# Nonregularity Proofs - PowerPoint PPT Presentation

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Nonregularity Proofs. Regular Languages: Grand Unification. (Parallel Simulation) (Rabin and Scott’s work). (Collapsing graphs; Structural Induction) (S. Kleene’s work). (Construction) (Solving linear equations). Role of various representations for Regular Languages.

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Nonregularity Proofs

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## Nonregularity Proofs

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### Regular Languages: Grand Unification

(Parallel Simulation)

(Rabin and Scott’s work)

(Collapsing graphs; Structural Induction)

(S. Kleene’s work)

(Construction)

(Solving linear equations)

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### Role of various representations for Regular Languages

• Closure under complementation. (DFAs)

• Closure under union, concatenation, and Kleene star. (NFA-ls, Regular expression.)

• Consequence:

Closure under intersection by De Morgan’s Laws.

• Relationship to context-free languages. (Regular Grammars.)

• Ease of specification. (Regular expression.)

• Building tokenizers/lexical analyzers. (DFAs)

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Consider pairs of strings:

IfL were regular,then there exists a DFA M

accepting L with the following property:

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CLAIM:

JUSTIFICATION: Otherwise, from the definition of DFA,

In order to satisfy

the machine M must have a unique state for every i.

Thus, M must have infinite number of states, if L

is regular. This violates the definition of DFA.

So, L must be non-regular.

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### Using Closure Properties

• Regular languages are closed under set-intersection.

• Note that regularity is a property of a collection, and not a property of an individual string in the collection.

L1=bit strings with even parity

L2=bit strings with number of 1’s divisible by 3

L=bit strings with number of 1’s a multiple of 6

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• If R is a regular language and C is context-free, then may not be regular.

• Proof:

• Show that

• is not regular.

• Proof: If L were regular, ought to be regular. However, is known to be non-regular. Hence, L cannot be regular.

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### DIGRESSION

• If R is a regular language and C is context-free, then can be regular.

• Proof:

• L = {/\, ab, ba, aabb, abab, abba, baba, bbaa, …}

• a*b* = {/\, a, b, aa, ab, bb,…, aabb,…}

• C = {/\, ab, aabb, …}

• intersect(a*b*, C) =intersect(L, a*b*) = C

• union(a*b*, C) = a*b* ; union(L, C) = L

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### Prelude to Pumping Lemma

• Is 46551 divisible by 46?

• Is 46554 divisible by 46?

• Is 46552 divisible by 46?

Necessary vs sufficient condition

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### Pumping Lemma for Regular Languages

• It is a necessary condition.

• Every regular language satisfies it.

• If a language violates it, it is not regular.

• RL => PL not PL => not RL

• It is not a sufficient condition.

• Not every non-regular language violates it.

• not RL =>? PL or not PL (no conclusion)

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b

a

q0

a

q1

b

b

a

q2

q3

a,b

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Note,

So,

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### Fundamental Observation

• Given a “sufficiently” long string, the states of a DFA must repeat in an accepting computation. These cycles can then be used to predict (generate) infinitely many other strings in (of) the language.

Pigeon-Hole Principle

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### Pumping Lemma (Theorem 7.3.3)

• Let L be a regular language that is accepted by a DFA M with k states. Let z be any string in L with . Then z can be decomposed as uvw with

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For all sufficiently long strings (z)

These exists non-null prefix (uv)

and substring (v)

For all repetitions of the substring (v),

we get strings in the language.

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### Proving non-regularity

• If there exists an arbitrarily long string s L, and for each decomposition s = uvw, there exists an i such that , then L is non-regular.

Negation of the necessary condition:

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