Proving of regular language

Question

Is this regular or not

L = {w1w^R | w ∈ {0,1}* (where for any word w ∈ {0,1})*, w^R denotes the reverse of w)

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score 0 · Answer 1 · answered Oct 01 '20 at 16:17

0

L is not a regular language.

We can contradict by pumping lemma for regular languages.

Let assume L is regular, so there exists an integer $ k $ of pummping lemma.

$ w = a^kb $

$ w^R = ba^k $

$ ww^R = a^kbba^k \in L $

$ w' = a^{k-r}bba^k \notin L , r>0 $

Contradiction to $ w' \in L $.

answered Oct 01 '20 at 16:17

Yakir

Actually in this case you can pump $bb$ and there is no problem. – user253751 Oct 01 '20 at 17:52