2 * tents.c: Puzzle involving placing tents next to trees subject to
3 * some confusing conditions.
7 * - it might be nice to make setter-provided tent/nontent clues
9 * * on the other hand, this would introduce considerable extra
10 * complexity and size into the game state; also inviolable
11 * clues would have to be marked as such somehow, in an
12 * intrusive and annoying manner. Since they're never
13 * generated by _my_ generator, I'm currently more inclined
16 * - more difficult levels at the top end?
17 * * for example, sometimes we can deduce that two BLANKs in
18 * the same row are each adjacent to the same unattached tree
19 * and to nothing else, implying that they can't both be
20 * tents; this enables us to rule out some extra combinations
21 * in the row-based deduction loop, and hence deduce more
22 * from the number in that row than we could otherwise do.
23 * * that by itself doesn't seem worth implementing a new
24 * difficulty level for, but if I can find a few more things
25 * like that then it might become worthwhile.
26 * * I wonder if there's a sensible heuristic for where to
27 * guess which would make a recursive solver viable?
44 * The rules of this puzzle as available on the WWW are poorly
45 * specified. The bits about tents having to be orthogonally
46 * adjacent to trees, tents not being even diagonally adjacent to
47 * one another, and the number of tents in each row and column
48 * being given are simple enough; the difficult bit is the
49 * tent-to-tree matching.
51 * Some sources use simplistic wordings such as `each tree is
52 * exactly connected to only one tent', which is extremely unclear:
53 * it's easy to read erroneously as `each tree is _orthogonally
54 * adjacent_ to exactly one tent', which is definitely incorrect.
55 * Even the most coherent sources I've found don't do a much better
56 * job of stating the rule.
58 * A more precise statement of the rule is that it must be possible
59 * to find a bijection f between tents and trees such that each
60 * tree T is orthogonally adjacent to the tent f(T), but that a
61 * tent is permitted to be adjacent to other trees in addition to
62 * its own. This slightly non-obvious criterion is what gives this
63 * puzzle most of its subtlety.
65 * However, there's a particularly subtle ambiguity left over. Is
66 * the bijection between tents and trees required to be _unique_?
67 * In other words, is that bijection conceptually something the
68 * player should be able to exhibit as part of the solution (even
69 * if they aren't actually required to do so)? Or is it sufficient
70 * to have a unique _placement_ of the tents which gives rise to at
71 * least one suitable bijection?
73 * The puzzle shown to the right of this .T. 2 *T* 2
74 * paragraph illustrates the problem. There T.T 0 -> T-T 0
75 * are two distinct bijections available. .T. 2 *T* 2
76 * The answer to the above question will
77 * determine whether it's a valid puzzle. 202 202
79 * This is an important question, because it affects both the
80 * player and the generator. Eventually I found all the instances
81 * of this puzzle I could Google up, solved them all by hand, and
82 * verified that in all cases the tree/tent matching was uniquely
83 * determined given the tree and tent positions. Therefore, the
84 * puzzle as implemented in this source file takes the following
87 * - When checking a user-supplied solution for correctness, only
88 * verify that there exists _at least_ one matching.
89 * - When generating a puzzle, enforce that there must be
92 * Algorithmic implications
93 * ------------------------
95 * Another way of phrasing the tree/tent matching criterion is to
96 * say that the bipartite adjacency graph between trees and tents
97 * has a perfect matching. That is, if you construct a graph which
98 * has a vertex per tree and a vertex per tent, and an edge between
99 * any tree and tent which are orthogonally adjacent, it is
100 * possible to find a set of N edges of that graph (where N is the
101 * number of trees and also the number of tents) which between them
102 * connect every tree to every tent.
104 * The most efficient known algorithms for finding such a matching
105 * given a graph, as far as I'm aware, are the Munkres assignment
106 * algorithm (also known as the Hungarian algorithm) and the
107 * Ford-Fulkerson algorithm (for finding optimal flows in
108 * networks). Each of these takes O(N^3) running time; so we're
109 * talking O(N^3) time to verify any candidate solution to this
110 * puzzle. That's just about OK if you're doing it once per mouse
111 * click (and in fact not even that, since the sensible thing to do
112 * is check all the _other_ puzzle criteria and only wade into this
113 * quagmire if none are violated); but if the solver had to keep
114 * doing N^3 work internally, then it would probably end up with
115 * more like N^5 or N^6 running time, and grid generation would
116 * become very clunky.
118 * Fortunately, I've been able to prove a very useful property of
119 * _unique_ perfect matchings, by adapting the proof of Hall's
120 * Marriage Theorem. For those unaware of Hall's Theorem, I'll
121 * recap it and its proof: it states that a bipartite graph
122 * contains a perfect matching iff every set of vertices on the
123 * left side of the graph have a neighbourhood _at least_ as big on
126 * This condition is obviously satisfied if a perfect matching does
127 * exist; each left-side node has a distinct right-side node which
128 * is the one assigned to it by the matching, and thus any set of n
129 * left vertices must have a combined neighbourhood containing at
130 * least the n corresponding right vertices, and possibly others
131 * too. Alternatively, imagine if you had (say) three left-side
132 * nodes all of which were connected to only two right-side nodes
133 * between them: any perfect matching would have to assign one of
134 * those two right nodes to each of the three left nodes, and still
135 * give the three left nodes a different right node each. This is
136 * of course impossible.
138 * To prove the converse (that if every subset of left vertices
139 * satisfies the Hall condition then a perfect matching exists),
140 * consider trying to find a proper subset of the left vertices
141 * which _exactly_ satisfies the Hall condition: that is, its right
142 * neighbourhood is precisely the same size as it. If we can find
143 * such a subset, then we can split the bipartite graph into two
144 * smaller ones: one consisting of the left subset and its right
145 * neighbourhood, the other consisting of everything else. Edges
146 * from the left side of the former graph to the right side of the
147 * latter do not exist, by construction; edges from the right side
148 * of the former to the left of the latter cannot be part of any
149 * perfect matching because otherwise the left subset would not be
150 * left with enough distinct right vertices to connect to (this is
151 * exactly the same deduction used in Solo's set analysis). You can
152 * then prove (left as an exercise) that both these smaller graphs
153 * still satisfy the Hall condition, and therefore the proof will
154 * follow by induction.
156 * There's one other possibility, which is the case where _no_
157 * proper subset of the left vertices has a right neighbourhood of
158 * exactly the same size. That is, every left subset has a strictly
159 * _larger_ right neighbourhood. In this situation, we can simply
160 * remove an _arbitrary_ edge from the graph. This cannot reduce
161 * the size of any left subset's right neighbourhood by more than
162 * one, so if all neighbourhoods were strictly bigger than they
163 * needed to be initially, they must now still be _at least as big_
164 * as they need to be. So we can keep throwing out arbitrary edges
165 * until we find a set which exactly satisfies the Hall condition,
166 * and then proceed as above. []
168 * That's Hall's theorem. I now build on this by examining the
169 * circumstances in which a bipartite graph can have a _unique_
170 * perfect matching. It is clear that in the second case, where no
171 * left subset exactly satisfies the Hall condition and so we can
172 * remove an arbitrary edge, there cannot be a unique perfect
173 * matching: given one perfect matching, we choose our arbitrary
174 * removed edge to be one of those contained in it, and then we can
175 * still find a perfect matching in the remaining graph, which will
176 * be a distinct perfect matching in the original.
178 * So it is a necessary condition for a unique perfect matching
179 * that there must be at least one proper left subset which
180 * _exactly_ satisfies the Hall condition. But now consider the
181 * smaller graph constructed by taking that left subset and its
182 * neighbourhood: if the graph as a whole had a unique perfect
183 * matching, then so must this smaller one, which means we can find
184 * a proper left subset _again_, and so on. Repeating this process
185 * must eventually reduce us to a graph with only one left-side
186 * vertex (so there are no proper subsets at all); this vertex must
187 * be connected to only one right-side vertex, and hence must be so
188 * in the original graph as well (by construction). So we can
189 * discard this vertex pair from the graph, and any other edges
190 * that involved it (which will by construction be from other left
191 * vertices only), and the resulting smaller graph still has a
192 * unique perfect matching which means we can do the same thing
195 * In other words, given any bipartite graph with a unique perfect
196 * matching, we can find that matching by the following extremely
199 * - Find a left-side vertex which is only connected to one
201 * - Assign those vertices to one another, and therefore discard
202 * any other edges connecting to that right vertex.
203 * - Repeat until all vertices have been matched.
205 * This algorithm can be run in O(V+E) time (where V is the number
206 * of vertices and E is the number of edges in the graph), and the
207 * only way it can fail is if there is not a unique perfect
208 * matching (either because there is no matching at all, or because
209 * it isn't unique; but it can't distinguish those cases).
211 * Thus, the internal solver in this source file can be confident
212 * that if the tree/tent matching is uniquely determined by the
213 * tree and tent positions, it can find it using only this kind of
214 * obvious and simple operation: assign a tree to a tent if it
215 * cannot possibly belong to any other tent, and vice versa. If the
216 * solver were _only_ trying to determine the matching, even that
217 * `vice versa' wouldn't be required; but it can come in handy when
218 * not all the tents have been placed yet. I can therefore be
219 * reasonably confident that as long as my solver doesn't need to
220 * cope with grids that have a non-unique matching, it will also
221 * not need to do anything complicated like set analysis between
226 * In standalone solver mode, `verbose' is a variable which can be
227 * set by command-line option; in debugging mode it's simply always
230 #if defined STANDALONE_SOLVER
231 #define SOLVER_DIAGNOSTICS
233 #elif defined SOLVER_DIAGNOSTICS
238 * Difficulty levels. I do some macro ickery here to ensure that my
239 * enum and the various forms of my name list always match up.
241 #define DIFFLIST(A) \
244 #define ENUM(upper,title,lower) DIFF_ ## upper,
245 #define TITLE(upper,title,lower) #title,
246 #define ENCODE(upper,title,lower) #lower
247 #define CONFIG(upper,title,lower) ":" #title
248 enum { DIFFLIST(ENUM
) DIFFCOUNT
};
249 static char const *const tents_diffnames
[] = { DIFFLIST(TITLE
) };
250 static char const tents_diffchars
[] = DIFFLIST(ENCODE
);
251 #define DIFFCONFIG DIFFLIST(CONFIG)
265 enum { BLANK
, TREE
, TENT
, NONTENT
, MAGIC
};
280 struct numbers
*numbers
;
281 int completed
, used_solve
;
284 static game_params
*default_params(void)
286 game_params
*ret
= snew(game_params
);
289 ret
->diff
= DIFF_EASY
;
294 static const struct game_params tents_presets
[] = {
298 {10, 10, DIFF_TRICKY
},
300 {15, 15, DIFF_TRICKY
},
303 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
308 if (i
< 0 || i
>= lenof(tents_presets
))
311 ret
= snew(game_params
);
312 *ret
= tents_presets
[i
];
314 sprintf(str
, "%dx%d %s", ret
->w
, ret
->h
, tents_diffnames
[ret
->diff
]);
321 static void free_params(game_params
*params
)
326 static game_params
*dup_params(game_params
*params
)
328 game_params
*ret
= snew(game_params
);
329 *ret
= *params
; /* structure copy */
333 static void decode_params(game_params
*params
, char const *string
)
335 params
->w
= params
->h
= atoi(string
);
336 while (*string
&& isdigit((unsigned char)*string
)) string
++;
337 if (*string
== 'x') {
339 params
->h
= atoi(string
);
340 while (*string
&& isdigit((unsigned char)*string
)) string
++;
342 if (*string
== 'd') {
345 for (i
= 0; i
< DIFFCOUNT
; i
++)
346 if (*string
== tents_diffchars
[i
])
348 if (*string
) string
++;
352 static char *encode_params(game_params
*params
, int full
)
356 sprintf(buf
, "%dx%d", params
->w
, params
->h
);
358 sprintf(buf
+ strlen(buf
), "d%c",
359 tents_diffchars
[params
->diff
]);
363 static config_item
*game_configure(game_params
*params
)
368 ret
= snewn(4, config_item
);
370 ret
[0].name
= "Width";
371 ret
[0].type
= C_STRING
;
372 sprintf(buf
, "%d", params
->w
);
373 ret
[0].sval
= dupstr(buf
);
376 ret
[1].name
= "Height";
377 ret
[1].type
= C_STRING
;
378 sprintf(buf
, "%d", params
->h
);
379 ret
[1].sval
= dupstr(buf
);
382 ret
[2].name
= "Difficulty";
383 ret
[2].type
= C_CHOICES
;
384 ret
[2].sval
= DIFFCONFIG
;
385 ret
[2].ival
= params
->diff
;
395 static game_params
*custom_params(config_item
*cfg
)
397 game_params
*ret
= snew(game_params
);
399 ret
->w
= atoi(cfg
[0].sval
);
400 ret
->h
= atoi(cfg
[1].sval
);
401 ret
->diff
= cfg
[2].ival
;
406 static char *validate_params(game_params
*params
, int full
)
409 * Generating anything under 4x4 runs into trouble of one kind
412 if (params
->w
< 4 || params
->h
< 4)
413 return "Width and height must both be at least four";
418 * Scratch space for solver.
420 enum { N
, U
, L
, R
, D
, MAXDIR
}; /* link directions */
421 #define dx(d) ( ((d)==R) - ((d)==L) )
422 #define dy(d) ( ((d)==D) - ((d)==U) )
423 #define F(d) ( U + D - (d) )
424 struct solver_scratch
{
425 char *links
; /* mapping between trees and tents */
427 char *place
, *mrows
, *trows
;
430 static struct solver_scratch
*new_scratch(int w
, int h
)
432 struct solver_scratch
*ret
= snew(struct solver_scratch
);
434 ret
->links
= snewn(w
*h
, char);
435 ret
->locs
= snewn(max(w
, h
), int);
436 ret
->place
= snewn(max(w
, h
), char);
437 ret
->mrows
= snewn(3 * max(w
, h
), char);
438 ret
->trows
= snewn(3 * max(w
, h
), char);
443 static void free_scratch(struct solver_scratch
*sc
)
454 * Solver. Returns 0 for impossibility, 1 for success, 2 for
455 * ambiguity or failure to converge.
457 static int tents_solve(int w
, int h
, const char *grid
, int *numbers
,
458 char *soln
, struct solver_scratch
*sc
, int diff
)
461 char *mrow
, *mrow1
, *mrow2
, *trow
, *trow1
, *trow2
;
464 * Set up solver data.
466 memset(sc
->links
, N
, w
*h
);
469 * Set up solution array.
471 memcpy(soln
, grid
, w
*h
);
477 int done_something
= FALSE
;
480 * Any tent which has only one unattached tree adjacent to
481 * it can be tied to that tree.
483 for (y
= 0; y
< h
; y
++)
484 for (x
= 0; x
< w
; x
++)
485 if (soln
[y
*w
+x
] == TENT
&& !sc
->links
[y
*w
+x
]) {
488 for (d
= 1; d
< MAXDIR
; d
++) {
489 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
490 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
491 soln
[y2
*w
+x2
] == TREE
&&
492 !sc
->links
[y2
*w
+x2
]) {
494 break; /* found more than one */
500 if (d
== MAXDIR
&& linkd
== 0) {
501 #ifdef SOLVER_DIAGNOSTICS
503 printf("tent at %d,%d cannot link to anything\n",
506 return 0; /* no solution exists */
507 } else if (d
== MAXDIR
) {
508 int x2
= x
+ dx(linkd
), y2
= y
+ dy(linkd
);
510 #ifdef SOLVER_DIAGNOSTICS
512 printf("tent at %d,%d can only link to tree at"
513 " %d,%d\n", x
, y
, x2
, y2
);
516 sc
->links
[y
*w
+x
] = linkd
;
517 sc
->links
[y2
*w
+x2
] = F(linkd
);
518 done_something
= TRUE
;
525 break; /* don't do anything else! */
528 * Mark a blank square as NONTENT if it is not orthogonally
529 * adjacent to any unmatched tree.
531 for (y
= 0; y
< h
; y
++)
532 for (x
= 0; x
< w
; x
++)
533 if (soln
[y
*w
+x
] == BLANK
) {
534 int can_be_tent
= FALSE
;
536 for (d
= 1; d
< MAXDIR
; d
++) {
537 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
538 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
539 soln
[y2
*w
+x2
] == TREE
&&
545 #ifdef SOLVER_DIAGNOSTICS
547 printf("%d,%d cannot be a tent (no adjacent"
548 " unmatched tree)\n", x
, y
);
550 soln
[y
*w
+x
] = NONTENT
;
551 done_something
= TRUE
;
559 * Mark a blank square as NONTENT if it is (perhaps
560 * diagonally) adjacent to any other tent.
562 for (y
= 0; y
< h
; y
++)
563 for (x
= 0; x
< w
; x
++)
564 if (soln
[y
*w
+x
] == BLANK
) {
565 int dx
, dy
, imposs
= FALSE
;
567 for (dy
= -1; dy
<= +1; dy
++)
568 for (dx
= -1; dx
<= +1; dx
++)
570 int x2
= x
+ dx
, y2
= y
+ dy
;
571 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
572 soln
[y2
*w
+x2
] == TENT
)
577 #ifdef SOLVER_DIAGNOSTICS
579 printf("%d,%d cannot be a tent (adjacent tent)\n",
582 soln
[y
*w
+x
] = NONTENT
;
583 done_something
= TRUE
;
591 * Any tree which has exactly one {unattached tent, BLANK}
592 * adjacent to it must have its tent in that square.
594 for (y
= 0; y
< h
; y
++)
595 for (x
= 0; x
< w
; x
++)
596 if (soln
[y
*w
+x
] == TREE
&& !sc
->links
[y
*w
+x
]) {
597 int linkd
= 0, linkd2
= 0, nd
= 0;
599 for (d
= 1; d
< MAXDIR
; d
++) {
600 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
601 if (!(x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
))
603 if (soln
[y2
*w
+x2
] == BLANK
||
604 (soln
[y2
*w
+x2
] == TENT
&& !sc
->links
[y2
*w
+x2
])) {
614 #ifdef SOLVER_DIAGNOSTICS
616 printf("tree at %d,%d cannot link to anything\n",
619 return 0; /* no solution exists */
620 } else if (nd
== 1) {
621 int x2
= x
+ dx(linkd
), y2
= y
+ dy(linkd
);
623 #ifdef SOLVER_DIAGNOSTICS
625 printf("tree at %d,%d can only link to tent at"
626 " %d,%d\n", x
, y
, x2
, y2
);
628 soln
[y2
*w
+x2
] = TENT
;
629 sc
->links
[y
*w
+x
] = linkd
;
630 sc
->links
[y2
*w
+x2
] = F(linkd
);
631 done_something
= TRUE
;
632 } else if (nd
== 2 && (!dx(linkd
) != !dx(linkd2
)) &&
633 diff
>= DIFF_TRICKY
) {
635 * If there are two possible places where
636 * this tree's tent can go, and they are
637 * diagonally separated rather than being
638 * on opposite sides of the tree, then the
639 * square (other than the tree square)
640 * which is adjacent to both of them must
643 int x2
= x
+ dx(linkd
) + dx(linkd2
);
644 int y2
= y
+ dy(linkd
) + dy(linkd2
);
645 assert(x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
);
646 if (soln
[y2
*w
+x2
] == BLANK
) {
647 #ifdef SOLVER_DIAGNOSTICS
649 printf("possible tent locations for tree at"
650 " %d,%d rule out tent at %d,%d\n",
653 soln
[y2
*w
+x2
] = NONTENT
;
654 done_something
= TRUE
;
663 * If localised deductions about the trees and tents
664 * themselves haven't helped us, it's time to resort to the
665 * numbers round the grid edge. For each row and column, we
666 * go through all possible combinations of locations for
667 * the unplaced tents, rule out any which have adjacent
668 * tents, and spot any square which is given the same state
669 * by all remaining combinations.
671 for (i
= 0; i
< w
+h
; i
++) {
672 int start
, step
, len
, start1
, start2
, n
, k
;
676 * This is the number for a column.
691 * This is the number for a row.
706 if (diff
< DIFF_TRICKY
) {
708 * In Easy mode, we don't look at the effect of one
709 * row on the next (i.e. ruling out a square if all
710 * possibilities for an adjacent row place a tent
713 start1
= start2
= -1;
719 * Count and store the locations of the free squares,
720 * and also count the number of tents already placed.
723 for (j
= 0; j
< len
; j
++) {
724 if (soln
[start
+j
*step
] == TENT
)
725 k
--; /* one fewer tent to place */
726 else if (soln
[start
+j
*step
] == BLANK
)
731 continue; /* nothing left to do here */
734 * Now we know we're placing k tents in n squares. Set
735 * up the first possibility.
737 for (j
= 0; j
< n
; j
++)
738 sc
->place
[j
] = (j
< k ? TENT
: NONTENT
);
741 * We're aiming to find squares in this row which are
742 * invariant over all valid possibilities. Thus, we
743 * maintain the current state of that invariance. We
744 * start everything off at MAGIC to indicate that it
745 * hasn't been set up yet.
748 mrow1
= sc
->mrows
+ len
;
749 mrow2
= sc
->mrows
+ 2*len
;
751 trow1
= sc
->trows
+ len
;
752 trow2
= sc
->trows
+ 2*len
;
753 memset(mrow
, MAGIC
, 3*len
);
756 * And iterate over all possibilities.
762 * See if this possibility is valid. The only way
763 * it can fail to be valid is if it contains two
764 * adjacent tents. (Other forms of invalidity, such
765 * as containing a tent adjacent to one already
766 * placed, will have been dealt with already by
767 * other parts of the solver.)
770 for (j
= 0; j
+1 < n
; j
++)
771 if (sc
->place
[j
] == TENT
&&
772 sc
->place
[j
+1] == TENT
&&
773 sc
->locs
[j
+1] == sc
->locs
[j
]+1) {
780 * Merge this valid combination into mrow.
782 memset(trow
, MAGIC
, len
);
783 memset(trow
+len
, BLANK
, 2*len
);
784 for (j
= 0; j
< n
; j
++) {
785 trow
[sc
->locs
[j
]] = sc
->place
[j
];
786 if (sc
->place
[j
] == TENT
) {
788 for (jj
= sc
->locs
[j
]-1; jj
<= sc
->locs
[j
]+1; jj
++)
789 if (jj
>= 0 && jj
< len
)
790 trow1
[jj
] = trow2
[jj
] = NONTENT
;
794 for (j
= 0; j
< 3*len
; j
++) {
795 if (trow
[j
] == MAGIC
)
797 if (mrow
[j
] == MAGIC
|| mrow
[j
] == trow
[j
]) {
799 * Either this is the first valid
800 * placement we've found at all, or
801 * this square's contents are
802 * consistent with every previous valid
808 * This square's contents fail to match
809 * what they were in a different
810 * combination, so we cannot deduce
811 * anything about this square.
819 * Find the next combination of k choices from n.
820 * We do this by finding the rightmost tent which
821 * can be moved one place right, doing so, and
822 * shunting all tents to the right of that as far
823 * left as they can go.
826 for (j
= n
-1; j
> 0; j
--) {
827 if (sc
->place
[j
] == TENT
)
829 if (sc
->place
[j
] == NONTENT
&& sc
->place
[j
-1] == TENT
) {
830 sc
->place
[j
-1] = NONTENT
;
833 sc
->place
[++j
] = TENT
;
835 sc
->place
[j
] = NONTENT
;
840 break; /* we've finished */
844 * It's just possible that _no_ placement was valid, in
845 * which case we have an internally inconsistent
848 if (mrow
[sc
->locs
[0]] == MAGIC
)
849 return 0; /* inconsistent */
852 * Now go through mrow and see if there's anything
853 * we've deduced which wasn't already mentioned in soln.
855 for (j
= 0; j
< len
; j
++) {
858 for (whichrow
= 0; whichrow
< 3; whichrow
++) {
859 char *mthis
= mrow
+ whichrow
* len
;
860 int tstart
= (whichrow
== 0 ? start
:
861 whichrow
== 1 ? start1
: start2
);
863 mthis
[j
] != MAGIC
&& mthis
[j
] != BLANK
&&
864 soln
[tstart
+j
*step
] == BLANK
) {
865 int pos
= tstart
+j
*step
;
867 #ifdef SOLVER_DIAGNOSTICS
869 printf("%s %d forces %s at %d,%d\n",
870 step
==1 ?
"row" : "column",
871 step
==1 ? start
/w
: start
,
872 mthis
[j
] == TENT ?
"tent" : "non-tent",
875 soln
[pos
] = mthis
[j
];
876 done_something
= TRUE
;
890 * The solver has nothing further it can do. Return 1 if both
891 * soln and sc->links are completely filled in, or 2 otherwise.
893 for (y
= 0; y
< h
; y
++)
894 for (x
= 0; x
< w
; x
++) {
895 if (soln
[y
*w
+x
] == BLANK
)
897 if (soln
[y
*w
+x
] != NONTENT
&& sc
->links
[y
*w
+x
] == 0)
904 static char *new_game_desc(game_params
*params
, random_state
*rs
,
905 char **aux
, int interactive
)
907 int w
= params
->w
, h
= params
->h
;
908 int ntrees
= w
* h
/ 5;
909 char *grid
= snewn(w
*h
, char);
910 char *puzzle
= snewn(w
*h
, char);
911 int *numbers
= snewn(w
+h
, int);
912 char *soln
= snewn(w
*h
, char);
913 int *temp
= snewn(2*w
*h
, int);
914 int maxedges
= ntrees
*4 + w
*h
;
915 int *edges
= snewn(2*maxedges
, int);
916 int *capacity
= snewn(maxedges
, int);
917 int *flow
= snewn(maxedges
, int);
918 struct solver_scratch
*sc
= new_scratch(w
, h
);
923 * Since this puzzle has many global deductions and doesn't
924 * permit limited clue sets, generating grids for this puzzle
925 * is hard enough that I see no better option than to simply
926 * generate a solution and see if it's unique and has the
927 * required difficulty. This turns out to be computationally
930 * We chose our tree count (hence also tent count) by dividing
931 * the total grid area by five above. Why five? Well, w*h/4 is
932 * the maximum number of tents you can _possibly_ fit into the
933 * grid without violating the separation criterion, and to
934 * achieve that you are constrained to a very small set of
935 * possible layouts (the obvious one with a tent at every
936 * (even,even) coordinate, and trivial variations thereon). So
937 * if we reduce the tent count a bit more, we enable more
938 * random-looking placement; 5 turns out to be a plausible
939 * figure which yields sensible puzzles. Increasing the tent
940 * count would give puzzles whose solutions were too regimented
941 * and could be solved by the use of that knowledge (and would
942 * also take longer to find a viable placement); decreasing it
943 * would make the grids emptier and more boring.
945 * Actually generating a grid is a matter of first placing the
946 * tents, and then placing the trees by the use of maxflow
947 * (finding a distinct square adjacent to every tent). We do it
948 * this way round because otherwise satisfying the tent
949 * separation condition would become onerous: most randomly
950 * chosen tent layouts do not satisfy this condition, so we'd
951 * have gone to a lot of work before finding that a candidate
952 * layout was unusable. Instead, we place the tents first and
953 * ensure they meet the separation criterion _before_ doing
954 * lots of computation; this works much better.
956 * The maxflow algorithm is not randomised, so employed naively
957 * it would give rise to grids with clear structure and
958 * directional bias. Hence, I assign the network nodes as seen
959 * by maxflow to be a _random_ permutation of the squares of
960 * the grid, so that any bias shown by maxflow towards
961 * low-numbered nodes is turned into a random bias.
963 * This generation strategy can fail at many points, including
964 * as early as tent placement (if you get a bad random order in
965 * which to greedily try the grid squares, you won't even
966 * manage to find enough mutually non-adjacent squares to put
967 * the tents in). Then it can fail if maxflow doesn't manage to
968 * find a good enough matching (i.e. the tent placements don't
969 * admit any adequate tree placements); and finally it can fail
970 * if the solver finds that the problem has the wrong
971 * difficulty (including being actually non-unique). All of
972 * these, however, are insufficiently frequent to cause
976 if (params
->diff
> DIFF_EASY
&& params
->w
<= 4 && params
->h
<= 4)
977 params
->diff
= DIFF_EASY
; /* downgrade to prevent tight loop */
981 * Arrange the grid squares into a random order.
983 for (i
= 0; i
< w
*h
; i
++)
985 shuffle(temp
, w
*h
, sizeof(*temp
), rs
);
988 * The first `ntrees' entries in temp which we can get
989 * without making two tents adjacent will be the tent
992 memset(grid
, BLANK
, w
*h
);
994 for (i
= 0; i
< w
*h
&& j
> 0; i
++) {
995 int x
= temp
[i
] % w
, y
= temp
[i
] / w
;
996 int dy
, dx
, ok
= TRUE
;
998 for (dy
= -1; dy
<= +1; dy
++)
999 for (dx
= -1; dx
<= +1; dx
++)
1000 if (x
+dx
>= 0 && x
+dx
< w
&&
1001 y
+dy
>= 0 && y
+dy
< h
&&
1002 grid
[(y
+dy
)*w
+(x
+dx
)] == TENT
)
1006 grid
[temp
[i
]] = TENT
;
1011 continue; /* couldn't place all the tents */
1014 * Now we build up the list of graph edges.
1017 for (i
= 0; i
< w
*h
; i
++) {
1018 if (grid
[temp
[i
]] == TENT
) {
1019 for (j
= 0; j
< w
*h
; j
++) {
1020 if (grid
[temp
[j
]] != TENT
) {
1021 int xi
= temp
[i
] % w
, yi
= temp
[i
] / w
;
1022 int xj
= temp
[j
] % w
, yj
= temp
[j
] / w
;
1023 if (abs(xi
-xj
) + abs(yi
-yj
) == 1) {
1024 edges
[nedges
*2] = i
;
1025 edges
[nedges
*2+1] = j
;
1026 capacity
[nedges
] = 1;
1033 * Special node w*h is the sink node; any non-tent node
1034 * has an edge going to it.
1036 edges
[nedges
*2] = i
;
1037 edges
[nedges
*2+1] = w
*h
;
1038 capacity
[nedges
] = 1;
1044 * Special node w*h+1 is the source node, with an edge going to
1047 for (i
= 0; i
< w
*h
; i
++) {
1048 if (grid
[temp
[i
]] == TENT
) {
1049 edges
[nedges
*2] = w
*h
+1;
1050 edges
[nedges
*2+1] = i
;
1051 capacity
[nedges
] = 1;
1056 assert(nedges
<= maxedges
);
1059 * Now we're ready to call the maxflow algorithm to place the
1062 j
= maxflow(w
*h
+2, w
*h
+1, w
*h
, nedges
, edges
, capacity
, flow
, NULL
);
1065 continue; /* couldn't place all the tents */
1068 * We've placed the trees. Now we need to work out _where_
1069 * we've placed them, which is a matter of reading back out
1070 * from the `flow' array.
1072 for (i
= 0; i
< nedges
; i
++) {
1073 if (edges
[2*i
] < w
*h
&& edges
[2*i
+1] < w
*h
&& flow
[i
] > 0)
1074 grid
[temp
[edges
[2*i
+1]]] = TREE
;
1078 * I think it looks ugly if there isn't at least one of
1079 * _something_ (tent or tree) in each row and each column
1080 * of the grid. This doesn't give any information away
1081 * since a completely empty row/column is instantly obvious
1082 * from the clues (it has no trees and a zero).
1084 for (i
= 0; i
< w
; i
++) {
1085 for (j
= 0; j
< h
; j
++) {
1086 if (grid
[j
*w
+i
] != BLANK
)
1087 break; /* found something in this column */
1090 break; /* found empty column */
1093 continue; /* a column was empty */
1095 for (j
= 0; j
< h
; j
++) {
1096 for (i
= 0; i
< w
; i
++) {
1097 if (grid
[j
*w
+i
] != BLANK
)
1098 break; /* found something in this row */
1101 break; /* found empty row */
1104 continue; /* a row was empty */
1107 * Now set up the numbers round the edge.
1109 for (i
= 0; i
< w
; i
++) {
1111 for (j
= 0; j
< h
; j
++)
1112 if (grid
[j
*w
+i
] == TENT
)
1116 for (i
= 0; i
< h
; i
++) {
1118 for (j
= 0; j
< w
; j
++)
1119 if (grid
[i
*w
+j
] == TENT
)
1125 * And now actually solve the puzzle, to see whether it's
1126 * unique and has the required difficulty.
1128 for (i
= 0; i
< w
*h
; i
++)
1129 puzzle
[i
] = grid
[i
] == TREE ? TREE
: BLANK
;
1130 i
= tents_solve(w
, h
, puzzle
, numbers
, soln
, sc
, params
->diff
-1);
1131 j
= tents_solve(w
, h
, puzzle
, numbers
, soln
, sc
, params
->diff
);
1134 * We expect solving with difficulty params->diff to have
1135 * succeeded (otherwise the problem is too hard), and
1136 * solving with diff-1 to have failed (otherwise it's too
1139 if (i
== 2 && j
== 1)
1144 * That's it. Encode as a game ID.
1146 ret
= snewn((w
+h
)*40 + ntrees
+ (w
*h
)/26 + 1, char);
1149 for (i
= 0; i
<= w
*h
; i
++) {
1150 int c
= (i
< w
*h ? grid
[i
] == TREE
: 1);
1152 *p
++ = (j
== 0 ?
'_' : j
-1 + 'a');
1162 for (i
= 0; i
< w
+h
; i
++)
1163 p
+= sprintf(p
, ",%d", numbers
[i
]);
1165 ret
= sresize(ret
, p
- ret
, char);
1168 * And encode the solution as an aux_info.
1170 *aux
= snewn(ntrees
* 40, char);
1173 for (i
= 0; i
< w
*h
; i
++)
1174 if (grid
[i
] == TENT
)
1175 p
+= sprintf(p
, ";T%d,%d", i
%w
, i
/w
);
1177 *aux
= sresize(*aux
, p
- *aux
, char);
1192 static char *validate_desc(game_params
*params
, char *desc
)
1194 int w
= params
->w
, h
= params
->h
;
1198 while (*desc
&& *desc
!= ',') {
1201 else if (*desc
>= 'a' && *desc
< 'z')
1202 area
+= *desc
- 'a' + 2;
1203 else if (*desc
== 'z')
1205 else if (*desc
== '!' || *desc
== '-')
1208 return "Invalid character in grid specification";
1213 for (i
= 0; i
< w
+h
; i
++) {
1215 return "Not enough numbers given after grid specification";
1216 else if (*desc
!= ',')
1217 return "Invalid character in number list";
1219 while (*desc
&& isdigit((unsigned char)*desc
)) desc
++;
1223 return "Unexpected additional data at end of game description";
1227 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1229 int w
= params
->w
, h
= params
->h
;
1230 game_state
*state
= snew(game_state
);
1233 state
->p
= *params
; /* structure copy */
1234 state
->grid
= snewn(w
*h
, char);
1235 state
->numbers
= snew(struct numbers
);
1236 state
->numbers
->refcount
= 1;
1237 state
->numbers
->numbers
= snewn(w
+h
, int);
1238 state
->completed
= state
->used_solve
= FALSE
;
1241 memset(state
->grid
, BLANK
, w
*h
);
1250 else if (*desc
>= 'a' && *desc
< 'z')
1251 run
= *desc
- ('a'-1);
1252 else if (*desc
== 'z') {
1256 assert(*desc
== '!' || *desc
== '-');
1258 type
= (*desc
== '!' ? TENT
: NONTENT
);
1264 assert(i
>= 0 && i
<= w
*h
);
1266 assert(type
== TREE
);
1270 state
->grid
[i
++] = type
;
1274 for (i
= 0; i
< w
+h
; i
++) {
1275 assert(*desc
== ',');
1277 state
->numbers
->numbers
[i
] = atoi(desc
);
1278 while (*desc
&& isdigit((unsigned char)*desc
)) desc
++;
1286 static game_state
*dup_game(game_state
*state
)
1288 int w
= state
->p
.w
, h
= state
->p
.h
;
1289 game_state
*ret
= snew(game_state
);
1291 ret
->p
= state
->p
; /* structure copy */
1292 ret
->grid
= snewn(w
*h
, char);
1293 memcpy(ret
->grid
, state
->grid
, w
*h
);
1294 ret
->numbers
= state
->numbers
;
1295 state
->numbers
->refcount
++;
1296 ret
->completed
= state
->completed
;
1297 ret
->used_solve
= state
->used_solve
;
1302 static void free_game(game_state
*state
)
1304 if (--state
->numbers
->refcount
<= 0) {
1305 sfree(state
->numbers
->numbers
);
1306 sfree(state
->numbers
);
1312 static char *solve_game(game_state
*state
, game_state
*currstate
,
1313 char *aux
, char **error
)
1315 int w
= state
->p
.w
, h
= state
->p
.h
;
1319 * If we already have the solution, save ourselves some
1324 struct solver_scratch
*sc
= new_scratch(w
, h
);
1330 soln
= snewn(w
*h
, char);
1331 ret
= tents_solve(w
, h
, state
->grid
, state
->numbers
->numbers
,
1332 soln
, sc
, DIFFCOUNT
-1);
1337 *error
= "This puzzle is not self-consistent";
1339 *error
= "Unable to find a unique solution for this puzzle";
1344 * Construct a move string which turns the current state
1345 * into the solved state.
1347 move
= snewn(w
*h
* 40, char);
1350 for (i
= 0; i
< w
*h
; i
++)
1351 if (soln
[i
] == TENT
)
1352 p
+= sprintf(p
, ";T%d,%d", i
%w
, i
/w
);
1354 move
= sresize(move
, p
- move
, char);
1362 static int game_can_format_as_text_now(game_params
*params
)
1367 static char *game_text_format(game_state
*state
)
1369 int w
= state
->p
.w
, h
= state
->p
.h
;
1374 * FIXME: We currently do not print the numbers round the edges
1375 * of the grid. I need to work out a sensible way of doing this
1376 * even when the column numbers exceed 9.
1378 * In the absence of those numbers, the result size is h lines
1379 * of w+1 characters each, plus a NUL.
1381 * This function is currently only used by the standalone
1382 * solver; until I make it look more sensible, I won't enable
1383 * it in the main game structure.
1385 ret
= snewn(h
*(w
+1) + 1, char);
1387 for (y
= 0; y
< h
; y
++) {
1388 for (x
= 0; x
< w
; x
++) {
1389 *p
= (state
->grid
[y
*w
+x
] == BLANK ?
'.' :
1390 state
->grid
[y
*w
+x
] == TREE ?
'T' :
1391 state
->grid
[y
*w
+x
] == TENT ?
'*' :
1392 state
->grid
[y
*w
+x
] == NONTENT ?
'-' : '?');
1403 int dsx
, dsy
; /* coords of drag start */
1404 int dex
, dey
; /* coords of drag end */
1405 int drag_button
; /* -1 for none, or a button code */
1406 int drag_ok
; /* dragged off the window, to cancel */
1408 int cx
, cy
, cdisp
; /* cursor position, and ?display. */
1411 static game_ui
*new_ui(game_state
*state
)
1413 game_ui
*ui
= snew(game_ui
);
1414 ui
->dsx
= ui
->dsy
= -1;
1415 ui
->dex
= ui
->dey
= -1;
1416 ui
->drag_button
= -1;
1417 ui
->drag_ok
= FALSE
;
1418 ui
->cx
= ui
->cy
= ui
->cdisp
= 0;
1422 static void free_ui(game_ui
*ui
)
1427 static char *encode_ui(game_ui
*ui
)
1432 static void decode_ui(game_ui
*ui
, char *encoding
)
1436 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
1437 game_state
*newstate
)
1441 struct game_drawstate
{
1445 int *drawn
, *numbersdrawn
;
1446 int cx
, cy
; /* last-drawn cursor pos, or (-1,-1) if absent. */
1449 #define PREFERRED_TILESIZE 32
1450 #define TILESIZE (ds->tilesize)
1451 #define TLBORDER (TILESIZE/2)
1452 #define BRBORDER (TILESIZE*3/2)
1453 #define COORD(x) ( (x) * TILESIZE + TLBORDER )
1454 #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 )
1456 #define FLASH_TIME 0.30F
1458 static int drag_xform(game_ui
*ui
, int x
, int y
, int v
)
1460 int xmin
, ymin
, xmax
, ymax
;
1462 xmin
= min(ui
->dsx
, ui
->dex
);
1463 xmax
= max(ui
->dsx
, ui
->dex
);
1464 ymin
= min(ui
->dsy
, ui
->dey
);
1465 ymax
= max(ui
->dsy
, ui
->dey
);
1468 * Left-dragging has no effect, so we treat a left-drag as a
1469 * single click on dsx,dsy.
1471 if (ui
->drag_button
== LEFT_BUTTON
) {
1472 xmin
= xmax
= ui
->dsx
;
1473 ymin
= ymax
= ui
->dsy
;
1476 if (x
< xmin
|| x
> xmax
|| y
< ymin
|| y
> ymax
)
1477 return v
; /* no change outside drag area */
1480 return v
; /* trees are inviolate always */
1482 if (xmin
== xmax
&& ymin
== ymax
) {
1484 * Results of a simple click. Left button sets blanks to
1485 * tents; right button sets blanks to non-tents; either
1486 * button clears a non-blank square.
1488 if (ui
->drag_button
== LEFT_BUTTON
)
1489 v
= (v
== BLANK ? TENT
: BLANK
);
1491 v
= (v
== BLANK ? NONTENT
: BLANK
);
1494 * Results of a drag. Left-dragging has no effect.
1495 * Right-dragging sets all blank squares to non-tents and
1496 * has no effect on anything else.
1498 if (ui
->drag_button
== RIGHT_BUTTON
)
1499 v
= (v
== BLANK ? NONTENT
: v
);
1507 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
1508 int x
, int y
, int button
)
1510 int w
= state
->p
.w
, h
= state
->p
.h
;
1513 if (button
== LEFT_BUTTON
|| button
== RIGHT_BUTTON
) {
1516 if (x
< 0 || y
< 0 || x
>= w
|| y
>= h
)
1519 ui
->drag_button
= button
;
1520 ui
->dsx
= ui
->dex
= x
;
1521 ui
->dsy
= ui
->dey
= y
;
1524 return ""; /* ui updated */
1527 if ((IS_MOUSE_DRAG(button
) || IS_MOUSE_RELEASE(button
)) &&
1528 ui
->drag_button
> 0) {
1529 int xmin
, ymin
, xmax
, ymax
;
1531 int buflen
, bufsize
, tmplen
;
1535 if (x
< 0 || y
< 0 || x
>= w
|| y
>= h
) {
1536 ui
->drag_ok
= FALSE
;
1539 * Drags are limited to one row or column. Hence, we
1540 * work out which coordinate is closer to the drag
1541 * start, and move it _to_ the drag start.
1543 if (abs(x
- ui
->dsx
) < abs(y
- ui
->dsy
))
1554 if (IS_MOUSE_DRAG(button
))
1555 return ""; /* ui updated */
1558 * The drag has been released. Enact it.
1561 ui
->drag_button
= -1;
1562 return ""; /* drag was just cancelled */
1565 xmin
= min(ui
->dsx
, ui
->dex
);
1566 xmax
= max(ui
->dsx
, ui
->dex
);
1567 ymin
= min(ui
->dsy
, ui
->dey
);
1568 ymax
= max(ui
->dsy
, ui
->dey
);
1569 assert(0 <= xmin
&& xmin
<= xmax
&& xmax
< w
);
1570 assert(0 <= ymin
&& ymin
<= ymax
&& ymax
< h
);
1574 buf
= snewn(bufsize
, char);
1576 for (y
= ymin
; y
<= ymax
; y
++)
1577 for (x
= xmin
; x
<= xmax
; x
++) {
1578 int v
= drag_xform(ui
, x
, y
, state
->grid
[y
*w
+x
]);
1579 if (state
->grid
[y
*w
+x
] != v
) {
1580 tmplen
= sprintf(tmpbuf
, "%s%c%d,%d", sep
,
1581 (int)(v
== BLANK ?
'B' :
1582 v
== TENT ?
'T' : 'N'),
1586 if (buflen
+ tmplen
>= bufsize
) {
1587 bufsize
= buflen
+ tmplen
+ 256;
1588 buf
= sresize(buf
, bufsize
, char);
1591 strcpy(buf
+buflen
, tmpbuf
);
1596 ui
->drag_button
= -1; /* drag is terminated */
1600 return ""; /* ui updated (drag was terminated) */
1607 if (IS_CURSOR_MOVE(button
)) {
1608 move_cursor(button
, &ui
->cx
, &ui
->cy
, w
, h
, 0);
1614 int v
= state
->grid
[ui
->cy
*w
+ui
->cx
];
1617 #ifdef SINGLE_CURSOR_SELECT
1618 if (button
== CURSOR_SELECT
)
1619 /* SELECT cycles T, N, B */
1620 rep
= v
== BLANK ?
'T' : v
== TENT ?
'N' : 'B';
1622 if (button
== CURSOR_SELECT
)
1623 rep
= v
== BLANK ?
'T' : 'B';
1624 else if (button
== CURSOR_SELECT2
)
1625 rep
= v
== BLANK ?
'N' : 'B';
1626 else if (button
== 'T' || button
== 'N' || button
== 'B')
1632 sprintf(tmpbuf
, "%c%d,%d", (int)rep
, ui
->cx
, ui
->cy
);
1633 return dupstr(tmpbuf
);
1635 } else if (IS_CURSOR_SELECT(button
)) {
1643 static game_state
*execute_move(game_state
*state
, char *move
)
1645 int w
= state
->p
.w
, h
= state
->p
.h
;
1647 int x
, y
, m
, n
, i
, j
;
1648 game_state
*ret
= dup_game(state
);
1654 ret
->used_solve
= TRUE
;
1656 * Set all non-tree squares to NONTENT. The rest of the
1657 * solve move will fill the tents in over the top.
1659 for (i
= 0; i
< w
*h
; i
++)
1660 if (ret
->grid
[i
] != TREE
)
1661 ret
->grid
[i
] = NONTENT
;
1663 } else if (c
== 'B' || c
== 'T' || c
== 'N') {
1665 if (sscanf(move
, "%d,%d%n", &x
, &y
, &n
) != 2 ||
1666 x
< 0 || y
< 0 || x
>= w
|| y
>= h
) {
1670 if (ret
->grid
[y
*w
+x
] == TREE
) {
1674 ret
->grid
[y
*w
+x
] = (c
== 'B' ? BLANK
: c
== 'T' ? TENT
: NONTENT
);
1689 * Check for completion.
1691 for (i
= n
= m
= 0; i
< w
*h
; i
++) {
1692 if (ret
->grid
[i
] == TENT
)
1694 else if (ret
->grid
[i
] == TREE
)
1698 int nedges
, maxedges
, *edges
, *capacity
, *flow
;
1701 * We have the right number of tents, which is a
1702 * precondition for the game being complete. Now check that
1703 * the numbers add up.
1705 for (i
= 0; i
< w
; i
++) {
1707 for (j
= 0; j
< h
; j
++)
1708 if (ret
->grid
[j
*w
+i
] == TENT
)
1710 if (ret
->numbers
->numbers
[i
] != n
)
1711 goto completion_check_done
;
1713 for (i
= 0; i
< h
; i
++) {
1715 for (j
= 0; j
< w
; j
++)
1716 if (ret
->grid
[i
*w
+j
] == TENT
)
1718 if (ret
->numbers
->numbers
[w
+i
] != n
)
1719 goto completion_check_done
;
1722 * Also, check that no two tents are adjacent.
1724 for (y
= 0; y
< h
; y
++)
1725 for (x
= 0; x
< w
; x
++) {
1727 ret
->grid
[y
*w
+x
] == TENT
&& ret
->grid
[y
*w
+x
+1] == TENT
)
1728 goto completion_check_done
;
1730 ret
->grid
[y
*w
+x
] == TENT
&& ret
->grid
[(y
+1)*w
+x
] == TENT
)
1731 goto completion_check_done
;
1732 if (x
+1 < w
&& y
+1 < h
) {
1733 if (ret
->grid
[y
*w
+x
] == TENT
&&
1734 ret
->grid
[(y
+1)*w
+(x
+1)] == TENT
)
1735 goto completion_check_done
;
1736 if (ret
->grid
[(y
+1)*w
+x
] == TENT
&&
1737 ret
->grid
[y
*w
+(x
+1)] == TENT
)
1738 goto completion_check_done
;
1743 * OK; we have the right number of tents, they match the
1744 * numeric clues, and they satisfy the non-adjacency
1745 * criterion. Finally, we need to verify that they can be
1746 * placed in a one-to-one matching with the trees such that
1747 * every tent is orthogonally adjacent to its tree.
1749 * This bit is where the hard work comes in: we have to do
1750 * it by finding such a matching using maxflow.
1752 * So we construct a network with one special source node,
1753 * one special sink node, one node per tent, and one node
1757 edges
= snewn(2 * maxedges
, int);
1758 capacity
= snewn(maxedges
, int);
1759 flow
= snewn(maxedges
, int);
1764 * 0..w*h trees/tents
1768 for (y
= 0; y
< h
; y
++)
1769 for (x
= 0; x
< w
; x
++)
1770 if (ret
->grid
[y
*w
+x
] == TREE
) {
1774 * Here we use the direction enum declared for
1775 * the solver. We make use of the fact that the
1776 * directions are declared in the order
1777 * U,L,R,D, meaning that we go through the four
1778 * neighbours of any square in numerically
1781 for (d
= 1; d
< MAXDIR
; d
++) {
1782 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
1783 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
1784 ret
->grid
[y2
*w
+x2
] == TENT
) {
1785 assert(nedges
< maxedges
);
1786 edges
[nedges
*2] = y
*w
+x
;
1787 edges
[nedges
*2+1] = y2
*w
+x2
;
1788 capacity
[nedges
] = 1;
1792 } else if (ret
->grid
[y
*w
+x
] == TENT
) {
1793 assert(nedges
< maxedges
);
1794 edges
[nedges
*2] = y
*w
+x
;
1795 edges
[nedges
*2+1] = w
*h
+1; /* edge going to sink */
1796 capacity
[nedges
] = 1;
1799 for (y
= 0; y
< h
; y
++)
1800 for (x
= 0; x
< w
; x
++)
1801 if (ret
->grid
[y
*w
+x
] == TREE
) {
1802 assert(nedges
< maxedges
);
1803 edges
[nedges
*2] = w
*h
; /* edge coming from source */
1804 edges
[nedges
*2+1] = y
*w
+x
;
1805 capacity
[nedges
] = 1;
1808 n
= maxflow(w
*h
+2, w
*h
, w
*h
+1, nedges
, edges
, capacity
, flow
, NULL
);
1815 goto completion_check_done
;
1818 * We haven't managed to fault the grid on any count. Score!
1820 ret
->completed
= TRUE
;
1822 completion_check_done
:
1827 /* ----------------------------------------------------------------------
1831 static void game_compute_size(game_params
*params
, int tilesize
,
1834 /* fool the macros */
1835 struct dummy
{ int tilesize
; } dummy
, *ds
= &dummy
;
1836 dummy
.tilesize
= tilesize
;
1838 *x
= TLBORDER
+ BRBORDER
+ TILESIZE
* params
->w
;
1839 *y
= TLBORDER
+ BRBORDER
+ TILESIZE
* params
->h
;
1842 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
1843 game_params
*params
, int tilesize
)
1845 ds
->tilesize
= tilesize
;
1848 static float *game_colours(frontend
*fe
, int *ncolours
)
1850 float *ret
= snewn(3 * NCOLOURS
, float);
1852 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
1854 ret
[COL_GRID
* 3 + 0] = 0.0F
;
1855 ret
[COL_GRID
* 3 + 1] = 0.0F
;
1856 ret
[COL_GRID
* 3 + 2] = 0.0F
;
1858 ret
[COL_GRASS
* 3 + 0] = 0.7F
;
1859 ret
[COL_GRASS
* 3 + 1] = 1.0F
;
1860 ret
[COL_GRASS
* 3 + 2] = 0.5F
;
1862 ret
[COL_TREETRUNK
* 3 + 0] = 0.6F
;
1863 ret
[COL_TREETRUNK
* 3 + 1] = 0.4F
;
1864 ret
[COL_TREETRUNK
* 3 + 2] = 0.0F
;
1866 ret
[COL_TREELEAF
* 3 + 0] = 0.0F
;
1867 ret
[COL_TREELEAF
* 3 + 1] = 0.7F
;
1868 ret
[COL_TREELEAF
* 3 + 2] = 0.0F
;
1870 ret
[COL_TENT
* 3 + 0] = 0.8F
;
1871 ret
[COL_TENT
* 3 + 1] = 0.7F
;
1872 ret
[COL_TENT
* 3 + 2] = 0.0F
;
1874 ret
[COL_ERROR
* 3 + 0] = 1.0F
;
1875 ret
[COL_ERROR
* 3 + 1] = 0.0F
;
1876 ret
[COL_ERROR
* 3 + 2] = 0.0F
;
1878 ret
[COL_ERRTEXT
* 3 + 0] = 1.0F
;
1879 ret
[COL_ERRTEXT
* 3 + 1] = 1.0F
;
1880 ret
[COL_ERRTEXT
* 3 + 2] = 1.0F
;
1882 *ncolours
= NCOLOURS
;
1886 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
1888 int w
= state
->p
.w
, h
= state
->p
.h
;
1889 struct game_drawstate
*ds
= snew(struct game_drawstate
);
1893 ds
->started
= FALSE
;
1894 ds
->p
= state
->p
; /* structure copy */
1895 ds
->drawn
= snewn(w
*h
, int);
1896 for (i
= 0; i
< w
*h
; i
++)
1897 ds
->drawn
[i
] = MAGIC
;
1898 ds
->numbersdrawn
= snewn(w
+h
, int);
1899 for (i
= 0; i
< w
+h
; i
++)
1900 ds
->numbersdrawn
[i
] = 2;
1901 ds
->cx
= ds
->cy
= -1;
1906 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
1909 sfree(ds
->numbersdrawn
);
1914 ERR_ADJ_TOPLEFT
= 4,
1925 static int *find_errors(game_state
*state
, char *grid
)
1927 int w
= state
->p
.w
, h
= state
->p
.h
;
1928 int *ret
= snewn(w
*h
+ w
+ h
, int);
1929 int *tmp
= snewn(w
*h
*2, int), *dsf
= tmp
+ w
*h
;
1933 * ret[0] through to ret[w*h-1] give error markers for the grid
1934 * squares. After that, ret[w*h] to ret[w*h+w-1] give error
1935 * markers for the column numbers, and ret[w*h+w] to
1936 * ret[w*h+w+h-1] for the row numbers.
1940 * Spot tent-adjacency violations.
1942 for (x
= 0; x
< w
*h
; x
++)
1944 for (y
= 0; y
< h
; y
++) {
1945 for (x
= 0; x
< w
; x
++) {
1946 if (y
+1 < h
&& x
+1 < w
&&
1947 ((grid
[y
*w
+x
] == TENT
&&
1948 grid
[(y
+1)*w
+(x
+1)] == TENT
) ||
1949 (grid
[(y
+1)*w
+x
] == TENT
&&
1950 grid
[y
*w
+(x
+1)] == TENT
))) {
1951 ret
[y
*w
+x
] |= 1 << ERR_ADJ_BOTRIGHT
;
1952 ret
[(y
+1)*w
+x
] |= 1 << ERR_ADJ_TOPRIGHT
;
1953 ret
[y
*w
+(x
+1)] |= 1 << ERR_ADJ_BOTLEFT
;
1954 ret
[(y
+1)*w
+(x
+1)] |= 1 << ERR_ADJ_TOPLEFT
;
1957 grid
[y
*w
+x
] == TENT
&&
1958 grid
[(y
+1)*w
+x
] == TENT
) {
1959 ret
[y
*w
+x
] |= 1 << ERR_ADJ_BOT
;
1960 ret
[(y
+1)*w
+x
] |= 1 << ERR_ADJ_TOP
;
1963 grid
[y
*w
+x
] == TENT
&&
1964 grid
[y
*w
+(x
+1)] == TENT
) {
1965 ret
[y
*w
+x
] |= 1 << ERR_ADJ_RIGHT
;
1966 ret
[y
*w
+(x
+1)] |= 1 << ERR_ADJ_LEFT
;
1972 * Spot numeric clue violations.
1974 for (x
= 0; x
< w
; x
++) {
1975 int tents
= 0, maybetents
= 0;
1976 for (y
= 0; y
< h
; y
++) {
1977 if (grid
[y
*w
+x
] == TENT
)
1979 else if (grid
[y
*w
+x
] == BLANK
)
1982 ret
[w
*h
+x
] = (tents
> state
->numbers
->numbers
[x
] ||
1983 tents
+ maybetents
< state
->numbers
->numbers
[x
]);
1985 for (y
= 0; y
< h
; y
++) {
1986 int tents
= 0, maybetents
= 0;
1987 for (x
= 0; x
< w
; x
++) {
1988 if (grid
[y
*w
+x
] == TENT
)
1990 else if (grid
[y
*w
+x
] == BLANK
)
1993 ret
[w
*h
+w
+y
] = (tents
> state
->numbers
->numbers
[w
+y
] ||
1994 tents
+ maybetents
< state
->numbers
->numbers
[w
+y
]);
1998 * Identify groups of tents with too few trees between them,
1999 * which we do by constructing the connected components of the
2000 * bipartite adjacency graph between tents and trees
2001 * ('bipartite' in the sense that we deliberately ignore
2002 * adjacency between tents or between trees), and highlighting
2003 * all the tents in any component which has a smaller tree
2007 /* Construct the equivalence classes. */
2008 for (y
= 0; y
< h
; y
++) {
2009 for (x
= 0; x
< w
-1; x
++) {
2010 if ((grid
[y
*w
+x
] == TREE
&& grid
[y
*w
+x
+1] == TENT
) ||
2011 (grid
[y
*w
+x
] == TENT
&& grid
[y
*w
+x
+1] == TREE
))
2012 dsf_merge(dsf
, y
*w
+x
, y
*w
+x
+1);
2015 for (y
= 0; y
< h
-1; y
++) {
2016 for (x
= 0; x
< w
; x
++) {
2017 if ((grid
[y
*w
+x
] == TREE
&& grid
[(y
+1)*w
+x
] == TENT
) ||
2018 (grid
[y
*w
+x
] == TENT
&& grid
[(y
+1)*w
+x
] == TREE
))
2019 dsf_merge(dsf
, y
*w
+x
, (y
+1)*w
+x
);
2022 /* Count up the tent/tree difference in each one. */
2023 for (x
= 0; x
< w
*h
; x
++)
2025 for (x
= 0; x
< w
*h
; x
++) {
2026 y
= dsf_canonify(dsf
, x
);
2027 if (grid
[x
] == TREE
)
2029 else if (grid
[x
] == TENT
)
2032 /* And highlight any tent belonging to an equivalence class with
2033 * a score less than zero. */
2034 for (x
= 0; x
< w
*h
; x
++) {
2035 y
= dsf_canonify(dsf
, x
);
2036 if (grid
[x
] == TENT
&& tmp
[y
] < 0)
2037 ret
[x
] |= 1 << ERR_OVERCOMMITTED
;
2041 * Identify groups of trees with too few tents between them.
2042 * This is done similarly, except that we now count BLANK as
2043 * equivalent to TENT, i.e. we only highlight such trees when
2044 * the user hasn't even left _room_ to provide tents for them
2045 * all. (Otherwise, we'd highlight all trees red right at the
2046 * start of the game, before the user had done anything wrong!)
2048 #define TENT(x) ((x)==TENT || (x)==BLANK)
2050 /* Construct the equivalence classes. */
2051 for (y
= 0; y
< h
; y
++) {
2052 for (x
= 0; x
< w
-1; x
++) {
2053 if ((grid
[y
*w
+x
] == TREE
&& TENT(grid
[y
*w
+x
+1])) ||
2054 (TENT(grid
[y
*w
+x
]) && grid
[y
*w
+x
+1] == TREE
))
2055 dsf_merge(dsf
, y
*w
+x
, y
*w
+x
+1);
2058 for (y
= 0; y
< h
-1; y
++) {
2059 for (x
= 0; x
< w
; x
++) {
2060 if ((grid
[y
*w
+x
] == TREE
&& TENT(grid
[(y
+1)*w
+x
])) ||
2061 (TENT(grid
[y
*w
+x
]) && grid
[(y
+1)*w
+x
] == TREE
))
2062 dsf_merge(dsf
, y
*w
+x
, (y
+1)*w
+x
);
2065 /* Count up the tent/tree difference in each one. */
2066 for (x
= 0; x
< w
*h
; x
++)
2068 for (x
= 0; x
< w
*h
; x
++) {
2069 y
= dsf_canonify(dsf
, x
);
2070 if (grid
[x
] == TREE
)
2072 else if (TENT(grid
[x
]))
2075 /* And highlight any tree belonging to an equivalence class with
2076 * a score more than zero. */
2077 for (x
= 0; x
< w
*h
; x
++) {
2078 y
= dsf_canonify(dsf
, x
);
2079 if (grid
[x
] == TREE
&& tmp
[y
] > 0)
2080 ret
[x
] |= 1 << ERR_OVERCOMMITTED
;
2088 static void draw_err_adj(drawing
*dr
, game_drawstate
*ds
, int x
, int y
)
2096 coords
[0] = x
- TILESIZE
*2/5;
2099 coords
[3] = y
- TILESIZE
*2/5;
2100 coords
[4] = x
+ TILESIZE
*2/5;
2103 coords
[7] = y
+ TILESIZE
*2/5;
2104 draw_polygon(dr
, coords
, 4, COL_ERROR
, COL_GRID
);
2107 * Draw an exclamation mark in the diamond. This turns out to
2108 * look unpleasantly off-centre if done via draw_text, so I do
2109 * it by hand on the basis that exclamation marks aren't that
2110 * difficult to draw...
2113 yext
= TILESIZE
*2/5 - (xext
*2+2);
2114 draw_rect(dr
, x
-xext
, y
-yext
, xext
*2+1, yext
*2+1 - (xext
*3),
2116 draw_rect(dr
, x
-xext
, y
+yext
-xext
*2+1, xext
*2+1, xext
*2, COL_ERRTEXT
);
2119 static void draw_tile(drawing
*dr
, game_drawstate
*ds
,
2120 int x
, int y
, int v
, int cur
, int printing
)
2123 int tx
= COORD(x
), ty
= COORD(y
);
2124 int cx
= tx
+ TILESIZE
/2, cy
= ty
+ TILESIZE
/2;
2129 clip(dr
, tx
, ty
, TILESIZE
, TILESIZE
);
2132 draw_rect(dr
, tx
, ty
, TILESIZE
, TILESIZE
, COL_GRID
);
2133 draw_rect(dr
, tx
+1, ty
+1, TILESIZE
-1, TILESIZE
-1,
2134 (v
== BLANK ? COL_BACKGROUND
: COL_GRASS
));
2140 (printing ? draw_rect_outline
: draw_rect
)
2141 (dr
, cx
-TILESIZE
/15, ty
+TILESIZE
*3/10,
2142 2*(TILESIZE
/15)+1, (TILESIZE
*9/10 - TILESIZE
*3/10),
2143 (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERROR
: COL_TREETRUNK
));
2145 for (i
= 0; i
< (printing ?
2 : 1); i
++) {
2146 int col
= (i
== 1 ? COL_BACKGROUND
:
2147 (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERROR
:
2149 int sub
= i
* (TILESIZE
/32);
2150 draw_circle(dr
, cx
, ty
+TILESIZE
*4/10, TILESIZE
/4 - sub
,
2152 draw_circle(dr
, cx
+TILESIZE
/5, ty
+TILESIZE
/4, TILESIZE
/8 - sub
,
2154 draw_circle(dr
, cx
-TILESIZE
/5, ty
+TILESIZE
/4, TILESIZE
/8 - sub
,
2156 draw_circle(dr
, cx
+TILESIZE
/4, ty
+TILESIZE
*6/13, TILESIZE
/8 - sub
,
2158 draw_circle(dr
, cx
-TILESIZE
/4, ty
+TILESIZE
*6/13, TILESIZE
/8 - sub
,
2161 } else if (v
== TENT
) {
2164 coords
[0] = cx
- TILESIZE
/3;
2165 coords
[1] = cy
+ TILESIZE
/3;
2166 coords
[2] = cx
+ TILESIZE
/3;
2167 coords
[3] = cy
+ TILESIZE
/3;
2169 coords
[5] = cy
- TILESIZE
/3;
2170 col
= (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERROR
: COL_TENT
);
2171 draw_polygon(dr
, coords
, 3, (printing ?
-1 : col
), col
);
2174 if (err
& (1 << ERR_ADJ_TOPLEFT
))
2175 draw_err_adj(dr
, ds
, tx
, ty
);
2176 if (err
& (1 << ERR_ADJ_TOP
))
2177 draw_err_adj(dr
, ds
, tx
+TILESIZE
/2, ty
);
2178 if (err
& (1 << ERR_ADJ_TOPRIGHT
))
2179 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
);
2180 if (err
& (1 << ERR_ADJ_LEFT
))
2181 draw_err_adj(dr
, ds
, tx
, ty
+TILESIZE
/2);
2182 if (err
& (1 << ERR_ADJ_RIGHT
))
2183 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
+TILESIZE
/2);
2184 if (err
& (1 << ERR_ADJ_BOTLEFT
))
2185 draw_err_adj(dr
, ds
, tx
, ty
+TILESIZE
);
2186 if (err
& (1 << ERR_ADJ_BOT
))
2187 draw_err_adj(dr
, ds
, tx
+TILESIZE
/2, ty
+TILESIZE
);
2188 if (err
& (1 << ERR_ADJ_BOTRIGHT
))
2189 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
+TILESIZE
);
2192 int coff
= TILESIZE
/8;
2193 draw_rect_outline(dr
, tx
+ coff
, ty
+ coff
,
2194 TILESIZE
- coff
*2 + 1, TILESIZE
- coff
*2 + 1,
2199 draw_update(dr
, tx
+1, ty
+1, TILESIZE
-1, TILESIZE
-1);
2203 * Internal redraw function, used for printing as well as drawing.
2205 static void int_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
2206 game_state
*state
, int dir
, game_ui
*ui
,
2207 float animtime
, float flashtime
, int printing
)
2209 int w
= state
->p
.w
, h
= state
->p
.h
;
2211 int cx
= -1, cy
= -1;
2217 if (ui
->cdisp
) { cx
= ui
->cx
; cy
= ui
->cy
; }
2218 if (cx
!= ds
->cx
|| cy
!= ds
->cy
) cmoved
= 1;
2221 if (printing
|| !ds
->started
) {
2224 game_compute_size(&state
->p
, TILESIZE
, &ww
, &wh
);
2225 draw_rect(dr
, 0, 0, ww
, wh
, COL_BACKGROUND
);
2226 draw_update(dr
, 0, 0, ww
, wh
);
2231 print_line_width(dr
, TILESIZE
/64);
2236 for (y
= 0; y
<= h
; y
++)
2237 draw_line(dr
, COORD(0), COORD(y
), COORD(w
), COORD(y
), COL_GRID
);
2238 for (x
= 0; x
<= w
; x
++)
2239 draw_line(dr
, COORD(x
), COORD(0), COORD(x
), COORD(h
), COL_GRID
);
2243 flashing
= (int)(flashtime
* 3 / FLASH_TIME
) != 1;
2248 * Find errors. For this we use _part_ of the information from a
2249 * currently active drag: we transform dsx,dsy but not anything
2250 * else. (This seems to strike a good compromise between having
2251 * the error highlights respond instantly to single clicks, but
2252 * not give constant feedback during a right-drag.)
2254 if (ui
&& ui
->drag_button
>= 0) {
2255 tmpgrid
= snewn(w
*h
, char);
2256 memcpy(tmpgrid
, state
->grid
, w
*h
);
2257 tmpgrid
[ui
->dsy
* w
+ ui
->dsx
] =
2258 drag_xform(ui
, ui
->dsx
, ui
->dsy
, tmpgrid
[ui
->dsy
* w
+ ui
->dsx
]);
2259 errors
= find_errors(state
, tmpgrid
);
2262 errors
= find_errors(state
, state
->grid
);
2268 for (y
= 0; y
< h
; y
++) {
2269 for (x
= 0; x
< w
; x
++) {
2270 int v
= state
->grid
[y
*w
+x
];
2274 * We deliberately do not take drag_ok into account
2275 * here, because user feedback suggests that it's
2276 * marginally nicer not to have the drag effects
2277 * flickering on and off disconcertingly.
2279 if (ui
&& ui
->drag_button
>= 0)
2280 v
= drag_xform(ui
, x
, y
, v
);
2282 if (flashing
&& (v
== TREE
|| v
== TENT
))
2286 if ((x
== cx
&& y
== cy
) ||
2287 (x
== ds
->cx
&& y
== ds
->cy
)) credraw
= 1;
2292 if (printing
|| ds
->drawn
[y
*w
+x
] != v
|| credraw
) {
2293 draw_tile(dr
, ds
, x
, y
, v
, (x
== cx
&& y
== cy
), printing
);
2295 ds
->drawn
[y
*w
+x
] = v
;
2301 * Draw (or redraw, if their error-highlighted state has
2302 * changed) the numbers.
2304 for (x
= 0; x
< w
; x
++) {
2305 if (ds
->numbersdrawn
[x
] != errors
[w
*h
+x
]) {
2307 draw_rect(dr
, COORD(x
), COORD(h
)+1, TILESIZE
, BRBORDER
-1,
2309 sprintf(buf
, "%d", state
->numbers
->numbers
[x
]);
2310 draw_text(dr
, COORD(x
) + TILESIZE
/2, COORD(h
+1),
2311 FONT_VARIABLE
, TILESIZE
/2, ALIGN_HCENTRE
|ALIGN_VNORMAL
,
2312 (errors
[w
*h
+x
] ? COL_ERROR
: COL_GRID
), buf
);
2313 draw_update(dr
, COORD(x
), COORD(h
)+1, TILESIZE
, BRBORDER
-1);
2314 ds
->numbersdrawn
[x
] = errors
[w
*h
+x
];
2317 for (y
= 0; y
< h
; y
++) {
2318 if (ds
->numbersdrawn
[w
+y
] != errors
[w
*h
+w
+y
]) {
2320 draw_rect(dr
, COORD(w
)+1, COORD(y
), BRBORDER
-1, TILESIZE
,
2322 sprintf(buf
, "%d", state
->numbers
->numbers
[w
+y
]);
2323 draw_text(dr
, COORD(w
+1), COORD(y
) + TILESIZE
/2,
2324 FONT_VARIABLE
, TILESIZE
/2, ALIGN_HRIGHT
|ALIGN_VCENTRE
,
2325 (errors
[w
*h
+w
+y
] ? COL_ERROR
: COL_GRID
), buf
);
2326 draw_update(dr
, COORD(w
)+1, COORD(y
), BRBORDER
-1, TILESIZE
);
2327 ds
->numbersdrawn
[w
+y
] = errors
[w
*h
+w
+y
];
2339 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
2340 game_state
*state
, int dir
, game_ui
*ui
,
2341 float animtime
, float flashtime
)
2343 int_redraw(dr
, ds
, oldstate
, state
, dir
, ui
, animtime
, flashtime
, FALSE
);
2346 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
2347 int dir
, game_ui
*ui
)
2352 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
2353 int dir
, game_ui
*ui
)
2355 if (!oldstate
->completed
&& newstate
->completed
&&
2356 !oldstate
->used_solve
&& !newstate
->used_solve
)
2362 static int game_timing_state(game_state
*state
, game_ui
*ui
)
2367 static void game_print_size(game_params
*params
, float *x
, float *y
)
2372 * I'll use 6mm squares by default.
2374 game_compute_size(params
, 600, &pw
, &ph
);
2379 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
2383 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
2384 game_drawstate ads
, *ds
= &ads
;
2385 game_set_size(dr
, ds
, NULL
, tilesize
);
2387 c
= print_mono_colour(dr
, 1); assert(c
== COL_BACKGROUND
);
2388 c
= print_mono_colour(dr
, 0); assert(c
== COL_GRID
);
2389 c
= print_mono_colour(dr
, 1); assert(c
== COL_GRASS
);
2390 c
= print_mono_colour(dr
, 0); assert(c
== COL_TREETRUNK
);
2391 c
= print_mono_colour(dr
, 0); assert(c
== COL_TREELEAF
);
2392 c
= print_mono_colour(dr
, 0); assert(c
== COL_TENT
);
2394 int_redraw(dr
, ds
, NULL
, state
, +1, NULL
, 0.0F
, 0.0F
, TRUE
);
2398 #define thegame tents
2401 const struct game thegame
= {
2402 "Tents", "games.tents", "tents",
2409 TRUE
, game_configure
, custom_params
,
2417 FALSE
, game_can_format_as_text_now
, game_text_format
,
2425 PREFERRED_TILESIZE
, game_compute_size
, game_set_size
,
2428 game_free_drawstate
,
2432 TRUE
, FALSE
, game_print_size
, game_print
,
2433 FALSE
, /* wants_statusbar */
2434 FALSE
, game_timing_state
,
2435 REQUIRE_RBUTTON
, /* flags */
2438 #ifdef STANDALONE_SOLVER
2442 int main(int argc
, char **argv
)
2446 char *id
= NULL
, *desc
, *err
;
2448 int ret
, diff
, really_verbose
= FALSE
;
2449 struct solver_scratch
*sc
;
2451 while (--argc
> 0) {
2453 if (!strcmp(p
, "-v")) {
2454 really_verbose
= TRUE
;
2455 } else if (!strcmp(p
, "-g")) {
2457 } else if (*p
== '-') {
2458 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
2466 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
2470 desc
= strchr(id
, ':');
2472 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
2477 p
= default_params();
2478 decode_params(p
, id
);
2479 err
= validate_desc(p
, desc
);
2481 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
2484 s
= new_game(NULL
, p
, desc
);
2485 s2
= new_game(NULL
, p
, desc
);
2487 sc
= new_scratch(p
->w
, p
->h
);
2490 * When solving an Easy puzzle, we don't want to bother the
2491 * user with Hard-level deductions. For this reason, we grade
2492 * the puzzle internally before doing anything else.
2494 ret
= -1; /* placate optimiser */
2495 for (diff
= 0; diff
< DIFFCOUNT
; diff
++) {
2496 ret
= tents_solve(p
->w
, p
->h
, s
->grid
, s
->numbers
->numbers
,
2497 s2
->grid
, sc
, diff
);
2502 if (diff
== DIFFCOUNT
) {
2504 printf("Difficulty rating: too hard to solve internally\n");
2506 printf("Unable to find a unique solution\n");
2510 printf("Difficulty rating: impossible (no solution exists)\n");
2512 printf("Difficulty rating: %s\n", tents_diffnames
[diff
]);
2514 verbose
= really_verbose
;
2515 ret
= tents_solve(p
->w
, p
->h
, s
->grid
, s
->numbers
->numbers
,
2516 s2
->grid
, sc
, diff
);
2518 printf("Puzzle is inconsistent\n");
2520 fputs(game_text_format(s2
), stdout
);
2529 /* vim: set shiftwidth=4 tabstop=8: */