Languages and Finite Automatasclab.yonsei.ac.kr/courses/05AM_old/lecture2.pdfFormal Languages and Automata - Yonsei CS 3 Contents For Models of Language Generation • Deterministic

Formal Languages and Automata - Yonsei CS 1

Finite Automata


Finite Accepter

Input

“Accept”or

“Reject”

String

FiniteAutomaton

Output


Contents

For Models of Language Generation

• Deterministic Finite Accepters(DFA)- What is DFA? transition graph- The languages which DFA accepts regular language

• Nondeterministic Finite Accepters (NFA)- What is NFA?- Equivalence of DFA & NFA- Reduction of Number of States in FA


Deterministic Accepters

statesfinal ofset a:stateinitial:

function) (totalfunction transition:alphabetinput ofset finite a:

statesinternal ofset finite a :),,,,(DFA.

0

0

QFQq

QQ

QFqQMDef

⊆∈

→∑×∑

∑=⋅

δ

δ

10 ),( qaq =δ

Finite Accepters: Deterministic in their operation

• At the initial time, q0 with input mechanism on the leftmost symbol of the input string.•When the end of the string, the string is accepted if the automaton is one of its final states. Otherwise, reject.•The transitions from one internal state to another by δ


DFA : Transition Graph

: states finalstate initial

) symbolinput : label (on transiti: edge

states :vertex

⋅⋅

⋅q1q0

aiq|Q|

a

To visualize and represent finite automata

10 ),( qaq =δ


Input Alphabet Σ

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

5qa a bb

ba,

{ }ba,=Σ

ba,

),,,,(DFA 0 FqQM δ∑=


Set of Internal States Q


5qa a bb

ba,

{ }543210 ,,,,, qqqqqqQ =

ba,



Initial State 0q

1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q



Set of Final States F

0q 1q 2q 3qa b b a

5qa a bb

ba,

{ }4qF =

ba,

4q



Transition Function δ


5qa a bb

ba,

QQ →Σ×:δ

ba,



( ) 10, qaq =δ

2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q 1q


( ) 50, qbq =δ

1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q



5qa a bb

ba,

ba,

( ) 32, qbq =δ


Transition Function δ


5qa a bb

ba,

δ a b0q1q2q3q4q5q

1q 5q5q 2q5q 3q4q 5q

ba,5q5q5q5q


Transition Graph

initialstate

finalstate

“accept”statetransition

abba -Finite Accepter


5qa a bb

ba,

ba,


Initial Configuration

1q 2q 3q 4qa b b a

5qa a bb

ba,

Input Stringa b b a

ba,

0q


Reading the Input


5qa a bb

ba,

a b b a

ba,



5qa a bb

ba,

a b b a

ba,



5qa a bb

ba,

a b b a

ba,



5qa a bb

ba,

a b b a

ba,



Output: “accept”

5qa a bb

ba,

a b b a

ba,

Input finished


Rejection

1q 2q 3q 4qa b b a

5qa a bb

ba,

a b a

ba,

0q



5qa a bb

ba,

a b a

ba,



5qa a bb

ba,

a b a

ba,



5qa a bb

ba,

a b a

ba,



5qa a bb

ba,Output:“reject”

a b a

ba,

Input finished


Another Rejection

1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q

λ


1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q

Output:“reject”

λ


Extended Transition Function *δ

QQ →Σ× *:*δ


5qa a bb

ba,

ba,

String rather than a symbol

Thus, its value gives the state the automaton will be in after reading that string.


( ) 20,* qabq =δ

3q 4qa b b a

5qa a bb

ba,

ba,

0q 1q 2q


( ) 40,* qabbaq =δ


5qa a bb

ba,

ba,


( ) 50,* qabbbaaq =δ

1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q


( ) qwq ′=,*δ

Observation: There is a walk from to with label

q q′w

q q′w

q q′kw σσσ L21=

1σ 2σ kσ


( ) 50,* qabbbaaq =δ

1q 2q 3q 4qa b b a

5qa a bb

ba,

ba,

0q

Example: There is a walk from to with label

0qabbbaa

5q


Recursive Definition

( )( ) )),,(*(,*

,*σδδσδ

λδwqwq

qq=

=

q q′σw1q

( )

qq

qwq

′=

′=

),(

,*

1 σδ

σδ

( )

( ) 1

1

,*

),(,*

qwq

qwq

=

=

δ

σδσδ

( ) )),,(*(,* σδδσδ wqwq =

Formally, can be defined recursively by*δ



5qa a bb

ba,

ba,

( )( )

( )( )( )( )( )

( )2

1

0

0

0

0

,,,

,,,*),,(*

,*

qbq

baqbaq

baqabq

=

=

=

=

=

δδδ

λδδδδδ

δ


Languages Accepted by DFAsTake a DFA

Definition:The language contains all input strings accepted by

= { strings that drive to a final state}

M

( )MLM

M( )ML


Example


5qa a bb

ba,

ba,

( ) { }abbaML = M

accept


Another Example


5qa a bb

ba,

ba,

( ) { }abbaabML ,,λ= M

acceptacceptaccept


FormallyFor a DFA

Language accepted by :

( )FqQM ,,,, 0δΣ=

M

( ) ( ){ }FwqwML ∈Σ∈= ,*:* 0δ

0q q′w Fq ∈′


ObservationLanguage rejected by :

( ) ( ){ }FwqwML ∉Σ∈= ,*:* 0δ

M

0q q′w Fq ∉′


More Examples

a

b ba,

ba,

0q 1q 2q

( ) }0:{ ≥= nbaML n

accept trap state

If b is the last symbol of the input, accept.Otherwise, the dfa goes into q2.


( )ML = { all strings with prefix }ab

a b

ba,

0q 1q 2q

accept

ba,3q

ab

Issue: After the first two symbols are read, no further decision:Solve the problem with an automaton with 4 states: initial, two for recognizing ab ending in a final trap state and one nonfinal trap state (1st is not a or 2nd is not b).


( )ML = { all strings without substring }001

λ 0 00 001

1

110

0 1,0

0

Find a dfa that accepts all the strings on {0,1} except those containing 001.

Need to remember if or not the current symbol has been preceded by on or two 0s. -> The state in which two 0s were the immediately preceding symbols can be labeled simply 00.


Regular Languages

A language is regular if there is a DFA such that

LM ( )MLL =

{ all strings with prefix }{ all strings without substring }

Examples of regular languages:

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

ab001


Another ExampleThe language is regular: - No explit way of testing the end of string: Simply putting

the dfa into a final state whenever the second a is encountered. But, if another b is found, it will take the dfa out of the final state.

{ }{ }*,: bawawaL ∈=

a

b

ba,

a

bb a

0q 2q 3q

4q

( )MLL =


IS a regular language?2LThen, find a dfa which accepts

{ }{ }*,:212 bawaaawawL ∈=

- The vertex q3 has to be modified: To recognize thesubstring (the same form), we replicate the states of the first part with q3- The dfa is found in Fig. 2.7.- Thus, if a language is regular, so are K,, 32 LL

Languages and Finite Automatasclab.yonsei.ac.kr/courses/05AM_old/lecture2.pdfFormal Languages and Automata - Yonsei CS 3 Contents For Models of Language Generation • Deterministic

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