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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 l.jpg

Nonregularity Proofs


Regular languages grand unification l.jpg

Regular Languages: Grand Unification

(Parallel Simulation)

(Rabin and Scott’s work)

(Collapsing graphs; Structural Induction)

(S. Kleene’s work)


(Solving linear equations)


Role of various representations for regular languages l.jpg

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)

  • L10PLemma

    Slide4 l.jpg

    Consider pairs of strings:

    IfL were regular,then there exists a DFA M

    accepting L with the following property:


    Slide5 l.jpg


    JUSTIFICATION: Otherwise, from the definition of DFA,

    which contradicts the earlier conclusion.

    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.


    Using closure properties l.jpg

    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


    Slide7 l.jpg

    • 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.


    Digression l.jpg


    • 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


    Pumping lemma for regular languages l.jpg

    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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    Basic Idea:













    Slide12 l.jpg




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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


    Pumping lemma theorem 7 3 3 l.jpg

    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


    Slide15 l.jpg

    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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