Finite Automata

Seungjin Choi

Department of Computer Science and EngineeringPohang University of Science and Technology

77 Cheongam-ro, Nam-gu, Pohang 37673, Koreaseungjin@postech.ac.kr

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Outline

I Examples of finite automata (=finite state machine, good modelsfor computers with extremely limited amount of memory)

I Deterministic finite automata (DFA)

I Nondeterministic finite automata (NFA)

I Equivalence between DFA and NFA

I More details and applications

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An Example of FA: Automatic Door

I Two states: ”open” or ”closed”

I Four inputs: front, rear, both, neither.

Door

Frontpad

Rearpad

(a) Automatic Door

open

front

neither

neitherbothrear front

rearboth

closed

(b) FA

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An Example of FA: e-Commerce

Protocol for e-commerce using e-money

Allowed events:

1. The customer can pay the store (=send he money-file to the store).

2. The customer can cancel the money (like putting a stop on a check).

3. The store can ship the goods to the customer.

4. The store can redeem the money (=cash the check).

5. The bank can transfer the money to the store.

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Deterministic Finite Automata (DFA)

Definition (DFA)A deterministic finite automaton is a quintuple (5-tuple),M = (Q,Σ, δ, q0,F ), where

I Q: a finite set of states,

I Σ: a finite set of input symbols (input alphabet),

I δ : Q × Σ→ Q: a transition function,

I q0 ∈ Q: a start state (an initial state),

I F ⊆ Q: a set of final or accepting states.

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q0 q1 q2

0 00

11

1

The DFA associated with this transition diagram is given by

M = ({q0, q1, q2}, {0, 1}, δ, q0, {q1}) ,

δ(q0, 0) = q0, δ(q0, 1) = q1,

δ(q1, 0) = q0, δ(q1, 1) = q2,

δ(q2, 0) = q2, δ(q2, 1) = q1.

M accepts 01, 101, 0111, 11001 and does not accept 00,100, 1100.

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Notations for DFA

I Transition diagram

I Transition table

I See Sec. 2.2.3

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Extended Transition Function

The extended transition function, δ∗ : Q × Σ∗ → Q, describes whathappens when we start in any state and follow any sequence of inputs.For example, given δ(q0, a) = q1 and δ(q1, b) = q2, we have

δ∗(q0, ab) = q2.

We can define δ∗ recursively by

δ∗(q, ε) = q,

δ∗(q,wa) = δ(δ∗(q,w), a), q ∈ Q,w ∈ Σ∗, a ∈ Σ.

For example, δ∗(q0, a) = δ(δ∗(q0, ε), a) = δ(q0, a) = q1. Thus,δ∗(q0, ab) = δ(δ∗(q0, a), b) = δ(q1, b) = q2.

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The Language of DFA

DefinitionThe language accepted by a DFA, M = (Q,Σ, δ, q0,F ), is the set of allstrings on Σ accepted by M,

L(M) = {w ∈ Σ∗ | δ∗(q0,w) ∈ F}.

Example: L = {0n1 | n ≥ 0}, the language accepted by the DFA shownbelow:

q0 q1 q2

0

1

0, 1

0, 1

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Example: Find a DFA that recognizes the set of all strings onΣ = {a, b} starting with the prefix ab.

Need 4 states: (1) a start state; (2) 2 states for recognizing ab ending ina final trap state; (3) one non-final trap state.

q0 q1 q2

q3

a ba, b

a, b

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Example: Find a DFA that accepts all the strings on Σ = {0, 1}, exceptthose containing the substring 001.

00100ǫ

0

0

0

01

1

1

0, 1

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Regular Language

Definition (Regular)A language L is called regular if and only if there exits a DFA M suchthat

L = L(M).

Regular language ⇐⇒ DFA

What you need to do to show that L is regular, is to find aDFA M such that L = L(M).

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Example: Show that the language, L = {awa |w ∈ {a, b}∗} is regular.

a,b

q0

q1

q2 q3

a

a

a

bb

b

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Example: Show that L2 is regular, whereL2 = {aw1aaw2a |w1,w2 ∈ {a, b}∗}.

a,b

q0

q1

q2 q3 q4 q5

a

aa

a a

bbb

b

b

If L is regular, so are L2, L3, . . .? (the answer is ”correct”)

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Nondeterministic Finite Automata (NFA)

Definition (NFA)A nondeterministic finite automaton is a quintuple (5-tuple),M = (Q,Σ, δ, q0,F ), where

I Q: a finite set of states,

I Σ: a finite set of input symbols (input alphabet),

I δ : Q × (Σ ∪ {ε})→ 2Q : a transition function,

I q0 ∈ Q: a start state (an initial state),

I F ⊆ Q: a set of final or accepting states.

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Differences between DFA and NFA

I In NFA, the range of the transition function δ is the power set 2Q ,implying that several different movements are allowed. For example,

δ(q1, a) = {q0, q2}.

I NFA can make a transition without consuming an input symbol. Forexample,

δ(q0, ε) = q1.

I The set δ(qi , a) may be empty. In other words, no transition may bedefined for a specific situation.

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An Example of NFA

ǫ

q0 q1 q2

0

1

0, 1

I Several edges with the same label originate from one vertex

I ε-transition

I δ(q2, 0) = φ

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NFA with Extended Transition Functions

ǫ

ǫ

q0 q1 q20

I For an NFA, the extended transition function is defined so thatδ∗(qi ,w) contains qj if and only if there is a walk in the transitiongraph from qi to qj labeled w .

I δ∗(q1, 0) = {q0, q1, q2} and δ∗(q2, ε) = {q0, q2}.

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The Language of NFADefinitionThe language L accepted by an NFA, M = (Q,Σ, δ, q0,F ), is defined by

L(M) = {w ∈ Σ∗ | δ∗(q0,w) ∩ F 6= φ} .

Example: L = {(10)n | n ≥ 0} is accepted by the NFA shown below:

ǫ

q0 q1 q2

0

1

0, 1

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Why Nondeterminism?

I Nondeterministic machines can serve as models ofsearch-and-backtrack algorithms.

I Sometimes helpful in solving problems easily.

I Certain results are more easily established for NFA’s than for DFA’s.

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Equivalence

DefinitionTwo finite automata M1 and M2 are said to be equivalent if

L(M1) = L(M2),

that is, they accept the same language.

ǫ

q0 q1 q2

0

1

0, 1 q0 q1 q2

0

0

1

1

0, 1

(a) NFA (b) DFA

These two FAs are equivalent.

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Equivalence of DFA and NFA

TheoremFor any NFA MN there exists a DFA MD such that

L(MD) = L(MN)

and vice versa.

I This involves the subset construction, an important example how anautomaton MB can be generically constructed from anotherautomaton MA.

I Given an NFA MN = (QN ,Σ, δN , q0,FN), we will construct a DFAMD = (QD ,Σ, δD , {q0},FD) such that L(MD) = L(MN).

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Subset Construction

I QD = {S |S ⊆ QN}, i.e., QD = 2QN .Note that |QD | = 2|QN |, although most states in QD are likely to begarbage.

I FD = {S ⊆ QN |S ∩ FN 6= φ}.I For every S ⊆ QN and a ∈ Σ,

δD(S , a) = ∪q∈S

δN(q, a).

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Example: Subset Construction

q0 q1 q20 1

0, 10 1

φ φ φ→ {q0} {q0, q1} {q0}{q1} φ {q2}∗{q2} φ φ{q0, q1} {q0, q1} {q0, q2}∗{q0, q2} {q0, q1} {q0}∗{q1, q2} φ {q2}

∗{q0, q1, q2} {q0, q1} {q0, q2}

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Example: An Equivalent DFA

Determine states accessible from the start state and draw a transitiongraph:

{q0} {q0, q1} {q0, q2}0

0

0

1

1

1

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