MELJUN CORTES Automata Lecture Non-context-free Languages 2

8/21/2019 MELJUN CORTES Automata Lecture Non-context-free Languages 2

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Theory of Computation (With Automata Theory)

* Property of STI

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Non-Context-Free Languages

NON-CONTEXT-FREE

LANGUAGES


Introduction

• Recall the following:

A regular language has either

a nondeterministic automatonthat recognizes it or a regular

expression that can generate

its strings.

A context-free language haseither a pushdown automaton

that recognizes it or a context-

free grammar that can

generate its strings.


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• Questions:

1. Are there non-context-free

languages?

2. How can a language beproven to be non-context-

free?

• The language L= {an

bn

c n

n ≥1} is non-context-free since

there is no PDA that can

recognize it.

• To prove that a language is

non-context-free, it must be

shown that there is no PDA or

CFG for that language (difficult

to do).


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• Recall that the pumpinglemma was used to prove that

a language is not regular.

The pumping lemma for

regular languages exploited aproperty of regular languages

to prove the non-regularity of

languages.

Strings of regular languageshave the property that if they

are long enough, then they

have a substring that can be

repeated an arbitrary number

of times and the resultingstrings will still be in language.

Languages which have strings

that do not have this property

are considered not regular.


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• To prove that a language is

non-context-free, there mustbe a pumping lemma for

context-free languages.

This pumping lemma should

also exploit a certain property

of context-free languages.

Languages whose strings do

not have this property areconsidered non-context-free.


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Context-Free Grammars Revisited

• Assume the following context-

free grammar G for language

L:

S → A1

A → 0A1 ε

• The leftmost derivation for the

string 0001111 is

S → A1

→ 0A11

→ 00A111

→ 000A1111

→ 0001111


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• The parse tree for the leftmostderivation for the string

0001111 is

A 10

ε

A 10

A 10

A 1

S


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• Notice that because the string

being derived is long, some

variables were used more than

once.

• In the pumping lemma for

regular languages, strings

whose length is greater than or

equal to the number of DFA

states (the pumping length)have a substring that can be

repeated an arbitrary number

of times, and the resulting

strings will still be in the

language.

• For the pumping lemma for

context-free languages, what

is then the pumping length?


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Deriving the Pumping Length

• Assume that for a certain

parse tree, the non-leaf nodes

have a maximum number of

children equal to c (this implies

that the right-hand side of the

rules of the grammar has a

string whose length is ≤ c ).

Assume also that in the

succeeding discussions c ≥ 2.

• Assume further that the height

of the parse tree for the

derivation of a certain string w

is h.

• A tree with height h can have a

maximum number of leaf

nodes equal to c h.


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• Since the leaf nodes of the

parse tree represent the

symbols of the string w being

derived, then this means that

w ≤ c h

• Assume now that the length of

a certain string w is

w ≥ c v + 1

where v is the number of

variables in the grammar.

This implies that the height of

the parse tree will be

h ≥ v + 1


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• If h ≥ v + 1, then there is a

path from the root node to a

leaf node that is greater than

or equal to v + 1.

This path will have at least v +

2 nodes. If the last node is a

terminal, then the remainingnodes are variables.

Since the derivation used

more than v + 1 variables, then

at least one variable was

repeated (by the pigeonhole

principle).


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• This is illustrated below:

S uAz uvAyz uvxyz

Assume that this parse tree

represents the shortest

derivation for the string S.

S

A. . . . . .

u z

A. . . . . .

v y

x


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• If the derivation A vAy is

repeated before the derivation

A x , then the parse tree

would look like:

S uAz uvAyz uvvAyyz

uvvxyyz uv 2 xy 2z

S

A. . . . . .

u z

A. . . . . .

v y

A. . . . . .

v y

x


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• If the derivation A vAy isrepeated several times, then

the string derived will look like

uv i xy i z

and this string would still

belong to the language

represented by the grammar.

• In the pumping lemma for regular languages, the input

string w can then be written as

w = xy i z

where the middle substring y

can be pumped several times

and the resulting string will still

belong to the language

concerned.


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context-free languages, theinput string w can then be

written as

w = uv i xy i z

where the 2nd and 4th

substrings v and y can be

pumped several times and the

resulting string will still belongto the language concerned.


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• In the pumping lemma for regular languages, the pumping

length p refers to the number of

states of the DFA recognizing

the language.

Hence, w ≥ p for the pumping

lemma to be used.


context-free languages, thepumping length p is equal to the

term c v + 1 where c is the length

of the longest right-hand side

string of any rule in the

grammar and v is the number of variables of the grammar.

Similarly, the length of the string

w being analyzed must be

greater than c v + 1.


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The Pumping Lemma for Context-

Free Languages

• If L is a context-free language,

then there is a number p (the

pump ing leng th ) where, if w

is any string in L whose length

is greater than or equal to p,

then w may be divided into five

parts, w = uvxyz , satisfying the

following conditions:

1. for each i ≥ 0, uv i xy iz

L,

2. vy ≥ 1, and

3. vxy p.


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• Major points:

1. The string w = uv i xy iz L

for each i ≥ 0.

This condition simply

states that strings

belonging to language L

that are longer than thepumping length can be

divided into 5 parts u, v , x ,

y , and z .

The 2nd and 4th parts can

be repeated several times

and the resulting string will

still belong to language L.


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2. vy ≥ 1 refers to the fact

that substrings v and y

cannot both be empty

strings.

If v = y = , then the

derivation A vAy

becomes A A. This is a

useless derivation and

therefore can be removed.

This, however, violates the

assumption that the parse

tree presented depicts theshortest derivation for the

string S.

Therefore vy ≥ 1.


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3. vxy p

Consider again the

following parse tree:

If w ≥ p, then the height

of the tree h ≥ v + 1.

S

A. . . . . .

u z

A. . . . . .

v y

x

h ≥ v + 1


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Consider now the following

subtree (rooted at the

upper variable A) of the

given parse tree:

A

A. . . . . .

v y

x

h ≤ v + 1

root of

the

subtree


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Take note that no other

variable is repeated below

the root node of this

subtree.

It follows that this subtree

has a height:

h ≤ v + 1

The string formed by this

subtree then has a length:

vxy c v + 1

vxy p


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• As in the pumping lemma for regular languages, proving a

language is non-context-free

will use proof by contradiction:

1. First, assume that thelanguage L is context-free.

2. Since L is context-free, the

pumping lemma states that

all strings of L that are long

enough can be pumped.

3. Find a string w that is long

enough that cannot be

pumped.

4. If there is such a string,

then the pumping lemma

was contradicted. Hence,

language L is non-context-

free.


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• Example 1: Prove that the

language L1 = {anbnc n n ≥ 1}

is not context-free.

Assume first that L1 is context-free.

Let p be the pumping length of

L1.

Let the string to be pumped be

w = a pb pc p.

Since w ≥ p, the string chosen

is long enough to be pumped.

Consider now the various

ways in which w can be

divided into the u, v , x , y , and z

substrings.


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Option 1: Let vxy be all a’s.

Pumping v and y would

create a string where there

are more a’s than b’s and

c ’s. Hence, the resulting

string will not be in L1.

This violates condition 1 of

the pumping lemma.

Option 2: Let vxy be somea’s followed by some b’s.



are more a’s and b’s thanc ’s. Hence, the resulting



the pumping lemma.


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Option 3: Let vxy be all b’s.



are more b’s than a’s and




the pumping lemma.

Option 4: Let vxy be someb’s followed by some c ’s.



are more b’s and c ’s thana’s. Hence, the resulting



the pumping lemma.


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Option 5: Let vxy be all c ’s.


create a string where thereare more c ’s than a’s and

b’s. Hence, the resulting



the pumping lemma.

Since L1 has a string that

cannot be pumped, then this

contradicts the original

assumption that L1 is context-

free. Therefore, L1 is non-

context-free.


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language L2 = {ww w {0,

1}*} is not context-free.



L2.


w = 0 p1 p0 p1 p.






substrings.


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Option 1: Let vxy be all 0s

from the first or second 0 p of the string.


create a string where the

first half is not equal to thesecond half. Hence, the

resulting string will not be

in L2. This violates

condition 1 of the pumping

lemma.

Option 2: Let vxy be all 1s

from the first or second 1 p of

the string.


create a string where the

first half is not equal to the

second half. Hence, the


in L2.


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Option 3: Let vxy be some 0s

followed by some 1s from the

first or second half of the

string.


create a string where thefirst half is not equal to the



in L2. This violates

condition 1 of the pumpinglemma.


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Option 4: Let vxy be some 1sfollowed by some 0s

somewhere in the middle

portion of the string.

Pumping v and y wouldcreate a string where the

first half is not equal to the



in L2. This violatescondition 1 of the pumping

lemma.



contradicts the original

assumption that L2 is context-


context-free.


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language L3 = {ai b j c k k ≥ j ≥ i ≥

0} is not context-free.



L3.


w = a pb pc p.






substrings.


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Option 1: Let vxy be all a’s.



are more a’s than b’s and




the pumping lemma.

Option 2: Let vxy be somea’s followed by some b’s.



are more a’s and b’s thanc ’s. Hence, the resulting



the pumping lemma.


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Option 3: Let vxy be all b’s.



are more b’s than c ’s.

Hence, the resulting stringwill not be in L3. This

violates condition 1 of the

pumping lemma.

Option 4: Let vxy be someb’s followed by some c ’s.

Pumping down v and y

would create a string

where there are more a’s

than b’s and c ’s. Hence,

the resulting string will not

be in L3. This violates


lemma.


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Option 5: Let vxy be all c ’s.

Pumping down v and y

would create a string

where there are more a’s

and b’s than c ’s. Hence,

the resulting string will not

be in L3. This violates


lemma.



contradicts the originalassumption that L3 is context-


context-free.

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MELJUN CORTES Automata Lecture Non-context-free Languages 2