 if (s
!= current_state
) {
1428 if (transitions
> 2)
1431 if (test_face
->dots
[i
] == starting_dot
&&
1432 current_face
== starting_face
)
1437 return (transitions
== 2) ? TRUE
: FALSE
;
1440 /* Count the number of neighbours of 'face', having colour 'colour' */
1441 static int face_num_neighbours(grid
*g
, char *board
, grid_face
*face
,
1442 enum face_colour colour
)
1444 int colour_count
= 0;
1448 for (i
= 0; i
< face
->order
; i
++) {
1450 f
= (e
->face1
== face
) ? e
->face2
: e
->face1
;
1451 if (FACE_COLOUR(f
) == colour
)
1454 return colour_count
;
1457 /* The 'score' of a face reflects its current desirability for selection
1458 * as the next face to colour white or black. We want to encourage moving
1459 * into grey areas and increasing loopiness, so we give scores according to
1460 * how many of the face's neighbours are currently coloured the same as the
1461 * proposed colour. */
1462 static int face_score(grid
*g
, char *board
, grid_face
*face
,
1463 enum face_colour colour
)
1465 /* Simple formula: score = 0 - num. same-coloured neighbours,
1466 * so a higher score means fewer same-coloured neighbours. */
1467 return -face_num_neighbours(g
, board
, face
, colour
);
1470 /* Generate a new complete set of clues for the given game_state.
1471 * The method is to generate a WHITE/BLACK colouring of all the faces,
1472 * such that the WHITE faces will define the inside of the path, and the
1473 * BLACK faces define the outside.
1474 * To do this, we initially colour all faces GREY. The infinite space outside
1475 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1476 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1477 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1478 * we avoid creating loops of a single colour, to preserve the topological
1479 * shape of the WHITE and BLACK regions.
1480 * We also try to make the boundary as loopy and twisty as possible, to avoid
1481 * generating paths that are uninteresting.
1482 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1483 * face that can be coloured with that colour (without violating the
1484 * topological shape of that region). It's not obvious, but I think this
1485 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1486 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1487 * regions can be grown.
1488 * This is checked using assert()ions, and I haven't seen any failures yet.
1490 * Hand-wavy proof: imagine what can go wrong...
1492 * Could the white faces get completely cut off by the black faces, and still
1493 * leave some grey faces remaining?
1494 * No, because then the black faces would form a loop around both the white
1495 * faces and the grey faces, which is disallowed because we continually
1496 * maintain the correct topological shape of the black region.
1497 * Similarly, the black faces can never get cut off by the white faces. That
1498 * means both the WHITE and BLACK regions always have some room to grow into
1500 * Could it be that we can't colour some GREY face, because there are too many
1501 * WHITE/BLACK transitions as we walk round the face? (see the
1502 * can_colour_face() function for details)
1503 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1504 * around the face. The two WHITE faces would be connected by a WHITE path,
1505 * and the BLACK faces would be connected by a BLACK path. These paths would
1506 * have to cross, which is impossible.
1507 * Another thing that could go wrong: perhaps we can't find any GREY face to
1508 * colour WHITE, because it would create a loop-violation or a corner-violation
1509 * with the other WHITE faces?
1510 * This is a little bit tricky to prove impossible. Imagine you have such a
1511 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1512 * or corner violation).
1513 * That would cut all the non-white area into two blobs. One of those blobs
1514 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1515 * So we have a connected GREY area, completely surrounded by WHITE
1516 * (including the GREY face we've tentatively coloured WHITE).
1517 * A well-known result in graph theory says that you can always find a GREY
1518 * face whose removal leaves the remaining GREY area connected. And it says
1519 * there are at least two such faces, so we can always choose the one that
1520 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1521 * everything nice and connected, including that "tentative" GREY face which
1522 * acts as a gateway to the rest of the non-WHITE grid.
1524 static void add_full_clues(game_state
*state
, random_state
*rs
)
1526 signed char *clues
= state
->clues
;
1528 grid
*g
= state
->game_grid
;
1530 int num_faces
= g
->num_faces
;
1531 struct face_score
*face_scores
; /* Array of face_score objects */
1532 struct face_score
*fs
; /* Points somewhere in the above list */
1533 struct grid_face
*cur_face
;
1534 tree234
*lightable_faces_sorted
;
1535 tree234
*darkable_faces_sorted
;
1539 board
= snewn(num_faces
, char);
1542 memset(board
, FACE_GREY
, num_faces
);
1544 /* Create and initialise the list of face_scores */
1545 face_scores
= snewn(num_faces
, struct face_score
);
1546 for (i
= 0; i
< num_faces
; i
++) {
1547 face_scores
[i
].random
= random_bits(rs
, 31);
1548 face_scores
[i
].black_score
= face_scores
[i
].white_score
= 0;
1551 /* Colour a random, finite face white. The infinite face is implicitly
1552 * coloured black. Together, they will seed the random growth process
1553 * for the black and white areas. */
1554 i
= random_upto(rs
, num_faces
);
1555 board
[i
] = FACE_WHITE
;
1557 /* We need a way of favouring faces that will increase our loopiness.
1558 * We do this by maintaining a list of all candidate faces sorted by
1559 * their score and choose randomly from that with appropriate skew.
1560 * In order to avoid consistently biasing towards particular faces, we
1561 * need the sort order _within_ each group of scores to be completely
1562 * random. But it would be abusing the hospitality of the tree234 data
1563 * structure if our comparison function were nondeterministic :-). So with
1564 * each face we associate a random number that does not change during a
1565 * particular run of the generator, and use that as a secondary sort key.
1566 * Yes, this means we will be biased towards particular random faces in
1567 * any one run but that doesn't actually matter. */
1569 lightable_faces_sorted
= newtree234(white_sort_cmpfn
);
1570 darkable_faces_sorted
= newtree234(black_sort_cmpfn
);
1572 /* Initialise the lists of lightable and darkable faces. This is
1573 * slightly different from the code inside the while-loop, because we need
1574 * to check every face of the board (the grid structure does not keep a
1575 * list of the infinite face's neighbours). */
1576 for (i
= 0; i
< num_faces
; i
++) {
1577 grid_face
*f
= g
->faces
+ i
;
1578 struct face_score
*fs
= face_scores
+ i
;
1579 if (board
[i
] != FACE_GREY
) continue;
1580 /* We need the full colourability check here, it's not enough simply
1581 * to check neighbourhood. On some grids, a neighbour of the infinite
1582 * face is not necessarily darkable. */
1583 if (can_colour_face(g
, board
, i
, FACE_BLACK
)) {
1584 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1585 add234(darkable_faces_sorted
, fs
);
1587 if (can_colour_face(g
, board
, i
, FACE_WHITE
)) {
1588 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1589 add234(lightable_faces_sorted
, fs
);
1593 /* Colour faces one at a time until no more faces are colourable. */
1596 enum face_colour colour
;
1597 struct face_score
*fs_white
, *fs_black
;
1598 int c_lightable
= count234(lightable_faces_sorted
);
1599 int c_darkable
= count234(darkable_faces_sorted
);
1600 if (c_lightable
== 0 && c_darkable
== 0) {
1601 /* No more faces we can use at all. */
1604 assert(c_lightable
!= 0 && c_darkable
!= 0);
1606 fs_white
= (struct face_score
*)index234(lightable_faces_sorted
, 0);
1607 fs_black
= (struct face_score
*)index234(darkable_faces_sorted
, 0);
1609 /* Choose a colour, and colour the best available face
1610 * with that colour. */
1611 colour
= random_upto(rs
, 2) ? FACE_WHITE
: FACE_BLACK
;
1613 if (colour
== FACE_WHITE
)
1618 i
= fs
- face_scores
;
1619 assert(board
[i
] == FACE_GREY
);
1622 /* Remove this newly-coloured face from the lists. These lists should
1623 * only contain grey faces. */
1624 del234(lightable_faces_sorted
, fs
);
1625 del234(darkable_faces_sorted
, fs
);
1627 /* Remember which face we've just coloured */
1628 cur_face
= g
->faces
+ i
;
1630 /* The face we've just coloured potentially affects the colourability
1631 * and the scores of any neighbouring faces (touching at a corner or
1632 * edge). So the search needs to be conducted around all faces
1633 * touching the one we've just lit. Iterate over its corners, then
1634 * over each corner's faces. For each such face, we remove it from
1635 * the lists, recalculate any scores, then add it back to the lists
1636 * (depending on whether it is lightable, darkable or both). */
1637 for (i
= 0; i
< cur_face
->order
; i
++) {
1638 grid_dot
*d
= cur_face
->dots
[i
];
1639 for (j
= 0; j
< d
->order
; j
++) {
1640 grid_face
*f
= d
->faces
[j
];
1641 int fi
; /* face index of f */
1648 /* If the face is already coloured, it won't be on our
1649 * lightable/darkable lists anyway, so we can skip it without
1650 * bothering with the removal step. */
1651 if (FACE_COLOUR(f
) != FACE_GREY
) continue;
1653 /* Find the face index and face_score* corresponding to f */
1655 fs
= face_scores
+ fi
;
1657 /* Remove from lightable list if it's in there. We do this,
1658 * even if it is still lightable, because the score might
1659 * be different, and we need to remove-then-add to maintain
1660 * correct sort order. */
1661 del234(lightable_faces_sorted
, fs
);
1662 if (can_colour_face(g
, board
, fi
, FACE_WHITE
)) {
1663 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1664 add234(lightable_faces_sorted
, fs
);
1666 /* Do the same for darkable list. */
1667 del234(darkable_faces_sorted
, fs
);
1668 if (can_colour_face(g
, board
, fi
, FACE_BLACK
)) {
1669 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1670 add234(darkable_faces_sorted
, fs
);
1677 freetree234(lightable_faces_sorted
);
1678 freetree234(darkable_faces_sorted
);
1681 /* The next step requires a shuffled list of all faces */
1682 face_list
= snewn(num_faces
, int);
1683 for (i
= 0; i
< num_faces
; ++i
) {
1686 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1688 /* The above loop-generation algorithm can often leave large clumps
1689 * of faces of one colour. In extreme cases, the resulting path can be
1690 * degenerate and not very satisfying to solve.
1691 * This next step alleviates this problem:
1692 * Go through the shuffled list, and flip the colour of any face we can
1693 * legally flip, and which is adjacent to only one face of the opposite
1694 * colour - this tends to grow 'tendrils' into any clumps.
1695 * Repeat until we can find no more faces to flip. This will
1696 * eventually terminate, because each flip increases the loop's
1697 * perimeter, which cannot increase for ever.
1698 * The resulting path will have maximal loopiness (in the sense that it
1699 * cannot be improved "locally". Unfortunately, this allows a player to
1700 * make some illicit deductions. To combat this (and make the path more
1701 * interesting), we do one final pass making random flips. */
1703 /* Set to TRUE for final pass */
1704 do_random_pass
= FALSE
;
1707 /* Remember whether a flip occurred during this pass */
1708 int flipped
= FALSE
;
1710 for (i
= 0; i
< num_faces
; ++i
) {
1711 int j
= face_list
[i
];
1712 enum face_colour opp
=
1713 (board
[j
] == FACE_WHITE
) ? FACE_BLACK
: FACE_WHITE
;
1714 if (can_colour_face(g
, board
, j
, opp
)) {
1715 grid_face
*face
= g
->faces
+j
;
1716 if (do_random_pass
) {
1717 /* final random pass */
1718 if (!random_upto(rs
, 10))
1721 /* normal pass - flip when neighbour count is 1 */
1722 if (face_num_neighbours(g
, board
, face
, opp
) == 1) {
1730 if (do_random_pass
) break;
1731 if (!flipped
) do_random_pass
= TRUE
;
1736 /* Fill out all the clues by initialising to 0, then iterating over
1737 * all edges and incrementing each clue as we find edges that border
1738 * between BLACK/WHITE faces. While we're at it, we verify that the
1739 * algorithm does work, and there aren't any GREY faces still there. */
1740 memset(clues
, 0, num_faces
);
1741 for (i
= 0; i
< g
->num_edges
; i
++) {
1742 grid_edge
*e
= g
->edges
+ i
;
1743 grid_face
*f1
= e
->face1
;
1744 grid_face
*f2
= e
->face2
;
1745 enum face_colour c1
= FACE_COLOUR(f1
);
1746 enum face_colour c2
= FACE_COLOUR(f2
);
1747 assert(c1
!= FACE_GREY
);
1748 assert(c2
!= FACE_GREY
);
1750 if (f1
) clues
[f1
- g
->faces
]++;
1751 if (f2
) clues
[f2
- g
->faces
]++;
1759 static int game_has_unique_soln(const game_state
*state
, int diff
)
1762 solver_state
*sstate_new
;
1763 solver_state
*sstate
= new_solver_state((game_state
*)state
, diff
);
1765 sstate_new
= solve_game_rec(sstate
);
1767 assert(sstate_new
->solver_status
!= SOLVER_MISTAKE
);
1768 ret
= (sstate_new
->solver_status
== SOLVER_SOLVED
);
1770 free_solver_state(sstate_new
);
1771 free_solver_state(sstate
);
1777 /* Remove clues one at a time at random. */
1778 static game_state
*remove_clues(game_state
*state
, random_state
*rs
,
1782 int num_faces
= state
->game_grid
->num_faces
;
1783 game_state
*ret
= dup_game(state
), *saved_ret
;
1786 /* We need to remove some clues. We'll do this by forming a list of all
1787 * available clues, shuffling it, then going along one at a
1788 * time clearing each clue in turn for which doing so doesn't render the
1789 * board unsolvable. */
1790 face_list
= snewn(num_faces
, int);
1791 for (n
= 0; n
< num_faces
; ++n
) {
1795 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1797 for (n
= 0; n
< num_faces
; ++n
) {
1798 saved_ret
= dup_game(ret
);
1799 ret
->clues
[face_list
[n
]] = -1;
1801 if (game_has_unique_soln(ret
, diff
)) {
1802 free_game(saved_ret
);
1814 static char *new_game_desc(game_params
*params
, random_state
*rs
,
1815 char **aux
, int interactive
)
1817 /* solution and description both use run-length encoding in obvious ways */
1820 game_state
*state
= snew(game_state
);
1821 game_state
*state_new
;
1822 params_generate_grid(params
);
1823 state
->game_grid
= g
= params
->game_grid
;
1825 state
->clues
= snewn(g
->num_faces
, signed char);
1826 state
->lines
= snewn(g
->num_edges
, char);
1827 state
->line_errors
= snewn(g
->num_edges
, unsigned char);
1829 state
->grid_type
= params
->type
;
1833 memset(state
->lines
, LINE_UNKNOWN
, g
->num_edges
);
1834 memset(state
->line_errors
, 0, g
->num_edges
);
1836 state
->solved
= state
->cheated
= FALSE
;
1838 /* Get a new random solvable board with all its clues filled in. Yes, this
1839 * can loop for ever if the params are suitably unfavourable, but
1840 * preventing games smaller than 4x4 seems to stop this happening */
1842 add_full_clues(state
, rs
);
1843 } while (!game_has_unique_soln(state
, params
->diff
));
1845 state_new
= remove_clues(state
, rs
, params
->diff
);
1850 if (params
->diff
> 0 && game_has_unique_soln(state
, params
->diff
-1)) {
1852 fprintf(stderr
, "Rejecting board, it is too easy\n");
1854 goto newboard_please
;
1857 retval
= state_to_text(state
);
1861 assert(!validate_desc(params
, retval
));
1866 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1869 game_state
*state
= snew(game_state
);
1870 int empties_to_make
= 0;
1872 const char *dp
= desc
;
1874 int num_faces
, num_edges
;
1876 params_generate_grid(params
);
1877 state
->game_grid
= g
= params
->game_grid
;
1879 num_faces
= g
->num_faces
;
1880 num_edges
= g
->num_edges
;
1882 state
->clues
= snewn(num_faces
, signed char);
1883 state
->lines
= snewn(num_edges
, char);
1884 state
->line_errors
= snewn(num_edges
, unsigned char);
1886 state
->solved
= state
->cheated
= FALSE
;
1888 state
->grid_type
= params
->type
;
1890 for (i
= 0; i
< num_faces
; i
++) {
1891 if (empties_to_make
) {
1893 state
->clues
[i
] = -1;
1899 if (n
>= 0 && n
< 10) {
1900 state
->clues
[i
] = n
;
1904 state
->clues
[i
] = -1;
1905 empties_to_make
= n
- 1;
1910 memset(state
->lines
, LINE_UNKNOWN
, num_edges
);
1911 memset(state
->line_errors
, 0, num_edges
);
1915 /* Calculates the line_errors data, and checks if the current state is a
1917 static int check_completion(game_state
*state
)
1919 grid
*g
= state
->game_grid
;
1921 int num_faces
= g
->num_faces
;
1923 int infinite_area
, finite_area
;
1924 int loops_found
= 0;
1925 int found_edge_not_in_loop
= FALSE
;
1927 memset(state
->line_errors
, 0, g
->num_edges
);
1929 /* LL implementation of SGT's idea:
1930 * A loop will partition the grid into an inside and an outside.
1931 * If there is more than one loop, the grid will be partitioned into
1932 * even more distinct regions. We can therefore track equivalence of
1933 * faces, by saying that two faces are equivalent when there is a non-YES
1934 * edge between them.
1935 * We could keep track of the number of connected components, by counting
1936 * the number of dsf-merges that aren't no-ops.
1937 * But we're only interested in 3 separate cases:
1938 * no loops, one loop, more than one loop.
1940 * No loops: all faces are equivalent to the infinite face.
1941 * One loop: only two equivalence classes - finite and infinite.
1942 * >= 2 loops: there are 2 distinct finite regions.
1944 * So we simply make two passes through all the edges.
1945 * In the first pass, we dsf-merge the two faces bordering each non-YES
1947 * In the second pass, we look for YES-edges bordering:
1948 * a) two non-equivalent faces.
1949 * b) two non-equivalent faces, and one of them is part of a different
1950 * finite area from the first finite area we've seen.
1952 * An occurrence of a) means there is at least one loop.
1953 * An occurrence of b) means there is more than one loop.
1954 * Edges satisfying a) are marked as errors.
1956 * While we're at it, we set a flag if we find a YES edge that is not
1958 * This information will help decide, if there's a single loop, whether it
1959 * is a candidate for being a solution (that is, all YES edges are part of
1962 * If there is a candidate loop, we then go through all clues and check
1963 * they are all satisfied. If so, we have found a solution and we can
1964 * unmark all line_errors.
1967 /* Infinite face is at the end - its index is num_faces.
1968 * This macro is just to make this obvious! */
1969 #define INF_FACE num_faces
1970 dsf
= snewn(num_faces
+ 1, int);
1971 dsf_init(dsf
, num_faces
+ 1);
1974 for (i
= 0; i
< g
->num_edges
; i
++) {
1975 grid_edge
*e
= g
->edges
+ i
;
1976 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1977 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1978 if (state
->lines
[i
] != LINE_YES
)
1979 dsf_merge(dsf
, f1
, f2
);
1983 infinite_area
= dsf_canonify(dsf
, INF_FACE
);
1985 for (i
= 0; i
< g
->num_edges
; i
++) {
1986 grid_edge
*e
= g
->edges
+ i
;
1987 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1988 int can1
= dsf_canonify(dsf
, f1
);
1989 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1990 int can2
= dsf_canonify(dsf
, f2
);
1991 if (state
->lines
[i
] != LINE_YES
) continue;
1994 /* Faces are equivalent, so this edge not part of a loop */
1995 found_edge_not_in_loop
= TRUE
;
1998 state
->line_errors
[i
] = TRUE
;
1999 if (loops_found
== 0) loops_found
= 1;
2001 /* Don't bother with further checks if we've already found 2 loops */
2002 if (loops_found
== 2) continue;
2004 if (finite_area
== -1) {
2005 /* Found our first finite area */
2006 if (can1
!= infinite_area
)
2012 /* Have we found a second area? */
2013 if (finite_area
!= -1) {
2014 if (can1
!= infinite_area
&& can1
!= finite_area
) {
2018 if (can2
!= infinite_area
&& can2
!= finite_area
) {
2025 printf("loops_found = %d\n", loops_found);
2026 printf("found_edge_not_in_loop = %s\n",
2027 found_edge_not_in_loop ? "TRUE" : "FALSE");
2030 sfree(dsf
); /* No longer need the dsf */
2032 /* Have we found a candidate loop? */
2033 if (loops_found
== 1 && !found_edge_not_in_loop
) {
2034 /* Yes, so check all clues are satisfied */
2035 int found_clue_violation
= FALSE
;
2036 for (i
= 0; i
< num_faces
; i
++) {
2037 int c
= state
->clues
[i
];
2039 if (face_order(state
, i
, LINE_YES
) != c
) {
2040 found_clue_violation
= TRUE
;
2046 if (!found_clue_violation
) {
2047 /* The loop is good */
2048 memset(state
->line_errors
, 0, g
->num_edges
);
2049 return TRUE
; /* No need to bother checking for dot violations */
2053 /* Check for dot violations */
2054 for (i
= 0; i
< g
->num_dots
; i
++) {
2055 int yes
= dot_order(state
, i
, LINE_YES
);
2056 int unknown
= dot_order(state
, i
, LINE_UNKNOWN
);
2057 if ((yes
== 1 && unknown
== 0) || (yes
>= 3)) {
2058 /* violation, so mark all YES edges as errors */
2059 grid_dot
*d
= g
->dots
+ i
;
2061 for (j
= 0; j
< d
->order
; j
++) {
2062 int e
= d
->edges
[j
] - g
->edges
;
2063 if (state
->lines
[e
] == LINE_YES
)
2064 state
->line_errors
[e
] = TRUE
;
2071 /* ----------------------------------------------------------------------
2074 * Our solver modes operate as follows. Each mode also uses the modes above it.
2077 * Just implement the rules of the game.
2079 * Normal and Tricky Modes
2080 * For each (adjacent) pair of lines through each dot we store a bit for
2081 * whether at least one of them is on and whether at most one is on. (If we
2082 * know both or neither is on that's already stored more directly.)
2085 * Use edsf data structure to make equivalence classes of lines that are
2086 * known identical to or opposite to one another.
2091 * For general grids, we consider "dlines" to be pairs of lines joined
2092 * at a dot. The lines must be adjacent around the dot, so we can think of
2093 * a dline as being a dot+face combination. Or, a dot+edge combination where
2094 * the second edge is taken to be the next clockwise edge from the dot.
2095 * Original loopy code didn't have this extra restriction of the lines being
2096 * adjacent. From my tests with square grids, this extra restriction seems to
2097 * take little, if anything, away from the quality of the puzzles.
2098 * A dline can be uniquely identified by an edge/dot combination, given that
2099 * a dline-pair always goes clockwise around its common dot. The edge/dot
2100 * combination can be represented by an edge/bool combination - if bool is
2101 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2102 * exactly twice the number of edges in the grid - although the dlines
2103 * spanning the infinite face are not all that useful to the solver.
2104 * Note that, by convention, a dline goes clockwise around its common dot,
2105 * which means the dline goes anti-clockwise around its common face.
2108 /* Helper functions for obtaining an index into an array of dlines, given
2109 * various information. We assume the grid layout conventions about how
2110 * the various lists are interleaved - see grid_make_consistent() for
2113 /* i points to the first edge of the dline pair, reading clockwise around
2115 static int dline_index_from_dot(grid
*g
, grid_dot
*d
, int i
)
2117 grid_edge
*e
= d
->edges
[i
];
2122 if (i2
== d
->order
) i2
= 0;
2125 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2127 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2128 (int)(d
- g
->dots
), i
, (int)(e
- g
->edges
),
2129 (int)(e2
- g
->edges
), ret
);
2133 /* i points to the second edge of the dline pair, reading clockwise around
2134 * the face. That is, the edges of the dline, starting at edge{i}, read
2135 * anti-clockwise around the face. By layout conventions, the common dot
2136 * of the dline will be f->dots[i] */
2137 static int dline_index_from_face(grid
*g
, grid_face
*f
, int i
)
2139 grid_edge
*e
= f
->edges
[i
];
2140 grid_dot
*d
= f
->dots
[i
];
2145 if (i2
< 0) i2
+= f
->order
;
2148 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2150 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2151 (int)(f
- g
->faces
), i
, (int)(e
- g
->edges
),
2152 (int)(e2
- g
->edges
), ret
);
2156 static int is_atleastone(const char *dline_array
, int index
)
2158 return BIT_SET(dline_array
[index
], 0);
2160 static int set_atleastone(char *dline_array
, int index
)
2162 return SET_BIT(dline_array
[index
], 0);
2164 static int is_atmostone(const char *dline_array
, int index
)
2166 return BIT_SET(dline_array
[index
], 1);
2168 static int set_atmostone(char *dline_array
, int index
)
2170 return SET_BIT(dline_array
[index
], 1);
2173 static void array_setall(char *array
, char from
, char to
, int len
)
2175 char *p
= array
, *p_old
= p
;
2176 int len_remaining
= len
;
2178 while ((p
= memchr(p
, from
, len_remaining
))) {
2180 len_remaining
-= p
- p_old
;
2185 /* Helper, called when doing dline dot deductions, in the case where we
2186 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2187 * them (because of dline atmostone/atleastone).
2188 * On entry, edge points to the first of these two UNKNOWNs. This function
2189 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2190 * and set their corresponding dline to atleastone. (Setting atmostone
2191 * already happens in earlier dline deductions) */
2192 static int dline_set_opp_atleastone(solver_state
*sstate
,
2193 grid_dot
*d
, int edge
)
2195 game_state
*state
= sstate
->state
;
2196 grid
*g
= state
->game_grid
;
2199 for (opp
= 0; opp
< N
; opp
++) {
2200 int opp_dline_index
;
2201 if (opp
== edge
|| opp
== edge
+1 || opp
== edge
-1)
2203 if (opp
== 0 && edge
== N
-1)
2205 if (opp
== N
-1 && edge
== 0)
2208 if (opp2
== N
) opp2
= 0;
2209 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2210 if (state
->lines
[d
->edges
[opp
] - g
->edges
] != LINE_UNKNOWN
)
2212 if (state
->lines
[d
->edges
[opp2
] - g
->edges
] != LINE_UNKNOWN
)
2214 /* Found opposite UNKNOWNS and they're next to each other */
2215 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2216 return set_atleastone(sstate
->dlines
, opp_dline_index
);
2222 /* Set pairs of lines around this face which are known to be identical, to
2223 * the given line_state */
2224 static int face_setall_identical(solver_state
*sstate
, int face_index
,
2225 enum line_state line_new
)
2227 /* can[dir] contains the canonical line associated with the line in
2228 * direction dir from the square in question. Similarly inv[dir] is
2229 * whether or not the line in question is inverse to its canonical
2232 game_state
*state
= sstate
->state
;
2233 grid
*g
= state
->game_grid
;
2234 grid_face
*f
= g
->faces
+ face_index
;
2237 int can1
, can2
, inv1
, inv2
;
2239 for (i
= 0; i
< N
; i
++) {
2240 int line1_index
= f
->edges
[i
] - g
->edges
;
2241 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2243 for (j
= i
+ 1; j
< N
; j
++) {
2244 int line2_index
= f
->edges
[j
] - g
->edges
;
2245 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2248 /* Found two UNKNOWNS */
2249 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2250 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2251 if (can1
== can2
&& inv1
== inv2
) {
2252 solver_set_line(sstate
, line1_index
, line_new
);
2253 solver_set_line(sstate
, line2_index
, line_new
);
2260 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2261 * return the edge indices into e. */
2262 static void find_unknowns(game_state
*state
,
2263 grid_edge
**edge_list
, /* Edge list to search (from a face or a dot) */
2264 int expected_count
, /* Number of UNKNOWNs (comes from solver's cache) */
2265 int *e
/* Returned edge indices */)
2268 grid
*g
= state
->game_grid
;
2269 while (c
< expected_count
) {
2270 int line_index
= *edge_list
- g
->edges
;
2271 if (state
->lines
[line_index
] == LINE_UNKNOWN
) {
2279 /* If we have a list of edges, and we know whether the number of YESs should
2280 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2281 * linedsf deductions. This can be used for both face and dot deductions.
2282 * Returns the difficulty level of the next solver that should be used,
2283 * or DIFF_MAX if no progress was made. */
2284 static int parity_deductions(solver_state
*sstate
,
2285 grid_edge
**edge_list
, /* Edge list (from a face or a dot) */
2286 int total_parity
, /* Expected number of YESs modulo 2 (either 0 or 1) */
2289 game_state
*state
= sstate
->state
;
2290 int diff
= DIFF_MAX
;
2291 int *linedsf
= sstate
->linedsf
;
2293 if (unknown_count
== 2) {
2294 /* Lines are known alike/opposite, depending on inv. */
2296 find_unknowns(state
, edge_list
, 2, e
);
2297 if (merge_lines(sstate
, e
[0], e
[1], total_parity
))
2298 diff
= min(diff
, DIFF_HARD
);
2299 } else if (unknown_count
== 3) {
2301 int can
[3]; /* canonical edges */
2302 int inv
[3]; /* whether can[x] is inverse to e[x] */
2303 find_unknowns(state
, edge_list
, 3, e
);
2304 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2305 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2306 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2307 if (can
[0] == can
[1]) {
2308 if (solver_set_line(sstate
, e
[2], (total_parity
^inv
[0]^inv
[1]) ?
2309 LINE_YES
: LINE_NO
))
2310 diff
= min(diff
, DIFF_EASY
);
2312 if (can
[0] == can
[2]) {
2313 if (solver_set_line(sstate
, e
[1], (total_parity
^inv
[0]^inv
[2]) ?
2314 LINE_YES
: LINE_NO
))
2315 diff
= min(diff
, DIFF_EASY
);
2317 if (can
[1] == can
[2]) {
2318 if (solver_set_line(sstate
, e
[0], (total_parity
^inv
[1]^inv
[2]) ?
2319 LINE_YES
: LINE_NO
))
2320 diff
= min(diff
, DIFF_EASY
);
2322 } else if (unknown_count
== 4) {
2324 int can
[4]; /* canonical edges */
2325 int inv
[4]; /* whether can[x] is inverse to e[x] */
2326 find_unknowns(state
, edge_list
, 4, e
);
2327 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2328 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2329 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2330 can
[3] = edsf_canonify(linedsf
, e
[3], inv
+3);
2331 if (can
[0] == can
[1]) {
2332 if (merge_lines(sstate
, e
[2], e
[3], total_parity
^inv
[0]^inv
[1]))
2333 diff
= min(diff
, DIFF_HARD
);
2334 } else if (can
[0] == can
[2]) {
2335 if (merge_lines(sstate
, e
[1], e
[3], total_parity
^inv
[0]^inv
[2]))
2336 diff
= min(diff
, DIFF_HARD
);
2337 } else if (can
[0] == can
[3]) {
2338 if (merge_lines(sstate
, e
[1], e
[2], total_parity
^inv
[0]^inv
[3]))
2339 diff
= min(diff
, DIFF_HARD
);
2340 } else if (can
[1] == can
[2]) {
2341 if (merge_lines(sstate
, e
[0], e
[3], total_parity
^inv
[1]^inv
[2]))
2342 diff
= min(diff
, DIFF_HARD
);
2343 } else if (can
[1] == can
[3]) {
2344 if (merge_lines(sstate
, e
[0], e
[2], total_parity
^inv
[1]^inv
[3]))
2345 diff
= min(diff
, DIFF_HARD
);
2346 } else if (can
[2] == can
[3]) {
2347 if (merge_lines(sstate
, e
[0], e
[1], total_parity
^inv
[2]^inv
[3]))
2348 diff
= min(diff
, DIFF_HARD
);
2356 * These are the main solver functions.
2358 * Their return values are diff values corresponding to the lowest mode solver
2359 * that would notice the work that they have done. For example if the normal
2360 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2361 * easy mode solver might be able to make progress using that. It doesn't make
2362 * sense for one of them to return a diff value higher than that of the
2365 * Each function returns the lowest value it can, as early as possible, in
2366 * order to try and pass as much work as possible back to the lower level
2367 * solvers which progress more quickly.
2370 /* PROPOSED NEW DESIGN:
2371 * We have a work queue consisting of 'events' notifying us that something has
2372 * happened that a particular solver mode might be interested in. For example
2373 * the hard mode solver might do something that helps the normal mode solver at
2374 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2375 * we pull events off the work queue, and hand each in turn to the solver that
2376 * is interested in them. If a solver reports that it failed we pass the same
2377 * event on to progressively more advanced solvers and the loop detector. Once
2378 * we've exhausted an event, or it has helped us progress, we drop it and
2379 * continue to the next one. The events are sorted first in order of solver
2380 * complexity (easy first) then order of insertion (oldest first).
2381 * Once we run out of events we loop over each permitted solver in turn
2382 * (easiest first) until either a deduction is made (and an event therefore
2383 * emerges) or no further deductions can be made (in which case we've failed).
2386 * * How do we 'loop over' a solver when both dots and squares are concerned.
2387 * Answer: first all squares then all dots.
2390 static int trivial_deductions(solver_state
*sstate
)
2392 int i
, current_yes
, current_no
;
2393 game_state
*state
= sstate
->state
;
2394 grid
*g
= state
->game_grid
;
2395 int diff
= DIFF_MAX
;
2397 /* Per-face deductions */
2398 for (i
= 0; i
< g
->num_faces
; i
++) {
2399 grid_face
*f
= g
->faces
+ i
;
2401 if (sstate
->face_solved
[i
])
2404 current_yes
= sstate
->face_yes_count
[i
];
2405 current_no
= sstate
->face_no_count
[i
];
2407 if (current_yes
+ current_no
== f
->order
) {
2408 sstate
->face_solved
[i
] = TRUE
;
2412 if (state
->clues
[i
] < 0)
2415 if (state
->clues
[i
] < current_yes
) {
2416 sstate
->solver_status
= SOLVER_MISTAKE
;
2419 if (state
->clues
[i
] == current_yes
) {
2420 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
))
2421 diff
= min(diff
, DIFF_EASY
);
2422 sstate
->face_solved
[i
] = TRUE
;
2426 if (f
->order
- state
->clues
[i
] < current_no
) {
2427 sstate
->solver_status
= SOLVER_MISTAKE
;
2430 if (f
->order
- state
->clues
[i
] == current_no
) {
2431 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
))
2432 diff
= min(diff
, DIFF_EASY
);
2433 sstate
->face_solved
[i
] = TRUE
;
2438 check_caches(sstate
);
2440 /* Per-dot deductions */
2441 for (i
= 0; i
< g
->num_dots
; i
++) {
2442 grid_dot
*d
= g
->dots
+ i
;
2443 int yes
, no
, unknown
;
2445 if (sstate
->dot_solved
[i
])
2448 yes
= sstate
->dot_yes_count
[i
];
2449 no
= sstate
->dot_no_count
[i
];
2450 unknown
= d
->order
- yes
- no
;
2454 sstate
->dot_solved
[i
] = TRUE
;
2455 } else if (unknown
== 1) {
2456 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2457 diff
= min(diff
, DIFF_EASY
);
2458 sstate
->dot_solved
[i
] = TRUE
;
2460 } else if (yes
== 1) {
2462 sstate
->solver_status
= SOLVER_MISTAKE
;
2464 } else if (unknown
== 1) {
2465 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
);
2466 diff
= min(diff
, DIFF_EASY
);
2468 } else if (yes
== 2) {
2470 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2471 diff
= min(diff
, DIFF_EASY
);
2473 sstate
->dot_solved
[i
] = TRUE
;
2475 sstate
->solver_status
= SOLVER_MISTAKE
;
2480 check_caches(sstate
);
2485 static int dline_deductions(solver_state
*sstate
)
2487 game_state
*state
= sstate
->state
;
2488 grid
*g
= state
->game_grid
;
2489 char *dlines
= sstate
->dlines
;
2491 int diff
= DIFF_MAX
;
2493 /* ------ Face deductions ------ */
2495 /* Given a set of dline atmostone/atleastone constraints, need to figure
2496 * out if we can deduce any further info. For more general faces than
2497 * squares, this turns out to be a tricky problem.
2498 * The approach taken here is to define (per face) NxN matrices:
2499 * "maxs" and "mins".
2500 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2501 * for the possible number of edges that are YES between positions j and k
2502 * going clockwise around the face. Can think of j and k as marking dots
2503 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2504 * edge1 joins dot1 to dot2 etc).
2505 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2506 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2507 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2508 * the dline atmostone/atleastone status for edges j and j+1.
2510 * Then we calculate the remaining entries recursively. We definitely
2512 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2513 * This is because any valid placement of YESs between j and k must give
2514 * a valid placement between j and u, and also between u and k.
2515 * I believe it's sufficient to use just the two values of u:
2516 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2517 * are rigorous, even if they might not be best-possible.
2519 * Once we have maxs and mins calculated, we can make inferences about
2520 * each dline{j,j+1} by looking at the possible complementary edge-counts
2521 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2522 * As well as dlines, we can make similar inferences about single edges.
2523 * For example, consider a pentagon with clue 3, and we know at most one
2524 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2525 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2526 * that final edge would have to be YES to make the count up to 3.
2529 /* Much quicker to allocate arrays on the stack than the heap, so
2530 * define the largest possible face size, and base our array allocations
2531 * on that. We check this with an assertion, in case someone decides to
2532 * make a grid which has larger faces than this. Note, this algorithm
2533 * could get quite expensive if there are many large faces. */
2534 #define MAX_FACE_SIZE 8
2536 for (i
= 0; i
< g
->num_faces
; i
++) {
2537 int maxs
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2538 int mins
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2539 grid_face
*f
= g
->faces
+ i
;
2542 int clue
= state
->clues
[i
];
2543 assert(N
<= MAX_FACE_SIZE
);
2544 if (sstate
->face_solved
[i
])
2546 if (clue
< 0) continue;
2548 /* Calculate the (j,j+1) entries */
2549 for (j
= 0; j
< N
; j
++) {
2550 int edge_index
= f
->edges
[j
] - g
->edges
;
2552 enum line_state line1
= state
->lines
[edge_index
];
2553 enum line_state line2
;
2557 maxs
[j
][k
] = (line1
== LINE_NO
) ?
0 : 1;
2558 mins
[j
][k
] = (line1
== LINE_YES
) ?
1 : 0;
2559 /* Calculate the (j,j+2) entries */
2560 dline_index
= dline_index_from_face(g
, f
, k
);
2561 edge_index
= f
->edges
[k
] - g
->edges
;
2562 line2
= state
->lines
[edge_index
];
2568 if (line1
== LINE_NO
) tmp
--;
2569 if (line2
== LINE_NO
) tmp
--;
2570 if (tmp
== 2 && is_atmostone(dlines
, dline_index
))
2576 if (line1
== LINE_YES
) tmp
++;
2577 if (line2
== LINE_YES
) tmp
++;
2578 if (tmp
== 0 && is_atleastone(dlines
, dline_index
))
2583 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2584 for (m
= 3; m
< N
; m
++) {
2585 for (j
= 0; j
< N
; j
++) {
2593 maxs
[j
][k
] = maxs
[j
][u
] + maxs
[u
][k
];
2594 mins
[j
][k
] = mins
[j
][u
] + mins
[u
][k
];
2595 tmp
= maxs
[j
][v
] + maxs
[v
][k
];
2596 maxs
[j
][k
] = min(maxs
[j
][k
], tmp
);
2597 tmp
= mins
[j
][v
] + mins
[v
][k
];
2598 mins
[j
][k
] = max(mins
[j
][k
], tmp
);
2602 /* See if we can make any deductions */
2603 for (j
= 0; j
< N
; j
++) {
2605 grid_edge
*e
= f
->edges
[j
];
2606 int line_index
= e
- g
->edges
;
2609 if (state
->lines
[line_index
] != LINE_UNKNOWN
)
2614 /* minimum YESs in the complement of this edge */
2615 if (mins
[k
][j
] > clue
) {
2616 sstate
->solver_status
= SOLVER_MISTAKE
;
2619 if (mins
[k
][j
] == clue
) {
2620 /* setting this edge to YES would make at least
2621 * (clue+1) edges - contradiction */
2622 solver_set_line(sstate
, line_index
, LINE_NO
);
2623 diff
= min(diff
, DIFF_EASY
);
2625 if (maxs
[k
][j
] < clue
- 1) {
2626 sstate
->solver_status
= SOLVER_MISTAKE
;
2629 if (maxs
[k
][j
] == clue
- 1) {
2630 /* Only way to satisfy the clue is to set edge{j} as YES */
2631 solver_set_line(sstate
, line_index
, LINE_YES
);
2632 diff
= min(diff
, DIFF_EASY
);
2635 /* More advanced deduction that allows propagation along diagonal
2636 * chains of faces connected by dots, for example, 3-2-...-2-3
2637 * in square grids. */
2638 if (sstate
->diff
>= DIFF_TRICKY
) {
2639 /* Now see if we can make dline deduction for edges{j,j+1} */
2641 if (state
->lines
[e
- g
->edges
] != LINE_UNKNOWN
)
2642 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2643 * Dlines where one of the edges is known, are handled in the
2647 dline_index
= dline_index_from_face(g
, f
, k
);
2651 /* minimum YESs in the complement of this dline */
2652 if (mins
[k
][j
] > clue
- 2) {
2653 /* Adding 2 YESs would break the clue */
2654 if (set_atmostone(dlines
, dline_index
))
2655 diff
= min(diff
, DIFF_NORMAL
);
2657 /* maximum YESs in the complement of this dline */
2658 if (maxs
[k
][j
] < clue
) {
2659 /* Adding 2 NOs would mean not enough YESs */
2660 if (set_atleastone(dlines
, dline_index
))
2661 diff
= min(diff
, DIFF_NORMAL
);
2667 if (diff
< DIFF_NORMAL
)
2670 /* ------ Dot deductions ------ */
2672 for (i
= 0; i
< g
->num_dots
; i
++) {
2673 grid_dot
*d
= g
->dots
+ i
;
2675 int yes
, no
, unknown
;
2677 if (sstate
->dot_solved
[i
])
2679 yes
= sstate
->dot_yes_count
[i
];
2680 no
= sstate
->dot_no_count
[i
];
2681 unknown
= N
- yes
- no
;
2683 for (j
= 0; j
< N
; j
++) {
2686 int line1_index
, line2_index
;
2687 enum line_state line1
, line2
;
2690 dline_index
= dline_index_from_dot(g
, d
, j
);
2691 line1_index
= d
->edges
[j
] - g
->edges
;
2692 line2_index
= d
->edges
[k
] - g
->edges
;
2693 line1
= state
->lines
[line1_index
];
2694 line2
= state
->lines
[line2_index
];
2696 /* Infer dline state from line state */
2697 if (line1
== LINE_NO
|| line2
== LINE_NO
) {
2698 if (set_atmostone(dlines
, dline_index
))
2699 diff
= min(diff
, DIFF_NORMAL
);
2701 if (line1
== LINE_YES
|| line2
== LINE_YES
) {
2702 if (set_atleastone(dlines
, dline_index
))
2703 diff
= min(diff
, DIFF_NORMAL
);
2705 /* Infer line state from dline state */
2706 if (is_atmostone(dlines
, dline_index
)) {
2707 if (line1
== LINE_YES
&& line2
== LINE_UNKNOWN
) {
2708 solver_set_line(sstate
, line2_index
, LINE_NO
);
2709 diff
= min(diff
, DIFF_EASY
);
2711 if (line2
== LINE_YES
&& line1
== LINE_UNKNOWN
) {
2712 solver_set_line(sstate
, line1_index
, LINE_NO
);
2713 diff
= min(diff
, DIFF_EASY
);
2716 if (is_atleastone(dlines
, dline_index
)) {
2717 if (line1
== LINE_NO
&& line2
== LINE_UNKNOWN
) {
2718 solver_set_line(sstate
, line2_index
, LINE_YES
);
2719 diff
= min(diff
, DIFF_EASY
);
2721 if (line2
== LINE_NO
&& line1
== LINE_UNKNOWN
) {
2722 solver_set_line(sstate
, line1_index
, LINE_YES
);
2723 diff
= min(diff
, DIFF_EASY
);
2726 /* Deductions that depend on the numbers of lines.
2727 * Only bother if both lines are UNKNOWN, otherwise the
2728 * easy-mode solver (or deductions above) would have taken
2730 if (line1
!= LINE_UNKNOWN
|| line2
!= LINE_UNKNOWN
)
2733 if (yes
== 0 && unknown
== 2) {
2734 /* Both these unknowns must be identical. If we know
2735 * atmostone or atleastone, we can make progress. */
2736 if (is_atmostone(dlines
, dline_index
)) {
2737 solver_set_line(sstate
, line1_index
, LINE_NO
);
2738 solver_set_line(sstate
, line2_index
, LINE_NO
);
2739 diff
= min(diff
, DIFF_EASY
);
2741 if (is_atleastone(dlines
, dline_index
)) {
2742 solver_set_line(sstate
, line1_index
, LINE_YES
);
2743 solver_set_line(sstate
, line2_index
, LINE_YES
);
2744 diff
= min(diff
, DIFF_EASY
);
2748 if (set_atmostone(dlines
, dline_index
))
2749 diff
= min(diff
, DIFF_NORMAL
);
2751 if (set_atleastone(dlines
, dline_index
))
2752 diff
= min(diff
, DIFF_NORMAL
);
2756 /* More advanced deduction that allows propagation along diagonal
2757 * chains of faces connected by dots, for example: 3-2-...-2-3
2758 * in square grids. */
2759 if (sstate
->diff
>= DIFF_TRICKY
) {
2760 /* If we have atleastone set for this dline, infer
2761 * atmostone for each "opposite" dline (that is, each
2762 * dline without edges in common with this one).
2763 * Again, this test is only worth doing if both these
2764 * lines are UNKNOWN. For if one of these lines were YES,
2765 * the (yes == 1) test above would kick in instead. */
2766 if (is_atleastone(dlines
, dline_index
)) {
2768 for (opp
= 0; opp
< N
; opp
++) {
2769 int opp_dline_index
;
2770 if (opp
== j
|| opp
== j
+1 || opp
== j
-1)
2772 if (j
== 0 && opp
== N
-1)
2774 if (j
== N
-1 && opp
== 0)
2776 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2777 if (set_atmostone(dlines
, opp_dline_index
))
2778 diff
= min(diff
, DIFF_NORMAL
);
2780 if (yes
== 0 && is_atmostone(dlines
, dline_index
)) {
2781 /* This dline has *exactly* one YES and there are no
2782 * other YESs. This allows more deductions. */
2784 /* Third unknown must be YES */
2785 for (opp
= 0; opp
< N
; opp
++) {
2787 if (opp
== j
|| opp
== k
)
2789 opp_index
= d
->edges
[opp
] - g
->edges
;
2790 if (state
->lines
[opp_index
] == LINE_UNKNOWN
) {
2791 solver_set_line(sstate
, opp_index
,
2793 diff
= min(diff
, DIFF_EASY
);
2796 } else if (unknown
== 4) {
2797 /* Exactly one of opposite UNKNOWNS is YES. We've
2798 * already set atmostone, so set atleastone as
2801 if (dline_set_opp_atleastone(sstate
, d
, j
))
2802 diff
= min(diff
, DIFF_NORMAL
);
2812 static int linedsf_deductions(solver_state
*sstate
)
2814 game_state
*state
= sstate
->state
;
2815 grid
*g
= state
->game_grid
;
2816 char *dlines
= sstate
->dlines
;
2818 int diff
= DIFF_MAX
;
2821 /* ------ Face deductions ------ */
2823 /* A fully-general linedsf deduction seems overly complicated
2824 * (I suspect the problem is NP-complete, though in practice it might just
2825 * be doable because faces are limited in size).
2826 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2827 * known to be identical. If setting them both to YES (or NO) would break
2828 * the clue, set them to NO (or YES). */
2830 for (i
= 0; i
< g
->num_faces
; i
++) {
2831 int N
, yes
, no
, unknown
;
2834 if (sstate
->face_solved
[i
])
2836 clue
= state
->clues
[i
];
2840 N
= g
->faces
[i
].order
;
2841 yes
= sstate
->face_yes_count
[i
];
2842 if (yes
+ 1 == clue
) {
2843 if (face_setall_identical(sstate
, i
, LINE_NO
))
2844 diff
= min(diff
, DIFF_EASY
);
2846 no
= sstate
->face_no_count
[i
];
2847 if (no
+ 1 == N
- clue
) {
2848 if (face_setall_identical(sstate
, i
, LINE_YES
))
2849 diff
= min(diff
, DIFF_EASY
);
2852 /* Reload YES count, it might have changed */
2853 yes
= sstate
->face_yes_count
[i
];
2854 unknown
= N
- no
- yes
;
2856 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2857 * parity of lines. */
2858 diff_tmp
= parity_deductions(sstate
, g
->faces
[i
].edges
,
2859 (clue
- yes
) % 2, unknown
);
2860 diff
= min(diff
, diff_tmp
);
2863 /* ------ Dot deductions ------ */
2864 for (i
= 0; i
< g
->num_dots
; i
++) {
2865 grid_dot
*d
= g
->dots
+ i
;
2868 int yes
, no
, unknown
;
2869 /* Go through dlines, and do any dline<->linedsf deductions wherever
2870 * we find two UNKNOWNS. */
2871 for (j
= 0; j
< N
; j
++) {
2872 int dline_index
= dline_index_from_dot(g
, d
, j
);
2875 int can1
, can2
, inv1
, inv2
;
2877 line1_index
= d
->edges
[j
] - g
->edges
;
2878 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2881 if (j2
== N
) j2
= 0;
2882 line2_index
= d
->edges
[j2
] - g
->edges
;
2883 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2885 /* Infer dline flags from linedsf */
2886 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2887 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2888 if (can1
== can2
&& inv1
!= inv2
) {
2889 /* These are opposites, so set dline atmostone/atleastone */
2890 if (set_atmostone(dlines
, dline_index
))
2891 diff
= min(diff
, DIFF_NORMAL
);
2892 if (set_atleastone(dlines
, dline_index
))
2893 diff
= min(diff
, DIFF_NORMAL
);
2896 /* Infer linedsf from dline flags */
2897 if (is_atmostone(dlines
, dline_index
)
2898 && is_atleastone(dlines
, dline_index
)) {
2899 if (merge_lines(sstate
, line1_index
, line2_index
, 1))
2900 diff
= min(diff
, DIFF_HARD
);
2904 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2905 * parity of lines. */
2906 yes
= sstate
->dot_yes_count
[i
];
2907 no
= sstate
->dot_no_count
[i
];
2908 unknown
= N
- yes
- no
;
2909 diff_tmp
= parity_deductions(sstate
, d
->edges
,
2911 diff
= min(diff
, diff_tmp
);
2914 /* ------ Edge dsf deductions ------ */
2916 /* If the state of a line is known, deduce the state of its canonical line
2917 * too, and vice versa. */
2918 for (i
= 0; i
< g
->num_edges
; i
++) {
2921 can
= edsf_canonify(sstate
->linedsf
, i
, &inv
);
2924 s
= sstate
->state
->lines
[can
];
2925 if (s
!= LINE_UNKNOWN
) {
2926 if (solver_set_line(sstate
, i
, inv ?
OPP(s
) : s
))
2927 diff
= min(diff
, DIFF_EASY
);
2929 s
= sstate
->state
->lines
[i
];
2930 if (s
!= LINE_UNKNOWN
) {
2931 if (solver_set_line(sstate
, can
, inv ?
OPP(s
) : s
))
2932 diff
= min(diff
, DIFF_EASY
);
2940 static int loop_deductions(solver_state
*sstate
)
2942 int edgecount
= 0, clues
= 0, satclues
= 0, sm1clues
= 0;
2943 game_state
*state
= sstate
->state
;
2944 grid
*g
= state
->game_grid
;
2945 int shortest_chainlen
= g
->num_dots
;
2946 int loop_found
= FALSE
;
2948 int progress
= FALSE
;
2952 * Go through the grid and update for all the new edges.
2953 * Since merge_dots() is idempotent, the simplest way to
2954 * do this is just to update for _all_ the edges.
2955 * Also, while we're here, we count the edges.
2957 for (i
= 0; i
< g
->num_edges
; i
++) {
2958 if (state
->lines
[i
] == LINE_YES
) {
2959 loop_found
|= merge_dots(sstate
, i
);
2965 * Count the clues, count the satisfied clues, and count the
2966 * satisfied-minus-one clues.
2968 for (i
= 0; i
< g
->num_faces
; i
++) {
2969 int c
= state
->clues
[i
];
2971 int o
= sstate
->face_yes_count
[i
];
2980 for (i
= 0; i
< g
->num_dots
; ++i
) {
2982 sstate
->looplen
[dsf_canonify(sstate
->dotdsf
, i
)];
2983 if (dots_connected
> 1)
2984 shortest_chainlen
= min(shortest_chainlen
, dots_connected
);
2987 assert(sstate
->solver_status
== SOLVER_INCOMPLETE
);
2989 if (satclues
== clues
&& shortest_chainlen
== edgecount
) {
2990 sstate
->solver_status
= SOLVER_SOLVED
;
2991 /* This discovery clearly counts as progress, even if we haven't
2992 * just added any lines or anything */
2994 goto finished_loop_deductionsing
;
2998 * Now go through looking for LINE_UNKNOWN edges which
2999 * connect two dots that are already in the same
3000 * equivalence class. If we find one, test to see if the
3001 * loop it would create is a solution.
3003 for (i
= 0; i
< g
->num_edges
; i
++) {
3004 grid_edge
*e
= g
->edges
+ i
;
3005 int d1
= e
->dot1
- g
->dots
;
3006 int d2
= e
->dot2
- g
->dots
;
3008 if (state
->lines
[i
] != LINE_UNKNOWN
)
3011 eqclass
= dsf_canonify(sstate
->dotdsf
, d1
);
3012 if (eqclass
!= dsf_canonify(sstate
->dotdsf
, d2
))
3015 val
= LINE_NO
; /* loop is bad until proven otherwise */
3018 * This edge would form a loop. Next
3019 * question: how long would the loop be?
3020 * Would it equal the total number of edges
3021 * (plus the one we'd be adding if we added
3024 if (sstate
->looplen
[eqclass
] == edgecount
+ 1) {
3028 * This edge would form a loop which
3029 * took in all the edges in the entire
3030 * grid. So now we need to work out
3031 * whether it would be a valid solution
3032 * to the puzzle, which means we have to
3033 * check if it satisfies all the clues.
3034 * This means that every clue must be
3035 * either satisfied or satisfied-minus-
3036 * 1, and also that the number of
3037 * satisfied-minus-1 clues must be at
3038 * most two and they must lie on either
3039 * side of this edge.
3043 int f
= e
->face1
- g
->faces
;
3044 int c
= state
->clues
[f
];
3045 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3049 int f
= e
->face2
- g
->faces
;
3050 int c
= state
->clues
[f
];
3051 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3054 if (sm1clues
== sm1_nearby
&&
3055 sm1clues
+ satclues
== clues
) {
3056 val
= LINE_YES
; /* loop is good! */
3061 * Right. Now we know that adding this edge
3062 * would form a loop, and we know whether
3063 * that loop would be a viable solution or
3066 * If adding this edge produces a solution,
3067 * then we know we've found _a_ solution but
3068 * we don't know that it's _the_ solution -
3069 * if it were provably the solution then
3070 * we'd have deduced this edge some time ago
3071 * without the need to do loop detection. So
3072 * in this state we return SOLVER_AMBIGUOUS,
3073 * which has the effect that hitting Solve
3074 * on a user-provided puzzle will fill in a
3075 * solution but using the solver to
3076 * construct new puzzles won't consider this
3077 * a reasonable deduction for the user to
3080 progress
= solver_set_line(sstate
, i
, val
);
3081 assert(progress
== TRUE
);
3082 if (val
== LINE_YES
) {
3083 sstate
->solver_status
= SOLVER_AMBIGUOUS
;
3084 goto finished_loop_deductionsing
;
3088 finished_loop_deductionsing
:
3089 return progress ? DIFF_EASY
: DIFF_MAX
;
3092 /* This will return a dynamically allocated solver_state containing the (more)
3094 static solver_state
*solve_game_rec(const solver_state
*sstate_start
)
3096 solver_state
*sstate
;
3098 /* Index of the solver we should call next. */
3101 /* As a speed-optimisation, we avoid re-running solvers that we know
3102 * won't make any progress. This happens when a high-difficulty
3103 * solver makes a deduction that can only help other high-difficulty
3105 * For example: if a new 'dline' flag is set by dline_deductions, the
3106 * trivial_deductions solver cannot do anything with this information.
3107 * If we've already run the trivial_deductions solver (because it's
3108 * earlier in the list), there's no point running it again.
3110 * Therefore: if a solver is earlier in the list than "threshold_index",
3111 * we don't bother running it if it's difficulty level is less than
3114 int threshold_diff
= 0;
3115 int threshold_index
= 0;
3117 sstate
= dup_solver_state(sstate_start
);
3119 check_caches(sstate
);
3121 while (i
< NUM_SOLVERS
) {
3122 if (sstate
->solver_status
== SOLVER_MISTAKE
)
3124 if (sstate
->solver_status
== SOLVER_SOLVED
||
3125 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3126 /* solver finished */
3130 if ((solver_diffs
[i
] >= threshold_diff
|| i
>= threshold_index
)
3131 && solver_diffs
[i
] <= sstate
->diff
) {
3132 /* current_solver is eligible, so use it */
3133 int next_diff
= solver_fns
[i
](sstate
);
3134 if (next_diff
!= DIFF_MAX
) {
3135 /* solver made progress, so use new thresholds and
3136 * start again at top of list. */
3137 threshold_diff
= next_diff
;
3138 threshold_index
= i
;
3143 /* current_solver is ineligible, or failed to make progress, so
3144 * go to the next solver in the list */
3148 if (sstate
->solver_status
== SOLVER_SOLVED
||
3149 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3150 /* s/LINE_UNKNOWN/LINE_NO/g */
3151 array_setall(sstate
->state
->lines
, LINE_UNKNOWN
, LINE_NO
,
3152 sstate
->state
->game_grid
->num_edges
);
3159 static char *solve_game(game_state
*state
, game_state
*currstate
,
3160 char *aux
, char **error
)
3163 solver_state
*sstate
, *new_sstate
;
3165 sstate
= new_solver_state(state
, DIFF_MAX
);
3166 new_sstate
= solve_game_rec(sstate
);
3168 if (new_sstate
->solver_status
== SOLVER_SOLVED
) {
3169 soln
= encode_solve_move(new_sstate
->state
);
3170 } else if (new_sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3171 soln
= encode_solve_move(new_sstate
->state
);
3172 /**error = "Solver found ambiguous solutions"; */
3174 soln
= encode_solve_move(new_sstate
->state
);
3175 /**error = "Solver failed"; */
3178 free_solver_state(new_sstate
);
3179 free_solver_state(sstate
);
3184 /* ----------------------------------------------------------------------
3185 * Drawing and mouse-handling
3188 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
3189 int x
, int y
, int button
)
3191 grid
*g
= state
->game_grid
;
3195 char button_char
= ' ';
3196 enum line_state old_state
;
3198 button
&= ~MOD_MASK
;
3200 /* Convert mouse-click (x,y) to grid coordinates */
3201 x
-= BORDER(ds
->tilesize
);
3202 y
-= BORDER(ds
->tilesize
);
3203 x
= x
* g
->tilesize
/ ds
->tilesize
;
3204 y
= y
* g
->tilesize
/ ds
->tilesize
;
3208 e
= grid_nearest_edge(g
, x
, y
);
3214 /* I think it's only possible to play this game with mouse clicks, sorry */
3215 /* Maybe will add mouse drag support some time */
3216 old_state
= state
->lines
[i
];
3220 switch (old_state
) {
3238 switch (old_state
) {
3257 sprintf(buf
, "%d%c", i
, (int)button_char
);
3263 static game_state
*execute_move(game_state
*state
, char *move
)
3266 game_state
*newstate
= dup_game(state
);
3268 if (move
[0] == 'S') {
3270 newstate
->cheated
= TRUE
;
3275 if (i
< 0 || i
>= newstate
->game_grid
->num_edges
)
3277 move
+= strspn(move
, "1234567890");
3278 switch (*(move
++)) {
3280 newstate
->lines
[i
] = LINE_YES
;
3283 newstate
->lines
[i
] = LINE_NO
;
3286 newstate
->lines
[i
] = LINE_UNKNOWN
;
3294 * Check for completion.
3296 if (check_completion(newstate
))
3297 newstate
->solved
= TRUE
;
3302 free_game(newstate
);
3306 /* ----------------------------------------------------------------------
3310 /* Convert from grid coordinates to screen coordinates */
3311 static void grid_to_screen(const game_drawstate
*ds
, const grid
*g
,
3312 int grid_x
, int grid_y
, int *x
, int *y
)
3314 *x
= grid_x
- g
->lowest_x
;
3315 *y
= grid_y
- g
->lowest_y
;
3316 *x
= *x
* ds
->tilesize
/ g
->tilesize
;
3317 *y
= *y
* ds
->tilesize
/ g
->tilesize
;
3318 *x
+= BORDER(ds
->tilesize
);
3319 *y
+= BORDER(ds
->tilesize
);
3322 /* Returns (into x,y) position of centre of face for rendering the text clue.
3324 static void face_text_pos(const game_drawstate
*ds
, const grid
*g
,
3325 const grid_face
*f
, int *x
, int *y
)
3329 /* Simplest solution is the centroid. Might not work in some cases. */
3331 /* Another algorithm to look into:
3332 * Find the midpoints of the sides, find the bounding-box,
3333 * then take the centre of that. */
3335 /* Best solution probably involves incentres (inscribed circles) */
3337 int sx
= 0, sy
= 0; /* sums */
3338 for (i
= 0; i
< f
->order
; i
++) {
3339 grid_dot
*d
= f
->dots
[i
];
3346 /* convert to screen coordinates */
3347 grid_to_screen(ds
, g
, sx
, sy
, x
, y
);
3350 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
3351 game_state
*state
, int dir
, game_ui
*ui
,
3352 float animtime
, float flashtime
)
3354 grid
*g
= state
->game_grid
;
3355 int border
= BORDER(ds
->tilesize
);
3358 int line_colour
, flash_changed
;
3364 * The initial contents of the window are not guaranteed and
3365 * can vary with front ends. To be on the safe side, all games
3366 * should start by drawing a big background-colour rectangle
3367 * covering the whole window.
3369 int grid_width
= g
->highest_x
- g
->lowest_x
;
3370 int grid_height
= g
->highest_y
- g
->lowest_y
;
3371 int w
= grid_width
* ds
->tilesize
/ g
->tilesize
;
3372 int h
= grid_height
* ds
->tilesize
/ g
->tilesize
;
3373 draw_rect(dr
, 0, 0, w
+ 2 * border
+ 1, h
+ 2 * border
+ 1,
3377 for (i
= 0; i
< g
->num_faces
; i
++) {
3381 c
[0] = CLUE2CHAR(state
->clues
[i
]);
3384 face_text_pos(ds
, g
, f
, &x
, &y
);
3385 draw_text(dr
, x
, y
, FONT_VARIABLE
, ds
->tilesize
/2,
3386 ALIGN_VCENTRE
| ALIGN_HCENTRE
, COL_FOREGROUND
, c
);
3388 draw_update(dr
, 0, 0, w
+ 2 * border
, h
+ 2 * border
);
3391 if (flashtime
> 0 &&
3392 (flashtime
<= FLASH_TIME
/3 ||
3393 flashtime
>= FLASH_TIME
*2/3)) {
3394 flash_changed
= !ds
->flashing
;
3395 ds
->flashing
= TRUE
;
3397 flash_changed
= ds
->flashing
;
3398 ds
->flashing
= FALSE
;
3401 /* Some platforms may perform anti-aliasing, which may prevent clean
3402 * repainting of lines when the colour is changed.
3403 * If a line needs to be over-drawn in a different colour, erase a
3404 * bounding-box around the line, then flag all nearby objects for redraw.
3407 const char redraw_flag
= (char)(1<<7);
3408 for (i
= 0; i
< g
->num_edges
; i
++) {
3409 char prev_ds
= (ds
->lines
[i
] & ~redraw_flag
);
3410 char new_ds
= state
->lines
[i
];
3411 if (state
->line_errors
[i
])
3412 new_ds
= DS_LINE_ERROR
;
3414 /* If we're changing state, AND
3415 * the previous state was a coloured line */
3416 if ((prev_ds
!= new_ds
) && (prev_ds
!= LINE_NO
)) {
3417 grid_edge
*e
= g
->edges
+ i
;
3418 int x1
= e
->dot1
->x
;
3419 int y1
= e
->dot1
->y
;
3420 int x2
= e
->dot2
->x
;
3421 int y2
= e
->dot2
->y
;
3422 int xmin
, xmax
, ymin
, ymax
;
3424 grid_to_screen(ds
, g
, x1
, y1
, &x1
, &y1
);
3425 grid_to_screen(ds
, g
, x2
, y2
, &x2
, &y2
);
3426 /* Allow extra margin for dots, and thickness of lines */
3427 xmin
= min(x1
, x2
) - 2;
3428 xmax
= max(x1
, x2
) + 2;
3429 ymin
= min(y1
, y2
) - 2;
3430 ymax
= max(y1
, y2
) + 2;
3431 /* For testing, I find it helpful to change COL_BACKGROUND
3432 * to COL_SATISFIED here. */
3433 draw_rect(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1,
3435 draw_update(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1);
3437 /* Mark nearby lines for redraw */
3438 for (j
= 0; j
< e
->dot1
->order
; j
++)
3439 ds
->lines
[e
->dot1
->edges
[j
] - g
->edges
] |= redraw_flag
;
3440 for (j
= 0; j
< e
->dot2
->order
; j
++)
3441 ds
->lines
[e
->dot2
->edges
[j
] - g
->edges
] |= redraw_flag
;
3442 /* Mark nearby clues for redraw. Use a value that is
3443 * neither TRUE nor FALSE for this. */
3445 ds
->clue_error
[e
->face1
- g
->faces
] = 2;
3447 ds
->clue_error
[e
->face2
- g
->faces
] = 2;
3452 /* Redraw clue colours if necessary */
3453 for (i
= 0; i
< g
->num_faces
; i
++) {
3454 grid_face
*f
= g
->faces
+ i
;
3455 int sides
= f
->order
;
3457 n
= state
->clues
[i
];
3461 c
[0] = CLUE2CHAR(n
);
3464 clue_mistake
= (face_order(state
, i
, LINE_YES
) > n
||
3465 face_order(state
, i
, LINE_NO
) > (sides
-n
));
3467 clue_satisfied
= (face_order(state
, i
, LINE_YES
) == n
&&
3468 face_order(state
, i
, LINE_NO
) == (sides
-n
));
3470 if (clue_mistake
!= ds
->clue_error
[i
]
3471 || clue_satisfied
!= ds
->clue_satisfied
[i
]) {
3473 face_text_pos(ds
, g
, f
, &x
, &y
);
3474 /* There seems to be a certain amount of trial-and-error
3475 * involved in working out the correct bounding-box for
3477 draw_rect(dr
, x
- ds
->tilesize
/4 - 1, y
- ds
->tilesize
/4 - 3,
3478 ds
->tilesize
/2 + 2, ds
->tilesize
/2 + 5,
3481 FONT_VARIABLE
, ds
->tilesize
/2,
3482 ALIGN_VCENTRE
| ALIGN_HCENTRE
,
3483 clue_mistake ? COL_MISTAKE
:
3484 clue_satisfied ? COL_SATISFIED
: COL_FOREGROUND
, c
);
3485 draw_update(dr
, x
- ds
->tilesize
/4 - 1, y
- ds
->tilesize
/4 - 3,
3486 ds
->tilesize
/2 + 2, ds
->tilesize
/2 + 5);
3488 ds
->clue_error
[i
] = clue_mistake
;
3489 ds
->clue_satisfied
[i
] = clue_satisfied
;
3491 /* Sometimes, the bounding-box encroaches into the surrounding
3492 * lines (particularly if the window is resized fairly small).
3493 * So redraw them. */
3494 for (j
= 0; j
< f
->order
; j
++)
3495 ds
->lines
[f
->edges
[j
] - g
->edges
] = -1;
3500 for (i
= 0; i
< g
->num_edges
; i
++) {
3501 grid_edge
*e
= g
->edges
+ i
;
3503 int xmin
, ymin
, xmax
, ymax
;
3504 char new_ds
, need_draw
;
3505 new_ds
= state
->lines
[i
];
3506 if (state
->line_errors
[i
])
3507 new_ds
= DS_LINE_ERROR
;
3508 need_draw
= (new_ds
!= ds
->lines
[i
]) ? TRUE
: FALSE
;
3509 if (flash_changed
&& (state
->lines
[i
] == LINE_YES
))
3512 need_draw
= TRUE
; /* draw everything at the start */
3513 ds
->lines
[i
] = new_ds
;
3516 if (state
->line_errors
[i
])
3517 line_colour
= COL_MISTAKE
;
3518 else if (state
->lines
[i
] == LINE_UNKNOWN
)
3519 line_colour
= COL_LINEUNKNOWN
;
3520 else if (state
->lines
[i
] == LINE_NO
)
3521 line_colour
= COL_FAINT
;
3522 else if (ds
->flashing
)
3523 line_colour
= COL_HIGHLIGHT
;
3525 line_colour
= COL_FOREGROUND
;
3527 /* Convert from grid to screen coordinates */
3528 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3529 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3536 if (line_colour
== COL_FAINT
) {
3537 static int draw_faint_lines
= -1;
3538 if (draw_faint_lines
< 0) {
3539 char *env
= getenv("LOOPY_FAINT_LINES");
3540 draw_faint_lines
= (!env
|| (env
[0] == 'y' ||
3543 if (draw_faint_lines
)
3544 draw_line(dr
, x1
, y1
, x2
, y2
, line_colour
);
3546 /* (dx, dy) points roughly from (x1, y1) to (x2, y2).
3547 * The line is then "fattened" in a (roughly) perpendicular
3548 * direction to create a thin rectangle. */
3549 int dx
= (x1
> x2
) ?
-1 : ((x1
< x2
) ?
1 : 0);
3550 int dy
= (y1
> y2
) ?
-1 : ((y1
< y2
) ?
1 : 0);
3552 points
[0] = x1
+ dy
;
3553 points
[1] = y1
- dx
;
3554 points
[2] = x1
- dy
;
3555 points
[3] = y1
+ dx
;
3556 points
[4] = x2
- dy
;
3557 points
[5] = y2
+ dx
;
3558 points
[6] = x2
+ dy
;
3559 points
[7] = y2
- dx
;
3560 draw_polygon(dr
, points
, 4, line_colour
, line_colour
);
3563 /* Draw dots at ends of the line */
3564 draw_circle(dr
, x1
, y1
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3565 draw_circle(dr
, x2
, y2
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3567 draw_update(dr
, xmin
-2, ymin
-2, xmax
- xmin
+ 4, ymax
- ymin
+ 4);
3572 for (i
= 0; i
< g
->num_dots
; i
++) {
3573 grid_dot
*d
= g
->dots
+ i
;
3575 grid_to_screen(ds
, g
, d
->x
, d
->y
, &x
, &y
);
3576 draw_circle(dr
, x
, y
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3582 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
3583 int dir
, game_ui
*ui
)
3585 if (!oldstate
->solved
&& newstate
->solved
&&
3586 !oldstate
->cheated
&& !newstate
->cheated
) {
3593 static void game_print_size(game_params
*params
, float *x
, float *y
)
3598 * I'll use 7mm "squares" by default.
3600 game_compute_size(params
, 700, &pw
, &ph
);
3605 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
3607 int ink
= print_mono_colour(dr
, 0);
3609 game_drawstate ads
, *ds
= &ads
;
3610 grid
*g
= state
->game_grid
;
3612 ds
->tilesize
= tilesize
;
3614 for (i
= 0; i
< g
->num_dots
; i
++) {
3616 grid_to_screen(ds
, g
, g
->dots
[i
].x
, g
->dots
[i
].y
, &x
, &y
);
3617 draw_circle(dr
, x
, y
, ds
->tilesize
/ 15, ink
, ink
);
3623 for (i
= 0; i
< g
->num_faces
; i
++) {
3624 grid_face
*f
= g
->faces
+ i
;
3625 int clue
= state
->clues
[i
];
3629 c
[0] = CLUE2CHAR(clue
);
3631 face_text_pos(ds
, g
, f
, &x
, &y
);
3633 FONT_VARIABLE
, ds
->tilesize
/ 2,
3634 ALIGN_VCENTRE
| ALIGN_HCENTRE
, ink
, c
);
3641 for (i
= 0; i
< g
->num_edges
; i
++) {
3642 int thickness
= (state
->lines
[i
] == LINE_YES
) ?
30 : 150;
3643 grid_edge
*e
= g
->edges
+ i
;
3645 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3646 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3647 if (state
->lines
[i
] == LINE_YES
)
3649 /* (dx, dy) points from (x1, y1) to (x2, y2).
3650 * The line is then "fattened" in a perpendicular
3651 * direction to create a thin rectangle. */
3652 double d
= sqrt(SQ((double)x1
- x2
) + SQ((double)y1
- y2
));
3653 double dx
= (x2
- x1
) / d
;
3654 double dy
= (y2
- y1
) / d
;
3657 dx
= (dx
* ds
->tilesize
) / thickness
;
3658 dy
= (dy
* ds
->tilesize
) / thickness
;
3659 points
[0] = x1
+ (int)dy
;
3660 points
[1] = y1
- (int)dx
;
3661 points
[2] = x1
- (int)dy
;
3662 points
[3] = y1
+ (int)dx
;
3663 points
[4] = x2
- (int)dy
;
3664 points
[5] = y2
+ (int)dx
;
3665 points
[6] = x2
+ (int)dy
;
3666 points
[7] = y2
- (int)dx
;
3667 draw_polygon(dr
, points
, 4, ink
, ink
);
3671 /* Draw a dotted line */
3674 for (j
= 1; j
< divisions
; j
++) {
3675 /* Weighted average */
3676 int x
= (x1
* (divisions
-j
) + x2
* j
) / divisions
;
3677 int y
= (y1
* (divisions
-j
) + y2
* j
) / divisions
;
3678 draw_circle(dr
, x
, y
, ds
->tilesize
/ thickness
, ink
, ink
);
3685 #define thegame loopy
3688 const struct game thegame
= {
3689 "Loopy", "games.loopy", "loopy",
3696 TRUE
, game_configure
, custom_params
,
3704 TRUE
, game_can_format_as_text_now
, game_text_format
,
3712 PREFERRED_TILE_SIZE
, game_compute_size
, game_set_size
,
3715 game_free_drawstate
,
3719 TRUE
, FALSE
, game_print_size
, game_print
,
3720 FALSE
/* wants_statusbar */,
3721 FALSE
, game_timing_state
,
3722 0, /* mouse_priorities */
3725 #ifdef STANDALONE_SOLVER
3728 * Half-hearted standalone solver. It can't output the solution to
3729 * anything but a square puzzle, and it can't log the deductions
3730 * it makes either. But it can solve square puzzles, and more
3731 * importantly it can use its solver to grade the difficulty of
3732 * any puzzle you give it.
3737 int main(int argc
, char **argv
)
3741 char *id
= NULL
, *desc
, *err
;
3744 #if 0 /* verbose solver not supported here (yet) */
3745 int really_verbose
= FALSE
;
3748 while (--argc
> 0) {
3750 #if 0 /* verbose solver not supported here (yet) */
3751 if (!strcmp(p
, "-v")) {
3752 really_verbose
= TRUE
;
3755 if (!strcmp(p
, "-g")) {
3757 } else if (*p
== '-') {
3758 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
3766 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
3770 desc
= strchr(id
, ':');
3772 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
3777 p
= default_params();
3778 decode_params(p
, id
);
3779 err
= validate_desc(p
, desc
);
3781 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
3784 s
= new_game(NULL
, p
, desc
);
3787 * When solving an Easy puzzle, we don't want to bother the
3788 * user with Hard-level deductions. For this reason, we grade
3789 * the puzzle internally before doing anything else.
3791 ret
= -1; /* placate optimiser */
3792 for (diff
= 0; diff
< DIFF_MAX
; diff
++) {
3793 solver_state
*sstate_new
;
3794 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3796 sstate_new
= solve_game_rec(sstate
);
3798 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3800 else if (sstate_new
->solver_status
== SOLVER_SOLVED
)
3805 free_solver_state(sstate_new
);
3806 free_solver_state(sstate
);
3812 if (diff
== DIFF_MAX
) {
3814 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3816 printf("Unable to find a unique solution\n");
3820 printf("Difficulty rating: impossible (no solution exists)\n");
3822 printf("Difficulty rating: %s\n", diffnames
[diff
]);
3824 solver_state
*sstate_new
;
3825 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3827 /* If we supported a verbose solver, we'd set verbosity here */
3829 sstate_new
= solve_game_rec(sstate
);
3831 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3832 printf("Puzzle is inconsistent\n");
3834 assert(sstate_new
->solver_status
== SOLVER_SOLVED
);
3835 if (s
->grid_type
== 0) {
3836 fputs(game_text_format(sstate_new
->state
), stdout
);
3838 printf("Unable to output non-square grids\n");
3842 free_solver_state(sstate_new
);
3843 free_solver_state(sstate
);