d083e766c9234aa7057025942c02a106baa31056
2 * Library code to divide up a rectangle into a number of equally
3 * sized ominoes, in a random fashion.
5 * Could use this for generating solved grids of
6 * http://www.nikoli.co.jp/ja/puzzles/block_puzzle/
7 * or for generating the playfield for Jigsaw Sudoku.
11 * Possible improvements which might cut the fail rate:
13 * - instead of picking one omino to extend in an iteration, try
14 * them all in succession (in a randomised order)
16 * - (for real rigour) instead of bfsing over ominoes, bfs over
17 * the space of possible _removed squares_. That way we aren't
18 * limited to randomly choosing a single square to remove from
19 * an omino and failing if that particular square doesn't
22 * However, I don't currently think it's neecss~|~
33 * Subroutine which implements a function used in computing both
34 * whether a square can safely be added to an omino, and whether
35 * it can safely be removed.
37 * We enumerate the eight squares 8-adjacent to this one, in
38 * cyclic order. We go round that loop and count the number of
39 * times we find a square owned by the target omino next to one
40 * not owned by it. We then return success iff that count is 2.
42 * When adding a square to an omino, this is precisely the
43 * criterion which tells us that adding the square won't leave a
44 * hole in the middle of the omino. (There's no explicit
45 * requirement in the statement of our problem that the ominoes be
46 * simply connected, but we do know they must be all of equal size
47 * and so it's clear that we must avoid leaving holes, since a
48 * hole would necessarily be smaller than the maximum omino size.)
50 * When removing a square from an omino, the _same_ criterion
51 * tells us that removing the square won't disconnect the omino.
53 static int addremcommon(int w
, int h
, int x
, int y
, int *own
, int val
)
58 for (dir
= 0; dir
< 8; dir
++) {
59 int dx
= ((dir
& 3) == 2 ?
0 : dir
> 2 && dir
< 6 ?
+1 : -1);
60 int dy
= ((dir
& 3) == 0 ?
0 : dir
< 4 ?
-1 : +1);
61 int sx
= x
+dx
, sy
= y
+dy
;
63 if (sx
< 0 || sx
>= w
|| sy
< 0 || sy
>= h
)
64 neighbours
[dir
] = -1; /* outside the grid */
66 neighbours
[dir
] = own
[sy
*w
+sx
];
70 * To begin with, check 4-adjacency.
72 if (neighbours
[0] != val
&& neighbours
[2] != val
&&
73 neighbours
[4] != val
&& neighbours
[6] != val
)
78 for (dir
= 0; dir
< 8; dir
++) {
79 int next
= (dir
+ 1) & 7;
80 int gotthis
= (neighbours
[dir
] == val
);
81 int gotnext
= (neighbours
[next
] == val
);
83 if (gotthis
!= gotnext
)
91 * w and h are the dimensions of the rectangle.
93 * k is the size of the required ominoes. (So k must divide w*h,
96 * The returned result is a w*h-sized dsf.
98 * In both of the above suggested use cases, the user would
99 * probably want w==h==k, but that isn't a requirement.
101 static int *divvy_internal(int w
, int h
, int k
, random_state
*rs
)
103 int *order
, *queue
, *tmp
, *own
, *sizes
, *addable
, *removable
, *retdsf
;
105 int i
, j
, n
, x
, y
, qhead
, qtail
;
110 order
= snewn(wh
, int);
111 tmp
= snewn(wh
, int);
112 own
= snewn(wh
, int);
113 sizes
= snewn(n
, int);
114 queue
= snewn(n
, int);
115 addable
= snewn(wh
*4, int);
116 removable
= snewn(wh
, int);
119 * Permute the grid squares into a random order, which will be
120 * used for iterating over the grid whenever we need to search
121 * for something. This prevents directional bias and arranges
122 * for the answer to be non-deterministic.
124 for (i
= 0; i
< wh
; i
++)
126 shuffle(order
, wh
, sizeof(*order
), rs
);
129 * Begin by choosing a starting square at random for each
132 for (i
= 0; i
< wh
; i
++) {
135 for (i
= 0; i
< n
; i
++) {
141 * Now repeatedly pick a random omino which isn't already at
142 * the target size, and find a way to expand it by one. This
143 * may involve stealing a square from another omino, in which
144 * case we then re-expand that omino, forming a chain of
145 * square-stealing which terminates in an as yet unclaimed
146 * square. Hence every successful iteration around this loop
147 * causes the number of unclaimed squares to drop by one, and
148 * so the process is bounded in duration.
152 #ifdef DIVVY_DIAGNOSTICS
155 printf("Top of loop. Current grid:\n");
156 for (y
= 0; y
< h
; y
++) {
157 for (x
= 0; x
< w
; x
++)
158 printf("%3d", own
[y
*w
+x
]);
165 * Go over the grid and figure out which squares can
166 * safely be added to, or removed from, each omino. We
167 * don't take account of other ominoes in this process, so
168 * we will often end up knowing that a square can be
169 * poached from one omino by another.
171 * For each square, there may be up to four ominoes to
172 * which it can be added (those to which it is
175 for (y
= 0; y
< h
; y
++) {
176 for (x
= 0; x
< w
; x
++) {
182 removable
[yx
] = FALSE
; /* can't remove if not owned! */
183 } else if (sizes
[curr
] == 1) {
184 removable
[yx
] = TRUE
; /* can always remove a singleton */
187 * See if this square can be removed from its
188 * omino without disconnecting it.
190 removable
[yx
] = addremcommon(w
, h
, x
, y
, own
, curr
);
193 for (dir
= 0; dir
< 4; dir
++) {
194 int dx
= (dir
== 0 ?
-1 : dir
== 1 ?
+1 : 0);
195 int dy
= (dir
== 2 ?
-1 : dir
== 3 ?
+1 : 0);
196 int sx
= x
+ dx
, sy
= y
+ dy
;
199 addable
[yx
*4+dir
] = -1;
201 if (sx
< 0 || sx
>= w
|| sy
< 0 || sy
>= h
)
202 continue; /* no omino here! */
204 continue; /* also no omino here */
205 if (own
[syx
] == own
[yx
])
206 continue; /* we already got one */
207 if (!addremcommon(w
, h
, x
, y
, own
, own
[syx
]))
208 continue; /* would non-simply connect the omino */
210 addable
[yx
*4+dir
] = own
[syx
];
215 for (i
= j
= 0; i
< n
; i
++)
219 break; /* all ominoes are complete! */
220 j
= tmp
[random_upto(rs
, j
)];
221 #ifdef DIVVY_DIAGNOSTICS
222 printf("Trying to extend %d\n", j
);
226 * So we're trying to expand omino j. We breadth-first
227 * search out from j across the space of ominoes.
229 * For bfs purposes, we use two elements of tmp per omino:
230 * tmp[2*i+0] tells us which omino we got to i from, and
231 * tmp[2*i+1] numbers the grid square that omino stole
234 * This requires that wh (the size of tmp) is at least 2n,
235 * i.e. k is at least 2. There would have been nothing to
236 * stop a user calling this function with k=1, but if they
237 * did then we wouldn't have got to _here_ in the code -
238 * we would have noticed above that all ominoes were
239 * already at their target sizes, and terminated :-)
242 for (i
= 0; i
< n
; i
++)
243 tmp
[2*i
] = tmp
[2*i
+1] = -1;
246 tmp
[2*j
] = tmp
[2*j
+1] = -2; /* special value: `starting point' */
248 while (qhead
< qtail
) {
254 * We wish to expand omino j. However, we might have
255 * got here by omino j having a square stolen from it,
256 * so first of all we must temporarily mark that
257 * square as not belonging to j, so that our adjacency
258 * calculations don't assume j _does_ belong to us.
262 assert(own
[tmpsq
] == j
);
267 * OK. Now begin by seeing if we can find any
268 * unclaimed square into which we can expand omino j.
269 * If we find one, the entire bfs terminates.
271 for (i
= 0; i
< wh
; i
++) {
274 if (own
[order
[i
]] != -1)
275 continue; /* this square is claimed */
278 * Special case: if our current omino was size 1
279 * and then had a square stolen from it, it's now
280 * size zero, which means it's valid to `expand'
281 * it into _any_ unclaimed square.
283 if (sizes
[j
] == 1 && tmpsq
>= 0)
287 * Failing that, we must do the full test for
290 for (dir
= 0; dir
< 4; dir
++)
291 if (addable
[order
[i
]*4+dir
] == j
) {
293 * We know this square is addable to this
294 * omino with the grid in the state it had
295 * at the top of the loop. However, we
296 * must now check that it's _still_
297 * addable to this omino when the omino is
298 * missing a square. To do this it's only
299 * necessary to re-check addremcommon.
301 if (!addremcommon(w
, h
, order
[i
]%w
, order
[i
]/w
,
307 continue; /* we can't add this square to j */
309 break; /* got one! */
315 * Restore the temporarily removed square _before_
316 * we start shifting ownerships about.
322 * We are done. We can add square i to omino j,
323 * and then backtrack along the trail in tmp
324 * moving squares between ominoes, ending up
325 * expanding our starting omino by one.
327 #ifdef DIVVY_DIAGNOSTICS
328 printf("(%d,%d)", i
%w
, i
/w
);
332 #ifdef DIVVY_DIAGNOSTICS
339 #ifdef DIVVY_DIAGNOSTICS
340 printf("; (%d,%d)", i
%w
, i
/w
);
343 #ifdef DIVVY_DIAGNOSTICS
348 * Increment the size of the starting omino.
353 * Terminate the bfs loop.
359 * If we get here, we haven't been able to expand
360 * omino j into an unclaimed square. So now we begin
361 * to investigate expanding it into squares which are
362 * claimed by ominoes the bfs has not yet visited.
364 for (i
= 0; i
< wh
; i
++) {
368 if (nj
< 0 || tmp
[2*nj
] != -1)
369 continue; /* unclaimed, or owned by wrong omino */
370 if (!removable
[order
[i
]])
371 continue; /* its omino won't let it go */
373 for (dir
= 0; dir
< 4; dir
++)
374 if (addable
[order
[i
]*4+dir
] == j
) {
376 * As above, re-check addremcommon.
378 if (!addremcommon(w
, h
, order
[i
]%w
, order
[i
]/w
,
383 * We have found a square we can use to
384 * expand omino j, at the expense of the
385 * as-yet unvisited omino nj. So add this
391 tmp
[2*nj
+1] = order
[i
];
394 * Now terminate the loop over dir, to
395 * ensure we don't accidentally add the
396 * same omino twice to the queue.
403 * Restore the temporarily removed square.
409 * Advance the queue head.
414 if (qhead
== qtail
) {
416 * We have finished the bfs and not found any way to
417 * expand omino j. Panic, and return failure.
419 * FIXME: or should we loop over all ominoes before we
422 #ifdef DIVVY_DIAGNOSTICS
430 #ifdef DIVVY_DIAGNOSTICS
433 printf("SUCCESS! Final grid:\n");
434 for (y
= 0; y
< h
; y
++) {
435 for (x
= 0; x
< w
; x
++)
436 printf("%3d", own
[y
*w
+x
]);
443 * Construct the output dsf.
445 for (i
= 0; i
< wh
; i
++) {
446 assert(own
[i
] >= 0 && own
[i
] < n
);
449 retdsf
= snew_dsf(wh
);
450 for (i
= 0; i
< wh
; i
++) {
451 dsf_merge(retdsf
, i
, tmp
[own
[i
]]);
455 * Construct the output dsf a different way, to verify that
456 * the ominoes really are k-ominoes and we haven't
457 * accidentally split one into two disconnected pieces.
460 for (y
= 0; y
< h
; y
++)
461 for (x
= 0; x
+1 < w
; x
++)
462 if (own
[y
*w
+x
] == own
[y
*w
+(x
+1)])
463 dsf_merge(tmp
, y
*w
+x
, y
*w
+(x
+1));
464 for (x
= 0; x
< w
; x
++)
465 for (y
= 0; y
+1 < h
; y
++)
466 if (own
[y
*w
+x
] == own
[(y
+1)*w
+x
])
467 dsf_merge(tmp
, y
*w
+x
, (y
+1)*w
+x
);
468 for (i
= 0; i
< wh
; i
++) {
469 j
= dsf_canonify(retdsf
, i
);
470 assert(dsf_canonify(tmp
, j
) == dsf_canonify(tmp
, i
));
476 * Free our temporary working space.
493 static int fail_counter
= 0;
496 int *divvy_rectangle(int w
, int h
, int k
, random_state
*rs
)
501 ret
= divvy_internal(w
, h
, k
, rs
);
516 * gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
520 * gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
523 int main(int argc
, char **argv
)
527 int w
= 9, h
= 4, k
= 6, tries
= 100;
530 rs
= random_new("123456", 6);
539 tries
= atoi(argv
[4]);
541 for (i
= 0; i
< tries
; i
++) {
544 dsf
= divvy_rectangle(w
, h
, k
, rs
);
547 for (y
= 0; y
<= 2*h
; y
++) {
548 for (x
= 0; x
<= 2*w
; x
++) {
549 int miny
= y
/2 - 1, maxy
= y
/2;
550 int minx
= x
/2 - 1, maxx
= x
/2;
551 int classes
[4], tx
, ty
;
552 for (ty
= 0; ty
< 2; ty
++)
553 for (tx
= 0; tx
< 2; tx
++) {
554 int cx
= minx
+tx
, cy
= miny
+ty
;
555 if (cx
< 0 || cx
>= w
|| cy
< 0 || cy
>= h
)
556 classes
[ty
*2+tx
] = -1;
558 classes
[ty
*2+tx
] = dsf_canonify(dsf
, cy
*w
+cx
);
560 switch (y
%2 * 2 + x
%2) {
563 * Cases for the corner:
565 * - if all four surrounding squares belong
566 * to the same omino, we print a space.
568 * - if the top two are the same and the
569 * bottom two are the same, we print a
572 * - if the left two are the same and the
573 * right two are the same, we print a
576 * - otherwise, we print a cross.
578 if (classes
[0] == classes
[1] &&
579 classes
[1] == classes
[2] &&
580 classes
[2] == classes
[3])
582 else if (classes
[0] == classes
[1] &&
583 classes
[2] == classes
[3])
585 else if (classes
[0] == classes
[2] &&
586 classes
[1] == classes
[3])
591 case 1: /* horiz edge */
592 if (classes
[1] == classes
[3])
597 case 2: /* vert edge */
598 if (classes
[2] == classes
[3])
603 case 3: /* square centre */
614 printf("%d retries needed for %d successes\n", fail_counter
, tries
);