Decidability of the language of DFAs accepting only odd-length strings 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 $L$:

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

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

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

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


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


Simplify $M$. Remove every unreachable state and transition from $M$ (disjoint from start state)
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.