Given a set of strings $S$ over $\{a,b\}$. How to determine whether there is an infinite sequence consisting of ${a, b}$ (i.e., a $\omega$-string), which doesn't have any string in $S$ as a substring?

The total length of all substrings in $S$ is at most $20000$.

  • $\begingroup$ Are you looking for an efficient algorithm, or for any algorithm? $\endgroup$ – Yuval Filmus Oct 19 '20 at 18:12
  • $\begingroup$ In general, I am looking for an efficient algorithm, but if you have any ideas on any algorithm, it may have been useful for me. $\endgroup$ – Kapa Oct 19 '20 at 18:24
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    $\begingroup$ Construct a DFA for the language of all finite strings not containing any of the given strings as a substring. Check whether there is a reachable state which belongs to a cycle containing only accepting states. This algorithm runs in exponential time. $\endgroup$ – Yuval Filmus Oct 20 '20 at 6:35
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    $\begingroup$ What's the motivation for this question? What's the context in which you encountered it? $\endgroup$ – D.W. Oct 20 '20 at 16:17

If every $\omega$-string contains a substring in $S$, then we say that $S$ is unavoidable or a universal hitting set (UHS). There is a classical algorithm for detecting whether a set of substrings is unavoidable, described for example in Lothaire's Algebraic combinatorics on words, Section 1.6.

Construct a graph $G$ as follows. The vertices are all prefixes of strings in $S$. For every prefix $p$ and symbol $\sigma$, we connect $p$ to the longest suffix of $p\sigma$ appearing in the graph. The set $S$ is unavoidable iff every cycle in $G$ passes through a vertex corresponding to a word in $S$.

To see this, suppose first that every cycle in $G$ passes through a vertex corresponding to a word in $S$. Consider some $\omega$-word $w$. Starting at the vertex $\epsilon$, "read" the symbols of $w$ one by one. Eventually you will hit a cycle (since $G$ is finite). By assumption, this cycle contains a vertex $v$ in $S$, and so $w$ contains $v$. In the other direction, tracing an $S$-less cycle (replacing each edge by the symbol $\sigma$ used to create it) results in a periodic $\omega$-word which avoids $S$.


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