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CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10 Mälardalen University 2012. Content The Pumping Lemma for CFL Applications of the Pumping Lemma for CFL Example of Midterm Exam 2 (CFL). The Pumping Lemma for Context-Free Languages.

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CDT314

FABER

Formal Languages, Automata and Models of Computation

Lecture 10

Mälardalen University

2012


ContentThe Pumping Lemma for CFLApplications of the Pumping Lemma for CFLExample of Midterm Exam 2 (CFL)


The pumping lemma for context free languages

The Pumping LemmaforContext-Free Languages

Based on C Busch, RPI, Models of Computation


Take an infinite context-free language.

It generates an infinite number of different strings:

Example:



string

Derivation tree


string

Derivation tree

repeated







Therefore, the string

is also generated by the grammar


We know:

We also know the following string is generated:


We know:

Therefore, the following string is also generated:


We know:

Therefore, the following string is also generated:


We know:

Therefore, the following string is also generated:


Therefore, knowing that

is generated by grammar

We also know that

is generated by


We are given an infinite

context-free grammar .

In general

Assume has no unit-productions

and no -productions.


Take a string

with length bigger than

(Number of productions) x

(Largest right side of a production)

>

Consequence:

Some variable must be repeated

in the derivation of .


string

Last repeated variable

repeated

stringsof terminals


Possible

derivations


We know:

Following string is also generated:


We know:

This string is also generated:

The original


We know:

This string is also generated:


We know:

This string is also generated:


We know:

This string is also generated:


Therefore, any string of the form

is generated by the grammar


Therefore

knowing that

we also know that


(Number of productions) x

(Largest right side of a production)

Observation:

Since is the last repeated variable

A string has length bigger than

>


Observation

Since there are no unit or productions


For infinite context-free language

there exists an integer such that

for any string

we can write

with lengths

and

The Pumping Lemma for CFL


Applications of the pumping lemma for cfl

Applicationsof The Pumping Lemma for CFL


Unrestricted grammarlanguages

Non-regular languages

Context-Free Languages

Regular Languages


Example

Theorem

The language

is not context free.

Proof

Use the Pumping Lemma

for context-free languages.


Assume thecontrary, that

is context-free.

Since is context-free and infinite

we can apply the pumping lemma.


Pumping Lemma gives a number

such that:

for any string with length

We can choose e.g.


We can write:

with lengths and



We examine all the possible locations

of string in


Case 1:

is within


Case 1:

and consist from only


Case 1:

Repeating and


Case 1:

From Pumping Lemma:


Case 1:

From Pumping Lemma:

However:

Contradiction!


Case 2:

is within


Case 2: Similar analysis to case 1


Case 3:

is within


Case 3: Similar analysis to case 1


Case 4:

overlaps and


Case 4:

contains only

Possibility 1:

contains only


Case 4:

Possibility 1:

contains only

contains only


Case 4:

From Pumping Lemma:


Case 4:

From Pumping Lemma:

However:

Contradiction!


Case 4:

Possibility 2:

contains and

contains only


Case 4:

Possibility 2:

contains and

contains only


Case 4:

From Pumping Lemma:


Case 4:

From Pumping Lemma:

However:

Contradiction!


Case 4:

Possibility 3:

contains only

contains and


Case 4: Possibility 3: contains only

contains and

Similar analysis with Possibility 2


Case 5:

overlaps and


Case 5: Similar analysis to case 4


(Since , string cannot

overlap , and at the same time)

There are no other cases to consider.


In all cases we obtain a contradiction.

Therefore:

The original assumption that

is context-free must be wrong.

Conclusion:

is not context-free.

END OF PROOF


Comparison between the pumping lemma for context free and regular languages

Comparison between the Pumping Lemma for Context Free and Regular Languages


What is the difference between context free languages and regular languages

......

......

What is the difference between Context Free Languages and Regular Languages?

In regular languages a single symbol/substring in the string w can be “pumped”.


The difference between context free languages and regular languages
The difference between Context Free Languages and Regular Languages

In CFL’s multiple symbols/substrings in the string w can be “pumped”.

Consider the language {an bn | n > 0}

No single symbol can be pumped and allow us to stay in the language.

However, there do exist pairs of symbols which can be pumped resulting in strings which stay in the language.

Thus a CFL pumping lemma applies.


CFG productions Languages


strings Languagesof terminals

String

Last repeated variable

repeated


Pumping conditions for rl and cfl

A language LanguagesL satisfies the RLpumping condition if:

there exists an integer m > 0 such that

for all strings w in L of length at least m

there exist strings x, y, zsuch that

w = xyzand

|xy| ≤ mand

|y| ≥ 1and

for all i ≥ 0, xyizis in L

|xy| ≤ m is in the beginning of the w and can be pumped within m symbols.

A language L satisfies theCFLpumping condition if:

there exists an integer m > 0such that

for all strings w in L of length at least m

there exist strings u, v, x, y, z such that

w = uvxyzand

|vxy| ≤ mand

|vy| ≥ 1and

for all i ≥ 0, uvixyiz is in L

|vxy| ≤ mand u comes first in the w and it can be arbitrarily long.

Pumping Conditions for RL and CFL


Pumping lemma

CFL’s Languages

“Pumping Languages”

All languages over {a,b}

Pumping Lemma

All CFL’s satisfy the CFL pumping condition

But some languages that satisfy CFL pumping condition are not CFL!


Implications
Implications Languages

CFL’s

“Pumping Languages”

All languages over {a,b}

We can use the pumping lemma to prove a language Lis not a CFL.

Show that L does not satisfy the CFL pumping condition.

We cannot use the pumping lemma to prove a language is CFL.

Showing L satisfies the pumping condition does not guarantee that L is context-free.


Pumping lemma1

Pumping Lemma Languages

What does it mean?


Pumping condition
Pumping Condition Languages

  • A language L satisfies the CFL pumping condition if:

    • there exists an integer m > 0such that

    • for all strings w in L of length at least m

    • there exist strings u, v, x, y, zsuch that

      • w = uvxyzand

      • |vxy| ≤ m and

      • |vy| ≥ 1 and

        Then for all i ≥ 0, uviwyizis in L


V and y can be pumped
v Languages and y can be pumped

1) w in L2) w = uvxyz3) for all i ≥ 0, uvixyiz is in L

  • Let w = abcdefgbe in L

  • Then there exist substrings v and y in w such that v and y can be repeated (pumped) and the resulting string is still in L:

    uvixyiz is in L for all i ≥ 0


For example Languagesw =abcdefg

v = cdand y = f

uv0xy0z = uxz =abegis in L

uv1xy1z = uvxyz = abcdefgis in L

uv2xy2z = uvvxyyz = abcdcdeffgis in L

uv3xy3z = uvvvxyyyz = abcdcdcdefffg is in L


What the other parts mean
What the other parts mean Languages

A language L satisfies the CFL pumping condition if:

there exists an integer m > 0 such that

for all strings w in L of length at least m

w must be in L


What the other parts mean1
What the other parts mean Languages

There exist strings u, v, x, y, z such that

  • w = uvxyz and

  • |vxy| ≤ m and

    v and y are contained within m characters of w

    Note: these are NOT necessarily the first m characters of w

    |vy| ≥ 1

    (v and y cannot both be , one of them might be , but not both)

  • For all i ≥ 0, uvixyiz is in L


Applying cfl pumping lemma to prove that a language l is not context free

APPLYING CFL PUMPING LEMMA Languages TO PROVE THAT A LANGUAGE LIS NOT CONTEXT-FREE


How we use the pumping lemma
How we use the Pumping Lemma Languages

  • We choose a specific language L.

  • We show that L does not satisfy the pumping condition.

  • We conclude that L is not context-free.


Showing that l does not pump

A language L Languagessatisfies the CFL pumping condition if:

there exists an integer m > 0 such that

for all strings w in L of length at least m

there exist strings u, v, x, y, z such that

w = uvxyz and

|vxy| ≤ m and

|vy| ≥ 1 and

for all i ≥ 0, uvixyiz is in L

A language L does not satisfy the CFL pumping condition if:

for all integers m of sufficient size

there exists a string w in L of length at least m such that

for all strings u, v, x, y, z where

w = uvxyz and

|vxy| ≤ m and

|vy| ≥ 1

there exists a i ≥ 0 such that uvixyiz is not in L

Showing that L “does not pump”


Unrestricted grammar Languageslanguages

Non-regular languages

Context-Free Languages

Regular Languages



For infinite context-free language Languages

there exists an integer such that

for any string

we can write

with lengths

then

The Pumping Lemma for CFL

and


Example Languages

Theorem

The language

is not context free.

Proof

Use the Pumping Lemma

for context-free languages.


Assume the Languagescontrary - that

is context-free.

Since is context-free and infinite

we can apply the pumping lemma.


Pumping Lemma gives a number Languages

such that we can

pick any string of

with length at least

So we pick:


and Languages

with lengths

We can write:

Pumping Lemma says:

for all


We examine Languagesall the possible locations

of string in


Case 1: Languages

is within the first


Case 1: Languages

is within the first


Case 1: Languages

is within the first


Case 1: Languages

is within the first

However, from Pumping Lemma:

Contradiction!


Case 2: Languages

is in the first

is in the first


Case 2: Languages

is in the first

is in the first


Case 2: Languages

is in the first

is in the first


Case 2: Languages

is in the first

is in the first

However, from Pumping Lemma:

Contradiction!


Case 3: Languages

overlaps the first

is in the first


Case 3: Languages

overlaps the first

is in the first


Case 3: Languages

overlaps the first

is in the first


Case 3: Languages

overlaps the first

is in the first

However, from Pumping Lemma:

Contradiction!


Case 4: Languages

in the first

overlaps the first

Analysis is similar to case 3


or Languages

or

Other cases:

is within

Analysis is similar to case 1:


More cases: Languages

overlaps

or

Analysis is similar to cases 2,3,4:


There are no other cases to consider. Languages

Since , it is impossible for

to overlap:

neither

nor

nor


is not context-free. Languages

In all cases we obtain a contradiction.

Therefore:

The original assumption that

is context-free must be wrong.

Conclusion:

END OF PROOF


Unrestricted grammar Languageslanguages

Non-regular languages

Context-Free Languages

Regular Languages


Example Languages

Theorem

The language

is not context free.

Proof

Use the Pumping Lemma

for context-free languages.


Since is context-free and infinite Languages

we can apply the pumping lemma.

Assume to the contrary that

is context-free.


Pumping Lemma gives a number Languages

such that we can:

pick any string of with length at least

so we pick:


We can write: Languages

with lengths

and

Pumping Lemma says:

for all


We examine Languagesall the possible locations

of the string in

There is only one case to consider.


Since Languages

for

we have:


However, from Pumping Lemma: Languages

Contradiction!


We obtained a Languagescontradiction

Therefore:

The original assumption that

is context-free must be wrong

Conclusion:

is not context-free

END OF PROOF


Unrestricted Languagesgrammarlanguages

Context-free languages

Regular Languages


Example Languages

Theorem

The language

is not context free

Proof

Use the Pumping Lemma

for Context-free languages


Since is context-free and infinite Languages

we can apply the pumping lemma.

Assume to the contrary that

is context-free.


Pumping Lemma gives a number Languages

such that we can:

pick any string of with length at least

so we choose:


We can write: Languages

with lengths and

Pumping Lemma says:

for all


We examine Languagesall the possible locations

of string in


The most complicated case: Languages

is in

is in


and Languages

The most complicated sub-case:


and Languages

The most complicated sub-case:


and Languages

The most complicated sub-case:


and Languages


However, from Pumping Lemma: Languages

Contradiction!


When we examine the rest of the cases Languages

we also obtain a contradiction.


is not context-free. Languages

Conclusion:

In all cases we obtain a contradiction.

Therefore:

The original assumption that

is context-free must be wrong.

END OF PROOF


Check your knowledge before the midterm 2 selected examples of context free language problems

Check your knowledge before the Midterm 2! LanguagesSelected ExamplesofContext Free Language Problems


Problem 1 find a cfg for the following language
Problem 1. Languages Find a CFG for the following language

Solution 1.

Let G be the grammar with productions:

Claim: L(G) = L


Find a cfg for the following language
Find a CFG for the following language: Languages

Proof:

Consider the following derivation:

(the first * applies S  aSc n times, the second * to B  bBc m times)

Since all words in L(G) must follow this pattern in their derivations, it is clear that L(G)  L


Find a cfg for the following language1
Find a CFG for the following language Languages

Considerw  L, w = anbmc(n + m)for some n, m  0

The derivation

S * anScn anBcn * anbmBcmcn anbmc(n + m)

clearly produces w for any n, m.

 L  L(G)

 L  L(G)

G is a CFG for L

END OF PROOF


Find a pda and cfg for the following language

Problem 2. Languages

Find a PDA and CFG for the following language

Solution 2.

Is the automaton deterministic? Yes.

It acts in a unique way in each state, no l-transitions.


CFG : Languages


Find a pda and cfg for the following language1

Problem 3. Languages

Find a PDA and CFG for the following language

Solution 3.

PDA


CFG : Languages


Problem 4. Languages

Prove that the language L is context-free

Consider the following two languages:

L1 ={w : w is made from a’s and b’s

and the length of w is a multiple of 10}

L2 = {an bn: n  0}

Solution 4.


L Languages1 ={w : w is made from a’s and b’s and the length of w is a multiple of ten}

L2 = {an bn: n  0}

LetL1cdenote the complement of L1.

We have that: L = L1c  L2.

L1is a regular language, since we can easily build a finite automaton with 10 states that accepts any string in this language.

L1cis regular too, since regular languages are closed under complement.


The language LanguagesL2is context-free.

The grammar is: S  aSb | 

Therefore, the language L = L1c  L2is also context-free,

since context-free languages are closed under regular intersection (Regular Closure).

END O PROOF


Find a pda and cfg for the following language2

Problem 5. Languages

Find a PDA and CFG for the following language

Solution 5.

CFG

Production ex.



PDA Languages


Find a pda and cfg for the following language3

Problem 6. Languages

Find a PDA and CFG for the following language:

Solution 6.

PDA


CFG, Languagesdirect construction

Strings start and finish with different symbols

Strings contain at least one more a than b

(we must have AA here as only one A just balances b)


Further Reading Languages

Famous Pushdown Automata Examples

http://www.liacs.nl/~hoogeboo/praatjes/tarragona/schoolpda-VIII.pdf

Computational and evolutionary aspects of language

http://www.nature.com/nature/journal/v417/n6889/full/nature00771.html


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