A special class of languages: “circular” languages. Is it known? Define the following class of "circular" languages over a finite alphabet Sigma. Actually, the name already exists to denote a different thing it seems, used in the field of DNA computing. AFAICT, that's a different class of languages.

A language L is circular iff for all words $w$ in $\Sigma^*$, we have:

$w$ belongs to L if and only if for all integers $k &gt; 0$, $w^k$ belongs to L.

Is this class of languages known? I am interested in the circular languages which are also regular and in particular in:


a name for them, if they are already known
decidability of the problem, given an automaton (in particular: a
DFA), whether the accepted language obeys to the above definition
In the first part, we show an exponential algorithm for deciding circularity. In the second part, we show that this the problem is coNP-hard. In the third part, we show that every circular language is a union of languages of the form $r^+$ (here $r$ could be the empty regexp); the union is not necessarily disjoint. In the fourth part, we exhibit a circular language which cannot be written as a disjoint sum $\sum r_i^+$.

Edit: Incorporated some corrections following Mark's comments. In particular, my earlier claims that circularity is coNP-complete or NP-hard are corrected.

Edit: Corrected normal form from $\sum r_i^*$ to $\sum r_i^+$. Exhibited an "inherently ambiguous" language.



Continuing Peter Taylor's comment, here's how to decide (extremely inefficiently) whether a language is circular given its DFA. Construct a new DFA whose states are $n$-tuples of the old states. This new DFA runs $n$ copies of the old DFA in parallel.

If the language is not circular then there is a word $w$ such that if we run it through the DFA repeatedly, starting with the initial state $s_0$, then we get states $s_1,\ldots,s_n$ such that $s_1$ is accepting but one of the other ones is not accepting (if all of them are accepting then then the sequence $s_0,\ldots,s_n$ must cycle so that $w^*$ is always in the language). In other words, we have a path from $s_0,\ldots,s_{n-1}$ to $s_1,\ldots,s_n$ where $s_1$ is accepting but one of the others is not accepting. Conversely, if the language is circular then that cannot happen.

So we've reduced the problem to a simple directed reachability test (just check all possible "bad" $n$-tuples).



The problem of circularity is coNP-hard. Suppose we're given a 3SAT instance with $n$ variables $\vec{x}$ and $m$ clauses $C_1,\ldots,C_m$. We can assume that $n = m$ (add dummy variables) and that $n$ is prime (otherwise find a prime between $n$ and $2n$ using AKS primality testing, and add dummy variables and clauses).

Consider the following language: "the input is not of the form $\vec{x}_1 \cdots \vec{x}_n$ where $\vec{x}_i$ is a satisfying assignment for $C_i$". It is easy to construct an $O(n^2)$ DFA for this language. If the language is not circular then there is a word $w$ in the language, some power of which is not in the language. Since the only words not in the language have length $n^2$, $w$ must be of length $1$ or $n$. If it is of length $1$, consider $w^n$ instead (it is still in the language), so that $w$ is in the language and $w^n$ is not in the language. The fact that $w^n$ is not in the language means that $w$ is a satisfying assignment.

Conversely, any satisfying assignment translates to a word proving the non-circularity of the language: the satisfying assignment $w$ belongs to the language but $w^n$ does not. Thus the language is circular iff the 3SAT instance is unsatisfiable.



In this part, we discuss a normal form for circular languages. Consider some DFA for a circular language $L$. A sequence $C = C_0,\ldots$ is real if $C_0 = s$ (the initial state), all other states are accepting, and $C_i = C_j$ implies $C_{i+1} = C_{j+1}$. Thus every real sequence is eventually periodic, and there are only finitely many real sequences (since the DFA has finitely many states).

We say that a word behaves according to $C$ if the word takes the DFA from state $c_i$ to state $c_{i+1}$, for all $i$. The set of all such words $E(C)$ is regular (the argument is similar to the first part of this answer). Note that $E(C)$ is a subset of $L$.

Given a real sequence $C$, define $C^k$ to be the sequence $C^k(t) = C(kt)$. The sequence $C^k$ is also real. Since there are only finitely many different sequences $C^k$, the language $D(C)$ which is the union of all $E(C^k)$ is also regular.

We claim that $D(C)$ has the property that if $x,y \in D(C)$ then $xy \in D(C)$. Indeed, suppose that $x \in C^k$ and $y \in C^l$. Then $xy \in C^{k+l}$. Thus $D(C) = D(C)^+$ can be written in the form $r^+$ for some regular expression $r$.

Every word $w$ in the language corresponds to some real sequence $C$, i.e. there exists a real sequence $C$ that $w$ behaves according to. Thus $L$ is the union of $D(C)$ over all real sequence $C$. Therefore every circular language has a representation of the form $\sum r_i^+$. Conversely, every such language is circular (trivially).



Consider the circular language $L$ of all words over $a,b$ that contain either an even number or $a$'s or an even number of $b$'s (or both). We show that it cannot be written as a disjoint sum $\sum r_i^+$; by "disjoint" we mean that $r_i^+ \cap r_j^+ = \varnothing$.

Let $N_i$ be the size of the some DFA for $r_i^+$, and $N &gt; \max N_i$ be some odd integer. Consider $x = a^N b^{N!}$. Since $x \in L$, $x \in r_i^+$ for some $i$. By the pumping lemma, we can pump a prefix of $x$ of length at most $N$. Thus $r_i^+$ generates $z = a^{N!} b^{N!}$. Similarly, $y = a^{N!} b^N$ is generated by some $r_j^+$, which also generates $z$. Note that $i \neq j$ since $xy \notin L$. Thus the representation cannot be disjoint.