Operations making non-context-free languages regular

Question

If $A$ is a non-context free language, is $A^*$ a regular language?
If $A$ is a non-context free language and $B$ is a regular language, is it possible that their concatenation is a regular language?

What do you think? What have you tried, and where did you get stuck? — Yuval Filmus, Apr 15 '19 at 18:07

rici · Answer 1 · 2019-04-11T04:36:35.920

1

Let $A$ be $\{a^p\mid p\text{ is prime}\}$. Let $B$ be $\{a\}^*$.

Both $A^*$ and $A\circ B$ are regular. $A^*$ is $\epsilon+aaa^*$ and $A\circ B$ is $aaa^*$.

edited Apr 11 '19 at 04:36

answered Apr 11 '19 at 03:35

rici

True, but it's still regular. I'll fix it. – rici Apr 11 '19 at 04:32
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Note that, if $A$ is a unary language (i.e., uses only a single letter alphabet) as in this case, then in fact the star $A^$ is always regular. See this question: Show that the Kleene star of any unary language is regular. It seems the same holds for $A \cdot {a}^$, because that language consists of all strings longer than the shortest in $A$. – Hendrik Jan Apr 11 '19 at 08:52

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