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I know that a DFA has to have exactly one transition for each symbol in the alphabet, but is it allowed to have two symbols on the same arrow? If, for example, I have a DFA with states $q_0$ and $q_1$, can I have one arrow from $q_0$ to $q_1$ with both $a$ and $b$?

This may be a stupid question, but I need to be completely sure that this is allowed (I believe it is).

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  • $\begingroup$ Yes. That's just the same thing as having two arrows, one labelled $a$ and one labelled $b$. $\endgroup$ – David Richerby Feb 6 '14 at 14:04
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The transition graph (as a drawing) is merely a representation of an Automaton, which is a well-defined model.

Formally, an DFA is a tuple $(Q,\Sigma,\delta,q_0,F)$, where the "type" of the transition function is $\delta:Q\times \Sigma\to Q$.

Thus, if you have $\delta(q_0,a)=\delta(q_0,b)=q_1$, that's fine. In the graphic representation, you will either have two arrows from $q_0$ to $q_1$, labeled $a$ and $b$, or you can just put both letters on the same arrow, it's not a formal thing anyway.

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As long as the starting and ending of respective states are same you can add as many symbols. enter image description here

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  • $\begingroup$ I don't understand what you mean. An arrow by definition goes from one state to another state. How could an arrow not have the same start and end state as itself? $\endgroup$ – David Richerby Feb 6 '14 at 14:07
  • $\begingroup$ I've edited it as 'respective states' $\endgroup$ – Terminal Feb 6 '14 at 14:07
  • $\begingroup$ I saw the edit. I still don't understand what you mean. $\endgroup$ – David Richerby Feb 6 '14 at 14:08

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