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.
18 * Subroutine which implements a function used in computing both
19 * whether a square can safely be added to an omino, and whether
20 * it can safely be removed.
22 * We enumerate the eight squares 8-adjacent to this one, in
23 * cyclic order. We go round that loop and count the number of
24 * times we find a square owned by the target omino next to one
25 * not owned by it. We then return success iff that count is 2.
27 * When adding a square to an omino, this is precisely the
28 * criterion which tells us that adding the square won't leave a
29 * hole in the middle of the omino. (There's no explicit
30 * requirement in the statement of our problem that the ominoes be
31 * simply connected, but we do know they must be all of equal size
32 * and so it's clear that we must avoid leaving holes, since a
33 * hole would necessarily be smaller than the maximum omino size.)
35 * When removing a square from an omino, the _same_ criterion
36 * tells us that removing the square won't disconnect the omino.
38 static int addremcommon(int w, int h, int x, int y, int *own, int val)
43 for (dir = 0; dir < 8; dir++) {
44 int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1);
45 int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1);
46 int sx = x+dx, sy = y+dy;
48 if (sx < 0 || sx >= w || sy < 0 || sy >= h)
49 neighbours[dir] = -1; /* outside the grid */
51 neighbours[dir] = own[sy*w+sx];
55 * To begin with, check 4-adjacency.
57 if (neighbours[0] != val && neighbours[2] != val &&
58 neighbours[4] != val && neighbours[6] != val)
63 for (dir = 0; dir < 8; dir++) {
64 int next = (dir + 1) & 7;
65 int gotthis = (neighbours[dir] == val);
66 int gotnext = (neighbours[next] == val);
68 if (gotthis != gotnext)
76 * w and h are the dimensions of the rectangle.
78 * k is the size of the required ominoes. (So k must divide w*h,
81 * The returned result is a w*h-sized dsf.
83 * In both of the above suggested use cases, the user would
84 * probably want w==h==k, but that isn't a requirement.
86 int *divvy_rectangle(int w, int h, int k, random_state *rs)
88 int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf;
90 int i, j, n, x, y, qhead, qtail;
95 order = snewn(wh, int);
98 sizes = snewn(n, int);
99 queue = snewn(n, int);
100 addable = snewn(wh*4, int);
101 removable = snewn(wh, int);
104 * Permute the grid squares into a random order, which will be
105 * used for iterating over the grid whenever we need to search
106 * for something. This prevents directional bias and arranges
107 * for the answer to be non-deterministic.
109 for (i = 0; i < wh; i++)
111 shuffle(order, wh, sizeof(*order), rs);
114 * Begin by choosing a starting square at random for each
117 for (i = 0; i < wh; i++) {
120 for (i = 0; i < n; i++) {
126 * Now repeatedly pick a random omino which isn't already at
127 * the target size, and find a way to expand it by one. This
128 * may involve stealing a square from another omino, in which
129 * case we then re-expand that omino, forming a chain of
130 * square-stealing which terminates in an as yet unclaimed
131 * square. Hence every successful iteration around this loop
132 * causes the number of unclaimed squares to drop by one, and
133 * so the process is bounded in duration.
137 #ifdef DIVVY_DIAGNOSTICS
140 printf("Top of loop. Current grid:\n");
141 for (y = 0; y < h; y++) {
142 for (x = 0; x < w; x++)
143 printf("%3d", own[y*w+x]);
150 * Go over the grid and figure out which squares can
151 * safely be added to, or removed from, each omino. We
152 * don't take account of other ominoes in this process, so
153 * we will often end up knowing that a square can be
154 * poached from one omino by another.
156 * For each square, there may be up to four ominoes to
157 * which it can be added (those to which it is
160 for (y = 0; y < h; y++) {
161 for (x = 0; x < w; x++) {
167 removable[yx] = 0; /* can't remove if it's not owned! */
170 * See if this square can be removed from its
171 * omino without disconnecting it.
173 removable[yx] = addremcommon(w, h, x, y, own, curr);
176 for (dir = 0; dir < 4; dir++) {
177 int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0);
178 int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0);
179 int sx = x + dx, sy = y + dy;
182 addable[yx*4+dir] = -1;
184 if (sx < 0 || sx >= w || sy < 0 || sy >= h)
185 continue; /* no omino here! */
187 continue; /* also no omino here */
188 if (own[syx] == own[yx])
189 continue; /* we already got one */
190 if (!addremcommon(w, h, x, y, own, own[syx]))
191 continue; /* would non-simply connect the omino */
193 addable[yx*4+dir] = own[syx];
198 for (i = j = 0; i < n; i++)
202 break; /* all ominoes are complete! */
203 j = tmp[random_upto(rs, j)];
206 * So we're trying to expand omino j. We breadth-first
207 * search out from j across the space of ominoes.
209 * For bfs purposes, we use two elements of tmp per omino:
210 * tmp[2*i+0] tells us which omino we got to i from, and
211 * tmp[2*i+1] numbers the grid square that omino stole
214 * This requires that wh (the size of tmp) is at least 2n,
215 * i.e. k is at least 2. There would have been nothing to
216 * stop a user calling this function with k=1, but if they
217 * did then we wouldn't have got to _here_ in the code -
218 * we would have noticed above that all ominoes were
219 * already at their target sizes, and terminated :-)
222 for (i = 0; i < n; i++)
223 tmp[2*i] = tmp[2*i+1] = -1;
226 tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */
228 while (qhead < qtail) {
234 * We wish to expand omino j. However, we might have
235 * got here by omino j having a square stolen from it,
236 * so first of all we must temporarily mark that
237 * square as not belonging to j, so that our adjacency
238 * calculations don't assume j _does_ belong to us.
242 assert(own[tmpsq] == j);
247 * OK. Now begin by seeing if we can find any
248 * unclaimed square into which we can expand omino j.
249 * If we find one, the entire bfs terminates.
251 for (i = 0; i < wh; i++) {
254 if (own[order[i]] >= 0)
255 continue; /* this square is claimed */
256 for (dir = 0; dir < 4; dir++)
257 if (addable[order[i]*4+dir] == j) {
259 * We know this square is addable to this
260 * omino with the grid in the state it had
261 * at the top of the loop. However, we
262 * must now check that it's _still_
263 * addable to this omino when the omino is
264 * missing a square. To do this it's only
265 * necessary to re-check addremcommon.
267 if (!addremcommon(w, h, order[i]%w, order[i]/w,
273 continue; /* we can't add this square to j */
274 break; /* got one! */
280 * We are done. We can add square i to omino j,
281 * and then backtrack along the trail in tmp
282 * moving squares between ominoes, ending up
283 * expanding our starting omino by one.
294 * Increment the size of the starting omino.
299 * Terminate the bfs loop.
305 * If we get here, we haven't been able to expand
306 * omino j into an unclaimed square. So now we begin
307 * to investigate expanding it into squares which are
308 * claimed by ominoes the bfs has not yet visited.
310 for (i = 0; i < wh; i++) {
314 if (nj < 0 || tmp[2*nj] != -1)
315 continue; /* unclaimed, or owned by wrong omino */
316 if (!removable[order[i]])
317 continue; /* its omino won't let it go */
319 for (dir = 0; dir < 4; dir++)
320 if (addable[order[i]*4+dir] == j) {
322 * As above, re-check addremcommon.
324 if (!addremcommon(w, h, order[i]%w, order[i]/w,
329 * We have found a square we can use to
330 * expand omino j, at the expense of the
331 * as-yet unvisited omino nj. So add this
337 tmp[2*nj+1] = order[i];
340 * Now terminate the loop over dir, to
341 * ensure we don't accidentally add the
342 * same omino twice to the queue.
349 * Restore the temporarily removed square.
355 * Advance the queue head.
360 if (qhead == qtail) {
362 * We have finished the bfs and not found any way to
363 * expand omino j. Panic, and return failure.
365 * FIXME: or should we loop over all ominoes before we
374 * Construct the output dsf.
376 for (i = 0; i < wh; i++) {
377 assert(own[i] >= 0 && own[i] < n);
380 retdsf = snew_dsf(wh);
381 for (i = 0; i < wh; i++) {
382 dsf_merge(retdsf, i, tmp[own[i]]);
386 * Construct the output dsf a different way, to verify that
387 * the ominoes really are k-ominoes and we haven't
388 * accidentally split one into two disconnected pieces.
391 for (y = 0; y < h; y++)
392 for (x = 0; x+1 < w; x++)
393 if (own[y*w+x] == own[y*w+(x+1)])
394 dsf_merge(tmp, y*w+x, y*w+(x+1));
395 for (x = 0; x < w; x++)
396 for (y = 0; y+1 < h; y++)
397 if (own[y*w+x] == own[(y+1)*w+x])
398 dsf_merge(tmp, y*w+x, (y+1)*w+x);
399 for (i = 0; i < wh; i++) {
400 j = dsf_canonify(retdsf, i);
401 assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i));
407 * Free our temporary working space.
426 * gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
430 * gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c
433 int main(int argc, char **argv)
437 int w = 9, h = 4, k = 6, tries = 100;
440 rs = random_new("123456", 6);
449 tries = atoi(argv[4]);
452 for (i = 0; i < tries; i++) {
453 dsf = divvy_rectangle(w, h, k, rs);
459 for (y = 0; y <= 2*h; y++) {
460 for (x = 0; x <= 2*w; x++) {
461 int miny = y/2 - 1, maxy = y/2;
462 int minx = x/2 - 1, maxx = x/2;
463 int classes[4], tx, ty;
464 for (ty = 0; ty < 2; ty++)
465 for (tx = 0; tx < 2; tx++) {
466 int cx = minx+tx, cy = miny+ty;
467 if (cx < 0 || cx >= w || cy < 0 || cy >= h)
468 classes[ty*2+tx] = -1;
470 classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx);
472 switch (y%2 * 2 + x%2) {
475 * Cases for the corner:
477 * - if all four surrounding squares
478 * belong to the same omino, we print a
481 * - if the top two are the same and the
482 * bottom two are the same, we print a
485 * - if the left two are the same and the
486 * right two are the same, we print a
489 * - otherwise, we print a cross.
491 if (classes[0] == classes[1] &&
492 classes[1] == classes[2] &&
493 classes[2] == classes[3])
495 else if (classes[0] == classes[1] &&
496 classes[2] == classes[3])
498 else if (classes[0] == classes[2] &&
499 classes[1] == classes[3])
504 case 1: /* horiz edge */
505 if (classes[1] == classes[3])
510 case 2: /* vert edge */
511 if (classes[2] == classes[3])
516 case 3: /* square centre */
528 printf("%d successes out of %d tries\n", successes, tries);