Consider the alphabet T={a}. Let M be the deterministic finite automaton (DFA) which has only one state q0, which is both the initial and also a final state. There is only one transition, going from q0 to q0, labeled 'a'. There are no other states or transitions in the automaton. Now, let L be the language accepted by the automaton M. Which one of the following is correct?

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BlackTom AI · Accepted Answer

Re-stating the problem in my own words: We have a DFA M with a single state q0, which is both the start state and a final state. The alphabet is T = {a}, and there is a single transition from q0 to q0 labeled 'a'. There are no other states or transitions. The language L is the set of strings accepted by M. We must evaluate the given options about L and its complement.

Option a: 'In a DFA, the initial state cannot be final. So, M is not a DFA.'
- This statement is incorrect. It is perfectly standard for a DFA to have the initial state also be a final (accepting) state. Deterministic finite automata are defined by having exactly one transition for each symbol from each state (where defined), a designated start state, and a set of accepting states. The fact that the initial state is also final does not violate the definition. Here, q0 is initial and final, and the transition on 'a' from q0 goes to q0, which satisfies the DFA requirements. The assertion that such a machine cannot be a DFA is simply false.

Option b: 'The complement of L is the empty language.'
- This requires analyzing L. With the given machine, the alphabet is {a}, so the set of all strings over this alphabet is a*, i.e., { ε, a, aa, aaa, ... }. The machine starts in a final state and loops on 'a', so it accepts ε (the empty string) and every string consisting of any number of a's. Thus L = a*. Therefore the complement of L, with respect to a*, is a* 
  − a* = ∅, the empty language. This statement is true under the standard interpretation of the complement relative to the complete set of strings over the alphabet. Keeping in mind that the complement is always taken with respect to the ambient universe a*, this option is correct.

Option c: 'L={a}'
- This is incorrect. As noted, the machine accepts ε as well (since the initial state is final), and it accepts every string consisting of any number of a's due to the loop on 'a' in the final state. Therefore L is not just {a}; it includes ε and aa, aaa, etc., i.e., L = a*. The proposed L = {a} excludes ε and excludes aa, which would not match the actual language accepted by M.

Option d: 'L is the empty language.'
- This is incorrect. The machine clearly accepts ε (empty string) because the start state q0 is final, so the empty string is accepted without consuming any input. Hence L is not empty. The empty language would contain no strings at all, which contradicts the behavior of M.

Summary of reasoning for each option:
- a is false because a DFA can have an initial state that is also final; the described machine satisfies DFA criteria.
- b is true because L = a*, given that q0 is final and the only transition on a loops back to q0, so every string of a's (including ε) is accepted; the complement relative to a* is indeed the empty language.
- c is false because L includes ε and all strings of a's, not just the single string 'a'.
- d is false because ε is accepted, so L is not empty.

Languages and Computation (COMP2049 UNNC) (SPC1 24-25) Lab Quiz 2 - Group A

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