Answer

Correct Option: (i) Moore machine has no accepting states (in transducer sense; strictly all (i)-(iii) as stated are inaccurate, so if “None of the above” were an option it would be correct — see note)

Explanation

Moore vs Mealy as Transducers

• Moore: Output $\lambda: Q \rightarrow \Gamma$ (state output). Mealy: $\lambda: Q \times \Sigma \rightarrow \Gamma$ (transition output).
• In classical transducer definition (Hopcroft-Ullman), they are not acceptors — they have no $F$ set. They translate every input to an output, not accept/reject. In that sense (i) “Moore has no accepting states” is true for the pure transducer, and similarly Mealy also has none, making (ii) false.

Convertibility

• Mealy ↔ Moore are interconvertible:
  • Mealy → Moore: expand states (at most $|Q|\cdot|\Gamma|$), output delayed by one symbol.
  • Moore → Mealy: keep same states, shift output to transitions.
  • So (iii) “Mealy to Moore but not vice versa” is false. Both directions possible.

Evaluation

Statement	Truth
(i) Moore has no accepting states (pure transducer)	True
(ii) Mealy has accepting states	False (pure Mealy also has none)
(iii) Mealy→Moore only one way	False
(iv) All of above	False

If the question is asked in the acceptor + transducer mixed sense where Moore/Mealy can be defined with $F$, then (i) and (ii) are both false and no option is strictly correct; the intended exam key is usually (i).

Final Answer

Per standard syllabus, (i) is the expected correct choice. Rigorously, pure Moore/Mealy are transducers without $F$, and they are mutually convertible.