Recently I had a question on one of my assignments asking to prove or disprove the following:

Let $L$ be a language. If $L^*$ is context-free then $L$ is context-free.

Now obviously this is false since we can take some non-context free language $L_1$ and the alphabet $\Sigma$, then make $(L_1 \cup \Sigma)^* = \Sigma^*$ is context-free and clearly $L_1\cup \Sigma$ is not.

Now I was thinking about a similar problem but now with respect to the words in the language, and was wondering if it is true or not. If $C$ is a context-free language then $C'$ is a context-free language where $w\in C'$ if and only if $\{w\}^*$ is contained in $C$.

My suspicion is that this is also false, but I can't come up with a counterexample. Somehow one needs to construct a CFL $C$ such that the subset of all the "periodic" words of $C$ together are not context-free.

  • $\begingroup$ Deleting my answer. Apparently I took a construction which works for NFAs and extrapolated it to PDAs rather recklessly. $\endgroup$ – dkaeae Mar 27 '19 at 11:43
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    $\begingroup$ What do you mean by "$w^*$ is contained in $C$"? $\{w^i\mid i\geq 0\}\subseteq C$? $\endgroup$ – David Richerby Mar 27 '19 at 12:20
  • $\begingroup$ @DavidRicherby I just updated the question. (This comment will be deleted shortly.) $\endgroup$ – John L. Mar 27 '19 at 19:20

Let me solve a simpler question first. Suppose we'd like to know

If $C$ is context-free, must $F(C)=\{w: ww \in C\}$ be context-free?

The answer is no: $F(\{a^i b^i c^j a^j b^k c^k\})=\{a^i b^i c^i\}$.

As for the original problem, consider the context-free language that contains words of the form $\{a^i b^i c^j d a^j b^k c^k d\}$ or those which don't contain exactly two $d$'s.

Define $D=\{w: w^\ast \subseteq C\}$. Suppose $D$ is context-free. Then $E=D\cap a^\ast b^\ast c^\ast d$ is context-free.

I claim $E=\{a^i b^i c^i d\}$, which is not context-free. Suppose $w = a^i b^j c^k d$. We have $w \in E$ exactly if $w^n \in C$ for every $n$. This condition is clearly satisfied when $n \neq 2$, because $w^n$ will not have exactly two $d$'s. For $n=2$, $w w=a^i b^j c^k d a^i b^j c^k d\in C$ which is equivalent to $i=j=k$.

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  • $\begingroup$ @AlexPatterson Your language works too (nitpick: you need to add $\epsilon$). The argument with $d$'s seemed to be easier to explain and I didn't like dealing with $a^{+} b^{+} c^{+}$ vs $a^{\ast} b^{\ast} c^{\ast}$ but it's a matter of preference. In the second question, you are right that $j=j'$ is unnecessary. However, an analogous argument can be said about $i=i'$ and $k=k'$; you can skip any one of the 3 equalities, but not two of them. For symmetry, I prefer to have all three instead of skipping an arbitrary one. $\endgroup$ – sdcvvc Mar 27 '19 at 17:30
  • $\begingroup$ @Apass.Jack It can be shown that if $C$ is regular, then $\{w:ww\in C\}$ and $\{w:w^\ast \subseteq C\}$ are also regular. $\endgroup$ – sdcvvc Mar 27 '19 at 18:31

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