4 * An implementation of the Nikoli game 'Loop the loop'.
5 * (c) Mike Pinna, 2005, 2006
6 * Substantially rewritten to allowing for more general types of grid.
7 * (c) Lambros Lambrou 2008
9 * vim: set shiftwidth=4 :set textwidth=80:
13 * Possible future solver enhancements:
15 * - There's an interesting deductive technique which makes use
16 * of topology rather than just graph theory. Each _face_ in
17 * the grid is either inside or outside the loop; you can tell
18 * that two faces are on the same side of the loop if they're
19 * separated by a LINE_NO (or, more generally, by a path
20 * crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
21 * and on the opposite side of the loop if they're separated by
22 * a LINE_YES (or an odd number of LINE_YESes and no
23 * LINE_UNKNOWNs). Oh, and any face separated from the outside
24 * of the grid by a LINE_YES or a LINE_NO is on the inside or
25 * outside respectively. So if you can track this for all
26 * faces, you figure out the state of the line between a pair
27 * once their relative insideness is known.
28 * + The way I envisage this working is simply to keep an edsf
29 * of all _faces_, which indicates whether they're on
30 * opposite sides of the loop from one another. We also
31 * include a special entry in the edsf for the infinite
33 * + So, the simple way to do this is to just go through the
34 * edges: every time we see an edge in a state other than
35 * LINE_UNKNOWN which separates two faces that aren't in the
36 * same edsf class, we can rectify that by merging the
37 * classes. Then, conversely, an edge in LINE_UNKNOWN state
38 * which separates two faces that _are_ in the same edsf
39 * class can immediately have its state determined.
40 * + But you can go one better, if you're prepared to loop
41 * over all _pairs_ of edges. Suppose we have edges A and B,
42 * which respectively separate faces A1,A2 and B1,B2.
43 * Suppose that A,B are in the same edge-edsf class and that
44 * A1,B1 (wlog) are in the same face-edsf class; then we can
45 * immediately place A2,B2 into the same face-edsf class (as
46 * each other, not as A1 and A2) one way round or the other.
47 * And conversely again, if A1,B1 are in the same face-edsf
48 * class and so are A2,B2, then we can put A,B into the same
50 * * Of course, this deduction requires a quadratic-time
51 * loop over all pairs of edges in the grid, so it should
52 * be reserved until there's nothing easier left to be
55 * - The generalised grid support has made me (SGT) notice a
56 * possible extension to the loop-avoidance code. When you have
57 * a path of connected edges such that no other edges at all
58 * are incident on any vertex in the middle of the path - or,
59 * alternatively, such that any such edges are already known to
60 * be LINE_NO - then you know those edges are either all
61 * LINE_YES or all LINE_NO. Hence you can mentally merge the
62 * entire path into a single long curly edge for the purposes
63 * of loop avoidance, and look directly at whether or not the
64 * extreme endpoints of the path are connected by some other
65 * route. I find this coming up fairly often when I play on the
66 * octagonal grid setting, so it might be worth implementing in
69 * - (Just a speed optimisation.) Consider some todo list queue where every
70 * time we modify something we mark it for consideration by other bits of
71 * the solver, to save iteration over things that have already been done.
86 /* Debugging options */
94 /* ----------------------------------------------------------------------
95 * Struct, enum and function declarations
112 /* Put -1 in a face that doesn't get a clue */
115 /* Array of line states, to store whether each line is
116 * YES, NO or UNKNOWN */
119 unsigned char *line_errors
;
124 /* Used in game_text_format(), so that it knows what type of
125 * grid it's trying to render as ASCII text. */
130 SOLVER_SOLVED
, /* This is the only solution the solver could find */
131 SOLVER_MISTAKE
, /* This is definitely not a solution */
132 SOLVER_AMBIGUOUS
, /* This _might_ be an ambiguous solution */
133 SOLVER_INCOMPLETE
/* This may be a partial solution */
136 /* ------ Solver state ------ */
137 typedef struct solver_state
{
139 enum solver_status solver_status
;
140 /* NB looplen is the number of dots that are joined together at a point, ie a
141 * looplen of 1 means there are no lines to a particular dot */
144 /* Difficulty level of solver. Used by solver functions that want to
145 * vary their behaviour depending on the requested difficulty level. */
151 char *face_yes_count
;
153 char *dot_solved
, *face_solved
;
156 /* Information for Normal level deductions:
157 * For each dline, store a bitmask for whether we know:
158 * (bit 0) at least one is YES
159 * (bit 1) at most one is YES */
162 /* Hard level information */
167 * Difficulty levels. I do some macro ickery here to ensure that my
168 * enum and the various forms of my name list always match up.
171 #define DIFFLIST(A) \
176 #define ENUM(upper,title,lower) DIFF_ ## upper,
177 #define TITLE(upper,title,lower) #title,
178 #define ENCODE(upper,title,lower) #lower
179 #define CONFIG(upper,title,lower) ":" #title
180 enum { DIFFLIST(ENUM
) DIFF_MAX
};
181 static char const *const diffnames
[] = { DIFFLIST(TITLE
) };
182 static char const diffchars
[] = DIFFLIST(ENCODE
);
183 #define DIFFCONFIG DIFFLIST(CONFIG)
186 * Solver routines, sorted roughly in order of computational cost.
187 * The solver will run the faster deductions first, and slower deductions are
188 * only invoked when the faster deductions are unable to make progress.
189 * Each function is associated with a difficulty level, so that the generated
190 * puzzles are solvable by applying only the functions with the chosen
191 * difficulty level or lower.
193 #define SOLVERLIST(A) \
194 A(trivial_deductions, DIFF_EASY) \
195 A(dline_deductions, DIFF_NORMAL) \
196 A(linedsf_deductions, DIFF_HARD) \
197 A(loop_deductions, DIFF_EASY)
198 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
199 #define SOLVER_FN(fn,diff) &fn,
200 #define SOLVER_DIFF(fn,diff) diff,
201 SOLVERLIST(SOLVER_FN_DECL
)
202 static int (*(solver_fns
[]))(solver_state
*) = { SOLVERLIST(SOLVER_FN
) };
203 static int const solver_diffs
[] = { SOLVERLIST(SOLVER_DIFF
) };
204 const int NUM_SOLVERS
= sizeof(solver_diffs
)/sizeof(*solver_diffs
);
211 /* Grid generation is expensive, so keep a (ref-counted) reference to the
212 * grid for these parameters, and only generate when required. */
216 /* line_drawstate is the same as line_state, but with the extra ERROR
217 * possibility. The drawing code copies line_state to line_drawstate,
218 * except in the case that the line is an error. */
219 enum line_state
{ LINE_YES
, LINE_UNKNOWN
, LINE_NO
};
220 enum line_drawstate
{ DS_LINE_YES
, DS_LINE_UNKNOWN
,
221 DS_LINE_NO
, DS_LINE_ERROR
};
223 #define OPP(line_state) \
227 struct game_drawstate
{
233 char *clue_satisfied
;
236 static char *validate_desc(game_params
*params
, char *desc
);
237 static int dot_order(const game_state
* state
, int i
, char line_type
);
238 static int face_order(const game_state
* state
, int i
, char line_type
);
239 static solver_state
*solve_game_rec(const solver_state
*sstate
);
242 static void check_caches(const solver_state
* sstate
);
244 #define check_caches(s)
247 /* ------- List of grid generators ------- */
248 #define GRIDLIST(A) \
249 A(Squares,grid_new_square,3,3) \
250 A(Triangular,grid_new_triangular,3,3) \
251 A(Honeycomb,grid_new_honeycomb,3,3) \
252 A(Snub-Square,grid_new_snubsquare,3,3) \
253 A(Cairo,grid_new_cairo,3,4) \
254 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
255 A(Octagonal,grid_new_octagonal,3,3) \
256 A(Kites,grid_new_kites,3,3) \
257 A(Floret,grid_new_floret,1,2)
259 #define GRID_NAME(title,fn,amin,omin) #title,
260 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
261 #define GRID_FN(title,fn,amin,omin) &fn,
262 #define GRID_SIZES(title,fn,amin,omin) \
264 "Width and height for this grid type must both be at least " #amin, \
265 "At least one of width and height for this grid type must be at least " #omin,},
266 static char const *const gridnames
[] = { GRIDLIST(GRID_NAME
) };
267 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
268 static grid
* (*(grid_fns
[]))(int w
, int h
) = { GRIDLIST(GRID_FN
) };
269 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
270 static const struct {
273 } grid_size_limits
[] = { GRIDLIST(GRID_SIZES
) };
275 /* Generates a (dynamically allocated) new grid, according to the
276 * type and size requested in params. Does nothing if the grid is already
277 * generated. The allocated grid is owned by the params object, and will be
278 * freed in free_params(). */
279 static void params_generate_grid(game_params
*params
)
281 if (!params
->game_grid
) {
282 params
->game_grid
= grid_fns
[params
->type
](params
->w
, params
->h
);
286 /* ----------------------------------------------------------------------
290 /* General constants */
291 #define PREFERRED_TILE_SIZE 32
292 #define BORDER(tilesize) ((tilesize) / 2)
293 #define FLASH_TIME 0.5F
295 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
297 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
298 ((field) |= (1<<(bit)), TRUE))
300 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
301 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
303 #define CLUE2CHAR(c) \
304 ((c < 0) ? ' ' : c + '0')
306 /* ----------------------------------------------------------------------
307 * General struct manipulation and other straightforward code
310 static game_state
*dup_game(game_state
*state
)
312 game_state
*ret
= snew(game_state
);
314 ret
->game_grid
= state
->game_grid
;
315 ret
->game_grid
->refcount
++;
317 ret
->solved
= state
->solved
;
318 ret
->cheated
= state
->cheated
;
320 ret
->clues
= snewn(state
->game_grid
->num_faces
, signed char);
321 memcpy(ret
->clues
, state
->clues
, state
->game_grid
->num_faces
);
323 ret
->lines
= snewn(state
->game_grid
->num_edges
, char);
324 memcpy(ret
->lines
, state
->lines
, state
->game_grid
->num_edges
);
326 ret
->line_errors
= snewn(state
->game_grid
->num_edges
, unsigned char);
327 memcpy(ret
->line_errors
, state
->line_errors
, state
->game_grid
->num_edges
);
329 ret
->grid_type
= state
->grid_type
;
333 static void free_game(game_state
*state
)
336 grid_free(state
->game_grid
);
339 sfree(state
->line_errors
);
344 static solver_state
*new_solver_state(game_state
*state
, int diff
) {
346 int num_dots
= state
->game_grid
->num_dots
;
347 int num_faces
= state
->game_grid
->num_faces
;
348 int num_edges
= state
->game_grid
->num_edges
;
349 solver_state
*ret
= snew(solver_state
);
351 ret
->state
= dup_game(state
);
353 ret
->solver_status
= SOLVER_INCOMPLETE
;
356 ret
->dotdsf
= snew_dsf(num_dots
);
357 ret
->looplen
= snewn(num_dots
, int);
359 for (i
= 0; i
< num_dots
; i
++) {
363 ret
->dot_solved
= snewn(num_dots
, char);
364 ret
->face_solved
= snewn(num_faces
, char);
365 memset(ret
->dot_solved
, FALSE
, num_dots
);
366 memset(ret
->face_solved
, FALSE
, num_faces
);
368 ret
->dot_yes_count
= snewn(num_dots
, char);
369 memset(ret
->dot_yes_count
, 0, num_dots
);
370 ret
->dot_no_count
= snewn(num_dots
, char);
371 memset(ret
->dot_no_count
, 0, num_dots
);
372 ret
->face_yes_count
= snewn(num_faces
, char);
373 memset(ret
->face_yes_count
, 0, num_faces
);
374 ret
->face_no_count
= snewn(num_faces
, char);
375 memset(ret
->face_no_count
, 0, num_faces
);
377 if (diff
< DIFF_NORMAL
) {
380 ret
->dlines
= snewn(2*num_edges
, char);
381 memset(ret
->dlines
, 0, 2*num_edges
);
384 if (diff
< DIFF_HARD
) {
387 ret
->linedsf
= snew_dsf(state
->game_grid
->num_edges
);
393 static void free_solver_state(solver_state
*sstate
) {
395 free_game(sstate
->state
);
396 sfree(sstate
->dotdsf
);
397 sfree(sstate
->looplen
);
398 sfree(sstate
->dot_solved
);
399 sfree(sstate
->face_solved
);
400 sfree(sstate
->dot_yes_count
);
401 sfree(sstate
->dot_no_count
);
402 sfree(sstate
->face_yes_count
);
403 sfree(sstate
->face_no_count
);
405 /* OK, because sfree(NULL) is a no-op */
406 sfree(sstate
->dlines
);
407 sfree(sstate
->linedsf
);
413 static solver_state
*dup_solver_state(const solver_state
*sstate
) {
414 game_state
*state
= sstate
->state
;
415 int num_dots
= state
->game_grid
->num_dots
;
416 int num_faces
= state
->game_grid
->num_faces
;
417 int num_edges
= state
->game_grid
->num_edges
;
418 solver_state
*ret
= snew(solver_state
);
420 ret
->state
= state
= dup_game(sstate
->state
);
422 ret
->solver_status
= sstate
->solver_status
;
423 ret
->diff
= sstate
->diff
;
425 ret
->dotdsf
= snewn(num_dots
, int);
426 ret
->looplen
= snewn(num_dots
, int);
427 memcpy(ret
->dotdsf
, sstate
->dotdsf
,
428 num_dots
* sizeof(int));
429 memcpy(ret
->looplen
, sstate
->looplen
,
430 num_dots
* sizeof(int));
432 ret
->dot_solved
= snewn(num_dots
, char);
433 ret
->face_solved
= snewn(num_faces
, char);
434 memcpy(ret
->dot_solved
, sstate
->dot_solved
, num_dots
);
435 memcpy(ret
->face_solved
, sstate
->face_solved
, num_faces
);
437 ret
->dot_yes_count
= snewn(num_dots
, char);
438 memcpy(ret
->dot_yes_count
, sstate
->dot_yes_count
, num_dots
);
439 ret
->dot_no_count
= snewn(num_dots
, char);
440 memcpy(ret
->dot_no_count
, sstate
->dot_no_count
, num_dots
);
442 ret
->face_yes_count
= snewn(num_faces
, char);
443 memcpy(ret
->face_yes_count
, sstate
->face_yes_count
, num_faces
);
444 ret
->face_no_count
= snewn(num_faces
, char);
445 memcpy(ret
->face_no_count
, sstate
->face_no_count
, num_faces
);
447 if (sstate
->dlines
) {
448 ret
->dlines
= snewn(2*num_edges
, char);
449 memcpy(ret
->dlines
, sstate
->dlines
,
455 if (sstate
->linedsf
) {
456 ret
->linedsf
= snewn(num_edges
, int);
457 memcpy(ret
->linedsf
, sstate
->linedsf
,
458 num_edges
* sizeof(int));
466 static game_params
*default_params(void)
468 game_params
*ret
= snew(game_params
);
477 ret
->diff
= DIFF_EASY
;
480 ret
->game_grid
= NULL
;
485 static game_params
*dup_params(game_params
*params
)
487 game_params
*ret
= snew(game_params
);
489 *ret
= *params
; /* structure copy */
490 if (ret
->game_grid
) {
491 ret
->game_grid
->refcount
++;
496 static const game_params presets
[] = {
498 { 7, 7, DIFF_EASY
, 0, NULL
},
499 { 7, 7, DIFF_NORMAL
, 0, NULL
},
500 { 7, 7, DIFF_HARD
, 0, NULL
},
501 { 7, 7, DIFF_HARD
, 1, NULL
},
502 { 7, 7, DIFF_HARD
, 2, NULL
},
503 { 5, 5, DIFF_HARD
, 3, NULL
},
504 { 7, 7, DIFF_HARD
, 4, NULL
},
505 { 5, 4, DIFF_HARD
, 5, NULL
},
506 { 5, 5, DIFF_HARD
, 6, NULL
},
507 { 5, 5, DIFF_HARD
, 7, NULL
},
508 { 3, 3, DIFF_HARD
, 8, NULL
},
510 { 7, 7, DIFF_EASY
, 0, NULL
},
511 { 10, 10, DIFF_EASY
, 0, NULL
},
512 { 7, 7, DIFF_NORMAL
, 0, NULL
},
513 { 10, 10, DIFF_NORMAL
, 0, NULL
},
514 { 7, 7, DIFF_HARD
, 0, NULL
},
515 { 10, 10, DIFF_HARD
, 0, NULL
},
516 { 10, 10, DIFF_HARD
, 1, NULL
},
517 { 12, 10, DIFF_HARD
, 2, NULL
},
518 { 7, 7, DIFF_HARD
, 3, NULL
},
519 { 9, 9, DIFF_HARD
, 4, NULL
},
520 { 5, 4, DIFF_HARD
, 5, NULL
},
521 { 7, 7, DIFF_HARD
, 6, NULL
},
522 { 5, 5, DIFF_HARD
, 7, NULL
},
523 { 5, 5, DIFF_HARD
, 8, NULL
},
527 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
532 if (i
< 0 || i
>= lenof(presets
))
535 tmppar
= snew(game_params
);
536 *tmppar
= presets
[i
];
538 sprintf(buf
, "%dx%d %s - %s", tmppar
->h
, tmppar
->w
,
539 gridnames
[tmppar
->type
], diffnames
[tmppar
->diff
]);
545 static void free_params(game_params
*params
)
547 if (params
->game_grid
) {
548 grid_free(params
->game_grid
);
553 static void decode_params(game_params
*params
, char const *string
)
555 if (params
->game_grid
) {
556 grid_free(params
->game_grid
);
557 params
->game_grid
= NULL
;
559 params
->h
= params
->w
= atoi(string
);
560 params
->diff
= DIFF_EASY
;
561 while (*string
&& isdigit((unsigned char)*string
)) string
++;
562 if (*string
== 'x') {
564 params
->h
= atoi(string
);
565 while (*string
&& isdigit((unsigned char)*string
)) string
++;
567 if (*string
== 't') {
569 params
->type
= atoi(string
);
570 while (*string
&& isdigit((unsigned char)*string
)) string
++;
572 if (*string
== 'd') {
575 for (i
= 0; i
< DIFF_MAX
; i
++)
576 if (*string
== diffchars
[i
])
578 if (*string
) string
++;
582 static char *encode_params(game_params
*params
, int full
)
585 sprintf(str
, "%dx%dt%d", params
->w
, params
->h
, params
->type
);
587 sprintf(str
+ strlen(str
), "d%c", diffchars
[params
->diff
]);
591 static config_item
*game_configure(game_params
*params
)
596 ret
= snewn(5, config_item
);
598 ret
[0].name
= "Width";
599 ret
[0].type
= C_STRING
;
600 sprintf(buf
, "%d", params
->w
);
601 ret
[0].sval
= dupstr(buf
);
604 ret
[1].name
= "Height";
605 ret
[1].type
= C_STRING
;
606 sprintf(buf
, "%d", params
->h
);
607 ret
[1].sval
= dupstr(buf
);
610 ret
[2].name
= "Grid type";
611 ret
[2].type
= C_CHOICES
;
612 ret
[2].sval
= GRID_CONFIGS
;
613 ret
[2].ival
= params
->type
;
615 ret
[3].name
= "Difficulty";
616 ret
[3].type
= C_CHOICES
;
617 ret
[3].sval
= DIFFCONFIG
;
618 ret
[3].ival
= params
->diff
;
628 static game_params
*custom_params(config_item
*cfg
)
630 game_params
*ret
= snew(game_params
);
632 ret
->w
= atoi(cfg
[0].sval
);
633 ret
->h
= atoi(cfg
[1].sval
);
634 ret
->type
= cfg
[2].ival
;
635 ret
->diff
= cfg
[3].ival
;
637 ret
->game_grid
= NULL
;
641 static char *validate_params(game_params
*params
, int full
)
643 if (params
->type
< 0 || params
->type
>= NUM_GRID_TYPES
)
644 return "Illegal grid type";
645 if (params
->w
< grid_size_limits
[params
->type
].amin
||
646 params
->h
< grid_size_limits
[params
->type
].amin
)
647 return grid_size_limits
[params
->type
].aerr
;
648 if (params
->w
< grid_size_limits
[params
->type
].omin
&&
649 params
->h
< grid_size_limits
[params
->type
].omin
)
650 return grid_size_limits
[params
->type
].oerr
;
653 * This shouldn't be able to happen at all, since decode_params
654 * and custom_params will never generate anything that isn't
657 assert(params
->diff
< DIFF_MAX
);
662 /* Returns a newly allocated string describing the current puzzle */
663 static char *state_to_text(const game_state
*state
)
665 grid
*g
= state
->game_grid
;
667 int num_faces
= g
->num_faces
;
668 char *description
= snewn(num_faces
+ 1, char);
669 char *dp
= description
;
673 for (i
= 0; i
< num_faces
; i
++) {
674 if (state
->clues
[i
] < 0) {
675 if (empty_count
> 25) {
676 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
682 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
685 dp
+= sprintf(dp
, "%c", (int)CLUE2CHAR(state
->clues
[i
]));
690 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
692 retval
= dupstr(description
);
698 /* We require that the params pass the test in validate_params and that the
699 * description fills the entire game area */
700 static char *validate_desc(game_params
*params
, char *desc
)
704 params_generate_grid(params
);
705 g
= params
->game_grid
;
707 for (; *desc
; ++desc
) {
708 if (*desc
>= '0' && *desc
<= '9') {
713 count
+= *desc
- 'a' + 1;
716 return "Unknown character in description";
719 if (count
< g
->num_faces
)
720 return "Description too short for board size";
721 if (count
> g
->num_faces
)
722 return "Description too long for board size";
727 /* Sums the lengths of the numbers in range [0,n) */
728 /* See equivalent function in solo.c for justification of this. */
729 static int len_0_to_n(int n
)
731 int len
= 1; /* Counting 0 as a bit of a special case */
734 for (i
= 1; i
< n
; i
*= 10) {
735 len
+= max(n
- i
, 0);
741 static char *encode_solve_move(const game_state
*state
)
746 int num_edges
= state
->game_grid
->num_edges
;
748 /* This is going to return a string representing the moves needed to set
749 * every line in a grid to be the same as the ones in 'state'. The exact
750 * length of this string is predictable. */
752 len
= 1; /* Count the 'S' prefix */
753 /* Numbers in all lines */
754 len
+= len_0_to_n(num_edges
);
755 /* For each line we also have a letter */
758 ret
= snewn(len
+ 1, char);
761 p
+= sprintf(p
, "S");
763 for (i
= 0; i
< num_edges
; i
++) {
764 switch (state
->lines
[i
]) {
766 p
+= sprintf(p
, "%dy", i
);
769 p
+= sprintf(p
, "%dn", i
);
774 /* No point in doing sums like that if they're going to be wrong */
775 assert(strlen(ret
) <= (size_t)len
);
779 static game_ui
*new_ui(game_state
*state
)
784 static void free_ui(game_ui
*ui
)
788 static char *encode_ui(game_ui
*ui
)
793 static void decode_ui(game_ui
*ui
, char *encoding
)
797 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
798 game_state
*newstate
)
802 static void game_compute_size(game_params
*params
, int tilesize
,
806 int grid_width
, grid_height
, rendered_width
, rendered_height
;
808 params_generate_grid(params
);
809 g
= params
->game_grid
;
810 grid_width
= g
->highest_x
- g
->lowest_x
;
811 grid_height
= g
->highest_y
- g
->lowest_y
;
812 /* multiply first to minimise rounding error on integer division */
813 rendered_width
= grid_width
* tilesize
/ g
->tilesize
;
814 rendered_height
= grid_height
* tilesize
/ g
->tilesize
;
815 *x
= rendered_width
+ 2 * BORDER(tilesize
) + 1;
816 *y
= rendered_height
+ 2 * BORDER(tilesize
) + 1;
819 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
820 game_params
*params
, int tilesize
)
822 ds
->tilesize
= tilesize
;
825 static float *game_colours(frontend
*fe
, int *ncolours
)
827 float *ret
= snewn(4 * NCOLOURS
, float);
829 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
831 ret
[COL_FOREGROUND
* 3 + 0] = 0.0F
;
832 ret
[COL_FOREGROUND
* 3 + 1] = 0.0F
;
833 ret
[COL_FOREGROUND
* 3 + 2] = 0.0F
;
835 ret
[COL_LINEUNKNOWN
* 3 + 0] = 0.8F
;
836 ret
[COL_LINEUNKNOWN
* 3 + 1] = 0.8F
;
837 ret
[COL_LINEUNKNOWN
* 3 + 2] = 0.0F
;
839 ret
[COL_HIGHLIGHT
* 3 + 0] = 1.0F
;
840 ret
[COL_HIGHLIGHT
* 3 + 1] = 1.0F
;
841 ret
[COL_HIGHLIGHT
* 3 + 2] = 1.0F
;
843 ret
[COL_MISTAKE
* 3 + 0] = 1.0F
;
844 ret
[COL_MISTAKE
* 3 + 1] = 0.0F
;
845 ret
[COL_MISTAKE
* 3 + 2] = 0.0F
;
847 ret
[COL_SATISFIED
* 3 + 0] = 0.0F
;
848 ret
[COL_SATISFIED
* 3 + 1] = 0.0F
;
849 ret
[COL_SATISFIED
* 3 + 2] = 0.0F
;
851 /* We want the faint lines to be a bit darker than the background.
852 * Except if the background is pretty dark already; then it ought to be a
853 * bit lighter. Oy vey.
855 ret
[COL_FAINT
* 3 + 0] = ret
[COL_BACKGROUND
* 3 + 0] * 0.9F
;
856 ret
[COL_FAINT
* 3 + 1] = ret
[COL_BACKGROUND
* 3 + 1] * 0.9F
;
857 ret
[COL_FAINT
* 3 + 2] = ret
[COL_BACKGROUND
* 3 + 2] * 0.9F
;
859 *ncolours
= NCOLOURS
;
863 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
865 struct game_drawstate
*ds
= snew(struct game_drawstate
);
866 int num_faces
= state
->game_grid
->num_faces
;
867 int num_edges
= state
->game_grid
->num_edges
;
871 ds
->lines
= snewn(num_edges
, char);
872 ds
->clue_error
= snewn(num_faces
, char);
873 ds
->clue_satisfied
= snewn(num_faces
, char);
876 memset(ds
->lines
, LINE_UNKNOWN
, num_edges
);
877 memset(ds
->clue_error
, 0, num_faces
);
878 memset(ds
->clue_satisfied
, 0, num_faces
);
883 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
885 sfree(ds
->clue_error
);
886 sfree(ds
->clue_satisfied
);
891 static int game_timing_state(game_state
*state
, game_ui
*ui
)
896 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
897 int dir
, game_ui
*ui
)
902 static int game_can_format_as_text_now(game_params
*params
)
904 if (params
->type
!= 0)
909 static char *game_text_format(game_state
*state
)
915 grid
*g
= state
->game_grid
;
918 assert(state
->grid_type
== 0);
920 /* Work out the basic size unit */
921 f
= g
->faces
; /* first face */
922 assert(f
->order
== 4);
923 /* The dots are ordered clockwise, so the two opposite
924 * corners are guaranteed to span the square */
925 cell_size
= abs(f
->dots
[0]->x
- f
->dots
[2]->x
);
927 w
= (g
->highest_x
- g
->lowest_x
) / cell_size
;
928 h
= (g
->highest_y
- g
->lowest_y
) / cell_size
;
930 /* Create a blank "canvas" to "draw" on */
933 ret
= snewn(W
* H
+ 1, char);
934 for (y
= 0; y
< H
; y
++) {
935 for (x
= 0; x
< W
-1; x
++) {
938 ret
[y
*W
+ W
-1] = '\n';
942 /* Fill in edge info */
943 for (i
= 0; i
< g
->num_edges
; i
++) {
944 grid_edge
*e
= g
->edges
+ i
;
945 /* Cell coordinates, from (0,0) to (w-1,h-1) */
946 int x1
= (e
->dot1
->x
- g
->lowest_x
) / cell_size
;
947 int x2
= (e
->dot2
->x
- g
->lowest_x
) / cell_size
;
948 int y1
= (e
->dot1
->y
- g
->lowest_y
) / cell_size
;
949 int y2
= (e
->dot2
->y
- g
->lowest_y
) / cell_size
;
950 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
951 * cell coordinates) */
954 switch (state
->lines
[i
]) {
956 ret
[y
*W
+ x
] = (y1
== y2
) ?
'-' : '|';
962 break; /* already a space */
964 assert(!"Illegal line state");
969 for (i
= 0; i
< g
->num_faces
; i
++) {
973 assert(f
->order
== 4);
974 /* Cell coordinates, from (0,0) to (w-1,h-1) */
975 x1
= (f
->dots
[0]->x
- g
->lowest_x
) / cell_size
;
976 x2
= (f
->dots
[2]->x
- g
->lowest_x
) / cell_size
;
977 y1
= (f
->dots
[0]->y
- g
->lowest_y
) / cell_size
;
978 y2
= (f
->dots
[2]->y
- g
->lowest_y
) / cell_size
;
979 /* Midpoint, in canvas coordinates */
982 ret
[y
*W
+ x
] = CLUE2CHAR(state
->clues
[i
]);
987 /* ----------------------------------------------------------------------
992 static void check_caches(const solver_state
* sstate
)
995 const game_state
*state
= sstate
->state
;
996 const grid
*g
= state
->game_grid
;
998 for (i
= 0; i
< g
->num_dots
; i
++) {
999 assert(dot_order(state
, i
, LINE_YES
) == sstate
->dot_yes_count
[i
]);
1000 assert(dot_order(state
, i
, LINE_NO
) == sstate
->dot_no_count
[i
]);
1003 for (i
= 0; i
< g
->num_faces
; i
++) {
1004 assert(face_order(state
, i
, LINE_YES
) == sstate
->face_yes_count
[i
]);
1005 assert(face_order(state
, i
, LINE_NO
) == sstate
->face_no_count
[i
]);
1010 #define check_caches(s) \
1012 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1016 #endif /* DEBUG_CACHES */
1018 /* ----------------------------------------------------------------------
1019 * Solver utility functions
1022 /* Sets the line (with index i) to the new state 'line_new', and updates
1023 * the cached counts of any affected faces and dots.
1024 * Returns TRUE if this actually changed the line's state. */
1025 static int solver_set_line(solver_state
*sstate
, int i
,
1026 enum line_state line_new
1028 , const char *reason
1032 game_state
*state
= sstate
->state
;
1036 assert(line_new
!= LINE_UNKNOWN
);
1038 check_caches(sstate
);
1040 if (state
->lines
[i
] == line_new
) {
1041 return FALSE
; /* nothing changed */
1043 state
->lines
[i
] = line_new
;
1046 fprintf(stderr
, "solver: set line [%d] to %s (%s)\n",
1047 i
, line_new
== LINE_YES ?
"YES" : "NO",
1051 g
= state
->game_grid
;
1054 /* Update the cache for both dots and both faces affected by this. */
1055 if (line_new
== LINE_YES
) {
1056 sstate
->dot_yes_count
[e
->dot1
- g
->dots
]++;
1057 sstate
->dot_yes_count
[e
->dot2
- g
->dots
]++;
1059 sstate
->face_yes_count
[e
->face1
- g
->faces
]++;
1062 sstate
->face_yes_count
[e
->face2
- g
->faces
]++;
1065 sstate
->dot_no_count
[e
->dot1
- g
->dots
]++;
1066 sstate
->dot_no_count
[e
->dot2
- g
->dots
]++;
1068 sstate
->face_no_count
[e
->face1
- g
->faces
]++;
1071 sstate
->face_no_count
[e
->face2
- g
->faces
]++;
1075 check_caches(sstate
);
1080 #define solver_set_line(a, b, c) \
1081 solver_set_line(a, b, c, __FUNCTION__)
1085 * Merge two dots due to the existence of an edge between them.
1086 * Updates the dsf tracking equivalence classes, and keeps track of
1087 * the length of path each dot is currently a part of.
1088 * Returns TRUE if the dots were already linked, ie if they are part of a
1089 * closed loop, and false otherwise.
1091 static int merge_dots(solver_state
*sstate
, int edge_index
)
1094 grid
*g
= sstate
->state
->game_grid
;
1095 grid_edge
*e
= g
->edges
+ edge_index
;
1097 i
= e
->dot1
- g
->dots
;
1098 j
= e
->dot2
- g
->dots
;
1100 i
= dsf_canonify(sstate
->dotdsf
, i
);
1101 j
= dsf_canonify(sstate
->dotdsf
, j
);
1106 len
= sstate
->looplen
[i
] + sstate
->looplen
[j
];
1107 dsf_merge(sstate
->dotdsf
, i
, j
);
1108 i
= dsf_canonify(sstate
->dotdsf
, i
);
1109 sstate
->looplen
[i
] = len
;
1114 /* Merge two lines because the solver has deduced that they must be either
1115 * identical or opposite. Returns TRUE if this is new information, otherwise
1117 static int merge_lines(solver_state
*sstate
, int i
, int j
, int inverse
1119 , const char *reason
1125 assert(i
< sstate
->state
->game_grid
->num_edges
);
1126 assert(j
< sstate
->state
->game_grid
->num_edges
);
1128 i
= edsf_canonify(sstate
->linedsf
, i
, &inv_tmp
);
1130 j
= edsf_canonify(sstate
->linedsf
, j
, &inv_tmp
);
1133 edsf_merge(sstate
->linedsf
, i
, j
, inverse
);
1137 fprintf(stderr
, "%s [%d] [%d] %s(%s)\n",
1139 inverse ?
"inverse " : "", reason
);
1146 #define merge_lines(a, b, c, d) \
1147 merge_lines(a, b, c, d, __FUNCTION__)
1150 /* Count the number of lines of a particular type currently going into the
1152 static int dot_order(const game_state
* state
, int dot
, char line_type
)
1155 grid
*g
= state
->game_grid
;
1156 grid_dot
*d
= g
->dots
+ dot
;
1159 for (i
= 0; i
< d
->order
; i
++) {
1160 grid_edge
*e
= d
->edges
[i
];
1161 if (state
->lines
[e
- g
->edges
] == line_type
)
1167 /* Count the number of lines of a particular type currently surrounding the
1169 static int face_order(const game_state
* state
, int face
, char line_type
)
1172 grid
*g
= state
->game_grid
;
1173 grid_face
*f
= g
->faces
+ face
;
1176 for (i
= 0; i
< f
->order
; i
++) {
1177 grid_edge
*e
= f
->edges
[i
];
1178 if (state
->lines
[e
- g
->edges
] == line_type
)
1184 /* Set all lines bordering a dot of type old_type to type new_type
1185 * Return value tells caller whether this function actually did anything */
1186 static int dot_setall(solver_state
*sstate
, int dot
,
1187 char old_type
, char new_type
)
1189 int retval
= FALSE
, r
;
1190 game_state
*state
= sstate
->state
;
1195 if (old_type
== new_type
)
1198 g
= state
->game_grid
;
1201 for (i
= 0; i
< d
->order
; i
++) {
1202 int line_index
= d
->edges
[i
] - g
->edges
;
1203 if (state
->lines
[line_index
] == old_type
) {
1204 r
= solver_set_line(sstate
, line_index
, new_type
);
1212 /* Set all lines bordering a face of type old_type to type new_type */
1213 static int face_setall(solver_state
*sstate
, int face
,
1214 char old_type
, char new_type
)
1216 int retval
= FALSE
, r
;
1217 game_state
*state
= sstate
->state
;
1222 if (old_type
== new_type
)
1225 g
= state
->game_grid
;
1226 f
= g
->faces
+ face
;
1228 for (i
= 0; i
< f
->order
; i
++) {
1229 int line_index
= f
->edges
[i
] - g
->edges
;
1230 if (state
->lines
[line_index
] == old_type
) {
1231 r
= solver_set_line(sstate
, line_index
, new_type
);
1239 /* ----------------------------------------------------------------------
1240 * Loop generation and clue removal
1243 /* We're going to store lists of current candidate faces for colouring black
1245 * Each face gets a 'score', which tells us how adding that face right
1246 * now would affect the curliness of the solution loop. We're trying to
1247 * maximise that quantity so will bias our random selection of faces to
1248 * colour those with high scores */
1252 unsigned long random
;
1253 /* No need to store a grid_face* here. The 'face_scores' array will
1254 * be a list of 'face_score' objects, one for each face of the grid, so
1255 * the position (index) within the 'face_scores' array will determine
1256 * which face corresponds to a particular face_score.
1257 * Having a single 'face_scores' array for all faces simplifies memory
1258 * management, and probably improves performance, because we don't have to
1259 * malloc/free each individual face_score, and we don't have to maintain
1260 * a mapping from grid_face* pointers to face_score* pointers.
1264 static int generic_sort_cmpfn(void *v1
, void *v2
, size_t offset
)
1266 struct face_score
*f1
= v1
;
1267 struct face_score
*f2
= v2
;
1270 r
= *(int *)((char *)f2
+ offset
) - *(int *)((char *)f1
+ offset
);
1275 if (f1
->random
< f2
->random
)
1277 else if (f1
->random
> f2
->random
)
1281 * It's _just_ possible that two faces might have been given
1282 * the same random value. In that situation, fall back to
1283 * comparing based on the positions within the face_scores list.
1284 * This introduces a tiny directional bias, but not a significant one.
1289 static int white_sort_cmpfn(void *v1
, void *v2
)
1291 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,white_score
));
1294 static int black_sort_cmpfn(void *v1
, void *v2
)
1296 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,black_score
));
1299 enum face_colour
{ FACE_WHITE
, FACE_GREY
, FACE_BLACK
};
1301 /* face should be of type grid_face* here. */
1302 #define FACE_COLOUR(face) \
1303 ( (face) == NULL ? FACE_BLACK : \
1304 board[(face) - g->faces] )
1306 /* 'board' is an array of these enums, indicating which faces are
1307 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1308 * Returns whether it's legal to colour the given face with this colour. */
1309 static int can_colour_face(grid
*g
, char* board
, int face_index
,
1310 enum face_colour colour
)
1313 grid_face
*test_face
= g
->faces
+ face_index
;
1314 grid_face
*starting_face
, *current_face
;
1315 grid_dot
*starting_dot
;
1317 int current_state
, s
; /* booleans: equal or not-equal to 'colour' */
1318 int found_same_coloured_neighbour
= FALSE
;
1319 assert(board
[face_index
] != colour
);
1321 /* Can only consider a face for colouring if it's adjacent to a face
1322 * with the same colour. */
1323 for (i
= 0; i
< test_face
->order
; i
++) {
1324 grid_edge
*e
= test_face
->edges
[i
];
1325 grid_face
*f
= (e
->face1
== test_face
) ? e
->face2
: e
->face1
;
1326 if (FACE_COLOUR(f
) == colour
) {
1327 found_same_coloured_neighbour
= TRUE
;
1331 if (!found_same_coloured_neighbour
)
1334 /* Need to avoid creating a loop of faces of this colour around some
1335 * differently-coloured faces.
1336 * Also need to avoid meeting a same-coloured face at a corner, with
1337 * other-coloured faces in between. Here's a simple test that (I believe)
1338 * takes care of both these conditions:
1340 * Take the circular path formed by this face's edges, and inflate it
1341 * slightly outwards. Imagine walking around this path and consider
1342 * the faces that you visit in sequence. This will include all faces
1343 * touching the given face, either along an edge or just at a corner.
1344 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1345 * you walk along the complete loop. This will obviously turn out to be
1347 * If 0, we're either in the middle of an "island" of this colour (should
1348 * be impossible as we're not supposed to create black or white loops),
1349 * or we're about to start a new island - also not allowed.
1350 * If 4 or greater, there are too many separate coloured regions touching
1351 * this face, and colouring it would create a loop or a corner-violation.
1352 * The only allowed case is when the count is exactly 2. */
1354 /* i points to a dot around the test face.
1355 * j points to a face around the i^th dot.
1356 * The current face will always be:
1357 * test_face->dots[i]->faces[j]
1358 * We assume dots go clockwise around the test face,
1359 * and faces go clockwise around dots. */
1362 * The end condition is slightly fiddly. In sufficiently strange
1363 * degenerate grids, our test face may be adjacent to the same
1364 * other face multiple times (typically if it's the exterior
1365 * face). Consider this, in particular:
1373 * The bottom left face there is adjacent to the exterior face
1374 * twice, so we can't just terminate our iteration when we reach
1375 * the same _face_ we started at. Furthermore, we can't
1376 * condition on having the same (i,j) pair either, because
1377 * several (i,j) pairs identify the bottom left contiguity with
1378 * the exterior face! We canonicalise the (i,j) pair by taking
1379 * one step around before we set the termination tracking.
1383 current_face
= test_face
->dots
[0]->faces
[0];
1384 if (current_face
== test_face
) {
1386 current_face
= test_face
->dots
[0]->faces
[1];
1389 current_state
= (FACE_COLOUR(current_face
) == colour
);
1390 starting_dot
= NULL
;
1391 starting_face
= NULL
;
1393 /* Advance to next face.
1394 * Need to loop here because it might take several goes to
1398 if (j
== test_face
->dots
[i
]->order
)
1401 if (test_face
->dots
[i
]->faces
[j
] == test_face
) {
1402 /* Advance to next dot round test_face, then
1403 * find current_face around new dot
1404 * and advance to the next face clockwise */
1406 if (i
== test_face
->order
)
1408 for (j
= 0; j
< test_face
->dots
[i
]->order
; j
++) {
1409 if (test_face
->dots
[i
]->faces
[j
] == current_face
)
1412 /* Must actually find current_face around new dot,
1413 * or else something's wrong with the grid. */
1414 assert(j
!= test_face
->dots
[i
]->order
);
1415 /* Found, so advance to next face and try again */
1420 /* (i,j) are now advanced to next face */
1421 current_face
= test_face
->dots
[i
]->faces
[j
];
1422 s
= (FACE_COLOUR(current_face
) == colour
);
1423 if (!starting_dot
) {
1424 starting_dot
= test_face
->dots
[i
];
1425 starting_face
= current_face
;
1428 if (s
!= current_state
) {
1431 if (transitions
> 2)
1434 if (test_face
->dots
[i
] == starting_dot
&&
1435 current_face
== starting_face
)
1440 return (transitions
== 2) ? TRUE
: FALSE
;
1443 /* Count the number of neighbours of 'face', having colour 'colour' */
1444 static int face_num_neighbours(grid
*g
, char *board
, grid_face
*face
,
1445 enum face_colour colour
)
1447 int colour_count
= 0;
1451 for (i
= 0; i
< face
->order
; i
++) {
1453 f
= (e
->face1
== face
) ? e
->face2
: e
->face1
;
1454 if (FACE_COLOUR(f
) == colour
)
1457 return colour_count
;
1460 /* The 'score' of a face reflects its current desirability for selection
1461 * as the next face to colour white or black. We want to encourage moving
1462 * into grey areas and increasing loopiness, so we give scores according to
1463 * how many of the face's neighbours are currently coloured the same as the
1464 * proposed colour. */
1465 static int face_score(grid
*g
, char *board
, grid_face
*face
,
1466 enum face_colour colour
)
1468 /* Simple formula: score = 0 - num. same-coloured neighbours,
1469 * so a higher score means fewer same-coloured neighbours. */
1470 return -face_num_neighbours(g
, board
, face
, colour
);
1473 /* Generate a new complete set of clues for the given game_state.
1474 * The method is to generate a WHITE/BLACK colouring of all the faces,
1475 * such that the WHITE faces will define the inside of the path, and the
1476 * BLACK faces define the outside.
1477 * To do this, we initially colour all faces GREY. The infinite space outside
1478 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1479 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1480 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1481 * we avoid creating loops of a single colour, to preserve the topological
1482 * shape of the WHITE and BLACK regions.
1483 * We also try to make the boundary as loopy and twisty as possible, to avoid
1484 * generating paths that are uninteresting.
1485 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1486 * face that can be coloured with that colour (without violating the
1487 * topological shape of that region). It's not obvious, but I think this
1488 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1489 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1490 * regions can be grown.
1491 * This is checked using assert()ions, and I haven't seen any failures yet.
1493 * Hand-wavy proof: imagine what can go wrong...
1495 * Could the white faces get completely cut off by the black faces, and still
1496 * leave some grey faces remaining?
1497 * No, because then the black faces would form a loop around both the white
1498 * faces and the grey faces, which is disallowed because we continually
1499 * maintain the correct topological shape of the black region.
1500 * Similarly, the black faces can never get cut off by the white faces. That
1501 * means both the WHITE and BLACK regions always have some room to grow into
1503 * Could it be that we can't colour some GREY face, because there are too many
1504 * WHITE/BLACK transitions as we walk round the face? (see the
1505 * can_colour_face() function for details)
1506 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1507 * around the face. The two WHITE faces would be connected by a WHITE path,
1508 * and the BLACK faces would be connected by a BLACK path. These paths would
1509 * have to cross, which is impossible.
1510 * Another thing that could go wrong: perhaps we can't find any GREY face to
1511 * colour WHITE, because it would create a loop-violation or a corner-violation
1512 * with the other WHITE faces?
1513 * This is a little bit tricky to prove impossible. Imagine you have such a
1514 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1515 * or corner violation).
1516 * That would cut all the non-white area into two blobs. One of those blobs
1517 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1518 * So we have a connected GREY area, completely surrounded by WHITE
1519 * (including the GREY face we've tentatively coloured WHITE).
1520 * A well-known result in graph theory says that you can always find a GREY
1521 * face whose removal leaves the remaining GREY area connected. And it says
1522 * there are at least two such faces, so we can always choose the one that
1523 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1524 * everything nice and connected, including that "tentative" GREY face which
1525 * acts as a gateway to the rest of the non-WHITE grid.
1527 static void add_full_clues(game_state
*state
, random_state
*rs
)
1529 signed char *clues
= state
->clues
;
1531 grid
*g
= state
->game_grid
;
1533 int num_faces
= g
->num_faces
;
1534 struct face_score
*face_scores
; /* Array of face_score objects */
1535 struct face_score
*fs
; /* Points somewhere in the above list */
1536 struct grid_face
*cur_face
;
1537 tree234
*lightable_faces_sorted
;
1538 tree234
*darkable_faces_sorted
;
1542 board
= snewn(num_faces
, char);
1545 memset(board
, FACE_GREY
, num_faces
);
1547 /* Create and initialise the list of face_scores */
1548 face_scores
= snewn(num_faces
, struct face_score
);
1549 for (i
= 0; i
< num_faces
; i
++) {
1550 face_scores
[i
].random
= random_bits(rs
, 31);
1551 face_scores
[i
].black_score
= face_scores
[i
].white_score
= 0;
1554 /* Colour a random, finite face white. The infinite face is implicitly
1555 * coloured black. Together, they will seed the random growth process
1556 * for the black and white areas. */
1557 i
= random_upto(rs
, num_faces
);
1558 board
[i
] = FACE_WHITE
;
1560 /* We need a way of favouring faces that will increase our loopiness.
1561 * We do this by maintaining a list of all candidate faces sorted by
1562 * their score and choose randomly from that with appropriate skew.
1563 * In order to avoid consistently biasing towards particular faces, we
1564 * need the sort order _within_ each group of scores to be completely
1565 * random. But it would be abusing the hospitality of the tree234 data
1566 * structure if our comparison function were nondeterministic :-). So with
1567 * each face we associate a random number that does not change during a
1568 * particular run of the generator, and use that as a secondary sort key.
1569 * Yes, this means we will be biased towards particular random faces in
1570 * any one run but that doesn't actually matter. */
1572 lightable_faces_sorted
= newtree234(white_sort_cmpfn
);
1573 darkable_faces_sorted
= newtree234(black_sort_cmpfn
);
1575 /* Initialise the lists of lightable and darkable faces. This is
1576 * slightly different from the code inside the while-loop, because we need
1577 * to check every face of the board (the grid structure does not keep a
1578 * list of the infinite face's neighbours). */
1579 for (i
= 0; i
< num_faces
; i
++) {
1580 grid_face
*f
= g
->faces
+ i
;
1581 struct face_score
*fs
= face_scores
+ i
;
1582 if (board
[i
] != FACE_GREY
) continue;
1583 /* We need the full colourability check here, it's not enough simply
1584 * to check neighbourhood. On some grids, a neighbour of the infinite
1585 * face is not necessarily darkable. */
1586 if (can_colour_face(g
, board
, i
, FACE_BLACK
)) {
1587 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1588 add234(darkable_faces_sorted
, fs
);
1590 if (can_colour_face(g
, board
, i
, FACE_WHITE
)) {
1591 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1592 add234(lightable_faces_sorted
, fs
);
1596 /* Colour faces one at a time until no more faces are colourable. */
1599 enum face_colour colour
;
1600 struct face_score
*fs_white
, *fs_black
;
1601 int c_lightable
= count234(lightable_faces_sorted
);
1602 int c_darkable
= count234(darkable_faces_sorted
);
1603 if (c_lightable
== 0 && c_darkable
== 0) {
1604 /* No more faces we can use at all. */
1607 assert(c_lightable
!= 0 && c_darkable
!= 0);
1609 fs_white
= (struct face_score
*)index234(lightable_faces_sorted
, 0);
1610 fs_black
= (struct face_score
*)index234(darkable_faces_sorted
, 0);
1612 /* Choose a colour, and colour the best available face
1613 * with that colour. */
1614 colour
= random_upto(rs
, 2) ? FACE_WHITE
: FACE_BLACK
;
1616 if (colour
== FACE_WHITE
)
1621 i
= fs
- face_scores
;
1622 assert(board
[i
] == FACE_GREY
);
1625 /* Remove this newly-coloured face from the lists. These lists should
1626 * only contain grey faces. */
1627 del234(lightable_faces_sorted
, fs
);
1628 del234(darkable_faces_sorted
, fs
);
1630 /* Remember which face we've just coloured */
1631 cur_face
= g
->faces
+ i
;
1633 /* The face we've just coloured potentially affects the colourability
1634 * and the scores of any neighbouring faces (touching at a corner or
1635 * edge). So the search needs to be conducted around all faces
1636 * touching the one we've just lit. Iterate over its corners, then
1637 * over each corner's faces. For each such face, we remove it from
1638 * the lists, recalculate any scores, then add it back to the lists
1639 * (depending on whether it is lightable, darkable or both). */
1640 for (i
= 0; i
< cur_face
->order
; i
++) {
1641 grid_dot
*d
= cur_face
->dots
[i
];
1642 for (j
= 0; j
< d
->order
; j
++) {
1643 grid_face
*f
= d
->faces
[j
];
1644 int fi
; /* face index of f */
1651 /* If the face is already coloured, it won't be on our
1652 * lightable/darkable lists anyway, so we can skip it without
1653 * bothering with the removal step. */
1654 if (FACE_COLOUR(f
) != FACE_GREY
) continue;
1656 /* Find the face index and face_score* corresponding to f */
1658 fs
= face_scores
+ fi
;
1660 /* Remove from lightable list if it's in there. We do this,
1661 * even if it is still lightable, because the score might
1662 * be different, and we need to remove-then-add to maintain
1663 * correct sort order. */
1664 del234(lightable_faces_sorted
, fs
);
1665 if (can_colour_face(g
, board
, fi
, FACE_WHITE
)) {
1666 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1667 add234(lightable_faces_sorted
, fs
);
1669 /* Do the same for darkable list. */
1670 del234(darkable_faces_sorted
, fs
);
1671 if (can_colour_face(g
, board
, fi
, FACE_BLACK
)) {
1672 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1673 add234(darkable_faces_sorted
, fs
);
1680 freetree234(lightable_faces_sorted
);
1681 freetree234(darkable_faces_sorted
);
1684 /* The next step requires a shuffled list of all faces */
1685 face_list
= snewn(num_faces
, int);
1686 for (i
= 0; i
< num_faces
; ++i
) {
1689 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1691 /* The above loop-generation algorithm can often leave large clumps
1692 * of faces of one colour. In extreme cases, the resulting path can be
1693 * degenerate and not very satisfying to solve.
1694 * This next step alleviates this problem:
1695 * Go through the shuffled list, and flip the colour of any face we can
1696 * legally flip, and which is adjacent to only one face of the opposite
1697 * colour - this tends to grow 'tendrils' into any clumps.
1698 * Repeat until we can find no more faces to flip. This will
1699 * eventually terminate, because each flip increases the loop's
1700 * perimeter, which cannot increase for ever.
1701 * The resulting path will have maximal loopiness (in the sense that it
1702 * cannot be improved "locally". Unfortunately, this allows a player to
1703 * make some illicit deductions. To combat this (and make the path more
1704 * interesting), we do one final pass making random flips. */
1706 /* Set to TRUE for final pass */
1707 do_random_pass
= FALSE
;
1710 /* Remember whether a flip occurred during this pass */
1711 int flipped
= FALSE
;
1713 for (i
= 0; i
< num_faces
; ++i
) {
1714 int j
= face_list
[i
];
1715 enum face_colour opp
=
1716 (board
[j
] == FACE_WHITE
) ? FACE_BLACK
: FACE_WHITE
;
1717 if (can_colour_face(g
, board
, j
, opp
)) {
1718 grid_face
*face
= g
->faces
+j
;
1719 if (do_random_pass
) {
1720 /* final random pass */
1721 if (!random_upto(rs
, 10))
1724 /* normal pass - flip when neighbour count is 1 */
1725 if (face_num_neighbours(g
, board
, face
, opp
) == 1) {
1733 if (do_random_pass
) break;
1734 if (!flipped
) do_random_pass
= TRUE
;
1739 /* Fill out all the clues by initialising to 0, then iterating over
1740 * all edges and incrementing each clue as we find edges that border
1741 * between BLACK/WHITE faces. While we're at it, we verify that the
1742 * algorithm does work, and there aren't any GREY faces still there. */
1743 memset(clues
, 0, num_faces
);
1744 for (i
= 0; i
< g
->num_edges
; i
++) {
1745 grid_edge
*e
= g
->edges
+ i
;
1746 grid_face
*f1
= e
->face1
;
1747 grid_face
*f2
= e
->face2
;
1748 enum face_colour c1
= FACE_COLOUR(f1
);
1749 enum face_colour c2
= FACE_COLOUR(f2
);
1750 assert(c1
!= FACE_GREY
);
1751 assert(c2
!= FACE_GREY
);
1753 if (f1
) clues
[f1
- g
->faces
]++;
1754 if (f2
) clues
[f2
- g
->faces
]++;
1762 static int game_has_unique_soln(const game_state
*state
, int diff
)
1765 solver_state
*sstate_new
;
1766 solver_state
*sstate
= new_solver_state((game_state
*)state
, diff
);
1768 sstate_new
= solve_game_rec(sstate
);
1770 assert(sstate_new
->solver_status
!= SOLVER_MISTAKE
);
1771 ret
= (sstate_new
->solver_status
== SOLVER_SOLVED
);
1773 free_solver_state(sstate_new
);
1774 free_solver_state(sstate
);
1780 /* Remove clues one at a time at random. */
1781 static game_state
*remove_clues(game_state
*state
, random_state
*rs
,
1785 int num_faces
= state
->game_grid
->num_faces
;
1786 game_state
*ret
= dup_game(state
), *saved_ret
;
1789 /* We need to remove some clues. We'll do this by forming a list of all
1790 * available clues, shuffling it, then going along one at a
1791 * time clearing each clue in turn for which doing so doesn't render the
1792 * board unsolvable. */
1793 face_list
= snewn(num_faces
, int);
1794 for (n
= 0; n
< num_faces
; ++n
) {
1798 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1800 for (n
= 0; n
< num_faces
; ++n
) {
1801 saved_ret
= dup_game(ret
);
1802 ret
->clues
[face_list
[n
]] = -1;
1804 if (game_has_unique_soln(ret
, diff
)) {
1805 free_game(saved_ret
);
1817 static char *new_game_desc(game_params
*params
, random_state
*rs
,
1818 char **aux
, int interactive
)
1820 /* solution and description both use run-length encoding in obvious ways */
1823 game_state
*state
= snew(game_state
);
1824 game_state
*state_new
;
1825 params_generate_grid(params
);
1826 state
->game_grid
= g
= params
->game_grid
;
1828 state
->clues
= snewn(g
->num_faces
, signed char);
1829 state
->lines
= snewn(g
->num_edges
, char);
1830 state
->line_errors
= snewn(g
->num_edges
, unsigned char);
1832 state
->grid_type
= params
->type
;
1836 memset(state
->lines
, LINE_UNKNOWN
, g
->num_edges
);
1837 memset(state
->line_errors
, 0, g
->num_edges
);
1839 state
->solved
= state
->cheated
= FALSE
;
1841 /* Get a new random solvable board with all its clues filled in. Yes, this
1842 * can loop for ever if the params are suitably unfavourable, but
1843 * preventing games smaller than 4x4 seems to stop this happening */
1845 add_full_clues(state
, rs
);
1846 } while (!game_has_unique_soln(state
, params
->diff
));
1848 state_new
= remove_clues(state
, rs
, params
->diff
);
1853 if (params
->diff
> 0 && game_has_unique_soln(state
, params
->diff
-1)) {
1855 fprintf(stderr
, "Rejecting board, it is too easy\n");
1857 goto newboard_please
;
1860 retval
= state_to_text(state
);
1864 assert(!validate_desc(params
, retval
));
1869 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1872 game_state
*state
= snew(game_state
);
1873 int empties_to_make
= 0;
1875 const char *dp
= desc
;
1877 int num_faces
, num_edges
;
1879 params_generate_grid(params
);
1880 state
->game_grid
= g
= params
->game_grid
;
1882 num_faces
= g
->num_faces
;
1883 num_edges
= g
->num_edges
;
1885 state
->clues
= snewn(num_faces
, signed char);
1886 state
->lines
= snewn(num_edges
, char);
1887 state
->line_errors
= snewn(num_edges
, unsigned char);
1889 state
->solved
= state
->cheated
= FALSE
;
1891 state
->grid_type
= params
->type
;
1893 for (i
= 0; i
< num_faces
; i
++) {
1894 if (empties_to_make
) {
1896 state
->clues
[i
] = -1;
1902 if (n
>= 0 && n
< 10) {
1903 state
->clues
[i
] = n
;
1907 state
->clues
[i
] = -1;
1908 empties_to_make
= n
- 1;
1913 memset(state
->lines
, LINE_UNKNOWN
, num_edges
);
1914 memset(state
->line_errors
, 0, num_edges
);
1918 /* Calculates the line_errors data, and checks if the current state is a
1920 static int check_completion(game_state
*state
)
1922 grid
*g
= state
->game_grid
;
1924 int num_faces
= g
->num_faces
;
1926 int infinite_area
, finite_area
;
1927 int loops_found
= 0;
1928 int found_edge_not_in_loop
= FALSE
;
1930 memset(state
->line_errors
, 0, g
->num_edges
);
1932 /* LL implementation of SGT's idea:
1933 * A loop will partition the grid into an inside and an outside.
1934 * If there is more than one loop, the grid will be partitioned into
1935 * even more distinct regions. We can therefore track equivalence of
1936 * faces, by saying that two faces are equivalent when there is a non-YES
1937 * edge between them.
1938 * We could keep track of the number of connected components, by counting
1939 * the number of dsf-merges that aren't no-ops.
1940 * But we're only interested in 3 separate cases:
1941 * no loops, one loop, more than one loop.
1943 * No loops: all faces are equivalent to the infinite face.
1944 * One loop: only two equivalence classes - finite and infinite.
1945 * >= 2 loops: there are 2 distinct finite regions.
1947 * So we simply make two passes through all the edges.
1948 * In the first pass, we dsf-merge the two faces bordering each non-YES
1950 * In the second pass, we look for YES-edges bordering:
1951 * a) two non-equivalent faces.
1952 * b) two non-equivalent faces, and one of them is part of a different
1953 * finite area from the first finite area we've seen.
1955 * An occurrence of a) means there is at least one loop.
1956 * An occurrence of b) means there is more than one loop.
1957 * Edges satisfying a) are marked as errors.
1959 * While we're at it, we set a flag if we find a YES edge that is not
1961 * This information will help decide, if there's a single loop, whether it
1962 * is a candidate for being a solution (that is, all YES edges are part of
1965 * If there is a candidate loop, we then go through all clues and check
1966 * they are all satisfied. If so, we have found a solution and we can
1967 * unmark all line_errors.
1970 /* Infinite face is at the end - its index is num_faces.
1971 * This macro is just to make this obvious! */
1972 #define INF_FACE num_faces
1973 dsf
= snewn(num_faces
+ 1, int);
1974 dsf_init(dsf
, num_faces
+ 1);
1977 for (i
= 0; i
< g
->num_edges
; i
++) {
1978 grid_edge
*e
= g
->edges
+ i
;
1979 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1980 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1981 if (state
->lines
[i
] != LINE_YES
)
1982 dsf_merge(dsf
, f1
, f2
);
1986 infinite_area
= dsf_canonify(dsf
, INF_FACE
);
1988 for (i
= 0; i
< g
->num_edges
; i
++) {
1989 grid_edge
*e
= g
->edges
+ i
;
1990 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
1991 int can1
= dsf_canonify(dsf
, f1
);
1992 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
1993 int can2
= dsf_canonify(dsf
, f2
);
1994 if (state
->lines
[i
] != LINE_YES
) continue;
1997 /* Faces are equivalent, so this edge not part of a loop */
1998 found_edge_not_in_loop
= TRUE
;
2001 state
->line_errors
[i
] = TRUE
;
2002 if (loops_found
== 0) loops_found
= 1;
2004 /* Don't bother with further checks if we've already found 2 loops */
2005 if (loops_found
== 2) continue;
2007 if (finite_area
== -1) {
2008 /* Found our first finite area */
2009 if (can1
!= infinite_area
)
2015 /* Have we found a second area? */
2016 if (finite_area
!= -1) {
2017 if (can1
!= infinite_area
&& can1
!= finite_area
) {
2021 if (can2
!= infinite_area
&& can2
!= finite_area
) {
2028 printf("loops_found = %d\n", loops_found);
2029 printf("found_edge_not_in_loop = %s\n",
2030 found_edge_not_in_loop ? "TRUE" : "FALSE");
2033 sfree(dsf
); /* No longer need the dsf */
2035 /* Have we found a candidate loop? */
2036 if (loops_found
== 1 && !found_edge_not_in_loop
) {
2037 /* Yes, so check all clues are satisfied */
2038 int found_clue_violation
= FALSE
;
2039 for (i
= 0; i
< num_faces
; i
++) {
2040 int c
= state
->clues
[i
];
2042 if (face_order(state
, i
, LINE_YES
) != c
) {
2043 found_clue_violation
= TRUE
;
2049 if (!found_clue_violation
) {
2050 /* The loop is good */
2051 memset(state
->line_errors
, 0, g
->num_edges
);
2052 return TRUE
; /* No need to bother checking for dot violations */
2056 /* Check for dot violations */
2057 for (i
= 0; i
< g
->num_dots
; i
++) {
2058 int yes
= dot_order(state
, i
, LINE_YES
);
2059 int unknown
= dot_order(state
, i
, LINE_UNKNOWN
);
2060 if ((yes
== 1 && unknown
== 0) || (yes
>= 3)) {
2061 /* violation, so mark all YES edges as errors */
2062 grid_dot
*d
= g
->dots
+ i
;
2064 for (j
= 0; j
< d
->order
; j
++) {
2065 int e
= d
->edges
[j
] - g
->edges
;
2066 if (state
->lines
[e
] == LINE_YES
)
2067 state
->line_errors
[e
] = TRUE
;
2074 /* ----------------------------------------------------------------------
2077 * Our solver modes operate as follows. Each mode also uses the modes above it.
2080 * Just implement the rules of the game.
2082 * Normal and Tricky Modes
2083 * For each (adjacent) pair of lines through each dot we store a bit for
2084 * whether at least one of them is on and whether at most one is on. (If we
2085 * know both or neither is on that's already stored more directly.)
2088 * Use edsf data structure to make equivalence classes of lines that are
2089 * known identical to or opposite to one another.
2094 * For general grids, we consider "dlines" to be pairs of lines joined
2095 * at a dot. The lines must be adjacent around the dot, so we can think of
2096 * a dline as being a dot+face combination. Or, a dot+edge combination where
2097 * the second edge is taken to be the next clockwise edge from the dot.
2098 * Original loopy code didn't have this extra restriction of the lines being
2099 * adjacent. From my tests with square grids, this extra restriction seems to
2100 * take little, if anything, away from the quality of the puzzles.
2101 * A dline can be uniquely identified by an edge/dot combination, given that
2102 * a dline-pair always goes clockwise around its common dot. The edge/dot
2103 * combination can be represented by an edge/bool combination - if bool is
2104 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2105 * exactly twice the number of edges in the grid - although the dlines
2106 * spanning the infinite face are not all that useful to the solver.
2107 * Note that, by convention, a dline goes clockwise around its common dot,
2108 * which means the dline goes anti-clockwise around its common face.
2111 /* Helper functions for obtaining an index into an array of dlines, given
2112 * various information. We assume the grid layout conventions about how
2113 * the various lists are interleaved - see grid_make_consistent() for
2116 /* i points to the first edge of the dline pair, reading clockwise around
2118 static int dline_index_from_dot(grid
*g
, grid_dot
*d
, int i
)
2120 grid_edge
*e
= d
->edges
[i
];
2125 if (i2
== d
->order
) i2
= 0;
2128 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2130 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2131 (int)(d
- g
->dots
), i
, (int)(e
- g
->edges
),
2132 (int)(e2
- g
->edges
), ret
);
2136 /* i points to the second edge of the dline pair, reading clockwise around
2137 * the face. That is, the edges of the dline, starting at edge{i}, read
2138 * anti-clockwise around the face. By layout conventions, the common dot
2139 * of the dline will be f->dots[i] */
2140 static int dline_index_from_face(grid
*g
, grid_face
*f
, int i
)
2142 grid_edge
*e
= f
->edges
[i
];
2143 grid_dot
*d
= f
->dots
[i
];
2148 if (i2
< 0) i2
+= f
->order
;
2151 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2153 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2154 (int)(f
- g
->faces
), i
, (int)(e
- g
->edges
),
2155 (int)(e2
- g
->edges
), ret
);
2159 static int is_atleastone(const char *dline_array
, int index
)
2161 return BIT_SET(dline_array
[index
], 0);
2163 static int set_atleastone(char *dline_array
, int index
)
2165 return SET_BIT(dline_array
[index
], 0);
2167 static int is_atmostone(const char *dline_array
, int index
)
2169 return BIT_SET(dline_array
[index
], 1);
2171 static int set_atmostone(char *dline_array
, int index
)
2173 return SET_BIT(dline_array
[index
], 1);
2176 static void array_setall(char *array
, char from
, char to
, int len
)
2178 char *p
= array
, *p_old
= p
;
2179 int len_remaining
= len
;
2181 while ((p
= memchr(p
, from
, len_remaining
))) {
2183 len_remaining
-= p
- p_old
;
2188 /* Helper, called when doing dline dot deductions, in the case where we
2189 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2190 * them (because of dline atmostone/atleastone).
2191 * On entry, edge points to the first of these two UNKNOWNs. This function
2192 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2193 * and set their corresponding dline to atleastone. (Setting atmostone
2194 * already happens in earlier dline deductions) */
2195 static int dline_set_opp_atleastone(solver_state
*sstate
,
2196 grid_dot
*d
, int edge
)
2198 game_state
*state
= sstate
->state
;
2199 grid
*g
= state
->game_grid
;
2202 for (opp
= 0; opp
< N
; opp
++) {
2203 int opp_dline_index
;
2204 if (opp
== edge
|| opp
== edge
+1 || opp
== edge
-1)
2206 if (opp
== 0 && edge
== N
-1)
2208 if (opp
== N
-1 && edge
== 0)
2211 if (opp2
== N
) opp2
= 0;
2212 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2213 if (state
->lines
[d
->edges
[opp
] - g
->edges
] != LINE_UNKNOWN
)
2215 if (state
->lines
[d
->edges
[opp2
] - g
->edges
] != LINE_UNKNOWN
)
2217 /* Found opposite UNKNOWNS and they're next to each other */
2218 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2219 return set_atleastone(sstate
->dlines
, opp_dline_index
);
2225 /* Set pairs of lines around this face which are known to be identical, to
2226 * the given line_state */
2227 static int face_setall_identical(solver_state
*sstate
, int face_index
,
2228 enum line_state line_new
)
2230 /* can[dir] contains the canonical line associated with the line in
2231 * direction dir from the square in question. Similarly inv[dir] is
2232 * whether or not the line in question is inverse to its canonical
2235 game_state
*state
= sstate
->state
;
2236 grid
*g
= state
->game_grid
;
2237 grid_face
*f
= g
->faces
+ face_index
;
2240 int can1
, can2
, inv1
, inv2
;
2242 for (i
= 0; i
< N
; i
++) {
2243 int line1_index
= f
->edges
[i
] - g
->edges
;
2244 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2246 for (j
= i
+ 1; j
< N
; j
++) {
2247 int line2_index
= f
->edges
[j
] - g
->edges
;
2248 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2251 /* Found two UNKNOWNS */
2252 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2253 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2254 if (can1
== can2
&& inv1
== inv2
) {
2255 solver_set_line(sstate
, line1_index
, line_new
);
2256 solver_set_line(sstate
, line2_index
, line_new
);
2263 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2264 * return the edge indices into e. */
2265 static void find_unknowns(game_state
*state
,
2266 grid_edge
**edge_list
, /* Edge list to search (from a face or a dot) */
2267 int expected_count
, /* Number of UNKNOWNs (comes from solver's cache) */
2268 int *e
/* Returned edge indices */)
2271 grid
*g
= state
->game_grid
;
2272 while (c
< expected_count
) {
2273 int line_index
= *edge_list
- g
->edges
;
2274 if (state
->lines
[line_index
] == LINE_UNKNOWN
) {
2282 /* If we have a list of edges, and we know whether the number of YESs should
2283 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2284 * linedsf deductions. This can be used for both face and dot deductions.
2285 * Returns the difficulty level of the next solver that should be used,
2286 * or DIFF_MAX if no progress was made. */
2287 static int parity_deductions(solver_state
*sstate
,
2288 grid_edge
**edge_list
, /* Edge list (from a face or a dot) */
2289 int total_parity
, /* Expected number of YESs modulo 2 (either 0 or 1) */
2292 game_state
*state
= sstate
->state
;
2293 int diff
= DIFF_MAX
;
2294 int *linedsf
= sstate
->linedsf
;
2296 if (unknown_count
== 2) {
2297 /* Lines are known alike/opposite, depending on inv. */
2299 find_unknowns(state
, edge_list
, 2, e
);
2300 if (merge_lines(sstate
, e
[0], e
[1], total_parity
))
2301 diff
= min(diff
, DIFF_HARD
);
2302 } else if (unknown_count
== 3) {
2304 int can
[3]; /* canonical edges */
2305 int inv
[3]; /* whether can[x] is inverse to e[x] */
2306 find_unknowns(state
, edge_list
, 3, e
);
2307 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2308 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2309 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2310 if (can
[0] == can
[1]) {
2311 if (solver_set_line(sstate
, e
[2], (total_parity
^inv
[0]^inv
[1]) ?
2312 LINE_YES
: LINE_NO
))
2313 diff
= min(diff
, DIFF_EASY
);
2315 if (can
[0] == can
[2]) {
2316 if (solver_set_line(sstate
, e
[1], (total_parity
^inv
[0]^inv
[2]) ?
2317 LINE_YES
: LINE_NO
))
2318 diff
= min(diff
, DIFF_EASY
);
2320 if (can
[1] == can
[2]) {
2321 if (solver_set_line(sstate
, e
[0], (total_parity
^inv
[1]^inv
[2]) ?
2322 LINE_YES
: LINE_NO
))
2323 diff
= min(diff
, DIFF_EASY
);
2325 } else if (unknown_count
== 4) {
2327 int can
[4]; /* canonical edges */
2328 int inv
[4]; /* whether can[x] is inverse to e[x] */
2329 find_unknowns(state
, edge_list
, 4, e
);
2330 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2331 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2332 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2333 can
[3] = edsf_canonify(linedsf
, e
[3], inv
+3);
2334 if (can
[0] == can
[1]) {
2335 if (merge_lines(sstate
, e
[2], e
[3], total_parity
^inv
[0]^inv
[1]))
2336 diff
= min(diff
, DIFF_HARD
);
2337 } else if (can
[0] == can
[2]) {
2338 if (merge_lines(sstate
, e
[1], e
[3], total_parity
^inv
[0]^inv
[2]))
2339 diff
= min(diff
, DIFF_HARD
);
2340 } else if (can
[0] == can
[3]) {
2341 if (merge_lines(sstate
, e
[1], e
[2], total_parity
^inv
[0]^inv
[3]))
2342 diff
= min(diff
, DIFF_HARD
);
2343 } else if (can
[1] == can
[2]) {
2344 if (merge_lines(sstate
, e
[0], e
[3], total_parity
^inv
[1]^inv
[2]))
2345 diff
= min(diff
, DIFF_HARD
);
2346 } else if (can
[1] == can
[3]) {
2347 if (merge_lines(sstate
, e
[0], e
[2], total_parity
^inv
[1]^inv
[3]))
2348 diff
= min(diff
, DIFF_HARD
);
2349 } else if (can
[2] == can
[3]) {
2350 if (merge_lines(sstate
, e
[0], e
[1], total_parity
^inv
[2]^inv
[3]))
2351 diff
= min(diff
, DIFF_HARD
);
2359 * These are the main solver functions.
2361 * Their return values are diff values corresponding to the lowest mode solver
2362 * that would notice the work that they have done. For example if the normal
2363 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2364 * easy mode solver might be able to make progress using that. It doesn't make
2365 * sense for one of them to return a diff value higher than that of the
2368 * Each function returns the lowest value it can, as early as possible, in
2369 * order to try and pass as much work as possible back to the lower level
2370 * solvers which progress more quickly.
2373 /* PROPOSED NEW DESIGN:
2374 * We have a work queue consisting of 'events' notifying us that something has
2375 * happened that a particular solver mode might be interested in. For example
2376 * the hard mode solver might do something that helps the normal mode solver at
2377 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2378 * we pull events off the work queue, and hand each in turn to the solver that
2379 * is interested in them. If a solver reports that it failed we pass the same
2380 * event on to progressively more advanced solvers and the loop detector. Once
2381 * we've exhausted an event, or it has helped us progress, we drop it and
2382 * continue to the next one. The events are sorted first in order of solver
2383 * complexity (easy first) then order of insertion (oldest first).
2384 * Once we run out of events we loop over each permitted solver in turn
2385 * (easiest first) until either a deduction is made (and an event therefore
2386 * emerges) or no further deductions can be made (in which case we've failed).
2389 * * How do we 'loop over' a solver when both dots and squares are concerned.
2390 * Answer: first all squares then all dots.
2393 static int trivial_deductions(solver_state
*sstate
)
2395 int i
, current_yes
, current_no
;
2396 game_state
*state
= sstate
->state
;
2397 grid
*g
= state
->game_grid
;
2398 int diff
= DIFF_MAX
;
2400 /* Per-face deductions */
2401 for (i
= 0; i
< g
->num_faces
; i
++) {
2402 grid_face
*f
= g
->faces
+ i
;
2404 if (sstate
->face_solved
[i
])
2407 current_yes
= sstate
->face_yes_count
[i
];
2408 current_no
= sstate
->face_no_count
[i
];
2410 if (current_yes
+ current_no
== f
->order
) {
2411 sstate
->face_solved
[i
] = TRUE
;
2415 if (state
->clues
[i
] < 0)
2418 if (state
->clues
[i
] < current_yes
) {
2419 sstate
->solver_status
= SOLVER_MISTAKE
;
2422 if (state
->clues
[i
] == current_yes
) {
2423 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
))
2424 diff
= min(diff
, DIFF_EASY
);
2425 sstate
->face_solved
[i
] = TRUE
;
2429 if (f
->order
- state
->clues
[i
] < current_no
) {
2430 sstate
->solver_status
= SOLVER_MISTAKE
;
2433 if (f
->order
- state
->clues
[i
] == current_no
) {
2434 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
))
2435 diff
= min(diff
, DIFF_EASY
);
2436 sstate
->face_solved
[i
] = TRUE
;
2441 check_caches(sstate
);
2443 /* Per-dot deductions */
2444 for (i
= 0; i
< g
->num_dots
; i
++) {
2445 grid_dot
*d
= g
->dots
+ i
;
2446 int yes
, no
, unknown
;
2448 if (sstate
->dot_solved
[i
])
2451 yes
= sstate
->dot_yes_count
[i
];
2452 no
= sstate
->dot_no_count
[i
];
2453 unknown
= d
->order
- yes
- no
;
2457 sstate
->dot_solved
[i
] = TRUE
;
2458 } else if (unknown
== 1) {
2459 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2460 diff
= min(diff
, DIFF_EASY
);
2461 sstate
->dot_solved
[i
] = TRUE
;
2463 } else if (yes
== 1) {
2465 sstate
->solver_status
= SOLVER_MISTAKE
;
2467 } else if (unknown
== 1) {
2468 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
);
2469 diff
= min(diff
, DIFF_EASY
);
2471 } else if (yes
== 2) {
2473 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2474 diff
= min(diff
, DIFF_EASY
);
2476 sstate
->dot_solved
[i
] = TRUE
;
2478 sstate
->solver_status
= SOLVER_MISTAKE
;
2483 check_caches(sstate
);
2488 static int dline_deductions(solver_state
*sstate
)
2490 game_state
*state
= sstate
->state
;
2491 grid
*g
= state
->game_grid
;
2492 char *dlines
= sstate
->dlines
;
2494 int diff
= DIFF_MAX
;
2496 /* ------ Face deductions ------ */
2498 /* Given a set of dline atmostone/atleastone constraints, need to figure
2499 * out if we can deduce any further info. For more general faces than
2500 * squares, this turns out to be a tricky problem.
2501 * The approach taken here is to define (per face) NxN matrices:
2502 * "maxs" and "mins".
2503 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2504 * for the possible number of edges that are YES between positions j and k
2505 * going clockwise around the face. Can think of j and k as marking dots
2506 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2507 * edge1 joins dot1 to dot2 etc).
2508 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2509 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2510 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2511 * the dline atmostone/atleastone status for edges j and j+1.
2513 * Then we calculate the remaining entries recursively. We definitely
2515 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2516 * This is because any valid placement of YESs between j and k must give
2517 * a valid placement between j and u, and also between u and k.
2518 * I believe it's sufficient to use just the two values of u:
2519 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2520 * are rigorous, even if they might not be best-possible.
2522 * Once we have maxs and mins calculated, we can make inferences about
2523 * each dline{j,j+1} by looking at the possible complementary edge-counts
2524 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2525 * As well as dlines, we can make similar inferences about single edges.
2526 * For example, consider a pentagon with clue 3, and we know at most one
2527 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2528 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2529 * that final edge would have to be YES to make the count up to 3.
2532 /* Much quicker to allocate arrays on the stack than the heap, so
2533 * define the largest possible face size, and base our array allocations
2534 * on that. We check this with an assertion, in case someone decides to
2535 * make a grid which has larger faces than this. Note, this algorithm
2536 * could get quite expensive if there are many large faces. */
2537 #define MAX_FACE_SIZE 8
2539 for (i
= 0; i
< g
->num_faces
; i
++) {
2540 int maxs
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2541 int mins
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2542 grid_face
*f
= g
->faces
+ i
;
2545 int clue
= state
->clues
[i
];
2546 assert(N
<= MAX_FACE_SIZE
);
2547 if (sstate
->face_solved
[i
])
2549 if (clue
< 0) continue;
2551 /* Calculate the (j,j+1) entries */
2552 for (j
= 0; j
< N
; j
++) {
2553 int edge_index
= f
->edges
[j
] - g
->edges
;
2555 enum line_state line1
= state
->lines
[edge_index
];
2556 enum line_state line2
;
2560 maxs
[j
][k
] = (line1
== LINE_NO
) ?
0 : 1;
2561 mins
[j
][k
] = (line1
== LINE_YES
) ?
1 : 0;
2562 /* Calculate the (j,j+2) entries */
2563 dline_index
= dline_index_from_face(g
, f
, k
);
2564 edge_index
= f
->edges
[k
] - g
->edges
;
2565 line2
= state
->lines
[edge_index
];
2571 if (line1
== LINE_NO
) tmp
--;
2572 if (line2
== LINE_NO
) tmp
--;
2573 if (tmp
== 2 && is_atmostone(dlines
, dline_index
))
2579 if (line1
== LINE_YES
) tmp
++;
2580 if (line2
== LINE_YES
) tmp
++;
2581 if (tmp
== 0 && is_atleastone(dlines
, dline_index
))
2586 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2587 for (m
= 3; m
< N
; m
++) {
2588 for (j
= 0; j
< N
; j
++) {
2596 maxs
[j
][k
] = maxs
[j
][u
] + maxs
[u
][k
];
2597 mins
[j
][k
] = mins
[j
][u
] + mins
[u
][k
];
2598 tmp
= maxs
[j
][v
] + maxs
[v
][k
];
2599 maxs
[j
][k
] = min(maxs
[j
][k
], tmp
);
2600 tmp
= mins
[j
][v
] + mins
[v
][k
];
2601 mins
[j
][k
] = max(mins
[j
][k
], tmp
);
2605 /* See if we can make any deductions */
2606 for (j
= 0; j
< N
; j
++) {
2608 grid_edge
*e
= f
->edges
[j
];
2609 int line_index
= e
- g
->edges
;
2612 if (state
->lines
[line_index
] != LINE_UNKNOWN
)
2617 /* minimum YESs in the complement of this edge */
2618 if (mins
[k
][j
] > clue
) {
2619 sstate
->solver_status
= SOLVER_MISTAKE
;
2622 if (mins
[k
][j
] == clue
) {
2623 /* setting this edge to YES would make at least
2624 * (clue+1) edges - contradiction */
2625 solver_set_line(sstate
, line_index
, LINE_NO
);
2626 diff
= min(diff
, DIFF_EASY
);
2628 if (maxs
[k
][j
] < clue
- 1) {
2629 sstate
->solver_status
= SOLVER_MISTAKE
;
2632 if (maxs
[k
][j
] == clue
- 1) {
2633 /* Only way to satisfy the clue is to set edge{j} as YES */
2634 solver_set_line(sstate
, line_index
, LINE_YES
);
2635 diff
= min(diff
, DIFF_EASY
);
2638 /* More advanced deduction that allows propagation along diagonal
2639 * chains of faces connected by dots, for example, 3-2-...-2-3
2640 * in square grids. */
2641 if (sstate
->diff
>= DIFF_TRICKY
) {
2642 /* Now see if we can make dline deduction for edges{j,j+1} */
2644 if (state
->lines
[e
- g
->edges
] != LINE_UNKNOWN
)
2645 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2646 * Dlines where one of the edges is known, are handled in the
2650 dline_index
= dline_index_from_face(g
, f
, k
);
2654 /* minimum YESs in the complement of this dline */
2655 if (mins
[k
][j
] > clue
- 2) {
2656 /* Adding 2 YESs would break the clue */
2657 if (set_atmostone(dlines
, dline_index
))
2658 diff
= min(diff
, DIFF_NORMAL
);
2660 /* maximum YESs in the complement of this dline */
2661 if (maxs
[k
][j
] < clue
) {
2662 /* Adding 2 NOs would mean not enough YESs */
2663 if (set_atleastone(dlines
, dline_index
))
2664 diff
= min(diff
, DIFF_NORMAL
);
2670 if (diff
< DIFF_NORMAL
)
2673 /* ------ Dot deductions ------ */
2675 for (i
= 0; i
< g
->num_dots
; i
++) {
2676 grid_dot
*d
= g
->dots
+ i
;
2678 int yes
, no
, unknown
;
2680 if (sstate
->dot_solved
[i
])
2682 yes
= sstate
->dot_yes_count
[i
];
2683 no
= sstate
->dot_no_count
[i
];
2684 unknown
= N
- yes
- no
;
2686 for (j
= 0; j
< N
; j
++) {
2689 int line1_index
, line2_index
;
2690 enum line_state line1
, line2
;
2693 dline_index
= dline_index_from_dot(g
, d
, j
);
2694 line1_index
= d
->edges
[j
] - g
->edges
;
2695 line2_index
= d
->edges
[k
] - g
->edges
;
2696 line1
= state
->lines
[line1_index
];
2697 line2
= state
->lines
[line2_index
];
2699 /* Infer dline state from line state */
2700 if (line1
== LINE_NO
|| line2
== LINE_NO
) {
2701 if (set_atmostone(dlines
, dline_index
))
2702 diff
= min(diff
, DIFF_NORMAL
);
2704 if (line1
== LINE_YES
|| line2
== LINE_YES
) {
2705 if (set_atleastone(dlines
, dline_index
))
2706 diff
= min(diff
, DIFF_NORMAL
);
2708 /* Infer line state from dline state */
2709 if (is_atmostone(dlines
, dline_index
)) {
2710 if (line1
== LINE_YES
&& line2
== LINE_UNKNOWN
) {
2711 solver_set_line(sstate
, line2_index
, LINE_NO
);
2712 diff
= min(diff
, DIFF_EASY
);
2714 if (line2
== LINE_YES
&& line1
== LINE_UNKNOWN
) {
2715 solver_set_line(sstate
, line1_index
, LINE_NO
);
2716 diff
= min(diff
, DIFF_EASY
);
2719 if (is_atleastone(dlines
, dline_index
)) {
2720 if (line1
== LINE_NO
&& line2
== LINE_UNKNOWN
) {
2721 solver_set_line(sstate
, line2_index
, LINE_YES
);
2722 diff
= min(diff
, DIFF_EASY
);
2724 if (line2
== LINE_NO
&& line1
== LINE_UNKNOWN
) {
2725 solver_set_line(sstate
, line1_index
, LINE_YES
);
2726 diff
= min(diff
, DIFF_EASY
);
2729 /* Deductions that depend on the numbers of lines.
2730 * Only bother if both lines are UNKNOWN, otherwise the
2731 * easy-mode solver (or deductions above) would have taken
2733 if (line1
!= LINE_UNKNOWN
|| line2
!= LINE_UNKNOWN
)
2736 if (yes
== 0 && unknown
== 2) {
2737 /* Both these unknowns must be identical. If we know
2738 * atmostone or atleastone, we can make progress. */
2739 if (is_atmostone(dlines
, dline_index
)) {
2740 solver_set_line(sstate
, line1_index
, LINE_NO
);
2741 solver_set_line(sstate
, line2_index
, LINE_NO
);
2742 diff
= min(diff
, DIFF_EASY
);
2744 if (is_atleastone(dlines
, dline_index
)) {
2745 solver_set_line(sstate
, line1_index
, LINE_YES
);
2746 solver_set_line(sstate
, line2_index
, LINE_YES
);
2747 diff
= min(diff
, DIFF_EASY
);
2751 if (set_atmostone(dlines
, dline_index
))
2752 diff
= min(diff
, DIFF_NORMAL
);
2754 if (set_atleastone(dlines
, dline_index
))
2755 diff
= min(diff
, DIFF_NORMAL
);
2759 /* More advanced deduction that allows propagation along diagonal
2760 * chains of faces connected by dots, for example: 3-2-...-2-3
2761 * in square grids. */
2762 if (sstate
->diff
>= DIFF_TRICKY
) {
2763 /* If we have atleastone set for this dline, infer
2764 * atmostone for each "opposite" dline (that is, each
2765 * dline without edges in common with this one).
2766 * Again, this test is only worth doing if both these
2767 * lines are UNKNOWN. For if one of these lines were YES,
2768 * the (yes == 1) test above would kick in instead. */
2769 if (is_atleastone(dlines
, dline_index
)) {
2771 for (opp
= 0; opp
< N
; opp
++) {
2772 int opp_dline_index
;
2773 if (opp
== j
|| opp
== j
+1 || opp
== j
-1)
2775 if (j
== 0 && opp
== N
-1)
2777 if (j
== N
-1 && opp
== 0)
2779 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2780 if (set_atmostone(dlines
, opp_dline_index
))
2781 diff
= min(diff
, DIFF_NORMAL
);
2783 if (yes
== 0 && is_atmostone(dlines
, dline_index
)) {
2784 /* This dline has *exactly* one YES and there are no
2785 * other YESs. This allows more deductions. */
2787 /* Third unknown must be YES */
2788 for (opp
= 0; opp
< N
; opp
++) {
2790 if (opp
== j
|| opp
== k
)
2792 opp_index
= d
->edges
[opp
] - g
->edges
;
2793 if (state
->lines
[opp_index
] == LINE_UNKNOWN
) {
2794 solver_set_line(sstate
, opp_index
,
2796 diff
= min(diff
, DIFF_EASY
);
2799 } else if (unknown
== 4) {
2800 /* Exactly one of opposite UNKNOWNS is YES. We've
2801 * already set atmostone, so set atleastone as
2804 if (dline_set_opp_atleastone(sstate
, d
, j
))
2805 diff
= min(diff
, DIFF_NORMAL
);
2815 static int linedsf_deductions(solver_state
*sstate
)
2817 game_state
*state
= sstate
->state
;
2818 grid
*g
= state
->game_grid
;
2819 char *dlines
= sstate
->dlines
;
2821 int diff
= DIFF_MAX
;
2824 /* ------ Face deductions ------ */
2826 /* A fully-general linedsf deduction seems overly complicated
2827 * (I suspect the problem is NP-complete, though in practice it might just
2828 * be doable because faces are limited in size).
2829 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2830 * known to be identical. If setting them both to YES (or NO) would break
2831 * the clue, set them to NO (or YES). */
2833 for (i
= 0; i
< g
->num_faces
; i
++) {
2834 int N
, yes
, no
, unknown
;
2837 if (sstate
->face_solved
[i
])
2839 clue
= state
->clues
[i
];
2843 N
= g
->faces
[i
].order
;
2844 yes
= sstate
->face_yes_count
[i
];
2845 if (yes
+ 1 == clue
) {
2846 if (face_setall_identical(sstate
, i
, LINE_NO
))
2847 diff
= min(diff
, DIFF_EASY
);
2849 no
= sstate
->face_no_count
[i
];
2850 if (no
+ 1 == N
- clue
) {
2851 if (face_setall_identical(sstate
, i
, LINE_YES
))
2852 diff
= min(diff
, DIFF_EASY
);
2855 /* Reload YES count, it might have changed */
2856 yes
= sstate
->face_yes_count
[i
];
2857 unknown
= N
- no
- yes
;
2859 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2860 * parity of lines. */
2861 diff_tmp
= parity_deductions(sstate
, g
->faces
[i
].edges
,
2862 (clue
- yes
) % 2, unknown
);
2863 diff
= min(diff
, diff_tmp
);
2866 /* ------ Dot deductions ------ */
2867 for (i
= 0; i
< g
->num_dots
; i
++) {
2868 grid_dot
*d
= g
->dots
+ i
;
2871 int yes
, no
, unknown
;
2872 /* Go through dlines, and do any dline<->linedsf deductions wherever
2873 * we find two UNKNOWNS. */
2874 for (j
= 0; j
< N
; j
++) {
2875 int dline_index
= dline_index_from_dot(g
, d
, j
);
2878 int can1
, can2
, inv1
, inv2
;
2880 line1_index
= d
->edges
[j
] - g
->edges
;
2881 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2884 if (j2
== N
) j2
= 0;
2885 line2_index
= d
->edges
[j2
] - g
->edges
;
2886 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2888 /* Infer dline flags from linedsf */
2889 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2890 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2891 if (can1
== can2
&& inv1
!= inv2
) {
2892 /* These are opposites, so set dline atmostone/atleastone */
2893 if (set_atmostone(dlines
, dline_index
))
2894 diff
= min(diff
, DIFF_NORMAL
);
2895 if (set_atleastone(dlines
, dline_index
))
2896 diff
= min(diff
, DIFF_NORMAL
);
2899 /* Infer linedsf from dline flags */
2900 if (is_atmostone(dlines
, dline_index
)
2901 && is_atleastone(dlines
, dline_index
)) {
2902 if (merge_lines(sstate
, line1_index
, line2_index
, 1))
2903 diff
= min(diff
, DIFF_HARD
);
2907 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2908 * parity of lines. */
2909 yes
= sstate
->dot_yes_count
[i
];
2910 no
= sstate
->dot_no_count
[i
];
2911 unknown
= N
- yes
- no
;
2912 diff_tmp
= parity_deductions(sstate
, d
->edges
,
2914 diff
= min(diff
, diff_tmp
);
2917 /* ------ Edge dsf deductions ------ */
2919 /* If the state of a line is known, deduce the state of its canonical line
2920 * too, and vice versa. */
2921 for (i
= 0; i
< g
->num_edges
; i
++) {
2924 can
= edsf_canonify(sstate
->linedsf
, i
, &inv
);
2927 s
= sstate
->state
->lines
[can
];
2928 if (s
!= LINE_UNKNOWN
) {
2929 if (solver_set_line(sstate
, i
, inv ?
OPP(s
) : s
))
2930 diff
= min(diff
, DIFF_EASY
);
2932 s
= sstate
->state
->lines
[i
];
2933 if (s
!= LINE_UNKNOWN
) {
2934 if (solver_set_line(sstate
, can
, inv ?
OPP(s
) : s
))
2935 diff
= min(diff
, DIFF_EASY
);
2943 static int loop_deductions(solver_state
*sstate
)
2945 int edgecount
= 0, clues
= 0, satclues
= 0, sm1clues
= 0;
2946 game_state
*state
= sstate
->state
;
2947 grid
*g
= state
->game_grid
;
2948 int shortest_chainlen
= g
->num_dots
;
2949 int loop_found
= FALSE
;
2951 int progress
= FALSE
;
2955 * Go through the grid and update for all the new edges.
2956 * Since merge_dots() is idempotent, the simplest way to
2957 * do this is just to update for _all_ the edges.
2958 * Also, while we're here, we count the edges.
2960 for (i
= 0; i
< g
->num_edges
; i
++) {
2961 if (state
->lines
[i
] == LINE_YES
) {
2962 loop_found
|= merge_dots(sstate
, i
);
2968 * Count the clues, count the satisfied clues, and count the
2969 * satisfied-minus-one clues.
2971 for (i
= 0; i
< g
->num_faces
; i
++) {
2972 int c
= state
->clues
[i
];
2974 int o
= sstate
->face_yes_count
[i
];
2983 for (i
= 0; i
< g
->num_dots
; ++i
) {
2985 sstate
->looplen
[dsf_canonify(sstate
->dotdsf
, i
)];
2986 if (dots_connected
> 1)
2987 shortest_chainlen
= min(shortest_chainlen
, dots_connected
);
2990 assert(sstate
->solver_status
== SOLVER_INCOMPLETE
);
2992 if (satclues
== clues
&& shortest_chainlen
== edgecount
) {
2993 sstate
->solver_status
= SOLVER_SOLVED
;
2994 /* This discovery clearly counts as progress, even if we haven't
2995 * just added any lines or anything */
2997 goto finished_loop_deductionsing
;
3001 * Now go through looking for LINE_UNKNOWN edges which
3002 * connect two dots that are already in the same
3003 * equivalence class. If we find one, test to see if the
3004 * loop it would create is a solution.
3006 for (i
= 0; i
< g
->num_edges
; i
++) {
3007 grid_edge
*e
= g
->edges
+ i
;
3008 int d1
= e
->dot1
- g
->dots
;
3009 int d2
= e
->dot2
- g
->dots
;
3011 if (state
->lines
[i
] != LINE_UNKNOWN
)
3014 eqclass
= dsf_canonify(sstate
->dotdsf
, d1
);
3015 if (eqclass
!= dsf_canonify(sstate
->dotdsf
, d2
))
3018 val
= LINE_NO
; /* loop is bad until proven otherwise */
3021 * This edge would form a loop. Next
3022 * question: how long would the loop be?
3023 * Would it equal the total number of edges
3024 * (plus the one we'd be adding if we added
3027 if (sstate
->looplen
[eqclass
] == edgecount
+ 1) {
3031 * This edge would form a loop which
3032 * took in all the edges in the entire
3033 * grid. So now we need to work out
3034 * whether it would be a valid solution
3035 * to the puzzle, which means we have to
3036 * check if it satisfies all the clues.
3037 * This means that every clue must be
3038 * either satisfied or satisfied-minus-
3039 * 1, and also that the number of
3040 * satisfied-minus-1 clues must be at
3041 * most two and they must lie on either
3042 * side of this edge.
3046 int f
= e
->face1
- g
->faces
;
3047 int c
= state
->clues
[f
];
3048 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3052 int f
= e
->face2
- g
->faces
;
3053 int c
= state
->clues
[f
];
3054 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3057 if (sm1clues
== sm1_nearby
&&
3058 sm1clues
+ satclues
== clues
) {
3059 val
= LINE_YES
; /* loop is good! */
3064 * Right. Now we know that adding this edge
3065 * would form a loop, and we know whether
3066 * that loop would be a viable solution or
3069 * If adding this edge produces a solution,
3070 * then we know we've found _a_ solution but
3071 * we don't know that it's _the_ solution -
3072 * if it were provably the solution then
3073 * we'd have deduced this edge some time ago
3074 * without the need to do loop detection. So
3075 * in this state we return SOLVER_AMBIGUOUS,
3076 * which has the effect that hitting Solve
3077 * on a user-provided puzzle will fill in a
3078 * solution but using the solver to
3079 * construct new puzzles won't consider this
3080 * a reasonable deduction for the user to
3083 progress
= solver_set_line(sstate
, i
, val
);
3084 assert(progress
== TRUE
);
3085 if (val
== LINE_YES
) {
3086 sstate
->solver_status
= SOLVER_AMBIGUOUS
;
3087 goto finished_loop_deductionsing
;
3091 finished_loop_deductionsing
:
3092 return progress ? DIFF_EASY
: DIFF_MAX
;
3095 /* This will return a dynamically allocated solver_state containing the (more)
3097 static solver_state
*solve_game_rec(const solver_state
*sstate_start
)
3099 solver_state
*sstate
;
3101 /* Index of the solver we should call next. */
3104 /* As a speed-optimisation, we avoid re-running solvers that we know
3105 * won't make any progress. This happens when a high-difficulty
3106 * solver makes a deduction that can only help other high-difficulty
3108 * For example: if a new 'dline' flag is set by dline_deductions, the
3109 * trivial_deductions solver cannot do anything with this information.
3110 * If we've already run the trivial_deductions solver (because it's
3111 * earlier in the list), there's no point running it again.
3113 * Therefore: if a solver is earlier in the list than "threshold_index",
3114 * we don't bother running it if it's difficulty level is less than
3117 int threshold_diff
= 0;
3118 int threshold_index
= 0;
3120 sstate
= dup_solver_state(sstate_start
);
3122 check_caches(sstate
);
3124 while (i
< NUM_SOLVERS
) {
3125 if (sstate
->solver_status
== SOLVER_MISTAKE
)
3127 if (sstate
->solver_status
== SOLVER_SOLVED
||
3128 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3129 /* solver finished */
3133 if ((solver_diffs
[i
] >= threshold_diff
|| i
>= threshold_index
)
3134 && solver_diffs
[i
] <= sstate
->diff
) {
3135 /* current_solver is eligible, so use it */
3136 int next_diff
= solver_fns
[i
](sstate
);
3137 if (next_diff
!= DIFF_MAX
) {
3138 /* solver made progress, so use new thresholds and
3139 * start again at top of list. */
3140 threshold_diff
= next_diff
;
3141 threshold_index
= i
;
3146 /* current_solver is ineligible, or failed to make progress, so
3147 * go to the next solver in the list */
3151 if (sstate
->solver_status
== SOLVER_SOLVED
||
3152 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3153 /* s/LINE_UNKNOWN/LINE_NO/g */
3154 array_setall(sstate
->state
->lines
, LINE_UNKNOWN
, LINE_NO
,
3155 sstate
->state
->game_grid
->num_edges
);
3162 static char *solve_game(game_state
*state
, game_state
*currstate
,
3163 char *aux
, char **error
)
3166 solver_state
*sstate
, *new_sstate
;
3168 sstate
= new_solver_state(state
, DIFF_MAX
);
3169 new_sstate
= solve_game_rec(sstate
);
3171 if (new_sstate
->solver_status
== SOLVER_SOLVED
) {
3172 soln
= encode_solve_move(new_sstate
->state
);
3173 } else if (new_sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3174 soln
= encode_solve_move(new_sstate
->state
);
3175 /**error = "Solver found ambiguous solutions"; */
3177 soln
= encode_solve_move(new_sstate
->state
);
3178 /**error = "Solver failed"; */
3181 free_solver_state(new_sstate
);
3182 free_solver_state(sstate
);
3187 /* ----------------------------------------------------------------------
3188 * Drawing and mouse-handling
3191 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
3192 int x
, int y
, int button
)
3194 grid
*g
= state
->game_grid
;
3198 char button_char
= ' ';
3199 enum line_state old_state
;
3201 button
&= ~MOD_MASK
;
3203 /* Convert mouse-click (x,y) to grid coordinates */
3204 x
-= BORDER(ds
->tilesize
);
3205 y
-= BORDER(ds
->tilesize
);
3206 x
= x
* g
->tilesize
/ ds
->tilesize
;
3207 y
= y
* g
->tilesize
/ ds
->tilesize
;
3211 e
= grid_nearest_edge(g
, x
, y
);
3217 /* I think it's only possible to play this game with mouse clicks, sorry */
3218 /* Maybe will add mouse drag support some time */
3219 old_state
= state
->lines
[i
];
3223 switch (old_state
) {
3241 switch (old_state
) {
3260 sprintf(buf
, "%d%c", i
, (int)button_char
);
3266 static game_state
*execute_move(game_state
*state
, char *move
)
3269 game_state
*newstate
= dup_game(state
);
3271 if (move
[0] == 'S') {
3273 newstate
->cheated
= TRUE
;
3278 if (i
< 0 || i
>= newstate
->game_grid
->num_edges
)
3280 move
+= strspn(move
, "1234567890");
3281 switch (*(move
++)) {
3283 newstate
->lines
[i
] = LINE_YES
;
3286 newstate
->lines
[i
] = LINE_NO
;
3289 newstate
->lines
[i
] = LINE_UNKNOWN
;
3297 * Check for completion.
3299 if (check_completion(newstate
))
3300 newstate
->solved
= TRUE
;
3305 free_game(newstate
);
3309 /* ----------------------------------------------------------------------
3313 /* Convert from grid coordinates to screen coordinates */
3314 static void grid_to_screen(const game_drawstate
*ds
, const grid
*g
,
3315 int grid_x
, int grid_y
, int *x
, int *y
)
3317 *x
= grid_x
- g
->lowest_x
;
3318 *y
= grid_y
- g
->lowest_y
;
3319 *x
= *x
* ds
->tilesize
/ g
->tilesize
;
3320 *y
= *y
* ds
->tilesize
/ g
->tilesize
;
3321 *x
+= BORDER(ds
->tilesize
);
3322 *y
+= BORDER(ds
->tilesize
);
3325 /* Returns (into x,y) position of centre of face for rendering the text clue.
3327 static void face_text_pos(const game_drawstate
*ds
, const grid
*g
,
3328 const grid_face
*f
, int *x
, int *y
)
3332 /* Simplest solution is the centroid. Might not work in some cases. */
3334 /* Another algorithm to look into:
3335 * Find the midpoints of the sides, find the bounding-box,
3336 * then take the centre of that. */
3338 /* Best solution probably involves incentres (inscribed circles) */
3340 int sx
= 0, sy
= 0; /* sums */
3341 for (i
= 0; i
< f
->order
; i
++) {
3342 grid_dot
*d
= f
->dots
[i
];
3349 /* convert to screen coordinates */
3350 grid_to_screen(ds
, g
, sx
, sy
, x
, y
);
3353 static void game_redraw_clue(drawing
*dr
, game_drawstate
*ds
,
3354 game_state
*state
, int i
)
3356 grid
*g
= state
->game_grid
;
3357 grid_face
*f
= g
->faces
+ i
;
3361 c
[0] = CLUE2CHAR(state
->clues
[i
]);
3364 face_text_pos(ds
, g
, f
, &x
, &y
);
3366 FONT_VARIABLE
, ds
->tilesize
/2,
3367 ALIGN_VCENTRE
| ALIGN_HCENTRE
,
3368 ds
->clue_error
[i
] ? COL_MISTAKE
:
3369 ds
->clue_satisfied
[i
] ? COL_SATISFIED
: COL_FOREGROUND
, c
);
3372 static void game_redraw_line(drawing
*dr
, game_drawstate
*ds
,
3373 game_state
*state
, int i
)
3375 grid
*g
= state
->game_grid
;
3376 grid_edge
*e
= g
->edges
+ i
;
3378 int xmin
, ymin
, xmax
, ymax
;
3381 if (state
->line_errors
[i
])
3382 line_colour
= COL_MISTAKE
;
3383 else if (state
->lines
[i
] == LINE_UNKNOWN
)
3384 line_colour
= COL_LINEUNKNOWN
;
3385 else if (state
->lines
[i
] == LINE_NO
)
3386 line_colour
= COL_FAINT
;
3387 else if (ds
->flashing
)
3388 line_colour
= COL_HIGHLIGHT
;
3390 line_colour
= COL_FOREGROUND
;
3392 /* Convert from grid to screen coordinates */
3393 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3394 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3401 if (line_colour
== COL_FAINT
) {
3402 static int draw_faint_lines
= -1;
3403 if (draw_faint_lines
< 0) {
3404 char *env
= getenv("LOOPY_FAINT_LINES");
3405 draw_faint_lines
= (!env
|| (env
[0] == 'y' ||
3408 if (draw_faint_lines
)
3409 draw_line(dr
, x1
, y1
, x2
, y2
, line_colour
);
3411 draw_thick_line(dr
, 3.0,
3418 static void game_redraw_dot(drawing
*dr
, game_drawstate
*ds
,
3419 game_state
*state
, int i
)
3421 grid
*g
= state
->game_grid
;
3422 grid_dot
*d
= g
->dots
+ i
;
3425 grid_to_screen(ds
, g
, d
->x
, d
->y
, &x
, &y
);
3426 draw_circle(dr
, x
, y
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3429 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
3430 game_state
*state
, int dir
, game_ui
*ui
,
3431 float animtime
, float flashtime
)
3433 #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
3435 grid
*g
= state
->game_grid
;
3436 int border
= BORDER(ds
->tilesize
);
3439 int redraw_everything
= FALSE
;
3441 int edges
[REDRAW_OBJECTS_LIMIT
], nedges
= 0;
3442 int faces
[REDRAW_OBJECTS_LIMIT
], nfaces
= 0;
3444 /* Redrawing is somewhat involved.
3446 * An update can theoretically affect an arbitrary number of edges
3447 * (consider, for example, completing or breaking a cycle which doesn't
3448 * satisfy all the clues -- we'll switch many edges between error and
3449 * normal states). On the other hand, redrawing the whole grid takes a
3450 * while, making the game feel sluggish, and many updates are actually
3451 * quite well localized.
3453 * This redraw algorithm attempts to cope with both situations gracefully
3454 * and correctly. For localized changes, we set a clip rectangle, fill
3455 * it with background, and then redraw (a plausible but conservative
3456 * guess at) the objects which intersect the rectangle; if several
3457 * objects need redrawing, we'll do them individually. However, if lots
3458 * of objects are affected, we'll just redraw everything.
3460 * The reason for all of this is that it's just not safe to do the redraw
3461 * piecemeal. If you try to draw an antialiased diagonal line over
3462 * itself, you get a slightly thicker antialiased diagonal line, which
3463 * looks rather ugly after a while.
3465 * So, we take two passes over the grid. The first attempts to work out
3466 * what needs doing, and the second actually does it.
3470 redraw_everything
= TRUE
;
3473 /* First, trundle through the faces. */
3474 for (i
= 0; i
< g
->num_faces
; i
++) {
3475 grid_face
*f
= g
->faces
+ i
;
3476 int sides
= f
->order
;
3479 int n
= state
->clues
[i
];
3483 clue_mistake
= (face_order(state
, i
, LINE_YES
) > n
||
3484 face_order(state
, i
, LINE_NO
) > (sides
-n
));
3485 clue_satisfied
= (face_order(state
, i
, LINE_YES
) == n
&&
3486 face_order(state
, i
, LINE_NO
) == (sides
-n
));
3488 if (clue_mistake
!= ds
->clue_error
[i
] ||
3489 clue_satisfied
!= ds
->clue_satisfied
[i
]) {
3490 ds
->clue_error
[i
] = clue_mistake
;
3491 ds
->clue_satisfied
[i
] = clue_satisfied
;
3492 if (nfaces
== REDRAW_OBJECTS_LIMIT
)
3493 redraw_everything
= TRUE
;
3495 faces
[nfaces
++] = i
;
3499 /* Work out what the flash state needs to be. */
3500 if (flashtime
> 0 &&
3501 (flashtime
<= FLASH_TIME
/3 ||
3502 flashtime
>= FLASH_TIME
*2/3)) {
3503 flash_changed
= !ds
->flashing
;
3504 ds
->flashing
= TRUE
;
3506 flash_changed
= ds
->flashing
;
3507 ds
->flashing
= FALSE
;
3510 /* Now, trundle through the edges. */
3511 for (i
= 0; i
< g
->num_edges
; i
++) {
3513 state
->line_errors
[i
] ? DS_LINE_ERROR
: state
->lines
[i
];
3514 if (new_ds
!= ds
->lines
[i
] ||
3515 (flash_changed
&& state
->lines
[i
] == LINE_YES
)) {
3516 ds
->lines
[i
] = new_ds
;
3517 if (nedges
== REDRAW_OBJECTS_LIMIT
)
3518 redraw_everything
= TRUE
;
3520 edges
[nedges
++] = i
;
3525 /* Pass one is now done. Now we do the actual drawing. */
3526 if (redraw_everything
) {
3528 /* This is the unsubtle version. */
3530 int grid_width
= g
->highest_x
- g
->lowest_x
;
3531 int grid_height
= g
->highest_y
- g
->lowest_y
;
3532 int w
= grid_width
* ds
->tilesize
/ g
->tilesize
;
3533 int h
= grid_height
* ds
->tilesize
/ g
->tilesize
;
3535 draw_rect(dr
, 0, 0, w
+ 2*border
+ 1, h
+ 2*border
+ 1,
3538 for (i
= 0; i
< g
->num_faces
; i
++)
3539 game_redraw_clue(dr
, ds
, state
, i
);
3540 for (i
= 0; i
< g
->num_edges
; i
++)
3541 game_redraw_line(dr
, ds
, state
, i
);
3542 for (i
= 0; i
< g
->num_dots
; i
++)
3543 game_redraw_dot(dr
, ds
, state
, i
);
3545 draw_update(dr
, 0, 0, w
+ 2*border
+ 1, h
+ 2*border
+ 1);
3548 /* Right. Now we roll up our sleeves. */
3550 for (i
= 0; i
< nfaces
; i
++) {
3551 grid_face
*f
= g
->faces
+ faces
[i
];
3556 /* There seems to be a certain amount of trial-and-error
3557 * involved in working out the correct bounding-box for
3559 face_text_pos(ds
, g
, f
, &xx
, &yy
);
3561 x
= xx
- ds
->tilesize
/4 - 1; w
= ds
->tilesize
/2 + 2;
3562 y
= yy
- ds
->tilesize
/4 - 3; h
= ds
->tilesize
/2 + 5;
3563 clip(dr
, x
, y
, w
, h
);
3564 draw_rect(dr
, x
, y
, w
, h
, COL_BACKGROUND
);
3566 game_redraw_clue(dr
, ds
, state
, faces
[i
]);
3567 for (j
= 0; j
< f
->order
; j
++)
3568 game_redraw_line(dr
, ds
, state
, f
->edges
[j
] - g
->edges
);
3569 for (j
= 0; j
< f
->order
; j
++)
3570 game_redraw_dot(dr
, ds
, state
, f
->dots
[j
] - g
->dots
);
3572 draw_update(dr
, x
, y
, w
, h
);
3575 for (i
= 0; i
< nedges
; i
++) {
3576 grid_edge
*e
= g
->edges
+ edges
[i
], *ee
;
3577 int x1
= e
->dot1
->x
;
3578 int y1
= e
->dot1
->y
;
3579 int x2
= e
->dot2
->x
;
3580 int y2
= e
->dot2
->y
;
3581 int xmin
, xmax
, ymin
, ymax
;
3584 grid_to_screen(ds
, g
, x1
, y1
, &x1
, &y1
);
3585 grid_to_screen(ds
, g
, x2
, y2
, &x2
, &y2
);
3586 /* Allow extra margin for dots, and thickness of lines */
3587 xmin
= min(x1
, x2
) - 2;
3588 xmax
= max(x1
, x2
) + 2;
3589 ymin
= min(y1
, y2
) - 2;
3590 ymax
= max(y1
, y2
) + 2;
3591 /* For testing, I find it helpful to change COL_BACKGROUND
3592 * to COL_SATISFIED here. */
3593 clip(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1);
3594 draw_rect(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1,
3598 game_redraw_clue(dr
, ds
, state
, e
->face1
- g
->faces
);
3600 game_redraw_clue(dr
, ds
, state
, e
->face2
- g
->faces
);
3602 game_redraw_line(dr
, ds
, state
, edges
[i
]);
3603 for (j
= 0; j
< e
->dot1
->order
; j
++) {
3604 ee
= e
->dot1
->edges
[j
];
3606 game_redraw_line(dr
, ds
, state
, ee
- g
->edges
);
3608 for (j
= 0; j
< e
->dot2
->order
; j
++) {
3609 ee
= e
->dot2
->edges
[j
];
3611 game_redraw_line(dr
, ds
, state
, ee
- g
->edges
);
3613 game_redraw_dot(dr
, ds
, state
, e
->dot1
- g
->dots
);
3614 game_redraw_dot(dr
, ds
, state
, e
->dot2
- g
->dots
);
3617 draw_update(dr
, xmin
, ymin
, xmax
- xmin
+ 1, ymax
- ymin
+ 1);
3624 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
3625 int dir
, game_ui
*ui
)
3627 if (!oldstate
->solved
&& newstate
->solved
&&
3628 !oldstate
->cheated
&& !newstate
->cheated
) {
3635 static void game_print_size(game_params
*params
, float *x
, float *y
)
3640 * I'll use 7mm "squares" by default.
3642 game_compute_size(params
, 700, &pw
, &ph
);
3647 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
3649 int ink
= print_mono_colour(dr
, 0);
3651 game_drawstate ads
, *ds
= &ads
;
3652 grid
*g
= state
->game_grid
;
3654 ds
->tilesize
= tilesize
;
3656 for (i
= 0; i
< g
->num_dots
; i
++) {
3658 grid_to_screen(ds
, g
, g
->dots
[i
].x
, g
->dots
[i
].y
, &x
, &y
);
3659 draw_circle(dr
, x
, y
, ds
->tilesize
/ 15, ink
, ink
);
3665 for (i
= 0; i
< g
->num_faces
; i
++) {
3666 grid_face
*f
= g
->faces
+ i
;
3667 int clue
= state
->clues
[i
];
3671 c
[0] = CLUE2CHAR(clue
);
3673 face_text_pos(ds
, g
, f
, &x
, &y
);
3675 FONT_VARIABLE
, ds
->tilesize
/ 2,
3676 ALIGN_VCENTRE
| ALIGN_HCENTRE
, ink
, c
);
3683 for (i
= 0; i
< g
->num_edges
; i
++) {
3684 int thickness
= (state
->lines
[i
] == LINE_YES
) ?
30 : 150;
3685 grid_edge
*e
= g
->edges
+ i
;
3687 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3688 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3689 if (state
->lines
[i
] == LINE_YES
)
3691 /* (dx, dy) points from (x1, y1) to (x2, y2).
3692 * The line is then "fattened" in a perpendicular
3693 * direction to create a thin rectangle. */
3694 double d
= sqrt(SQ((double)x1
- x2
) + SQ((double)y1
- y2
));
3695 double dx
= (x2
- x1
) / d
;
3696 double dy
= (y2
- y1
) / d
;
3699 dx
= (dx
* ds
->tilesize
) / thickness
;
3700 dy
= (dy
* ds
->tilesize
) / thickness
;
3701 points
[0] = x1
+ (int)dy
;
3702 points
[1] = y1
- (int)dx
;
3703 points
[2] = x1
- (int)dy
;
3704 points
[3] = y1
+ (int)dx
;
3705 points
[4] = x2
- (int)dy
;
3706 points
[5] = y2
+ (int)dx
;
3707 points
[6] = x2
+ (int)dy
;
3708 points
[7] = y2
- (int)dx
;
3709 draw_polygon(dr
, points
, 4, ink
, ink
);
3713 /* Draw a dotted line */
3716 for (j
= 1; j
< divisions
; j
++) {
3717 /* Weighted average */
3718 int x
= (x1
* (divisions
-j
) + x2
* j
) / divisions
;
3719 int y
= (y1
* (divisions
-j
) + y2
* j
) / divisions
;
3720 draw_circle(dr
, x
, y
, ds
->tilesize
/ thickness
, ink
, ink
);
3727 #define thegame loopy
3730 const struct game thegame
= {
3731 "Loopy", "games.loopy", "loopy",
3738 TRUE
, game_configure
, custom_params
,
3746 TRUE
, game_can_format_as_text_now
, game_text_format
,
3754 PREFERRED_TILE_SIZE
, game_compute_size
, game_set_size
,
3757 game_free_drawstate
,
3761 TRUE
, FALSE
, game_print_size
, game_print
,
3762 FALSE
/* wants_statusbar */,
3763 FALSE
, game_timing_state
,
3764 0, /* mouse_priorities */
3767 #ifdef STANDALONE_SOLVER
3770 * Half-hearted standalone solver. It can't output the solution to
3771 * anything but a square puzzle, and it can't log the deductions
3772 * it makes either. But it can solve square puzzles, and more
3773 * importantly it can use its solver to grade the difficulty of
3774 * any puzzle you give it.
3779 int main(int argc
, char **argv
)
3783 char *id
= NULL
, *desc
, *err
;
3786 #if 0 /* verbose solver not supported here (yet) */
3787 int really_verbose
= FALSE
;
3790 while (--argc
> 0) {
3792 #if 0 /* verbose solver not supported here (yet) */
3793 if (!strcmp(p
, "-v")) {
3794 really_verbose
= TRUE
;
3797 if (!strcmp(p
, "-g")) {
3799 } else if (*p
== '-') {
3800 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
3808 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
3812 desc
= strchr(id
, ':');
3814 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
3819 p
= default_params();
3820 decode_params(p
, id
);
3821 err
= validate_desc(p
, desc
);
3823 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
3826 s
= new_game(NULL
, p
, desc
);
3829 * When solving an Easy puzzle, we don't want to bother the
3830 * user with Hard-level deductions. For this reason, we grade
3831 * the puzzle internally before doing anything else.
3833 ret
= -1; /* placate optimiser */
3834 for (diff
= 0; diff
< DIFF_MAX
; diff
++) {
3835 solver_state
*sstate_new
;
3836 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3838 sstate_new
= solve_game_rec(sstate
);
3840 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3842 else if (sstate_new
->solver_status
== SOLVER_SOLVED
)
3847 free_solver_state(sstate_new
);
3848 free_solver_state(sstate
);
3854 if (diff
== DIFF_MAX
) {
3856 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3858 printf("Unable to find a unique solution\n");
3862 printf("Difficulty rating: impossible (no solution exists)\n");
3864 printf("Difficulty rating: %s\n", diffnames
[diff
]);
3866 solver_state
*sstate_new
;
3867 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
3869 /* If we supported a verbose solver, we'd set verbosity here */
3871 sstate_new
= solve_game_rec(sstate
);
3873 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
3874 printf("Puzzle is inconsistent\n");
3876 assert(sstate_new
->solver_status
== SOLVER_SOLVED
);
3877 if (s
->grid_type
== 0) {
3878 fputs(game_text_format(sstate_new
->state
), stdout
);
3880 printf("Unable to output non-square grids\n");
3884 free_solver_state(sstate_new
);
3885 free_solver_state(sstate
);