A deterministic finite automaton (or DFA) is a deterministic automaton with a finite input alphabet and a finite number of states. It can be formally defined as a 5-tuple $(S, \Sigma, \delta, q_0, F)$ , where

• $S$ is a non-empty finite set of states,
• $\Sigma$ is the alphabet (defining what set of input strings the automaton operates on),
• $\delta : S\times\Sigma\rightarrow S$ is the transition function,
• $q_0 \in S$ is the starting state, and
• $F \subseteq S$ is a set of final (or accepting states).

DFAs represent regular languages, and can be used to test whether any string in $\Sigma^*$ is in the language it represents. Consider the following regular language over the alphabet $\Sigma := \left\{ \verb.a., \verb.b. \right\}$ (represented by the regular expression aa*b):

\begin{eqnarray*} \verb.<.S\verb.>. & \verb.::=. & \verb=a=\,A \\ \verb.<.A\verb.>. & \verb.::=. & b\,\verb.|.\,\verb=a=\,A \\ \end{eqnarray*} This language can be represented by the DFA with the following state diagram:

The vertex 0 is the initial state $q_0$ , and the vertex 2 is the only state in $F$ . Note that for every vertex there is an edge leading away from it with a label for each symbol in $\Sigma$ . This is a requirement of DFAs, which guarantees that operation is well-defined for any finite string.

If given the string aaab as input, operation of the DFA above is as follows. The first a is removed from the input string, so the edge from 0 to 1 is followed. The resulting input string is aab. For each of the next two as, the edge is followed from 1 to itself. Finally, b is read from the input string and the edge from 1 to 2 is followed. Since the input string is now $\lambda$ , the operation of the DFA halts. Since it has halted in the accepting state 2, the string aaab is accepted as a sentence in the regular language implemented by this DFA.

Now let us trace operation on the string aaaba. Execution is as above, until state 2 is reached with a remaining in the input string. The edge from 2 to 3 is then followed and the operation of the DFA halts. Since 3 is not an accepting state for this DFA, aaaba is not accepted.

Remarks.

• A DFA can be modified to include $\epsilon$ -transitionsEpsilonTransitions. But the resulting DFA can be simulated by another DFA (without any epsilon transitions).
• Although the operation of a DFA is much easier to compute than that of a non-deterministic automaton, it is non-trivial to directly generate a DFA from a regular grammar. It is much easier to generate a non-deterministic finite automaton from the regular grammar, and then transform the non-deterministic finite automaton into a DFA.