1 votes 1 votes L={ W(W^r) , W ∈ (a,b)* } W^r is reverse of W. Is it a regular language ? please Explain. Diksha Aswal asked Sep 13, 2017 Diksha Aswal 413 views answer comment Share Follow See 1 comment See all 1 1 comment reply Rishabh Gupta 2 commented Sep 13, 2017 reply Follow Share No. You can check it using pumping lemma. It is CFL. A nondeterministic pda can accept it. 1 votes 1 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes wwR is CFL. NPDA can simulate it. It cannot be simulated by DPDA. It is not regular. NB : wxwR is regular sh!va answered Sep 13, 2017 • selected Sep 14, 2017 by srestha sh!va comment Share Follow See all 3 Comments See all 3 3 Comments reply Warlock lord commented Sep 13, 2017 reply Follow Share 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. 0 votes 0 votes Hemant Parihar commented Sep 13, 2017 reply Follow Share @warlock lord Check here more such questions. http://gatecse.in/identify-the-class-of-a-given-language/ 0 votes 0 votes Warlock lord commented Sep 13, 2017 reply Follow Share 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 0 votes 0 votes Please log in or register to add a comment.