Nonregularity Proofs

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Nonregularity Proofs
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Regular Languages Grand Unification
(Parallel Simulation) (Rabin and Scotts
work)
(Collapsing graphs Structural Induction) (S.
Kleenes 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 Morgans Laws.
• Relationship to context-free languages. (Regular
  Grammars.)
• Ease of specification. (Regular expression.)
• Building tokenizers/lexical analyzers. (DFAs)

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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.

L1bit strings with even parity L2bit strings
with number of 1s divisible by 3 Lbit strings
with number of 1s 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,
• ab /\, a, b, aa, ab, bb,, aabb,
• C /\, ab, aabb,
• intersect(ab, C) intersect(L, ab) C
• union(ab, C) ab 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 gt PL not PL gt not RL
• It is not a sufficient condition.
• Not every non-regular language violates it.
• not RL gt? PL or not PL (no conclusion)

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Basic Idea
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