applications of regular closure

1

Applications of

Regular Closure

2

The intersection of a context-free language and a regular languageis a context-free language

1L context free

2L regular

21 LL

context-free

Regular Closure

3

An Application of Regular Closure

Prove that: }100:{ nbaL nn

is context-free

4

}{ nnbaWe know:

is context-free

5

}{ 1001001 baL is regular

}{}){( 100100*1 babaL is regular

We also know:

6

}{}){( 100100*1 babaL

regular

}{ nnbacontext-free

1}{ Lba nn is context-free

Lnba

Lbann

nn

}100:{

}{ 1

context-free

(regular closure)

7

Another Application of Regular Closure

Prove that: }:{ cba nnnwL

is not context-free

8

}:{ cba nnnwL

}{*}**{ nnn cbacbaL

context-free regular context-free

If is context-free

Then

Impossible!!!

Therefore, is not context free L

(regular closure)

9

Decidable Propertiesof

Context-Free Languages

10

Membership Question:

for context-free grammarfind if string

G)(GLw

Membership Algorithms: Parsers

• Exhaustive search parser

• CYK parsing algorithm

11

Empty Language Question:

for context-free grammar find if

G)(GL

Algorithm:

S

1. Remove useless variables

2. Check if start variable is useless

12

Infinite Language Question:

for context-free grammar find if is infinite

G)(GL

Algorithm:

1. Remove useless variables

2. Remove unit and productions

3. Create dependency graph for variables

4. If there is a loop in the dependency graph then the language is infinite

13

Example:

cBSC

bbbBB

aaCbA

ABS

|

|

S

A

B

C

Dependency graph Infinite language

14

cBSC

bbbBB

aaCbA

ABS

|

|

acbbSbbbacBSbBaCbBABS

ii bbbSacbb

bbbSacbbacbbSbbbS

)()(

)()( 22

15

The Pumping Lemmafor

Context-Free Languages

16

Take an infinite context-free language

Example:

bB

SbB

aBbA

ABS

Generates an infinite numberof different strings

17abbabbbbabbabbBb

abbaBbBbabbABbabbSb

abbBaBbBABS

bB

SbB

aBbA

ABS

A derivation:

18

S

A B

bBa

b

bS

A B

bBa

b

b

Derivation tree string abbabbbb

19

S

A B

bBa

b

bS

A B

bBa

b

b

repeated

Derivation tree string abbabbbb

20

B

bS

A B

bBa

b

b

aBbbbaBbBb

ABbSbB

bB

21

B

bS

A B

bBa b

aBbbbB

Repeated Part

22

B

bS

A B

ba bB

bS

A B

bBa b

aaBbbbbbbaBbbbB

Another possible derivation

aBbbbB

23

S

A B

bBa

b

bS

A B

ba bB

abbaBbbbS

aBbbbB

24bbabbaaBbbbbabbaBbbbS

S

A B

bBa

b

bS

A B

ba bB

bS

A B

bBa b

aBbbbB

25bbabbaabbbbbbbabbaaBbbbbS

S

A B

bBa

b

bS

A B

ba bB

bS

A B

bBa b bBb

26

bbabbaabbbbbS

Therefore, the string

bbabbaabbbbb

is also generated by the grammar

27

abbaBbbbS

aBbbbB bBWe know:

We also know this string is generated:

abbaabbbb

abbaBbbbS

28

abbaBbbbS

aBbbbB bBWe know:

Therefore, this string is also generated:

bbabbaabbbbb

bbabbaaBbbbb

abbaBbbbS

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abbaBbbbS

aBbbbB bBWe know:


bbbbbbbaabba

bbbbbbBaabba

bbbbbbBaabba

abbaBbbbS

22

22

)()(

)()(

)()(

30


bbbbbbbaabba

bbbbbbBaabba

abbaBbbbS

ii

ii

)()(

)()(

abbaBbbbS

aBbbbB bBWe know:

31

Therefore, knowing that

abbabbbb

is generated by grammar , we also know that

bbbbbbbaabba ii )()(

is generated by

G

G

32

In general:

We are given an infinite context-free grammar G

Assume has no unit-productions no -productions

G

33

Take a string with length bigger than

Some variable must be repeatedin the derivation of

)(GLw

(Number of productions) x(Largest right side of a production)

m >

w

Consequence:

34

S

A

A

x

Last repeated variable

u z

v y

uvxyzw

terminalsof strings :,,,, zyxvu

String

repeated

35

S

A

A

x

u z

v y

uAzS

vAyA

xA

Possiblederivations:

36

uAzS vAyA

xA

We know:

This string is also generated:

uxzuAzS*

zxyuv 00

37


uvxyzuvAyzuAzS**

The original zxyuvw 11

uAzS vAyA

xA

We know:

38


uvvxyyzuvvAyyzuvAyzuAzS***

zxyuv 22

uAzS vAyA

xA

We know:

39


uvvvxyyyzuvvvAyyyz

uvvAyyzuvAyzuAzS**

**

zxyuv 33

uAzS vAyA

xA

We know:

40


yyyzvxyuvvv

yyyzvAyuvvv

uvvvAyyyz

uvvAyyzuvAyzuAzS

zxyuv ii

uAzS vAyA

xA

We know:

*

***

*****

41

Therefore, any string of the form

zxyuv ii

is generated by the grammar G

0i

42

knowing that )(GLuvxyz

we also know that )(GLzxyuv ii

Therefore,

43

S

A

A

x

u z

v y

mvxy ||Observation:

Since is the last repeated variableA

44

S

A

A

x

u z

v y

1|| vyObservation:

Since there are no unit or productions

45

The Pumping Lemma:

there exists an integer such that m

for any string mwLw || ,

we can write

For infinite context-free language L

uvxyzw

with lengths 1||and || vymvxy

and it must be:

0 allfor , iLzxyuv ii

46

Applicationsof

The Pumping Lemma

47

Context-free languages

}0:{ nba nn

Non-context free languages

}0:{ ncba nnn

48

Theorem:The language

}0:{ ncbaL nnn

is not context free

Proof: Use the Pumping Lemmafor context-free languages

49

}0:{ ncbaL nnn

Assume for contradiction that

is context-free

Since is context-free and infinitewe can apply the pumping lemma

L

L

50

Pumping Lemma gives a magic numbersuch that:

m

Pick any string with length Lw mw ||

We pick:mmm cbaw

}0:{ ncbaL nnn

51

}0:{ ncbaL nnn

mmm cbaw

We can write: uvxyzw

with lengths andmvxy || 1|| vy

52

}0:{ ncbaL nnn

Pumping Lemma says:

Lzxyuv ii for all 0i

mmm cbawuvxyzw mvxy || 1|| vy

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}0:{ ncbaL nnn

We examine all the possible locationsof string in vxy w


54

}0:{ ncbaL nnn

Case 1: vxy is withinma

ccccccbbbbbbaaaaaa .........

vxy

m m m

u z


55

}0:{ ncbaL nnn

Case 1: and consist from only v y a


vxy

m m m

u z


56

}0:{ ncbaL nnn

Case 1:

ccccccbbbbbbaaaaaaaaaaaa .........22xyv

km m m

u z

Repeating andv y


1k

57

}0:{ ncbaL nnn

Case 1:

ccccccbbbbbbaaaaaaaaaaaa .........22xyv

km m m

u z

From Pumping Lemma: Lzxyuv 22


1k

58

}0:{ ncbaL nnn

Case 1:

Lcbazxyuv mmkm 22


However:

Contradiction!!!


1k

59

}0:{ ncbaL nnn

Case 2: vxy is withinmb


vxy

m m m

u z


60

}0:{ ncbaL nnn

Case 2:


vxy

m m m

u z

Similar analysis with case 1


61

}0:{ ncbaL nnn

Case 3: vxy is withinmc


vxy

m m m

u z


62

}0:{ ncbaL nnn

Case 3:


vxy

m m m

u z



63

}0:{ ncbaL nnn

Case 4: vxy overlaps and ma


vxy

m m m

u z

mb


64

}0:{ ncbaL nnn

Case 4:


vxy

m m m

u z

Possibility 1: contains onlycontains onlyvy

ab


65

}0:{ ncbaL nnn

ccccccbbbbbbbbbbaaaaaaaaaa .........22xyv

1km 2km m

u z


Case 4:Possibility 1: contains onlycontains onlyvy

ab121 kk

66

}0:{ ncbaL nnn

Case 4:

ccccccbbbbbbbbbbaaaaaaaaaa .........22xyv

1km 2km m

u z



121 kk

67

}0:{ ncbaL nnn

Case 4:From Pumping Lemma: Lzxyuv 22

Lcbazxyuv mkmkm 2122However:

Contradiction!!!


121 kk

68

}0:{ ncbaL nnn

Case 4:


vxy

m m m

u z

Possibility 2: contains andcontains onlyvy

ab


b

69

}0:{ ncbaL nnn

Case 4:

ccccccbbbbbbbbbbbaaaaabbaabaaa .........

22xyv

m km m

u z

Possibility 2: contains andcontains onlyvy

ab


b

1k 2k

121 kkk

70

}0:{ ncbaL nnn



ccccccbbbbbbbbbbbaaaaabbaabaaa .........

22xyv

m km m

u z

1k 2k

121 kkk

71

}0:{ ncbaL nnn



Lcbabazxyuv mkmkkm 2122

However:

Contradiction!!!

121 kkk

72

}0:{ ncbaL nnn

Case 4:


vxy

m m m

u z

Possibility 3: contains onlycontains andvy

ab


a

73

}0:{ ncbaL nnn

Case 4:Possibility 3: contains onlycontains andvy

ab


a

Similar analysis with Possibility 2

74

}0:{ ncbaL nnn

Case 5: vxy overlaps and mb


vxy

m m m

u z

mc


75

}0:{ ncbaL nnn

Case 5:




vxy

m m m

u z

76

There are no other cases to consider

(since , string cannot overlap , and at the same time)

vxymvxy ||

ma mb mc

77

In all cases we obtained a contradiction

Therefore: The original assumption that

}0:{ ncbaL nnn

is context-free must be wrong

Conclusion: is not context-freeL

applications of regular closure

Documents

contextfree grammarfind

infinite contextfree

context freeproof

pumping lemmapumping

productionsthe pumping

possible locationsof

infinite language question

infinite example