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Finite AutomataFinite Automata

FiniteFinite AutomatonAutomatonInput

String

Output

String

FiniteAutomaton

A string over a given alphabet written on an input file

A control unit consisting of a finite number of internal states

The internal state at the next time step is determined by the transition function

FiniteFinite AccepterAccepter

“ Accept” or“Reject”

String

FiniteAutomaton

Output

Input String of

Internal states

““ Accept”Accept”

““Reject”Reject”

Initial state

final state

FormalitiesFormalitiesDeterministic Finite Accepter (DFA)

(p.38)

FqQM ,,,, 0Q

: set of internal states

: input alphabet

: Q × Q transition function

: initial state

: set of final states

configuration

Input String

Internal states

a ab b

Leftmost position

initial state

: a particular status of the input file and internal stateInitial

Internal states

a ab b 10 , qaq Transition function

current state current input symbol

move to the next state

Next state

At q0 now check for current symbol-

Internal states

a ab b 21, qbq Transition function

Next state

move to the next state

Internal states

a ab b 32 , qbq Transition function

Internal states

a ab b 43 , qaq Transition function

Internal states

a ab b No symbol left and it stops at final state

The string abba is accepted by the machine.

Transition functions are the most important part for a finite accepter.

We use a transition graph to represent a finite accepter such that Node state, and Edge with label transition function

(current symbol)

Transition Graph Transition Graph GGMM

32 , qbq Transition Graph forTransition Graph for

Next state

q2 q3b

Initial state

Final state qf

Transition Graph

initialstate

“accept”statetransition

abba –Finite Accepter

0q 1q 2q 3q 4qa b b a

a a bb

32 , qbq It is important to have a clear and intuitive picture to work with DFA. Transition graph: vertices > states, edges > transitions

Initial ConfigurationInitial Configuration

Input Stringa b b a

1q 2q 3q 4qa b b a

a a bb

Reading the InputReading the Input

a a bb

One at a time from left to right

a b b a

a a bb

a b b a

a a bb

a b b a

a a bb

a b b a

Output: “accept”

a b b a

a a bb

Input finished

a b b a

Another Example: Another Example: abaaba

1q 2q 3q 4qa b b a

a a bb

1q 2q 3q 4qa b b a

a a bb

Initial ConfigurationInitial Configuration

a a bb

Output:“reject”

a a bb

Input finished

Another Input StringAnother Input String

1q 2q 3q 4qa b b a

a a bb

1q 2q 3q 4qa b b a

a a bb

Output:“ reject”

End of input!

a a bb

A Trap StateA Trap State

It is important to have a clear and intuitive picture to work with DFA. Transition graph: vertices > states, edges > transitions

What are Q, , , q0, F?

A Deterministic Finite Accepter or DFA is defined as (Def. 2.1 p.38)

FqQM ,,,, 0

a a bb

InputInput AlphabetAlphabet

a a bb

SetSet ofof StatesStates

543210 ,,,,, qqqqqqQ

a a bb

InitialInitial StateState

1q 2q 3q 4qa b b a

a a bb

SetSet ofof FinalFinal StatesStates

0q 1q 2q 3qa b b a

a a bb

TransitionTransition FunctionFunction

a a bb

By Transition Graph

What does “deterministic” mean?What does “deterministic” mean?

Other than using Other than using transition graph transition graph GGMM to represent a to represent a

transition function, there are two common ways to transition function, there are two common ways to represent a transition function.represent a transition function.

is a total function

a a bb

FqQM ,,,, 0

a a bb

10 , qaq 50 , qbq

51, qaq 21, qbq

52 , qaq 32 , qbq

43 , qaq 53 , qbq

54 , qaq 54 , qbq

55 , qaq 55 , qbq

a a bb

5q5q5q5q

ExtendedExtended Transition Transition Function Function

*QQ *:*

QQ :Transition Transition functionfunction

including including

a a bb

20 ,* qabq

3q 4qa b b a

a a bb

0q 1q 2q

QQ *:*

40 ,* qabbaq

a a bb

1q 2q 3q 4qa b b a

a a bb

50 ,* qabbbaaq

Extended Transition Extended Transition Function Function

)),,(*(),(*

awqwaq

Recursive DefinitionRecursive Definition

for all for all qq Q, w Q, w*, a*, a

QQ *:*

RecursiveRecursive DefinitionDefinition

)),,(*(,*

awqwaq

)),,(*(,* awqwaq

qwaq ,*If

then a state, say q1, such that

aq qw1q

a a bb

, ,ba0q , 0 ,,* baq

* b 0 ),,( aq * ab

)),,(*(,*

qwq ,*kw ...21

In the DFA M, there iswith

q q1 2 k

There is a walk in GM

from to with labelq q

kw ...21

q q1 2 k

kw ...21

qwq ,*In the DFA M, there iswith kw ...21

q q1 2 k

kw ...21

qwq ,*In the DFA M, there iswith kw ...21

Theorem 2.1Theorem 2.1

50 ,* qabbbaaq

1q 2q 3q 4qa b b a

a a bb

There is a walk from to with label

0qabbbaa

If and only if

LanguagesLanguages AcceptedAccepted byby DFAsDFAs

Take DFA

Definition: (p.40)

The language contains all input strings accepted by

= { strings that drive to a final state}

= { = { w w * : * : *(*(qq00, , ww)) FF } }

FqQM ,,,, 0

ExampleExample

a a bb

abbaML

acceptaccept

back to complement Ex. p.51

AnotherAnother ExampleExample

abbaabML ,,

acceptaccept

a a bb

acceptacceptacceptaccept

Formally, for a DFA Formally, for a DFA

Since M is DFA, every w*, corresponding to a unique walk in the transition graph of M.

qwq ,* 0

FqQM ,,,, 0

0q qwFq )(MLw

Fq )(MLw

ObservationObservation Language rejected by :

FwqwML ,*:* 0

What is the complement of language on example (p.48)?Example on p.48

Examples on Constructing Examples on Constructing DFAsDFAs

}0:{ nbaML n

0q 1q 2q

accept trap state

= { all strings with prefix }ab ML

accept

0q 1q 2q

= { all strings without substring 001} ML

001 0 00

What if consider DFA for {strings with What if consider DFA for {strings with 001}?001}?

RegularRegular LanguagesLanguages

A language is regular if there is a DFA such that

All regular languages form a language family

LM MLL

abba abbaab,,

}0:{ nban

{ all strings with prefix }{ all strings with prefix }ab

{ all strings without substring }{ all strings without substring }001

Examples of regular languages:

There exist automata that accept theseLanguages (see previous slides).

abbaIs the complement of also regular?

How can you show that the complement of a regular language is regular? (referring to Hw# 3, 4 on p.47)

The language is regular

*,: bawawaL

0q 2q 3q

since we can find a DFA M such that

L*=L {}

The language is regular. How about L2 ?

*,: bawawaL

Can you show LL* is a regular language?Try Hw # 20, 21, 22

Is Is L?L?

If there is a DFA M for L, can you tell if L?

If L is regular, then how about L-{}?

Try hw #17 & 18 p.48

How can you construct a DFA for L-{How can you construct a DFA for L-{} } fromfrom the DFA of L? the DFA of L? How can you construct a DFA for LHow can you construct a DFA for L{{} } from from the DFA of L? the DFA of L?

Homework 2.1 Homework 2.1 p.47

Try: 1 ~7, 9, 11~25 Hand in: 2bd, 4, 6, 7e, 9bf, 18

For hw18, state the general method then use the example to illustrate the method where L {a,b}*L= {w: w= or w contains a substring “ab”}

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