CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 10

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CD5560

FABER

Formal Languages, Automata and Models of Computation

Lecture 10

Mälardalen University

2010

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Content

The Pumping Lemma for CFLApplications of the Pumping Lemma for CFL

Midterm Exam 2: Context-Free Languages

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Pumping Lemma for CFL’s

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Comparison to Regular Language Pumping Lemma/Condition

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What’s Difference between CFL’s and Regular Languages?

In regular languages, a single substring “pumps”

...... ......q

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What’s Difference between CFL’s and Regular Languages?

In CFL’s, multiple substrings can be “pumped”– Consider the language {anbn | n > 0}

– No single substring can be pumped and allow us to stay in the language

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

Thus, a modified pumping lemma applies.

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Pumping Conditions for RL and CFL

A language L satisfies the RL pumping condition if: – there exists an integer m > 0

such that– for all strings x in L of length at

least m– there exist strings u, v, w such

that• x = uvw and• |uv| ≤ m and• |v| ≥ 1 and• For all i ≥ 0, uviw is in L

A language L satisfies the CFL pumping condition if: – there exists an integer m > 0

such that– for all strings x in L of length at

least m– there exist strings u, v, w, y, z

such that• x = uvwyz and• |vwy| ≤ m and• |vy| ≥ 1 and• For all i ≥ 0, uviwyiz is in L

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Pumping Lemma

All CFL’s satisfy the CFL pumping condition

All languages over {a,b}

“Pumping Languages”

CFL’s

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Implications

• We can use the pumping lemma to prove a language L is not a CFL– Show 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

All languages over {a,b}

“Pumping Languages”

CFL’s

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Pumping Lemma

What does it mean?

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Pumping Condition

• A language L satisfies the CFL pumping condition if: – there exists an integer m > 0 such that– for all strings x in L of length at least m– there exist strings u, v, w, y, z such that

• x = uvwyz and• |vwy| ≤ m and• |vy| ≥ 1 and• For all i ≥ 0, uviwyiz is in L

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v and y can be pumped

• Let x = abcdefg be in L• Then there exist substrings v and y in x such

that v and y can be repeated (pumped) and the resulting string is still in L– uviwyiz is in L for all i ≥ 0

1) x in L2) x = uvwyz3) For all i ≥ 0, uviwyiz is in L

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For example x = abcdefg

v = cd and y = f

uv0wy0z = uwz = abeg is in L

uv1wy1z = uvwyz = abcdefg is in L

uv2wy2z = uvvwyyz = abcdcdeffg is in L

uv3wy3z = uvvvwyyyz = abcdcdcdefffg is in L

…

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What the other parts mean

A language L satisfies the CFL pumping condition if: – there exists an integer m > 0 such that– for all strings x in L of length at least m

• x must be in L and have sufficient length

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What the other parts mean

– there exist strings u, v, w, y, z such that• x = uvwyz and• |vwy| ≤ m and

–v and y are contained within m characters of x–Note: these are NOT necessarily the first m

characters of x• |vy| ≥ 1 and

–v and y cannot both be l,–One of them might be l, but not both

• For all i ≥ 0, uviwyiz is in L

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Pumping Lemma

Applying it to prove a specific language L is not context-free

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How we use the Pumping Lemma

• We choose a specific language L

For example {aj bj cj | j > 0}

• We show that L does not satisfy the pumping condition

• We conclude that L is not context-free

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Showing L “does not pump”

A language L satisfies the CFL pumping condition if: – there exists an integer n >

0 such that– for all strings x in L of

length at least m– there exist strings u, v, w,

y, z such that• x = uvwyz and• |vwy| ≤ m and• |vy| ≥ 1 and• For all i ≥ 0, uviwyiz 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 x in L of

length at least m such that– for all strings u, v, w, y, z

such that• x = uvwyz and• |vwy| ≤ m and• |vy| ≥ 1

– There exists a i ≥ 0 such that uviwyiz is not in L

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Example Languages

• TWOCOPIES = {ww | w is in {a,b}* }– abbabb is in TWOCOPIES but abaabb is not

• EQUAL3 = {the set of strings over {a, b, c} such that the number of a’s equals the number of b’s equals the number of c’s}

• {ai bj ck | i < j < k}

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More Pumping Lemma Applications

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The Pumping Lemma for CFL

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 0 allfor , iLzxyuv ii

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The Pumping Lemma for CFL

There exists an integer such that m

mwGLw || ),(

can be written

Let be a context free grammar. G

uvxyzw

with lengths 1||and || vymvxy

and 0 allfor , iLzxyuv ii

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Regular Languages

}{ nnba }{ RwwContext-Free Languages

Non-regular languages

Unrestricted grammar languages

}0:{ ncba nnn

**ba

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Theorem The language

}*},{:{ bawwwL

is not context free

Proof Use the Pumping Lemma

for context-free languages

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Assume the contrary - that

is context-free

Since is context-free and infinite

we can apply the pumping lemma

L

L

}*},{:{ bawwwL

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Pumping Lemma gives a number

such that:

m

we pick: Lbaba mmmm

}*},{:{ bawwwL

Pick any string of

with length at least mL

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We can write:

}*},{:{ bawwwL

uvxyzbaba mmmm

Pumping Lemma says:

Lzxyuv ii for all 0i

with lengths mvxy || 1|| vyand

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We examine all the possible locations

of string in vxy

mvxy || 1|| vyuvxyzbaba mmmm

}*},{:{ bawwwL

mmmm baba

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Case 1: vxy is within the firstma

bbaabbaa ........................v

m m m

u z


}*},{:{ bawwwL

x y

m

1kav 2kay 121 kk

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bbaabbaa ..................................2v

21 kkm m m

u z


}*},{:{ bawwwL

x 2y

m


1kav 2kay 121 kk

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}*},{:{ bawwwL


121 kk

Lzxyuvbaba mmmkkm 2221

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}*},{:{ bawwwL


Contradiction!

Lzxyuvbaba mmmkkm 2221

Lzxyuv 22However, from Pumping Lemma:

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}*},{:{ bawwwL

bbaabbaa ........................v

m m m

u zx y

m

is in the first

is in the firstCase 2: mamb

vy

1kav 2kby 121 kk

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}*},{:{ bawwwL

bbaabbaa ....................................2v

1km 2km m

u zx 2y

m

is in the first


vy

1kav 2kby 121 kk

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}*},{:{ bawwwL

is in the first


vy

121 kk

Lzxyuvbaba mmkmkm 2221

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}*},{:{ bawwwL

is in the first


vy

Lzxyuvbaba mmkmkm 2221

Contradiction!


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}*},{:{ bawwwL

bbaabbaa ........................v

m m m

u zx y

m

is in the first

overlaps the firstCase 3: mmbamb

vy

21 kk bav 3kby 1, 21 kk

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}*},{:{ bawwwL

is in the first


vy

21 kk bav 3kby 1, 21 kk

2v zx 2y

bbaabbbbaaaa .................................m 2k 3km

u

m1k m

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}*},{:{ bawwwL

is in the first


vy

Lzxyuvbabbaa mmkmkkm 22321

1, 21 kk

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}*},{:{ bawwwL

is in the first


vy

Lzxyuvbabbaa mmkkkm 22321

Contradiction!


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Overlaps the first

in the firstCase 4:

bbaabbaa ........................v

m m m

u z


}*},{:{ bawwwL

x y

m

mammba

vy

Analysis is similar to case 3

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Other cases: vxy is within

mmmm babaor

or

mmmm baba

mmmm baba

Analysis is similar to case 1: mmmm baba

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More cases: vxy overlaps

mmmm baba ormmmm baba

Analysis is similar to cases 2,3,4:

mmmm baba

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Since , it is impossible for

to overlap:

There are no other cases to consider

mvxy ||vxy

mmmm baba

nor normmmm baba mmmm baba

neither

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In all cases we obtained a contradiction

Therefore: The original assumption that

is context-free must be wrong

Conclusion: is not context-freeL

}*},{:{ bawwwL

END OF PROOF

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Regular Languages

}{ nnba }{ RwwContext-Free Languages

Non-regular languages

Unrestricted grammar languages

}0:{ ncba nnn

**ba

}{ww

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is not context free



}0:{ ! naL n

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L

Assume to the contrary that

is context-free

L

}0:{ ! naL n

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Pumping Lemma gives a magic number

such that:

m

we pick: Lam !

Pick any string of with length at least mL

}0:{ ! naL n

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We can write: uvxyzam !

Pumping Lemma says:


}0:{ ! naL n

with lengths mvxy || 1|| vy and

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of string in vxy !ma

mvxy || 1|| vyuvxyzam !

There is only one case to consider

}0:{ ! naL n

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v

!m

u zx y

1kav 2kay mkk 211

aa ...............


}0:{ ! naL n

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2v

21! kkm

u zx 2y

1kav 2kay

aa ...........................


}0:{ ! naL n

mkk 211

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2v

km !

u zx 2y

1kav 2kay

aa ...........................


}0:{ ! naL n

mk 1

21 kkk

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}0:{ ! naL n

mk 1

zxyuva km 22!

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)!1(

)1(!

!!

!!

m

mm

mmm

mmkm

)!1(!! mkmm

mk 1Since

2m we have:for

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}0:{ ! naL n

Lzxyuva km 22!

)!1(!! mkmm

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}0:{ ! naL n

Lzxyuva km 22!

Lzxyuv 22

Contradiction!

However, from Pumping Lemma:

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We obtained a contradiction




}0:{ ! naL n

END OF PROOF

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Regular Languages

**ba

Context-free languages

Unrestricted grammar languages}0:{ ncba nnn

}0:{ nba nn

}},{:{ bawww

}*},{:{ bawwwR

}0:{ ! nan }0:{2

nba nn

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is not context free



}0:{2

nbaL nn

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L

Assume to the contrary that

is context-free

L

}0:{2

nbaL nn

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Pumping Lemma gives a number

such that:

m

we pick: Lba mm 2

Pick any string of with length at least mL

}0:{2

nbaL nn

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We can write:

with lengths andmvxy || 1|| vy

uvxyzba mm 2

Pumping Lemma says:


}0:{2

nbaL nn

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of string in vxy

mvxy || 1|| vyuvxyzba mm 2

mm ba2

}0:{2

nbaL nn

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Most complicated case:

bbaa ...........................v

2m m

u zx y

is in

is inmamb

vy

}0:{2

nbaL nn


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bbaa ...........................v

2m m

u zx y

1kav 2kby

}0:{2

nbaL nn

mkk 211


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bbaa ...........................v

2m m

u zx y

1kav 2kby

Most complicated sub-case: 01 k 02 k

}0:{2

nbaL nn

mkk 211

and


690v

12 km 2km

u zx 0y

bbaa ..................

}0:{2

nbaL nn

1kav 2kby


mkk 211

and


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}0:{2

nbaL nn

1kav 2kby


mkk 211

and

zxyuvba kmkm 00212


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12

2

222

12

)1()(

km

mm

mkm

01 k 02 k mkk 211and

221

2 )( kmkm

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}0:{2

nbaL nn

221

2 )( kmkm

Lzxyuvba kmkm 00212


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}0:{2

nbaL nn

Lzxyuvba kmkm 00212

However, from Pumping Lemma:

Lzxyuv 00

Contradiction!


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When we examine the rest of the cases

we also obtain a contradiction

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In all cases we obtained a contradiction




}0:{2

nbaL nn

END OF PROOF

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Check your knowledge before midterm exam!

Selected Examplesof

CF Language Problems

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Let G be the grammar with productions:

Claim: L(G) = L

||

bBcB

BaScS

}:{ mnkcbaL kmn Find a CFG for the following language

Example

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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

||

bBcB

BaScS

}:{ mnkcbaL kmn

)(

**

mncmbna

ncmcBmbnancBnancSnaS

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Consider w 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

Find a CFG for the following language

}:{ mnkcbaL kmn

END OF PROOF

||

bBcB

BaScS

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}:{ 32 NnbaL nn Find a PDA and CFG for the following language

Example

Is the automaton deterministic? Yes. It acts in a unique way in each state, no -transitions.

fqiq

/, 23 ab22 /, aa

/, 23 ab

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CFG : 32| SbaS

}:{ 32 NnbaL nn

fqiq

/, 23 ab22 /, aa

/, 23 ab

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}2:},{{ ba nnbaxL Find a PDA and CFG for the following language

Example

bb

aa

/,

/,

/,

/,

/,

/,

aab

aba

aab

baa

PDA

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CFG :

SSbSaSaaSaSbaSbSaS ||||

}2:},{{ ba nnbaxL bb

aa

/,

/,

/,

/,

/,

/,

aab

aba

aab

baa

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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 = {anbn: n 0}

}:{ tenofmultipleNOTisnnbnaL

Prove that the language L is context-free

Example

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L1 ={w : w is made from a’s and b’s and the length of w is a multiple of ten}

L2 = {anbn: n 0}

Let L1 denote the complement of L1. We have that L = L1 L2.

L1 is a regular language, since we can easily build a finite automaton with 10 states that accepts any string in this language.

L1 is regular too, since regular languages are closed under complement.

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The language L2 is context-free.

The grammar is: S aSb |

Therefore, the language L = L1 L2 is also context-free,

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

END O PROOF

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}||,},{,:{ nxbaxNnaxL n Find a PDA and CFG for the following language

Example

CFG

babAaAA

AASaS

|||

||

ababaaabaAaaabAaa

AAaaAASaaASaS

Production ex.

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}||,},{,:{ nxbaxNnaxL n

PDA

fqiq

babAaAA

AASaS

|||/,

||/,

/,

/,

bb

aa

S/,

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}different are symbols finishing theand starting the

,:},{{ ba nnbaxL

Find a PDA and CFG for the following language

Example

PDA /,a

/,

/,

/,

/,

ba

bb

ab

aa

bb /,

/, ab /,a

/, a

/,

/,

/,

/,

ba

bb

ab

aa

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}different are symbols finishing theand starting the

,:},{{ ba nnbaxL

/,a

/,

/,

/,

/,

ba

bb

ab

aa

bb /,

/, ab /,a

/, a

/,

/,

/,

/,

ba

bb

ab

aa

CFG, direct construction

•Strings start and finish with different symbols

•Strings contain at least one more a than b

bAaaAbS |

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

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

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