Proving $E_{DFA}$ is decidable by running $A_{DFA}$ several times

Question

I am trying to prove that language $E_{DFA}$ is decidable using multiple executions of $A_{DFA}$ (not using the proof in Sipser's book "Introduction to the Theory of Computation").

Can I just use the given $M$ as a decider?

$M =$ "On input $\langle B,w \rangle$, where $B$ is a DFA and $w$ is an empty string ($L(w)=\emptyset$):

1. Mark the start of the DFA as $q_0$.
2. Simulate B on input w a finite number of times:
   1. First check if the encoding is correct, if $L(w)=\emptyset$, if not - reject.
   2. Recursively: mark states that can be obtained within a finite $δ$ operations from any marked states.
3. If no accept state marked - accept. else, reject."

Is this correct?

Definitions: $$\begin{align*} A_{DFA} &= \left\{ \langle B,w \rangle \mid \text{$B$ is a DFA that accepts input string $w$} \right\} \\ E_{DFA} &= \left\{ \langle A \rangle \mid \text{$A$ is a DFA and $L\left(A\right) = \varnothing$} \right\} \end{align*}$$

Emphasis: I am trying to prove $E_{DFA}$ by running the proof of $A_{DFA}$ several times (finite), so I thought the correct thing would be to use the input $w$ as empty set.

Answer 1

If you want to reduce $E_{DFA}$ to $A_{DFA}$, you should use the following theorem:

If a DFA with $n$ states accepts some word, then it accepts some word of length less than $n$.

But a much more efficient algorithm would just run DFS from the initial state, and check whether some accepting state is reachable from it.