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