DFA Equivalence via Minimization and Isomorphism | Generated by AI

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Question: Give some hints to solve problem 1671 “History of Languages” (determine if two DFAs accept the same language)

Answer:

Here are the most practical and commonly used hints/approaches to solve this problem efficiently:

Core Idea

Two DFAs accept the same language ⇔ their minimized versions are isomorphic (same number of states, same structure up to state renaming).

Since N ≤ 2000 per automaton and T ≤ 26, direct minimization + isomorphism check is feasible but needs to be implemented carefully.

Useful Hints & Strategies

  1. Handle dead/unreachable states first
    • Many input DFAs contain unreachable states or states with no outgoing transitions.
    • Run a DFS/BFS from state 0 on each DFA separately and only keep reachable states.
    • This often reduces the effective number of states significantly.
  2. Treat missing transitions (−1) correctly
    • The standard way is to create an explicit trap (dead) state.
    • Add one extra state (say index N) that is:
      • non-accepting
      • transitions to itself on every symbol
    • Replace every −1 transition with → this trap state.
    • This makes the DFA complete (total transition function).

  3. Minimize each DFA (most important step) The standard efficient algorithm for your constraints is Hopcroft’s algorithm (O(|S| × |Σ| log |S|)), which is fast enough for N=2000, Σ=26.

    Alternative (simpler to code but slower):
    Use partition refinement with the basic table-filling / equivalence-class method:

    • Start with two groups: accepting vs non-accepting states
    • Iteratively split groups until no more distinctions can be made
    • Works well in practice for N ≤ 2000
  4. After minimization, check if the two minimal DFAs are isomorphic
    • They must have the same number of states after minimization.
    • There must exist a bijection between states that preserves:
      • start state → start state
      • accepting ↔ accepting
      • transitions: δ₁(p,a) → q ⇔ δ₂(f(p),a) → f(q)
    • You can do this by:
      • Renumber states of one DFA starting from 0
      • Try to build the mapping using BFS from the start states (state 0 of both)
      • Check that the mapping is total, injective, and preserves transitions and acceptance
  5. Quick necessary (but not sufficient) checks (can early-reject)
    • Same number of reachable states?
    • Same number of accepting reachable states?
    • Same number of states reachable in exactly k steps for small k?
    • Same acceptance behavior on some random / short strings? (These are fast filters before doing full minimization.)
  6. Practical implementation order that works well ``` For each DFA:
    1. Read N, transitions, accepting flags
    2. Replace -1 → trap state (add state N if needed)
    3. Compute reachable states from 0 (BFS/DFS)
    4. Build the subgraph of only reachable states
    5. Minimize this DFA (Hopcroft or partition refinement)
    6. Get the number of states in minimal DFA + canonical state numbering

    Finally:

    • If number of states in min-DFA A ≠ min-DFA B → No
    • Else try to check isomorphism by BFS-matching from start state ```
  7. Very useful observation for contest setting
    • If both DFAs are already minimal and have ≤ 2000 states, isomorphism check alone is often enough.
    • But in practice most test cases require minimization — many problems deliberately give non-minimal DFAs.
  8. Edge cases to watch for
    • One or both automata accept nothing (only trap state reachable and non-accepting)
    • One accepts everything (after minimization becomes 1 accepting state looping everywhere)
    • Disconnected components (but you already filter unreachable)
    • Start state not accepting vs accepting
    • Empty string acceptance (check if state 0 is accepting)

Following steps 1–4 carefully (especially minimization + isomorphism) solves the problem correctly.

Good luck!

References:

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