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Given L is regular, Prove/Disprove that the following language is regular or not. $EXP = \{w| w^{|w|} ∈L\}$

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    $\begingroup$ What do you think? What have you tried, and where did you get stuck? $\endgroup$ – Yuval Filmus Feb 4 '18 at 16:17
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  • $\begingroup$ Oh, I did to try to find and used others to solve this. I'm sorry I should've mentioned my thoughts. Thanks! $\endgroup$ – Jatin Arora Feb 5 '18 at 17:39
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Suppose that $L$ is accepted by a DFA with states $Q$, initial state $q_0$, accepting states $F$, and transition function $\delta$.

With every word $w$ we can associate the function $\delta_w\colon Q \to Q$ given by $\delta_w(q) = \delta(q,w)$. Note that there are finitely many such function. It is not hard to check that for any such function $\delta_w$, the language $L_{\delta_w} = \{ x : \delta_x = \delta_w \}$ is regular.

We can also associate with $w$ a set $N_w = \{ n \in \mathbb{N} : \delta(q_0,w^n) \in F \}$. Note that $N_w$ depends only on $\delta_w$, and so there are finitely many such sets. Moreover, each of them is eventually periodic (why?), and so the language $L_{N_w} = \{ x : |x| \in N_w \}$ is regular.

Combining the two observations, we see that $$ EXP = \bigcup_{w \in \Sigma^*} (L_{\delta_w} \cap L_{N_w}). $$ While the sum is infinite, there are only finitely many different summands, and we deduce that $EXP$ is regular.

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