# Complexity of Proper Prefix-Convex Regular Languages

Conference paper
Part of the Lecture Notes in Computer Science book series (LNCS, volume 10329)

## Abstract

A language L over an alphabet $$\varSigma$$ is prefix-convex if, for any words $$x,y,z\in \varSigma ^*$$, whenever x and xyz are in L, then so is xy. Prefix-convex languages include right-ideal, prefix-closed, and prefix-free languages, which were studied elsewhere. Here we concentrate on prefix-convex languages that do not belong to any one of these classes; we call such languages proper. We exhibit most complex proper prefix-convex languages, which meet the bounds for the size of the syntactic semigroup, reversal, complexity of atoms, star, product, and Boolean operations.

### Keywords

Atom Most complex Prefix-convex Proper Quotient complexity Regular language State complexity Syntactic semigroup

## 1 Introduction

Prefix-Convex Languages. We examine the complexity properties of a class of regular languages that has never been studied before: the class of proper prefix-convex languages [7]. Let $$\varSigma$$ be a finite alphabet; if $$w=xy$$, for $$x,y\in \varSigma ^*$$, then x is a prefix of w. A language $$L\subseteq \varSigma ^*$$ is prefix-convex [1, 16] if whenever x and xyz are in L, then so is xy. Prefix-convex languages include three special cases:
1. 1.

A language $$L\subseteq \varSigma$$ is a right ideal if it is non-empty and satisfies $$L=L\varSigma ^*$$. Right ideals appear in pattern matching [11]: $$L\varSigma ^*$$ is the set of all words in some text (word in $$\varSigma ^*$$) beginning with words in L.

2. 2.

A language is prefix-closed [6] if whenever w is in L, then so is every prefix of w. The set of allowed sequences to any system is prefix-closed. Every prefix-closed language other than $$\varSigma ^*$$ is the complement of a right ideal [1].

3. 3.

A language is prefix-free if $$w\in L$$ implies that no prefix of w other than w is in L. Prefix-free languages other than $$\{\varepsilon \}$$, where $$\varepsilon$$ is the empty word, are prefix codes and are of considerable importance in coding theory [2].

The complexities of these three special prefix-convex languages were studied in [8]. We now turn to the “real” prefix-convex languages that do not belong to any of the three special classes.

Omitted proofs can be found in [7].

Complexities of Operations. If $$L\subseteq \varSigma ^*$$ is a language, the (left) quotient of L by a word $$w\in \varSigma ^*$$ is $$w^{-1}L=\{x\mid wx\in L\}$$. A language is regular if and only if it has a finite number of distinct quotients. So the number of quotients of L, the quotient complexity [3] $$\kappa (L)$$ of L, is a natural measure of complexity for L. An equivalent concept is the state complexity [15, 17, 18] of L, which is the number of states in a complete minimal deterministic finite automaton (DFA) over $$\varSigma$$ recognizing L. We refer to quotient/state complexity simply as complexity.

If $$L_n$$ is a regular language of complexity n, and $$\circ$$ is a unary operation, the complexity of$$\circ$$ is the maximal value of $$\kappa (L_n^\circ )$$, expressed as a function of n, as $$L_n$$ ranges over all languages of complexity n. If $$L'_m$$ and $$L_n$$ are regular languages of complexities m and n respectively, and $$\circ$$ is a binary operation, the complexity of$$\circ$$ is the maximal value of $$\kappa (L'_m \circ L_n)$$, expressed as a function of m and n, as $$L'_m$$ and $$L_n$$ range over all languages of complexities m and n. The complexity of an operation is a lower bound on its time and space complexities. The operations reversal, (Kleene) star, product (concatenation), and binary boolean operations are considered “common”, and their complexities are known; see [4, 17, 18].

Witnesses. To find the complexity of a unary operation we find an upper bound on this complexity, and languages that meet this bound. We require a language $$L_n$$ for each n, that is, a sequence, $$(L_k, L_{k+1}, \dots )$$, called a stream of languages, where k is a small integer, because the bound may not hold for small values of n. For a binary operation we need two streams. The same stream cannot always be used for both operands, but for all common binary operations the second stream can be a “dialect” of the first, that is it can “differ only slightly” from the first [4]. Let $$\varSigma =\{a_1,\dots ,a_k\}$$ be an alphabet ordered as shown; if $$L\subseteq \varSigma ^*$$, we denote it by $$L(a_1,\dots ,a_k)$$. A dialect of L is obtained by deleting letters of $$\varSigma$$ in the words of L, or replacing them by letters of another alphabet $$\varSigma '$$. More precisely, for an injective partial map $$\pi :\varSigma \mapsto \varSigma '$$, we get a dialect of L by replacing each letter $$a \in \varSigma$$ by $$\pi (a)$$ in every word of L, or deleting the word if $$\pi (a)$$ is undefined. We write $$L(\pi (a_1),\dots , \pi (a_k))$$ to denote the dialect of $$L(a_1,\dots ,a_k)$$ given by $$\pi$$, and we denote undefined values of $$\pi$$ by “−”. Undefined values for letters at the end of the alphabet are omitted; for example, $$L(a,c,-,-)$$ is written as L(ac). Our definition of dialect is more general than that of [5], where only the case $$\varSigma '=\varSigma$$ was allowed.

Finite Automata. A deterministic finite automaton (DFA) is a quintuple $${{\mathcal {D}}}=(Q, \varSigma , \delta , q_0,F)$$, where Q is a finite non-empty set of states, $$\varSigma$$ is a finite non-empty alphabet, $$\delta :Q\times \varSigma \rightarrow Q$$ 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 a function $$\delta :Q\times \varSigma ^*\rightarrow Q$$ as usual. A DFA $${{\mathcal {D}}}$$accepts a word $$w \in \varSigma ^*$$ if $${\delta }(q_0,w)\in F$$. The set of all words accepted by $${{\mathcal {D}}}$$ is the language of$${{\mathcal {D}}}$$. If $$q\in Q$$, then the language$$L_q$$ofq is the language accepted by the DFA $$(Q,\varSigma ,\delta ,q,F)$$. A state is empty or dead or a sink if its language is empty. Two states p and q of $${{\mathcal {D}}}$$ are equivalent if $$L_p = L_q$$. A state q is reachable if there exists $$w\in \varSigma ^*$$ such that $$\delta (q_0,w)=q$$. A DFA is minimal if all of its states are reachable and no two states are equivalent. A nondeterministic finite automaton (NFA) is a quintuple $${{\mathcal {D}}}=(Q, \varSigma , \delta , I,F)$$, where Q, $$\varSigma$$ and F are defined as in a DFA, $$\delta :Q\times \varSigma \rightarrow 2^Q$$ is the transition function, and $$I\subseteq Q$$ is the set of initial states. An $$\varepsilon$$-NFA is an NFA in which transitions under the empty word $$\varepsilon$$ are also permitted.

Transformations. We use $$Q_n=\{0,\dots ,n-1\}$$ as the set of states of every DFA with n states. A transformation of $$Q_n$$ is a mapping $$t:Q_n\rightarrow Q_n$$. The image of $$q\in Q_n$$ under t is qt. In any DFA, each letter $$a\in \varSigma$$ induces a transformation $$\delta _a$$ of the set $$Q_n$$ defined by $$q\delta _a=\delta (q,a)$$; we denote this by $$a:\delta _a$$. Often we use the letter a to denote the transformation it induces; thus we write qa instead of $$q\delta _a$$. We extend the notation to sets: if $$P\subseteq Q_n$$, then $$Pa=\{pa\mid p\in P\}$$. We also write $$P\mathop {\longrightarrow }\limits ^{a} Pa$$ to indicate that the image of P under a is Pa. If st are transformations of $$Q_n$$, their composition is (qs)t.

For $$k\geqslant 2$$, a transformation (permutation) t of a set $$P=\{q_0,q_1,\ldots ,q_{k-1}\} \subseteq Q_{n}$$ is a k-cycle if $$q_0t=q_1, q_1t=q_2,\ldots ,q_{k-2}t=q_{k-1},q_{k-1}t=q_0$$. This k-cycle is denoted by $$(q_0,q_1,\ldots ,q_{k-1})$$. A 2-cycle $$(q_0,q_1)$$ is called a transposition. A transformation that sends all the states of P to q and acts as the identity on the other states is denoted by $$(P \rightarrow q)$$, and $$(Q_n\rightarrow p)$$ is called a constant transformation. If $$P=\{p\}$$ we write $$(p\rightarrow q)$$ for $$(\{p\} \rightarrow q)$$. The identity transformation is denoted by $$\mathbbm {1}$$. Also, $$(_i^j \; q\rightarrow q+1)$$ is a transformation that sends q to $$q+1$$ for $$i\leqslant q\leqslant j$$ and is the identity for the remaining states; $$(_i^j \; q\rightarrow q-1)$$ is defined similarly.

Semigroups. The syntactic congruence of $$L\subseteq \varSigma ^*$$ is defined on $$\varSigma ^+$$: For $$x, y \in \varSigma ^+$$, $$x\, {\mathbin {\approx _L}}\, y \text { if and only if } wxz\in L \Leftrightarrow wyz\in L\text { for all } w,z \in \varSigma ^*$$. The quotient set $$\varSigma ^+/ {\mathbin {\approx _L}}$$ of equivalence classes of $${\mathbin {\approx _L}}$$ is the syntactic semigroup of L. Let $${{\mathcal {D}}}_n = (Q_n, \varSigma , \delta , q_0, F)$$ be a DFA, and let $$L_n=L({{\mathcal {D}}}_n)$$. For each word $$w \in \varSigma ^*$$, 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 $$T_{{{\mathcal {D}}}_n}$$ of all such transformations by non-empty words is a semigroup under composition called the transition semigroup of $${{\mathcal {D}}}_n$$. If $${{\mathcal {D}}}_n$$ is a minimal DFA of $$L_n$$, then $$T_{{{\mathcal {D}}}_n}$$ is isomorphic to the syntactic semigroup $$T_{L_n}$$ of $$L_n$$, and we represent elements of $$T_{L_n}$$ by transformations in $$T_{{{\mathcal {D}}}_n}$$. The size of the syntactic semigroup has been used as a measure of complexity for regular languages [4, 10, 12, 14].

Atoms. are defined by a left congruence, where two words x and y are equivalent if $$ux\in L$$ if and only if $$uy\in L$$ for all $$u\in \varSigma ^*$$. Thus x and y are equivalent if $$x\in u^{-1}L$$ if and only if $$y\in u^{-1}L$$. An equivalence class of this relation is an atom of L [9, 13].

One can conclude that an atom is a non-empty intersection of complemented and uncomplemented quotients of L. That is, every atom of a language with quotients $$K_0, K_1, \dots , K_{n-1}$$ can be written as $$A_S = \bigcap _{i \in S} K_i \cap \bigcap _{i \in \overline{S}} \overline{K_i}$$ for some set $$S \subseteq Q_n$$. The number of atoms and their complexities were suggested as possible measures of complexity [4], because all the quotients of a language and the quotients of its atoms are unions of atoms [9].

Most Complex Regular Stream. The stream $$({{\mathcal {D}}}_n(a,b,c) \mid n\geqslant 3)$$ of Definition 1 and Fig. 1 will be used as a component in the class of proper prefix-convex languages. This stream together with some dialects meets the complexity bounds for reversal, star, product, and all binary boolean operations [7, 8]. Moreover, it has the maximal syntactic semigroup and most complex atoms, making it a most complex regular stream.

### Definition 1

For $$n\geqslant 3$$, let $${{\mathcal {D}}}_n={{\mathcal {D}}}_n(a,b,c)=(Q_n,\varSigma ,\delta _n, 0, \{n-1\})$$, where $$\varSigma =\{a,b,c\}$$, and $$\delta _n$$ is defined by $$a:(0,\dots ,n-1)$$, $$b:(0,1)$$, $$c:(1 \rightarrow 0)$$.

Most complex streams are useful in systems dealing with regular languages and finite automata. To know the maximal sizes of automata that can be handled by a system it suffices to use the most complex stream to test all the operations.

## 2 Proper Prefix-Convex Languages

We begin with some properties of prefix-convex languages that will be used frequently in this section. The following lemma and propositions characterize the classes of prefix-convex languages in terms of their minimal DFAs.

### Lemma 1

Let L be a prefix-convex language over $$\varSigma$$. Either L is a right ideal or L has an empty quotient.

### Proposition 1

Let $$L_n$$ be a regular language of complexity n, and let $${{\mathcal {D}}}_n = (Q_n, \varSigma , \delta , 0, F)$$ be a minimal DFA recognizing $$L_n$$. The following are equivalent:
1. 1.

$$L_n$$ is prefix-convex.

2. 2.

For all $$p,q,r \in Q_n$$, if p and r are final, q is reachable from p, and r is reachable from q, then q is final.

3. 3.

Every state reachable in $${{\mathcal {D}}}_n$$ from any final state is either final or empty.

### Proposition 2

Let $$L_n$$ be a non-empty prefix-convex language of complexity n, and let $${{\mathcal {D}}}_n = (Q_n, \varSigma , \delta , 0, F)$$ be a minimal DFA recognizing $$L_n$$.

1. 1.

$$L_n$$ is prefix-closed if and only if $$0 \in F$$.

2. 2.

$$L_n$$ is prefix-free if and only if $${{\mathcal {D}}}_n$$ has a unique final state p and an empty state $$p'$$ such that $$\delta (p, a) = p'$$ for all $$a \in \varSigma$$.

3. 3.

$$L_n$$ is a right ideal if and only if $${{\mathcal {D}}}_n$$ has a unique final state p and $$\delta (p, a) = p$$ for all $$a \in \varSigma$$.

A prefix-convex language L is proper if it is not a right ideal and it is neither prefix-closed nor prefix-free. We say it is k-proper if it has k final states, $$1\leqslant k\leqslant n-2$$. Every minimal DFA for a k-proper language with complexity n has the same general structure: there are $$n-1-k$$ non-final, non-empty states, k final states, and one empty state. Every letter fixes the empty state and, by Proposition 1, no letter sends a final state to a non-final, non-empty state.

Next we define a stream of k-proper DFAs and languages, which we will show to be most complex.

### Definition 2

For $$n \geqslant 3$$, $$1 \leqslant k \leqslant n-2$$, let $${\mathcal {D}}_{n,k}(\varSigma ) = (Q_n, \varSigma , \delta _{n,k}, 0, F_{n,k})$$ where $$\varSigma = \{a, b, c_1, c_2, d_1,d_2,e\}$$, $$F_{n,k} = \{n-1-k, \dots , n-2\}$$, and $$\delta _{n,k}$$ is given by the transformations
\begin{aligned} a&:{\left\{ \begin{array}{ll} (1, \dots , n-2-k)(n-1-k, n-k), &{} \mathrm {\,\, if\,\, } n-1-k \mathrm {\,\, is\,\, even\,\, and\,\, } k \geqslant 2; \\ (0, \dots , n-2-k)(n-1-k, n-k), &{}\mathrm {\,\,if\,\, } n-1-k \mathrm {\,\, is\,\, odd\,\, and\,\,} k \geqslant 2; \\ (1, \dots , n-2-k), &{}\mathrm {\,\,if\,\,} n-1-k \mathrm {\,\,is\,\,even\,\,and\,\,} k = 1; \\ (0, \dots , n-2-k), &{} \mathrm {\,\, if\,\, } n-1-k \mathrm {\,\,is\,\,odd\,\,and\,\,} k = 1. \\ \end{array}\right. }\\ b&:{\left\{ \begin{array}{ll} (n-k, \dots , n-2)(0, 1), &{} \mathrm {\,\, if\,\, } k \mathrm {\,\,is\,\,even\,\,and\,\,} n-1-k \geqslant 2; \\ (n-1-k, \dots , n-2)(0, 1), &{} \mathrm {\,\, if\,\, } k \mathrm {\,\,is\,\,odd\,\,and\,\,} n-1-k \geqslant 2; \\ (n-k, \dots , n-2), &{} \mathrm {\,\, if\,\, }k \mathrm {\,\,is\,\,even\,\,and\,\,} n-1-k = 1; \\ (n-1-k, \dots , n-2), &{} \mathrm {\,\, if\,\, } k \mathrm {\,\,is\,\,odd\,\,and\,\,} n-1-k = 1. \\ \end{array}\right. }\\ c_1&:{\left\{ \begin{array}{ll}(1 \rightarrow 0), &{} \mathrm {\,\, if\,\, } n-1-k \geqslant 2; \\ \quad ~ \mathbbm {1}, &{} \mathrm {\,\, if\,\, } n-1-k = 1. \end{array}\right. }\\ c_2&:{\left\{ \begin{array}{ll}(n-k \rightarrow n-1-k), &{} \mathrm {\,\, if\,\, } k \geqslant 2; \\ \quad \quad \quad ~ \mathbbm {1}, &{}\mathrm {\,\, if\,\, } k = 1. \end{array}\right. } \\ d_1&:(n-2-k \rightarrow n-1)(_0^{n-3-k} \;\; q\rightarrow q+1).\\ d_2&:(_{n-1-k}^{n-2} \;\; q\rightarrow q+1).\\ e&:(0 \rightarrow n-1-k). \end{aligned}
Also, let $$E_{n,k} = \{0, \dots , n-2-k\}$$; it is useful to partition $$Q_n$$ into $$E_{n,k}$$, $$F_{n,k}$$, and $$\{n-1\}$$. Letters a and b have complementary behaviours on $$E_{n,k}$$ and $$F_{n,k}$$, depending on the parities of n and k. Letters $$c_1$$ and $$d_1$$ act on $$E_{n,k}$$ in exactly the same way as $$c_2$$ and $$d_2$$ act on $$F_{n,k}$$. In addition, $$d_1$$ and $$d_2$$ send states $$n-2-k$$ and $$n-2$$, respectively, to state $$n-1$$, and letter e connects the two parts of the DFA. The structure of $$\mathcal {D}_n(\varSigma )$$ is shown in Figs. 2 and 3 for certain parities of $$n-1-k$$ and k. Let $$L_{n,k}(\varSigma )$$ be the language recognized by $${\mathcal {D}}_{n,k}(\varSigma )$$.

### Theorem 1

(Proper Prefix-Convex Languages). For $$n\geqslant 3$$ and $$1 \leqslant k \leqslant n-2$$, the DFA $$\mathcal {D}_{n,k}(\varSigma )$$ of Definition 2 is minimal and $$L_{n,k}(\varSigma )$$ is a k-proper language of complexity n. The bounds below are maximal for k-proper prefix-convex languages. At least seven letters are required to meet these bounds.

1. 1.

The syntactic semigroup of $$L_{n,k}(\varSigma )$$ has cardinality $$n^{n-1-k}(k+1)^k$$; the maximal value $$n(n-1)^{n-2}$$ is reached only when $$k=n-2$$.

2. 2.

The non-empty, non-final quotients of $$L_{n,k}(a, b, -, -, -, d_2, e)$$ have complexity n, the final quotients have complexity $$k+1$$, and $$\emptyset$$ has complexity 1.

3. 3.

The reverse of $$L_{n,k}(a,b,-,-,-,d_2,e)$$ has complexity $$2^{n-1}$$; moreover, the language $$L_{n,k}(a,b,-,-,-,d_2,e)$$ has $$2^{n-1}$$ atoms for all k.

4. 4.
For each atom $$A_S$$ of $$L_{n,k}(\varSigma )$$, write $$S = X_1 \cup X_2$$, where $$X_1 \subseteq E_{n,k}$$ and $$X_2 \subseteq F_{n,k}$$. Let $$\overline{X_1} = E_{n,k}\setminus X_1$$ and $$\overline{X_2} = F_{n,k}\setminus X_2$$. If $$X_2 \not = \emptyset$$, then $$\kappa (A_S) =$$
\begin{aligned}1 + \sum _{x_1=0}^{|X_1|}\sum _{x_2=1}^{|X_1| + |X_2| - x_1}\sum _{y_1=0}^{|\overline{X_1}|}\sum _{y_2=0}^{|\overline{X_1}|+|\overline{X_2}| -y_1}\left( {\begin{array}{c}n-1-k\\ x_1\end{array}}\right) \left( {\begin{array}{c}k\\ x_2\end{array}}\right) \left( {\begin{array}{c}n-1-k-x_1\\ y_1\end{array}}\right) \left( {\begin{array}{c}k-x_2\\ y_2\end{array}}\right) .\end{aligned}
If $$X_1 \not = \emptyset$$ and $$X_2 = \emptyset$$, then $$\kappa (A_S) =$$
\begin{aligned} 1 + \sum _{x_1=0}^{|X_1|}\sum _{x_2=0}^{|X_1| - x_1}\sum _{y_1=0}^{|\overline{X_1}|}\sum _{y_2=0}^{k}\left( {\begin{array}{c}n-1-k\\ x_1\end{array}}\right) \left( {\begin{array}{c}k\\ x_2\end{array}}\right) \left( {\begin{array}{c}n-1-k-x_1\\ y_1\end{array}}\right) \left( {\begin{array}{c}k-x_2\\ y_2\end{array}}\right) -2^k\sum _{y=0}^{|\overline{X_1}|}\left( {\begin{array}{c}n-1-k\\ y\end{array}}\right) . \end{aligned}
Otherwise, $$S = \emptyset$$ and $$\kappa (A_S) = 2^{n-1}$$.

5. 5.

The star of $$L_{n,k}(a,b,-,-,d_1,d_2, e)$$ has complexity $$2^{n-2}+2^{n-2-k}+1$$. The maximal value $$2^{n-2}+2^{n-3}+1$$ is reached only when $$k=1$$.

6. 6.

$$L'_{m,j}(a,b,c_1,-, d_1, d_2, e) L_{n,k}(a,d_2, c_1,-,d_1, b, e)$$ has complexity $$m-1-j +j2^{n-2}+2^{n-1}$$. The maximal value $$m 2^{n-2} + 1$$ is reached only when $$j=m-2$$.

7. 7.
For $$m,n\geqslant 3$$, $$1 \leqslant j \leqslant m-2$$, and $$1 \leqslant k \leqslant n-2$$, define the languages $$L'_{m,j} = L'_{m,j}(a, b, c_1, -, d_1, d_2, e)$$ and $$L_{n,k} = L_{n,k}(a, b, e, -, d_2, d_1, c_1)$$. For any proper binary boolean function $$\circ$$, the complexity of $$L'_{m,j} \circ L_{n,k}$$ is maximal. In particular,
1. (a)

$$L'_{m,j} \cup L_{n,k}$$ and $$L'_{m,j} \oplus L_{n,k}$$ have complexity mn.

2. (b)

$$L'_{m,j} \setminus L_{n,k}$$ has complexity $$mn-(n-1)$$.

3. (c)

$$L'_{m,j} \cap L_{n,k}$$ has complexity $$mn-(m+n-2)$$.

### Proof

The remainder of this paper is an outline of the proof of this theorem. The longer parts of the proof are separated into individual propositions and lemmas.

DFA $$\mathcal {D}_{n,k}(a,b,-,-,-,d_2,e)$$ is easily seen to be minimal. Language $$L_{n,k}(\varSigma )$$ is k-proper by Propositions 1 and 2.

1. 1.

See Lemma 2 and Proposition 3.

2. 2.

If the initial state of $$\mathcal {D}_{n,k}(a, b, -, -, -, d_2, e)$$ is changed to $$q \in E_{n,k}$$, the new DFA accepts a quotient of $$L_{n,k}$$ and is still minimal; hence the complexity of that quotient is n. If the initial state is changed to $$q \in F_{n,k}$$ then states in $$E_{n,k}$$ are unreachable, but the DFA on $$\{n-1-k, \dots , n-1\}$$ is minimal; hence the complexity of that quotient is $$k+1$$. The remaining quotient is empty, and hence has complexity 1. By Proposition 1, these are maximal.

3. 3.

See Proposition 4 for the reverse. It was shown in [9] that the number of atoms is equal to the complexity of the reverse.

4. 4.

See [7].

5. 5.

See Proposition 5.

6. 6.

See [7].

7. 7.

By [3, Theorem 2], all boolean operations on regular languages have the upper bound mn, which gives the bound for (a). The bounds for (b) and (c) follow from [3, Theorem 5]. The proof that all these bounds are tight for $$L'_{m,j} \circ L_{n,k}$$ can be found in [7].   $$\square$$

### Lemma 2

Let $$n \geqslant 1$$ and $$1 \leqslant k \leqslant n-2$$. For any permutation t of $$Q_n$$ such that $$E_{n,k}t = E_{n,k}$$, $$F_{n,k}t = F_{n,k}$$, and $$(n-1)t = n-1$$, there is a word $$w \in \{a,b\}^*$$ that induces t on $$\mathcal {D}_{n,k}$$.

### Proof

Only a and b induce permutations of $$Q_n$$; every other letter induces a properly injective map. Furthermore, a and b permute $$E_{n,k}$$ and $$F_{n,k}$$ separately, and both fix $$n-1$$. Hence every $$w \in \{a,b\}^*$$ induces a permutation on $$Q_n$$ such that $$E_{n,k}w = E_{n,k}$$, $$F_{n,k} w = F_{n,k}$$, and $$(n-1)w = n-1$$. Each such permutation naturally corresponds to an element of $$S_{n-1-k} \times S_k$$, where $$S_m$$ denotes the symmetric group on m elements. To be consistent with the DFA, assume $$S_{n-1-k}$$ contains permutations of $$\{0, \dots , n-2-k\}$$ and $$S_k$$ contains permutations of $$\{n-1-k, \dots , n-2\}$$. Let $$s_a$$ and $$s_b$$ denote the group elements corresponding to the transformations induced by a and b respectively. We show that $$s_a$$ and $$s_b$$ generate $$S_{n-1-k} \times S_k$$.

It is well known that $$(0, \dots , m-1)$$, and (0, 1) generate the symmetric group on $$\{0, \dots , m-1\}$$ for any $$m \ge 2$$. Note that $$(1, \dots , m-1)$$ and (0, 1) are also generators, since $$(0, 1) (1, \dots , m-1) = (0, \dots , m-1)$$.

If $$n-1-k=1$$ and $$k=1$$, then $$S_{n-1-k} \times S_k$$ is the trivial group. If $$n-1-k = 1$$ and $$k \geqslant 2$$, then $$s_a = (\mathbbm {1}, (n-1-k,n-k))$$ and $$s_b$$ is either $$(\mathbbm {1}, (n-1-k, \dots , n-2))$$ or $$(\mathbbm {1}, (n-k, \dots , n-2))$$, and either pair generates the group. There is a similar argument when $$k = 1$$.

Assume now $$n-1-k \geqslant 2$$ and $$k \geqslant 2$$. If $$n-1-k$$ is odd then $$s_a = ((0,\dots , n-2-k), (n-1-k,n-k))$$, and hence $$s_a^{n-1-k} = ((0,\dots , n-2-k)^{n-1-k}, (n-1-k,n-k)^{n-1-k}) = (\mathbbm {1}, (n-1-k,n-k))$$. Similarly if $$n-1-k$$ is even then $$s_a = ((1,\dots , n-2-k), (n-1-k,n-k))$$, and hence $$s_a^{n-2-k}= (\mathbbm {1}, (n-1-k,n-k))$$. Therefore $$(\mathbbm {1}, (n-1-k,n-k))$$ is always generated by $$s_a$$. By symmetry, $$((0,1), \mathbbm {1})$$ is always generated by $$s_b$$ regardless of the parity of k.

Since we can isolate the transposition component of $$s_a$$, we can isolate the other component as well: $$(\mathbbm {1}, (n-1-k,n-k))s_a$$ is either $$((0, \dots , n-2-k),\mathbbm {1})$$ or $$((1, \dots , n-2-k),\mathbbm {1})$$. Paired with $$((0,1), \mathbbm {1})$$, either element is sufficient to generate $$S_{n-1-k} \times \{\mathbbm {1}\}$$. Similarly, $$s_a$$ and $$s_b$$ generate $$\{\mathbbm {1}\} \times S_k$$. Therefore $$s_a$$ and $$s_b$$ generate $$S_{n-1-k} \times S_k$$. It follows that a and b generate all permutations t of $$Q_n$$ such that $$E_{n,k}t = E_{n,k}$$, $$F_{n,k}t = F_{n,k}$$, and $$(n-1)t = n-1$$.    $$\square$$

### Proposition 3

(Syntactic Semigroup). The syntactic semigroup of $$L_{n,k}(\varSigma )$$ has cardinality $$n^{n-1-k}(k+1)^k$$, which is maximal for a k-proper language. Furthermore, seven letters are required to meet this bound. The maximum value $$n(n-1)^{n-2}$$ is reached only when $$k=n-2$$.

### Proof

Let L be a k-proper language of complexity n and let $$\mathcal {D}$$ be a minimal DFA recognizing L. By Lemma 1, $$\mathcal {D}$$ has an empty state. By Proposition 1, the only states that can be reached from one of the k final states are either final or empty. Thus, a transformation in the transition semigroup of $$\mathcal {D}$$ may map each final state to one of $$k+1$$ possible states, while each non-final, non-empty state may be mapped to any of the n states. Since the empty state can only be mapped to itself, we are left with $$n^{n-1-k}(k+1)^k$$ possible transformations in the transition semigroup. Therefore the syntactic semigroup of any k-proper language has size at most $$n^{n-1-k}(k+1)^k$$.

Now consider the transition semigroup of $$\mathcal {D}_{n,k}(\varSigma )$$. Every transformation t in the semigroup must satisfy $$F_{n,k}t \subseteq F_{n,k} \cup \{n-1\}$$ and $$(n-1)t = n-1$$, since any other transformation would violate prefix-convexity. We show that the semigroup contains every such transformation, and hence the syntactic semigroup of $$L_{n,k}(\varSigma )$$ is maximal.

First, consider the transformations t such that $$E_{n,k}t \subseteq E_{n,k} \cup \{n-1\}$$ and $$qt =q$$ for all $$q \in F_{n,k} \cup \{n-1\}$$. By Lemma 2, a and b generate every permutation of $$E_{n,k}$$. When t is not a permutation, we can use $$c_1$$ to combine any states p and q: apply a permutation on $$E_{n,k}$$ so that $$p \rightarrow 0$$ and $$q \rightarrow 1$$, and then apply $$c_1$$ so that $$1 \rightarrow 0$$. Repeat this method to combine any set of states, and further apply permutations to induce the desired transformation while leaving the states of $$F_{n,k} \cup \{n-1\}$$ in place. The same idea applies with $$d_1$$; apply permutations and $$d_1$$ to send any states of $$E_{n,k}$$ to $$n-1$$. Hence a, b, $$c_1$$, and $$d_1$$ generate every transformation t such that $$E_{n,k}t \subseteq E_{n,k}\cup \{n-1\}$$ and $$qt = q$$ for all $$q \in F_{n,k} \cup \{n-1\}$$.

We can make the same argument for transformations that act only on $$F_{n,k}$$ and fix every other state. Since $$c_2$$ and $$d_2$$ act on $$F_{n,k}$$ exactly as $$c_1$$ and $$d_1$$ act on $$E_{n,k}$$, the letters a, b, $$c_2$$, and $$d_2$$ generate every transformation t such that $$F_{n,k}t \subseteq F_{n,k} \cup \{n-1\}$$ and $$qt = q$$ for all $$q \in E_{n,k}\cup \{n-1\}$$. It follows that a, b, $$c_1$$, $$c_2$$, $$d_1$$, and $$d_2$$ generate every transformation t such that $$E_{n,k}t \subseteq E_{n,k} \cup \{n-1\}$$, $$F_{n,k}t \subseteq F_{n,k} \cup \{n-1\}$$, and $$(n-1)t = n-1$$.

Note the similarity between this DFA restricted to the states $$E_{n,k} \cup \{n-1\}$$ (or $$F_{n,k} \cup \{n-1\}$$) and the witness for right ideals introduced in [7]. The argument for the size of the syntactic semigroup of right ideals is similar to this; see [10].

Finally, consider an arbitrary transformation t such that $$F_{n,k}t \subseteq F_{n,k} \cup \{n-1\}$$ and $$(n-1)t = n-1$$. Let $$j_t$$ be the number of states $$p \in E_{n,k}$$ such that $$pt \in F_{n,k}$$. We show by induction on $$j_t$$ that t is in the transition semigroup of $$\mathcal {D}$$. If $$j_t = 0$$, then t is generated by $$\varSigma \setminus \{e\}$$. If $$j_t \geqslant 1$$, there exist $$p,q \in E_{n,k}$$ such that $$pt \in F_{n,k}$$ and q is not in the image of t. Consider the transformations $$s_1$$ and $$s_2$$ defined by $$qs_1 = pt$$ and $$rs_1 = r$$ for $$r\not =q$$, and $$ps_2 = q$$ and $$rs_2 = rt$$ for $$r\not =p$$. Then $$(rs_2)s_1 = rt$$ for all $$r \in Q_n$$. Notice that $$j_{s_2} = j_t -1$$, and hence $$\varSigma$$ generates $$s_2$$ by inductive assumption. One can verify that $$s_1 = (n-1-k,pt) (0,q) (0 \rightarrow n-1-k) (0,q) (n-1-k,pt)$$. From this expression, we see that $$s_1$$ is the composition of transpositions induced by words in $$\{a,b\}^*$$ and the transformation $$(0 \rightarrow n-1-k)$$ induced by e, and hence $$s_1$$ is generated by $$\varSigma$$. Thus, t is in the transition semigroup. By induction on $$j_t$$, it follows that the syntactic semigroup of $$L_{n,k}$$ is maximal.

Now we show that seven letters are required to meet this bound. Two letters (like a and b) are required to generate the permutations, since clearly one letter is not sufficient. Every other letter will induce a properly injective map. A letter (like $$c_1$$) that induces a properly injective map on $$E_{n,k}$$ and permutes $$F_{n,k}$$ is required. Similarly, a letter (like $$c_2$$) that permutes $$E_{n,k}$$ and induces a properly injective map on $$F_{n,k}$$ is required. A letter (like $$d_1$$) that sends a state in $$E_{n,k}$$ to $$n-1$$ and permutes $$F_{n,k}$$ is required. Similarly, a letter (like $$d_2$$) that sends a state in $$F_{n,k}$$ to $$n-1$$ and permutes $$E_{n,k}$$ is required. Finally, a letter (like e) that connects $$E_{n,k}$$ and $$F_{n,k}$$ is required.

For a fixed n, we may want to know which $$k \in \{1 , \dots , n-2\}$$ maximizes $$s_n(k) = n^{n-1-k}(k+1)^k$$; this corresponds to the largest syntactic semigroup of a proper prefix-convex language with n quotients. We show that $$s_n(k)$$ is largest at $$k = n-2$$. Consider the ratio $$\frac{s_n(k+1)}{s_n(k)} = \frac{(k+2)^{k+1}}{n(k+1)^k}$$. Notice this ratio is increasing with k, and hence $$s_n$$ is a convex function on $$\{1, \dots ,n-2\}$$. It follows that the maximum value of $$s_n$$ must occur at one the endpoints, 1 and $$n-2$$.

Now we show that $$s_n(n-2) \geqslant s_n(1)$$ for all $$n \geqslant 3$$. We can check this explicitly for $$n = 3, 4, 5$$. When $$n \geqslant 6$$, $$s_n(n-2)/s_n(1) = \frac{n}{2} \left( \frac{n-1}{n}\right) ^{n-2} \geqslant 3 \left( 1/e\right) > 1$$; so the largest syntactic semigroup of $$L_{n,k}(\varSigma )$$ occurs only at $$k = n-2$$ for all $$n \geqslant 3$$.    $$\square$$

### Proposition 4

(Reverse). For any regular language L of complexity n with an empty quotient, the reversal has complexity at most $$2^{n-1}$$. Moreover, the reverse of $$L_{n,k}(a,b,-,-,-,d_2, e)$$ has complexity $$2^{n-1}$$ for $$n \geqslant 3$$ and $$1 \leqslant k \leqslant n-2$$.

### Proof

The first claim is left for the reader to verify. For the second claim, let $$\mathcal {D}_{n,k} = (Q_n, \{a,b,d_2,e\}, \delta _{n,k}, 0, F_{n,k})$$ denote the DFA $$\mathcal {D}_{n,k}(a,b,-,-,-,d_2,e)$$ in Definition 2 and let $$L_{n,k} = L(D_{n,k})$$. Construct an NFA $${\mathcal {N}}$$ recognizing the reverse of $$L_{n,k}$$ by reversing each transition, letting the initial state 0 be the unique final state, and letting the final states in $$F_{ n,k}$$ be the initial states. Applying the subset construction to $${\mathcal {N}}$$ yields a DFA $$\mathcal {D}^R$$ whose states are subsets of $$Q_{n-1}$$, with initial state $$F_{n,k}$$ and final states $$\{U \subseteq Q_{n-1} \mid 0 \in U\}$$. We show that $$\mathcal {D}^R$$ is minimal, and hence the reverse of $$L_{n,k}$$ has complexity $$2^{n-1}$$.

Recall from Lemma 2 that a and b generate all permutations of $$E_{n,k}$$ and $$F_{n,k}$$ in $$\mathcal {D}_{n,k}$$ and, although the transitions are reversed in $$\mathcal {D}^R$$, they still generate all such permutations. Let $$u_1, u_2 \in \{a,b\}^*$$ be such that $$u_1$$ induces $$(0, \dots , n-2-k)$$ and $$u_2$$ induces $$(n-1-k, \dots , n-2)$$ in $$\mathcal {D}^R$$.

Consider a state $$U = \{q_1, \dots , q_h, n-1-k, \dots , n-2\}$$ where $$0 \leqslant q_1< q_2< \dots < q_h \leqslant n-2-k$$. If $$h = 0$$, then U is the initial state. When $$h \geqslant 1$$, $$\{q_2 - q_1, q_3 - q_1, \dots , q_h - q_1, n-1-k, \dots , n-2\}eu_1^{q_1} = U$$. By induction, all such states are reachable.

Now we show that any state $$U = \{q_1, \dots , q_h, p_1, \dots , p_i\}$$ where $$0 \leqslant q_1< q_2< \dots < q_h \leqslant n-2-k$$ and $$n-1-k \leqslant p_1< p_2< \dots < p_i \leqslant n-2$$ is reachable. If $$i = k$$, then $$U = \{q_1, \dots , q_h, n-1-k, \dots , n-2\}$$ is reachable by the argument above. When $$0 \leqslant i < k$$, choose $$p \in F_{n,k} \setminus U$$ and see that U is reached from $$U \cup \{p\}$$ by $$u_2^{n-1-p}d_2u_2^{p - (n-2-k)}$$. By induction, every state is reachable.

To prove distinguishability, consider distinct states U and V. Choose $$q \in U \oplus V$$. If $$q \in E_{n,k}$$, then U and V are distinguished by $$u_1^{n-1-k-q}$$. When $$q \in F_{n,k}$$, they are distinguished by $$u_2^{n -1-q} e$$. So $$\mathcal {D}^R$$ is minimal.    $$\square$$

### Proposition 5

(Star). Let L be a regular language with $$n \geqslant 2$$ quotients, including $$k \geqslant 1$$ final quotients and one empty quotient. Then $$\kappa (L^*) \leqslant 2^{n-2} + 2^{n-2-k} + 1$$. This bound is tight for prefix-convex languages; in particular, the language $$( L_{n,k}(a,b,-,-,d_1,d_2, e))^*$$ meets this bound for $$n \geqslant 3$$ and $$1 \leqslant k \leqslant n-2$$.

Table 1.

Complexities of prefix-convex languages

Right-ideal

Prefix-closed

Prefix-free

Proper

SeGr

$$\mathbf {n^{n-1}}$$

$$\mathbf {n^{n-1}}$$

$$n^{n-2}$$

$$n^{n-1-k}(k+1)^k$$

Rev

$$\mathbf {2^{n-1}}$$

$$\mathbf {2^{n-1}}$$

$$2^{n-2} +1$$

$$\mathbf {2^{n-1} }$$

Star

$$n+1$$

$$2^{n-2}+1$$

n

$$\mathbf {2^{n-2} +2^{n-2-k}+1}$$

Prod

$$m+2^{n-2}$$

$$(m+1)2^{n-2}$$

$$m+n-2$$

$$\mathbf {m-1-j + j2^{n-2} +2^{n-1}}$$

$$\cup$$

$$mn-(m+n-2)$$

$$\mathbf {mn}$$

$$mn-2$$

$$\mathbf {mn}$$

$$\oplus$$

$$\mathbf {mn}$$

$$\mathbf {mn}$$

$$mn-2$$

$$\mathbf {mn}$$

$$\setminus$$

$$\mathbf {mn-(m-1)}$$

$$\mathbf {mn-(n-1)}$$

$$mn-(m+2n-4)$$

$$\mathbf {mn-(n-1)}$$

$$\cap$$

$$\mathbf {mn}$$

$$mn-(m+n-2)$$

$$mn-2(m+n-3)$$

$$mn -(m+n-2)$$

### Proof

Since L has an empty quotient, let $$n-1$$ be the empty state of its minimal DFA $$\mathcal {D}$$. To obtain an $$\varepsilon$$-NFA for $$L^*$$, we add a new initial state $$0'$$ which is final and has the same transitions as 0. We then add an $$\varepsilon$$-transition from every state in F to 0. Applying the subset construction to this $$\varepsilon$$-NFA yields a DFA $${{\mathcal {D}}}' = (Q', \varSigma , \delta ', \{0'\}, F')$$ recognizing $$L^*$$, in which $$Q'$$ contains non-empty subsets of $$Q_n \cup \{0'\}$$.

Many of the states of $$Q'$$ are unreachable or indistinguishable from other states. Since there is no transition in the $$\varepsilon$$-NFA to $$0'$$, the only reachable state in $$Q'$$ containing $$0'$$ is $$\{0'\}$$. As well, any reachable final state $$U \not = \{0'\}$$ must contain 0 because of the $$\varepsilon$$-transitions. Finally, for any $$U \in Q'$$, we have $$U \in F'$$ if and only if $$U \cup \{n-1\} \in F'$$, and since $$\delta '(U \cup \{n-1\}, w) = \delta '(U, w) \cup \{n-1\}$$ for all $$w \in \varSigma ^*$$, the states U and $$U \cup \{n-1\}$$ are equivalent in $$D'$$.

Hence $${\mathcal {D}}'$$ is equivalent to a DFA with the states $$\{\{0'\}\} \cup \{U \subseteq Q_{n-1} \mid U \cap F = \emptyset \} \cup \{U \subseteq Q_{n-1} \mid \hbox {} 0 \in U \text { and } U \cap F \not = \emptyset \}$$. This DFA has $$1 + 2^{n-1-k} + (2^{n-2} - 2^{n-2-k}) = 2^{n-2} + 2^{n-2-k} + 1$$ states. Thus, $$\kappa (L^*) \leqslant 2^{n-2} + 2^{n-2-k} + 1$$.

This bound applies when L is a prefix-convex language and $$n \geqslant 3$$. By Lemma 1, L is either a right ideal or has an empty state. If L is a right ideal, then $$\kappa (L^*) \leqslant n+1$$, which is at most $$2^{n-2}+2^{n-2-k}+1$$ for $$n \geqslant 3$$.

For the last claim, let $$\mathcal {D}_{n,k}(a,b,-,-,d_1,d_2,e)$$ of Definition 2 be denoted by $$\mathcal {D}_{n,k} = (Q_n, \{a, b, d_1, d_2, e\}, \delta _{n,k}, 0, F_{n,k})$$ and let $$L_{n,k} = L(D_{n,k})$$. We apply the same construction and reduction as before to obtain a DFA $$\mathcal {D}_{n,k}'$$ recognizing $$L_{n,k}^*$$ with states $$Q' = \{\{0'\}\} \cup \{U \subseteq E_{n,k}\} \cup \{U \subseteq Q_{n-1} \mid \hbox {} 0 \in U \hbox { and } U \cap F_{n,k} \not = \emptyset \}$$. We show that the states of $$Q'$$ are reachable and pairwise distinguishable.

By Lemma 2, a and b generate all permutations of $$E_{n,k}$$ and $$F_{n,k}$$ in $$\mathcal {D}_{n,k}$$. Choose $$u_1, u_2 \in \{a,b\}^*$$ such that $$u_1$$ induces $$(0, \dots , n-2-k)$$ and $$u_2$$ induces $$(n-1-k, \dots , n-2)$$ in $$\mathcal {D}_{n,k}$$.

For reachability, we consider three cases. (1) State $$\{0'\}$$ is reachable by $$\varepsilon$$. (2) Let $$U \subseteq E_{n,k}$$. For any $$q \in E_{n,k}$$, we can reach $$U \setminus \{q\}$$ by $$u_1^{n-2-k-q}d_1u_1^q$$; hence if U is reachable, then every subset of U is reachable. Observe that state $$E_{n,k}$$ is reachable by $$eu_1^{n-2-k}d_2^k$$, and we can reach any subset of this state. Therefore, all non-final states are reachable. (3) If $$U \cap F_{n,k} \not = \emptyset$$, then $$U = \{0, q_1, q_2, \dots , q_h, r_1, \dots , r_i\}$$ where $$0< q_1< \dots < q_h \leqslant n-2-k$$ and $$n-1-k \leqslant r_1< \dots< r_i < n-1$$ and $$i \geqslant 1$$. We prove that U is reachable by induction on i. If $$i = 0$$, then U is reachable by (2). For any $$i \geqslant 1$$, we can reach U from $$\{0, q_1, \dots , q_h, r_2 - (r_1 - (n-1-k)), \dots , r_i - (r_1 - (n-1-k)) \}$$ by $$eu_2^{r_1 - (n-1-k)}$$. Therefore, all states of this form are reachable.

Now we show that the states are pairwise distinguishable. (1) The initial state $$\{0'\}$$ is distinguishable from any other final state U since $$\{0'\}u_1$$ is non-final and $$Uu_1$$ is final. (2) If U and V are distinct subsets of $$E_{n,k}$$, then there is some $$q \in U \oplus V$$. We distinguish U and V by $$u_1^{n-1-k-q}e$$. (3) If U and V are distinct and final and neither one is $$\{0'\}$$, then there is some $$q \in U \oplus V$$. If $$q \in E_{n,k}$$, then $$Ud_2^k = U \setminus F_{n,k}$$ and $$Vd_2^k = V \setminus F_{n,k}$$ are distinct, non-final states as in (2). Otherwise, $$q \in F_{n,k}$$ and we distinguish U and V by $$u_2^{n-1-q}d_2^{k-1}$$.    $$\square$$

## 3 Conclusions

The bounds for prefix-convex languages (see also [8]) are summarized in Table 1. The largest bounds are shown in boldface type, and they are reached either in the class of right-ideal languages or the class of proper languages. Recall that for regular languages we have the following results: semigroup $$n^n$$, reverse $$2^n$$, star $$2^{n-1}+2^{n-2}$$, product $$m2^n-2^{n-1}$$, boolean operations mn.

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