Limitations of Context Free Languages - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

Download Limitations of Context Free Languages - Automata and Complexity Theory - Lecture Slides and more Slides Theory of Automata in PDF only on Docsity!

Limitations of context-free languages

Non context-free languages

• Recall the pumping lemma for regular languages

allows us to show some languages are not regular

Are these languages context-free?

L

1

= {a

n

b

n

: n ≥ 0}

L

2

= {x: x has same number of as and bs}

L

3

n

: n is prime}

L

4

= {a

n

b

n

c

n

: n ≥ 0}

L

5

= {x#x

R

: x ∈ {0, 1}*}

L

6

= {x#x: x ∈ {0, 1}*}

More intuition

• Suppose we could construct some CFG for L

4

, e.g.

• We do some derivations

of “long” strings

S → BC

B → CS | b

C → SB | a

S ⇒ BC

⇒ CSC

⇒ aSC

⇒ aBCC

⇒ abCC

⇒ abaC

⇒ abaSB

⇒ abaBCB

⇒ ababCB

⇒ ababaB

⇒ ababab

More intuition

• If derivation is long enough, some variable must

appear twice on same path in parse tree

S ⇒ BC

⇒ CSC

⇒ aSC

⇒ aBCC

⇒ abCC

⇒ abaC

⇒ abaSB

⇒ abaBCB

⇒ ababCB

⇒ ababaB

⇒ ababab

S

B C

C S S B

B C B C

a

b a b a b

More intuition

• We can repeat this many times

• Every sufficiently large derivation will have a

part that

can be repeated indefinitely

• This is caused by cycles in the grammar

ababab

ababbabb

ababbbabbb

abab

n

ab

n

bb

General picture

u

u

v

v

v

w

x

y

y

u

v

v

v

w

x

y

uvwxy uv

3

wx

3

y

A

A

A

A

A

A

A

A

A

x

x

x

x

w

xvvwxxy

Pumping lemma for context-free

languages

• Theorem: For every context-free language L

There exists a number n such that for every

string z in L, we can write z = uvwxy where

 |vwx| ≤ n

 |vx| ≥ 1

 For every i ≥ 0, the string uv

i

wx

i

y is in L.

u w y

x v

Pumping lemma for context-free

languages

• So to prove L is not context-free, it is enough that

For every n there exists z in L, such that for

every way of writing z = uvwxy where

 |vwx| ≤ n and  |vx| ≥ 1, the string uv

i

wx

i

y

is

not in L for some i ≥ 0.

u w y

x v

Example

adversary

choose n

write z = uvwxy

(|vwx| ≤ n,|vx| ≥ 1)

you

choose z ∈ L

choose i

you win if uv

i

wx

i

y ∉ L

1

2

adversary

n

write z = uvwxy

you

z = a

n

b

n

c

n

i =?

1

2

L

4

= {a

n

b

n

c

n

: n ≥ 0}

u w y

x v

a a a ... a a b b b ... b b c c c ... c c

Example

• Case 1: v or x contains two kinds of symbols

Then uv

2

wx

2

y not in L because pattern is wrong

• Case 2: v and x both contain one kind of

symbol

Then uv

2

wx

2

y does not have same number of

x

v

a a a ... a a b b b ... b b c c c ... c c

x

v

a a a ... a a b b b ... b b c c c ... c c