Solution: Equivalent Rooms in a Dungeon¶
Codex (GPT-6): models obstacle behavior as a deterministic finite automaton and applies Hopcroft partition refinement, with an implementation and proof for this problem. Automaton minimization is an established method.
Experiments as an automaton¶
Add an absorbing state 0 whose transitions all return to 0. Rooms are accepting states, while 0 is rejecting. A command word ends in a room exactly when no obstacle has been encountered. The requested relation is therefore automaton state equivalence.
Start with blocks \(\{0\}\) and \(\{1,\ldots,n\}\) . Split a block whenever some command \(j\) sends its members to different blocks. A stable partition sends members of the same block into the same block under every command.
In a stable partition, same-block states agree on the empty experiment. For a nonempty word, their first transitions enter the same block, and induction on the remaining length proves equal results. Conversely, truly equivalent states are never separated: their initial accepting status agrees, and their successors under a common command are equivalent. The final blocks are exactly the equivalence classes. Exclude the obstacle block from the count.
Refine with smaller blocks¶
Maintain a queue of blocks to process. For a dequeued block \(A\) and each command \(j\) , use reverse edges to find
Split each touched block \(Y\) into \(Y\cap X\) and \(Y\setminus X\) , discarding empty parts.
Initially only the obstacle block needs to be queued. The automaton is complete and the other block is its complement, so refining by that complement adds no distinction. After a split, keep the larger half's block identifier and enqueue the smaller half under a new identifier. If the old identifier is already queued, it will still be processed. If it has been processed, the smaller half suffices: membership in a set and one of its halves determines membership in the other half.
A block may split again while its different commands are being processed. Copy its members when dequeuing it and use that snapshot for every command in this round, so the round consistently processes the same set.
Move only touched states¶
A permutation of states and its inverse represent the partition; each block occupies a contiguous interval. While scanning \(X\) , swap each encountered state to the front of its block and count the hits. Update block identifiers only for the smaller resulting half. The number of moved states is at most the number of discovered predecessors, avoiding a scan of the untouched large half.
Whenever a state enters a new queued smaller block, its block size at least halves. A retained identifier is dequeued at most once. Thus each state participates in reverse-edge scans \(O(\log(n+1))\) times. A state has exactly one outgoing edge per command, so a command's predecessor scan cannot hit it twice.
Total time is \(O(nk\log(n+1))\) and space is \(O(nk+n)\) . The product constraint bounds the reverse-edge arrays. Repeatedly rebuilding every transition signature can require linearly many rounds on a chain; it is useful as a small independent model and is implemented in the generator. The full reference uses smaller-block refinement.
See the standard program, test recipes, and measurements.