If a language is accepted by a finite-state automaton, then it is also decided by a deterministic finite-state automaton that has no transition rules

Figure 2.3.3 A deterministic finite-state automaton.

M₁ has two transition rules, and M₂ has one.

A configuration , or an instantaneous description, of the finite-state automaton is a singleton uqv, where q is a state in Q, and uv is a string in *. The configuration is said to be an initial configuration if u = and q is the initial state. The configuration is said to be an accepting , or final, configuration if v = and q is an accepting state. With no loss of generality it is assumed that Q and are mutually disjoint.

Other definitions, like those of _M , , _M *, *, and acceptance, recognition, and decidability of a language by a finite-state automaton, are similar to those given for finite-state transducers.

Nondeterminism versus Determinism in Finite-State Automata

By the following theorem, nondeterminism does not add to the recognition power of finite-state automata, even though it might add to their succinctness. The proof of the theorem provides an algorithm for

constructing, from any given n-state finite-state automaton, an equivalent deterministic finite-state automaton of at most 2ⁿ states.

Theorem 2.3.1 If a language is accepted by a finite-state automaton, then it is also decided by a deterministic finite-state automaton that has no transition rules.

Proof Consider any finite-state automaton M = <Q, , , q₀, F>. Let A_x denote the set of all the states

that M can reach from its initial state q₀, by the sequences of moves that consume the string x, that is, the set { q | q₀x * xq }. Then an input w is accepted by M if and only if A_w contains an accepting state.

The proof relies on the observation that A_xa contains exactly those states that can be reached from the states in A_x, by the sequences of transition rules that consume a, that is, A_xa = { p | q is in A_x, and qa *

Figure 2.3.4 Sequences of transition rules that consume xa.

Consequently, q is in A_x and the subsequence _i+1, . . . , _t of transition rules takes M from state q to state p while consuming a.

On the other hand, if q is in A_x and if p is a state that is reachable from state q by a sequence ₁, . . . , _s of transition rules that consumes a, then the state p is in A_xa. In such a case, if '₁, . . . , '_r is a sequence of transition rules that takes M from the initial state q₀ to state q while consuming x, then M can reach the state p from state q₀ by the sequence '₁, . . . , '_r, ₁, . . . , _s of transition rules that consumes xa (see Therefore, a deterministic finite-state automaton M' of the following form decides the language that is accepted by M.

The set of states of M' is equal to { A | A is a subset of Q, and A = A_x for some x in * }. Since Q is finite, it follows that Q has only a finite number of subsets A, and consequently M' has also only a finite number of states. The initial state of M' is the subset of Q that is equal to A . The accepting states of M' are those states of M' that contain at least one accepting state of M. The transition table of M' is the set { (A, a, A') | A and A' are states of M', a is in , and A' is the set of states that the finite-state automaton M can reach by consuming a from those states that are in A }.

By definition, M' has no transition rules. Moreover, M' is deterministic because, for each x in * and each a in , the set A_xa is uniquely defined from the set A_x and the symbol a.

Example 2.3.3 Let M be the finite-state automaton whose transition diagram is given in Figure 2.3.2.

The transition diagram in Figure 2.3.5

Figure 2.3.5 A transition diagram of an -free, deterministic finite-state automaton that is equivalent to the finite-state automaton whose transition diagram is given in Figure 2.3.2.

represents an -free, deterministic finite-state automaton that is equivalent to M. Using the terminology of the proof of Theorem 2.3.1 A = {q₀, q₁, q₄}, A₀ = {q₀, q₁, q₂, q₄}, and A₀₀ = A₀₀₀ = = A_{0 0} = {q₀, q₁, q₂, q₃, q₄}.

A is the set of all the states that M can reach without reading any input. q₀ is in A because it is the initial state of M. q₁ and q₂ are in A because M has transition rules that leave the initial state q₀ and enter states q₁ and q₂, respectively.

A₀ is the set of all the states that M can reach just by reading 0 from those states that are in A . q₀ is in A₀ because q₀ is in A and M has the transition rule (q₀, 0, q₀). q₁ is in A₀ because q₀ is in A and M can use

the pair (q₀, 0, q₀) and (q₀, , q₁) of transition rules to reach q₁ from q₀ just by reading 0. q₂ is in A₀ because q₀ is in A and M can use the pair (q₀, , q₁) and (q₁, 0, q₂) of transition rules to reach q₂ from q₀ just by reading 0.

The result of the last theorem cannot be generalized to finite-state transducers, because deterministic finite-state transducers can only compute functions, whereas nondeterministic finite-state transducers can also compute relations which are not functions, for example, the relation {(a, b), (a, c)}. In fact, there are also functions that can be computed by nondeterministic finite-state transducers but that cannot be

computed by deterministic finite-state transducers. R = { (x0, 0^|x|) | x is a string in {0, 1}* } { (x1, 1^|x|) | x is a string in {0, 1}* } is an example of such a function. The function cannot be computed by a

deterministic finite-state transducer because each deterministic finite-state transducer M satisfies the following condition, which is not shared by the function: if x₁ is a prefix of x₂ and M accepts x₁ and x₂, then the output of M on input x₁ is a prefix of the output of M on input x₂ (Exercise 2.2.5).

Finite-State Automata and Type 3 Grammars

The following two results imply that a language is accepted by a finite-state automaton if and only if it is a Type 3 language. The proof of the first result shows how Type 3 grammars can simulate the

computations of finite-state automata.