CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10

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CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10

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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 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:

A derivation

string

Derivation tree

string

Derivation tree

repeated

Repeated part

Another possible derivation

derivation

derivation

derivation

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

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

Pumping Lemma says:

for all

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

......

......

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

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

stringsof terminals

String

Last repeated variable

repeated

A language L 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.

CFL’s

“Pumping Languages”

All languages over {a,b}

All CFL’s satisfy the CFL pumping condition

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

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 Lemma

What does it mean?

- 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

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 w =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

…

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

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 LIS NOT CONTEXT-FREE

- We choose a specific language L.
- We show that L does not satisfy the pumping condition.
- We conclude that L is not context-free.

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

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

Unrestricted grammarlanguages

Non-regular languages

Context-Free Languages

Regular Languages

More Pumping Lemma Applications

For infinite context-free language

there exists an integer such that

for any string

we can write

with lengths

then

The Pumping Lemma for CFL

and

Example

Theorem

The language

is not context free.

Proof

Use the Pumping Lemma

for context-free languages.

Assume the contrary - that

is context-free.

Since is context-free and infinite

we can apply the pumping lemma.

Pumping Lemma gives a number

such that we can

pick any string of

with length at least

So we pick:

and

with lengths

We can write:

Pumping Lemma says:

for all

We examine all the possible locations

of string in

Case 1:

is within the first

Case 1:

is within the first

Case 1:

is within the first

Case 1:

is within the first

However, from Pumping Lemma:

Contradiction!

Case 2:

is in the first

is in the first

Case 2:

is in the first

is in the first

Case 2:

is in the first

is in the first

Case 2:

is in the first

is in the first

However, from Pumping Lemma:

Contradiction!

Case 3:

overlaps the first

is in the first

Case 3:

overlaps the first

is in the first

Case 3:

overlaps the first

is in the first

Case 3:

overlaps the first

is in the first

However, from Pumping Lemma:

Contradiction!

Case 4:

in the first

overlaps the first

Analysis is similar to case 3

or

or

Other cases:

is within

Analysis is similar to case 1:

More cases:

overlaps

or

Analysis is similar to cases 2,3,4:

There are no other cases to consider.

Since , it is impossible for

to overlap:

neither

nor

nor

is not context-free.

In all cases we obtain a contradiction.

Therefore:

The original assumption that

is context-free must be wrong.

Conclusion:

END OF PROOF

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.

Since is context-free and infinite

we can apply the pumping lemma.

Assume to the contrary that

is context-free.

Pumping Lemma gives a number

such that we can:

pick any string of with length at least

so we pick:

We can write:

with lengths

and

Pumping Lemma says:

for all

We examine all the possible locations

of the string in

There is only one case to consider.

Since

for

we have:

However, from Pumping Lemma:

Contradiction!

We obtained a contradiction

Therefore:

The original assumption that

is context-free must be wrong

Conclusion:

is not context-free

END OF PROOF

Unrestrictedgrammarlanguages

Context-free languages

Regular Languages

Example

Theorem

The language

is not context free

Proof

Use the Pumping Lemma

for Context-free languages

Since is context-free and infinite

we can apply the pumping lemma.

Assume to the contrary that

is context-free.

Pumping Lemma gives a number

such that we can:

pick any string of with length at least

so we choose:

We can write:

with lengths and

Pumping Lemma says:

for all

We examine all the possible locations

of string in

The most complicated case:

is in

is in

and

The most complicated sub-case:

and

The most complicated sub-case:

and

The most complicated sub-case:

and

However, from Pumping Lemma:

Contradiction!

When we examine the rest of the cases

we also obtain a contradiction.

is not context-free.

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 ExamplesofContext Free Language Problems

Solution 1.

Let G be the grammar with productions:

Claim: L(G) = L

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

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

Problem 2.

Solution 2.

Is the automaton deterministic? Yes.

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

CFG :

Problem 3.

Solution 3.

PDA

CFG :

Problem 4.

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.

L1 ={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 L2is 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

Problem 5.

Solution 5.

CFG

Production ex.

Remember converting grammar to NPDA, example:

Grammar

NPDA

PDA

Problem 6.

Solution 6.

PDA

CFG, direct 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

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