Given a language $L$, let $Double(L)=\{w: ww\in L\}$ and $NotDouble(L)=\{w: ww\notin L\}$
If L is regular, are $Double(L)$ and $NotDouble(L)$ regular?
I tried using the closure properties of $L_{reg}$ but I figured out that $Double(L) \cup NotDouble(L)$ are not complementing of $L$.
I tried thinking about the regular expression of $L$ but I couldn't get anywhere with it.
I saw this question but I couldn't understand the answers. Is there a way to prove this using regular expressions, automata and closure properties only?
If $L$ is regular, prove that $\sqrt{L}=\left\{ w : ww\in L\right\}$ is regular
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Yinon Eliraz
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The accepted answer to the previous question is in terms of automata. It is not very heavy on intuition, though. – hmakholm left over Monica Jan 30 '16 at 12:42
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Why do you think they not complementing? In fact, for each $w$ over some alphabet, either $ww \in L$, or $ww \not \in L$. So $\sqrt{L}$ should solve everything.
Jára Cimrman
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