I know that there are non-regular languages, so that $L^*$ is regular, but all examples I can find are context-sensitive but not context free.
In case there are none how do you prove it?
Asked By : Simon S
Answered By : Gilles
$L = \{a^n b^n \mid n\in\mathbb{N}\}$ is context-free but not regular (classical example). So is $L' = \{a^n b^n \mid n\in\mathbb{N}\} \cup \{a,b\}$.
$L'^\ast = \{a,b\}^\ast$ is regular.
Best Answer from StackOverflow
Question Source : http://cs.stackexchange.com/questions/2081
0 comments:
Post a Comment
Let us know your responses and feedback