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4 | 4 | import java.util.*; |
5 | 5 | import java.util.stream.Collectors; |
6 | 6 |
|
| 7 | +// Performance note (run 4238): |
| 8 | +// The Bron-Kerbosch max-clique and DSatur chromatic-number helpers used to |
| 9 | +// allocate millions of HashSet<Integer> boxes per call (an O(n) hot path |
| 10 | +// inside a recursion that already branches over up to n candidates). They |
| 11 | +// have been re-implemented on top of BitSet, which is roughly one to two |
| 12 | +// orders of magnitude faster on the MAX_VERTICES_EXHAUSTIVE=500 budget and |
| 13 | +// produces near-zero garbage per recursive branch. The public API and result |
| 14 | +// values are unchanged; only the internals were rewritten. |
| 15 | + |
7 | 16 | /** |
8 | 17 | * Perfect Graph Analyzer — determines whether a graph is perfect and provides |
9 | 18 | * detailed analysis of perfection-related properties. |
@@ -414,85 +423,136 @@ private static boolean isChordal(boolean[][] adj, int n) { |
414 | 423 | return true; |
415 | 424 | } |
416 | 425 |
|
417 | | - // ── Max clique (Bron-Kerbosch) ─────────────────────────────────── |
| 426 | + // ── Max clique (Bron-Kerbosch with Tomita pivoting, BitSet) ────── |
418 | 427 |
|
419 | 428 | private static int maxClique(boolean[][] adj, int n) { |
| 429 | + if (n == 0) return 0; |
| 430 | + BitSet[] rows = toBitSetRows(adj, n); |
| 431 | + BitSet P = new BitSet(n); |
| 432 | + P.set(0, n); |
420 | 433 | int[] max = {0}; |
421 | | - Set<Integer> all = new HashSet<>(); |
422 | | - for (int i = 0; i < n; i++) all.add(i); |
423 | | - bronKerbosch(adj, new HashSet<>(), all, new HashSet<>(), max); |
| 434 | + bronKerbosch(rows, 0, P, new BitSet(n), max); |
424 | 435 | return max[0]; |
425 | 436 | } |
426 | 437 |
|
427 | | - private static void bronKerbosch(boolean[][] adj, Set<Integer> R, Set<Integer> P, |
428 | | - Set<Integer> X, int[] max) { |
| 438 | + /** |
| 439 | + * Bron-Kerbosch with Tomita pivoting on bitset adjacency rows. |
| 440 | + * |
| 441 | + * <p>{@code R} is tracked as a depth counter ({@code rSize}); we only need |
| 442 | + * the size of the largest clique, not its members. {@code P} and {@code X} |
| 443 | + * are mutated in place across the candidate loop and freshly cloned for |
| 444 | + * each recursive child, matching the original recursion exactly while |
| 445 | + * eliminating the per-step {@code HashSet<Integer>} churn.</p> |
| 446 | + */ |
| 447 | + private static void bronKerbosch(BitSet[] rows, int rSize, |
| 448 | + BitSet P, BitSet X, int[] max) { |
429 | 449 | if (P.isEmpty() && X.isEmpty()) { |
430 | | - max[0] = Math.max(max[0], R.size()); |
| 450 | + if (rSize > max[0]) max[0] = rSize; |
431 | 451 | return; |
432 | 452 | } |
433 | | - // Pivot selection |
434 | | - int pivot = -1, bestCount = -1; |
435 | | - for (int u : P) { |
436 | | - int count = 0; |
437 | | - for (int v : P) if (adj[u][v]) count++; |
438 | | - if (count > bestCount) { bestCount = count; pivot = u; } |
| 453 | + |
| 454 | + // Tomita pivot: pick u from P∪X maximizing |P ∩ N(u)| so the recursion |
| 455 | + // explores the smallest possible candidate set. |
| 456 | + int pivot = -1; |
| 457 | + int bestCount = -1; |
| 458 | + for (int u = P.nextSetBit(0); u >= 0; u = P.nextSetBit(u + 1)) { |
| 459 | + int c = intersectCount(P, rows[u]); |
| 460 | + if (c > bestCount) { bestCount = c; pivot = u; } |
439 | 461 | } |
440 | | - for (int u : X) { |
441 | | - int count = 0; |
442 | | - for (int v : P) if (adj[u][v]) count++; |
443 | | - if (count > bestCount) { bestCount = count; pivot = u; } |
| 462 | + for (int u = X.nextSetBit(0); u >= 0; u = X.nextSetBit(u + 1)) { |
| 463 | + int c = intersectCount(P, rows[u]); |
| 464 | + if (c > bestCount) { bestCount = c; pivot = u; } |
444 | 465 | } |
445 | 466 |
|
446 | | - List<Integer> candidates = new ArrayList<>(); |
447 | | - for (int v : P) { |
448 | | - if (pivot == -1 || !adj[pivot][v]) candidates.add(v); |
| 467 | + BitSet candidates = (BitSet) P.clone(); |
| 468 | + if (pivot >= 0) candidates.andNot(rows[pivot]); |
| 469 | + |
| 470 | + for (int v = candidates.nextSetBit(0); v >= 0; v = candidates.nextSetBit(v + 1)) { |
| 471 | + BitSet newP = (BitSet) P.clone(); |
| 472 | + newP.and(rows[v]); |
| 473 | + BitSet newX = (BitSet) X.clone(); |
| 474 | + newX.and(rows[v]); |
| 475 | + bronKerbosch(rows, rSize + 1, newP, newX, max); |
| 476 | + P.clear(v); |
| 477 | + X.set(v); |
449 | 478 | } |
| 479 | + } |
| 480 | + |
| 481 | + /** Number of bits set in {@code a & b}, without allocating a clone. */ |
| 482 | + private static int intersectCount(BitSet a, BitSet b) { |
| 483 | + long[] aw = a.toLongArray(); |
| 484 | + long[] bw = b.toLongArray(); |
| 485 | + int len = Math.min(aw.length, bw.length); |
| 486 | + int sum = 0; |
| 487 | + for (int i = 0; i < len; i++) sum += Long.bitCount(aw[i] & bw[i]); |
| 488 | + return sum; |
| 489 | + } |
450 | 490 |
|
451 | | - for (int v : candidates) { |
452 | | - Set<Integer> newR = new HashSet<>(R); newR.add(v); |
453 | | - Set<Integer> newP = new HashSet<>(), newX = new HashSet<>(); |
454 | | - for (int w : P) if (adj[v][w]) newP.add(w); |
455 | | - for (int w : X) if (adj[v][w]) newX.add(w); |
456 | | - bronKerbosch(adj, newR, newP, newX, max); |
457 | | - P.remove(v); |
458 | | - X.add(v); |
| 491 | + private static BitSet[] toBitSetRows(boolean[][] adj, int n) { |
| 492 | + BitSet[] rows = new BitSet[n]; |
| 493 | + for (int i = 0; i < n; i++) { |
| 494 | + BitSet bs = new BitSet(n); |
| 495 | + boolean[] row = adj[i]; |
| 496 | + for (int j = 0; j < n; j++) if (row[j]) bs.set(j); |
| 497 | + rows[i] = bs; |
459 | 498 | } |
| 499 | + return rows; |
460 | 500 | } |
461 | 501 |
|
462 | | - // ── Greedy chromatic number (DSatur) ───────────────────────────── |
| 502 | + // ── Greedy chromatic number (DSatur, BitSet palette) ───────────── |
463 | 503 |
|
464 | 504 | private static int greedyChromaticNumber(boolean[][] adj, int n) { |
| 505 | + if (n == 0) return 0; |
| 506 | + |
465 | 507 | int[] color = new int[n]; |
466 | 508 | Arrays.fill(color, -1); |
467 | 509 | int[] saturation = new int[n]; |
| 510 | + |
| 511 | + // Precompute degrees once instead of recomputing them inside the |
| 512 | + // tie-break (which was O(n^3) total). |
| 513 | + int[] degree = new int[n]; |
| 514 | + for (int v = 0; v < n; v++) { |
| 515 | + int d = 0; |
| 516 | + boolean[] row = adj[v]; |
| 517 | + for (int w = 0; w < n; w++) if (row[w]) d++; |
| 518 | + degree[v] = d; |
| 519 | + } |
| 520 | + |
| 521 | + // Per-vertex bitset of colors already used by its (colored) neighbors. |
| 522 | + // The smallest available color is then just nextClearBit(0). |
| 523 | + BitSet[] neighborColors = new BitSet[n]; |
| 524 | + for (int v = 0; v < n; v++) neighborColors[v] = new BitSet(); |
| 525 | + |
468 | 526 | int maxColor = 0; |
469 | 527 |
|
470 | 528 | for (int step = 0; step < n; step++) { |
471 | | - // Pick uncolored vertex with highest saturation, break ties by degree |
| 529 | + // Pick uncolored vertex with highest saturation; break ties by |
| 530 | + // (precomputed) degree. |
472 | 531 | int best = -1; |
| 532 | + int bestSat = -1; |
| 533 | + int bestDeg = -1; |
473 | 534 | for (int v = 0; v < n; v++) { |
474 | 535 | if (color[v] != -1) continue; |
475 | | - if (best == -1 || saturation[v] > saturation[best]) best = v; |
476 | | - else if (saturation[v] == saturation[best]) { |
477 | | - int dv = 0, db = 0; |
478 | | - for (int w = 0; w < n; w++) { if (adj[v][w]) dv++; if (adj[best][w]) db++; } |
479 | | - if (dv > db) best = v; |
| 536 | + int sv = saturation[v]; |
| 537 | + if (sv > bestSat || (sv == bestSat && degree[v] > bestDeg)) { |
| 538 | + best = v; |
| 539 | + bestSat = sv; |
| 540 | + bestDeg = degree[v]; |
480 | 541 | } |
481 | 542 | } |
482 | 543 |
|
483 | | - // Find smallest available color |
484 | | - Set<Integer> usedColors = new HashSet<>(); |
485 | | - for (int w = 0; w < n; w++) { |
486 | | - if (adj[best][w] && color[w] != -1) usedColors.add(color[w]); |
487 | | - } |
488 | | - int c = 0; |
489 | | - while (usedColors.contains(c)) c++; |
| 544 | + int c = neighborColors[best].nextClearBit(0); |
490 | 545 | color[best] = c; |
491 | | - maxColor = Math.max(maxColor, c); |
| 546 | + if (c > maxColor) maxColor = c; |
492 | 547 |
|
493 | | - // Update saturation |
| 548 | + // Propagate saturation to uncolored neighbors. |
| 549 | + boolean[] row = adj[best]; |
494 | 550 | for (int w = 0; w < n; w++) { |
495 | | - if (adj[best][w] && color[w] == -1) saturation[w]++; |
| 551 | + if (!row[w] || color[w] != -1) continue; |
| 552 | + if (!neighborColors[w].get(c)) { |
| 553 | + neighborColors[w].set(c); |
| 554 | + saturation[w]++; |
| 555 | + } |
496 | 556 | } |
497 | 557 | } |
498 | 558 | return maxColor + 1; |
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