Consider the language L over the alphabet {0, 1}, given below: \(L \)= {\(w…

Answer: 4 states. A minimal DFA for the language requires four states.

Key idea: the automaton only needs to remember how many consecutive 1s have just been seen (0, 1, or 2). If three consecutive 1s occur, we move to a rejecting trap state.

• State S0 (start, accepting): no trailing 1s. On input 0 stay in S0; on input 1 go to S1.

• State S1 (accepting): exactly one trailing 1. On input 0 go to S0; on input 1 go to S2.

• State S2 (accepting): exactly two trailing 1s. On input 0 go to S0; on input 1 go to SD (trap).

• State SD (rejecting trap): three or more consecutive 1s have occurred. On both inputs 0 and 1 stay in SD.

Minimality (why 4 states are necessary):

• Consider the four strings: ε (empty), 1, 11, and 111. For any two different strings among these, there exists a continuation that makes one accepted and the other rejected, so they belong to distinct Myhill–Nerode equivalence classes.

• Examples of distinguishing continuations:

• ε vs 1: append 11. ε·11 = 11 (accepted) but 1·11 = 111 (rejected).

• 1 vs 11: append 1. 1·1 = 11 (accepted) but 11·1 = 111 (rejected).

• Any of ε, 1, or 11 vs 111: append the empty string. 111 is already rejected while the others are accepted.

Because there are at least four pairwise distinguishable equivalence classes, any DFA must have at least four states. The construction above with four states shows four states suffice, so the minimum number of states is 4.