Decidability of the language of DFAs accepting only odd-length strings [closed]

Question

Let $L$ be the language $\{\langle M \rangle : M\text{ is a DFA that accepts only odd-length strings}\}$.

Prove that $L$ is decidable.

How wrong is my answer?

Create a TM $T$ that decides $L$:

1. Create an equivalent input string $v$ that is in the language $L(M)$.

2. Give encoding to $T = \langle M, v \rangle$.

3. TM $T$ simulates $M$ on input $v$.

4. If TM $T$ halts and accepts, accept. Otherwise, reject.

I would like to know how off I am with my answer.

Answer 1

Every DFA can be simplified by a Turing Machine, so you can simplify the DFA you receive as input and compare it to the simplified DFA that accepts odd-length strings.

On input $M$:

1. Simplify $M$. Remove every unreachable state and transition from $M$ (disjoint from start state)
2. If the start state of $M$ is reject, and the only other state in $M$ is accept, and for every $w \in \Gamma, \exists$ a $w-arrow$ from accept to reject and from reject to accept, then accept. Otherwise reject.

Answer 2

Forget TMs here, all you need is a decision algorithm, so do this

1. Create a DFA, $N$ that accepts only odd-length strings (a two-state DFA will do it).
2. On input $M$, determine whether $L(M)=L(N)$, using known results about regular language closure properties (for example, for any sets $A, B$ we know $A=B$ if and only if their symmetric difference, $(A\cap \overline{B})\cup (\overline{A}\cap B)$ is empty). If $A, B$ are regular, then their symmetric difference is regular (by closure properties of regular languages).
3. Test the symmetric difference for emptiness. There's a well-known result about how to do that for regular languages.
4. If the difference is empty, answer "yes" else answer "no".

There's your decider.