Nonregularity proofs l.jpg
This presentation is the property of its rightful owner.
Sponsored Links
1 / 16

Nonregularity Proofs PowerPoint PPT Presentation


  • 70 Views
  • Uploaded on
  • Presentation posted in: General

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.

Download Presentation

Nonregularity Proofs

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Nonregularity proofs l.jpg

Nonregularity Proofs

L10PLemma


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)

(Construction)

(Solving linear equations)

L10PLemma


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:

    L10PLemma


    Slide5 l.jpg

    CLAIM:

    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.

    L10PLemma


    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

    L10PLemma


    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.

    L10PLemma


    Digression l.jpg

    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

    L10PLemma


    Prelude to pumping lemma l.jpg

    Prelude to Pumping Lemma

    • Is 46551 divisible by 46?

    • Is 46554 divisible by 46?

    • Is 46552 divisible by 46?

      Necessary vs sufficient condition

    L10PLemma


    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)

    L10PLemma


    Basic idea l.jpg

    Basic Idea:

    b

    a

    q0

    a

    q1

    b

    b

    a

    q2

    q3

    a,b

    L10PLemma


    Slide12 l.jpg

    Note,

    So,

    L10PLemma


    Fundamental observation l.jpg

    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

    L10PLemma


    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

    L10PLemma


    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.

    L10PLemma


    Proving non regularity l.jpg

    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:

    L10PLemma


  • Login