¿Cómo encuentro la representación más corta para un subconjunto de un conjunto de potencia?

Estoy buscando un algoritmo eficiente para el siguiente problema o una prueba de dureza NP.

Sea $\Sigma$ un conjunto y $A\subseteq\mathcal{P}(\Sigma)$ un conjunto de subconjuntos de $\Sigma$ . Encuentre una secuencia $w\in \Sigma^*$ de menor longitud tal que para cada $L\in A$ , haya una $k\in\mathbb{N}$ tal que { w k + i ∣ 0 ≤ i < | L | } = L .$\{ w_{k+i} \mid 0\leq i < |L| \} = L$

Por ejemplo, para $A = \{\{a,b\},\{a,c\}\}$ , la palabra $w = bac$ es una solución al problema, ya que para $\{a,b\}$ hay $k=0$ , para $\{a,c\}$ hay $k=1$ .

En cuanto a mi motivación, estoy tratando de representar el conjunto de bordes de un autómata finito, donde cada borde puede ser etiquetado por un conjunto de letras del alfabeto de entrada. Me gustaría almacenar una sola cadena y luego mantener un par de punteros a esa cadena en cada borde. Mi objetivo es minimizar la longitud de esa cuerda.

I believe I found a reduction from Hamiltonian path, thus proving the problem NP-hard.

Call the word $w\in\Sigma^*$ a witness for $A$, if it satisfies the condition from the question (for each $L\in A$, there's $m\geq 1$ such that $\{w_{m+i}\mid 0\leq i<|L|\} = L$).

Consider the decision version of the original problem, i.e. decide whether for some $A$ and $k\geq 0$, there's a witness for $A$ of length at most $k$. This problem can be solved using the original problem as an oracle in polynomial time (find the shortest witness, then compare its length to $k$).

Now for the core of the reduction. Let $G=(V,E)$ be a simple, undirected, connected graph. For each $v\in V$, let $L_v=\{v\}\cup\{e\in E\mid v\in e\}$ be the set containing the vertex $v$ and all of its adjacent edges. Set $\Sigma=E$ and $A=\{L_v\mid v\in V\}$. Then $G$ has a Hamiltonian path if and only if there is a witness for $A$ of length at most $2|E|+1$.

Proof. Let $v_1e_1v_2\ldots e_{n-1}v_n$ be a Hamiltonian path in $G$ and $H=\{e_1, e_2, \ldots, e_{n-1}\}$ the set of all edges on the path. For each vertex $v$, define the set $U_v=L_v\setminus H$. Choose an arbitrary ordering $\alpha_v$ for each $U_v$. The word $w=\alpha_{v_1}e_1\alpha_{v_2}e_2\ldots e_{n-1}\alpha_{v_n}$ is a witness for $A$, since $L_{v_1}$ is represented by the substring $\alpha_1e_1$, $L_{v_n}$ by $e_{n-1}\alpha_n$, and for each $v_i$, $i\notin\{1, n\}$, $L_{v_i}$ is represented by $e_{i-1}u_{v_i}e_i$. Furthermore, each edge in $E$ occurs twice in $w$ with the exception of $|V|-1$ edges in $H$, which occur once, and each vertex in $V$ occurs once, giving $|w|=2|E|+1$.

For the other direction, let $w$ be an arbitrary witness for $A$ of length at most $2|E|+1$. Clearly, each $e\in E$ and $v\in V$ occurs in $w$ at least once. Without loss of generality, assume that each $e\in E$ occurs in $w$ at most twice and each $v\in V$ occurs exactly once; otherwise a shorter witness can be found by removing elements from $w$. Let $H\subseteq E$ be the set of all edges occurring in $w$ exactly once. Given the assumptions above, it holds that $|w|=2|E|-|H|+|V|$.

Consider a contiguous substring of $w$ of the form $ue_1e_2\ldots e_kv$, where $u,v\in V$, $e_i\in E$. We say that $u,v$ are adjacent. Notice that if $e_i\in H$, then $e_i=\{u,v\}$, because $e_i$ occurs only once, yet it is adjacent to two vertices in $G$. Therefore, at most one of $e_i$ can be in $H$. Similarly, no edge in $H$ can occur in $w$ before the first vertex or after the last vertex.

Now, there are $|V|$ vertices, therefore $|H|\leq |V|-1$. From there, it follows that $|w|\geq 2|E|+1$. Since we assume $|w|\leq 2|E|+1$, we get equality. From there we get $|H|=|V|-1$. By pigeonhole principle, there is an edge from $H$ between each pair of vertices adjacent in $w$. Denote $h_1h_2\ldots h_{n-1}$ all elements from $H$ in the order they appear in $w$. It follows that $v_1h_1v_2h_2\ldots h_{n-1}v_n$ is a Hamiltonian path in $G$. $\square$

Since the problem of deciding the existence of Hamiltonian path is NP-hard and the above reduction is polynomial, the original problem is NP-hard too.