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Consider the language defined as L = { $a^pb^qa^r$ | p = q or q = r} .

L complement is

  1. Regular
  2. CFL but not regular
  3. CSL but not CSL
  4. DCFL
in Theory of Computation by (83 points)
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Complement shall be p!=q and q!=r

CSL it must be

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Actually the language L is CFL, and we know that CFL are not closed under Compliment and Intersection.

So compliment of given language will be CSL.
by Junior (865 points)
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L is Non-deterministic context free right?
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@balchandar reddy san Yes L is Non-Deterministic Context Free Language.

We generally use NDCFL as CFL only and Deterministic Context Free Language as DCFL.

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