language recognition (12.4) longin jan latecki temple university
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Language Recognition (12.4) Longin Jan Latecki Temple University. Based on slides by Costas Busch from the course http://www.cs.rpi.edu/courses/spring05/modcomp/ and …. Three Equivalent Representations. Regular expressions. Each can describe the others. Regular languages. - PowerPoint PPT PresentationTRANSCRIPT
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Language Recognition (12.4) Longin Jan LateckiTemple University
Based on slides by Costas Busch from the coursehttp://www.cs.rpi.edu/courses/spring05/modcomp/and …
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Three Equivalent Representations
Finite automata
Regularexpressions
Regular languages
Each can
describethe others
Kleene’s Theorem: For every regular expression, there is a deterministic finite-state automaton that defines the same language, and vice versa.
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EXAMPLE 1
Consider the language { ambn | m, n N}, which is represented by the regular expression a*b*.
A regular grammar for this language can
be written as follows:
S | aS | B B b | bB.
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Regular Expression
Regular Grammar
a* S | aS(a+b)* S | aS | bSa* + b* S | A | B
A a | aAB b | bB
a*b S b | aSba* S bA
A | aA(ab)* S | abS
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NFAs Regular grammarsThus, the language recognized by FSA
is a regular language Every NFA can be converted into a corresponding regular grammar and vice versa.Each symbol A of the grammar is associated with a non-terminal node of the NFA sA, in particular, start symbol
S is associated with the start state sS.
Every transition is associated with a grammar production: T(sA,a) = sB A aB.
Every production B is associated with final state sB.
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Equivalent FSA and regular grammar, Ex. 4, p. 772.
G=(V,T,S,P)V={S, A, B, 0, 1} withS=s0, A=s1, and B=s2,
T={0,1}, and productions areS 0A | 1B | 1 | λA 0A | 1B | 1 B 0A | 1B | 1 | λ
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Kleene’s Theorem
LanguagesGenerated byRegular Expressions
LanguagesRecognizedby FSA
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LanguagesGenerated byRegular Expressions
LanguagesRecognizedby FSA
LanguagesGenerated byRegular Expressions
LanguagesRecognizedby FSA
We will show:
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Proof - Part 1
r)(rL
For any regular expression the language is recognized by FSA (= is a regular language)
LanguagesGenerated byRegular Expressions
LanguagesRecognizedby FSA
Proof by induction on the size of r
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Induction BasisPrimitive Regular Expressions: a,,
NFAs
)()( 1 LML
)(}{)( 2 LML
)(}{)( 3 aLaML
regularlanguages
a
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Inductive Hypothesis
Assume for regular expressions andthat and are regular languages
1r 2r
)( 1rL )( 2rL
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Inductive StepWe will prove:
1
1
21
21
*
rL
rL
rrL
rrL
Are regular Languages
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By definition of regular expressions:
11
11
2121
2121
**
rLrL
rLrL
rLrLrrL
rLrLrrL
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)( 1rL )( 2rLBy inductive hypothesis we know: and are regular languages
Regular languages are closed under: *1
21
21
rLrLrL
rLrL Union Concatenation
Star
We need to show:
This fact is illustrated in Fig. 2 on p. 821.
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Therefore:
** 11
2121
2121
rLrL
rLrLrrL
rLrLrrL
Are regularlanguages
And trivially: ))(( 1rL is a regular language
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Proof - Part 2
LanguagesGenerated byRegular Expressions
LanguagesRecognizedby FSA
Lr LrL )(
For any regular language there is a regular expression with
Proof by construction of regular expression
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Since is regular take the NFA that accepts it
LM
LML )(
Single final state
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From construct the equivalentGeneralized Transition Graph in which transition labels are regular
expressions
M
Example:
a
ba,
cM
a
ba
c
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Another Example:
ba a
b
b0q 1q 2q
ba,a
b
b0q 1q 2q
b
b
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Reducing the states:ba
ab
b0q 1q 2q
b
0q 2q
babb*
)(* babb
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Resulting Regular Expression:
0q 2q
babb*
)(* babb
*)(**)*( bbabbabbr
LMLrL )()(
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In GeneralRemoving states:
iq q jqa b
cde
iq jq
dae* bce*dce*
bae*
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The final transition graph:
0q fq
1r
2r
3r4r
*)*(* 213421 rrrrrrr
LMLrL )()(
The resulting regular expression:
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Three Equivalent Representations
Finite automata
Regularexpressions
Regular languages
Each can
describethe others