Most Complex Deterministic Union-Free Regular Languages^††thanks: This work was supported by the Natural Sciences and Engineering Research Council of Canada grant No. OGP0000871.

Formal definitions are postponed until Section 2.

The class of regular languages over a finite alphabet $\Sigma$ is the smallest class of languages containing the empty language $\emptyset$, the language $\{\varepsilon\}$, where $\varepsilon$ is the empty word, and the letter languages $\{a\}$ for each $a\in\Sigma$, and closed under the operations of union, concatenation, and (Kleene) star. Hence each regular language can be written as a finite expression involving the above basic languages and operations. An expression defining a regular language in this way is called a regular expression. Because regular languages are also closed under complementation, we may also consider regular expressions that allow complementation, which are called extended regular expressions. In this paper we deal exclusively with regular languages.

A natural question is: what kind of languages are defined if one of the operations in the definitions given above is missing? If the star operation is removed from the extended regular expressions we get the well known star-free languages [8, 18, 23], which have been extensively studied. Less attention was given to classes defined by removing an operation from ordinary regular expressions, but recently language classes defined without union or concatenation have been studied.

If we remove some operations from regular expressions, we obtain the following classes of languages:

Union only

subsets of $\{\varepsilon\}\cup\Sigma$.

Concatenation only

$\emptyset$ and $\{w\}$ for each $w\in\Sigma^{*}$.

Star only

$\emptyset$, $\{\varepsilon\}$, $\{a\}$ for each $a\in\Sigma$, and $\{a\}^{*}$ for each $a\in\Sigma$.

Union and Concatenation

Finite languages.

Concatenation and Star

These are the union-free languages that constitute the main topic of this paper.

Union and Star

These are the concatenation-free languages that were studied in [11, 13, 17].

Union-free regular languages were first considered by Brzozowski [3] in 1962 under the name star-dot regular languages, where dot stands for concatenation. He proved that every regular language is a union of union-free languages [3, p. 216, Theorem 9.5]¹¹1Terminology changed to that of the present paper.. Much more recently, in 2001, Crvenković, Dolinka and Ésik [11] studied equations satisfied by union-free regular languages, and proved that the class of these languages cannot be axiomatized by a finite set of equations. This is also known to be true for the class of all regular languages. In 2006 Nagy studied union-free languages in detail and characterized them in terms of nondeterministic finite automata (NFAs) recognizing them [19], which he called one-cycle-free-path NFAs. In 2009 minimal union-free decompositions of regular languages were studied in [1] by Afonin and Golomazov. They also presented a new algorithm for deciding whether a given deterministic finite automaton (DFA) accepts a union-free language. Decompositions of regular languages in terms of union-free languages were further studied by Nagy in 2010 [20]. The state complexities of operations on union-free languages were examined in 2011 by Jirásková and Masopust [15], who proved that the state complexities of basic operations on these languages are the same as those in the class of all regular languages. It was shown in [15] that the class of languages defined by DFAs with the one-cycle-free-path property is a proper subclass of that defined by one-cycle-free-path NFAs; the former class is called the class of deterministic union-free languages. In 2012 Jirásková and Nagy [16] proved that the class of finite unions of deterministic union-free languages is a proper subclass of the class of regular languages. They also showed that every deterministic union-free language is accepted by a special kind of a one-cycle-free-path DFA called a balloon DFA. A summary of the properties of union-free languages was presented in 2017 in [13].

Let $L$ be a regular language. We define the alphabet of $L$ to be the set of letters which appear at least once in a word of $L$. For example, consider the language $L=\{a,ab,ac\}$ and the subset $K=\{a,ac\}$; we say $L$ has alphabet $\{a,b,c\}$ and $K$ has alphabet $\{a,c\}$.

A deterministic finite automaton (DFA) is a quintuple ${\mathcal{D}}=(Q,\Sigma,\delta,q_{0},F)$, where $Q$ is a finite non-empty set of states, $\Sigma$ is a finite non-empty alphabet, $\delta\colon Q\times\Sigma\to Q$ is the transition function, $q_{0}\in Q$ is the initial state, and $F\subseteq Q$ is the set of final states. We extend $\delta$ to functions $\delta\colon Q\times\Sigma^{*}\to Q$ and $\delta\colon 2^{Q}\times\Sigma^{*}\to 2^{Q}$ as usual (where $2^{Q}$ denotes the set of all subsets of $Q$). A DFA ${\mathcal{D}}$ accepts a word $w\in\Sigma^{*}$ if ${\delta}(q_{0},w)\in F$. The language accepted by ${\mathcal{D}}$ is the set of all words accepted by ${\mathcal{D}}$, and is denoted by $L({\mathcal{D}})$. If $q$ is a state of ${\mathcal{D}}$, then the language $L_{q}({\mathcal{D}})$ of $q$ is the language accepted by the DFA $(Q,\Sigma,\delta,q,F)$. A state is empty (or dead or a sink state) if its language is empty. Two states $p$ and $q$ of ${\mathcal{D}}$ are equivalent if $L_{p}({\mathcal{D}})=L_{q}({\mathcal{D}})$. A state $q$ is reachable if there exists $w\in\Sigma^{*}$ such that $\delta(q_{0},w)=q$. A DFA ${\mathcal{D}}$ is minimal if it has the smallest number of states among all DFAs accepting $L({\mathcal{D}})$. We say a DFA has a minimal alphabet if its alphabet is equal to the alphabet of $L({\mathcal{D}})$. It is well known that a DFA with a minimal alphabet is minimal if and only if all of its states are reachable and no two states are equivalent.

A nondeterministic finite automaton (NFA) is a quintuple $\mathcal{N}=(Q,\Sigma,\delta,I,F)$, where $Q$, $\Sigma$ and $F$ are as in a DFA, $\delta\colon Q\times\Sigma\to 2^{Q}$, and $I\subseteq Q$ is the set of initial states. Each triple $(p,a,q)$ with $p,q\in Q$, $a\in\Sigma$ is a transition if $q\in\delta(p,a)$. A sequence $((p_{0},a_{0},q_{0}),(p_{1},a_{1},q_{1}),\dots,(p_{k-1},a_{k-1},q_{k-1}))$ of transitions, where $p_{i+1}=q_{i}$ for $i=0,\dots,k-2$ is a path in ${\mathcal{N}}$. The word $a_{0}a_{1}\cdots a_{k-1}$ is the word spelled by the path. A word $w$ is accepted by ${\mathcal{N}}$ is there exists a path with $p_{0}\in I$ and $q_{k-1}\in F$ that spells $w$. If $q\in\delta(p,a)$ we also use the notation $p\xrightarrow{a}q$. We extend this notation also to words, and write $p\xrightarrow{w}q$ for $w\in\Sigma^{*}$.

The state complexity of a regular language $L$, denoted by $\kappa(L)$, is the number of states in the minimal DFA accepting $L$. Henceforth we frequently refer to state complexity as simply complexity, and we denote a language of complexity $n$ by $L_{n}$, and a DFA with $n$ states by ${\mathcal{D}}_{n}$.

The state complexity of a regularity-preserving unary operation $\circ$ on regular languages is the maximal value of $\kappa(L^{\circ})$, expressed as a function of one parameter $n$, where $L$ varies over all regular languages with complexity at most $n$. For example, the state complexity of the reversal operation is $2^{n}$; it is known that if $L$ has complexity at most $n$, then $\kappa(L^{R})\leqslant 2^{n}$, and furthermore this upper bound is tight in the sense that for each $n\geqslant 1$ there exists a language $L_{n}$ such that $\kappa(L_{n}^{R})=2^{n}$. In general, to show that an upper bound on $\kappa(L^{\circ})$ is tight, we need to exhibit a sequence $(L_{n}\mid n\geqslant k)=(L_{k},L_{k+1},\dots)$, called a stream, of languages of each complexity $n\geqslant k$ (for some small constant $k$) that meet this upper bound. Often we are not interested in the special-case behaviour of the operation that may occur at very small values of $n$; the parameter $k$ allows us to ignore these small values and simplify the statements of results.

The state complexity of a regularity-preserving binary operation $\circ$ on regular languages is the maximal value of $\kappa(L^{\prime}\circ L)$, epxressed as a function of two parameters $m$ and $n$, where $L^{\prime}$ varies over all regular languages of complexity at most $m$ and $L$ varies over all regular languages of complexity at most $n$. In this case, to show an upper bound on the state complexity is tight, we need to exhibit two classes $(L^{\prime}_{m,n}\mid m\geqslant h,n\geqslant k)$ and $(L_{m,n}\mid m\geqslant h,n\geqslant k)$ of languages meeting the bound; the notation $L^{\prime}_{m,n}$ and $L_{m,n}$ implies that $L^{\prime}_{m,n}$ and $L_{m,n}$ depend on both $m$ and $n$. However, in most cases studied in the literature, it is enough to use witness streams $(L^{\prime}_{m}\mid m\geqslant h)$ and $(L_{n}\mid n\geqslant k)$, where $L^{\prime}_{m}$ is independent of $n$ and $L_{n}$ is independent of $m$.

For binary operations we consider two types of state complexity: restricted and unrestricted state complexity. For restricted state complexity the operands of the binary operations are required to have the same alphabet. For unrestricted state complexity the alphabets of the operands may differ. See [7] for more details.

Sometimes the same stream can be used for both operands of a binary operation, but this is not always possible. For example, for boolean operations when $m=n$, the state complexity of $L_{n}\cup L_{n}=L_{n}$ is $n$, whereas the upper bound is $mn=n^{2}$. However, in many cases the second language is a "dialect" of the first, that is, it “differs only slightly” from the first. A dialect of $L_{n}(\Sigma)$ is a language obtained from $L_{n}(\Sigma)$ by deleting some letters of $\Sigma$ in the words of $L_{n}(\Sigma)$ – by this we mean that words containing these letters are deleted – or replacing them by letters of another alphabet $\Sigma^{\prime}$. In this paper we consider only the cases where $\Sigma=\Sigma^{\prime}$, and we encounter only two types of dialects:

1. 1.

   A dialect in which some letters were deleted; for example, $L_{n}(a,b)$ is a dialect of $L_{n}(a,b,c)$ with $c$ deleted, and $L_{n}(a,-,c)$ is a dialect with $b$ deleted.

2. 2.

   A dialect in which the roles of two letters are exchanged; for example, $L_{n}(b,a)$ is such a dialect of $L_{n}(a,b)$.

These two types of dialects can be combined, for example, in $L_{n}(a,-,b)$ the letter $c$ is deleted, and $b$ plays the role that $c$ played originally. The notion of dialects also extends to DFAs; for example, if ${\mathcal{D}}_{n}(a,b,c)$ recognizes $L_{n}(a,b,c)$ then ${\mathcal{D}}_{n}(a,-,b)$ recognizes the dialect $L_{n}(a,-,b)$.

We use $Q_{n}=\{0,\dots,n-1\}$ as our basic set with $n$ elements. A transformation of $Q_{n}$ is a mapping $t\colon Q_{n}\to Q_{n}$. The image of $q\in Q_{n}$ under $t$ is denoted by $qt$, and this notation is extended to subsets of $Q_{n}$. The preimage of $q\in Q_{n}$ under $t$ the set $qt^{-1}=\{p\in Q_{n}:pt=q\}$, and this notation is extended to subsets of $Q_{n}$ as follows: $St^{-1}=\{p\in Q_{n}:pt\in S\}$. The rank of a transformation $t$ is the cardinality of $Q_{n}t$. If $s$ and $t$ are transformations of $Q_{n}$, their composition is denoted $st$ and we have $q(st)=(qs)t$ for $q\in Q_{n}$. The $k$-fold composition $tt\dotsb t$ (with $k$ occurences of $t$) is denoted $t^{k}$, and for $S\subseteq Q_{n}$ we define $St^{-k}=S(t^{k})^{-1}$. Let $\mathcal{T}_{Q_{n}}$ be the set of all $n^{n}$ transformations of $Q_{n}$; then $\mathcal{T}_{Q_{n}}$ is a monoid under composition.

For $k\geqslant 2$, a transformation $t$ of a set $P=\{q_{0},q_{1},\ldots,q_{k-1}\}\subseteq Q_{n}$ is a $k$-cycle if $q_{0}t=q_{1},q_{1}t=q_{2},\ldots,q_{k-2}t=q_{k-1},q_{k-1}t=q_{0}$. This $k$-cycle is denoted by $(q_{0},q_{1},\ldots,q_{k-1})$, and leaves the states in $Q_{n}\setminus P$ unchanged. A 2-cycle $(q_{0},q_{1})$ is called a transposition. A transformation that sends state $p$ to $q$ and acts as the identity on the remaining states is denoted by $(p\to q)$. The identity transformation is denoted by 1.

Let $\mathcal{D}=(Q_{n},\Sigma,\delta,0,F)$ be a DFA. For each word $w\in\Sigma^{*}$, the transition function induces a transformation $\delta_{w}$ of $Q_{n}$ by $w$: for all $q\in Q_{n}$, $q\delta_{w}=\delta(q,w).$ The set $T_{\mathcal{D}}$ of all such transformations by non-empty words is the transition semigroup of $\mathcal{D}$ under composition. Often we use the word $w$ to denote the transformation $t$ it induces; thus we write $qw$ instead of $q\delta_{w}$. We also write $w\colon t$ to mean that $w$ induces the transformation $t$.

The size of the syntactic semigroup of a regular language is another measure of the complexity of the language [4]. Write $\Sigma^{+}$ for $\Sigma^{*}\setminus\{\varepsilon\}$. The syntactic congruence of a language $L\subseteq\Sigma^{*}$ is defined on $\Sigma^{+}$ as follows: For $x,y\in\Sigma^{+},x\,{\mathbin{\approx_{L}}}\,y$ if and only if $wxz\in L\Leftrightarrow wyz\in L$ for all $w,z\in\Sigma^{*}.$ The quotient set $\Sigma^{+}/{\mathbin{\approx_{L}}}$ of equivalence classes of ${\mathbin{\approx_{L}}}$ is a semigroup, the syntactic semigroup $T_{L}$ of $L$. The syntactic semigroup is isomorphic to the transition semigroup of the minimal DFA of $L$ [21].

The (left) quotient of $L\subseteq\Sigma^{*}$ by a word $w\in\Sigma^{*}$ is the language $w^{-1}L=\{x:wx\in L\}$. It is well known that the number of quotients of a regular language is finite and equal to the state complexity of the language.

The atoms of a regular language are defined by a left congruence, where two words $x$ and $y$ are congruent whenever $ux\in L$ if and only if $uy\in L$ for all $u\in\Sigma^{*}$. Thus $x$ and $y$ are congruent whenever $x\in u^{-1}L$ if and only if $y\in u^{-1}L$ for all $u\in\Sigma^{*}$. An equivalence class of this relation is an atom of $L$ [10]. Atoms can be expressed as non-empty intersections of complemented and uncomplemented quotients of $L$. The number of atoms and their state complexities were suggested as measures of complexity of regular languages [4] because all quotients of a language and all quotients of its atoms are unions of atoms [9, 10, 14].

The automata described in [19] that characterize union-free languages are called there one-cycle-free-path automata. They are defined by the property that there is only one final state and a unique cycle-free path from each state to the final state. We are now ready to define a most complex deterministic one-cycle-free-path DFA and its most complex deterministic union-free language.

The most complex stream below meets all of our complexity bounds. However, our witness uses three letters for restricted product whereas [15] uses binary witnesses. The same shortcoming of most complex streams occurs in the case of regular languages [4]; that seems to be the price of getting a witness for all operations rather than minimizing the alphabet for each operation.

The DFA of Definition 1 bears some similarities to the DFA for reversal in Fig. 6 in [15, p. 1650]. It is evident that it is a one-cycle-free-path DFA. Let $E=(a(b\cup c\cup d)^{*})^{n-2}a$. One verifies that

Noting that $(E_{1}\cup E_{2}\cup\cdots\cup E_{k})^{*}=(E_{1}^{*}E_{2}^{*}\cdots E_{k}^{*})^{*}$ for all regular expressions $E_{i}$, $i=1,\dots,k$, we obtain a union-free expression for $L_{n}$.

We have exhibited a single ternary language stream that is a witness for the maximal state complexities of star and reversal of union-free languages. Together with some dialects it also constitutes a witness for union, intersection, difference, symmetric difference, and product in case the alphabets of the two operands are the same. As was shown in [15] these bounds are the same as those for regular languages. We prove that our witness also has the largest syntactic semigroup and most complex atoms, and that these complexities are again the same as those for arbitrary regular languages. By adding a fourth input inducing the identity transformation to our witness we obtain witnesses for unrestricted binary operations, where the alphabets of the operands are not the same. The bounds here are again the same as those for regular languages. In summary, this shows that the complexity measures proposed in [4] do not distinguish union-free languages from regular languages.