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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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regular languages grand unification
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
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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slide4

Consider pairs of strings:

IfL were regular,then there exists a DFA M

accepting L with the following property:

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slide5

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.

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using closure properties
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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slide7
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
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
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
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
Basic Idea:

b

a

q0

a

q1

b

b

a

q2

q3

a,b

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slide12

Note,

So,

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fundamental observation
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
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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slide15
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
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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