Why is E (dfa) a solvable language?

Question

I do not understand why the Turing machine T ACCEPTS, when the state is not accepted, marked and rejected, when the state of acceptance is noted:

E (dfa) = {| A is DFA and L (A) = empty set (do not have a character)}

E (dfa) is a decidable language.

Evidence. DFA accepts some line, if it is possible to achieve the acceptance state from the start state,> move along the arrows of the DFA. To verify this condition, we can design> TM T, which uses a labeling algorithm similar to that used in Example 3.23.

T = "At the input, where A is DFA: 1. Mark the initial state of A. 2. Repeat until new states are noted: 3. Mark any state that has a transition entering it from any state that is already marked. 4. If the acceptance status is not accepted, accept; otherwise reject. "

It seems to me the opposite. Can anyone explain this?

Thank.

Answer 1

, "" "" . , T, , DFA, A, .

L (T) {A | L (A) }. , E (dfa), , L (T) , , , .

, :

• L (T) A , L (A) .
• , L (A) ? L (A) , A .
• , A? .

:

• , A, accept, .
• A, L (A) .
• L (A) , L (T) A.

, T A , , . , , T A iff A .