Minimal deterministic finite automaton for the language \(L=\{0^n \mid n \geq…

Answer: The minimal DFA has 6 states with 5 accepting (final) states.

Reasoning:

• Distinguishability for lengths 0–4: For each n in {0,1,2,3,4}, the string of length n can be turned into length 4 by appending the appropriate number of zeros, so these five lengths are pairwise distinguishable and require distinct states.

• All lengths ≥5 are equivalent: Appending any number of zeros to a string already of length at least 5 never produces length 4, so all such strings behave the same with respect to future input and form a single equivalence class.

• Equivalence classes and states: The classes are length 0, 1, 2, 3, 4, and ≥5, giving 6 Myhill–Nerode classes, hence a minimal DFA with 6 states.

• Accepting states: All classes except the class for length 4 are accepted (the language excludes length 4 only), so there are 5 accepting states.

• Brief transition description: From the start state (length 0) each input 0 moves to the next length state 1,2,3,4, then to the ≥5 state which loops on 0.