Finishing the Festival

Propagate-then-split, run to the end, finds every legal festival schedule — three — from eight surviving candidates, almost entirely by deletion.

Propagation left Anvil with 2 and 3, Brook with 2 and 3, Cinder fixed at 3, Dune with 1 and 2. Split on Anvil. Case one, Anvil = 2: the opener rule demands Dune earlier than 2, collapsing Dune to 1; the drummer rule strikes Brook's 2, collapsing Brook to 3. Every domain is now a single value — Dune 1, Anvil 2, Brook 3, Cinder 3 — a full solution, produced by deletion alone, and identical to the one backtracking sweated for. Case two, Anvil = 3: the drummer rule collapses Brook to 2; the opener rule trims nothing further, leaving Dune with 1 and 2. One more split, on Dune, and both branches survive every check: Dune 1 or Dune 2, with Anvil 3, Brook 2, Cinder 3. Count the harvest: three legal schedules in total, found by a case tree that touched a handful of nodes — against the 81 conceivable assignments the napkin started with. That ratio is the trademark of constraint solving: propagation does the demolition, splitting handles the remainder, and the exhaustive tour that backtracking took through the doomed drawer never happens at all. One final audit then closes the run: check each schedule against all four rules — drummer, fireworks, opener, van — and each passes; check that no fourth schedule survived, and the case tree itself says so, because every branch was either solved or split and nothing was dropped.