Regularity of the exact middle of words from a regular language

Question

Let $L$ be a regular language.
Is the language $L_2 = \{y : \exists x,z\ \ s.t.|x|=|z|\ and\ xyz \in L \}$ regular?

I know it's very similar to the question here, but the catch is that it's not a simple substring of a word in a regular language, but rather an "exact middle" - we have to count the prefix and suffix length.

Therefore, I assume it's not regular, but I couldn't find a way to prove it. I also couldn't think of any way to modify the NFA of $L$ to accept $L_2$.

Answer 1

Hint: Consider some DFA for $L$. For every $n \geq 0$, let $A_n$ be the set of states $s$ such that there is some word of length $n$ which leads the DFA from the initial state to $s$. Let $B_n$ be the set of states $t$ such that there is some word of length $n$ which leads the DFA from $t$ to an accepting state. Finally, for any two states $s,t$, let $R_{s,t}$ be the (regular) set of words leading the DFA from $s$ to $t$. We have $$ L_2 = \bigcup_{n \geq 0} \bigcup_{\substack{s \in A_n \\ t \in B_n}} R_{s,t}. $$ Since there are only finitely many possibilities for $s,t$, the union is in fact finite, and so regular.

Answer 2

Let $D = (Q, Σ, δ, q_0, F)$ be a DFA for $L$. Without loss of generality, assume $q_S, q_F \notin Q$. We construct a ε-NFA $N = (Q ∪ \{q_S, q_F\}, Σ, Δ, q_S, \{q_F\})$ for $L_2$ the following way:

Find each path in $D$ from $q_0$ to any $f ∈ F$. For each such path $p_k: q_0 = q_{k,0} \xrightarrow{α_{k,1}} q_{k,1} \xrightarrow{α_{k,2}} … \xrightarrow{α_{k,i}} q_{k,i} \xrightarrow{α_{k,i+1}} … \xrightarrow{α_{k,n_k}} q_{k,n_k}$ construct the paths $p_k^{(i)} : q_{k,i} \xrightarrow{α_{k,i+1}} q_{k,i+1} \xrightarrow{α_{k,i+2}} … \xrightarrow{α_{k,n_k-i}} q_{k,n_k-i}$ for $0 \le i \le \frac{n_k}2$ (i.e. constuct all “middle parts“ of the path). This can be done effectively. Construct $Δ$ by combining all these paths, together with:

• $(q_S, ε, q_{k,i})$ for all $i$ as above
• $(q_{k, n_k-i}, ε, q_F)$ for all $i$ as above

$L(N)$ is regular by construction.

Proof sketch that $L(N) = L_2$: Let $w ∈ L(N)$. By construction we know that $w$ must match at least on of the paths $p_k^{(i)}$ above. Each of this path belongs to a path in $D$, which contains an additional prefix and suffix of length $i$. Choose $x$ as the word described by this prefix and $y$ the one described by the suffix. We find that $xwy ∈ L$, with $|x| = |y| = i$. With similar reasoning we find for each $w ∈ L_2$ a path in $N$. Let $i$ be the length of $x$ and $y$ belonging to $w$. $p_k^{(i)}$ for some $k$ forms $w$.

Thus $L(N) = L_2$.