If I have a DFA with only one state that is not an accept state, it accepts only the empty set.

I get confused when if the DFA of one state is an accept state. Does this mean it accepts everything? Or could you have it only accept "a" and not "b". I guess I don't understand if not having an arrow out implies there is a second "invisible state"

In other words, for a DFA of one state, are there 2 or 4 distinct machines?



First let $\Sigma = \{a,b\}$.

Assuming that you're taking a strict approach, there has to be a transition from every state for every symbol in the alphabet, then you have only two possible DFAs (that accept $\emptyset$ and $\Sigma^{\ast}$ respectively).

If you allow a partial transition function, where missing transitions imply rejection (note that this can still be interpreted deterministically), then there are five different languages that can be accepted (strictly there's seven automata): $\emptyset$, $\{\varepsilon\}$, $a^{\ast}$, $b^{\ast}$ and $\Sigma^{\ast}$.

With a larger alphabet, in the partial function case, you get more of course. With $|\Sigma| = n$ there are $2^{n} + 1$ possible languages, from $2^{n+1}$ different automata (half produce the same language). You get $2^{n}$ from each possible choice of which symbols will have a transition in the case where the state is accepting, and the $+ 1$ from the case where the state is non-accepting (it doesn't matter what transitions you have then, the languages are all $\emptyset$).

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    $\begingroup$ @DavidRicherby, no epsilon transitions, just a partial function for the transition function instead of a [total] function. I don't particularly like this version, but some people do. You just get implicit early rejection. $\endgroup$ Sep 22 '15 at 6:45
  • $\begingroup$ Gotcha. I somehow attributed the "+1" to $\epsilon$ transitions, rather than to the state being rejecting. $\endgroup$ Sep 22 '15 at 7:21
  • $\begingroup$ @DavidRicherby, ahh, actually that's worth explaining in the answer, ta. $\endgroup$ Sep 22 '15 at 8:53

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