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L={ W(W^r) , W ∈ (a,b)* }

W^r is reverse of W.

Is it a regular language ?

please Explain.
in Theory of Computation by Junior (779 points) | 58 views
+1
No. You can check it using pumping lemma.
It is CFL. A nondeterministic pda can accept it.

1 Answer

+2 votes
Best answer

wwR is CFL.

  • NPDA can simulate it.
  • It cannot be simulated by DPDA. 
  • It is not regular.

NB : wxwR is regular

by Boss (33k points)
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wxwR is regular only if w,x both belong to (a,b)*, right? If x belongs to some other element group then it will not be regular.

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@warlock lord Check here more such questions.

http://gatecse.in/identify-the-class-of-a-given-language/

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Yes that is what I meant to ask if w belongs to (a,b)* but x belongs to (c,d)*.. will it still be regular?

I'm just pointing out that "wxwR" is not simply regular. It has to belong to the same symbol group
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