11d273f7 |
1 | /* |
2 | * Library code to divide up a rectangle into a number of equally |
3 | * sized ominoes, in a random fashion. |
4 | * |
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. |
8 | */ |
9 | |
9d36cbd7 |
10 | /* |
82ed4789 |
11 | * This code is restricted to simply connected solutions: that is, |
12 | * no single polyomino may completely surround another (not even |
13 | * with a corner visible to the outside world, in the sense that a |
14 | * 7-omino can `surround' a single square). |
15 | * |
16 | * It's tempting to think that this is a natural consequence of |
17 | * all the ominoes being the same size - after all, a division of |
18 | * anything into 7-ominoes must necessarily have all of them |
19 | * simply connected, because if one was not then the 1-square |
20 | * space in the middle could not be part of any 7-omino - but in |
21 | * fact, for sufficiently large k, it is perfectly possible for a |
22 | * k-omino to completely surround another k-omino. A simple |
23 | * example is this one with two 25-ominoes: |
24 | * |
25 | * +--+--+--+--+--+--+--+ |
26 | * | | |
27 | * + +--+--+--+--+--+ + |
28 | * | | | | |
29 | * + + + + |
30 | * | | | | |
31 | * + + + +--+ |
32 | * | | | | |
33 | * + + + +--+ |
34 | * | | | | |
35 | * + + + + |
36 | * | | | | |
37 | * + +--+--+--+--+--+ + |
38 | * | | |
39 | * +--+--+--+--+--+--+--+ |
40 | * |
41 | * I believe the smallest k which can manage this is 23, which I |
42 | * derive by considering the largest _rectangle_ a k-omino can |
43 | * surround. Consider: to surround an m x n rectangle, a polyomino |
44 | * must have 2m squares along the two m-sides of the rectangle, 2n |
45 | * squares along the n-sides, and fill in three of the corners. So |
46 | * m+n must be at most (k-3)/2. Hence, to find the largest area of |
47 | * such a rectangle, we must find m so as to maximise the product |
48 | * m((k-3)/2-m). This is achieved when m is as close as possible |
49 | * to half of (k-3)/2; so the largest rectangle surroundable by a |
50 | * k-omino is equal to floor(k'/2)*ceil(k'/2), with k'=(k-3)/2. |
51 | * The smallest k for which this is at least k is 23: a 23-omino |
52 | * can surround a 5x5 rectangle, whereas the best a 22-omino can |
53 | * manage is a 5x4. |
54 | * |
55 | * (I don't have a definite argument to show that a k-omino cannot |
56 | * surround a larger area in non-rectangular than rectangular |
57 | * form, but it seems intuitively obvious to me that it can't. I |
58 | * may update this with a more rigorous proof if I think of one.) |
59 | * |
60 | * Anyway: the point is, although constructions of this type are |
61 | * possible for sufficiently large k, divvy_rectangle() will never |
62 | * generate them. This could be considered a weakness for some |
63 | * purposes, in the sense that we can't generate all possible |
64 | * divisions. However, there are many divisions which we are |
65 | * highly unlikely to generate anyway, so in practice it probably |
66 | * isn't _too_ bad. |
67 | * |
68 | * If I wanted to fix this issue, I would have to make the rules |
69 | * more complicated for determining when a square can safely be |
70 | * _removed_ from a polyomino. Adding one becomes easier (a square |
71 | * may be added to a polyomino iff it is 4-adjacent to any square |
72 | * currently part of the polyomino, and the current test for loop |
73 | * formation may be dispensed with), but to determine which |
74 | * squares may be removed we must now resort to analysis of the |
75 | * overall structure of the polyomino rather than the simple local |
76 | * properties we can currently get away with measuring. |
77 | */ |
78 | |
79 | /* |
9d36cbd7 |
80 | * Possible improvements which might cut the fail rate: |
81 | * |
82 | * - instead of picking one omino to extend in an iteration, try |
83 | * them all in succession (in a randomised order) |
84 | * |
85 | * - (for real rigour) instead of bfsing over ominoes, bfs over |
86 | * the space of possible _removed squares_. That way we aren't |
87 | * limited to randomly choosing a single square to remove from |
88 | * an omino and failing if that particular square doesn't |
89 | * happen to work. |
90 | * |
4d31c526 |
91 | * However, I don't currently think it's necessary to do either of |
92 | * these, because the failure rate is already low enough to be |
93 | * easily tolerable, under all circumstances I've been able to |
94 | * think of. |
9d36cbd7 |
95 | */ |
96 | |
11d273f7 |
97 | #include <assert.h> |
98 | #include <stdio.h> |
99 | #include <stdlib.h> |
100 | #include <stddef.h> |
101 | |
102 | #include "puzzles.h" |
103 | |
104 | /* |
105 | * Subroutine which implements a function used in computing both |
106 | * whether a square can safely be added to an omino, and whether |
107 | * it can safely be removed. |
108 | * |
109 | * We enumerate the eight squares 8-adjacent to this one, in |
110 | * cyclic order. We go round that loop and count the number of |
111 | * times we find a square owned by the target omino next to one |
112 | * not owned by it. We then return success iff that count is 2. |
113 | * |
114 | * When adding a square to an omino, this is precisely the |
115 | * criterion which tells us that adding the square won't leave a |
a9ea89c5 |
116 | * hole in the middle of the omino. (If it did, then things get |
117 | * more complicated; see above.) |
11d273f7 |
118 | * |
119 | * When removing a square from an omino, the _same_ criterion |
120 | * tells us that removing the square won't disconnect the omino. |
a9ea89c5 |
121 | * (This only works _because_ we've ensured the omino is simply |
122 | * connected.) |
11d273f7 |
123 | */ |
124 | static int addremcommon(int w, int h, int x, int y, int *own, int val) |
125 | { |
126 | int neighbours[8]; |
127 | int dir, count; |
128 | |
129 | for (dir = 0; dir < 8; dir++) { |
130 | int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1); |
131 | int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1); |
132 | int sx = x+dx, sy = y+dy; |
133 | |
134 | if (sx < 0 || sx >= w || sy < 0 || sy >= h) |
135 | neighbours[dir] = -1; /* outside the grid */ |
136 | else |
137 | neighbours[dir] = own[sy*w+sx]; |
138 | } |
139 | |
140 | /* |
141 | * To begin with, check 4-adjacency. |
142 | */ |
143 | if (neighbours[0] != val && neighbours[2] != val && |
144 | neighbours[4] != val && neighbours[6] != val) |
145 | return FALSE; |
146 | |
147 | count = 0; |
148 | |
149 | for (dir = 0; dir < 8; dir++) { |
150 | int next = (dir + 1) & 7; |
151 | int gotthis = (neighbours[dir] == val); |
152 | int gotnext = (neighbours[next] == val); |
153 | |
154 | if (gotthis != gotnext) |
155 | count++; |
156 | } |
157 | |
158 | return (count == 2); |
159 | } |
160 | |
161 | /* |
162 | * w and h are the dimensions of the rectangle. |
163 | * |
164 | * k is the size of the required ominoes. (So k must divide w*h, |
165 | * of course.) |
166 | * |
167 | * The returned result is a w*h-sized dsf. |
168 | * |
169 | * In both of the above suggested use cases, the user would |
170 | * probably want w==h==k, but that isn't a requirement. |
171 | */ |
9d36cbd7 |
172 | static int *divvy_internal(int w, int h, int k, random_state *rs) |
11d273f7 |
173 | { |
174 | int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf; |
175 | int wh = w*h; |
176 | int i, j, n, x, y, qhead, qtail; |
177 | |
178 | n = wh / k; |
179 | assert(wh == k*n); |
180 | |
181 | order = snewn(wh, int); |
182 | tmp = snewn(wh, int); |
183 | own = snewn(wh, int); |
184 | sizes = snewn(n, int); |
185 | queue = snewn(n, int); |
186 | addable = snewn(wh*4, int); |
187 | removable = snewn(wh, int); |
188 | |
189 | /* |
190 | * Permute the grid squares into a random order, which will be |
191 | * used for iterating over the grid whenever we need to search |
192 | * for something. This prevents directional bias and arranges |
193 | * for the answer to be non-deterministic. |
194 | */ |
195 | for (i = 0; i < wh; i++) |
196 | order[i] = i; |
197 | shuffle(order, wh, sizeof(*order), rs); |
198 | |
199 | /* |
200 | * Begin by choosing a starting square at random for each |
201 | * omino. |
202 | */ |
203 | for (i = 0; i < wh; i++) { |
204 | own[i] = -1; |
205 | } |
206 | for (i = 0; i < n; i++) { |
207 | own[order[i]] = i; |
208 | sizes[i] = 1; |
209 | } |
210 | |
211 | /* |
212 | * Now repeatedly pick a random omino which isn't already at |
213 | * the target size, and find a way to expand it by one. This |
214 | * may involve stealing a square from another omino, in which |
215 | * case we then re-expand that omino, forming a chain of |
216 | * square-stealing which terminates in an as yet unclaimed |
217 | * square. Hence every successful iteration around this loop |
218 | * causes the number of unclaimed squares to drop by one, and |
219 | * so the process is bounded in duration. |
220 | */ |
221 | while (1) { |
222 | |
223 | #ifdef DIVVY_DIAGNOSTICS |
224 | { |
225 | int x, y; |
226 | printf("Top of loop. Current grid:\n"); |
227 | for (y = 0; y < h; y++) { |
228 | for (x = 0; x < w; x++) |
229 | printf("%3d", own[y*w+x]); |
230 | printf("\n"); |
231 | } |
232 | } |
233 | #endif |
234 | |
235 | /* |
236 | * Go over the grid and figure out which squares can |
237 | * safely be added to, or removed from, each omino. We |
238 | * don't take account of other ominoes in this process, so |
239 | * we will often end up knowing that a square can be |
240 | * poached from one omino by another. |
241 | * |
242 | * For each square, there may be up to four ominoes to |
243 | * which it can be added (those to which it is |
244 | * 4-adjacent). |
245 | */ |
246 | for (y = 0; y < h; y++) { |
247 | for (x = 0; x < w; x++) { |
248 | int yx = y*w+x; |
249 | int curr = own[yx]; |
250 | int dir; |
251 | |
252 | if (curr < 0) { |
9d36cbd7 |
253 | removable[yx] = FALSE; /* can't remove if not owned! */ |
254 | } else if (sizes[curr] == 1) { |
255 | removable[yx] = TRUE; /* can always remove a singleton */ |
11d273f7 |
256 | } else { |
257 | /* |
258 | * See if this square can be removed from its |
259 | * omino without disconnecting it. |
260 | */ |
261 | removable[yx] = addremcommon(w, h, x, y, own, curr); |
262 | } |
263 | |
264 | for (dir = 0; dir < 4; dir++) { |
265 | int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0); |
266 | int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0); |
267 | int sx = x + dx, sy = y + dy; |
268 | int syx = sy*w+sx; |
269 | |
270 | addable[yx*4+dir] = -1; |
271 | |
272 | if (sx < 0 || sx >= w || sy < 0 || sy >= h) |
273 | continue; /* no omino here! */ |
274 | if (own[syx] < 0) |
275 | continue; /* also no omino here */ |
276 | if (own[syx] == own[yx]) |
277 | continue; /* we already got one */ |
278 | if (!addremcommon(w, h, x, y, own, own[syx])) |
279 | continue; /* would non-simply connect the omino */ |
280 | |
281 | addable[yx*4+dir] = own[syx]; |
282 | } |
283 | } |
284 | } |
285 | |
286 | for (i = j = 0; i < n; i++) |
287 | if (sizes[i] < k) |
288 | tmp[j++] = i; |
289 | if (j == 0) |
290 | break; /* all ominoes are complete! */ |
291 | j = tmp[random_upto(rs, j)]; |
f40d37bd |
292 | #ifdef DIVVY_DIAGNOSTICS |
293 | printf("Trying to extend %d\n", j); |
294 | #endif |
11d273f7 |
295 | |
296 | /* |
297 | * So we're trying to expand omino j. We breadth-first |
298 | * search out from j across the space of ominoes. |
299 | * |
300 | * For bfs purposes, we use two elements of tmp per omino: |
301 | * tmp[2*i+0] tells us which omino we got to i from, and |
302 | * tmp[2*i+1] numbers the grid square that omino stole |
303 | * from us. |
304 | * |
305 | * This requires that wh (the size of tmp) is at least 2n, |
306 | * i.e. k is at least 2. There would have been nothing to |
307 | * stop a user calling this function with k=1, but if they |
308 | * did then we wouldn't have got to _here_ in the code - |
309 | * we would have noticed above that all ominoes were |
310 | * already at their target sizes, and terminated :-) |
311 | */ |
312 | assert(wh >= 2*n); |
313 | for (i = 0; i < n; i++) |
314 | tmp[2*i] = tmp[2*i+1] = -1; |
315 | qhead = qtail = 0; |
316 | queue[qtail++] = j; |
317 | tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */ |
318 | |
319 | while (qhead < qtail) { |
320 | int tmpsq; |
321 | |
322 | j = queue[qhead]; |
323 | |
324 | /* |
325 | * We wish to expand omino j. However, we might have |
326 | * got here by omino j having a square stolen from it, |
327 | * so first of all we must temporarily mark that |
328 | * square as not belonging to j, so that our adjacency |
329 | * calculations don't assume j _does_ belong to us. |
330 | */ |
331 | tmpsq = tmp[2*j+1]; |
332 | if (tmpsq >= 0) { |
333 | assert(own[tmpsq] == j); |
9d36cbd7 |
334 | own[tmpsq] = -3; |
11d273f7 |
335 | } |
336 | |
337 | /* |
338 | * OK. Now begin by seeing if we can find any |
339 | * unclaimed square into which we can expand omino j. |
340 | * If we find one, the entire bfs terminates. |
341 | */ |
342 | for (i = 0; i < wh; i++) { |
343 | int dir; |
344 | |
9d36cbd7 |
345 | if (own[order[i]] != -1) |
11d273f7 |
346 | continue; /* this square is claimed */ |
9d36cbd7 |
347 | |
348 | /* |
349 | * Special case: if our current omino was size 1 |
350 | * and then had a square stolen from it, it's now |
351 | * size zero, which means it's valid to `expand' |
352 | * it into _any_ unclaimed square. |
353 | */ |
354 | if (sizes[j] == 1 && tmpsq >= 0) |
355 | break; /* got one */ |
356 | |
357 | /* |
358 | * Failing that, we must do the full test for |
359 | * addability. |
360 | */ |
11d273f7 |
361 | for (dir = 0; dir < 4; dir++) |
362 | if (addable[order[i]*4+dir] == j) { |
363 | /* |
364 | * We know this square is addable to this |
365 | * omino with the grid in the state it had |
366 | * at the top of the loop. However, we |
367 | * must now check that it's _still_ |
368 | * addable to this omino when the omino is |
369 | * missing a square. To do this it's only |
370 | * necessary to re-check addremcommon. |
f40d37bd |
371 | */ |
11d273f7 |
372 | if (!addremcommon(w, h, order[i]%w, order[i]/w, |
373 | own, j)) |
374 | continue; |
375 | break; |
376 | } |
377 | if (dir == 4) |
378 | continue; /* we can't add this square to j */ |
9d36cbd7 |
379 | |
11d273f7 |
380 | break; /* got one! */ |
381 | } |
382 | if (i < wh) { |
383 | i = order[i]; |
384 | |
385 | /* |
9d36cbd7 |
386 | * Restore the temporarily removed square _before_ |
387 | * we start shifting ownerships about. |
388 | */ |
389 | if (tmpsq >= 0) |
390 | own[tmpsq] = j; |
391 | |
392 | /* |
11d273f7 |
393 | * We are done. We can add square i to omino j, |
394 | * and then backtrack along the trail in tmp |
395 | * moving squares between ominoes, ending up |
396 | * expanding our starting omino by one. |
397 | */ |
f40d37bd |
398 | #ifdef DIVVY_DIAGNOSTICS |
399 | printf("(%d,%d)", i%w, i/w); |
400 | #endif |
11d273f7 |
401 | while (1) { |
402 | own[i] = j; |
f40d37bd |
403 | #ifdef DIVVY_DIAGNOSTICS |
404 | printf(" -> %d", j); |
405 | #endif |
11d273f7 |
406 | if (tmp[2*j] == -2) |
407 | break; |
408 | i = tmp[2*j+1]; |
409 | j = tmp[2*j]; |
f40d37bd |
410 | #ifdef DIVVY_DIAGNOSTICS |
411 | printf("; (%d,%d)", i%w, i/w); |
412 | #endif |
11d273f7 |
413 | } |
f40d37bd |
414 | #ifdef DIVVY_DIAGNOSTICS |
415 | printf("\n"); |
416 | #endif |
11d273f7 |
417 | |
418 | /* |
419 | * Increment the size of the starting omino. |
420 | */ |
421 | sizes[j]++; |
422 | |
423 | /* |
424 | * Terminate the bfs loop. |
425 | */ |
426 | break; |
427 | } |
428 | |
429 | /* |
430 | * If we get here, we haven't been able to expand |
431 | * omino j into an unclaimed square. So now we begin |
432 | * to investigate expanding it into squares which are |
433 | * claimed by ominoes the bfs has not yet visited. |
434 | */ |
435 | for (i = 0; i < wh; i++) { |
436 | int dir, nj; |
437 | |
438 | nj = own[order[i]]; |
439 | if (nj < 0 || tmp[2*nj] != -1) |
440 | continue; /* unclaimed, or owned by wrong omino */ |
441 | if (!removable[order[i]]) |
442 | continue; /* its omino won't let it go */ |
443 | |
444 | for (dir = 0; dir < 4; dir++) |
445 | if (addable[order[i]*4+dir] == j) { |
446 | /* |
447 | * As above, re-check addremcommon. |
448 | */ |
449 | if (!addremcommon(w, h, order[i]%w, order[i]/w, |
450 | own, j)) |
451 | continue; |
452 | |
453 | /* |
454 | * We have found a square we can use to |
455 | * expand omino j, at the expense of the |
456 | * as-yet unvisited omino nj. So add this |
457 | * to the bfs queue. |
458 | */ |
459 | assert(qtail < n); |
460 | queue[qtail++] = nj; |
461 | tmp[2*nj] = j; |
462 | tmp[2*nj+1] = order[i]; |
463 | |
464 | /* |
465 | * Now terminate the loop over dir, to |
466 | * ensure we don't accidentally add the |
467 | * same omino twice to the queue. |
468 | */ |
469 | break; |
470 | } |
471 | } |
472 | |
473 | /* |
474 | * Restore the temporarily removed square. |
475 | */ |
476 | if (tmpsq >= 0) |
477 | own[tmpsq] = j; |
478 | |
479 | /* |
480 | * Advance the queue head. |
481 | */ |
482 | qhead++; |
483 | } |
484 | |
485 | if (qhead == qtail) { |
486 | /* |
487 | * We have finished the bfs and not found any way to |
488 | * expand omino j. Panic, and return failure. |
489 | * |
490 | * FIXME: or should we loop over all ominoes before we |
491 | * give up? |
492 | */ |
f40d37bd |
493 | #ifdef DIVVY_DIAGNOSTICS |
494 | printf("FAIL!\n"); |
495 | #endif |
11d273f7 |
496 | retdsf = NULL; |
497 | goto cleanup; |
498 | } |
499 | } |
500 | |
f40d37bd |
501 | #ifdef DIVVY_DIAGNOSTICS |
502 | { |
503 | int x, y; |
504 | printf("SUCCESS! Final grid:\n"); |
505 | for (y = 0; y < h; y++) { |
506 | for (x = 0; x < w; x++) |
507 | printf("%3d", own[y*w+x]); |
508 | printf("\n"); |
509 | } |
510 | } |
511 | #endif |
512 | |
11d273f7 |
513 | /* |
514 | * Construct the output dsf. |
515 | */ |
516 | for (i = 0; i < wh; i++) { |
517 | assert(own[i] >= 0 && own[i] < n); |
518 | tmp[own[i]] = i; |
519 | } |
520 | retdsf = snew_dsf(wh); |
521 | for (i = 0; i < wh; i++) { |
522 | dsf_merge(retdsf, i, tmp[own[i]]); |
523 | } |
524 | |
525 | /* |
526 | * Construct the output dsf a different way, to verify that |
527 | * the ominoes really are k-ominoes and we haven't |
528 | * accidentally split one into two disconnected pieces. |
529 | */ |
530 | dsf_init(tmp, wh); |
531 | for (y = 0; y < h; y++) |
532 | for (x = 0; x+1 < w; x++) |
533 | if (own[y*w+x] == own[y*w+(x+1)]) |
534 | dsf_merge(tmp, y*w+x, y*w+(x+1)); |
535 | for (x = 0; x < w; x++) |
536 | for (y = 0; y+1 < h; y++) |
537 | if (own[y*w+x] == own[(y+1)*w+x]) |
538 | dsf_merge(tmp, y*w+x, (y+1)*w+x); |
539 | for (i = 0; i < wh; i++) { |
540 | j = dsf_canonify(retdsf, i); |
541 | assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i)); |
542 | } |
543 | |
544 | cleanup: |
545 | |
546 | /* |
547 | * Free our temporary working space. |
548 | */ |
549 | sfree(order); |
550 | sfree(tmp); |
551 | sfree(own); |
552 | sfree(sizes); |
553 | sfree(queue); |
554 | sfree(addable); |
555 | sfree(removable); |
556 | |
557 | /* |
558 | * And we're done. |
559 | */ |
560 | return retdsf; |
561 | } |
562 | |
563 | #ifdef TESTMODE |
9d36cbd7 |
564 | static int fail_counter = 0; |
565 | #endif |
566 | |
567 | int *divvy_rectangle(int w, int h, int k, random_state *rs) |
568 | { |
569 | int *ret; |
570 | |
571 | do { |
572 | ret = divvy_internal(w, h, k, rs); |
573 | |
574 | #ifdef TESTMODE |
575 | if (!ret) |
576 | fail_counter++; |
577 | #endif |
578 | |
579 | } while (!ret); |
580 | |
581 | return ret; |
582 | } |
583 | |
584 | #ifdef TESTMODE |
11d273f7 |
585 | |
586 | /* |
587 | * gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c |
588 | * |
589 | * or to debug |
590 | * |
591 | * gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c |
592 | */ |
593 | |
594 | int main(int argc, char **argv) |
595 | { |
596 | int *dsf; |
9d36cbd7 |
597 | int i; |
11d273f7 |
598 | int w = 9, h = 4, k = 6, tries = 100; |
599 | random_state *rs; |
600 | |
601 | rs = random_new("123456", 6); |
602 | |
603 | if (argc > 1) |
604 | w = atoi(argv[1]); |
605 | if (argc > 2) |
606 | h = atoi(argv[2]); |
607 | if (argc > 3) |
608 | k = atoi(argv[3]); |
609 | if (argc > 4) |
610 | tries = atoi(argv[4]); |
611 | |
11d273f7 |
612 | for (i = 0; i < tries; i++) { |
9d36cbd7 |
613 | int x, y; |
11d273f7 |
614 | |
9d36cbd7 |
615 | dsf = divvy_rectangle(w, h, k, rs); |
616 | assert(dsf); |
617 | |
618 | for (y = 0; y <= 2*h; y++) { |
619 | for (x = 0; x <= 2*w; x++) { |
620 | int miny = y/2 - 1, maxy = y/2; |
621 | int minx = x/2 - 1, maxx = x/2; |
622 | int classes[4], tx, ty; |
623 | for (ty = 0; ty < 2; ty++) |
624 | for (tx = 0; tx < 2; tx++) { |
625 | int cx = minx+tx, cy = miny+ty; |
626 | if (cx < 0 || cx >= w || cy < 0 || cy >= h) |
627 | classes[ty*2+tx] = -1; |
11d273f7 |
628 | else |
9d36cbd7 |
629 | classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx); |
11d273f7 |
630 | } |
9d36cbd7 |
631 | switch (y%2 * 2 + x%2) { |
632 | case 0: /* corner */ |
633 | /* |
634 | * Cases for the corner: |
635 | * |
636 | * - if all four surrounding squares belong |
637 | * to the same omino, we print a space. |
638 | * |
639 | * - if the top two are the same and the |
640 | * bottom two are the same, we print a |
641 | * horizontal line. |
642 | * |
643 | * - if the left two are the same and the |
644 | * right two are the same, we print a |
645 | * vertical line. |
646 | * |
647 | * - otherwise, we print a cross. |
648 | */ |
649 | if (classes[0] == classes[1] && |
650 | classes[1] == classes[2] && |
651 | classes[2] == classes[3]) |
652 | printf(" "); |
653 | else if (classes[0] == classes[1] && |
654 | classes[2] == classes[3]) |
655 | printf("-"); |
656 | else if (classes[0] == classes[2] && |
657 | classes[1] == classes[3]) |
658 | printf("|"); |
659 | else |
660 | printf("+"); |
661 | break; |
662 | case 1: /* horiz edge */ |
663 | if (classes[1] == classes[3]) |
664 | printf(" "); |
665 | else |
666 | printf("--"); |
667 | break; |
668 | case 2: /* vert edge */ |
669 | if (classes[2] == classes[3]) |
670 | printf(" "); |
671 | else |
672 | printf("|"); |
673 | break; |
674 | case 3: /* square centre */ |
675 | printf(" "); |
676 | break; |
11d273f7 |
677 | } |
11d273f7 |
678 | } |
679 | printf("\n"); |
11d273f7 |
680 | } |
9d36cbd7 |
681 | printf("\n"); |
682 | sfree(dsf); |
11d273f7 |
683 | } |
684 | |
9d36cbd7 |
685 | printf("%d retries needed for %d successes\n", fail_counter, tries); |
11d273f7 |
686 | |
687 | return 0; |
688 | } |
689 | |
690 | #endif |