I'm having trouble with proving the following, and my attempt and confusion is mentioned further below. **PROBLEM STATEMENT:** > For a string $w = a_1 a_2 a_3 a_4 a_5 a_6 a_7 \dots$, define $third(w) = a_3 a_6 a_9 \dots$ > Then, for a language $L$, define $third(L) = \{third(w) : w ∈ L\}$. > Show that if $L$ is regular, then $third(L)$ is also regular. > (Hint: Construct an ϵ-NFA from the DFA for $L$.) **TEACHER'S SOLUTION:** If $L$ is regular, it is accepted by some DFA, say $A = (Q,Σ,δ,s_0,F)$. We will construct an ϵ-NFA $B$ such that $L(B) = third(L(A))$. Here, you need four copies of $A$. Formally, $B = (Q × \{1,2,3,4\}, Σ, ρ, F × \{2,3,4\})$, where $ρ =$ ${\{(⟨p, 1⟩, ϵ, ⟨q, 2⟩) : (p, a, q) ∈ δ}$, for some ${a ∈ Σ\}}$ ${∪}$ ${\{(⟨p, 2⟩, ϵ, ⟨q, 3⟩) : (p, a, q) ∈ δ}$, for some ${a ∈ Σ\}}$ ${∪}$ ${\{(⟨p, 3⟩, a, ⟨q, 4⟩) : (p, a, q) ∈ δ\}}$ ${∪}$ ${\{(⟨p, 4⟩, ϵ, ⟨p, 1⟩) : p ∈ Q\}}$. (Do not confuse ρ and p.) **WHAT I DO UNDERSTAND:** • I understand how Cartesian products work. **WHAT I MAY OR MAY NOT UNDERSTAND:** • Are the ${(p, a, q) ∈ δ}$ parts another way of saying ${δ(p,a) = q}$? • I suspect that the gist of the proof is to modify the initial DFA of L with ϵ transitions, so that there are only non-ϵ transitions for every third symbol in any particular string of L, w, being analyzed. And, I suspect that that's the purpose of the stuff with the angle brackets. • I suspect that the purpose of taking the Cartesian product of Q and ${\{1,2,3,4\}}$ and the purpose of taking the Cartesian product of F with ${\{2,3,4\}}$ is to have a state for each copy of A, where ${⟨p, 1⟩}$ represents the state p of the first copy of A (but is technically a state of B, not A). • Should ${⟨s_0, 1⟩}$ be the start state of B, such that B's n-tuple is a 5-tuple, rather than a 4-tuple, and that that 5-tuple is ${B = (Q × \{1,2,3,4\}, Σ, ρ, ⟨s_0, 1⟩, F × \{2,3,4\})}$? **WHAT I DON'T UNDERSTAND:** • I don't understand what the transitions regarding the four copies (defined by B's transition relation, ρ) achieve exactly. • Also, why does one need four copies of A? Can't one just add those additional transitions to a single copy of A? Any input would be GREATLY appreciated!