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 \(a:(0,\dots ,n-1)\), \(b:(0,1)\), \(c:(1 \rightarrow 0)\).

Fig. 1.

Minimal DFA of a most complex regular language.

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 \({\mathcal {D}}_{n,k}(\varSigma )\).
Fig. 2.

DFA \(\mathcal {D}_{n,k}(a,b,c_1,c_2, d_1, d_2,e)\) of Definition 2 when \(n-1-k\) is odd, k is even, and both are at least 2; missing transitions are self-loops.

Fig. 3.

DFA \(\mathcal {D}_{n,k}(a,b,c_1,c_2, d_1, d_2,e)\) of Definition 2 when \(n-1-k\) is even, k is odd, and both are at least 2; missing transitions are self-loops.

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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Copyright information

© Springer International Publishing AG 2017

Authors and Affiliations

  1. 1.David R. Cheriton School of Computer ScienceUniversity of WaterlooWaterlooCanada

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