| 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 | |
| 10 | /* |
| 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 claim the smallest k which can manage this is 23. More |
| 42 | * formally: |
| 43 | * |
| 44 | * If a k-omino P is completely surrounded by another k-omino Q, |
| 45 | * such that every edge of P borders on Q, then k >= 23. |
| 46 | * |
| 47 | * Proof: |
| 48 | * |
| 49 | * It's relatively simple to find the largest _rectangle_ a |
| 50 | * k-omino can enclose. So I'll construct my proof in two parts: |
| 51 | * firstly, show that no 22-omino or smaller can enclose a |
| 52 | * rectangle as large as itself, and secondly, show that no |
| 53 | * polyomino can enclose a larger non-rectangle than a rectangle. |
| 54 | * |
| 55 | * The first of those claims: |
| 56 | * |
| 57 | * To surround an m x n rectangle, a polyomino must have 2m |
| 58 | * squares along the two m-sides of the rectangle, 2n squares |
| 59 | * along the two n-sides, and must fill in at least three of the |
| 60 | * corners in order to be connected. Thus, 2(m+n)+3 <= k. We wish |
| 61 | * to find the largest value of mn subject to that constraint, and |
| 62 | * it's clear that this is achieved when m and n are as close to |
| 63 | * equal as possible. (If they aren't, WLOG suppose m < n; then |
| 64 | * (m+1)(n-1) = mn + n - m - 1 >= mn, with equality only when |
| 65 | * m=n-1.) |
| 66 | * |
| 67 | * So the area of the largest rectangle which can be enclosed by a |
| 68 | * k-omino is given by floor(k'/2) * ceil(k'/2), where k' = |
| 69 | * (k-3)/2. This is a monotonic function in k, so there will be a |
| 70 | * unique point at which it goes from being smaller than k to |
| 71 | * being larger than k. That point is between 22 (maximum area 20) |
| 72 | * and 23 (maximum area 25). |
| 73 | * |
| 74 | * The second claim: |
| 75 | * |
| 76 | * Suppose we have an inner polyomino P surrounded by an outer |
| 77 | * polyomino Q. I seek to show that if P is non-rectangular, then |
| 78 | * P is also non-maximal, in the sense that we can transform P and |
| 79 | * Q into a new pair of polyominoes in which P is larger and Q is |
| 80 | * at most the same size. |
| 81 | * |
| 82 | * Consider walking along the boundary of P in a clockwise |
| 83 | * direction. (We may assume, of course, that there is only _one_ |
| 84 | * boundary of P, i.e. P has no hole in the middle. If it does |
| 85 | * have a hole in the middle, it's _trivially_ non-maximal because |
| 86 | * we can just fill the hole in!) Our walk will take us along many |
| 87 | * edges between squares; sometimes we might turn left, and |
| 88 | * certainly sometimes we will turn right. Always there will be a |
| 89 | * square of P on our right, and a square of Q on our left. |
| 90 | * |
| 91 | * The net angle through which we turn during the entire walk must |
| 92 | * add up to 360 degrees rightwards. So if there are no left |
| 93 | * turns, then we must turn right exactly four times, meaning we |
| 94 | * have described a rectangle. Hence, if P is _not_ rectangular, |
| 95 | * then there must have been a left turn at some point. A left |
| 96 | * turn must mean we walk along two edges of the same square of Q. |
| 97 | * |
| 98 | * Thus, there is some square X in Q which is adjacent to two |
| 99 | * diagonally separated squares in P. Let us call those two |
| 100 | * squares N and E; let us refer to the other two neighbours of X |
| 101 | * as S and W; let us refer to the other mutual neighbour of S and |
| 102 | * W as D; and let us refer to the other mutual neighbour of S and |
| 103 | * E as Y. In other words, we have named seven squares, arranged |
| 104 | * thus: |
| 105 | * |
| 106 | * N |
| 107 | * W X E |
| 108 | * D S Y |
| 109 | * |
| 110 | * where N and E are in P, and X is in Q. |
| 111 | * |
| 112 | * Clearly at least one of W and S must be in Q (because otherwise |
| 113 | * X would not be connected to any other square in Q, and would |
| 114 | * hence have to be the whole of Q; and evidently if Q were a |
| 115 | * 1-omino it could not enclose _anything_). So we divide into |
| 116 | * cases: |
| 117 | * |
| 118 | * If both W and S are in Q, then we take X out of Q and put it in |
| 119 | * P, which does not expose any edge of P. If this disconnects Q, |
| 120 | * then we can reconnect it by adding D to Q. |
| 121 | * |
| 122 | * If only one of W and S is in Q, then wlog let it be W. If S is |
| 123 | * in _P_, then we have a particularly easy case: we can simply |
| 124 | * take X out of Q and add it to P, and this cannot disconnect X |
| 125 | * since X was a leaf square of Q. |
| 126 | * |
| 127 | * Our remaining case is that W is in Q and S is in neither P nor |
| 128 | * Q. Again we take X out of Q and put it in P; we also add S to |
| 129 | * Q. This ensures we do not expose an edge of P, but we must now |
| 130 | * prove that S is adjacent to some other existing square of Q so |
| 131 | * that we haven't disconnected Q by adding it. |
| 132 | * |
| 133 | * To do this, we recall that we walked along the edge XE, and |
| 134 | * then turned left to walk along XN. So just before doing all |
| 135 | * that, we must have reached the corner XSE, and we must have |
| 136 | * done it by walking along one of the three edges meeting at that |
| 137 | * corner which are _not_ XE. It can't have been SY, since S would |
| 138 | * then have been on our left and it isn't in Q; and it can't have |
| 139 | * been XS, since S would then have been on our right and it isn't |
| 140 | * in P. So it must have been YE, in which case Y was on our left, |
| 141 | * and hence is in Q. |
| 142 | * |
| 143 | * So in all cases we have shown that we can take X out of Q and |
| 144 | * add it to P, and add at most one square to Q to restore the |
| 145 | * containment and connectedness properties. Hence, we can keep |
| 146 | * doing this until we run out of left turns and P becomes |
| 147 | * rectangular. [] |
| 148 | * |
| 149 | * ------------ |
| 150 | * |
| 151 | * Anyway, that entire proof was a bit of a sidetrack. The point |
| 152 | * is, although constructions of this type are possible for |
| 153 | * sufficiently large k, divvy_rectangle() will never generate |
| 154 | * them. This could be considered a weakness for some purposes, in |
| 155 | * the sense that we can't generate all possible divisions. |
| 156 | * However, there are many divisions which we are highly unlikely |
| 157 | * to generate anyway, so in practice it probably isn't _too_ bad. |
| 158 | * |
| 159 | * If I wanted to fix this issue, I would have to make the rules |
| 160 | * more complicated for determining when a square can safely be |
| 161 | * _removed_ from a polyomino. Adding one becomes easier (a square |
| 162 | * may be added to a polyomino iff it is 4-adjacent to any square |
| 163 | * currently part of the polyomino, and the current test for loop |
| 164 | * formation may be dispensed with), but to determine which |
| 165 | * squares may be removed we must now resort to analysis of the |
| 166 | * overall structure of the polyomino rather than the simple local |
| 167 | * properties we can currently get away with measuring. |
| 168 | */ |
| 169 | |
| 170 | /* |
| 171 | * Possible improvements which might cut the fail rate: |
| 172 | * |
| 173 | * - instead of picking one omino to extend in an iteration, try |
| 174 | * them all in succession (in a randomised order) |
| 175 | * |
| 176 | * - (for real rigour) instead of bfsing over ominoes, bfs over |
| 177 | * the space of possible _removed squares_. That way we aren't |
| 178 | * limited to randomly choosing a single square to remove from |
| 179 | * an omino and failing if that particular square doesn't |
| 180 | * happen to work. |
| 181 | * |
| 182 | * However, I don't currently think it's necessary to do either of |
| 183 | * these, because the failure rate is already low enough to be |
| 184 | * easily tolerable, under all circumstances I've been able to |
| 185 | * think of. |
| 186 | */ |
| 187 | |
| 188 | #include <assert.h> |
| 189 | #include <stdio.h> |
| 190 | #include <stdlib.h> |
| 191 | #include <stddef.h> |
| 192 | |
| 193 | #include "puzzles.h" |
| 194 | |
| 195 | /* |
| 196 | * Subroutine which implements a function used in computing both |
| 197 | * whether a square can safely be added to an omino, and whether |
| 198 | * it can safely be removed. |
| 199 | * |
| 200 | * We enumerate the eight squares 8-adjacent to this one, in |
| 201 | * cyclic order. We go round that loop and count the number of |
| 202 | * times we find a square owned by the target omino next to one |
| 203 | * not owned by it. We then return success iff that count is 2. |
| 204 | * |
| 205 | * When adding a square to an omino, this is precisely the |
| 206 | * criterion which tells us that adding the square won't leave a |
| 207 | * hole in the middle of the omino. (If it did, then things get |
| 208 | * more complicated; see above.) |
| 209 | * |
| 210 | * When removing a square from an omino, the _same_ criterion |
| 211 | * tells us that removing the square won't disconnect the omino. |
| 212 | * (This only works _because_ we've ensured the omino is simply |
| 213 | * connected.) |
| 214 | */ |
| 215 | static int addremcommon(int w, int h, int x, int y, int *own, int val) |
| 216 | { |
| 217 | int neighbours[8]; |
| 218 | int dir, count; |
| 219 | |
| 220 | for (dir = 0; dir < 8; dir++) { |
| 221 | int dx = ((dir & 3) == 2 ? 0 : dir > 2 && dir < 6 ? +1 : -1); |
| 222 | int dy = ((dir & 3) == 0 ? 0 : dir < 4 ? -1 : +1); |
| 223 | int sx = x+dx, sy = y+dy; |
| 224 | |
| 225 | if (sx < 0 || sx >= w || sy < 0 || sy >= h) |
| 226 | neighbours[dir] = -1; /* outside the grid */ |
| 227 | else |
| 228 | neighbours[dir] = own[sy*w+sx]; |
| 229 | } |
| 230 | |
| 231 | /* |
| 232 | * To begin with, check 4-adjacency. |
| 233 | */ |
| 234 | if (neighbours[0] != val && neighbours[2] != val && |
| 235 | neighbours[4] != val && neighbours[6] != val) |
| 236 | return FALSE; |
| 237 | |
| 238 | count = 0; |
| 239 | |
| 240 | for (dir = 0; dir < 8; dir++) { |
| 241 | int next = (dir + 1) & 7; |
| 242 | int gotthis = (neighbours[dir] == val); |
| 243 | int gotnext = (neighbours[next] == val); |
| 244 | |
| 245 | if (gotthis != gotnext) |
| 246 | count++; |
| 247 | } |
| 248 | |
| 249 | return (count == 2); |
| 250 | } |
| 251 | |
| 252 | /* |
| 253 | * w and h are the dimensions of the rectangle. |
| 254 | * |
| 255 | * k is the size of the required ominoes. (So k must divide w*h, |
| 256 | * of course.) |
| 257 | * |
| 258 | * The returned result is a w*h-sized dsf. |
| 259 | * |
| 260 | * In both of the above suggested use cases, the user would |
| 261 | * probably want w==h==k, but that isn't a requirement. |
| 262 | */ |
| 263 | static int *divvy_internal(int w, int h, int k, random_state *rs) |
| 264 | { |
| 265 | int *order, *queue, *tmp, *own, *sizes, *addable, *removable, *retdsf; |
| 266 | int wh = w*h; |
| 267 | int i, j, n, x, y, qhead, qtail; |
| 268 | |
| 269 | n = wh / k; |
| 270 | assert(wh == k*n); |
| 271 | |
| 272 | order = snewn(wh, int); |
| 273 | tmp = snewn(wh, int); |
| 274 | own = snewn(wh, int); |
| 275 | sizes = snewn(n, int); |
| 276 | queue = snewn(n, int); |
| 277 | addable = snewn(wh*4, int); |
| 278 | removable = snewn(wh, int); |
| 279 | |
| 280 | /* |
| 281 | * Permute the grid squares into a random order, which will be |
| 282 | * used for iterating over the grid whenever we need to search |
| 283 | * for something. This prevents directional bias and arranges |
| 284 | * for the answer to be non-deterministic. |
| 285 | */ |
| 286 | for (i = 0; i < wh; i++) |
| 287 | order[i] = i; |
| 288 | shuffle(order, wh, sizeof(*order), rs); |
| 289 | |
| 290 | /* |
| 291 | * Begin by choosing a starting square at random for each |
| 292 | * omino. |
| 293 | */ |
| 294 | for (i = 0; i < wh; i++) { |
| 295 | own[i] = -1; |
| 296 | } |
| 297 | for (i = 0; i < n; i++) { |
| 298 | own[order[i]] = i; |
| 299 | sizes[i] = 1; |
| 300 | } |
| 301 | |
| 302 | /* |
| 303 | * Now repeatedly pick a random omino which isn't already at |
| 304 | * the target size, and find a way to expand it by one. This |
| 305 | * may involve stealing a square from another omino, in which |
| 306 | * case we then re-expand that omino, forming a chain of |
| 307 | * square-stealing which terminates in an as yet unclaimed |
| 308 | * square. Hence every successful iteration around this loop |
| 309 | * causes the number of unclaimed squares to drop by one, and |
| 310 | * so the process is bounded in duration. |
| 311 | */ |
| 312 | while (1) { |
| 313 | |
| 314 | #ifdef DIVVY_DIAGNOSTICS |
| 315 | { |
| 316 | int x, y; |
| 317 | printf("Top of loop. Current grid:\n"); |
| 318 | for (y = 0; y < h; y++) { |
| 319 | for (x = 0; x < w; x++) |
| 320 | printf("%3d", own[y*w+x]); |
| 321 | printf("\n"); |
| 322 | } |
| 323 | } |
| 324 | #endif |
| 325 | |
| 326 | /* |
| 327 | * Go over the grid and figure out which squares can |
| 328 | * safely be added to, or removed from, each omino. We |
| 329 | * don't take account of other ominoes in this process, so |
| 330 | * we will often end up knowing that a square can be |
| 331 | * poached from one omino by another. |
| 332 | * |
| 333 | * For each square, there may be up to four ominoes to |
| 334 | * which it can be added (those to which it is |
| 335 | * 4-adjacent). |
| 336 | */ |
| 337 | for (y = 0; y < h; y++) { |
| 338 | for (x = 0; x < w; x++) { |
| 339 | int yx = y*w+x; |
| 340 | int curr = own[yx]; |
| 341 | int dir; |
| 342 | |
| 343 | if (curr < 0) { |
| 344 | removable[yx] = FALSE; /* can't remove if not owned! */ |
| 345 | } else if (sizes[curr] == 1) { |
| 346 | removable[yx] = TRUE; /* can always remove a singleton */ |
| 347 | } else { |
| 348 | /* |
| 349 | * See if this square can be removed from its |
| 350 | * omino without disconnecting it. |
| 351 | */ |
| 352 | removable[yx] = addremcommon(w, h, x, y, own, curr); |
| 353 | } |
| 354 | |
| 355 | for (dir = 0; dir < 4; dir++) { |
| 356 | int dx = (dir == 0 ? -1 : dir == 1 ? +1 : 0); |
| 357 | int dy = (dir == 2 ? -1 : dir == 3 ? +1 : 0); |
| 358 | int sx = x + dx, sy = y + dy; |
| 359 | int syx = sy*w+sx; |
| 360 | |
| 361 | addable[yx*4+dir] = -1; |
| 362 | |
| 363 | if (sx < 0 || sx >= w || sy < 0 || sy >= h) |
| 364 | continue; /* no omino here! */ |
| 365 | if (own[syx] < 0) |
| 366 | continue; /* also no omino here */ |
| 367 | if (own[syx] == own[yx]) |
| 368 | continue; /* we already got one */ |
| 369 | if (!addremcommon(w, h, x, y, own, own[syx])) |
| 370 | continue; /* would non-simply connect the omino */ |
| 371 | |
| 372 | addable[yx*4+dir] = own[syx]; |
| 373 | } |
| 374 | } |
| 375 | } |
| 376 | |
| 377 | for (i = j = 0; i < n; i++) |
| 378 | if (sizes[i] < k) |
| 379 | tmp[j++] = i; |
| 380 | if (j == 0) |
| 381 | break; /* all ominoes are complete! */ |
| 382 | j = tmp[random_upto(rs, j)]; |
| 383 | #ifdef DIVVY_DIAGNOSTICS |
| 384 | printf("Trying to extend %d\n", j); |
| 385 | #endif |
| 386 | |
| 387 | /* |
| 388 | * So we're trying to expand omino j. We breadth-first |
| 389 | * search out from j across the space of ominoes. |
| 390 | * |
| 391 | * For bfs purposes, we use two elements of tmp per omino: |
| 392 | * tmp[2*i+0] tells us which omino we got to i from, and |
| 393 | * tmp[2*i+1] numbers the grid square that omino stole |
| 394 | * from us. |
| 395 | * |
| 396 | * This requires that wh (the size of tmp) is at least 2n, |
| 397 | * i.e. k is at least 2. There would have been nothing to |
| 398 | * stop a user calling this function with k=1, but if they |
| 399 | * did then we wouldn't have got to _here_ in the code - |
| 400 | * we would have noticed above that all ominoes were |
| 401 | * already at their target sizes, and terminated :-) |
| 402 | */ |
| 403 | assert(wh >= 2*n); |
| 404 | for (i = 0; i < n; i++) |
| 405 | tmp[2*i] = tmp[2*i+1] = -1; |
| 406 | qhead = qtail = 0; |
| 407 | queue[qtail++] = j; |
| 408 | tmp[2*j] = tmp[2*j+1] = -2; /* special value: `starting point' */ |
| 409 | |
| 410 | while (qhead < qtail) { |
| 411 | int tmpsq; |
| 412 | |
| 413 | j = queue[qhead]; |
| 414 | |
| 415 | /* |
| 416 | * We wish to expand omino j. However, we might have |
| 417 | * got here by omino j having a square stolen from it, |
| 418 | * so first of all we must temporarily mark that |
| 419 | * square as not belonging to j, so that our adjacency |
| 420 | * calculations don't assume j _does_ belong to us. |
| 421 | */ |
| 422 | tmpsq = tmp[2*j+1]; |
| 423 | if (tmpsq >= 0) { |
| 424 | assert(own[tmpsq] == j); |
| 425 | own[tmpsq] = -3; |
| 426 | } |
| 427 | |
| 428 | /* |
| 429 | * OK. Now begin by seeing if we can find any |
| 430 | * unclaimed square into which we can expand omino j. |
| 431 | * If we find one, the entire bfs terminates. |
| 432 | */ |
| 433 | for (i = 0; i < wh; i++) { |
| 434 | int dir; |
| 435 | |
| 436 | if (own[order[i]] != -1) |
| 437 | continue; /* this square is claimed */ |
| 438 | |
| 439 | /* |
| 440 | * Special case: if our current omino was size 1 |
| 441 | * and then had a square stolen from it, it's now |
| 442 | * size zero, which means it's valid to `expand' |
| 443 | * it into _any_ unclaimed square. |
| 444 | */ |
| 445 | if (sizes[j] == 1 && tmpsq >= 0) |
| 446 | break; /* got one */ |
| 447 | |
| 448 | /* |
| 449 | * Failing that, we must do the full test for |
| 450 | * addability. |
| 451 | */ |
| 452 | for (dir = 0; dir < 4; dir++) |
| 453 | if (addable[order[i]*4+dir] == j) { |
| 454 | /* |
| 455 | * We know this square is addable to this |
| 456 | * omino with the grid in the state it had |
| 457 | * at the top of the loop. However, we |
| 458 | * must now check that it's _still_ |
| 459 | * addable to this omino when the omino is |
| 460 | * missing a square. To do this it's only |
| 461 | * necessary to re-check addremcommon. |
| 462 | */ |
| 463 | if (!addremcommon(w, h, order[i]%w, order[i]/w, |
| 464 | own, j)) |
| 465 | continue; |
| 466 | break; |
| 467 | } |
| 468 | if (dir == 4) |
| 469 | continue; /* we can't add this square to j */ |
| 470 | |
| 471 | break; /* got one! */ |
| 472 | } |
| 473 | if (i < wh) { |
| 474 | i = order[i]; |
| 475 | |
| 476 | /* |
| 477 | * Restore the temporarily removed square _before_ |
| 478 | * we start shifting ownerships about. |
| 479 | */ |
| 480 | if (tmpsq >= 0) |
| 481 | own[tmpsq] = j; |
| 482 | |
| 483 | /* |
| 484 | * We are done. We can add square i to omino j, |
| 485 | * and then backtrack along the trail in tmp |
| 486 | * moving squares between ominoes, ending up |
| 487 | * expanding our starting omino by one. |
| 488 | */ |
| 489 | #ifdef DIVVY_DIAGNOSTICS |
| 490 | printf("(%d,%d)", i%w, i/w); |
| 491 | #endif |
| 492 | while (1) { |
| 493 | own[i] = j; |
| 494 | #ifdef DIVVY_DIAGNOSTICS |
| 495 | printf(" -> %d", j); |
| 496 | #endif |
| 497 | if (tmp[2*j] == -2) |
| 498 | break; |
| 499 | i = tmp[2*j+1]; |
| 500 | j = tmp[2*j]; |
| 501 | #ifdef DIVVY_DIAGNOSTICS |
| 502 | printf("; (%d,%d)", i%w, i/w); |
| 503 | #endif |
| 504 | } |
| 505 | #ifdef DIVVY_DIAGNOSTICS |
| 506 | printf("\n"); |
| 507 | #endif |
| 508 | |
| 509 | /* |
| 510 | * Increment the size of the starting omino. |
| 511 | */ |
| 512 | sizes[j]++; |
| 513 | |
| 514 | /* |
| 515 | * Terminate the bfs loop. |
| 516 | */ |
| 517 | break; |
| 518 | } |
| 519 | |
| 520 | /* |
| 521 | * If we get here, we haven't been able to expand |
| 522 | * omino j into an unclaimed square. So now we begin |
| 523 | * to investigate expanding it into squares which are |
| 524 | * claimed by ominoes the bfs has not yet visited. |
| 525 | */ |
| 526 | for (i = 0; i < wh; i++) { |
| 527 | int dir, nj; |
| 528 | |
| 529 | nj = own[order[i]]; |
| 530 | if (nj < 0 || tmp[2*nj] != -1) |
| 531 | continue; /* unclaimed, or owned by wrong omino */ |
| 532 | if (!removable[order[i]]) |
| 533 | continue; /* its omino won't let it go */ |
| 534 | |
| 535 | for (dir = 0; dir < 4; dir++) |
| 536 | if (addable[order[i]*4+dir] == j) { |
| 537 | /* |
| 538 | * As above, re-check addremcommon. |
| 539 | */ |
| 540 | if (!addremcommon(w, h, order[i]%w, order[i]/w, |
| 541 | own, j)) |
| 542 | continue; |
| 543 | |
| 544 | /* |
| 545 | * We have found a square we can use to |
| 546 | * expand omino j, at the expense of the |
| 547 | * as-yet unvisited omino nj. So add this |
| 548 | * to the bfs queue. |
| 549 | */ |
| 550 | assert(qtail < n); |
| 551 | queue[qtail++] = nj; |
| 552 | tmp[2*nj] = j; |
| 553 | tmp[2*nj+1] = order[i]; |
| 554 | |
| 555 | /* |
| 556 | * Now terminate the loop over dir, to |
| 557 | * ensure we don't accidentally add the |
| 558 | * same omino twice to the queue. |
| 559 | */ |
| 560 | break; |
| 561 | } |
| 562 | } |
| 563 | |
| 564 | /* |
| 565 | * Restore the temporarily removed square. |
| 566 | */ |
| 567 | if (tmpsq >= 0) |
| 568 | own[tmpsq] = j; |
| 569 | |
| 570 | /* |
| 571 | * Advance the queue head. |
| 572 | */ |
| 573 | qhead++; |
| 574 | } |
| 575 | |
| 576 | if (qhead == qtail) { |
| 577 | /* |
| 578 | * We have finished the bfs and not found any way to |
| 579 | * expand omino j. Panic, and return failure. |
| 580 | * |
| 581 | * FIXME: or should we loop over all ominoes before we |
| 582 | * give up? |
| 583 | */ |
| 584 | #ifdef DIVVY_DIAGNOSTICS |
| 585 | printf("FAIL!\n"); |
| 586 | #endif |
| 587 | retdsf = NULL; |
| 588 | goto cleanup; |
| 589 | } |
| 590 | } |
| 591 | |
| 592 | #ifdef DIVVY_DIAGNOSTICS |
| 593 | { |
| 594 | int x, y; |
| 595 | printf("SUCCESS! Final grid:\n"); |
| 596 | for (y = 0; y < h; y++) { |
| 597 | for (x = 0; x < w; x++) |
| 598 | printf("%3d", own[y*w+x]); |
| 599 | printf("\n"); |
| 600 | } |
| 601 | } |
| 602 | #endif |
| 603 | |
| 604 | /* |
| 605 | * Construct the output dsf. |
| 606 | */ |
| 607 | for (i = 0; i < wh; i++) { |
| 608 | assert(own[i] >= 0 && own[i] < n); |
| 609 | tmp[own[i]] = i; |
| 610 | } |
| 611 | retdsf = snew_dsf(wh); |
| 612 | for (i = 0; i < wh; i++) { |
| 613 | dsf_merge(retdsf, i, tmp[own[i]]); |
| 614 | } |
| 615 | |
| 616 | /* |
| 617 | * Construct the output dsf a different way, to verify that |
| 618 | * the ominoes really are k-ominoes and we haven't |
| 619 | * accidentally split one into two disconnected pieces. |
| 620 | */ |
| 621 | dsf_init(tmp, wh); |
| 622 | for (y = 0; y < h; y++) |
| 623 | for (x = 0; x+1 < w; x++) |
| 624 | if (own[y*w+x] == own[y*w+(x+1)]) |
| 625 | dsf_merge(tmp, y*w+x, y*w+(x+1)); |
| 626 | for (x = 0; x < w; x++) |
| 627 | for (y = 0; y+1 < h; y++) |
| 628 | if (own[y*w+x] == own[(y+1)*w+x]) |
| 629 | dsf_merge(tmp, y*w+x, (y+1)*w+x); |
| 630 | for (i = 0; i < wh; i++) { |
| 631 | j = dsf_canonify(retdsf, i); |
| 632 | assert(dsf_canonify(tmp, j) == dsf_canonify(tmp, i)); |
| 633 | } |
| 634 | |
| 635 | cleanup: |
| 636 | |
| 637 | /* |
| 638 | * Free our temporary working space. |
| 639 | */ |
| 640 | sfree(order); |
| 641 | sfree(tmp); |
| 642 | sfree(own); |
| 643 | sfree(sizes); |
| 644 | sfree(queue); |
| 645 | sfree(addable); |
| 646 | sfree(removable); |
| 647 | |
| 648 | /* |
| 649 | * And we're done. |
| 650 | */ |
| 651 | return retdsf; |
| 652 | } |
| 653 | |
| 654 | #ifdef TESTMODE |
| 655 | static int fail_counter = 0; |
| 656 | #endif |
| 657 | |
| 658 | int *divvy_rectangle(int w, int h, int k, random_state *rs) |
| 659 | { |
| 660 | int *ret; |
| 661 | |
| 662 | do { |
| 663 | ret = divvy_internal(w, h, k, rs); |
| 664 | |
| 665 | #ifdef TESTMODE |
| 666 | if (!ret) |
| 667 | fail_counter++; |
| 668 | #endif |
| 669 | |
| 670 | } while (!ret); |
| 671 | |
| 672 | return ret; |
| 673 | } |
| 674 | |
| 675 | #ifdef TESTMODE |
| 676 | |
| 677 | /* |
| 678 | * gcc -g -O0 -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c |
| 679 | * |
| 680 | * or to debug |
| 681 | * |
| 682 | * gcc -g -O0 -DDIVVY_DIAGNOSTICS -DTESTMODE -I.. -o divvy divvy.c ../random.c ../malloc.c ../dsf.c ../misc.c ../nullfe.c |
| 683 | */ |
| 684 | |
| 685 | int main(int argc, char **argv) |
| 686 | { |
| 687 | int *dsf; |
| 688 | int i; |
| 689 | int w = 9, h = 4, k = 6, tries = 100; |
| 690 | random_state *rs; |
| 691 | |
| 692 | rs = random_new("123456", 6); |
| 693 | |
| 694 | if (argc > 1) |
| 695 | w = atoi(argv[1]); |
| 696 | if (argc > 2) |
| 697 | h = atoi(argv[2]); |
| 698 | if (argc > 3) |
| 699 | k = atoi(argv[3]); |
| 700 | if (argc > 4) |
| 701 | tries = atoi(argv[4]); |
| 702 | |
| 703 | for (i = 0; i < tries; i++) { |
| 704 | int x, y; |
| 705 | |
| 706 | dsf = divvy_rectangle(w, h, k, rs); |
| 707 | assert(dsf); |
| 708 | |
| 709 | for (y = 0; y <= 2*h; y++) { |
| 710 | for (x = 0; x <= 2*w; x++) { |
| 711 | int miny = y/2 - 1, maxy = y/2; |
| 712 | int minx = x/2 - 1, maxx = x/2; |
| 713 | int classes[4], tx, ty; |
| 714 | for (ty = 0; ty < 2; ty++) |
| 715 | for (tx = 0; tx < 2; tx++) { |
| 716 | int cx = minx+tx, cy = miny+ty; |
| 717 | if (cx < 0 || cx >= w || cy < 0 || cy >= h) |
| 718 | classes[ty*2+tx] = -1; |
| 719 | else |
| 720 | classes[ty*2+tx] = dsf_canonify(dsf, cy*w+cx); |
| 721 | } |
| 722 | switch (y%2 * 2 + x%2) { |
| 723 | case 0: /* corner */ |
| 724 | /* |
| 725 | * Cases for the corner: |
| 726 | * |
| 727 | * - if all four surrounding squares belong |
| 728 | * to the same omino, we print a space. |
| 729 | * |
| 730 | * - if the top two are the same and the |
| 731 | * bottom two are the same, we print a |
| 732 | * horizontal line. |
| 733 | * |
| 734 | * - if the left two are the same and the |
| 735 | * right two are the same, we print a |
| 736 | * vertical line. |
| 737 | * |
| 738 | * - otherwise, we print a cross. |
| 739 | */ |
| 740 | if (classes[0] == classes[1] && |
| 741 | classes[1] == classes[2] && |
| 742 | classes[2] == classes[3]) |
| 743 | printf(" "); |
| 744 | else if (classes[0] == classes[1] && |
| 745 | classes[2] == classes[3]) |
| 746 | printf("-"); |
| 747 | else if (classes[0] == classes[2] && |
| 748 | classes[1] == classes[3]) |
| 749 | printf("|"); |
| 750 | else |
| 751 | printf("+"); |
| 752 | break; |
| 753 | case 1: /* horiz edge */ |
| 754 | if (classes[1] == classes[3]) |
| 755 | printf(" "); |
| 756 | else |
| 757 | printf("--"); |
| 758 | break; |
| 759 | case 2: /* vert edge */ |
| 760 | if (classes[2] == classes[3]) |
| 761 | printf(" "); |
| 762 | else |
| 763 | printf("|"); |
| 764 | break; |
| 765 | case 3: /* square centre */ |
| 766 | printf(" "); |
| 767 | break; |
| 768 | } |
| 769 | } |
| 770 | printf("\n"); |
| 771 | } |
| 772 | printf("\n"); |
| 773 | sfree(dsf); |
| 774 | } |
| 775 | |
| 776 | printf("%d retries needed for %d successes\n", fail_counter, tries); |
| 777 | |
| 778 | return 0; |
| 779 | } |
| 780 | |
| 781 | #endif |