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

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

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

### The Pumping LemmaforContext-Free Languages

### Applicationsof The Pumping Lemma for CFL

### Comparison between the Pumping Lemma for Context Free and Regular Languages

### Pumping Lemma Languages

### More Pumping Lemma Applications Languages

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

FABER

Formal Languages, Automata and Models of Computation

Lecture 10

Mälardalen University

2012

Based on C Busch, RPI, Models of Computation

Take an infinite context-free language.

It generates an infinite number of different strings:

Example:

Derivation tree

is also generated by the grammar

We also know the following string is generated:

Therefore, the following string is also generated:

Therefore, the following string is also generated:

Therefore, the following string is also generated:

context-free grammar .

In general

Assume has no unit-productions

and no -productions.

with length bigger than

(Number of productions) x

(Largest right side of a production)

>

Consequence:

Some variable must be repeated

in the derivation of .

derivations

Following string is also generated:

This string is also generated:

This string is also generated:

This string is also generated:

Therefore, any string of the form

is generated by the grammar

(Largest right side of a production)

Observation:

Since is the last repeated variable

A string has length bigger than

>

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

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.

with lengths and

for all

We examine all the possible locations

of string in

is within

and consist from only

Repeating and

From Pumping Lemma:

is within

Case 2: Similar analysis to case 1

is within

Case 3: Similar analysis to case 1

overlaps and

From Pumping Lemma:

From Pumping Lemma:

overlaps and

Case 5: Similar analysis to case 4

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

......

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

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

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 CFLCFL’s Languages

“Pumping Languages”

All languages over {a,b}

Pumping LemmaAll CFL’s satisfy the CFL pumping condition

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

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.

What does it mean?

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 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 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 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 Languages TO PROVE THAT A LANGUAGE LIS NOT CONTEXT-FREE

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.

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:

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

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:

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

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 examine Languagesall the possible locations

of string 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

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: 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 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

Problem 2. Languages

Find a PDA and CFG for the following languageSolution 2.

Is the automaton deterministic? Yes.

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

CFG : Languages

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

PDA Languages

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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