[PDF] [PDF] More Applications of The Pumping Lemma

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Here is the Pumping Lemma If L is a regular language, then there is an integer n > 0 with the property that: (*) for any string x ∈ L where x ≥ n, there are strings u, v, w such that (i) x = uvw, (ii) v = ǫ, (iii) uv ≤ n, (iv) uvkw ∈ L for all k ∈ N

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1 CS 301 - Lecture 17 Pumping Lemma for Context Free Grammars

Fall 2008

Review

• Languages and Grammars - Alphabets, strings, languages • Regular Languages

- Deterministic Finite and Nondeterministic Automata - Equivalence of NFA and DFA - Regular Expressions - Regular Grammars - Properties of Regular Languages - Languages that are not regular and the pumping lemma

• Context Free Languages

- Context Free Grammars - Derivations: leftmost, rightmost and derivation trees - Parsing and ambiguity - Simplifications and Normal Forms - Nondeterministic Pushdown Automata - Pushdown Automata and Context Free Grammars - Deterministic Pushdown Automata

• Today: - Pumping Lemma for context free grammars

More Applications of The Pumping Lemma

The Pumping Lemma: there exists an integer such that m for any string mwLw≥∈|| , we can write For infinite context-free language L uvxyzw= with lengths and it must be:

0 allfor ,≥∈iLzxyuv

ii 2

Context-free languages

}0:{≥nba nn

Non-context free languages

}0:{≥ncba nnn }*},{:{bawww R }},{:{bavvv∈

Theorem: The language

}*},{:{bavvvL∈= is not context free

Proof: Use the Pumping Lemma

for context-free languages Assume for contradiction that is context-free Since is context-free and infinite we can apply the pumping lemma LL }*},{:{bavvvL∈=

Pumping Lemma gives a magic number

such that: m

Pick any string of with length at least

m we pick: Lbaba mmmm L }*},{:{bavvvL∈= 3 We can write: with lengths and uvxyzbaba mmmm

Pumping Lemma says:

Lzxyuv

ii for all

0≥i

}*},{:{bavvvL∈=

We examine all the possible locations

of string in vxy uvxyzbaba mmmm mmmm baba }*},{:{bavvvL∈=

Case 1:

vxy is within the first m a bbaabbaa........................ vmmmu z uvxyzbaba mmmm xym 1 k av= 2 k ay= 1 21
≥+kk }*},{:{bavvvL∈= 2 v 21
kkm++ mmu z uvxyzbaba mmmm x 2 y m

Case 1:

vxy is within the first m a 1 k av= 2 k ay= 1 21
≥+kk }*},{:{bavvvL∈= 4 uvxyzbaba mmmm

Case 1:

vxy is within the first m a 1 21
≥+kk

Lzxyuvbaba

mmm kkm 22
21
}*},{:{bavvvL∈= uvxyzbaba mmmm

Case 1:

vxy is within the first m aLzxyuv∈ 22

Contradiction!!!

Lzxyuvbaba

mmm kkm 22
21

However, from Pumping Lemma:

}*},{:{bavvvL∈= is in the first is in the first

Case 2:

bbaabbaa........................ vmmmu z uvxyzbaba mmmm xym m a m b vy 1 k av= 2 k by= 1 21
≥+kk }*},{:{bavvvL∈= is in the first is in the first

Case 2:

2 v 1 km+ 2 km+ mu z uvxyzbaba mmmm x 2 y m m a m b vy 1 k av= 2 k by= 1 21
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