Can we transfer every DFA to DFAs with start state having no in edge?

The start state cannot have any "in edge" (an arrow point directly to the start state) and only out edge is possible for the start state. Other states except the start state are free of limit in DFA (deterministic finite automaton).

I have doubt between two awnsers which are:

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It depends on the language and is not possible for every DFA.

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Yes we can.

If the answer is positive then how ?

It is possible. You just need to duplicate the start state, one copy becoming the starting state with no in-edge, the other copy becoming a state that can have in-edges.

Formally, if $A = (Q, \delta, q_0, F)$ is a DFA, then consider $B = (Q', \delta', q_0', F')$ with:

• $Q' = Q\cup \{q_0'\}$;

• $F' = F$ if $q_0\notin F$, or $F' = F\cup \{q_0'\}$ otherwise;

• for $q\in Q$ and $a\in \Sigma$, $\delta'(q, a) = \delta(q, a)$;

• for $a\in \Sigma$, $\delta'(q_0', a) = \delta(q_0, a)$.

StackExchange Computer Science Q#158966, answer score: 15