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computations of programs are driven by their inputs. The outputs are just the results of the

computations, and they have no influence on the course that the computations take.

Consequently, it seems that much can be studied about finite-state transducers, or equivalently,about finite-memory programs even when their outputs are ignored. The advantage of

conducting a study of such stripped-down finite-state transducers is in the simplified

argumentation that they allow.

Finite-State Automata

A finite-state transducer whose output components are ignored is called a finite-state automaton.

Formally, afinite-state automaton M is a tuple , where Q, , q0, and F are

defined as for finite-state transducers, and the transition table is a relation from Q ( { }) toQ.

Transition diagrams similar to those used for representing finite-state transducers can also beused to represent finite-state automata. The only difference is that in the case of finite-state

automata, an edge that corresponds to a transition rule (p, , p) is labeled by the string .

Example 2.3.1 The finite-state automaton that is induced by the finite-state transducer of

Figure2.2.2is , where Q = {q0, q1}, = {a, b}, = {(q0, a, q1), (q0, b, q1), (q1, b,

q1), (q1, a, q0)}, and F = {q0}.

The transition diagram in Figure2.3.1represents the finite-state automaton.

Figure

2.3.1

A finite-state automaton that corresponds to the finite-state transducer of

Figure2.2.2.

The finite-state automaton M is said to be deterministic if, for each state q in Q and for each

input symbol a in , the union (q, a) (q, ) is a multiset that contains at most one element.

The finite-state automaton is said to be nondeterministic if it is not a deterministic finite-stateautomaton.

A transition rule (q, , p) of the finite-state automaton is said to be an transition rule if = . Afinite-state automaton with no transition rules is said to be an -free finite-state automaton.

Example 2.3.2 Consider the finite-state automaton M1 =


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(q4, 1, q5), (q5, 0, q5), (q5, 1, q6), (q6, 1, q6), (q6, 0, q4)}, q0, {q0, q3, q6}>. The transition diagram

of M1 is given in Figure2.3.2.

Figure 2.3.2A nondeterministic finite-state automaton.

M1 is nondeterministic owing to the transition rules that originate at state q0. One of thetransition rules requires that an input value be read, whereas the other two transition rules require

that no input value be read. Moreover, M1 is also nondeterministic when the transition rule (q0, 0,q0) is ignored, because M1 cannot determine locally which of the other transition rules to followon the moves that originate at state q0.

The finite-state automaton M2 in Figure2.3.3is a deterministic finite-state automaton.

Figure 2.3.3A 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, orfinal, configuration if v = and q is an accepting state. With no loss of generality it is assumed

that Q and are mutually disjoint.
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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-stateautomata, 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 bya deterministic finite-state automaton that has no transition rules.

Proof Consider any finite-state automaton M = . Let Ax denote the set of all

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

contains an accepting state.

The proof relies on the observation that Axa contains exactly those states that can be reached

from the states in Ax, by the sequences of transition rules that consume a, that is, Axa = { p | q is

in Ax, and qa * ap }.

Specifically, if p is a state in Axa, then by definition there is a sequence of transition rules 1, . . ., t that takes M from the initial state q0 to state p while consuming xa. This sequence must have

a prefix 1, . . . , i that takes M from q0 to some state q while consuming x (see Figure2.3.4(a)).

Figure 2.3.4Sequences of transition rules that consume xa.

Consequently, q is in Ax and the subsequence i+1, . . . , t of transition rules takes M from stateq to state p while consuming a.

On the other hand, if q is in Ax 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 Axa. In such a case, if '1, . . . , 'ris 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 oftransition rules that consumes xa (see Figure2.3.4(b)).

As a result, to determine if a1 an is accepted by M, one needs only to follow the sequence A ,Aa1, Aa1a2, . . . , Aa1 an of sets of states, where each Aa1 ai+1 is uniquely determined from Aa1 ai
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and ai+1. 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 = Ax 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 . Theaccepting 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 inA }.

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

Example 2.3.3 Let M be the finite-state automaton whose transition diagram is given inFigure2.3.2. The transition diagram in Figure2.3.5

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

Figure2.3.2.

represents an -free, deterministic finite-state automaton that is equivalent to M. Using the

terminology of the proof of Theorem2.3.1A = {q0, q1, q4}, A0 = {q0, q1, q2, q4}, and A00 = A000= = A0 0 = {q0, q1, q2, q3, q4}.
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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 Aand 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 nondeterministicfinite-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 transducerbecause 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 oninput x1 is a prefix of the output of M on input x2 (Exercise2.2.5).

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 cansimulate the computations of finite-state automata.

Theorem 2.3.2 Finite-state automata accept only Type 3 languages.

Proof Consider any finite-state automaton M = . By Theorem2.3.1it can beassumed that M is an -free, finite-state automaton. With no loss of generality, it can also be

assumed that no transition rule takes M to its initial state when that state is an accepting one. (If

such is not the case, then one can add a new state q'0 to Q, make the new state q'0 both an initial

and an accepting state, and add a new transition rule (q'0, , q) to for each transition rule of theform (q0, , q) that is in .)

Let G = be a Type 3 grammar, where N has a nonterminal symbol [q] for eachstate q in Q and P has the following production rules.

a. A production rule of the form [q] a[p] for each transition rule (q, a, p) in the transitiontable .

b. A production rule of the form [q] a for each transition rule (q, a, p) in such that p isan accepting state in F.

c. A production rule of the form [q0] if the initial state q0 is an accepting state in F.
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The grammar G is constructed to simulate the computations of the finite-state automaton M. G

records the states of M through the nonterminal symbols. In particular, G uses its start symbol

[q0] to initiate a simulation of M at state q0. G uses a production rule of the form [q] a[p] tosimulate a move of M from state q to state p. In using such a production rule, G generates the

symbol a that M reads in the corresponding move. G uses a production rule of the form [q] a

instead of the production rule of the form [q] a[p], when it wants to terminate a simulation atan accepting state p.

By induction on n it follows that a string a1a2 an has a derivation in G of the form [q] a1[q1]a1a2[q2] a1a2 an-1[qn-1] a1a2 an if and only if M has a sequence of moves of the form

qa1a2 an a1q1a2 an a1a2q2a3 an a1 an-1qn-1an a1a2 anqn for some accepting state qn. In

particular the correspondence above holds for q = q0. Therefore L(G) = L(M).

Example 2.3.4 The finite-state automaton M1, whose transition diagram is given in

Figure2.3.6(b),

Figure 2.3.6Two equivalent finite-state automata.

is an -free, deterministic finite-state automaton. M1 is not suitable for a direct simulation by a

Type 3 grammar because its initial state q0 is both an accepting state and a destination of atransition rule. Without modifications to M1 the algorithm that constructs the grammar G will

produce the production rule [q0] because q0 is an accepting state, and the production rule [q1]

b[q0] because of the transition rule (q1, b, q0). Such a pair of production rules cannot coexist ina Type 3 grammar.

M1 is equivalent to the finite-state automaton M2, whose transition diagram is given inFigure2.3.6(a). The Type 3 grammar G =


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The accepting computation q'0abaa aq1baa abq0aa abaq1a abaaq2 of M2 on input abaa is

simulated by the derivation [q'0] a[q1] ab[q0] aba[q1] abaa of the grammar.

The production rule [q1] a[q2] can be eliminated from the grammar without affecting the

generated language.

The next theorem shows that the converse of Theorem2.3.2also holds. The proof shows how

finite-state automata can trace the derivations of Type 3 grammars.

Theorem 2.3.3 Each Type 3 language is accepted by a finite-state automaton.

Proof Consider any Type 3 grammar G = . The finite-state automaton M = accepts the language that G generates if Q, , qS, and F are as defined below.

M has a state qA in Q for each nonterminal symbol A in N. In addition, Q also has a

distinguished state named qf. The state qS of M, which corresponds to the start symbol S, is

designated as the initial state of M. The state qfof M is designated to be the only accepting state

of M, that is, F = {qf}.

M has a transition rule in if and only if the transition rule corresponds to a production rule of G.Each transition rule of the form (qA, a, qB) in corresponds to a production rule of the form A

aB in G. Each transition rule of the form (qA, a, qf) in corresponds to a production rule of the

form A a in G. Each transition rule of the form (qS, , qf) in corresponds to a production rule

of the form S in G.

The finite-state automaton M is constructed so as to trace the derivations of the grammar G in its

computations. M uses its states to keep track of the nonterminal symbols in use in the sententialforms of G. M uses its transition rules to consume the input symbols that G generates in the

direct derivations that use the corresponding production rules.

By induction on n, the constructed finite-state automaton M has a sequence qA0x u1qA1v1

u2qA2v2 un-1qAn-1vn-1 xqAn of n moves if and only if the grammar G has a derivation of

length n of the form A0 u1A1 u2A2 un-1An-1 x. In particular, such correspondenceholds for A0 = S. Consequently, x is in L(M) if and only if it is in L(G).

Example 2.3.5 Consider the Type 3 grammar G = , where P consistsof the following transition rules.
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The transition diagram in Figure2.3.7

Figure

2.3.7

A finite-state automaton that accepts L(G), where G is the grammar of

Example2.3.5.

represents a finite-state automaton that accepts the language L(G). The derivation S aA aaA

aab in G is traced by the computation qSaab aqAab aaqAb aabqfof M.

It turns out that finite-state automata and Type 3 grammars are quite similar mathematicalsystems. The states in the automata play a role similar to the nonterminal symbols in the

grammars, and the transition rules in the automata play a role similar to the production rules inthe grammars.

Type 3 Grammars and Regular Grammars

Type 3 grammars seem to be minimal in the sense that placing further meaningful restrictions on

them results in grammars that cannot generate all the Type 3 languages. On the other hand, someof the restrictions placed on Type 3 grammars can be relaxed without increasing the class of

languages that they can generate.
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Specifically, a grammar G = is said to be a right-linear grammarif each of its

production rules is either of the form A xB or of the form A x, where A and B are

nonterminal symbols in N and x is a string of terminal symbols in *.

The grammar is said to be a left-linear grammarif each of its production rules is either of the

form A Bx or of the form A x, where A and B are nonterminal symbols in N and x is a stringof terminal symbols in *.

The grammar is said to be a regular grammarif it is either a right-linear grammar or a left-lineargrammar. A language is said to be a regular language if it is generated by a regular grammar.

By Exercise2.3.5a language is a Type 3 language if and only if it is regular.

Regular Languages and Regular Expressions

Regular languages can also be defined, from the empty set and from some finite number of

singleton sets, by the operations
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