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Assume we apply the following operation on the regular language $L$:

$$OP(L)=\{a_2a_1a_4a_3\dots a_{2n}a_{2n-1}:a_1a_2a_3\dots a_{2n}\in L\}$$

Why $OP(L)$ remains regular?

I think this is as the same as shuffle but I have no idea to show that.

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Make a DFA $D$ for language $L$.

Now make a copy $D'$ of this DFA, but only of the states, removing the transitions.

To add back transitions we check for all possible symbols $a, b$ for all possible pairs of states $x, y$ if $\delta(\delta(x, a), b) = y$ in $D$. If so, we add a new non-accepting state $s'$ to $D'$, and the transitions $\delta(x', b) = s'$, $\delta(s', a) = y'$.

We've now ensured that $D'$ recognizes $a_2a_1a_4a_3\dots$ iff $D$ recognizes $a_1a_2a_3a_4$, thus $OP(L)$ must also be regular.

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