2 * tents.c: Puzzle involving placing tents next to trees subject to
3 * some confusing conditions.
7 * - my technique for highlighting errors in the tent/tree matching
8 * is not perfect. It currently works by finding the connected
9 * components of the bipartite adjacency graph between tents and
10 * trees, and highlighting red all the tents in such a component
11 * if they outnumber the trees (the red meaning "these tents have
12 * too few trees between them") and vice versa if the trees
13 * outnumber the tents (but this time considering BLANKs as
14 * potential tents as yet unplaced, to avoid highlighting
15 * 'errors' from the word go before the player has actually made
16 * any mistake). However, something more subtle can go wrong
17 * within a component: consider, for instance, the setup
23 * in which there is one connected component containing equal
24 * numbers of trees and tents, but nonetheless there is no
25 * perfect matching that can link the two sensibly. This will be
26 * rejected by the rigorous solution checker, but the error
27 * highlighter won't currently spot it.
29 * Well, the _matching_ error highlighter won't spot it, anyway.
30 * In that diagram, there are two pairs of diagonally adjacent
31 * tents, which will be flagged as erroneous because that's much
32 * easier. So if I could prove that _all_ such setups require
33 * diagonally adjacent tents, I could safely ignore this problem.
34 * If not, however, then a proper treatment will require running
35 * the maxflow matcher over each component once I've identified
38 * - it might be nice to make setter-provided tent/nontent clues
40 * * on the other hand, this would introduce considerable extra
41 * complexity and size into the game state; also inviolable
42 * clues would have to be marked as such somehow, in an
43 * intrusive and annoying manner. Since they're never
44 * generated by _my_ generator, I'm currently more inclined
47 * - more difficult levels at the top end?
48 * * for example, sometimes we can deduce that two BLANKs in
49 * the same row are each adjacent to the same unattached tree
50 * and to nothing else, implying that they can't both be
51 * tents; this enables us to rule out some extra combinations
52 * in the row-based deduction loop, and hence deduce more
53 * from the number in that row than we could otherwise do.
54 * * that by itself doesn't seem worth implementing a new
55 * difficulty level for, but if I can find a few more things
56 * like that then it might become worthwhile.
57 * * I wonder if there's a sensible heuristic for where to
58 * guess which would make a recursive solver viable?
75 * The rules of this puzzle as available on the WWW are poorly
76 * specified. The bits about tents having to be orthogonally
77 * adjacent to trees, tents not being even diagonally adjacent to
78 * one another, and the number of tents in each row and column
79 * being given are simple enough; the difficult bit is the
80 * tent-to-tree matching.
82 * Some sources use simplistic wordings such as `each tree is
83 * exactly connected to only one tent', which is extremely unclear:
84 * it's easy to read erroneously as `each tree is _orthogonally
85 * adjacent_ to exactly one tent', which is definitely incorrect.
86 * Even the most coherent sources I've found don't do a much better
87 * job of stating the rule.
89 * A more precise statement of the rule is that it must be possible
90 * to find a bijection f between tents and trees such that each
91 * tree T is orthogonally adjacent to the tent f(T), but that a
92 * tent is permitted to be adjacent to other trees in addition to
93 * its own. This slightly non-obvious criterion is what gives this
94 * puzzle most of its subtlety.
96 * However, there's a particularly subtle ambiguity left over. Is
97 * the bijection between tents and trees required to be _unique_?
98 * In other words, is that bijection conceptually something the
99 * player should be able to exhibit as part of the solution (even
100 * if they aren't actually required to do so)? Or is it sufficient
101 * to have a unique _placement_ of the tents which gives rise to at
102 * least one suitable bijection?
104 * The puzzle shown to the right of this .T. 2 *T* 2
105 * paragraph illustrates the problem. There T.T 0 -> T-T 0
106 * are two distinct bijections available. .T. 2 *T* 2
107 * The answer to the above question will
108 * determine whether it's a valid puzzle. 202 202
110 * This is an important question, because it affects both the
111 * player and the generator. Eventually I found all the instances
112 * of this puzzle I could Google up, solved them all by hand, and
113 * verified that in all cases the tree/tent matching was uniquely
114 * determined given the tree and tent positions. Therefore, the
115 * puzzle as implemented in this source file takes the following
118 * - When checking a user-supplied solution for correctness, only
119 * verify that there exists _at least_ one matching.
120 * - When generating a puzzle, enforce that there must be
123 * Algorithmic implications
124 * ------------------------
126 * Another way of phrasing the tree/tent matching criterion is to
127 * say that the bipartite adjacency graph between trees and tents
128 * has a perfect matching. That is, if you construct a graph which
129 * has a vertex per tree and a vertex per tent, and an edge between
130 * any tree and tent which are orthogonally adjacent, it is
131 * possible to find a set of N edges of that graph (where N is the
132 * number of trees and also the number of tents) which between them
133 * connect every tree to every tent.
135 * The most efficient known algorithms for finding such a matching
136 * given a graph, as far as I'm aware, are the Munkres assignment
137 * algorithm (also known as the Hungarian algorithm) and the
138 * Ford-Fulkerson algorithm (for finding optimal flows in
139 * networks). Each of these takes O(N^3) running time; so we're
140 * talking O(N^3) time to verify any candidate solution to this
141 * puzzle. That's just about OK if you're doing it once per mouse
142 * click (and in fact not even that, since the sensible thing to do
143 * is check all the _other_ puzzle criteria and only wade into this
144 * quagmire if none are violated); but if the solver had to keep
145 * doing N^3 work internally, then it would probably end up with
146 * more like N^5 or N^6 running time, and grid generation would
147 * become very clunky.
149 * Fortunately, I've been able to prove a very useful property of
150 * _unique_ perfect matchings, by adapting the proof of Hall's
151 * Marriage Theorem. For those unaware of Hall's Theorem, I'll
152 * recap it and its proof: it states that a bipartite graph
153 * contains a perfect matching iff every set of vertices on the
154 * left side of the graph have a neighbourhood _at least_ as big on
157 * This condition is obviously satisfied if a perfect matching does
158 * exist; each left-side node has a distinct right-side node which
159 * is the one assigned to it by the matching, and thus any set of n
160 * left vertices must have a combined neighbourhood containing at
161 * least the n corresponding right vertices, and possibly others
162 * too. Alternatively, imagine if you had (say) three left-side
163 * nodes all of which were connected to only two right-side nodes
164 * between them: any perfect matching would have to assign one of
165 * those two right nodes to each of the three left nodes, and still
166 * give the three left nodes a different right node each. This is
167 * of course impossible.
169 * To prove the converse (that if every subset of left vertices
170 * satisfies the Hall condition then a perfect matching exists),
171 * consider trying to find a proper subset of the left vertices
172 * which _exactly_ satisfies the Hall condition: that is, its right
173 * neighbourhood is precisely the same size as it. If we can find
174 * such a subset, then we can split the bipartite graph into two
175 * smaller ones: one consisting of the left subset and its right
176 * neighbourhood, the other consisting of everything else. Edges
177 * from the left side of the former graph to the right side of the
178 * latter do not exist, by construction; edges from the right side
179 * of the former to the left of the latter cannot be part of any
180 * perfect matching because otherwise the left subset would not be
181 * left with enough distinct right vertices to connect to (this is
182 * exactly the same deduction used in Solo's set analysis). You can
183 * then prove (left as an exercise) that both these smaller graphs
184 * still satisfy the Hall condition, and therefore the proof will
185 * follow by induction.
187 * There's one other possibility, which is the case where _no_
188 * proper subset of the left vertices has a right neighbourhood of
189 * exactly the same size. That is, every left subset has a strictly
190 * _larger_ right neighbourhood. In this situation, we can simply
191 * remove an _arbitrary_ edge from the graph. This cannot reduce
192 * the size of any left subset's right neighbourhood by more than
193 * one, so if all neighbourhoods were strictly bigger than they
194 * needed to be initially, they must now still be _at least as big_
195 * as they need to be. So we can keep throwing out arbitrary edges
196 * until we find a set which exactly satisfies the Hall condition,
197 * and then proceed as above. []
199 * That's Hall's theorem. I now build on this by examining the
200 * circumstances in which a bipartite graph can have a _unique_
201 * perfect matching. It is clear that in the second case, where no
202 * left subset exactly satisfies the Hall condition and so we can
203 * remove an arbitrary edge, there cannot be a unique perfect
204 * matching: given one perfect matching, we choose our arbitrary
205 * removed edge to be one of those contained in it, and then we can
206 * still find a perfect matching in the remaining graph, which will
207 * be a distinct perfect matching in the original.
209 * So it is a necessary condition for a unique perfect matching
210 * that there must be at least one proper left subset which
211 * _exactly_ satisfies the Hall condition. But now consider the
212 * smaller graph constructed by taking that left subset and its
213 * neighbourhood: if the graph as a whole had a unique perfect
214 * matching, then so must this smaller one, which means we can find
215 * a proper left subset _again_, and so on. Repeating this process
216 * must eventually reduce us to a graph with only one left-side
217 * vertex (so there are no proper subsets at all); this vertex must
218 * be connected to only one right-side vertex, and hence must be so
219 * in the original graph as well (by construction). So we can
220 * discard this vertex pair from the graph, and any other edges
221 * that involved it (which will by construction be from other left
222 * vertices only), and the resulting smaller graph still has a
223 * unique perfect matching which means we can do the same thing
226 * In other words, given any bipartite graph with a unique perfect
227 * matching, we can find that matching by the following extremely
230 * - Find a left-side vertex which is only connected to one
232 * - Assign those vertices to one another, and therefore discard
233 * any other edges connecting to that right vertex.
234 * - Repeat until all vertices have been matched.
236 * This algorithm can be run in O(V+E) time (where V is the number
237 * of vertices and E is the number of edges in the graph), and the
238 * only way it can fail is if there is not a unique perfect
239 * matching (either because there is no matching at all, or because
240 * it isn't unique; but it can't distinguish those cases).
242 * Thus, the internal solver in this source file can be confident
243 * that if the tree/tent matching is uniquely determined by the
244 * tree and tent positions, it can find it using only this kind of
245 * obvious and simple operation: assign a tree to a tent if it
246 * cannot possibly belong to any other tent, and vice versa. If the
247 * solver were _only_ trying to determine the matching, even that
248 * `vice versa' wouldn't be required; but it can come in handy when
249 * not all the tents have been placed yet. I can therefore be
250 * reasonably confident that as long as my solver doesn't need to
251 * cope with grids that have a non-unique matching, it will also
252 * not need to do anything complicated like set analysis between
257 * In standalone solver mode, `verbose' is a variable which can be
258 * set by command-line option; in debugging mode it's simply always
261 #if defined STANDALONE_SOLVER
262 #define SOLVER_DIAGNOSTICS
264 #elif defined SOLVER_DIAGNOSTICS
269 * Difficulty levels. I do some macro ickery here to ensure that my
270 * enum and the various forms of my name list always match up.
272 #define DIFFLIST(A) \
275 #define ENUM(upper,title,lower) DIFF_ ## upper,
276 #define TITLE(upper,title,lower) #title,
277 #define ENCODE(upper,title,lower) #lower
278 #define CONFIG(upper,title,lower) ":" #title
279 enum { DIFFLIST(ENUM
) DIFFCOUNT
};
280 static char const *const tents_diffnames
[] = { DIFFLIST(TITLE
) };
281 static char const tents_diffchars
[] = DIFFLIST(ENCODE
);
282 #define DIFFCONFIG DIFFLIST(CONFIG)
297 enum { BLANK
, TREE
, TENT
, NONTENT
, MAGIC
};
312 struct numbers
*numbers
;
313 int completed
, used_solve
;
316 static game_params
*default_params(void)
318 game_params
*ret
= snew(game_params
);
321 ret
->diff
= DIFF_EASY
;
326 static const struct game_params tents_presets
[] = {
330 {10, 10, DIFF_TRICKY
},
332 {15, 15, DIFF_TRICKY
},
335 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
340 if (i
< 0 || i
>= lenof(tents_presets
))
343 ret
= snew(game_params
);
344 *ret
= tents_presets
[i
];
346 sprintf(str
, "%dx%d %s", ret
->w
, ret
->h
, tents_diffnames
[ret
->diff
]);
353 static void free_params(game_params
*params
)
358 static game_params
*dup_params(game_params
*params
)
360 game_params
*ret
= snew(game_params
);
361 *ret
= *params
; /* structure copy */
365 static void decode_params(game_params
*params
, char const *string
)
367 params
->w
= params
->h
= atoi(string
);
368 while (*string
&& isdigit((unsigned char)*string
)) string
++;
369 if (*string
== 'x') {
371 params
->h
= atoi(string
);
372 while (*string
&& isdigit((unsigned char)*string
)) string
++;
374 if (*string
== 'd') {
377 for (i
= 0; i
< DIFFCOUNT
; i
++)
378 if (*string
== tents_diffchars
[i
])
380 if (*string
) string
++;
384 static char *encode_params(game_params
*params
, int full
)
388 sprintf(buf
, "%dx%d", params
->w
, params
->h
);
390 sprintf(buf
+ strlen(buf
), "d%c",
391 tents_diffchars
[params
->diff
]);
395 static config_item
*game_configure(game_params
*params
)
400 ret
= snewn(4, config_item
);
402 ret
[0].name
= "Width";
403 ret
[0].type
= C_STRING
;
404 sprintf(buf
, "%d", params
->w
);
405 ret
[0].sval
= dupstr(buf
);
408 ret
[1].name
= "Height";
409 ret
[1].type
= C_STRING
;
410 sprintf(buf
, "%d", params
->h
);
411 ret
[1].sval
= dupstr(buf
);
414 ret
[2].name
= "Difficulty";
415 ret
[2].type
= C_CHOICES
;
416 ret
[2].sval
= DIFFCONFIG
;
417 ret
[2].ival
= params
->diff
;
427 static game_params
*custom_params(config_item
*cfg
)
429 game_params
*ret
= snew(game_params
);
431 ret
->w
= atoi(cfg
[0].sval
);
432 ret
->h
= atoi(cfg
[1].sval
);
433 ret
->diff
= cfg
[2].ival
;
438 static char *validate_params(game_params
*params
, int full
)
441 * Generating anything under 4x4 runs into trouble of one kind
444 if (params
->w
< 4 || params
->h
< 4)
445 return "Width and height must both be at least four";
450 * Scratch space for solver.
452 enum { N
, U
, L
, R
, D
, MAXDIR
}; /* link directions */
453 #define dx(d) ( ((d)==R) - ((d)==L) )
454 #define dy(d) ( ((d)==D) - ((d)==U) )
455 #define F(d) ( U + D - (d) )
456 struct solver_scratch
{
457 char *links
; /* mapping between trees and tents */
459 char *place
, *mrows
, *trows
;
462 static struct solver_scratch
*new_scratch(int w
, int h
)
464 struct solver_scratch
*ret
= snew(struct solver_scratch
);
466 ret
->links
= snewn(w
*h
, char);
467 ret
->locs
= snewn(max(w
, h
), int);
468 ret
->place
= snewn(max(w
, h
), char);
469 ret
->mrows
= snewn(3 * max(w
, h
), char);
470 ret
->trows
= snewn(3 * max(w
, h
), char);
475 static void free_scratch(struct solver_scratch
*sc
)
486 * Solver. Returns 0 for impossibility, 1 for success, 2 for
487 * ambiguity or failure to converge.
489 static int tents_solve(int w
, int h
, const char *grid
, int *numbers
,
490 char *soln
, struct solver_scratch
*sc
, int diff
)
493 char *mrow
, *mrow1
, *mrow2
, *trow
, *trow1
, *trow2
;
496 * Set up solver data.
498 memset(sc
->links
, N
, w
*h
);
501 * Set up solution array.
503 memcpy(soln
, grid
, w
*h
);
509 int done_something
= FALSE
;
512 * Any tent which has only one unattached tree adjacent to
513 * it can be tied to that tree.
515 for (y
= 0; y
< h
; y
++)
516 for (x
= 0; x
< w
; x
++)
517 if (soln
[y
*w
+x
] == TENT
&& !sc
->links
[y
*w
+x
]) {
520 for (d
= 1; d
< MAXDIR
; d
++) {
521 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
522 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
523 soln
[y2
*w
+x2
] == TREE
&&
524 !sc
->links
[y2
*w
+x2
]) {
526 break; /* found more than one */
532 if (d
== MAXDIR
&& linkd
== 0) {
533 #ifdef SOLVER_DIAGNOSTICS
535 printf("tent at %d,%d cannot link to anything\n",
538 return 0; /* no solution exists */
539 } else if (d
== MAXDIR
) {
540 int x2
= x
+ dx(linkd
), y2
= y
+ dy(linkd
);
542 #ifdef SOLVER_DIAGNOSTICS
544 printf("tent at %d,%d can only link to tree at"
545 " %d,%d\n", x
, y
, x2
, y2
);
548 sc
->links
[y
*w
+x
] = linkd
;
549 sc
->links
[y2
*w
+x2
] = F(linkd
);
550 done_something
= TRUE
;
557 break; /* don't do anything else! */
560 * Mark a blank square as NONTENT if it is not orthogonally
561 * adjacent to any unmatched tree.
563 for (y
= 0; y
< h
; y
++)
564 for (x
= 0; x
< w
; x
++)
565 if (soln
[y
*w
+x
] == BLANK
) {
566 int can_be_tent
= FALSE
;
568 for (d
= 1; d
< MAXDIR
; d
++) {
569 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
570 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
571 soln
[y2
*w
+x2
] == TREE
&&
577 #ifdef SOLVER_DIAGNOSTICS
579 printf("%d,%d cannot be a tent (no adjacent"
580 " unmatched tree)\n", x
, y
);
582 soln
[y
*w
+x
] = NONTENT
;
583 done_something
= TRUE
;
591 * Mark a blank square as NONTENT if it is (perhaps
592 * diagonally) adjacent to any other tent.
594 for (y
= 0; y
< h
; y
++)
595 for (x
= 0; x
< w
; x
++)
596 if (soln
[y
*w
+x
] == BLANK
) {
597 int dx
, dy
, imposs
= FALSE
;
599 for (dy
= -1; dy
<= +1; dy
++)
600 for (dx
= -1; dx
<= +1; dx
++)
602 int x2
= x
+ dx
, y2
= y
+ dy
;
603 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
604 soln
[y2
*w
+x2
] == TENT
)
609 #ifdef SOLVER_DIAGNOSTICS
611 printf("%d,%d cannot be a tent (adjacent tent)\n",
614 soln
[y
*w
+x
] = NONTENT
;
615 done_something
= TRUE
;
623 * Any tree which has exactly one {unattached tent, BLANK}
624 * adjacent to it must have its tent in that square.
626 for (y
= 0; y
< h
; y
++)
627 for (x
= 0; x
< w
; x
++)
628 if (soln
[y
*w
+x
] == TREE
&& !sc
->links
[y
*w
+x
]) {
629 int linkd
= 0, linkd2
= 0, nd
= 0;
631 for (d
= 1; d
< MAXDIR
; d
++) {
632 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
633 if (!(x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
))
635 if (soln
[y2
*w
+x2
] == BLANK
||
636 (soln
[y2
*w
+x2
] == TENT
&& !sc
->links
[y2
*w
+x2
])) {
646 #ifdef SOLVER_DIAGNOSTICS
648 printf("tree at %d,%d cannot link to anything\n",
651 return 0; /* no solution exists */
652 } else if (nd
== 1) {
653 int x2
= x
+ dx(linkd
), y2
= y
+ dy(linkd
);
655 #ifdef SOLVER_DIAGNOSTICS
657 printf("tree at %d,%d can only link to tent at"
658 " %d,%d\n", x
, y
, x2
, y2
);
660 soln
[y2
*w
+x2
] = TENT
;
661 sc
->links
[y
*w
+x
] = linkd
;
662 sc
->links
[y2
*w
+x2
] = F(linkd
);
663 done_something
= TRUE
;
664 } else if (nd
== 2 && (!dx(linkd
) != !dx(linkd2
)) &&
665 diff
>= DIFF_TRICKY
) {
667 * If there are two possible places where
668 * this tree's tent can go, and they are
669 * diagonally separated rather than being
670 * on opposite sides of the tree, then the
671 * square (other than the tree square)
672 * which is adjacent to both of them must
675 int x2
= x
+ dx(linkd
) + dx(linkd2
);
676 int y2
= y
+ dy(linkd
) + dy(linkd2
);
677 assert(x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
);
678 if (soln
[y2
*w
+x2
] == BLANK
) {
679 #ifdef SOLVER_DIAGNOSTICS
681 printf("possible tent locations for tree at"
682 " %d,%d rule out tent at %d,%d\n",
685 soln
[y2
*w
+x2
] = NONTENT
;
686 done_something
= TRUE
;
695 * If localised deductions about the trees and tents
696 * themselves haven't helped us, it's time to resort to the
697 * numbers round the grid edge. For each row and column, we
698 * go through all possible combinations of locations for
699 * the unplaced tents, rule out any which have adjacent
700 * tents, and spot any square which is given the same state
701 * by all remaining combinations.
703 for (i
= 0; i
< w
+h
; i
++) {
704 int start
, step
, len
, start1
, start2
, n
, k
;
708 * This is the number for a column.
723 * This is the number for a row.
738 if (diff
< DIFF_TRICKY
) {
740 * In Easy mode, we don't look at the effect of one
741 * row on the next (i.e. ruling out a square if all
742 * possibilities for an adjacent row place a tent
745 start1
= start2
= -1;
751 * Count and store the locations of the free squares,
752 * and also count the number of tents already placed.
755 for (j
= 0; j
< len
; j
++) {
756 if (soln
[start
+j
*step
] == TENT
)
757 k
--; /* one fewer tent to place */
758 else if (soln
[start
+j
*step
] == BLANK
)
763 continue; /* nothing left to do here */
766 * Now we know we're placing k tents in n squares. Set
767 * up the first possibility.
769 for (j
= 0; j
< n
; j
++)
770 sc
->place
[j
] = (j
< k ? TENT
: NONTENT
);
773 * We're aiming to find squares in this row which are
774 * invariant over all valid possibilities. Thus, we
775 * maintain the current state of that invariance. We
776 * start everything off at MAGIC to indicate that it
777 * hasn't been set up yet.
780 mrow1
= sc
->mrows
+ len
;
781 mrow2
= sc
->mrows
+ 2*len
;
783 trow1
= sc
->trows
+ len
;
784 trow2
= sc
->trows
+ 2*len
;
785 memset(mrow
, MAGIC
, 3*len
);
788 * And iterate over all possibilities.
794 * See if this possibility is valid. The only way
795 * it can fail to be valid is if it contains two
796 * adjacent tents. (Other forms of invalidity, such
797 * as containing a tent adjacent to one already
798 * placed, will have been dealt with already by
799 * other parts of the solver.)
802 for (j
= 0; j
+1 < n
; j
++)
803 if (sc
->place
[j
] == TENT
&&
804 sc
->place
[j
+1] == TENT
&&
805 sc
->locs
[j
+1] == sc
->locs
[j
]+1) {
812 * Merge this valid combination into mrow.
814 memset(trow
, MAGIC
, len
);
815 memset(trow
+len
, BLANK
, 2*len
);
816 for (j
= 0; j
< n
; j
++) {
817 trow
[sc
->locs
[j
]] = sc
->place
[j
];
818 if (sc
->place
[j
] == TENT
) {
820 for (jj
= sc
->locs
[j
]-1; jj
<= sc
->locs
[j
]+1; jj
++)
821 if (jj
>= 0 && jj
< len
)
822 trow1
[jj
] = trow2
[jj
] = NONTENT
;
826 for (j
= 0; j
< 3*len
; j
++) {
827 if (trow
[j
] == MAGIC
)
829 if (mrow
[j
] == MAGIC
|| mrow
[j
] == trow
[j
]) {
831 * Either this is the first valid
832 * placement we've found at all, or
833 * this square's contents are
834 * consistent with every previous valid
840 * This square's contents fail to match
841 * what they were in a different
842 * combination, so we cannot deduce
843 * anything about this square.
851 * Find the next combination of k choices from n.
852 * We do this by finding the rightmost tent which
853 * can be moved one place right, doing so, and
854 * shunting all tents to the right of that as far
855 * left as they can go.
858 for (j
= n
-1; j
> 0; j
--) {
859 if (sc
->place
[j
] == TENT
)
861 if (sc
->place
[j
] == NONTENT
&& sc
->place
[j
-1] == TENT
) {
862 sc
->place
[j
-1] = NONTENT
;
865 sc
->place
[++j
] = TENT
;
867 sc
->place
[j
] = NONTENT
;
872 break; /* we've finished */
876 * It's just possible that _no_ placement was valid, in
877 * which case we have an internally inconsistent
880 if (mrow
[sc
->locs
[0]] == MAGIC
)
881 return 0; /* inconsistent */
884 * Now go through mrow and see if there's anything
885 * we've deduced which wasn't already mentioned in soln.
887 for (j
= 0; j
< len
; j
++) {
890 for (whichrow
= 0; whichrow
< 3; whichrow
++) {
891 char *mthis
= mrow
+ whichrow
* len
;
892 int tstart
= (whichrow
== 0 ? start
:
893 whichrow
== 1 ? start1
: start2
);
895 mthis
[j
] != MAGIC
&& mthis
[j
] != BLANK
&&
896 soln
[tstart
+j
*step
] == BLANK
) {
897 int pos
= tstart
+j
*step
;
899 #ifdef SOLVER_DIAGNOSTICS
901 printf("%s %d forces %s at %d,%d\n",
902 step
==1 ?
"row" : "column",
903 step
==1 ? start
/w
: start
,
904 mthis
[j
] == TENT ?
"tent" : "non-tent",
907 soln
[pos
] = mthis
[j
];
908 done_something
= TRUE
;
922 * The solver has nothing further it can do. Return 1 if both
923 * soln and sc->links are completely filled in, or 2 otherwise.
925 for (y
= 0; y
< h
; y
++)
926 for (x
= 0; x
< w
; x
++) {
927 if (soln
[y
*w
+x
] == BLANK
)
929 if (soln
[y
*w
+x
] != NONTENT
&& sc
->links
[y
*w
+x
] == 0)
936 static char *new_game_desc(game_params
*params
, random_state
*rs
,
937 char **aux
, int interactive
)
939 int w
= params
->w
, h
= params
->h
;
940 int ntrees
= w
* h
/ 5;
941 char *grid
= snewn(w
*h
, char);
942 char *puzzle
= snewn(w
*h
, char);
943 int *numbers
= snewn(w
+h
, int);
944 char *soln
= snewn(w
*h
, char);
945 int *temp
= snewn(2*w
*h
, int);
946 int maxedges
= ntrees
*4 + w
*h
;
947 int *edges
= snewn(2*maxedges
, int);
948 int *capacity
= snewn(maxedges
, int);
949 int *flow
= snewn(maxedges
, int);
950 struct solver_scratch
*sc
= new_scratch(w
, h
);
955 * Since this puzzle has many global deductions and doesn't
956 * permit limited clue sets, generating grids for this puzzle
957 * is hard enough that I see no better option than to simply
958 * generate a solution and see if it's unique and has the
959 * required difficulty. This turns out to be computationally
962 * We chose our tree count (hence also tent count) by dividing
963 * the total grid area by five above. Why five? Well, w*h/4 is
964 * the maximum number of tents you can _possibly_ fit into the
965 * grid without violating the separation criterion, and to
966 * achieve that you are constrained to a very small set of
967 * possible layouts (the obvious one with a tent at every
968 * (even,even) coordinate, and trivial variations thereon). So
969 * if we reduce the tent count a bit more, we enable more
970 * random-looking placement; 5 turns out to be a plausible
971 * figure which yields sensible puzzles. Increasing the tent
972 * count would give puzzles whose solutions were too regimented
973 * and could be solved by the use of that knowledge (and would
974 * also take longer to find a viable placement); decreasing it
975 * would make the grids emptier and more boring.
977 * Actually generating a grid is a matter of first placing the
978 * tents, and then placing the trees by the use of maxflow
979 * (finding a distinct square adjacent to every tent). We do it
980 * this way round because otherwise satisfying the tent
981 * separation condition would become onerous: most randomly
982 * chosen tent layouts do not satisfy this condition, so we'd
983 * have gone to a lot of work before finding that a candidate
984 * layout was unusable. Instead, we place the tents first and
985 * ensure they meet the separation criterion _before_ doing
986 * lots of computation; this works much better.
988 * The maxflow algorithm is not randomised, so employed naively
989 * it would give rise to grids with clear structure and
990 * directional bias. Hence, I assign the network nodes as seen
991 * by maxflow to be a _random_ permutation of the squares of
992 * the grid, so that any bias shown by maxflow towards
993 * low-numbered nodes is turned into a random bias.
995 * This generation strategy can fail at many points, including
996 * as early as tent placement (if you get a bad random order in
997 * which to greedily try the grid squares, you won't even
998 * manage to find enough mutually non-adjacent squares to put
999 * the tents in). Then it can fail if maxflow doesn't manage to
1000 * find a good enough matching (i.e. the tent placements don't
1001 * admit any adequate tree placements); and finally it can fail
1002 * if the solver finds that the problem has the wrong
1003 * difficulty (including being actually non-unique). All of
1004 * these, however, are insufficiently frequent to cause
1008 if (params
->diff
> DIFF_EASY
&& params
->w
<= 4 && params
->h
<= 4)
1009 params
->diff
= DIFF_EASY
; /* downgrade to prevent tight loop */
1013 * Arrange the grid squares into a random order.
1015 for (i
= 0; i
< w
*h
; i
++)
1017 shuffle(temp
, w
*h
, sizeof(*temp
), rs
);
1020 * The first `ntrees' entries in temp which we can get
1021 * without making two tents adjacent will be the tent
1024 memset(grid
, BLANK
, w
*h
);
1026 for (i
= 0; i
< w
*h
&& j
> 0; i
++) {
1027 int x
= temp
[i
] % w
, y
= temp
[i
] / w
;
1028 int dy
, dx
, ok
= TRUE
;
1030 for (dy
= -1; dy
<= +1; dy
++)
1031 for (dx
= -1; dx
<= +1; dx
++)
1032 if (x
+dx
>= 0 && x
+dx
< w
&&
1033 y
+dy
>= 0 && y
+dy
< h
&&
1034 grid
[(y
+dy
)*w
+(x
+dx
)] == TENT
)
1038 grid
[temp
[i
]] = TENT
;
1043 continue; /* couldn't place all the tents */
1046 * Now we build up the list of graph edges.
1049 for (i
= 0; i
< w
*h
; i
++) {
1050 if (grid
[temp
[i
]] == TENT
) {
1051 for (j
= 0; j
< w
*h
; j
++) {
1052 if (grid
[temp
[j
]] != TENT
) {
1053 int xi
= temp
[i
] % w
, yi
= temp
[i
] / w
;
1054 int xj
= temp
[j
] % w
, yj
= temp
[j
] / w
;
1055 if (abs(xi
-xj
) + abs(yi
-yj
) == 1) {
1056 edges
[nedges
*2] = i
;
1057 edges
[nedges
*2+1] = j
;
1058 capacity
[nedges
] = 1;
1065 * Special node w*h is the sink node; any non-tent node
1066 * has an edge going to it.
1068 edges
[nedges
*2] = i
;
1069 edges
[nedges
*2+1] = w
*h
;
1070 capacity
[nedges
] = 1;
1076 * Special node w*h+1 is the source node, with an edge going to
1079 for (i
= 0; i
< w
*h
; i
++) {
1080 if (grid
[temp
[i
]] == TENT
) {
1081 edges
[nedges
*2] = w
*h
+1;
1082 edges
[nedges
*2+1] = i
;
1083 capacity
[nedges
] = 1;
1088 assert(nedges
<= maxedges
);
1091 * Now we're ready to call the maxflow algorithm to place the
1094 j
= maxflow(w
*h
+2, w
*h
+1, w
*h
, nedges
, edges
, capacity
, flow
, NULL
);
1097 continue; /* couldn't place all the tents */
1100 * We've placed the trees. Now we need to work out _where_
1101 * we've placed them, which is a matter of reading back out
1102 * from the `flow' array.
1104 for (i
= 0; i
< nedges
; i
++) {
1105 if (edges
[2*i
] < w
*h
&& edges
[2*i
+1] < w
*h
&& flow
[i
] > 0)
1106 grid
[temp
[edges
[2*i
+1]]] = TREE
;
1110 * I think it looks ugly if there isn't at least one of
1111 * _something_ (tent or tree) in each row and each column
1112 * of the grid. This doesn't give any information away
1113 * since a completely empty row/column is instantly obvious
1114 * from the clues (it has no trees and a zero).
1116 for (i
= 0; i
< w
; i
++) {
1117 for (j
= 0; j
< h
; j
++) {
1118 if (grid
[j
*w
+i
] != BLANK
)
1119 break; /* found something in this column */
1122 break; /* found empty column */
1125 continue; /* a column was empty */
1127 for (j
= 0; j
< h
; j
++) {
1128 for (i
= 0; i
< w
; i
++) {
1129 if (grid
[j
*w
+i
] != BLANK
)
1130 break; /* found something in this row */
1133 break; /* found empty row */
1136 continue; /* a row was empty */
1139 * Now set up the numbers round the edge.
1141 for (i
= 0; i
< w
; i
++) {
1143 for (j
= 0; j
< h
; j
++)
1144 if (grid
[j
*w
+i
] == TENT
)
1148 for (i
= 0; i
< h
; i
++) {
1150 for (j
= 0; j
< w
; j
++)
1151 if (grid
[i
*w
+j
] == TENT
)
1157 * And now actually solve the puzzle, to see whether it's
1158 * unique and has the required difficulty.
1160 for (i
= 0; i
< w
*h
; i
++)
1161 puzzle
[i
] = grid
[i
] == TREE ? TREE
: BLANK
;
1162 i
= tents_solve(w
, h
, puzzle
, numbers
, soln
, sc
, params
->diff
-1);
1163 j
= tents_solve(w
, h
, puzzle
, numbers
, soln
, sc
, params
->diff
);
1166 * We expect solving with difficulty params->diff to have
1167 * succeeded (otherwise the problem is too hard), and
1168 * solving with diff-1 to have failed (otherwise it's too
1171 if (i
== 2 && j
== 1)
1176 * That's it. Encode as a game ID.
1178 ret
= snewn((w
+h
)*40 + ntrees
+ (w
*h
)/26 + 1, char);
1181 for (i
= 0; i
<= w
*h
; i
++) {
1182 int c
= (i
< w
*h ? grid
[i
] == TREE
: 1);
1184 *p
++ = (j
== 0 ?
'_' : j
-1 + 'a');
1194 for (i
= 0; i
< w
+h
; i
++)
1195 p
+= sprintf(p
, ",%d", numbers
[i
]);
1197 ret
= sresize(ret
, p
- ret
, char);
1200 * And encode the solution as an aux_info.
1202 *aux
= snewn(ntrees
* 40, char);
1205 for (i
= 0; i
< w
*h
; i
++)
1206 if (grid
[i
] == TENT
)
1207 p
+= sprintf(p
, ";T%d,%d", i
%w
, i
/w
);
1209 *aux
= sresize(*aux
, p
- *aux
, char);
1224 static char *validate_desc(game_params
*params
, char *desc
)
1226 int w
= params
->w
, h
= params
->h
;
1230 while (*desc
&& *desc
!= ',') {
1233 else if (*desc
>= 'a' && *desc
< 'z')
1234 area
+= *desc
- 'a' + 2;
1235 else if (*desc
== 'z')
1237 else if (*desc
== '!' || *desc
== '-')
1240 return "Invalid character in grid specification";
1245 for (i
= 0; i
< w
+h
; i
++) {
1247 return "Not enough numbers given after grid specification";
1248 else if (*desc
!= ',')
1249 return "Invalid character in number list";
1251 while (*desc
&& isdigit((unsigned char)*desc
)) desc
++;
1255 return "Unexpected additional data at end of game description";
1259 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
1261 int w
= params
->w
, h
= params
->h
;
1262 game_state
*state
= snew(game_state
);
1265 state
->p
= *params
; /* structure copy */
1266 state
->grid
= snewn(w
*h
, char);
1267 state
->numbers
= snew(struct numbers
);
1268 state
->numbers
->refcount
= 1;
1269 state
->numbers
->numbers
= snewn(w
+h
, int);
1270 state
->completed
= state
->used_solve
= FALSE
;
1273 memset(state
->grid
, BLANK
, w
*h
);
1282 else if (*desc
>= 'a' && *desc
< 'z')
1283 run
= *desc
- ('a'-1);
1284 else if (*desc
== 'z') {
1288 assert(*desc
== '!' || *desc
== '-');
1290 type
= (*desc
== '!' ? TENT
: NONTENT
);
1296 assert(i
>= 0 && i
<= w
*h
);
1298 assert(type
== TREE
);
1302 state
->grid
[i
++] = type
;
1306 for (i
= 0; i
< w
+h
; i
++) {
1307 assert(*desc
== ',');
1309 state
->numbers
->numbers
[i
] = atoi(desc
);
1310 while (*desc
&& isdigit((unsigned char)*desc
)) desc
++;
1318 static game_state
*dup_game(game_state
*state
)
1320 int w
= state
->p
.w
, h
= state
->p
.h
;
1321 game_state
*ret
= snew(game_state
);
1323 ret
->p
= state
->p
; /* structure copy */
1324 ret
->grid
= snewn(w
*h
, char);
1325 memcpy(ret
->grid
, state
->grid
, w
*h
);
1326 ret
->numbers
= state
->numbers
;
1327 state
->numbers
->refcount
++;
1328 ret
->completed
= state
->completed
;
1329 ret
->used_solve
= state
->used_solve
;
1334 static void free_game(game_state
*state
)
1336 if (--state
->numbers
->refcount
<= 0) {
1337 sfree(state
->numbers
->numbers
);
1338 sfree(state
->numbers
);
1344 static char *solve_game(game_state
*state
, game_state
*currstate
,
1345 char *aux
, char **error
)
1347 int w
= state
->p
.w
, h
= state
->p
.h
;
1351 * If we already have the solution, save ourselves some
1356 struct solver_scratch
*sc
= new_scratch(w
, h
);
1362 soln
= snewn(w
*h
, char);
1363 ret
= tents_solve(w
, h
, state
->grid
, state
->numbers
->numbers
,
1364 soln
, sc
, DIFFCOUNT
-1);
1369 *error
= "This puzzle is not self-consistent";
1371 *error
= "Unable to find a unique solution for this puzzle";
1376 * Construct a move string which turns the current state
1377 * into the solved state.
1379 move
= snewn(w
*h
* 40, char);
1382 for (i
= 0; i
< w
*h
; i
++)
1383 if (soln
[i
] == TENT
)
1384 p
+= sprintf(p
, ";T%d,%d", i
%w
, i
/w
);
1386 move
= sresize(move
, p
- move
, char);
1394 static int game_can_format_as_text_now(game_params
*params
)
1399 static char *game_text_format(game_state
*state
)
1401 int w
= state
->p
.w
, h
= state
->p
.h
;
1406 * FIXME: We currently do not print the numbers round the edges
1407 * of the grid. I need to work out a sensible way of doing this
1408 * even when the column numbers exceed 9.
1410 * In the absence of those numbers, the result size is h lines
1411 * of w+1 characters each, plus a NUL.
1413 * This function is currently only used by the standalone
1414 * solver; until I make it look more sensible, I won't enable
1415 * it in the main game structure.
1417 ret
= snewn(h
*(w
+1) + 1, char);
1419 for (y
= 0; y
< h
; y
++) {
1420 for (x
= 0; x
< w
; x
++) {
1421 *p
= (state
->grid
[y
*w
+x
] == BLANK ?
'.' :
1422 state
->grid
[y
*w
+x
] == TREE ?
'T' :
1423 state
->grid
[y
*w
+x
] == TENT ?
'*' :
1424 state
->grid
[y
*w
+x
] == NONTENT ?
'-' : '?');
1435 int dsx
, dsy
; /* coords of drag start */
1436 int dex
, dey
; /* coords of drag end */
1437 int drag_button
; /* -1 for none, or a button code */
1438 int drag_ok
; /* dragged off the window, to cancel */
1440 int cx
, cy
, cdisp
; /* cursor position, and ?display. */
1443 static game_ui
*new_ui(game_state
*state
)
1445 game_ui
*ui
= snew(game_ui
);
1446 ui
->dsx
= ui
->dsy
= -1;
1447 ui
->dex
= ui
->dey
= -1;
1448 ui
->drag_button
= -1;
1449 ui
->drag_ok
= FALSE
;
1450 ui
->cx
= ui
->cy
= ui
->cdisp
= 0;
1454 static void free_ui(game_ui
*ui
)
1459 static char *encode_ui(game_ui
*ui
)
1464 static void decode_ui(game_ui
*ui
, char *encoding
)
1468 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
1469 game_state
*newstate
)
1473 struct game_drawstate
{
1477 int *drawn
, *numbersdrawn
;
1478 int cx
, cy
; /* last-drawn cursor pos, or (-1,-1) if absent. */
1481 #define PREFERRED_TILESIZE 32
1482 #define TILESIZE (ds->tilesize)
1483 #define TLBORDER (TILESIZE/2)
1484 #define BRBORDER (TILESIZE*3/2)
1485 #define COORD(x) ( (x) * TILESIZE + TLBORDER )
1486 #define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 )
1488 #define FLASH_TIME 0.30F
1490 static int drag_xform(game_ui
*ui
, int x
, int y
, int v
)
1492 int xmin
, ymin
, xmax
, ymax
;
1494 xmin
= min(ui
->dsx
, ui
->dex
);
1495 xmax
= max(ui
->dsx
, ui
->dex
);
1496 ymin
= min(ui
->dsy
, ui
->dey
);
1497 ymax
= max(ui
->dsy
, ui
->dey
);
1500 * Left-dragging has no effect, so we treat a left-drag as a
1501 * single click on dsx,dsy.
1503 if (ui
->drag_button
== LEFT_BUTTON
) {
1504 xmin
= xmax
= ui
->dsx
;
1505 ymin
= ymax
= ui
->dsy
;
1508 if (x
< xmin
|| x
> xmax
|| y
< ymin
|| y
> ymax
)
1509 return v
; /* no change outside drag area */
1512 return v
; /* trees are inviolate always */
1514 if (xmin
== xmax
&& ymin
== ymax
) {
1516 * Results of a simple click. Left button sets blanks to
1517 * tents; right button sets blanks to non-tents; either
1518 * button clears a non-blank square.
1520 if (ui
->drag_button
== LEFT_BUTTON
)
1521 v
= (v
== BLANK ? TENT
: BLANK
);
1523 v
= (v
== BLANK ? NONTENT
: BLANK
);
1526 * Results of a drag. Left-dragging has no effect.
1527 * Right-dragging sets all blank squares to non-tents and
1528 * has no effect on anything else.
1530 if (ui
->drag_button
== RIGHT_BUTTON
)
1531 v
= (v
== BLANK ? NONTENT
: v
);
1539 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
1540 int x
, int y
, int button
)
1542 int w
= state
->p
.w
, h
= state
->p
.h
;
1545 if (button
== LEFT_BUTTON
|| button
== RIGHT_BUTTON
) {
1548 if (x
< 0 || y
< 0 || x
>= w
|| y
>= h
)
1551 ui
->drag_button
= button
;
1552 ui
->dsx
= ui
->dex
= x
;
1553 ui
->dsy
= ui
->dey
= y
;
1556 return ""; /* ui updated */
1559 if ((IS_MOUSE_DRAG(button
) || IS_MOUSE_RELEASE(button
)) &&
1560 ui
->drag_button
> 0) {
1561 int xmin
, ymin
, xmax
, ymax
;
1563 int buflen
, bufsize
, tmplen
;
1567 if (x
< 0 || y
< 0 || x
>= w
|| y
>= h
) {
1568 ui
->drag_ok
= FALSE
;
1571 * Drags are limited to one row or column. Hence, we
1572 * work out which coordinate is closer to the drag
1573 * start, and move it _to_ the drag start.
1575 if (abs(x
- ui
->dsx
) < abs(y
- ui
->dsy
))
1586 if (IS_MOUSE_DRAG(button
))
1587 return ""; /* ui updated */
1590 * The drag has been released. Enact it.
1593 ui
->drag_button
= -1;
1594 return ""; /* drag was just cancelled */
1597 xmin
= min(ui
->dsx
, ui
->dex
);
1598 xmax
= max(ui
->dsx
, ui
->dex
);
1599 ymin
= min(ui
->dsy
, ui
->dey
);
1600 ymax
= max(ui
->dsy
, ui
->dey
);
1601 assert(0 <= xmin
&& xmin
<= xmax
&& xmax
< w
);
1602 assert(0 <= ymin
&& ymin
<= ymax
&& ymax
< h
);
1606 buf
= snewn(bufsize
, char);
1608 for (y
= ymin
; y
<= ymax
; y
++)
1609 for (x
= xmin
; x
<= xmax
; x
++) {
1610 int v
= drag_xform(ui
, x
, y
, state
->grid
[y
*w
+x
]);
1611 if (state
->grid
[y
*w
+x
] != v
) {
1612 tmplen
= sprintf(tmpbuf
, "%s%c%d,%d", sep
,
1613 (int)(v
== BLANK ?
'B' :
1614 v
== TENT ?
'T' : 'N'),
1618 if (buflen
+ tmplen
>= bufsize
) {
1619 bufsize
= buflen
+ tmplen
+ 256;
1620 buf
= sresize(buf
, bufsize
, char);
1623 strcpy(buf
+buflen
, tmpbuf
);
1628 ui
->drag_button
= -1; /* drag is terminated */
1632 return ""; /* ui updated (drag was terminated) */
1639 if (IS_CURSOR_MOVE(button
)) {
1640 move_cursor(button
, &ui
->cx
, &ui
->cy
, w
, h
, 0);
1646 int v
= state
->grid
[ui
->cy
*w
+ui
->cx
];
1649 #ifdef SINGLE_CURSOR_SELECT
1650 if (button
== CURSOR_SELECT
)
1651 /* SELECT cycles T, N, B */
1652 rep
= v
== BLANK ?
'T' : v
== TENT ?
'N' : 'B';
1654 if (button
== CURSOR_SELECT
)
1655 rep
= v
== BLANK ?
'T' : 'B';
1656 else if (button
== CURSOR_SELECT2
)
1657 rep
= v
== BLANK ?
'N' : 'B';
1658 else if (button
== 'T' || button
== 'N' || button
== 'B')
1664 sprintf(tmpbuf
, "%c%d,%d", (int)rep
, ui
->cx
, ui
->cy
);
1665 return dupstr(tmpbuf
);
1667 } else if (IS_CURSOR_SELECT(button
)) {
1675 static game_state
*execute_move(game_state
*state
, char *move
)
1677 int w
= state
->p
.w
, h
= state
->p
.h
;
1679 int x
, y
, m
, n
, i
, j
;
1680 game_state
*ret
= dup_game(state
);
1686 ret
->used_solve
= TRUE
;
1688 * Set all non-tree squares to NONTENT. The rest of the
1689 * solve move will fill the tents in over the top.
1691 for (i
= 0; i
< w
*h
; i
++)
1692 if (ret
->grid
[i
] != TREE
)
1693 ret
->grid
[i
] = NONTENT
;
1695 } else if (c
== 'B' || c
== 'T' || c
== 'N') {
1697 if (sscanf(move
, "%d,%d%n", &x
, &y
, &n
) != 2 ||
1698 x
< 0 || y
< 0 || x
>= w
|| y
>= h
) {
1702 if (ret
->grid
[y
*w
+x
] == TREE
) {
1706 ret
->grid
[y
*w
+x
] = (c
== 'B' ? BLANK
: c
== 'T' ? TENT
: NONTENT
);
1721 * Check for completion.
1723 for (i
= n
= m
= 0; i
< w
*h
; i
++) {
1724 if (ret
->grid
[i
] == TENT
)
1726 else if (ret
->grid
[i
] == TREE
)
1730 int nedges
, maxedges
, *edges
, *capacity
, *flow
;
1733 * We have the right number of tents, which is a
1734 * precondition for the game being complete. Now check that
1735 * the numbers add up.
1737 for (i
= 0; i
< w
; i
++) {
1739 for (j
= 0; j
< h
; j
++)
1740 if (ret
->grid
[j
*w
+i
] == TENT
)
1742 if (ret
->numbers
->numbers
[i
] != n
)
1743 goto completion_check_done
;
1745 for (i
= 0; i
< h
; i
++) {
1747 for (j
= 0; j
< w
; j
++)
1748 if (ret
->grid
[i
*w
+j
] == TENT
)
1750 if (ret
->numbers
->numbers
[w
+i
] != n
)
1751 goto completion_check_done
;
1754 * Also, check that no two tents are adjacent.
1756 for (y
= 0; y
< h
; y
++)
1757 for (x
= 0; x
< w
; x
++) {
1759 ret
->grid
[y
*w
+x
] == TENT
&& ret
->grid
[y
*w
+x
+1] == TENT
)
1760 goto completion_check_done
;
1762 ret
->grid
[y
*w
+x
] == TENT
&& ret
->grid
[(y
+1)*w
+x
] == TENT
)
1763 goto completion_check_done
;
1764 if (x
+1 < w
&& y
+1 < h
) {
1765 if (ret
->grid
[y
*w
+x
] == TENT
&&
1766 ret
->grid
[(y
+1)*w
+(x
+1)] == TENT
)
1767 goto completion_check_done
;
1768 if (ret
->grid
[(y
+1)*w
+x
] == TENT
&&
1769 ret
->grid
[y
*w
+(x
+1)] == TENT
)
1770 goto completion_check_done
;
1775 * OK; we have the right number of tents, they match the
1776 * numeric clues, and they satisfy the non-adjacency
1777 * criterion. Finally, we need to verify that they can be
1778 * placed in a one-to-one matching with the trees such that
1779 * every tent is orthogonally adjacent to its tree.
1781 * This bit is where the hard work comes in: we have to do
1782 * it by finding such a matching using maxflow.
1784 * So we construct a network with one special source node,
1785 * one special sink node, one node per tent, and one node
1789 edges
= snewn(2 * maxedges
, int);
1790 capacity
= snewn(maxedges
, int);
1791 flow
= snewn(maxedges
, int);
1796 * 0..w*h trees/tents
1800 for (y
= 0; y
< h
; y
++)
1801 for (x
= 0; x
< w
; x
++)
1802 if (ret
->grid
[y
*w
+x
] == TREE
) {
1806 * Here we use the direction enum declared for
1807 * the solver. We make use of the fact that the
1808 * directions are declared in the order
1809 * U,L,R,D, meaning that we go through the four
1810 * neighbours of any square in numerically
1813 for (d
= 1; d
< MAXDIR
; d
++) {
1814 int x2
= x
+ dx(d
), y2
= y
+ dy(d
);
1815 if (x2
>= 0 && x2
< w
&& y2
>= 0 && y2
< h
&&
1816 ret
->grid
[y2
*w
+x2
] == TENT
) {
1817 assert(nedges
< maxedges
);
1818 edges
[nedges
*2] = y
*w
+x
;
1819 edges
[nedges
*2+1] = y2
*w
+x2
;
1820 capacity
[nedges
] = 1;
1824 } else if (ret
->grid
[y
*w
+x
] == TENT
) {
1825 assert(nedges
< maxedges
);
1826 edges
[nedges
*2] = y
*w
+x
;
1827 edges
[nedges
*2+1] = w
*h
+1; /* edge going to sink */
1828 capacity
[nedges
] = 1;
1831 for (y
= 0; y
< h
; y
++)
1832 for (x
= 0; x
< w
; x
++)
1833 if (ret
->grid
[y
*w
+x
] == TREE
) {
1834 assert(nedges
< maxedges
);
1835 edges
[nedges
*2] = w
*h
; /* edge coming from source */
1836 edges
[nedges
*2+1] = y
*w
+x
;
1837 capacity
[nedges
] = 1;
1840 n
= maxflow(w
*h
+2, w
*h
, w
*h
+1, nedges
, edges
, capacity
, flow
, NULL
);
1847 goto completion_check_done
;
1850 * We haven't managed to fault the grid on any count. Score!
1852 ret
->completed
= TRUE
;
1854 completion_check_done
:
1859 /* ----------------------------------------------------------------------
1863 static void game_compute_size(game_params
*params
, int tilesize
,
1866 /* fool the macros */
1867 struct dummy
{ int tilesize
; } dummy
, *ds
= &dummy
;
1868 dummy
.tilesize
= tilesize
;
1870 *x
= TLBORDER
+ BRBORDER
+ TILESIZE
* params
->w
;
1871 *y
= TLBORDER
+ BRBORDER
+ TILESIZE
* params
->h
;
1874 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
1875 game_params
*params
, int tilesize
)
1877 ds
->tilesize
= tilesize
;
1880 static float *game_colours(frontend
*fe
, int *ncolours
)
1882 float *ret
= snewn(3 * NCOLOURS
, float);
1884 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
1886 ret
[COL_GRID
* 3 + 0] = 0.0F
;
1887 ret
[COL_GRID
* 3 + 1] = 0.0F
;
1888 ret
[COL_GRID
* 3 + 2] = 0.0F
;
1890 ret
[COL_GRASS
* 3 + 0] = 0.7F
;
1891 ret
[COL_GRASS
* 3 + 1] = 1.0F
;
1892 ret
[COL_GRASS
* 3 + 2] = 0.5F
;
1894 ret
[COL_TREETRUNK
* 3 + 0] = 0.6F
;
1895 ret
[COL_TREETRUNK
* 3 + 1] = 0.4F
;
1896 ret
[COL_TREETRUNK
* 3 + 2] = 0.0F
;
1898 ret
[COL_TREELEAF
* 3 + 0] = 0.0F
;
1899 ret
[COL_TREELEAF
* 3 + 1] = 0.7F
;
1900 ret
[COL_TREELEAF
* 3 + 2] = 0.0F
;
1902 ret
[COL_TENT
* 3 + 0] = 0.8F
;
1903 ret
[COL_TENT
* 3 + 1] = 0.7F
;
1904 ret
[COL_TENT
* 3 + 2] = 0.0F
;
1906 ret
[COL_ERROR
* 3 + 0] = 1.0F
;
1907 ret
[COL_ERROR
* 3 + 1] = 0.0F
;
1908 ret
[COL_ERROR
* 3 + 2] = 0.0F
;
1910 ret
[COL_ERRTEXT
* 3 + 0] = 1.0F
;
1911 ret
[COL_ERRTEXT
* 3 + 1] = 1.0F
;
1912 ret
[COL_ERRTEXT
* 3 + 2] = 1.0F
;
1914 ret
[COL_ERRTRUNK
* 3 + 0] = 0.6F
;
1915 ret
[COL_ERRTRUNK
* 3 + 1] = 0.0F
;
1916 ret
[COL_ERRTRUNK
* 3 + 2] = 0.0F
;
1918 *ncolours
= NCOLOURS
;
1922 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
1924 int w
= state
->p
.w
, h
= state
->p
.h
;
1925 struct game_drawstate
*ds
= snew(struct game_drawstate
);
1929 ds
->started
= FALSE
;
1930 ds
->p
= state
->p
; /* structure copy */
1931 ds
->drawn
= snewn(w
*h
, int);
1932 for (i
= 0; i
< w
*h
; i
++)
1933 ds
->drawn
[i
] = MAGIC
;
1934 ds
->numbersdrawn
= snewn(w
+h
, int);
1935 for (i
= 0; i
< w
+h
; i
++)
1936 ds
->numbersdrawn
[i
] = 2;
1937 ds
->cx
= ds
->cy
= -1;
1942 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
1945 sfree(ds
->numbersdrawn
);
1950 ERR_ADJ_TOPLEFT
= 4,
1961 static int *find_errors(game_state
*state
, char *grid
)
1963 int w
= state
->p
.w
, h
= state
->p
.h
;
1964 int *ret
= snewn(w
*h
+ w
+ h
, int);
1965 int *tmp
= snewn(w
*h
*2, int), *dsf
= tmp
+ w
*h
;
1969 * ret[0] through to ret[w*h-1] give error markers for the grid
1970 * squares. After that, ret[w*h] to ret[w*h+w-1] give error
1971 * markers for the column numbers, and ret[w*h+w] to
1972 * ret[w*h+w+h-1] for the row numbers.
1976 * Spot tent-adjacency violations.
1978 for (x
= 0; x
< w
*h
; x
++)
1980 for (y
= 0; y
< h
; y
++) {
1981 for (x
= 0; x
< w
; x
++) {
1982 if (y
+1 < h
&& x
+1 < w
&&
1983 ((grid
[y
*w
+x
] == TENT
&&
1984 grid
[(y
+1)*w
+(x
+1)] == TENT
) ||
1985 (grid
[(y
+1)*w
+x
] == TENT
&&
1986 grid
[y
*w
+(x
+1)] == TENT
))) {
1987 ret
[y
*w
+x
] |= 1 << ERR_ADJ_BOTRIGHT
;
1988 ret
[(y
+1)*w
+x
] |= 1 << ERR_ADJ_TOPRIGHT
;
1989 ret
[y
*w
+(x
+1)] |= 1 << ERR_ADJ_BOTLEFT
;
1990 ret
[(y
+1)*w
+(x
+1)] |= 1 << ERR_ADJ_TOPLEFT
;
1993 grid
[y
*w
+x
] == TENT
&&
1994 grid
[(y
+1)*w
+x
] == TENT
) {
1995 ret
[y
*w
+x
] |= 1 << ERR_ADJ_BOT
;
1996 ret
[(y
+1)*w
+x
] |= 1 << ERR_ADJ_TOP
;
1999 grid
[y
*w
+x
] == TENT
&&
2000 grid
[y
*w
+(x
+1)] == TENT
) {
2001 ret
[y
*w
+x
] |= 1 << ERR_ADJ_RIGHT
;
2002 ret
[y
*w
+(x
+1)] |= 1 << ERR_ADJ_LEFT
;
2008 * Spot numeric clue violations.
2010 for (x
= 0; x
< w
; x
++) {
2011 int tents
= 0, maybetents
= 0;
2012 for (y
= 0; y
< h
; y
++) {
2013 if (grid
[y
*w
+x
] == TENT
)
2015 else if (grid
[y
*w
+x
] == BLANK
)
2018 ret
[w
*h
+x
] = (tents
> state
->numbers
->numbers
[x
] ||
2019 tents
+ maybetents
< state
->numbers
->numbers
[x
]);
2021 for (y
= 0; y
< h
; y
++) {
2022 int tents
= 0, maybetents
= 0;
2023 for (x
= 0; x
< w
; x
++) {
2024 if (grid
[y
*w
+x
] == TENT
)
2026 else if (grid
[y
*w
+x
] == BLANK
)
2029 ret
[w
*h
+w
+y
] = (tents
> state
->numbers
->numbers
[w
+y
] ||
2030 tents
+ maybetents
< state
->numbers
->numbers
[w
+y
]);
2034 * Identify groups of tents with too few trees between them,
2035 * which we do by constructing the connected components of the
2036 * bipartite adjacency graph between tents and trees
2037 * ('bipartite' in the sense that we deliberately ignore
2038 * adjacency between tents or between trees), and highlighting
2039 * all the tents in any component which has a smaller tree
2043 /* Construct the equivalence classes. */
2044 for (y
= 0; y
< h
; y
++) {
2045 for (x
= 0; x
< w
-1; x
++) {
2046 if ((grid
[y
*w
+x
] == TREE
&& grid
[y
*w
+x
+1] == TENT
) ||
2047 (grid
[y
*w
+x
] == TENT
&& grid
[y
*w
+x
+1] == TREE
))
2048 dsf_merge(dsf
, y
*w
+x
, y
*w
+x
+1);
2051 for (y
= 0; y
< h
-1; y
++) {
2052 for (x
= 0; x
< w
; x
++) {
2053 if ((grid
[y
*w
+x
] == TREE
&& grid
[(y
+1)*w
+x
] == TENT
) ||
2054 (grid
[y
*w
+x
] == TENT
&& grid
[(y
+1)*w
+x
] == TREE
))
2055 dsf_merge(dsf
, y
*w
+x
, (y
+1)*w
+x
);
2058 /* Count up the tent/tree difference in each one. */
2059 for (x
= 0; x
< w
*h
; x
++)
2061 for (x
= 0; x
< w
*h
; x
++) {
2062 y
= dsf_canonify(dsf
, x
);
2063 if (grid
[x
] == TREE
)
2065 else if (grid
[x
] == TENT
)
2068 /* And highlight any tent belonging to an equivalence class with
2069 * a score less than zero. */
2070 for (x
= 0; x
< w
*h
; x
++) {
2071 y
= dsf_canonify(dsf
, x
);
2072 if (grid
[x
] == TENT
&& tmp
[y
] < 0)
2073 ret
[x
] |= 1 << ERR_OVERCOMMITTED
;
2077 * Identify groups of trees with too few tents between them.
2078 * This is done similarly, except that we now count BLANK as
2079 * equivalent to TENT, i.e. we only highlight such trees when
2080 * the user hasn't even left _room_ to provide tents for them
2081 * all. (Otherwise, we'd highlight all trees red right at the
2082 * start of the game, before the user had done anything wrong!)
2084 #define TENT(x) ((x)==TENT || (x)==BLANK)
2086 /* Construct the equivalence classes. */
2087 for (y
= 0; y
< h
; y
++) {
2088 for (x
= 0; x
< w
-1; x
++) {
2089 if ((grid
[y
*w
+x
] == TREE
&& TENT(grid
[y
*w
+x
+1])) ||
2090 (TENT(grid
[y
*w
+x
]) && grid
[y
*w
+x
+1] == TREE
))
2091 dsf_merge(dsf
, y
*w
+x
, y
*w
+x
+1);
2094 for (y
= 0; y
< h
-1; y
++) {
2095 for (x
= 0; x
< w
; x
++) {
2096 if ((grid
[y
*w
+x
] == TREE
&& TENT(grid
[(y
+1)*w
+x
])) ||
2097 (TENT(grid
[y
*w
+x
]) && grid
[(y
+1)*w
+x
] == TREE
))
2098 dsf_merge(dsf
, y
*w
+x
, (y
+1)*w
+x
);
2101 /* Count up the tent/tree difference in each one. */
2102 for (x
= 0; x
< w
*h
; x
++)
2104 for (x
= 0; x
< w
*h
; x
++) {
2105 y
= dsf_canonify(dsf
, x
);
2106 if (grid
[x
] == TREE
)
2108 else if (TENT(grid
[x
]))
2111 /* And highlight any tree belonging to an equivalence class with
2112 * a score more than zero. */
2113 for (x
= 0; x
< w
*h
; x
++) {
2114 y
= dsf_canonify(dsf
, x
);
2115 if (grid
[x
] == TREE
&& tmp
[y
] > 0)
2116 ret
[x
] |= 1 << ERR_OVERCOMMITTED
;
2124 static void draw_err_adj(drawing
*dr
, game_drawstate
*ds
, int x
, int y
)
2132 coords
[0] = x
- TILESIZE
*2/5;
2135 coords
[3] = y
- TILESIZE
*2/5;
2136 coords
[4] = x
+ TILESIZE
*2/5;
2139 coords
[7] = y
+ TILESIZE
*2/5;
2140 draw_polygon(dr
, coords
, 4, COL_ERROR
, COL_GRID
);
2143 * Draw an exclamation mark in the diamond. This turns out to
2144 * look unpleasantly off-centre if done via draw_text, so I do
2145 * it by hand on the basis that exclamation marks aren't that
2146 * difficult to draw...
2149 yext
= TILESIZE
*2/5 - (xext
*2+2);
2150 draw_rect(dr
, x
-xext
, y
-yext
, xext
*2+1, yext
*2+1 - (xext
*3),
2152 draw_rect(dr
, x
-xext
, y
+yext
-xext
*2+1, xext
*2+1, xext
*2, COL_ERRTEXT
);
2155 static void draw_tile(drawing
*dr
, game_drawstate
*ds
,
2156 int x
, int y
, int v
, int cur
, int printing
)
2159 int tx
= COORD(x
), ty
= COORD(y
);
2160 int cx
= tx
+ TILESIZE
/2, cy
= ty
+ TILESIZE
/2;
2165 clip(dr
, tx
, ty
, TILESIZE
, TILESIZE
);
2168 draw_rect(dr
, tx
, ty
, TILESIZE
, TILESIZE
, COL_GRID
);
2169 draw_rect(dr
, tx
+1, ty
+1, TILESIZE
-1, TILESIZE
-1,
2170 (v
== BLANK ? COL_BACKGROUND
: COL_GRASS
));
2176 (printing ? draw_rect_outline
: draw_rect
)
2177 (dr
, cx
-TILESIZE
/15, ty
+TILESIZE
*3/10,
2178 2*(TILESIZE
/15)+1, (TILESIZE
*9/10 - TILESIZE
*3/10),
2179 (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERRTRUNK
: COL_TREETRUNK
));
2181 for (i
= 0; i
< (printing ?
2 : 1); i
++) {
2182 int col
= (i
== 1 ? COL_BACKGROUND
:
2183 (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERROR
:
2185 int sub
= i
* (TILESIZE
/32);
2186 draw_circle(dr
, cx
, ty
+TILESIZE
*4/10, TILESIZE
/4 - sub
,
2188 draw_circle(dr
, cx
+TILESIZE
/5, ty
+TILESIZE
/4, TILESIZE
/8 - sub
,
2190 draw_circle(dr
, cx
-TILESIZE
/5, ty
+TILESIZE
/4, TILESIZE
/8 - sub
,
2192 draw_circle(dr
, cx
+TILESIZE
/4, ty
+TILESIZE
*6/13, TILESIZE
/8 - sub
,
2194 draw_circle(dr
, cx
-TILESIZE
/4, ty
+TILESIZE
*6/13, TILESIZE
/8 - sub
,
2197 } else if (v
== TENT
) {
2200 coords
[0] = cx
- TILESIZE
/3;
2201 coords
[1] = cy
+ TILESIZE
/3;
2202 coords
[2] = cx
+ TILESIZE
/3;
2203 coords
[3] = cy
+ TILESIZE
/3;
2205 coords
[5] = cy
- TILESIZE
/3;
2206 col
= (err
& (1<<ERR_OVERCOMMITTED
) ? COL_ERROR
: COL_TENT
);
2207 draw_polygon(dr
, coords
, 3, (printing ?
-1 : col
), col
);
2210 if (err
& (1 << ERR_ADJ_TOPLEFT
))
2211 draw_err_adj(dr
, ds
, tx
, ty
);
2212 if (err
& (1 << ERR_ADJ_TOP
))
2213 draw_err_adj(dr
, ds
, tx
+TILESIZE
/2, ty
);
2214 if (err
& (1 << ERR_ADJ_TOPRIGHT
))
2215 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
);
2216 if (err
& (1 << ERR_ADJ_LEFT
))
2217 draw_err_adj(dr
, ds
, tx
, ty
+TILESIZE
/2);
2218 if (err
& (1 << ERR_ADJ_RIGHT
))
2219 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
+TILESIZE
/2);
2220 if (err
& (1 << ERR_ADJ_BOTLEFT
))
2221 draw_err_adj(dr
, ds
, tx
, ty
+TILESIZE
);
2222 if (err
& (1 << ERR_ADJ_BOT
))
2223 draw_err_adj(dr
, ds
, tx
+TILESIZE
/2, ty
+TILESIZE
);
2224 if (err
& (1 << ERR_ADJ_BOTRIGHT
))
2225 draw_err_adj(dr
, ds
, tx
+TILESIZE
, ty
+TILESIZE
);
2228 int coff
= TILESIZE
/8;
2229 draw_rect_outline(dr
, tx
+ coff
, ty
+ coff
,
2230 TILESIZE
- coff
*2 + 1, TILESIZE
- coff
*2 + 1,
2235 draw_update(dr
, tx
+1, ty
+1, TILESIZE
-1, TILESIZE
-1);
2239 * Internal redraw function, used for printing as well as drawing.
2241 static void int_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
2242 game_state
*state
, int dir
, game_ui
*ui
,
2243 float animtime
, float flashtime
, int printing
)
2245 int w
= state
->p
.w
, h
= state
->p
.h
;
2247 int cx
= -1, cy
= -1;
2253 if (ui
->cdisp
) { cx
= ui
->cx
; cy
= ui
->cy
; }
2254 if (cx
!= ds
->cx
|| cy
!= ds
->cy
) cmoved
= 1;
2257 if (printing
|| !ds
->started
) {
2260 game_compute_size(&state
->p
, TILESIZE
, &ww
, &wh
);
2261 draw_rect(dr
, 0, 0, ww
, wh
, COL_BACKGROUND
);
2262 draw_update(dr
, 0, 0, ww
, wh
);
2267 print_line_width(dr
, TILESIZE
/64);
2272 for (y
= 0; y
<= h
; y
++)
2273 draw_line(dr
, COORD(0), COORD(y
), COORD(w
), COORD(y
), COL_GRID
);
2274 for (x
= 0; x
<= w
; x
++)
2275 draw_line(dr
, COORD(x
), COORD(0), COORD(x
), COORD(h
), COL_GRID
);
2279 flashing
= (int)(flashtime
* 3 / FLASH_TIME
) != 1;
2284 * Find errors. For this we use _part_ of the information from a
2285 * currently active drag: we transform dsx,dsy but not anything
2286 * else. (This seems to strike a good compromise between having
2287 * the error highlights respond instantly to single clicks, but
2288 * not give constant feedback during a right-drag.)
2290 if (ui
&& ui
->drag_button
>= 0) {
2291 tmpgrid
= snewn(w
*h
, char);
2292 memcpy(tmpgrid
, state
->grid
, w
*h
);
2293 tmpgrid
[ui
->dsy
* w
+ ui
->dsx
] =
2294 drag_xform(ui
, ui
->dsx
, ui
->dsy
, tmpgrid
[ui
->dsy
* w
+ ui
->dsx
]);
2295 errors
= find_errors(state
, tmpgrid
);
2298 errors
= find_errors(state
, state
->grid
);
2304 for (y
= 0; y
< h
; y
++) {
2305 for (x
= 0; x
< w
; x
++) {
2306 int v
= state
->grid
[y
*w
+x
];
2310 * We deliberately do not take drag_ok into account
2311 * here, because user feedback suggests that it's
2312 * marginally nicer not to have the drag effects
2313 * flickering on and off disconcertingly.
2315 if (ui
&& ui
->drag_button
>= 0)
2316 v
= drag_xform(ui
, x
, y
, v
);
2318 if (flashing
&& (v
== TREE
|| v
== TENT
))
2322 if ((x
== cx
&& y
== cy
) ||
2323 (x
== ds
->cx
&& y
== ds
->cy
)) credraw
= 1;
2328 if (printing
|| ds
->drawn
[y
*w
+x
] != v
|| credraw
) {
2329 draw_tile(dr
, ds
, x
, y
, v
, (x
== cx
&& y
== cy
), printing
);
2331 ds
->drawn
[y
*w
+x
] = v
;
2337 * Draw (or redraw, if their error-highlighted state has
2338 * changed) the numbers.
2340 for (x
= 0; x
< w
; x
++) {
2341 if (ds
->numbersdrawn
[x
] != errors
[w
*h
+x
]) {
2343 draw_rect(dr
, COORD(x
), COORD(h
)+1, TILESIZE
, BRBORDER
-1,
2345 sprintf(buf
, "%d", state
->numbers
->numbers
[x
]);
2346 draw_text(dr
, COORD(x
) + TILESIZE
/2, COORD(h
+1),
2347 FONT_VARIABLE
, TILESIZE
/2, ALIGN_HCENTRE
|ALIGN_VNORMAL
,
2348 (errors
[w
*h
+x
] ? COL_ERROR
: COL_GRID
), buf
);
2349 draw_update(dr
, COORD(x
), COORD(h
)+1, TILESIZE
, BRBORDER
-1);
2350 ds
->numbersdrawn
[x
] = errors
[w
*h
+x
];
2353 for (y
= 0; y
< h
; y
++) {
2354 if (ds
->numbersdrawn
[w
+y
] != errors
[w
*h
+w
+y
]) {
2356 draw_rect(dr
, COORD(w
)+1, COORD(y
), BRBORDER
-1, TILESIZE
,
2358 sprintf(buf
, "%d", state
->numbers
->numbers
[w
+y
]);
2359 draw_text(dr
, COORD(w
+1), COORD(y
) + TILESIZE
/2,
2360 FONT_VARIABLE
, TILESIZE
/2, ALIGN_HRIGHT
|ALIGN_VCENTRE
,
2361 (errors
[w
*h
+w
+y
] ? COL_ERROR
: COL_GRID
), buf
);
2362 draw_update(dr
, COORD(w
)+1, COORD(y
), BRBORDER
-1, TILESIZE
);
2363 ds
->numbersdrawn
[w
+y
] = errors
[w
*h
+w
+y
];
2375 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
2376 game_state
*state
, int dir
, game_ui
*ui
,
2377 float animtime
, float flashtime
)
2379 int_redraw(dr
, ds
, oldstate
, state
, dir
, ui
, animtime
, flashtime
, FALSE
);
2382 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
2383 int dir
, game_ui
*ui
)
2388 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
2389 int dir
, game_ui
*ui
)
2391 if (!oldstate
->completed
&& newstate
->completed
&&
2392 !oldstate
->used_solve
&& !newstate
->used_solve
)
2398 static int game_timing_state(game_state
*state
, game_ui
*ui
)
2403 static void game_print_size(game_params
*params
, float *x
, float *y
)
2408 * I'll use 6mm squares by default.
2410 game_compute_size(params
, 600, &pw
, &ph
);
2415 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
2419 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
2420 game_drawstate ads
, *ds
= &ads
;
2421 game_set_size(dr
, ds
, NULL
, tilesize
);
2423 c
= print_mono_colour(dr
, 1); assert(c
== COL_BACKGROUND
);
2424 c
= print_mono_colour(dr
, 0); assert(c
== COL_GRID
);
2425 c
= print_mono_colour(dr
, 1); assert(c
== COL_GRASS
);
2426 c
= print_mono_colour(dr
, 0); assert(c
== COL_TREETRUNK
);
2427 c
= print_mono_colour(dr
, 0); assert(c
== COL_TREELEAF
);
2428 c
= print_mono_colour(dr
, 0); assert(c
== COL_TENT
);
2430 int_redraw(dr
, ds
, NULL
, state
, +1, NULL
, 0.0F
, 0.0F
, TRUE
);
2434 #define thegame tents
2437 const struct game thegame
= {
2438 "Tents", "games.tents", "tents",
2445 TRUE
, game_configure
, custom_params
,
2453 FALSE
, game_can_format_as_text_now
, game_text_format
,
2461 PREFERRED_TILESIZE
, game_compute_size
, game_set_size
,
2464 game_free_drawstate
,
2468 TRUE
, FALSE
, game_print_size
, game_print
,
2469 FALSE
, /* wants_statusbar */
2470 FALSE
, game_timing_state
,
2471 REQUIRE_RBUTTON
, /* flags */
2474 #ifdef STANDALONE_SOLVER
2478 int main(int argc
, char **argv
)
2482 char *id
= NULL
, *desc
, *err
;
2484 int ret
, diff
, really_verbose
= FALSE
;
2485 struct solver_scratch
*sc
;
2487 while (--argc
> 0) {
2489 if (!strcmp(p
, "-v")) {
2490 really_verbose
= TRUE
;
2491 } else if (!strcmp(p
, "-g")) {
2493 } else if (*p
== '-') {
2494 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
2502 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
2506 desc
= strchr(id
, ':');
2508 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
2513 p
= default_params();
2514 decode_params(p
, id
);
2515 err
= validate_desc(p
, desc
);
2517 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
2520 s
= new_game(NULL
, p
, desc
);
2521 s2
= new_game(NULL
, p
, desc
);
2523 sc
= new_scratch(p
->w
, p
->h
);
2526 * When solving an Easy puzzle, we don't want to bother the
2527 * user with Hard-level deductions. For this reason, we grade
2528 * the puzzle internally before doing anything else.
2530 ret
= -1; /* placate optimiser */
2531 for (diff
= 0; diff
< DIFFCOUNT
; diff
++) {
2532 ret
= tents_solve(p
->w
, p
->h
, s
->grid
, s
->numbers
->numbers
,
2533 s2
->grid
, sc
, diff
);
2538 if (diff
== DIFFCOUNT
) {
2540 printf("Difficulty rating: too hard to solve internally\n");
2542 printf("Unable to find a unique solution\n");
2546 printf("Difficulty rating: impossible (no solution exists)\n");
2548 printf("Difficulty rating: %s\n", tents_diffnames
[diff
]);
2550 verbose
= really_verbose
;
2551 ret
= tents_solve(p
->w
, p
->h
, s
->grid
, s
->numbers
->numbers
,
2552 s2
->grid
, sc
, diff
);
2554 printf("Puzzle is inconsistent\n");
2556 fputs(game_text_format(s2
), stdout
);
2565 /* vim: set shiftwidth=4 tabstop=8: */