4 * An implementation of the Nikoli game 'Loop the loop'.
5 * (c) Mike Pinna, 2005, 2006
6 * Substantially rewritten to allowing for more general types of grid.
7 * (c) Lambros Lambrou 2008
9 * vim: set shiftwidth=4 :set textwidth=80:
13 * Possible future solver enhancements:
15 * - There's an interesting deductive technique which makes use
16 * of topology rather than just graph theory. Each _face_ in
17 * the grid is either inside or outside the loop; you can tell
18 * that two faces are on the same side of the loop if they're
19 * separated by a LINE_NO (or, more generally, by a path
20 * crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
21 * and on the opposite side of the loop if they're separated by
22 * a LINE_YES (or an odd number of LINE_YESes and no
23 * LINE_UNKNOWNs). Oh, and any face separated from the outside
24 * of the grid by a LINE_YES or a LINE_NO is on the inside or
25 * outside respectively. So if you can track this for all
26 * faces, you figure out the state of the line between a pair
27 * once their relative insideness is known.
28 * + The way I envisage this working is simply to keep an edsf
29 * of all _faces_, which indicates whether they're on
30 * opposite sides of the loop from one another. We also
31 * include a special entry in the edsf for the infinite
33 * + So, the simple way to do this is to just go through the
34 * edges: every time we see an edge in a state other than
35 * LINE_UNKNOWN which separates two faces that aren't in the
36 * same edsf class, we can rectify that by merging the
37 * classes. Then, conversely, an edge in LINE_UNKNOWN state
38 * which separates two faces that _are_ in the same edsf
39 * class can immediately have its state determined.
40 * + But you can go one better, if you're prepared to loop
41 * over all _pairs_ of edges. Suppose we have edges A and B,
42 * which respectively separate faces A1,A2 and B1,B2.
43 * Suppose that A,B are in the same edge-edsf class and that
44 * A1,B1 (wlog) are in the same face-edsf class; then we can
45 * immediately place A2,B2 into the same face-edsf class (as
46 * each other, not as A1 and A2) one way round or the other.
47 * And conversely again, if A1,B1 are in the same face-edsf
48 * class and so are A2,B2, then we can put A,B into the same
50 * * Of course, this deduction requires a quadratic-time
51 * loop over all pairs of edges in the grid, so it should
52 * be reserved until there's nothing easier left to be
55 * - The generalised grid support has made me (SGT) notice a
56 * possible extension to the loop-avoidance code. When you have
57 * a path of connected edges such that no other edges at all
58 * are incident on any vertex in the middle of the path - or,
59 * alternatively, such that any such edges are already known to
60 * be LINE_NO - then you know those edges are either all
61 * LINE_YES or all LINE_NO. Hence you can mentally merge the
62 * entire path into a single long curly edge for the purposes
63 * of loop avoidance, and look directly at whether or not the
64 * extreme endpoints of the path are connected by some other
65 * route. I find this coming up fairly often when I play on the
66 * octagonal grid setting, so it might be worth implementing in
69 * - (Just a speed optimisation.) Consider some todo list queue where every
70 * time we modify something we mark it for consideration by other bits of
71 * the solver, to save iteration over things that have already been done.
86 /* Debugging options */
94 /* ----------------------------------------------------------------------
95 * Struct, enum and function declarations
112 /* Put -1 in a face that doesn't get a clue */
115 /* Array of line states, to store whether each line is
116 * YES, NO or UNKNOWN */
119 unsigned char *line_errors
;
124 /* Used in game_text_format(), so that it knows what type of
125 * grid it's trying to render as ASCII text. */
130 SOLVER_SOLVED
, /* This is the only solution the solver could find */
131 SOLVER_MISTAKE
, /* This is definitely not a solution */
132 SOLVER_AMBIGUOUS
, /* This _might_ be an ambiguous solution */
133 SOLVER_INCOMPLETE
/* This may be a partial solution */
136 /* ------ Solver state ------ */
137 typedef struct solver_state
{
139 enum solver_status solver_status
;
140 /* NB looplen is the number of dots that are joined together at a point, ie a
141 * looplen of 1 means there are no lines to a particular dot */
144 /* Difficulty level of solver. Used by solver functions that want to
145 * vary their behaviour depending on the requested difficulty level. */
151 char *face_yes_count
;
153 char *dot_solved
, *face_solved
;
156 /* Information for Normal level deductions:
157 * For each dline, store a bitmask for whether we know:
158 * (bit 0) at least one is YES
159 * (bit 1) at most one is YES */
162 /* Hard level information */
167 * Difficulty levels. I do some macro ickery here to ensure that my
168 * enum and the various forms of my name list always match up.
171 #define DIFFLIST(A) \
176 #define ENUM(upper,title,lower) DIFF_ ## upper,
177 #define TITLE(upper,title,lower) #title,
178 #define ENCODE(upper,title,lower) #lower
179 #define CONFIG(upper,title,lower) ":" #title
180 enum { DIFFLIST(ENUM
) DIFF_MAX
};
181 static char const *const diffnames
[] = { DIFFLIST(TITLE
) };
182 static char const diffchars
[] = DIFFLIST(ENCODE
);
183 #define DIFFCONFIG DIFFLIST(CONFIG)
186 * Solver routines, sorted roughly in order of computational cost.
187 * The solver will run the faster deductions first, and slower deductions are
188 * only invoked when the faster deductions are unable to make progress.
189 * Each function is associated with a difficulty level, so that the generated
190 * puzzles are solvable by applying only the functions with the chosen
191 * difficulty level or lower.
193 #define SOLVERLIST(A) \
194 A(trivial_deductions, DIFF_EASY) \
195 A(dline_deductions, DIFF_NORMAL) \
196 A(linedsf_deductions, DIFF_HARD) \
197 A(loop_deductions, DIFF_EASY)
198 #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
199 #define SOLVER_FN(fn,diff) &fn,
200 #define SOLVER_DIFF(fn,diff) diff,
201 SOLVERLIST(SOLVER_FN_DECL
)
202 static int (*(solver_fns
[]))(solver_state
*) = { SOLVERLIST(SOLVER_FN
) };
203 static int const solver_diffs
[] = { SOLVERLIST(SOLVER_DIFF
) };
204 const int NUM_SOLVERS
= sizeof(solver_diffs
)/sizeof(*solver_diffs
);
211 /* Grid generation is expensive, so keep a (ref-counted) reference to the
212 * grid for these parameters, and only generate when required. */
216 /* line_drawstate is the same as line_state, but with the extra ERROR
217 * possibility. The drawing code copies line_state to line_drawstate,
218 * except in the case that the line is an error. */
219 enum line_state
{ LINE_YES
, LINE_UNKNOWN
, LINE_NO
};
220 enum line_drawstate
{ DS_LINE_YES
, DS_LINE_UNKNOWN
,
221 DS_LINE_NO
, DS_LINE_ERROR
};
223 #define OPP(line_state) \
227 struct game_drawstate
{
234 char *clue_satisfied
;
237 static char *validate_desc(game_params
*params
, char *desc
);
238 static int dot_order(const game_state
* state
, int i
, char line_type
);
239 static int face_order(const game_state
* state
, int i
, char line_type
);
240 static solver_state
*solve_game_rec(const solver_state
*sstate
);
243 static void check_caches(const solver_state
* sstate
);
245 #define check_caches(s)
248 /* ------- List of grid generators ------- */
249 #define GRIDLIST(A) \
250 A(Squares,grid_new_square,3,3) \
251 A(Triangular,grid_new_triangular,3,3) \
252 A(Honeycomb,grid_new_honeycomb,3,3) \
253 A(Snub-Square,grid_new_snubsquare,3,3) \
254 A(Cairo,grid_new_cairo,3,4) \
255 A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
256 A(Octagonal,grid_new_octagonal,3,3) \
257 A(Kites,grid_new_kites,3,3) \
258 A(Floret,grid_new_floret,1,2) \
259 A(Dodecagonal,grid_new_dodecagonal,2,2) \
260 A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2)
262 #define GRID_NAME(title,fn,amin,omin) #title,
263 #define GRID_CONFIG(title,fn,amin,omin) ":" #title
264 #define GRID_FN(title,fn,amin,omin) &fn,
265 #define GRID_SIZES(title,fn,amin,omin) \
267 "Width and height for this grid type must both be at least " #amin, \
268 "At least one of width and height for this grid type must be at least " #omin,},
269 static char const *const gridnames
[] = { GRIDLIST(GRID_NAME
) };
270 #define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
271 static grid
* (*(grid_fns
[]))(int w
, int h
) = { GRIDLIST(GRID_FN
) };
272 #define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
273 static const struct {
276 } grid_size_limits
[] = { GRIDLIST(GRID_SIZES
) };
278 /* Generates a (dynamically allocated) new grid, according to the
279 * type and size requested in params. Does nothing if the grid is already
280 * generated. The allocated grid is owned by the params object, and will be
281 * freed in free_params(). */
282 static void params_generate_grid(game_params
*params
)
284 if (!params
->game_grid
) {
285 params
->game_grid
= grid_fns
[params
->type
](params
->w
, params
->h
);
289 /* ----------------------------------------------------------------------
293 /* General constants */
294 #define PREFERRED_TILE_SIZE 32
295 #define BORDER(tilesize) ((tilesize) / 2)
296 #define FLASH_TIME 0.5F
298 #define BIT_SET(field, bit) ((field) & (1<<(bit)))
300 #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
301 ((field) |= (1<<(bit)), TRUE))
303 #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
304 ((field) &= ~(1<<(bit)), TRUE) : FALSE)
306 #define CLUE2CHAR(c) \
307 ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
309 /* ----------------------------------------------------------------------
310 * General struct manipulation and other straightforward code
313 static game_state
*dup_game(game_state
*state
)
315 game_state
*ret
= snew(game_state
);
317 ret
->game_grid
= state
->game_grid
;
318 ret
->game_grid
->refcount
++;
320 ret
->solved
= state
->solved
;
321 ret
->cheated
= state
->cheated
;
323 ret
->clues
= snewn(state
->game_grid
->num_faces
, signed char);
324 memcpy(ret
->clues
, state
->clues
, state
->game_grid
->num_faces
);
326 ret
->lines
= snewn(state
->game_grid
->num_edges
, char);
327 memcpy(ret
->lines
, state
->lines
, state
->game_grid
->num_edges
);
329 ret
->line_errors
= snewn(state
->game_grid
->num_edges
, unsigned char);
330 memcpy(ret
->line_errors
, state
->line_errors
, state
->game_grid
->num_edges
);
332 ret
->grid_type
= state
->grid_type
;
336 static void free_game(game_state
*state
)
339 grid_free(state
->game_grid
);
342 sfree(state
->line_errors
);
347 static solver_state
*new_solver_state(game_state
*state
, int diff
) {
349 int num_dots
= state
->game_grid
->num_dots
;
350 int num_faces
= state
->game_grid
->num_faces
;
351 int num_edges
= state
->game_grid
->num_edges
;
352 solver_state
*ret
= snew(solver_state
);
354 ret
->state
= dup_game(state
);
356 ret
->solver_status
= SOLVER_INCOMPLETE
;
359 ret
->dotdsf
= snew_dsf(num_dots
);
360 ret
->looplen
= snewn(num_dots
, int);
362 for (i
= 0; i
< num_dots
; i
++) {
366 ret
->dot_solved
= snewn(num_dots
, char);
367 ret
->face_solved
= snewn(num_faces
, char);
368 memset(ret
->dot_solved
, FALSE
, num_dots
);
369 memset(ret
->face_solved
, FALSE
, num_faces
);
371 ret
->dot_yes_count
= snewn(num_dots
, char);
372 memset(ret
->dot_yes_count
, 0, num_dots
);
373 ret
->dot_no_count
= snewn(num_dots
, char);
374 memset(ret
->dot_no_count
, 0, num_dots
);
375 ret
->face_yes_count
= snewn(num_faces
, char);
376 memset(ret
->face_yes_count
, 0, num_faces
);
377 ret
->face_no_count
= snewn(num_faces
, char);
378 memset(ret
->face_no_count
, 0, num_faces
);
380 if (diff
< DIFF_NORMAL
) {
383 ret
->dlines
= snewn(2*num_edges
, char);
384 memset(ret
->dlines
, 0, 2*num_edges
);
387 if (diff
< DIFF_HARD
) {
390 ret
->linedsf
= snew_dsf(state
->game_grid
->num_edges
);
396 static void free_solver_state(solver_state
*sstate
) {
398 free_game(sstate
->state
);
399 sfree(sstate
->dotdsf
);
400 sfree(sstate
->looplen
);
401 sfree(sstate
->dot_solved
);
402 sfree(sstate
->face_solved
);
403 sfree(sstate
->dot_yes_count
);
404 sfree(sstate
->dot_no_count
);
405 sfree(sstate
->face_yes_count
);
406 sfree(sstate
->face_no_count
);
408 /* OK, because sfree(NULL) is a no-op */
409 sfree(sstate
->dlines
);
410 sfree(sstate
->linedsf
);
416 static solver_state
*dup_solver_state(const solver_state
*sstate
) {
417 game_state
*state
= sstate
->state
;
418 int num_dots
= state
->game_grid
->num_dots
;
419 int num_faces
= state
->game_grid
->num_faces
;
420 int num_edges
= state
->game_grid
->num_edges
;
421 solver_state
*ret
= snew(solver_state
);
423 ret
->state
= state
= dup_game(sstate
->state
);
425 ret
->solver_status
= sstate
->solver_status
;
426 ret
->diff
= sstate
->diff
;
428 ret
->dotdsf
= snewn(num_dots
, int);
429 ret
->looplen
= snewn(num_dots
, int);
430 memcpy(ret
->dotdsf
, sstate
->dotdsf
,
431 num_dots
* sizeof(int));
432 memcpy(ret
->looplen
, sstate
->looplen
,
433 num_dots
* sizeof(int));
435 ret
->dot_solved
= snewn(num_dots
, char);
436 ret
->face_solved
= snewn(num_faces
, char);
437 memcpy(ret
->dot_solved
, sstate
->dot_solved
, num_dots
);
438 memcpy(ret
->face_solved
, sstate
->face_solved
, num_faces
);
440 ret
->dot_yes_count
= snewn(num_dots
, char);
441 memcpy(ret
->dot_yes_count
, sstate
->dot_yes_count
, num_dots
);
442 ret
->dot_no_count
= snewn(num_dots
, char);
443 memcpy(ret
->dot_no_count
, sstate
->dot_no_count
, num_dots
);
445 ret
->face_yes_count
= snewn(num_faces
, char);
446 memcpy(ret
->face_yes_count
, sstate
->face_yes_count
, num_faces
);
447 ret
->face_no_count
= snewn(num_faces
, char);
448 memcpy(ret
->face_no_count
, sstate
->face_no_count
, num_faces
);
450 if (sstate
->dlines
) {
451 ret
->dlines
= snewn(2*num_edges
, char);
452 memcpy(ret
->dlines
, sstate
->dlines
,
458 if (sstate
->linedsf
) {
459 ret
->linedsf
= snewn(num_edges
, int);
460 memcpy(ret
->linedsf
, sstate
->linedsf
,
461 num_edges
* sizeof(int));
469 static game_params
*default_params(void)
471 game_params
*ret
= snew(game_params
);
480 ret
->diff
= DIFF_EASY
;
483 ret
->game_grid
= NULL
;
488 static game_params
*dup_params(game_params
*params
)
490 game_params
*ret
= snew(game_params
);
492 *ret
= *params
; /* structure copy */
493 if (ret
->game_grid
) {
494 ret
->game_grid
->refcount
++;
499 static const game_params presets
[] = {
501 { 7, 7, DIFF_EASY
, 0, NULL
},
502 { 7, 7, DIFF_NORMAL
, 0, NULL
},
503 { 7, 7, DIFF_HARD
, 0, NULL
},
504 { 7, 7, DIFF_HARD
, 1, NULL
},
505 { 7, 7, DIFF_HARD
, 2, NULL
},
506 { 5, 5, DIFF_HARD
, 3, NULL
},
507 { 7, 7, DIFF_HARD
, 4, NULL
},
508 { 5, 4, DIFF_HARD
, 5, NULL
},
509 { 5, 5, DIFF_HARD
, 6, NULL
},
510 { 5, 5, DIFF_HARD
, 7, NULL
},
511 { 3, 3, DIFF_HARD
, 8, NULL
},
512 { 3, 3, DIFF_HARD
, 9, NULL
},
513 { 3, 3, DIFF_HARD
, 10, NULL
},
515 { 7, 7, DIFF_EASY
, 0, NULL
},
516 { 10, 10, DIFF_EASY
, 0, NULL
},
517 { 7, 7, DIFF_NORMAL
, 0, NULL
},
518 { 10, 10, DIFF_NORMAL
, 0, NULL
},
519 { 7, 7, DIFF_HARD
, 0, NULL
},
520 { 10, 10, DIFF_HARD
, 0, NULL
},
521 { 10, 10, DIFF_HARD
, 1, NULL
},
522 { 12, 10, DIFF_HARD
, 2, NULL
},
523 { 7, 7, DIFF_HARD
, 3, NULL
},
524 { 9, 9, DIFF_HARD
, 4, NULL
},
525 { 5, 4, DIFF_HARD
, 5, NULL
},
526 { 7, 7, DIFF_HARD
, 6, NULL
},
527 { 5, 5, DIFF_HARD
, 7, NULL
},
528 { 5, 5, DIFF_HARD
, 8, NULL
},
529 { 5, 4, DIFF_HARD
, 9, NULL
},
530 { 5, 4, DIFF_HARD
, 10, NULL
},
534 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
539 if (i
< 0 || i
>= lenof(presets
))
542 tmppar
= snew(game_params
);
543 *tmppar
= presets
[i
];
545 sprintf(buf
, "%dx%d %s - %s", tmppar
->h
, tmppar
->w
,
546 gridnames
[tmppar
->type
], diffnames
[tmppar
->diff
]);
552 static void free_params(game_params
*params
)
554 if (params
->game_grid
) {
555 grid_free(params
->game_grid
);
560 static void decode_params(game_params
*params
, char const *string
)
562 if (params
->game_grid
) {
563 grid_free(params
->game_grid
);
564 params
->game_grid
= NULL
;
566 params
->h
= params
->w
= atoi(string
);
567 params
->diff
= DIFF_EASY
;
568 while (*string
&& isdigit((unsigned char)*string
)) string
++;
569 if (*string
== 'x') {
571 params
->h
= atoi(string
);
572 while (*string
&& isdigit((unsigned char)*string
)) string
++;
574 if (*string
== 't') {
576 params
->type
= atoi(string
);
577 while (*string
&& isdigit((unsigned char)*string
)) string
++;
579 if (*string
== 'd') {
582 for (i
= 0; i
< DIFF_MAX
; i
++)
583 if (*string
== diffchars
[i
])
585 if (*string
) string
++;
589 static char *encode_params(game_params
*params
, int full
)
592 sprintf(str
, "%dx%dt%d", params
->w
, params
->h
, params
->type
);
594 sprintf(str
+ strlen(str
), "d%c", diffchars
[params
->diff
]);
598 static config_item
*game_configure(game_params
*params
)
603 ret
= snewn(5, config_item
);
605 ret
[0].name
= "Width";
606 ret
[0].type
= C_STRING
;
607 sprintf(buf
, "%d", params
->w
);
608 ret
[0].sval
= dupstr(buf
);
611 ret
[1].name
= "Height";
612 ret
[1].type
= C_STRING
;
613 sprintf(buf
, "%d", params
->h
);
614 ret
[1].sval
= dupstr(buf
);
617 ret
[2].name
= "Grid type";
618 ret
[2].type
= C_CHOICES
;
619 ret
[2].sval
= GRID_CONFIGS
;
620 ret
[2].ival
= params
->type
;
622 ret
[3].name
= "Difficulty";
623 ret
[3].type
= C_CHOICES
;
624 ret
[3].sval
= DIFFCONFIG
;
625 ret
[3].ival
= params
->diff
;
635 static game_params
*custom_params(config_item
*cfg
)
637 game_params
*ret
= snew(game_params
);
639 ret
->w
= atoi(cfg
[0].sval
);
640 ret
->h
= atoi(cfg
[1].sval
);
641 ret
->type
= cfg
[2].ival
;
642 ret
->diff
= cfg
[3].ival
;
644 ret
->game_grid
= NULL
;
648 static char *validate_params(game_params
*params
, int full
)
650 if (params
->type
< 0 || params
->type
>= NUM_GRID_TYPES
)
651 return "Illegal grid type";
652 if (params
->w
< grid_size_limits
[params
->type
].amin
||
653 params
->h
< grid_size_limits
[params
->type
].amin
)
654 return grid_size_limits
[params
->type
].aerr
;
655 if (params
->w
< grid_size_limits
[params
->type
].omin
&&
656 params
->h
< grid_size_limits
[params
->type
].omin
)
657 return grid_size_limits
[params
->type
].oerr
;
660 * This shouldn't be able to happen at all, since decode_params
661 * and custom_params will never generate anything that isn't
664 assert(params
->diff
< DIFF_MAX
);
669 /* Returns a newly allocated string describing the current puzzle */
670 static char *state_to_text(const game_state
*state
)
672 grid
*g
= state
->game_grid
;
674 int num_faces
= g
->num_faces
;
675 char *description
= snewn(num_faces
+ 1, char);
676 char *dp
= description
;
680 for (i
= 0; i
< num_faces
; i
++) {
681 if (state
->clues
[i
] < 0) {
682 if (empty_count
> 25) {
683 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
689 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
692 dp
+= sprintf(dp
, "%c", (int)CLUE2CHAR(state
->clues
[i
]));
697 dp
+= sprintf(dp
, "%c", (int)(empty_count
+ 'a' - 1));
699 retval
= dupstr(description
);
705 /* We require that the params pass the test in validate_params and that the
706 * description fills the entire game area */
707 static char *validate_desc(game_params
*params
, char *desc
)
711 params_generate_grid(params
);
712 g
= params
->game_grid
;
714 for (; *desc
; ++desc
) {
715 if ((*desc
>= '0' && *desc
<= '9') || (*desc
>= 'A' && *desc
<= 'Z')) {
720 count
+= *desc
- 'a' + 1;
723 return "Unknown character in description";
726 if (count
< g
->num_faces
)
727 return "Description too short for board size";
728 if (count
> g
->num_faces
)
729 return "Description too long for board size";
734 /* Sums the lengths of the numbers in range [0,n) */
735 /* See equivalent function in solo.c for justification of this. */
736 static int len_0_to_n(int n
)
738 int len
= 1; /* Counting 0 as a bit of a special case */
741 for (i
= 1; i
< n
; i
*= 10) {
742 len
+= max(n
- i
, 0);
748 static char *encode_solve_move(const game_state
*state
)
753 int num_edges
= state
->game_grid
->num_edges
;
755 /* This is going to return a string representing the moves needed to set
756 * every line in a grid to be the same as the ones in 'state'. The exact
757 * length of this string is predictable. */
759 len
= 1; /* Count the 'S' prefix */
760 /* Numbers in all lines */
761 len
+= len_0_to_n(num_edges
);
762 /* For each line we also have a letter */
765 ret
= snewn(len
+ 1, char);
768 p
+= sprintf(p
, "S");
770 for (i
= 0; i
< num_edges
; i
++) {
771 switch (state
->lines
[i
]) {
773 p
+= sprintf(p
, "%dy", i
);
776 p
+= sprintf(p
, "%dn", i
);
781 /* No point in doing sums like that if they're going to be wrong */
782 assert(strlen(ret
) <= (size_t)len
);
786 static game_ui
*new_ui(game_state
*state
)
791 static void free_ui(game_ui
*ui
)
795 static char *encode_ui(game_ui
*ui
)
800 static void decode_ui(game_ui
*ui
, char *encoding
)
804 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
805 game_state
*newstate
)
809 static void game_compute_size(game_params
*params
, int tilesize
,
813 int grid_width
, grid_height
, rendered_width
, rendered_height
;
815 params_generate_grid(params
);
816 g
= params
->game_grid
;
817 grid_width
= g
->highest_x
- g
->lowest_x
;
818 grid_height
= g
->highest_y
- g
->lowest_y
;
819 /* multiply first to minimise rounding error on integer division */
820 rendered_width
= grid_width
* tilesize
/ g
->tilesize
;
821 rendered_height
= grid_height
* tilesize
/ g
->tilesize
;
822 *x
= rendered_width
+ 2 * BORDER(tilesize
) + 1;
823 *y
= rendered_height
+ 2 * BORDER(tilesize
) + 1;
826 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
827 game_params
*params
, int tilesize
)
829 ds
->tilesize
= tilesize
;
832 static float *game_colours(frontend
*fe
, int *ncolours
)
834 float *ret
= snewn(4 * NCOLOURS
, float);
836 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
838 ret
[COL_FOREGROUND
* 3 + 0] = 0.0F
;
839 ret
[COL_FOREGROUND
* 3 + 1] = 0.0F
;
840 ret
[COL_FOREGROUND
* 3 + 2] = 0.0F
;
843 * We want COL_LINEUNKNOWN to be a yellow which is a bit darker
844 * than the background. (I previously set it to 0.8,0.8,0, but
845 * found that this went badly with the 0.8,0.8,0.8 favoured as a
846 * background by the Java frontend.)
848 ret
[COL_LINEUNKNOWN
* 3 + 0] = ret
[COL_BACKGROUND
* 3 + 0] * 0.9F
;
849 ret
[COL_LINEUNKNOWN
* 3 + 1] = ret
[COL_BACKGROUND
* 3 + 1] * 0.9F
;
850 ret
[COL_LINEUNKNOWN
* 3 + 2] = 0.0F
;
852 ret
[COL_HIGHLIGHT
* 3 + 0] = 1.0F
;
853 ret
[COL_HIGHLIGHT
* 3 + 1] = 1.0F
;
854 ret
[COL_HIGHLIGHT
* 3 + 2] = 1.0F
;
856 ret
[COL_MISTAKE
* 3 + 0] = 1.0F
;
857 ret
[COL_MISTAKE
* 3 + 1] = 0.0F
;
858 ret
[COL_MISTAKE
* 3 + 2] = 0.0F
;
860 ret
[COL_SATISFIED
* 3 + 0] = 0.0F
;
861 ret
[COL_SATISFIED
* 3 + 1] = 0.0F
;
862 ret
[COL_SATISFIED
* 3 + 2] = 0.0F
;
864 /* We want the faint lines to be a bit darker than the background.
865 * Except if the background is pretty dark already; then it ought to be a
866 * bit lighter. Oy vey.
868 ret
[COL_FAINT
* 3 + 0] = ret
[COL_BACKGROUND
* 3 + 0] * 0.9F
;
869 ret
[COL_FAINT
* 3 + 1] = ret
[COL_BACKGROUND
* 3 + 1] * 0.9F
;
870 ret
[COL_FAINT
* 3 + 2] = ret
[COL_BACKGROUND
* 3 + 2] * 0.9F
;
872 *ncolours
= NCOLOURS
;
876 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
878 struct game_drawstate
*ds
= snew(struct game_drawstate
);
879 int num_faces
= state
->game_grid
->num_faces
;
880 int num_edges
= state
->game_grid
->num_edges
;
885 ds
->lines
= snewn(num_edges
, char);
886 ds
->clue_error
= snewn(num_faces
, char);
887 ds
->clue_satisfied
= snewn(num_faces
, char);
888 ds
->textx
= snewn(num_faces
, int);
889 ds
->texty
= snewn(num_faces
, int);
892 memset(ds
->lines
, LINE_UNKNOWN
, num_edges
);
893 memset(ds
->clue_error
, 0, num_faces
);
894 memset(ds
->clue_satisfied
, 0, num_faces
);
895 for (i
= 0; i
< num_faces
; i
++)
896 ds
->textx
[i
] = ds
->texty
[i
] = -1;
901 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
903 sfree(ds
->clue_error
);
904 sfree(ds
->clue_satisfied
);
909 static int game_timing_state(game_state
*state
, game_ui
*ui
)
914 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
915 int dir
, game_ui
*ui
)
920 static int game_can_format_as_text_now(game_params
*params
)
922 if (params
->type
!= 0)
927 static char *game_text_format(game_state
*state
)
933 grid
*g
= state
->game_grid
;
936 assert(state
->grid_type
== 0);
938 /* Work out the basic size unit */
939 f
= g
->faces
; /* first face */
940 assert(f
->order
== 4);
941 /* The dots are ordered clockwise, so the two opposite
942 * corners are guaranteed to span the square */
943 cell_size
= abs(f
->dots
[0]->x
- f
->dots
[2]->x
);
945 w
= (g
->highest_x
- g
->lowest_x
) / cell_size
;
946 h
= (g
->highest_y
- g
->lowest_y
) / cell_size
;
948 /* Create a blank "canvas" to "draw" on */
951 ret
= snewn(W
* H
+ 1, char);
952 for (y
= 0; y
< H
; y
++) {
953 for (x
= 0; x
< W
-1; x
++) {
956 ret
[y
*W
+ W
-1] = '\n';
960 /* Fill in edge info */
961 for (i
= 0; i
< g
->num_edges
; i
++) {
962 grid_edge
*e
= g
->edges
+ i
;
963 /* Cell coordinates, from (0,0) to (w-1,h-1) */
964 int x1
= (e
->dot1
->x
- g
->lowest_x
) / cell_size
;
965 int x2
= (e
->dot2
->x
- g
->lowest_x
) / cell_size
;
966 int y1
= (e
->dot1
->y
- g
->lowest_y
) / cell_size
;
967 int y2
= (e
->dot2
->y
- g
->lowest_y
) / cell_size
;
968 /* Midpoint, in canvas coordinates (canvas coordinates are just twice
969 * cell coordinates) */
972 switch (state
->lines
[i
]) {
974 ret
[y
*W
+ x
] = (y1
== y2
) ?
'-' : '|';
980 break; /* already a space */
982 assert(!"Illegal line state");
987 for (i
= 0; i
< g
->num_faces
; i
++) {
991 assert(f
->order
== 4);
992 /* Cell coordinates, from (0,0) to (w-1,h-1) */
993 x1
= (f
->dots
[0]->x
- g
->lowest_x
) / cell_size
;
994 x2
= (f
->dots
[2]->x
- g
->lowest_x
) / cell_size
;
995 y1
= (f
->dots
[0]->y
- g
->lowest_y
) / cell_size
;
996 y2
= (f
->dots
[2]->y
- g
->lowest_y
) / cell_size
;
997 /* Midpoint, in canvas coordinates */
1000 ret
[y
*W
+ x
] = CLUE2CHAR(state
->clues
[i
]);
1005 /* ----------------------------------------------------------------------
1010 static void check_caches(const solver_state
* sstate
)
1013 const game_state
*state
= sstate
->state
;
1014 const grid
*g
= state
->game_grid
;
1016 for (i
= 0; i
< g
->num_dots
; i
++) {
1017 assert(dot_order(state
, i
, LINE_YES
) == sstate
->dot_yes_count
[i
]);
1018 assert(dot_order(state
, i
, LINE_NO
) == sstate
->dot_no_count
[i
]);
1021 for (i
= 0; i
< g
->num_faces
; i
++) {
1022 assert(face_order(state
, i
, LINE_YES
) == sstate
->face_yes_count
[i
]);
1023 assert(face_order(state
, i
, LINE_NO
) == sstate
->face_no_count
[i
]);
1028 #define check_caches(s) \
1030 fprintf(stderr, "check_caches at line %d\n", __LINE__); \
1034 #endif /* DEBUG_CACHES */
1036 /* ----------------------------------------------------------------------
1037 * Solver utility functions
1040 /* Sets the line (with index i) to the new state 'line_new', and updates
1041 * the cached counts of any affected faces and dots.
1042 * Returns TRUE if this actually changed the line's state. */
1043 static int solver_set_line(solver_state
*sstate
, int i
,
1044 enum line_state line_new
1046 , const char *reason
1050 game_state
*state
= sstate
->state
;
1054 assert(line_new
!= LINE_UNKNOWN
);
1056 check_caches(sstate
);
1058 if (state
->lines
[i
] == line_new
) {
1059 return FALSE
; /* nothing changed */
1061 state
->lines
[i
] = line_new
;
1064 fprintf(stderr
, "solver: set line [%d] to %s (%s)\n",
1065 i
, line_new
== LINE_YES ?
"YES" : "NO",
1069 g
= state
->game_grid
;
1072 /* Update the cache for both dots and both faces affected by this. */
1073 if (line_new
== LINE_YES
) {
1074 sstate
->dot_yes_count
[e
->dot1
- g
->dots
]++;
1075 sstate
->dot_yes_count
[e
->dot2
- g
->dots
]++;
1077 sstate
->face_yes_count
[e
->face1
- g
->faces
]++;
1080 sstate
->face_yes_count
[e
->face2
- g
->faces
]++;
1083 sstate
->dot_no_count
[e
->dot1
- g
->dots
]++;
1084 sstate
->dot_no_count
[e
->dot2
- g
->dots
]++;
1086 sstate
->face_no_count
[e
->face1
- g
->faces
]++;
1089 sstate
->face_no_count
[e
->face2
- g
->faces
]++;
1093 check_caches(sstate
);
1098 #define solver_set_line(a, b, c) \
1099 solver_set_line(a, b, c, __FUNCTION__)
1103 * Merge two dots due to the existence of an edge between them.
1104 * Updates the dsf tracking equivalence classes, and keeps track of
1105 * the length of path each dot is currently a part of.
1106 * Returns TRUE if the dots were already linked, ie if they are part of a
1107 * closed loop, and false otherwise.
1109 static int merge_dots(solver_state
*sstate
, int edge_index
)
1112 grid
*g
= sstate
->state
->game_grid
;
1113 grid_edge
*e
= g
->edges
+ edge_index
;
1115 i
= e
->dot1
- g
->dots
;
1116 j
= e
->dot2
- g
->dots
;
1118 i
= dsf_canonify(sstate
->dotdsf
, i
);
1119 j
= dsf_canonify(sstate
->dotdsf
, j
);
1124 len
= sstate
->looplen
[i
] + sstate
->looplen
[j
];
1125 dsf_merge(sstate
->dotdsf
, i
, j
);
1126 i
= dsf_canonify(sstate
->dotdsf
, i
);
1127 sstate
->looplen
[i
] = len
;
1132 /* Merge two lines because the solver has deduced that they must be either
1133 * identical or opposite. Returns TRUE if this is new information, otherwise
1135 static int merge_lines(solver_state
*sstate
, int i
, int j
, int inverse
1137 , const char *reason
1143 assert(i
< sstate
->state
->game_grid
->num_edges
);
1144 assert(j
< sstate
->state
->game_grid
->num_edges
);
1146 i
= edsf_canonify(sstate
->linedsf
, i
, &inv_tmp
);
1148 j
= edsf_canonify(sstate
->linedsf
, j
, &inv_tmp
);
1151 edsf_merge(sstate
->linedsf
, i
, j
, inverse
);
1155 fprintf(stderr
, "%s [%d] [%d] %s(%s)\n",
1157 inverse ?
"inverse " : "", reason
);
1164 #define merge_lines(a, b, c, d) \
1165 merge_lines(a, b, c, d, __FUNCTION__)
1168 /* Count the number of lines of a particular type currently going into the
1170 static int dot_order(const game_state
* state
, int dot
, char line_type
)
1173 grid
*g
= state
->game_grid
;
1174 grid_dot
*d
= g
->dots
+ dot
;
1177 for (i
= 0; i
< d
->order
; i
++) {
1178 grid_edge
*e
= d
->edges
[i
];
1179 if (state
->lines
[e
- g
->edges
] == line_type
)
1185 /* Count the number of lines of a particular type currently surrounding the
1187 static int face_order(const game_state
* state
, int face
, char line_type
)
1190 grid
*g
= state
->game_grid
;
1191 grid_face
*f
= g
->faces
+ face
;
1194 for (i
= 0; i
< f
->order
; i
++) {
1195 grid_edge
*e
= f
->edges
[i
];
1196 if (state
->lines
[e
- g
->edges
] == line_type
)
1202 /* Set all lines bordering a dot of type old_type to type new_type
1203 * Return value tells caller whether this function actually did anything */
1204 static int dot_setall(solver_state
*sstate
, int dot
,
1205 char old_type
, char new_type
)
1207 int retval
= FALSE
, r
;
1208 game_state
*state
= sstate
->state
;
1213 if (old_type
== new_type
)
1216 g
= state
->game_grid
;
1219 for (i
= 0; i
< d
->order
; i
++) {
1220 int line_index
= d
->edges
[i
] - g
->edges
;
1221 if (state
->lines
[line_index
] == old_type
) {
1222 r
= solver_set_line(sstate
, line_index
, new_type
);
1230 /* Set all lines bordering a face of type old_type to type new_type */
1231 static int face_setall(solver_state
*sstate
, int face
,
1232 char old_type
, char new_type
)
1234 int retval
= FALSE
, r
;
1235 game_state
*state
= sstate
->state
;
1240 if (old_type
== new_type
)
1243 g
= state
->game_grid
;
1244 f
= g
->faces
+ face
;
1246 for (i
= 0; i
< f
->order
; i
++) {
1247 int line_index
= f
->edges
[i
] - g
->edges
;
1248 if (state
->lines
[line_index
] == old_type
) {
1249 r
= solver_set_line(sstate
, line_index
, new_type
);
1257 /* ----------------------------------------------------------------------
1258 * Loop generation and clue removal
1261 /* We're going to store lists of current candidate faces for colouring black
1263 * Each face gets a 'score', which tells us how adding that face right
1264 * now would affect the curliness of the solution loop. We're trying to
1265 * maximise that quantity so will bias our random selection of faces to
1266 * colour those with high scores */
1270 unsigned long random
;
1271 /* No need to store a grid_face* here. The 'face_scores' array will
1272 * be a list of 'face_score' objects, one for each face of the grid, so
1273 * the position (index) within the 'face_scores' array will determine
1274 * which face corresponds to a particular face_score.
1275 * Having a single 'face_scores' array for all faces simplifies memory
1276 * management, and probably improves performance, because we don't have to
1277 * malloc/free each individual face_score, and we don't have to maintain
1278 * a mapping from grid_face* pointers to face_score* pointers.
1282 static int generic_sort_cmpfn(void *v1
, void *v2
, size_t offset
)
1284 struct face_score
*f1
= v1
;
1285 struct face_score
*f2
= v2
;
1288 r
= *(int *)((char *)f2
+ offset
) - *(int *)((char *)f1
+ offset
);
1293 if (f1
->random
< f2
->random
)
1295 else if (f1
->random
> f2
->random
)
1299 * It's _just_ possible that two faces might have been given
1300 * the same random value. In that situation, fall back to
1301 * comparing based on the positions within the face_scores list.
1302 * This introduces a tiny directional bias, but not a significant one.
1307 static int white_sort_cmpfn(void *v1
, void *v2
)
1309 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,white_score
));
1312 static int black_sort_cmpfn(void *v1
, void *v2
)
1314 return generic_sort_cmpfn(v1
, v2
, offsetof(struct face_score
,black_score
));
1317 enum face_colour
{ FACE_WHITE
, FACE_GREY
, FACE_BLACK
};
1319 /* face should be of type grid_face* here. */
1320 #define FACE_COLOUR(face) \
1321 ( (face) == NULL ? FACE_BLACK : \
1322 board[(face) - g->faces] )
1324 /* 'board' is an array of these enums, indicating which faces are
1325 * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
1326 * Returns whether it's legal to colour the given face with this colour. */
1327 static int can_colour_face(grid
*g
, char* board
, int face_index
,
1328 enum face_colour colour
)
1331 grid_face
*test_face
= g
->faces
+ face_index
;
1332 grid_face
*starting_face
, *current_face
;
1333 grid_dot
*starting_dot
;
1335 int current_state
, s
; /* booleans: equal or not-equal to 'colour' */
1336 int found_same_coloured_neighbour
= FALSE
;
1337 assert(board
[face_index
] != colour
);
1339 /* Can only consider a face for colouring if it's adjacent to a face
1340 * with the same colour. */
1341 for (i
= 0; i
< test_face
->order
; i
++) {
1342 grid_edge
*e
= test_face
->edges
[i
];
1343 grid_face
*f
= (e
->face1
== test_face
) ? e
->face2
: e
->face1
;
1344 if (FACE_COLOUR(f
) == colour
) {
1345 found_same_coloured_neighbour
= TRUE
;
1349 if (!found_same_coloured_neighbour
)
1352 /* Need to avoid creating a loop of faces of this colour around some
1353 * differently-coloured faces.
1354 * Also need to avoid meeting a same-coloured face at a corner, with
1355 * other-coloured faces in between. Here's a simple test that (I believe)
1356 * takes care of both these conditions:
1358 * Take the circular path formed by this face's edges, and inflate it
1359 * slightly outwards. Imagine walking around this path and consider
1360 * the faces that you visit in sequence. This will include all faces
1361 * touching the given face, either along an edge or just at a corner.
1362 * Count the number of 'colour'/not-'colour' transitions you encounter, as
1363 * you walk along the complete loop. This will obviously turn out to be
1365 * If 0, we're either in the middle of an "island" of this colour (should
1366 * be impossible as we're not supposed to create black or white loops),
1367 * or we're about to start a new island - also not allowed.
1368 * If 4 or greater, there are too many separate coloured regions touching
1369 * this face, and colouring it would create a loop or a corner-violation.
1370 * The only allowed case is when the count is exactly 2. */
1372 /* i points to a dot around the test face.
1373 * j points to a face around the i^th dot.
1374 * The current face will always be:
1375 * test_face->dots[i]->faces[j]
1376 * We assume dots go clockwise around the test face,
1377 * and faces go clockwise around dots. */
1380 * The end condition is slightly fiddly. In sufficiently strange
1381 * degenerate grids, our test face may be adjacent to the same
1382 * other face multiple times (typically if it's the exterior
1383 * face). Consider this, in particular:
1391 * The bottom left face there is adjacent to the exterior face
1392 * twice, so we can't just terminate our iteration when we reach
1393 * the same _face_ we started at. Furthermore, we can't
1394 * condition on having the same (i,j) pair either, because
1395 * several (i,j) pairs identify the bottom left contiguity with
1396 * the exterior face! We canonicalise the (i,j) pair by taking
1397 * one step around before we set the termination tracking.
1401 current_face
= test_face
->dots
[0]->faces
[0];
1402 if (current_face
== test_face
) {
1404 current_face
= test_face
->dots
[0]->faces
[1];
1407 current_state
= (FACE_COLOUR(current_face
) == colour
);
1408 starting_dot
= NULL
;
1409 starting_face
= NULL
;
1411 /* Advance to next face.
1412 * Need to loop here because it might take several goes to
1416 if (j
== test_face
->dots
[i
]->order
)
1419 if (test_face
->dots
[i
]->faces
[j
] == test_face
) {
1420 /* Advance to next dot round test_face, then
1421 * find current_face around new dot
1422 * and advance to the next face clockwise */
1424 if (i
== test_face
->order
)
1426 for (j
= 0; j
< test_face
->dots
[i
]->order
; j
++) {
1427 if (test_face
->dots
[i
]->faces
[j
] == current_face
)
1430 /* Must actually find current_face around new dot,
1431 * or else something's wrong with the grid. */
1432 assert(j
!= test_face
->dots
[i
]->order
);
1433 /* Found, so advance to next face and try again */
1438 /* (i,j) are now advanced to next face */
1439 current_face
= test_face
->dots
[i
]->faces
[j
];
1440 s
= (FACE_COLOUR(current_face
) == colour
);
1441 if (!starting_dot
) {
1442 starting_dot
= test_face
->dots
[i
];
1443 starting_face
= current_face
;
1446 if (s
!= current_state
) {
1449 if (transitions
> 2)
1452 if (test_face
->dots
[i
] == starting_dot
&&
1453 current_face
== starting_face
)
1458 return (transitions
== 2) ? TRUE
: FALSE
;
1461 /* Count the number of neighbours of 'face', having colour 'colour' */
1462 static int face_num_neighbours(grid
*g
, char *board
, grid_face
*face
,
1463 enum face_colour colour
)
1465 int colour_count
= 0;
1469 for (i
= 0; i
< face
->order
; i
++) {
1471 f
= (e
->face1
== face
) ? e
->face2
: e
->face1
;
1472 if (FACE_COLOUR(f
) == colour
)
1475 return colour_count
;
1478 /* The 'score' of a face reflects its current desirability for selection
1479 * as the next face to colour white or black. We want to encourage moving
1480 * into grey areas and increasing loopiness, so we give scores according to
1481 * how many of the face's neighbours are currently coloured the same as the
1482 * proposed colour. */
1483 static int face_score(grid
*g
, char *board
, grid_face
*face
,
1484 enum face_colour colour
)
1486 /* Simple formula: score = 0 - num. same-coloured neighbours,
1487 * so a higher score means fewer same-coloured neighbours. */
1488 return -face_num_neighbours(g
, board
, face
, colour
);
1491 /* Generate a new complete set of clues for the given game_state.
1492 * The method is to generate a WHITE/BLACK colouring of all the faces,
1493 * such that the WHITE faces will define the inside of the path, and the
1494 * BLACK faces define the outside.
1495 * To do this, we initially colour all faces GREY. The infinite space outside
1496 * the grid is coloured BLACK, and we choose a random face to colour WHITE.
1497 * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
1498 * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
1499 * we avoid creating loops of a single colour, to preserve the topological
1500 * shape of the WHITE and BLACK regions.
1501 * We also try to make the boundary as loopy and twisty as possible, to avoid
1502 * generating paths that are uninteresting.
1503 * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
1504 * face that can be coloured with that colour (without violating the
1505 * topological shape of that region). It's not obvious, but I think this
1506 * algorithm is guaranteed to terminate without leaving any GREY faces behind.
1507 * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
1508 * regions can be grown.
1509 * This is checked using assert()ions, and I haven't seen any failures yet.
1511 * Hand-wavy proof: imagine what can go wrong...
1513 * Could the white faces get completely cut off by the black faces, and still
1514 * leave some grey faces remaining?
1515 * No, because then the black faces would form a loop around both the white
1516 * faces and the grey faces, which is disallowed because we continually
1517 * maintain the correct topological shape of the black region.
1518 * Similarly, the black faces can never get cut off by the white faces. That
1519 * means both the WHITE and BLACK regions always have some room to grow into
1521 * Could it be that we can't colour some GREY face, because there are too many
1522 * WHITE/BLACK transitions as we walk round the face? (see the
1523 * can_colour_face() function for details)
1524 * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
1525 * around the face. The two WHITE faces would be connected by a WHITE path,
1526 * and the BLACK faces would be connected by a BLACK path. These paths would
1527 * have to cross, which is impossible.
1528 * Another thing that could go wrong: perhaps we can't find any GREY face to
1529 * colour WHITE, because it would create a loop-violation or a corner-violation
1530 * with the other WHITE faces?
1531 * This is a little bit tricky to prove impossible. Imagine you have such a
1532 * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
1533 * or corner violation).
1534 * That would cut all the non-white area into two blobs. One of those blobs
1535 * must be free of BLACK faces (because the BLACK stuff is a connected blob).
1536 * So we have a connected GREY area, completely surrounded by WHITE
1537 * (including the GREY face we've tentatively coloured WHITE).
1538 * A well-known result in graph theory says that you can always find a GREY
1539 * face whose removal leaves the remaining GREY area connected. And it says
1540 * there are at least two such faces, so we can always choose the one that
1541 * isn't the "tentative" GREY face. Colouring that face WHITE leaves
1542 * everything nice and connected, including that "tentative" GREY face which
1543 * acts as a gateway to the rest of the non-WHITE grid.
1545 static void add_full_clues(game_state
*state
, random_state
*rs
)
1547 signed char *clues
= state
->clues
;
1549 grid
*g
= state
->game_grid
;
1551 int num_faces
= g
->num_faces
;
1552 struct face_score
*face_scores
; /* Array of face_score objects */
1553 struct face_score
*fs
; /* Points somewhere in the above list */
1554 struct grid_face
*cur_face
;
1555 tree234
*lightable_faces_sorted
;
1556 tree234
*darkable_faces_sorted
;
1560 board
= snewn(num_faces
, char);
1563 memset(board
, FACE_GREY
, num_faces
);
1565 /* Create and initialise the list of face_scores */
1566 face_scores
= snewn(num_faces
, struct face_score
);
1567 for (i
= 0; i
< num_faces
; i
++) {
1568 face_scores
[i
].random
= random_bits(rs
, 31);
1569 face_scores
[i
].black_score
= face_scores
[i
].white_score
= 0;
1572 /* Colour a random, finite face white. The infinite face is implicitly
1573 * coloured black. Together, they will seed the random growth process
1574 * for the black and white areas. */
1575 i
= random_upto(rs
, num_faces
);
1576 board
[i
] = FACE_WHITE
;
1578 /* We need a way of favouring faces that will increase our loopiness.
1579 * We do this by maintaining a list of all candidate faces sorted by
1580 * their score and choose randomly from that with appropriate skew.
1581 * In order to avoid consistently biasing towards particular faces, we
1582 * need the sort order _within_ each group of scores to be completely
1583 * random. But it would be abusing the hospitality of the tree234 data
1584 * structure if our comparison function were nondeterministic :-). So with
1585 * each face we associate a random number that does not change during a
1586 * particular run of the generator, and use that as a secondary sort key.
1587 * Yes, this means we will be biased towards particular random faces in
1588 * any one run but that doesn't actually matter. */
1590 lightable_faces_sorted
= newtree234(white_sort_cmpfn
);
1591 darkable_faces_sorted
= newtree234(black_sort_cmpfn
);
1593 /* Initialise the lists of lightable and darkable faces. This is
1594 * slightly different from the code inside the while-loop, because we need
1595 * to check every face of the board (the grid structure does not keep a
1596 * list of the infinite face's neighbours). */
1597 for (i
= 0; i
< num_faces
; i
++) {
1598 grid_face
*f
= g
->faces
+ i
;
1599 struct face_score
*fs
= face_scores
+ i
;
1600 if (board
[i
] != FACE_GREY
) continue;
1601 /* We need the full colourability check here, it's not enough simply
1602 * to check neighbourhood. On some grids, a neighbour of the infinite
1603 * face is not necessarily darkable. */
1604 if (can_colour_face(g
, board
, i
, FACE_BLACK
)) {
1605 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1606 add234(darkable_faces_sorted
, fs
);
1608 if (can_colour_face(g
, board
, i
, FACE_WHITE
)) {
1609 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1610 add234(lightable_faces_sorted
, fs
);
1614 /* Colour faces one at a time until no more faces are colourable. */
1617 enum face_colour colour
;
1618 struct face_score
*fs_white
, *fs_black
;
1619 int c_lightable
= count234(lightable_faces_sorted
);
1620 int c_darkable
= count234(darkable_faces_sorted
);
1621 if (c_lightable
== 0 && c_darkable
== 0) {
1622 /* No more faces we can use at all. */
1625 assert(c_lightable
!= 0 && c_darkable
!= 0);
1627 fs_white
= (struct face_score
*)index234(lightable_faces_sorted
, 0);
1628 fs_black
= (struct face_score
*)index234(darkable_faces_sorted
, 0);
1630 /* Choose a colour, and colour the best available face
1631 * with that colour. */
1632 colour
= random_upto(rs
, 2) ? FACE_WHITE
: FACE_BLACK
;
1634 if (colour
== FACE_WHITE
)
1639 i
= fs
- face_scores
;
1640 assert(board
[i
] == FACE_GREY
);
1643 /* Remove this newly-coloured face from the lists. These lists should
1644 * only contain grey faces. */
1645 del234(lightable_faces_sorted
, fs
);
1646 del234(darkable_faces_sorted
, fs
);
1648 /* Remember which face we've just coloured */
1649 cur_face
= g
->faces
+ i
;
1651 /* The face we've just coloured potentially affects the colourability
1652 * and the scores of any neighbouring faces (touching at a corner or
1653 * edge). So the search needs to be conducted around all faces
1654 * touching the one we've just lit. Iterate over its corners, then
1655 * over each corner's faces. For each such face, we remove it from
1656 * the lists, recalculate any scores, then add it back to the lists
1657 * (depending on whether it is lightable, darkable or both). */
1658 for (i
= 0; i
< cur_face
->order
; i
++) {
1659 grid_dot
*d
= cur_face
->dots
[i
];
1660 for (j
= 0; j
< d
->order
; j
++) {
1661 grid_face
*f
= d
->faces
[j
];
1662 int fi
; /* face index of f */
1669 /* If the face is already coloured, it won't be on our
1670 * lightable/darkable lists anyway, so we can skip it without
1671 * bothering with the removal step. */
1672 if (FACE_COLOUR(f
) != FACE_GREY
) continue;
1674 /* Find the face index and face_score* corresponding to f */
1676 fs
= face_scores
+ fi
;
1678 /* Remove from lightable list if it's in there. We do this,
1679 * even if it is still lightable, because the score might
1680 * be different, and we need to remove-then-add to maintain
1681 * correct sort order. */
1682 del234(lightable_faces_sorted
, fs
);
1683 if (can_colour_face(g
, board
, fi
, FACE_WHITE
)) {
1684 fs
->white_score
= face_score(g
, board
, f
, FACE_WHITE
);
1685 add234(lightable_faces_sorted
, fs
);
1687 /* Do the same for darkable list. */
1688 del234(darkable_faces_sorted
, fs
);
1689 if (can_colour_face(g
, board
, fi
, FACE_BLACK
)) {
1690 fs
->black_score
= face_score(g
, board
, f
, FACE_BLACK
);
1691 add234(darkable_faces_sorted
, fs
);
1698 freetree234(lightable_faces_sorted
);
1699 freetree234(darkable_faces_sorted
);
1702 /* The next step requires a shuffled list of all faces */
1703 face_list
= snewn(num_faces
, int);
1704 for (i
= 0; i
< num_faces
; ++i
) {
1707 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1709 /* The above loop-generation algorithm can often leave large clumps
1710 * of faces of one colour. In extreme cases, the resulting path can be
1711 * degenerate and not very satisfying to solve.
1712 * This next step alleviates this problem:
1713 * Go through the shuffled list, and flip the colour of any face we can
1714 * legally flip, and which is adjacent to only one face of the opposite
1715 * colour - this tends to grow 'tendrils' into any clumps.
1716 * Repeat until we can find no more faces to flip. This will
1717 * eventually terminate, because each flip increases the loop's
1718 * perimeter, which cannot increase for ever.
1719 * The resulting path will have maximal loopiness (in the sense that it
1720 * cannot be improved "locally". Unfortunately, this allows a player to
1721 * make some illicit deductions. To combat this (and make the path more
1722 * interesting), we do one final pass making random flips. */
1724 /* Set to TRUE for final pass */
1725 do_random_pass
= FALSE
;
1728 /* Remember whether a flip occurred during this pass */
1729 int flipped
= FALSE
;
1731 for (i
= 0; i
< num_faces
; ++i
) {
1732 int j
= face_list
[i
];
1733 enum face_colour opp
=
1734 (board
[j
] == FACE_WHITE
) ? FACE_BLACK
: FACE_WHITE
;
1735 if (can_colour_face(g
, board
, j
, opp
)) {
1736 grid_face
*face
= g
->faces
+j
;
1737 if (do_random_pass
) {
1738 /* final random pass */
1739 if (!random_upto(rs
, 10))
1742 /* normal pass - flip when neighbour count is 1 */
1743 if (face_num_neighbours(g
, board
, face
, opp
) == 1) {
1751 if (do_random_pass
) break;
1752 if (!flipped
) do_random_pass
= TRUE
;
1757 /* Fill out all the clues by initialising to 0, then iterating over
1758 * all edges and incrementing each clue as we find edges that border
1759 * between BLACK/WHITE faces. While we're at it, we verify that the
1760 * algorithm does work, and there aren't any GREY faces still there. */
1761 memset(clues
, 0, num_faces
);
1762 for (i
= 0; i
< g
->num_edges
; i
++) {
1763 grid_edge
*e
= g
->edges
+ i
;
1764 grid_face
*f1
= e
->face1
;
1765 grid_face
*f2
= e
->face2
;
1766 enum face_colour c1
= FACE_COLOUR(f1
);
1767 enum face_colour c2
= FACE_COLOUR(f2
);
1768 assert(c1
!= FACE_GREY
);
1769 assert(c2
!= FACE_GREY
);
1771 if (f1
) clues
[f1
- g
->faces
]++;
1772 if (f2
) clues
[f2
- g
->faces
]++;
1780 static int game_has_unique_soln(const game_state
*state
, int diff
)
1783 solver_state
*sstate_new
;
1784 solver_state
*sstate
= new_solver_state((game_state
*)state
, diff
);
1786 sstate_new
= solve_game_rec(sstate
);
1788 assert(sstate_new
->solver_status
!= SOLVER_MISTAKE
);
1789 ret
= (sstate_new
->solver_status
== SOLVER_SOLVED
);
1791 free_solver_state(sstate_new
);
1792 free_solver_state(sstate
);
1798 /* Remove clues one at a time at random. */
1799 static game_state
*remove_clues(game_state
*state
, random_state
*rs
,
1803 int num_faces
= state
->game_grid
->num_faces
;
1804 game_state
*ret
= dup_game(state
), *saved_ret
;
1807 /* We need to remove some clues. We'll do this by forming a list of all
1808 * available clues, shuffling it, then going along one at a
1809 * time clearing each clue in turn for which doing so doesn't render the
1810 * board unsolvable. */
1811 face_list
= snewn(num_faces
, int);
1812 for (n
= 0; n
< num_faces
; ++n
) {
1816 shuffle(face_list
, num_faces
, sizeof(int), rs
);
1818 for (n
= 0; n
< num_faces
; ++n
) {
1819 saved_ret
= dup_game(ret
);
1820 ret
->clues
[face_list
[n
]] = -1;
1822 if (game_has_unique_soln(ret
, diff
)) {
1823 free_game(saved_ret
);
1835 static char *new_game_desc(game_params
*params
, random_state
*rs
,
1836 char **aux
, int interactive
)
1838 /* solution and description both use run-length encoding in obvious ways */
1841 game_state
*state
= snew(game_state
);
1842 game_state
*state_new
;
1843 params_generate_grid(params
);
1844 state
->game_grid
= g
= params
->game_grid
;
1846 state
->clues
= snewn(g
->num_faces
, signed char);
1847 state
->lines
= snewn(g
->num_edges
, char);
1848 state
->line_errors
= snewn(g
->num_edges
, unsigned char);
1850 state
->grid_type
= params
->type
;
1854 memset(state
->lines
, LINE_UNKNOWN
, g
->num_edges
);
1855 memset(state
->line_errors
, 0, g
->num_edges
);
1857 state
->solved
= state
->cheated
= FALSE
;
1859 /* Get a new random solvable board with all its clues filled in. Yes, this
1860 * can loop for ever if the params are suitably unfavourable, but
1861 * preventing games smaller than 4x4 seems to stop this happening */
1863 add_full_clues(state
, rs
);
1864 } while (!game_has_unique_soln(state
, params
->diff
));
1866 state_new
= remove_clues(state
, rs
, params
->diff
);
1871 if (params
->diff
> 0 && game_has_unique_soln(state
, params
->diff
-1)) {
1873 fprintf(stderr
, "Rejecting board, it is too easy\n");
1875 goto newboard_please
;
1878 retval
= state_to_text(state
);
1882 assert(!validate_desc(params
, retval
));
1887 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1890 game_state
*state
= snew(game_state
);
1891 int empties_to_make
= 0;
1893 const char *dp
= desc
;
1895 int num_faces
, num_edges
;
1897 params_generate_grid(params
);
1898 state
->game_grid
= g
= params
->game_grid
;
1900 num_faces
= g
->num_faces
;
1901 num_edges
= g
->num_edges
;
1903 state
->clues
= snewn(num_faces
, signed char);
1904 state
->lines
= snewn(num_edges
, char);
1905 state
->line_errors
= snewn(num_edges
, unsigned char);
1907 state
->solved
= state
->cheated
= FALSE
;
1909 state
->grid_type
= params
->type
;
1911 for (i
= 0; i
< num_faces
; i
++) {
1912 if (empties_to_make
) {
1914 state
->clues
[i
] = -1;
1920 n2
= *dp
- 'A' + 10;
1921 if (n
>= 0 && n
< 10) {
1922 state
->clues
[i
] = n
;
1923 } else if (n2
>= 10 && n2
< 36) {
1924 state
->clues
[i
] = n2
;
1928 state
->clues
[i
] = -1;
1929 empties_to_make
= n
- 1;
1934 memset(state
->lines
, LINE_UNKNOWN
, num_edges
);
1935 memset(state
->line_errors
, 0, num_edges
);
1939 /* Calculates the line_errors data, and checks if the current state is a
1941 static int check_completion(game_state
*state
)
1943 grid
*g
= state
->game_grid
;
1945 int num_faces
= g
->num_faces
;
1947 int infinite_area
, finite_area
;
1948 int loops_found
= 0;
1949 int found_edge_not_in_loop
= FALSE
;
1951 memset(state
->line_errors
, 0, g
->num_edges
);
1953 /* LL implementation of SGT's idea:
1954 * A loop will partition the grid into an inside and an outside.
1955 * If there is more than one loop, the grid will be partitioned into
1956 * even more distinct regions. We can therefore track equivalence of
1957 * faces, by saying that two faces are equivalent when there is a non-YES
1958 * edge between them.
1959 * We could keep track of the number of connected components, by counting
1960 * the number of dsf-merges that aren't no-ops.
1961 * But we're only interested in 3 separate cases:
1962 * no loops, one loop, more than one loop.
1964 * No loops: all faces are equivalent to the infinite face.
1965 * One loop: only two equivalence classes - finite and infinite.
1966 * >= 2 loops: there are 2 distinct finite regions.
1968 * So we simply make two passes through all the edges.
1969 * In the first pass, we dsf-merge the two faces bordering each non-YES
1971 * In the second pass, we look for YES-edges bordering:
1972 * a) two non-equivalent faces.
1973 * b) two non-equivalent faces, and one of them is part of a different
1974 * finite area from the first finite area we've seen.
1976 * An occurrence of a) means there is at least one loop.
1977 * An occurrence of b) means there is more than one loop.
1978 * Edges satisfying a) are marked as errors.
1980 * While we're at it, we set a flag if we find a YES edge that is not
1982 * This information will help decide, if there's a single loop, whether it
1983 * is a candidate for being a solution (that is, all YES edges are part of
1986 * If there is a candidate loop, we then go through all clues and check
1987 * they are all satisfied. If so, we have found a solution and we can
1988 * unmark all line_errors.
1991 /* Infinite face is at the end - its index is num_faces.
1992 * This macro is just to make this obvious! */
1993 #define INF_FACE num_faces
1994 dsf
= snewn(num_faces
+ 1, int);
1995 dsf_init(dsf
, num_faces
+ 1);
1998 for (i
= 0; i
< g
->num_edges
; i
++) {
1999 grid_edge
*e
= g
->edges
+ i
;
2000 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
2001 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
2002 if (state
->lines
[i
] != LINE_YES
)
2003 dsf_merge(dsf
, f1
, f2
);
2007 infinite_area
= dsf_canonify(dsf
, INF_FACE
);
2009 for (i
= 0; i
< g
->num_edges
; i
++) {
2010 grid_edge
*e
= g
->edges
+ i
;
2011 int f1
= e
->face1 ? e
->face1
- g
->faces
: INF_FACE
;
2012 int can1
= dsf_canonify(dsf
, f1
);
2013 int f2
= e
->face2 ? e
->face2
- g
->faces
: INF_FACE
;
2014 int can2
= dsf_canonify(dsf
, f2
);
2015 if (state
->lines
[i
] != LINE_YES
) continue;
2018 /* Faces are equivalent, so this edge not part of a loop */
2019 found_edge_not_in_loop
= TRUE
;
2022 state
->line_errors
[i
] = TRUE
;
2023 if (loops_found
== 0) loops_found
= 1;
2025 /* Don't bother with further checks if we've already found 2 loops */
2026 if (loops_found
== 2) continue;
2028 if (finite_area
== -1) {
2029 /* Found our first finite area */
2030 if (can1
!= infinite_area
)
2036 /* Have we found a second area? */
2037 if (finite_area
!= -1) {
2038 if (can1
!= infinite_area
&& can1
!= finite_area
) {
2042 if (can2
!= infinite_area
&& can2
!= finite_area
) {
2049 printf("loops_found = %d\n", loops_found);
2050 printf("found_edge_not_in_loop = %s\n",
2051 found_edge_not_in_loop ? "TRUE" : "FALSE");
2054 sfree(dsf
); /* No longer need the dsf */
2056 /* Have we found a candidate loop? */
2057 if (loops_found
== 1 && !found_edge_not_in_loop
) {
2058 /* Yes, so check all clues are satisfied */
2059 int found_clue_violation
= FALSE
;
2060 for (i
= 0; i
< num_faces
; i
++) {
2061 int c
= state
->clues
[i
];
2063 if (face_order(state
, i
, LINE_YES
) != c
) {
2064 found_clue_violation
= TRUE
;
2070 if (!found_clue_violation
) {
2071 /* The loop is good */
2072 memset(state
->line_errors
, 0, g
->num_edges
);
2073 return TRUE
; /* No need to bother checking for dot violations */
2077 /* Check for dot violations */
2078 for (i
= 0; i
< g
->num_dots
; i
++) {
2079 int yes
= dot_order(state
, i
, LINE_YES
);
2080 int unknown
= dot_order(state
, i
, LINE_UNKNOWN
);
2081 if ((yes
== 1 && unknown
== 0) || (yes
>= 3)) {
2082 /* violation, so mark all YES edges as errors */
2083 grid_dot
*d
= g
->dots
+ i
;
2085 for (j
= 0; j
< d
->order
; j
++) {
2086 int e
= d
->edges
[j
] - g
->edges
;
2087 if (state
->lines
[e
] == LINE_YES
)
2088 state
->line_errors
[e
] = TRUE
;
2095 /* ----------------------------------------------------------------------
2098 * Our solver modes operate as follows. Each mode also uses the modes above it.
2101 * Just implement the rules of the game.
2103 * Normal and Tricky Modes
2104 * For each (adjacent) pair of lines through each dot we store a bit for
2105 * whether at least one of them is on and whether at most one is on. (If we
2106 * know both or neither is on that's already stored more directly.)
2109 * Use edsf data structure to make equivalence classes of lines that are
2110 * known identical to or opposite to one another.
2115 * For general grids, we consider "dlines" to be pairs of lines joined
2116 * at a dot. The lines must be adjacent around the dot, so we can think of
2117 * a dline as being a dot+face combination. Or, a dot+edge combination where
2118 * the second edge is taken to be the next clockwise edge from the dot.
2119 * Original loopy code didn't have this extra restriction of the lines being
2120 * adjacent. From my tests with square grids, this extra restriction seems to
2121 * take little, if anything, away from the quality of the puzzles.
2122 * A dline can be uniquely identified by an edge/dot combination, given that
2123 * a dline-pair always goes clockwise around its common dot. The edge/dot
2124 * combination can be represented by an edge/bool combination - if bool is
2125 * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
2126 * exactly twice the number of edges in the grid - although the dlines
2127 * spanning the infinite face are not all that useful to the solver.
2128 * Note that, by convention, a dline goes clockwise around its common dot,
2129 * which means the dline goes anti-clockwise around its common face.
2132 /* Helper functions for obtaining an index into an array of dlines, given
2133 * various information. We assume the grid layout conventions about how
2134 * the various lists are interleaved - see grid_make_consistent() for
2137 /* i points to the first edge of the dline pair, reading clockwise around
2139 static int dline_index_from_dot(grid
*g
, grid_dot
*d
, int i
)
2141 grid_edge
*e
= d
->edges
[i
];
2146 if (i2
== d
->order
) i2
= 0;
2149 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2151 printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
2152 (int)(d
- g
->dots
), i
, (int)(e
- g
->edges
),
2153 (int)(e2
- g
->edges
), ret
);
2157 /* i points to the second edge of the dline pair, reading clockwise around
2158 * the face. That is, the edges of the dline, starting at edge{i}, read
2159 * anti-clockwise around the face. By layout conventions, the common dot
2160 * of the dline will be f->dots[i] */
2161 static int dline_index_from_face(grid
*g
, grid_face
*f
, int i
)
2163 grid_edge
*e
= f
->edges
[i
];
2164 grid_dot
*d
= f
->dots
[i
];
2169 if (i2
< 0) i2
+= f
->order
;
2172 ret
= 2 * (e
- g
->edges
) + ((e
->dot1
== d
) ?
1 : 0);
2174 printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
2175 (int)(f
- g
->faces
), i
, (int)(e
- g
->edges
),
2176 (int)(e2
- g
->edges
), ret
);
2180 static int is_atleastone(const char *dline_array
, int index
)
2182 return BIT_SET(dline_array
[index
], 0);
2184 static int set_atleastone(char *dline_array
, int index
)
2186 return SET_BIT(dline_array
[index
], 0);
2188 static int is_atmostone(const char *dline_array
, int index
)
2190 return BIT_SET(dline_array
[index
], 1);
2192 static int set_atmostone(char *dline_array
, int index
)
2194 return SET_BIT(dline_array
[index
], 1);
2197 static void array_setall(char *array
, char from
, char to
, int len
)
2199 char *p
= array
, *p_old
= p
;
2200 int len_remaining
= len
;
2202 while ((p
= memchr(p
, from
, len_remaining
))) {
2204 len_remaining
-= p
- p_old
;
2209 /* Helper, called when doing dline dot deductions, in the case where we
2210 * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
2211 * them (because of dline atmostone/atleastone).
2212 * On entry, edge points to the first of these two UNKNOWNs. This function
2213 * will find the opposite UNKNOWNS (if they are adjacent to one another)
2214 * and set their corresponding dline to atleastone. (Setting atmostone
2215 * already happens in earlier dline deductions) */
2216 static int dline_set_opp_atleastone(solver_state
*sstate
,
2217 grid_dot
*d
, int edge
)
2219 game_state
*state
= sstate
->state
;
2220 grid
*g
= state
->game_grid
;
2223 for (opp
= 0; opp
< N
; opp
++) {
2224 int opp_dline_index
;
2225 if (opp
== edge
|| opp
== edge
+1 || opp
== edge
-1)
2227 if (opp
== 0 && edge
== N
-1)
2229 if (opp
== N
-1 && edge
== 0)
2232 if (opp2
== N
) opp2
= 0;
2233 /* Check if opp, opp2 point to LINE_UNKNOWNs */
2234 if (state
->lines
[d
->edges
[opp
] - g
->edges
] != LINE_UNKNOWN
)
2236 if (state
->lines
[d
->edges
[opp2
] - g
->edges
] != LINE_UNKNOWN
)
2238 /* Found opposite UNKNOWNS and they're next to each other */
2239 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2240 return set_atleastone(sstate
->dlines
, opp_dline_index
);
2246 /* Set pairs of lines around this face which are known to be identical, to
2247 * the given line_state */
2248 static int face_setall_identical(solver_state
*sstate
, int face_index
,
2249 enum line_state line_new
)
2251 /* can[dir] contains the canonical line associated with the line in
2252 * direction dir from the square in question. Similarly inv[dir] is
2253 * whether or not the line in question is inverse to its canonical
2256 game_state
*state
= sstate
->state
;
2257 grid
*g
= state
->game_grid
;
2258 grid_face
*f
= g
->faces
+ face_index
;
2261 int can1
, can2
, inv1
, inv2
;
2263 for (i
= 0; i
< N
; i
++) {
2264 int line1_index
= f
->edges
[i
] - g
->edges
;
2265 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2267 for (j
= i
+ 1; j
< N
; j
++) {
2268 int line2_index
= f
->edges
[j
] - g
->edges
;
2269 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2272 /* Found two UNKNOWNS */
2273 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2274 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2275 if (can1
== can2
&& inv1
== inv2
) {
2276 solver_set_line(sstate
, line1_index
, line_new
);
2277 solver_set_line(sstate
, line2_index
, line_new
);
2284 /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
2285 * return the edge indices into e. */
2286 static void find_unknowns(game_state
*state
,
2287 grid_edge
**edge_list
, /* Edge list to search (from a face or a dot) */
2288 int expected_count
, /* Number of UNKNOWNs (comes from solver's cache) */
2289 int *e
/* Returned edge indices */)
2292 grid
*g
= state
->game_grid
;
2293 while (c
< expected_count
) {
2294 int line_index
= *edge_list
- g
->edges
;
2295 if (state
->lines
[line_index
] == LINE_UNKNOWN
) {
2303 /* If we have a list of edges, and we know whether the number of YESs should
2304 * be odd or even, and there are only a few UNKNOWNs, we can do some simple
2305 * linedsf deductions. This can be used for both face and dot deductions.
2306 * Returns the difficulty level of the next solver that should be used,
2307 * or DIFF_MAX if no progress was made. */
2308 static int parity_deductions(solver_state
*sstate
,
2309 grid_edge
**edge_list
, /* Edge list (from a face or a dot) */
2310 int total_parity
, /* Expected number of YESs modulo 2 (either 0 or 1) */
2313 game_state
*state
= sstate
->state
;
2314 int diff
= DIFF_MAX
;
2315 int *linedsf
= sstate
->linedsf
;
2317 if (unknown_count
== 2) {
2318 /* Lines are known alike/opposite, depending on inv. */
2320 find_unknowns(state
, edge_list
, 2, e
);
2321 if (merge_lines(sstate
, e
[0], e
[1], total_parity
))
2322 diff
= min(diff
, DIFF_HARD
);
2323 } else if (unknown_count
== 3) {
2325 int can
[3]; /* canonical edges */
2326 int inv
[3]; /* whether can[x] is inverse to e[x] */
2327 find_unknowns(state
, edge_list
, 3, e
);
2328 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2329 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2330 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2331 if (can
[0] == can
[1]) {
2332 if (solver_set_line(sstate
, e
[2], (total_parity
^inv
[0]^inv
[1]) ?
2333 LINE_YES
: LINE_NO
))
2334 diff
= min(diff
, DIFF_EASY
);
2336 if (can
[0] == can
[2]) {
2337 if (solver_set_line(sstate
, e
[1], (total_parity
^inv
[0]^inv
[2]) ?
2338 LINE_YES
: LINE_NO
))
2339 diff
= min(diff
, DIFF_EASY
);
2341 if (can
[1] == can
[2]) {
2342 if (solver_set_line(sstate
, e
[0], (total_parity
^inv
[1]^inv
[2]) ?
2343 LINE_YES
: LINE_NO
))
2344 diff
= min(diff
, DIFF_EASY
);
2346 } else if (unknown_count
== 4) {
2348 int can
[4]; /* canonical edges */
2349 int inv
[4]; /* whether can[x] is inverse to e[x] */
2350 find_unknowns(state
, edge_list
, 4, e
);
2351 can
[0] = edsf_canonify(linedsf
, e
[0], inv
);
2352 can
[1] = edsf_canonify(linedsf
, e
[1], inv
+1);
2353 can
[2] = edsf_canonify(linedsf
, e
[2], inv
+2);
2354 can
[3] = edsf_canonify(linedsf
, e
[3], inv
+3);
2355 if (can
[0] == can
[1]) {
2356 if (merge_lines(sstate
, e
[2], e
[3], total_parity
^inv
[0]^inv
[1]))
2357 diff
= min(diff
, DIFF_HARD
);
2358 } else if (can
[0] == can
[2]) {
2359 if (merge_lines(sstate
, e
[1], e
[3], total_parity
^inv
[0]^inv
[2]))
2360 diff
= min(diff
, DIFF_HARD
);
2361 } else if (can
[0] == can
[3]) {
2362 if (merge_lines(sstate
, e
[1], e
[2], total_parity
^inv
[0]^inv
[3]))
2363 diff
= min(diff
, DIFF_HARD
);
2364 } else if (can
[1] == can
[2]) {
2365 if (merge_lines(sstate
, e
[0], e
[3], total_parity
^inv
[1]^inv
[2]))
2366 diff
= min(diff
, DIFF_HARD
);
2367 } else if (can
[1] == can
[3]) {
2368 if (merge_lines(sstate
, e
[0], e
[2], total_parity
^inv
[1]^inv
[3]))
2369 diff
= min(diff
, DIFF_HARD
);
2370 } else if (can
[2] == can
[3]) {
2371 if (merge_lines(sstate
, e
[0], e
[1], total_parity
^inv
[2]^inv
[3]))
2372 diff
= min(diff
, DIFF_HARD
);
2380 * These are the main solver functions.
2382 * Their return values are diff values corresponding to the lowest mode solver
2383 * that would notice the work that they have done. For example if the normal
2384 * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
2385 * easy mode solver might be able to make progress using that. It doesn't make
2386 * sense for one of them to return a diff value higher than that of the
2389 * Each function returns the lowest value it can, as early as possible, in
2390 * order to try and pass as much work as possible back to the lower level
2391 * solvers which progress more quickly.
2394 /* PROPOSED NEW DESIGN:
2395 * We have a work queue consisting of 'events' notifying us that something has
2396 * happened that a particular solver mode might be interested in. For example
2397 * the hard mode solver might do something that helps the normal mode solver at
2398 * dot [x,y] in which case it will enqueue an event recording this fact. Then
2399 * we pull events off the work queue, and hand each in turn to the solver that
2400 * is interested in them. If a solver reports that it failed we pass the same
2401 * event on to progressively more advanced solvers and the loop detector. Once
2402 * we've exhausted an event, or it has helped us progress, we drop it and
2403 * continue to the next one. The events are sorted first in order of solver
2404 * complexity (easy first) then order of insertion (oldest first).
2405 * Once we run out of events we loop over each permitted solver in turn
2406 * (easiest first) until either a deduction is made (and an event therefore
2407 * emerges) or no further deductions can be made (in which case we've failed).
2410 * * How do we 'loop over' a solver when both dots and squares are concerned.
2411 * Answer: first all squares then all dots.
2414 static int trivial_deductions(solver_state
*sstate
)
2416 int i
, current_yes
, current_no
;
2417 game_state
*state
= sstate
->state
;
2418 grid
*g
= state
->game_grid
;
2419 int diff
= DIFF_MAX
;
2421 /* Per-face deductions */
2422 for (i
= 0; i
< g
->num_faces
; i
++) {
2423 grid_face
*f
= g
->faces
+ i
;
2425 if (sstate
->face_solved
[i
])
2428 current_yes
= sstate
->face_yes_count
[i
];
2429 current_no
= sstate
->face_no_count
[i
];
2431 if (current_yes
+ current_no
== f
->order
) {
2432 sstate
->face_solved
[i
] = TRUE
;
2436 if (state
->clues
[i
] < 0)
2440 * This code checks whether the numeric clue on a face is so
2441 * large as to permit all its remaining LINE_UNKNOWNs to be
2442 * filled in as LINE_YES, or alternatively so small as to
2443 * permit them all to be filled in as LINE_NO.
2446 if (state
->clues
[i
] < current_yes
) {
2447 sstate
->solver_status
= SOLVER_MISTAKE
;
2450 if (state
->clues
[i
] == current_yes
) {
2451 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
))
2452 diff
= min(diff
, DIFF_EASY
);
2453 sstate
->face_solved
[i
] = TRUE
;
2457 if (f
->order
- state
->clues
[i
] < current_no
) {
2458 sstate
->solver_status
= SOLVER_MISTAKE
;
2461 if (f
->order
- state
->clues
[i
] == current_no
) {
2462 if (face_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
))
2463 diff
= min(diff
, DIFF_EASY
);
2464 sstate
->face_solved
[i
] = TRUE
;
2468 if (f
->order
- state
->clues
[i
] == current_no
+ 1 &&
2469 f
->order
- current_yes
- current_no
> 2) {
2471 * One small refinement to the above: we also look for any
2472 * adjacent pair of LINE_UNKNOWNs around the face with
2473 * some LINE_YES incident on it from elsewhere. If we find
2474 * one, then we know that pair of LINE_UNKNOWNs can't
2475 * _both_ be LINE_YES, and hence that pushes us one line
2476 * closer to being able to determine all the rest.
2478 int j
, k
, e1
, e2
, e
, d
;
2480 for (j
= 0; j
< f
->order
; j
++) {
2481 e1
= f
->edges
[j
] - g
->edges
;
2482 e2
= f
->edges
[j
+1 < f
->order ? j
+1 : 0] - g
->edges
;
2484 if (g
->edges
[e1
].dot1
== g
->edges
[e2
].dot1
||
2485 g
->edges
[e1
].dot1
== g
->edges
[e2
].dot2
) {
2486 d
= g
->edges
[e1
].dot1
- g
->dots
;
2488 assert(g
->edges
[e1
].dot2
== g
->edges
[e2
].dot1
||
2489 g
->edges
[e1
].dot2
== g
->edges
[e2
].dot2
);
2490 d
= g
->edges
[e1
].dot2
- g
->dots
;
2493 if (state
->lines
[e1
] == LINE_UNKNOWN
&&
2494 state
->lines
[e2
] == LINE_UNKNOWN
) {
2495 for (k
= 0; k
< g
->dots
[d
].order
; k
++) {
2496 int e
= g
->dots
[d
].edges
[k
] - g
->edges
;
2497 if (state
->lines
[e
] == LINE_YES
)
2498 goto found
; /* multi-level break */
2506 * If we get here, we've found such a pair of edges, and
2507 * they're e1 and e2.
2509 for (j
= 0; j
< f
->order
; j
++) {
2510 e
= f
->edges
[j
] - g
->edges
;
2511 if (state
->lines
[e
] == LINE_UNKNOWN
&& e
!= e1
&& e
!= e2
) {
2512 int r
= solver_set_line(sstate
, e
, LINE_YES
);
2514 diff
= min(diff
, DIFF_EASY
);
2520 check_caches(sstate
);
2522 /* Per-dot deductions */
2523 for (i
= 0; i
< g
->num_dots
; i
++) {
2524 grid_dot
*d
= g
->dots
+ i
;
2525 int yes
, no
, unknown
;
2527 if (sstate
->dot_solved
[i
])
2530 yes
= sstate
->dot_yes_count
[i
];
2531 no
= sstate
->dot_no_count
[i
];
2532 unknown
= d
->order
- yes
- no
;
2536 sstate
->dot_solved
[i
] = TRUE
;
2537 } else if (unknown
== 1) {
2538 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2539 diff
= min(diff
, DIFF_EASY
);
2540 sstate
->dot_solved
[i
] = TRUE
;
2542 } else if (yes
== 1) {
2544 sstate
->solver_status
= SOLVER_MISTAKE
;
2546 } else if (unknown
== 1) {
2547 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_YES
);
2548 diff
= min(diff
, DIFF_EASY
);
2550 } else if (yes
== 2) {
2552 dot_setall(sstate
, i
, LINE_UNKNOWN
, LINE_NO
);
2553 diff
= min(diff
, DIFF_EASY
);
2555 sstate
->dot_solved
[i
] = TRUE
;
2557 sstate
->solver_status
= SOLVER_MISTAKE
;
2562 check_caches(sstate
);
2567 static int dline_deductions(solver_state
*sstate
)
2569 game_state
*state
= sstate
->state
;
2570 grid
*g
= state
->game_grid
;
2571 char *dlines
= sstate
->dlines
;
2573 int diff
= DIFF_MAX
;
2575 /* ------ Face deductions ------ */
2577 /* Given a set of dline atmostone/atleastone constraints, need to figure
2578 * out if we can deduce any further info. For more general faces than
2579 * squares, this turns out to be a tricky problem.
2580 * The approach taken here is to define (per face) NxN matrices:
2581 * "maxs" and "mins".
2582 * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
2583 * for the possible number of edges that are YES between positions j and k
2584 * going clockwise around the face. Can think of j and k as marking dots
2585 * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
2586 * edge1 joins dot1 to dot2 etc).
2587 * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
2588 * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
2589 * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
2590 * the dline atmostone/atleastone status for edges j and j+1.
2592 * Then we calculate the remaining entries recursively. We definitely
2594 * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
2595 * This is because any valid placement of YESs between j and k must give
2596 * a valid placement between j and u, and also between u and k.
2597 * I believe it's sufficient to use just the two values of u:
2598 * j+1 and j+2. Seems to work well in practice - the bounds we compute
2599 * are rigorous, even if they might not be best-possible.
2601 * Once we have maxs and mins calculated, we can make inferences about
2602 * each dline{j,j+1} by looking at the possible complementary edge-counts
2603 * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
2604 * As well as dlines, we can make similar inferences about single edges.
2605 * For example, consider a pentagon with clue 3, and we know at most one
2606 * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
2607 * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
2608 * that final edge would have to be YES to make the count up to 3.
2611 /* Much quicker to allocate arrays on the stack than the heap, so
2612 * define the largest possible face size, and base our array allocations
2613 * on that. We check this with an assertion, in case someone decides to
2614 * make a grid which has larger faces than this. Note, this algorithm
2615 * could get quite expensive if there are many large faces. */
2616 #define MAX_FACE_SIZE 12
2618 for (i
= 0; i
< g
->num_faces
; i
++) {
2619 int maxs
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2620 int mins
[MAX_FACE_SIZE
][MAX_FACE_SIZE
];
2621 grid_face
*f
= g
->faces
+ i
;
2624 int clue
= state
->clues
[i
];
2625 assert(N
<= MAX_FACE_SIZE
);
2626 if (sstate
->face_solved
[i
])
2628 if (clue
< 0) continue;
2630 /* Calculate the (j,j+1) entries */
2631 for (j
= 0; j
< N
; j
++) {
2632 int edge_index
= f
->edges
[j
] - g
->edges
;
2634 enum line_state line1
= state
->lines
[edge_index
];
2635 enum line_state line2
;
2639 maxs
[j
][k
] = (line1
== LINE_NO
) ?
0 : 1;
2640 mins
[j
][k
] = (line1
== LINE_YES
) ?
1 : 0;
2641 /* Calculate the (j,j+2) entries */
2642 dline_index
= dline_index_from_face(g
, f
, k
);
2643 edge_index
= f
->edges
[k
] - g
->edges
;
2644 line2
= state
->lines
[edge_index
];
2650 if (line1
== LINE_NO
) tmp
--;
2651 if (line2
== LINE_NO
) tmp
--;
2652 if (tmp
== 2 && is_atmostone(dlines
, dline_index
))
2658 if (line1
== LINE_YES
) tmp
++;
2659 if (line2
== LINE_YES
) tmp
++;
2660 if (tmp
== 0 && is_atleastone(dlines
, dline_index
))
2665 /* Calculate the (j,j+m) entries for m between 3 and N-1 */
2666 for (m
= 3; m
< N
; m
++) {
2667 for (j
= 0; j
< N
; j
++) {
2675 maxs
[j
][k
] = maxs
[j
][u
] + maxs
[u
][k
];
2676 mins
[j
][k
] = mins
[j
][u
] + mins
[u
][k
];
2677 tmp
= maxs
[j
][v
] + maxs
[v
][k
];
2678 maxs
[j
][k
] = min(maxs
[j
][k
], tmp
);
2679 tmp
= mins
[j
][v
] + mins
[v
][k
];
2680 mins
[j
][k
] = max(mins
[j
][k
], tmp
);
2684 /* See if we can make any deductions */
2685 for (j
= 0; j
< N
; j
++) {
2687 grid_edge
*e
= f
->edges
[j
];
2688 int line_index
= e
- g
->edges
;
2691 if (state
->lines
[line_index
] != LINE_UNKNOWN
)
2696 /* minimum YESs in the complement of this edge */
2697 if (mins
[k
][j
] > clue
) {
2698 sstate
->solver_status
= SOLVER_MISTAKE
;
2701 if (mins
[k
][j
] == clue
) {
2702 /* setting this edge to YES would make at least
2703 * (clue+1) edges - contradiction */
2704 solver_set_line(sstate
, line_index
, LINE_NO
);
2705 diff
= min(diff
, DIFF_EASY
);
2707 if (maxs
[k
][j
] < clue
- 1) {
2708 sstate
->solver_status
= SOLVER_MISTAKE
;
2711 if (maxs
[k
][j
] == clue
- 1) {
2712 /* Only way to satisfy the clue is to set edge{j} as YES */
2713 solver_set_line(sstate
, line_index
, LINE_YES
);
2714 diff
= min(diff
, DIFF_EASY
);
2717 /* More advanced deduction that allows propagation along diagonal
2718 * chains of faces connected by dots, for example, 3-2-...-2-3
2719 * in square grids. */
2720 if (sstate
->diff
>= DIFF_TRICKY
) {
2721 /* Now see if we can make dline deduction for edges{j,j+1} */
2723 if (state
->lines
[e
- g
->edges
] != LINE_UNKNOWN
)
2724 /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
2725 * Dlines where one of the edges is known, are handled in the
2729 dline_index
= dline_index_from_face(g
, f
, k
);
2733 /* minimum YESs in the complement of this dline */
2734 if (mins
[k
][j
] > clue
- 2) {
2735 /* Adding 2 YESs would break the clue */
2736 if (set_atmostone(dlines
, dline_index
))
2737 diff
= min(diff
, DIFF_NORMAL
);
2739 /* maximum YESs in the complement of this dline */
2740 if (maxs
[k
][j
] < clue
) {
2741 /* Adding 2 NOs would mean not enough YESs */
2742 if (set_atleastone(dlines
, dline_index
))
2743 diff
= min(diff
, DIFF_NORMAL
);
2749 if (diff
< DIFF_NORMAL
)
2752 /* ------ Dot deductions ------ */
2754 for (i
= 0; i
< g
->num_dots
; i
++) {
2755 grid_dot
*d
= g
->dots
+ i
;
2757 int yes
, no
, unknown
;
2759 if (sstate
->dot_solved
[i
])
2761 yes
= sstate
->dot_yes_count
[i
];
2762 no
= sstate
->dot_no_count
[i
];
2763 unknown
= N
- yes
- no
;
2765 for (j
= 0; j
< N
; j
++) {
2768 int line1_index
, line2_index
;
2769 enum line_state line1
, line2
;
2772 dline_index
= dline_index_from_dot(g
, d
, j
);
2773 line1_index
= d
->edges
[j
] - g
->edges
;
2774 line2_index
= d
->edges
[k
] - g
->edges
;
2775 line1
= state
->lines
[line1_index
];
2776 line2
= state
->lines
[line2_index
];
2778 /* Infer dline state from line state */
2779 if (line1
== LINE_NO
|| line2
== LINE_NO
) {
2780 if (set_atmostone(dlines
, dline_index
))
2781 diff
= min(diff
, DIFF_NORMAL
);
2783 if (line1
== LINE_YES
|| line2
== LINE_YES
) {
2784 if (set_atleastone(dlines
, dline_index
))
2785 diff
= min(diff
, DIFF_NORMAL
);
2787 /* Infer line state from dline state */
2788 if (is_atmostone(dlines
, dline_index
)) {
2789 if (line1
== LINE_YES
&& line2
== LINE_UNKNOWN
) {
2790 solver_set_line(sstate
, line2_index
, LINE_NO
);
2791 diff
= min(diff
, DIFF_EASY
);
2793 if (line2
== LINE_YES
&& line1
== LINE_UNKNOWN
) {
2794 solver_set_line(sstate
, line1_index
, LINE_NO
);
2795 diff
= min(diff
, DIFF_EASY
);
2798 if (is_atleastone(dlines
, dline_index
)) {
2799 if (line1
== LINE_NO
&& line2
== LINE_UNKNOWN
) {
2800 solver_set_line(sstate
, line2_index
, LINE_YES
);
2801 diff
= min(diff
, DIFF_EASY
);
2803 if (line2
== LINE_NO
&& line1
== LINE_UNKNOWN
) {
2804 solver_set_line(sstate
, line1_index
, LINE_YES
);
2805 diff
= min(diff
, DIFF_EASY
);
2808 /* Deductions that depend on the numbers of lines.
2809 * Only bother if both lines are UNKNOWN, otherwise the
2810 * easy-mode solver (or deductions above) would have taken
2812 if (line1
!= LINE_UNKNOWN
|| line2
!= LINE_UNKNOWN
)
2815 if (yes
== 0 && unknown
== 2) {
2816 /* Both these unknowns must be identical. If we know
2817 * atmostone or atleastone, we can make progress. */
2818 if (is_atmostone(dlines
, dline_index
)) {
2819 solver_set_line(sstate
, line1_index
, LINE_NO
);
2820 solver_set_line(sstate
, line2_index
, LINE_NO
);
2821 diff
= min(diff
, DIFF_EASY
);
2823 if (is_atleastone(dlines
, dline_index
)) {
2824 solver_set_line(sstate
, line1_index
, LINE_YES
);
2825 solver_set_line(sstate
, line2_index
, LINE_YES
);
2826 diff
= min(diff
, DIFF_EASY
);
2830 if (set_atmostone(dlines
, dline_index
))
2831 diff
= min(diff
, DIFF_NORMAL
);
2833 if (set_atleastone(dlines
, dline_index
))
2834 diff
= min(diff
, DIFF_NORMAL
);
2838 /* More advanced deduction that allows propagation along diagonal
2839 * chains of faces connected by dots, for example: 3-2-...-2-3
2840 * in square grids. */
2841 if (sstate
->diff
>= DIFF_TRICKY
) {
2842 /* If we have atleastone set for this dline, infer
2843 * atmostone for each "opposite" dline (that is, each
2844 * dline without edges in common with this one).
2845 * Again, this test is only worth doing if both these
2846 * lines are UNKNOWN. For if one of these lines were YES,
2847 * the (yes == 1) test above would kick in instead. */
2848 if (is_atleastone(dlines
, dline_index
)) {
2850 for (opp
= 0; opp
< N
; opp
++) {
2851 int opp_dline_index
;
2852 if (opp
== j
|| opp
== j
+1 || opp
== j
-1)
2854 if (j
== 0 && opp
== N
-1)
2856 if (j
== N
-1 && opp
== 0)
2858 opp_dline_index
= dline_index_from_dot(g
, d
, opp
);
2859 if (set_atmostone(dlines
, opp_dline_index
))
2860 diff
= min(diff
, DIFF_NORMAL
);
2862 if (yes
== 0 && is_atmostone(dlines
, dline_index
)) {
2863 /* This dline has *exactly* one YES and there are no
2864 * other YESs. This allows more deductions. */
2866 /* Third unknown must be YES */
2867 for (opp
= 0; opp
< N
; opp
++) {
2869 if (opp
== j
|| opp
== k
)
2871 opp_index
= d
->edges
[opp
] - g
->edges
;
2872 if (state
->lines
[opp_index
] == LINE_UNKNOWN
) {
2873 solver_set_line(sstate
, opp_index
,
2875 diff
= min(diff
, DIFF_EASY
);
2878 } else if (unknown
== 4) {
2879 /* Exactly one of opposite UNKNOWNS is YES. We've
2880 * already set atmostone, so set atleastone as
2883 if (dline_set_opp_atleastone(sstate
, d
, j
))
2884 diff
= min(diff
, DIFF_NORMAL
);
2894 static int linedsf_deductions(solver_state
*sstate
)
2896 game_state
*state
= sstate
->state
;
2897 grid
*g
= state
->game_grid
;
2898 char *dlines
= sstate
->dlines
;
2900 int diff
= DIFF_MAX
;
2903 /* ------ Face deductions ------ */
2905 /* A fully-general linedsf deduction seems overly complicated
2906 * (I suspect the problem is NP-complete, though in practice it might just
2907 * be doable because faces are limited in size).
2908 * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
2909 * known to be identical. If setting them both to YES (or NO) would break
2910 * the clue, set them to NO (or YES). */
2912 for (i
= 0; i
< g
->num_faces
; i
++) {
2913 int N
, yes
, no
, unknown
;
2916 if (sstate
->face_solved
[i
])
2918 clue
= state
->clues
[i
];
2922 N
= g
->faces
[i
].order
;
2923 yes
= sstate
->face_yes_count
[i
];
2924 if (yes
+ 1 == clue
) {
2925 if (face_setall_identical(sstate
, i
, LINE_NO
))
2926 diff
= min(diff
, DIFF_EASY
);
2928 no
= sstate
->face_no_count
[i
];
2929 if (no
+ 1 == N
- clue
) {
2930 if (face_setall_identical(sstate
, i
, LINE_YES
))
2931 diff
= min(diff
, DIFF_EASY
);
2934 /* Reload YES count, it might have changed */
2935 yes
= sstate
->face_yes_count
[i
];
2936 unknown
= N
- no
- yes
;
2938 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2939 * parity of lines. */
2940 diff_tmp
= parity_deductions(sstate
, g
->faces
[i
].edges
,
2941 (clue
- yes
) % 2, unknown
);
2942 diff
= min(diff
, diff_tmp
);
2945 /* ------ Dot deductions ------ */
2946 for (i
= 0; i
< g
->num_dots
; i
++) {
2947 grid_dot
*d
= g
->dots
+ i
;
2950 int yes
, no
, unknown
;
2951 /* Go through dlines, and do any dline<->linedsf deductions wherever
2952 * we find two UNKNOWNS. */
2953 for (j
= 0; j
< N
; j
++) {
2954 int dline_index
= dline_index_from_dot(g
, d
, j
);
2957 int can1
, can2
, inv1
, inv2
;
2959 line1_index
= d
->edges
[j
] - g
->edges
;
2960 if (state
->lines
[line1_index
] != LINE_UNKNOWN
)
2963 if (j2
== N
) j2
= 0;
2964 line2_index
= d
->edges
[j2
] - g
->edges
;
2965 if (state
->lines
[line2_index
] != LINE_UNKNOWN
)
2967 /* Infer dline flags from linedsf */
2968 can1
= edsf_canonify(sstate
->linedsf
, line1_index
, &inv1
);
2969 can2
= edsf_canonify(sstate
->linedsf
, line2_index
, &inv2
);
2970 if (can1
== can2
&& inv1
!= inv2
) {
2971 /* These are opposites, so set dline atmostone/atleastone */
2972 if (set_atmostone(dlines
, dline_index
))
2973 diff
= min(diff
, DIFF_NORMAL
);
2974 if (set_atleastone(dlines
, dline_index
))
2975 diff
= min(diff
, DIFF_NORMAL
);
2978 /* Infer linedsf from dline flags */
2979 if (is_atmostone(dlines
, dline_index
)
2980 && is_atleastone(dlines
, dline_index
)) {
2981 if (merge_lines(sstate
, line1_index
, line2_index
, 1))
2982 diff
= min(diff
, DIFF_HARD
);
2986 /* Deductions with small number of LINE_UNKNOWNs, based on overall
2987 * parity of lines. */
2988 yes
= sstate
->dot_yes_count
[i
];
2989 no
= sstate
->dot_no_count
[i
];
2990 unknown
= N
- yes
- no
;
2991 diff_tmp
= parity_deductions(sstate
, d
->edges
,
2993 diff
= min(diff
, diff_tmp
);
2996 /* ------ Edge dsf deductions ------ */
2998 /* If the state of a line is known, deduce the state of its canonical line
2999 * too, and vice versa. */
3000 for (i
= 0; i
< g
->num_edges
; i
++) {
3003 can
= edsf_canonify(sstate
->linedsf
, i
, &inv
);
3006 s
= sstate
->state
->lines
[can
];
3007 if (s
!= LINE_UNKNOWN
) {
3008 if (solver_set_line(sstate
, i
, inv ?
OPP(s
) : s
))
3009 diff
= min(diff
, DIFF_EASY
);
3011 s
= sstate
->state
->lines
[i
];
3012 if (s
!= LINE_UNKNOWN
) {
3013 if (solver_set_line(sstate
, can
, inv ?
OPP(s
) : s
))
3014 diff
= min(diff
, DIFF_EASY
);
3022 static int loop_deductions(solver_state
*sstate
)
3024 int edgecount
= 0, clues
= 0, satclues
= 0, sm1clues
= 0;
3025 game_state
*state
= sstate
->state
;
3026 grid
*g
= state
->game_grid
;
3027 int shortest_chainlen
= g
->num_dots
;
3028 int loop_found
= FALSE
;
3030 int progress
= FALSE
;
3034 * Go through the grid and update for all the new edges.
3035 * Since merge_dots() is idempotent, the simplest way to
3036 * do this is just to update for _all_ the edges.
3037 * Also, while we're here, we count the edges.
3039 for (i
= 0; i
< g
->num_edges
; i
++) {
3040 if (state
->lines
[i
] == LINE_YES
) {
3041 loop_found
|= merge_dots(sstate
, i
);
3047 * Count the clues, count the satisfied clues, and count the
3048 * satisfied-minus-one clues.
3050 for (i
= 0; i
< g
->num_faces
; i
++) {
3051 int c
= state
->clues
[i
];
3053 int o
= sstate
->face_yes_count
[i
];
3062 for (i
= 0; i
< g
->num_dots
; ++i
) {
3064 sstate
->looplen
[dsf_canonify(sstate
->dotdsf
, i
)];
3065 if (dots_connected
> 1)
3066 shortest_chainlen
= min(shortest_chainlen
, dots_connected
);
3069 assert(sstate
->solver_status
== SOLVER_INCOMPLETE
);
3071 if (satclues
== clues
&& shortest_chainlen
== edgecount
) {
3072 sstate
->solver_status
= SOLVER_SOLVED
;
3073 /* This discovery clearly counts as progress, even if we haven't
3074 * just added any lines or anything */
3076 goto finished_loop_deductionsing
;
3080 * Now go through looking for LINE_UNKNOWN edges which
3081 * connect two dots that are already in the same
3082 * equivalence class. If we find one, test to see if the
3083 * loop it would create is a solution.
3085 for (i
= 0; i
< g
->num_edges
; i
++) {
3086 grid_edge
*e
= g
->edges
+ i
;
3087 int d1
= e
->dot1
- g
->dots
;
3088 int d2
= e
->dot2
- g
->dots
;
3090 if (state
->lines
[i
] != LINE_UNKNOWN
)
3093 eqclass
= dsf_canonify(sstate
->dotdsf
, d1
);
3094 if (eqclass
!= dsf_canonify(sstate
->dotdsf
, d2
))
3097 val
= LINE_NO
; /* loop is bad until proven otherwise */
3100 * This edge would form a loop. Next
3101 * question: how long would the loop be?
3102 * Would it equal the total number of edges
3103 * (plus the one we'd be adding if we added
3106 if (sstate
->looplen
[eqclass
] == edgecount
+ 1) {
3110 * This edge would form a loop which
3111 * took in all the edges in the entire
3112 * grid. So now we need to work out
3113 * whether it would be a valid solution
3114 * to the puzzle, which means we have to
3115 * check if it satisfies all the clues.
3116 * This means that every clue must be
3117 * either satisfied or satisfied-minus-
3118 * 1, and also that the number of
3119 * satisfied-minus-1 clues must be at
3120 * most two and they must lie on either
3121 * side of this edge.
3125 int f
= e
->face1
- g
->faces
;
3126 int c
= state
->clues
[f
];
3127 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3131 int f
= e
->face2
- g
->faces
;
3132 int c
= state
->clues
[f
];
3133 if (c
>= 0 && sstate
->face_yes_count
[f
] == c
- 1)
3136 if (sm1clues
== sm1_nearby
&&
3137 sm1clues
+ satclues
== clues
) {
3138 val
= LINE_YES
; /* loop is good! */
3143 * Right. Now we know that adding this edge
3144 * would form a loop, and we know whether
3145 * that loop would be a viable solution or
3148 * If adding this edge produces a solution,
3149 * then we know we've found _a_ solution but
3150 * we don't know that it's _the_ solution -
3151 * if it were provably the solution then
3152 * we'd have deduced this edge some time ago
3153 * without the need to do loop detection. So
3154 * in this state we return SOLVER_AMBIGUOUS,
3155 * which has the effect that hitting Solve
3156 * on a user-provided puzzle will fill in a
3157 * solution but using the solver to
3158 * construct new puzzles won't consider this
3159 * a reasonable deduction for the user to
3162 progress
= solver_set_line(sstate
, i
, val
);
3163 assert(progress
== TRUE
);
3164 if (val
== LINE_YES
) {
3165 sstate
->solver_status
= SOLVER_AMBIGUOUS
;
3166 goto finished_loop_deductionsing
;
3170 finished_loop_deductionsing
:
3171 return progress ? DIFF_EASY
: DIFF_MAX
;
3174 /* This will return a dynamically allocated solver_state containing the (more)
3176 static solver_state
*solve_game_rec(const solver_state
*sstate_start
)
3178 solver_state
*sstate
;
3180 /* Index of the solver we should call next. */
3183 /* As a speed-optimisation, we avoid re-running solvers that we know
3184 * won't make any progress. This happens when a high-difficulty
3185 * solver makes a deduction that can only help other high-difficulty
3187 * For example: if a new 'dline' flag is set by dline_deductions, the
3188 * trivial_deductions solver cannot do anything with this information.
3189 * If we've already run the trivial_deductions solver (because it's
3190 * earlier in the list), there's no point running it again.
3192 * Therefore: if a solver is earlier in the list than "threshold_index",
3193 * we don't bother running it if it's difficulty level is less than
3196 int threshold_diff
= 0;
3197 int threshold_index
= 0;
3199 sstate
= dup_solver_state(sstate_start
);
3201 check_caches(sstate
);
3203 while (i
< NUM_SOLVERS
) {
3204 if (sstate
->solver_status
== SOLVER_MISTAKE
)
3206 if (sstate
->solver_status
== SOLVER_SOLVED
||
3207 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3208 /* solver finished */
3212 if ((solver_diffs
[i
] >= threshold_diff
|| i
>= threshold_index
)
3213 && solver_diffs
[i
] <= sstate
->diff
) {
3214 /* current_solver is eligible, so use it */
3215 int next_diff
= solver_fns
[i
](sstate
);
3216 if (next_diff
!= DIFF_MAX
) {
3217 /* solver made progress, so use new thresholds and
3218 * start again at top of list. */
3219 threshold_diff
= next_diff
;
3220 threshold_index
= i
;
3225 /* current_solver is ineligible, or failed to make progress, so
3226 * go to the next solver in the list */
3230 if (sstate
->solver_status
== SOLVER_SOLVED
||
3231 sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3232 /* s/LINE_UNKNOWN/LINE_NO/g */
3233 array_setall(sstate
->state
->lines
, LINE_UNKNOWN
, LINE_NO
,
3234 sstate
->state
->game_grid
->num_edges
);
3241 static char *solve_game(game_state
*state
, game_state
*currstate
,
3242 char *aux
, char **error
)
3245 solver_state
*sstate
, *new_sstate
;
3247 sstate
= new_solver_state(state
, DIFF_MAX
);
3248 new_sstate
= solve_game_rec(sstate
);
3250 if (new_sstate
->solver_status
== SOLVER_SOLVED
) {
3251 soln
= encode_solve_move(new_sstate
->state
);
3252 } else if (new_sstate
->solver_status
== SOLVER_AMBIGUOUS
) {
3253 soln
= encode_solve_move(new_sstate
->state
);
3254 /**error = "Solver found ambiguous solutions"; */
3256 soln
= encode_solve_move(new_sstate
->state
);
3257 /**error = "Solver failed"; */
3260 free_solver_state(new_sstate
);
3261 free_solver_state(sstate
);
3266 /* ----------------------------------------------------------------------
3267 * Drawing and mouse-handling
3270 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
3271 int x
, int y
, int button
)
3273 grid
*g
= state
->game_grid
;
3277 char button_char
= ' ';
3278 enum line_state old_state
;
3280 button
&= ~MOD_MASK
;
3282 /* Convert mouse-click (x,y) to grid coordinates */
3283 x
-= BORDER(ds
->tilesize
);
3284 y
-= BORDER(ds
->tilesize
);
3285 x
= x
* g
->tilesize
/ ds
->tilesize
;
3286 y
= y
* g
->tilesize
/ ds
->tilesize
;
3290 e
= grid_nearest_edge(g
, x
, y
);
3296 /* I think it's only possible to play this game with mouse clicks, sorry */
3297 /* Maybe will add mouse drag support some time */
3298 old_state
= state
->lines
[i
];
3302 switch (old_state
) {
3320 switch (old_state
) {
3339 sprintf(buf
, "%d%c", i
, (int)button_char
);
3345 static game_state
*execute_move(game_state
*state
, char *move
)
3348 game_state
*newstate
= dup_game(state
);
3350 if (move
[0] == 'S') {
3352 newstate
->cheated
= TRUE
;
3357 if (i
< 0 || i
>= newstate
->game_grid
->num_edges
)
3359 move
+= strspn(move
, "1234567890");
3360 switch (*(move
++)) {
3362 newstate
->lines
[i
] = LINE_YES
;
3365 newstate
->lines
[i
] = LINE_NO
;
3368 newstate
->lines
[i
] = LINE_UNKNOWN
;
3376 * Check for completion.
3378 if (check_completion(newstate
))
3379 newstate
->solved
= TRUE
;
3384 free_game(newstate
);
3388 /* ----------------------------------------------------------------------
3392 /* Convert from grid coordinates to screen coordinates */
3393 static void grid_to_screen(const game_drawstate
*ds
, const grid
*g
,
3394 int grid_x
, int grid_y
, int *x
, int *y
)
3396 *x
= grid_x
- g
->lowest_x
;
3397 *y
= grid_y
- g
->lowest_y
;
3398 *x
= *x
* ds
->tilesize
/ g
->tilesize
;
3399 *y
= *y
* ds
->tilesize
/ g
->tilesize
;
3400 *x
+= BORDER(ds
->tilesize
);
3401 *y
+= BORDER(ds
->tilesize
);
3404 /* Returns (into x,y) position of centre of face for rendering the text clue.
3406 static void face_text_pos(const game_drawstate
*ds
, const grid
*g
,
3407 const grid_face
*f
, int *xret
, int *yret
)
3409 int x
, y
, x0
, y0
, x1
, y1
, xbest
, ybest
, i
, shift
;
3411 int faceindex
= f
- g
->faces
;
3414 * Return the cached position for this face, if we've already
3417 if (ds
->textx
[faceindex
] >= 0) {
3418 *xret
= ds
->textx
[faceindex
];
3419 *yret
= ds
->texty
[faceindex
];
3424 * Otherwise, try to find the point in the polygon with the
3425 * maximum distance to any edge or corner.
3427 * Start by working out the face's bounding box, in grid
3430 x0
= x1
= f
->dots
[0]->x
;
3431 y0
= y1
= f
->dots
[0]->y
;
3432 for (i
= 1; i
< f
->order
; i
++) {
3433 if (x0
> f
->dots
[i
]->x
) x0
= f
->dots
[i
]->x
;
3434 if (x1
< f
->dots
[i
]->x
) x1
= f
->dots
[i
]->x
;
3435 if (y0
> f
->dots
[i
]->y
) y0
= f
->dots
[i
]->y
;
3436 if (y1
< f
->dots
[i
]->y
) y1
= f
->dots
[i
]->y
;
3440 * If the grid is at excessive resolution, decide on a scaling
3441 * factor to bring it within reasonable bounds so we don't have to
3442 * think too hard or suffer integer overflow.
3445 while (x1
- x0
> 128 || y1
- y0
> 128) {
3454 * Now iterate over every point in that bounding box.
3458 for (y
= y0
; y
<= y1
; y
++) {
3459 for (x
= x0
; x
<= x1
; x
++) {
3461 * First, disqualify the point if it's not inside the
3462 * polygon, which we work out by counting the edges to the
3463 * right of the point. (For tiebreaking purposes when
3464 * edges start or end on our y-coordinate or go right
3465 * through it, we consider our point to be offset by a
3466 * small _positive_ epsilon in both the x- and
3470 for (i
= 0; i
< f
->order
; i
++) {
3471 int xs
= f
->edges
[i
]->dot1
->x
>> shift
;
3472 int xe
= f
->edges
[i
]->dot2
->x
>> shift
;
3473 int ys
= f
->edges
[i
]->dot1
->y
>> shift
;
3474 int ye
= f
->edges
[i
]->dot2
->y
>> shift
;
3475 if ((y
>= ys
&& y
< ye
) || (y
>= ye
&& y
< ys
)) {
3477 * The line goes past our y-position. Now we need
3478 * to know if its x-coordinate when it does so is
3481 * The x-coordinate in question is mathematically
3482 * (y - ys) * (xe - xs) / (ye - ys), and we want
3483 * to know whether (x - xs) >= that. Of course we
3484 * avoid the division, so we can work in integers;
3485 * to do this we must multiply both sides of the
3486 * inequality by ye - ys, which means we must
3487 * first check that's not negative.
3489 int num
= xe
- xs
, denom
= ye
- ys
;
3494 if ((x
- xs
) * denom
>= (y
- ys
) * num
)
3500 long mindist
= LONG_MAX
;
3503 * This point is inside the polygon, so now we check
3504 * its minimum distance to every edge and corner.
3505 * First the corners ...
3507 for (i
= 0; i
< f
->order
; i
++) {
3508 int xp
= f
->dots
[i
]->x
>> shift
;
3509 int yp
= f
->dots
[i
]->y
>> shift
;
3510 int dx
= x
- xp
, dy
= y
- yp
;
3511 long dist
= (long)dx
*dx
+ (long)dy
*dy
;
3517 * ... and now also check the perpendicular distance
3518 * to every edge, if the perpendicular lies between
3519 * the edge's endpoints.
3521 for (i
= 0; i
< f
->order
; i
++) {
3522 int xs
= f
->edges
[i
]->dot1
->x
>> shift
;
3523 int xe
= f
->edges
[i
]->dot2
->x
>> shift
;
3524 int ys
= f
->edges
[i
]->dot1
->y
>> shift
;
3525 int ye
= f
->edges
[i
]->dot2
->y
>> shift
;
3528 * If s and e are our endpoints, and p our
3529 * candidate circle centre, the foot of a
3530 * perpendicular from p to the line se lies
3531 * between s and e if and only if (p-s).(e-s) lies
3532 * strictly between 0 and (e-s).(e-s).
3534 int edx
= xe
- xs
, edy
= ye
- ys
;
3535 int pdx
= x
- xs
, pdy
= y
- ys
;
3536 long pde
= (long)pdx
* edx
+ (long)pdy
* edy
;
3537 long ede
= (long)edx
* edx
+ (long)edy
* edy
;
3538 if (0 < pde
&& pde
< ede
) {
3540 * Yes, the nearest point on this edge is
3541 * closer than either endpoint, so we must
3542 * take it into account by measuring the
3543 * perpendicular distance to the edge and
3544 * checking its square against mindist.
3547 long pdre
= (long)pdx
* edy
- (long)pdy
* edx
;
3548 long sqlen
= pdre
* pdre
/ ede
;
3550 if (mindist
> sqlen
)
3556 * Right. Now we know the biggest circle around this
3557 * point, so we can check it against bestdist.
3559 if (bestdist
< mindist
) {
3568 assert(bestdist
>= 0);
3570 /* convert to screen coordinates */
3571 grid_to_screen(ds
, g
, xbest
<< shift
, ybest
<< shift
,
3572 &ds
->textx
[faceindex
], &ds
->texty
[faceindex
]);
3574 *xret
= ds
->textx
[faceindex
];
3575 *yret
= ds
->texty
[faceindex
];
3578 static void face_text_bbox(game_drawstate
*ds
, grid
*g
, grid_face
*f
,
3579 int *x
, int *y
, int *w
, int *h
)
3582 face_text_pos(ds
, g
, f
, &xx
, &yy
);
3584 /* There seems to be a certain amount of trial-and-error involved
3585 * in working out the correct bounding-box for the text. */
3587 *x
= xx
- ds
->tilesize
/4 - 1;
3588 *y
= yy
- ds
->tilesize
/4 - 3;
3589 *w
= ds
->tilesize
/2 + 2;
3590 *h
= ds
->tilesize
/2 + 5;
3593 static void game_redraw_clue(drawing
*dr
, game_drawstate
*ds
,
3594 game_state
*state
, int i
)
3596 grid
*g
= state
->game_grid
;
3597 grid_face
*f
= g
->faces
+ i
;
3601 if (state
->clues
[i
] < 10) {
3602 c
[0] = CLUE2CHAR(state
->clues
[i
]);
3605 sprintf(c
, "%d", state
->clues
[i
]);
3608 face_text_pos(ds
, g
, f
, &x
, &y
);
3610 FONT_VARIABLE
, ds
->tilesize
/2,
3611 ALIGN_VCENTRE
| ALIGN_HCENTRE
,
3612 ds
->clue_error
[i
] ? COL_MISTAKE
:
3613 ds
->clue_satisfied
[i
] ? COL_SATISFIED
: COL_FOREGROUND
, c
);
3616 static void edge_bbox(game_drawstate
*ds
, grid
*g
, grid_edge
*e
,
3617 int *x
, int *y
, int *w
, int *h
)
3619 int x1
= e
->dot1
->x
;
3620 int y1
= e
->dot1
->y
;
3621 int x2
= e
->dot2
->x
;
3622 int y2
= e
->dot2
->y
;
3623 int xmin
, xmax
, ymin
, ymax
;
3625 grid_to_screen(ds
, g
, x1
, y1
, &x1
, &y1
);
3626 grid_to_screen(ds
, g
, x2
, y2
, &x2
, &y2
);
3627 /* Allow extra margin for dots, and thickness of lines */
3628 xmin
= min(x1
, x2
) - 2;
3629 xmax
= max(x1
, x2
) + 2;
3630 ymin
= min(y1
, y2
) - 2;
3631 ymax
= max(y1
, y2
) + 2;
3635 *w
= xmax
- xmin
+ 1;
3636 *h
= ymax
- ymin
+ 1;
3639 static void dot_bbox(game_drawstate
*ds
, grid
*g
, grid_dot
*d
,
3640 int *x
, int *y
, int *w
, int *h
)
3644 grid_to_screen(ds
, g
, d
->x
, d
->y
, &x1
, &y1
);
3652 static const int loopy_line_redraw_phases
[] = {
3653 COL_FAINT
, COL_LINEUNKNOWN
, COL_FOREGROUND
, COL_HIGHLIGHT
, COL_MISTAKE
3655 #define NPHASES lenof(loopy_line_redraw_phases)
3657 static void game_redraw_line(drawing
*dr
, game_drawstate
*ds
,
3658 game_state
*state
, int i
, int phase
)
3660 grid
*g
= state
->game_grid
;
3661 grid_edge
*e
= g
->edges
+ i
;
3663 int xmin
, ymin
, xmax
, ymax
;
3666 if (state
->line_errors
[i
])
3667 line_colour
= COL_MISTAKE
;
3668 else if (state
->lines
[i
] == LINE_UNKNOWN
)
3669 line_colour
= COL_LINEUNKNOWN
;
3670 else if (state
->lines
[i
] == LINE_NO
)
3671 line_colour
= COL_FAINT
;
3672 else if (ds
->flashing
)
3673 line_colour
= COL_HIGHLIGHT
;
3675 line_colour
= COL_FOREGROUND
;
3676 if (line_colour
!= loopy_line_redraw_phases
[phase
])
3679 /* Convert from grid to screen coordinates */
3680 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3681 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3688 if (line_colour
== COL_FAINT
) {
3689 static int draw_faint_lines
= -1;
3690 if (draw_faint_lines
< 0) {
3691 char *env
= getenv("LOOPY_FAINT_LINES");
3692 draw_faint_lines
= (!env
|| (env
[0] == 'y' ||
3695 if (draw_faint_lines
)
3696 draw_line(dr
, x1
, y1
, x2
, y2
, line_colour
);
3698 draw_thick_line(dr
, 3.0,
3705 static void game_redraw_dot(drawing
*dr
, game_drawstate
*ds
,
3706 game_state
*state
, int i
)
3708 grid
*g
= state
->game_grid
;
3709 grid_dot
*d
= g
->dots
+ i
;
3712 grid_to_screen(ds
, g
, d
->x
, d
->y
, &x
, &y
);
3713 draw_circle(dr
, x
, y
, 2, COL_FOREGROUND
, COL_FOREGROUND
);
3716 static int boxes_intersect(int x0
, int y0
, int w0
, int h0
,
3717 int x1
, int y1
, int w1
, int h1
)
3720 * Two intervals intersect iff neither is wholly on one side of
3721 * the other. Two boxes intersect iff their horizontal and
3722 * vertical intervals both intersect.
3724 return (x0
< x1
+w1
&& x1
< x0
+w0
&& y0
< y1
+h1
&& y1
< y0
+h0
);
3727 static void game_redraw_in_rect(drawing
*dr
, game_drawstate
*ds
,
3728 game_state
*state
, int x
, int y
, int w
, int h
)
3730 grid
*g
= state
->game_grid
;
3734 clip(dr
, x
, y
, w
, h
);
3735 draw_rect(dr
, x
, y
, w
, h
, COL_BACKGROUND
);
3737 for (i
= 0; i
< g
->num_faces
; i
++) {
3738 if (state
->clues
[i
] >= 0) {
3739 face_text_bbox(ds
, g
, &g
->faces
[i
], &bx
, &by
, &bw
, &bh
);
3740 if (boxes_intersect(x
, y
, w
, h
, bx
, by
, bw
, bh
))
3741 game_redraw_clue(dr
, ds
, state
, i
);
3744 for (phase
= 0; phase
< NPHASES
; phase
++) {
3745 for (i
= 0; i
< g
->num_edges
; i
++) {
3746 edge_bbox(ds
, g
, &g
->edges
[i
], &bx
, &by
, &bw
, &bh
);
3747 if (boxes_intersect(x
, y
, w
, h
, bx
, by
, bw
, bh
))
3748 game_redraw_line(dr
, ds
, state
, i
, phase
);
3751 for (i
= 0; i
< g
->num_dots
; i
++) {
3752 dot_bbox(ds
, g
, &g
->dots
[i
], &bx
, &by
, &bw
, &bh
);
3753 if (boxes_intersect(x
, y
, w
, h
, bx
, by
, bw
, bh
))
3754 game_redraw_dot(dr
, ds
, state
, i
);
3758 draw_update(dr
, x
, y
, w
, h
);
3761 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
3762 game_state
*state
, int dir
, game_ui
*ui
,
3763 float animtime
, float flashtime
)
3765 #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
3767 grid
*g
= state
->game_grid
;
3768 int border
= BORDER(ds
->tilesize
);
3771 int redraw_everything
= FALSE
;
3773 int edges
[REDRAW_OBJECTS_LIMIT
], nedges
= 0;
3774 int faces
[REDRAW_OBJECTS_LIMIT
], nfaces
= 0;
3776 /* Redrawing is somewhat involved.
3778 * An update can theoretically affect an arbitrary number of edges
3779 * (consider, for example, completing or breaking a cycle which doesn't
3780 * satisfy all the clues -- we'll switch many edges between error and
3781 * normal states). On the other hand, redrawing the whole grid takes a
3782 * while, making the game feel sluggish, and many updates are actually
3783 * quite well localized.
3785 * This redraw algorithm attempts to cope with both situations gracefully
3786 * and correctly. For localized changes, we set a clip rectangle, fill
3787 * it with background, and then redraw (a plausible but conservative
3788 * guess at) the objects which intersect the rectangle; if several
3789 * objects need redrawing, we'll do them individually. However, if lots
3790 * of objects are affected, we'll just redraw everything.
3792 * The reason for all of this is that it's just not safe to do the redraw
3793 * piecemeal. If you try to draw an antialiased diagonal line over
3794 * itself, you get a slightly thicker antialiased diagonal line, which
3795 * looks rather ugly after a while.
3797 * So, we take two passes over the grid. The first attempts to work out
3798 * what needs doing, and the second actually does it.
3802 redraw_everything
= TRUE
;
3805 /* First, trundle through the faces. */
3806 for (i
= 0; i
< g
->num_faces
; i
++) {
3807 grid_face
*f
= g
->faces
+ i
;
3808 int sides
= f
->order
;
3811 int n
= state
->clues
[i
];
3815 clue_mistake
= (face_order(state
, i
, LINE_YES
) > n
||
3816 face_order(state
, i
, LINE_NO
) > (sides
-n
));
3817 clue_satisfied
= (face_order(state
, i
, LINE_YES
) == n
&&
3818 face_order(state
, i
, LINE_NO
) == (sides
-n
));
3820 if (clue_mistake
!= ds
->clue_error
[i
] ||
3821 clue_satisfied
!= ds
->clue_satisfied
[i
]) {
3822 ds
->clue_error
[i
] = clue_mistake
;
3823 ds
->clue_satisfied
[i
] = clue_satisfied
;
3824 if (nfaces
== REDRAW_OBJECTS_LIMIT
)
3825 redraw_everything
= TRUE
;
3827 faces
[nfaces
++] = i
;
3831 /* Work out what the flash state needs to be. */
3832 if (flashtime
> 0 &&
3833 (flashtime
<= FLASH_TIME
/3 ||
3834 flashtime
>= FLASH_TIME
*2/3)) {
3835 flash_changed
= !ds
->flashing
;
3836 ds
->flashing
= TRUE
;
3838 flash_changed
= ds
->flashing
;
3839 ds
->flashing
= FALSE
;
3842 /* Now, trundle through the edges. */
3843 for (i
= 0; i
< g
->num_edges
; i
++) {
3845 state
->line_errors
[i
] ? DS_LINE_ERROR
: state
->lines
[i
];
3846 if (new_ds
!= ds
->lines
[i
] ||
3847 (flash_changed
&& state
->lines
[i
] == LINE_YES
)) {
3848 ds
->lines
[i
] = new_ds
;
3849 if (nedges
== REDRAW_OBJECTS_LIMIT
)
3850 redraw_everything
= TRUE
;
3852 edges
[nedges
++] = i
;
3857 /* Pass one is now done. Now we do the actual drawing. */
3858 if (redraw_everything
) {
3859 int grid_width
= g
->highest_x
- g
->lowest_x
;
3860 int grid_height
= g
->highest_y
- g
->lowest_y
;
3861 int w
= grid_width
* ds
->tilesize
/ g
->tilesize
;
3862 int h
= grid_height
* ds
->tilesize
/ g
->tilesize
;
3864 game_redraw_in_rect(dr
, ds
, state
,
3865 0, 0, w
+ 2*border
+ 1, h
+ 2*border
+ 1);
3868 /* Right. Now we roll up our sleeves. */
3870 for (i
= 0; i
< nfaces
; i
++) {
3871 grid_face
*f
= g
->faces
+ faces
[i
];
3874 face_text_bbox(ds
, g
, f
, &x
, &y
, &w
, &h
);
3875 game_redraw_in_rect(dr
, ds
, state
, x
, y
, w
, h
);
3878 for (i
= 0; i
< nedges
; i
++) {
3879 grid_edge
*e
= g
->edges
+ edges
[i
];
3882 edge_bbox(ds
, g
, e
, &x
, &y
, &w
, &h
);
3883 game_redraw_in_rect(dr
, ds
, state
, x
, y
, w
, h
);
3890 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
3891 int dir
, game_ui
*ui
)
3893 if (!oldstate
->solved
&& newstate
->solved
&&
3894 !oldstate
->cheated
&& !newstate
->cheated
) {
3901 static int game_is_solved(game_state
*state
)
3903 return state
->solved
;
3906 static void game_print_size(game_params
*params
, float *x
, float *y
)
3911 * I'll use 7mm "squares" by default.
3913 game_compute_size(params
, 700, &pw
, &ph
);
3918 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
3920 int ink
= print_mono_colour(dr
, 0);
3922 game_drawstate ads
, *ds
= &ads
;
3923 grid
*g
= state
->game_grid
;
3925 ds
->tilesize
= tilesize
;
3927 for (i
= 0; i
< g
->num_dots
; i
++) {
3929 grid_to_screen(ds
, g
, g
->dots
[i
].x
, g
->dots
[i
].y
, &x
, &y
);
3930 draw_circle(dr
, x
, y
, ds
->tilesize
/ 15, ink
, ink
);
3936 for (i
= 0; i
< g
->num_faces
; i
++) {
3937 grid_face
*f
= g
->faces
+ i
;
3938 int clue
= state
->clues
[i
];
3942 c
[0] = CLUE2CHAR(clue
);
3944 face_text_pos(ds
, g
, f
, &x
, &y
);
3946 FONT_VARIABLE
, ds
->tilesize
/ 2,
3947 ALIGN_VCENTRE
| ALIGN_HCENTRE
, ink
, c
);
3954 for (i
= 0; i
< g
->num_edges
; i
++) {
3955 int thickness
= (state
->lines
[i
] == LINE_YES
) ?
30 : 150;
3956 grid_edge
*e
= g
->edges
+ i
;
3958 grid_to_screen(ds
, g
, e
->dot1
->x
, e
->dot1
->y
, &x1
, &y1
);
3959 grid_to_screen(ds
, g
, e
->dot2
->x
, e
->dot2
->y
, &x2
, &y2
);
3960 if (state
->lines
[i
] == LINE_YES
)
3962 /* (dx, dy) points from (x1, y1) to (x2, y2).
3963 * The line is then "fattened" in a perpendicular
3964 * direction to create a thin rectangle. */
3965 double d
= sqrt(SQ((double)x1
- x2
) + SQ((double)y1
- y2
));
3966 double dx
= (x2
- x1
) / d
;
3967 double dy
= (y2
- y1
) / d
;
3970 dx
= (dx
* ds
->tilesize
) / thickness
;
3971 dy
= (dy
* ds
->tilesize
) / thickness
;
3972 points
[0] = x1
+ (int)dy
;
3973 points
[1] = y1
- (int)dx
;
3974 points
[2] = x1
- (int)dy
;
3975 points
[3] = y1
+ (int)dx
;
3976 points
[4] = x2
- (int)dy
;
3977 points
[5] = y2
+ (int)dx
;
3978 points
[6] = x2
+ (int)dy
;
3979 points
[7] = y2
- (int)dx
;
3980 draw_polygon(dr
, points
, 4, ink
, ink
);
3984 /* Draw a dotted line */
3987 for (j
= 1; j
< divisions
; j
++) {
3988 /* Weighted average */
3989 int x
= (x1
* (divisions
-j
) + x2
* j
) / divisions
;
3990 int y
= (y1
* (divisions
-j
) + y2
* j
) / divisions
;
3991 draw_circle(dr
, x
, y
, ds
->tilesize
/ thickness
, ink
, ink
);
3998 #define thegame loopy
4001 const struct game thegame
= {
4002 "Loopy", "games.loopy", "loopy",
4009 TRUE
, game_configure
, custom_params
,
4017 TRUE
, game_can_format_as_text_now
, game_text_format
,
4025 PREFERRED_TILE_SIZE
, game_compute_size
, game_set_size
,
4028 game_free_drawstate
,
4033 TRUE
, FALSE
, game_print_size
, game_print
,
4034 FALSE
/* wants_statusbar */,
4035 FALSE
, game_timing_state
,
4036 0, /* mouse_priorities */
4039 #ifdef STANDALONE_SOLVER
4042 * Half-hearted standalone solver. It can't output the solution to
4043 * anything but a square puzzle, and it can't log the deductions
4044 * it makes either. But it can solve square puzzles, and more
4045 * importantly it can use its solver to grade the difficulty of
4046 * any puzzle you give it.
4051 int main(int argc
, char **argv
)
4055 char *id
= NULL
, *desc
, *err
;
4058 #if 0 /* verbose solver not supported here (yet) */
4059 int really_verbose
= FALSE
;
4062 while (--argc
> 0) {
4064 #if 0 /* verbose solver not supported here (yet) */
4065 if (!strcmp(p
, "-v")) {
4066 really_verbose
= TRUE
;
4069 if (!strcmp(p
, "-g")) {
4071 } else if (*p
== '-') {
4072 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
4080 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
4084 desc
= strchr(id
, ':');
4086 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
4091 p
= default_params();
4092 decode_params(p
, id
);
4093 err
= validate_desc(p
, desc
);
4095 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
4098 s
= new_game(NULL
, p
, desc
);
4101 * When solving an Easy puzzle, we don't want to bother the
4102 * user with Hard-level deductions. For this reason, we grade
4103 * the puzzle internally before doing anything else.
4105 ret
= -1; /* placate optimiser */
4106 for (diff
= 0; diff
< DIFF_MAX
; diff
++) {
4107 solver_state
*sstate_new
;
4108 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
4110 sstate_new
= solve_game_rec(sstate
);
4112 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
4114 else if (sstate_new
->solver_status
== SOLVER_SOLVED
)
4119 free_solver_state(sstate_new
);
4120 free_solver_state(sstate
);
4126 if (diff
== DIFF_MAX
) {
4128 printf("Difficulty rating: harder than Hard, or ambiguous\n");
4130 printf("Unable to find a unique solution\n");
4134 printf("Difficulty rating: impossible (no solution exists)\n");
4136 printf("Difficulty rating: %s\n", diffnames
[diff
]);
4138 solver_state
*sstate_new
;
4139 solver_state
*sstate
= new_solver_state((game_state
*)s
, diff
);
4141 /* If we supported a verbose solver, we'd set verbosity here */
4143 sstate_new
= solve_game_rec(sstate
);
4145 if (sstate_new
->solver_status
== SOLVER_MISTAKE
)
4146 printf("Puzzle is inconsistent\n");
4148 assert(sstate_new
->solver_status
== SOLVER_SOLVED
);
4149 if (s
->grid_type
== 0) {
4150 fputs(game_text_format(sstate_new
->state
), stdout
);
4152 printf("Unable to output non-square grids\n");
4156 free_solver_state(sstate_new
);
4157 free_solver_state(sstate
);