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Let $\Sigma=\left\{0,1\right\}, L = \Sigma^*$ and $R=\left\{0^n1^n \mid n > 0\right\} $ then the languages $L \cup R$ and $R$ are respectively

  1. regular, regular
  2. not regular, regular
  3. regular, not regular
  4. not regular, not regular
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5 Answers

Best answer
29 votes
29 votes

Answer is (C). $L \cup R$ is nothing but $L$ as $R$ is a subset of $L$ and hence regular. $R$ is deterministic context-free but not regular as we require a stack to keep the count of 0's to match that of $1$'s.

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2 votes
2 votes
It will be D.

Since L is regular and R is CFL. (L U R) is regular union... which results in the language union with regular. Since R is not regular, so its result will also be not regular.
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