Consider the regular grammar: S → Xa | Ya X → Za Z → Sa | ϵ Y → Wa W → Sa…

Key insight: derive the set of strings generated by the start symbol S.

• From Z → ε | S a we get L(Z) = { ε } ∪ { w a | w ∈ L(S) }.

• Then X → Z a gives L(X) = { a } ∪ { w a a | w ∈ L(S) }.

• Also W → S a and Y → W a yield L(Y) = { w a a | w ∈ L(S) }.

• Since S → X a | Y a, combine the above to get L(S) = { a a } ∪ { w a^3 | w ∈ L(S) }.

• Therefore L(S) = { a^{3k+2} | k ≥ 0 } (strings of a whose length is 2 modulo 3).

Minimal DFA reasoning:

• Recognizing a unary language that depends only on length modulo 3 requires distinguishing three residue classes (0, 1, 2 mod 3).

• A DFA with three states arranged in a 3-cycle on input 'a' can track these residues and accept exactly the residue 2 class.

• Two states are insufficient (they can only track parity), and any larger DFA is not minimal.

Conclusion: the minimum number of states required for the DFA is 3.