Publications (121)35.81 Total impact
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ABSTRACT: We prove that $f(m,n)=2^{mn1} + 2^{(m1)(n1)}(2^{m1}1)(2^{n1}1)$ is an upper bound on the state complexity of the shuffle operation on two regular languages having the same alphabet and state complexities $m$ and $n$, respectively. We also state partial results about the tightness of this bound: We show that there exist witness languages meeting the bound for every $m \le 5$ and any $n$, and also for $m=n=6$. Moreover, we prove that in the subset automaton of the NFA accepting the shuffle, all $2^{mn}$ states can be distinguishable, and an alphabet of size three suffices for that. It follows that the bound can be met if all $f(m,n)$ states are reachable. We know that an alphabet of size at least $mn$ is required (provided $m,n \ge 2$). The question of reachability, and hence also of the tightness of the bound $f(m,n)$ in general, remains open. 
Article: Most Complex Regular Ideals
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ABSTRACT: A right ideal (left ideal, twosided ideal) is a nonempty language $L$ over an alphabet $\Sigma$ that satisfies $L=L\Sigma^*$ ($L=\Sigma^*L$, $L=\Sigma^*L\Sigma^*$). Let $k=3$ for right ideals, 4 for left ideals and 5 for twosided ideals. We show that there exist sequences ($L_n \mid n \ge k $) of right, left, and twosided regular ideals, where $L_n$ has quotient complexity (state complexity) $n$, such that $L_n$ is most complex in its class under the following measures of complexity: the size of the syntactic semigroup, the quotient complexities of the left quotients of $L_n$, the number of atoms (intersections of complemented and uncomplemented left quotients), the quotient complexities of the atoms, and the quotient complexities of reversal, star, product (concatenation), and all binary boolean operations. In that sense, these ideals are "most complex" languages in their classes, or "universal witnesses" to the complexity of the various operations.  [Show abstract] [Hide abstract]
ABSTRACT: The state complexity of a regular language is the number of states in a minimal deterministic finite automaton accepting the language. The syntactic complexity of a regular language is the cardinality of its syntactic semigroup. The syntactic complexity of a subclass of regular languages is the worstcase syntactic complexity taken as a function of the state complexity $n$ of languages in that class. We prove that $n^{n1}$, $n^{n1}+n1$, and $n^{n2}+(n2)2^{n2}+1$ are tight upper bounds on the syntactic complexities of right ideals and prefixclosed languages, left ideals and suffixclosed languages, and twosided ideals and factorclosed languages, respectively. Moreover, we show that the transition semigroups meeting the upper bounds for all three types of ideals are unique, and the numbers of generators (4, 5, and 6, respectively) cannot be reduced.  [Show abstract] [Hide abstract]
ABSTRACT: A sequence \((L_k,L_{k+1} \dots )\) of regular languages in some class \({\mathcal C}\), where n is the state complexity of \(L_n\), is called a stream. A stream is most complex in class \({\mathcal C}\) if its languages together with their dialects (that is, languages that differ only very slightly from the languages in the stream) meet the state complexity bounds for boolean operations, product (concatenation), star, and reversal, have the largest syntactic semigroups, and have the maximal numbers of atoms, each of which has maximal state complexity. It is known that there exist such most complex streams in the class of regular languages, and also in the classes of right, left, and twosided ideals. In contrast to this, we prove that there does not exist a most complex stream in the class of suffixfree regular languages. However, we do exhibit one ternary suffixfree stream that meets the bound for product and whose restrictions to binary alphabets meet the bounds for star and boolean operations. We also exhibit a quinary stream that meets the bounds for boolean operations, reversal, size of syntactic semigroup, and atom complexities. Moreover, we solve an open problem about the bound for the product of two languages of state complexities m and n in the binary case by showing that it can be met for infinitely many m and n. Two transition semigroups play an important role for suffixfree languages: semigroup \(\mathbf {T}^{\leqslant 5}(n)\) is the largest suffixfree semigroup for \(n\leqslant 5\), while semigroup \(\mathbf {T}^{\geqslant 6}(n)\) is largest for \(n=2,3\) and \(n\geqslant 6\). We prove that all witnesses meeting the bounds for the star and the second witness in a product must have transition semigroups in \(\mathbf {T}^{\leqslant 5}(n)\). On the other hand, witnesses meeting the bounds for reversal, size of syntactic semigroup and the complexity of atoms must have semigroups in \(\mathbf {T}^{\geqslant 6}(n)\).  [Show abstract] [Hide abstract]
ABSTRACT: A (left) quotient of a language $L$ by a word $w$ is the language $w^{1}L=\{x\mid wx\in L\}$. The quotient complexity of a regular language $L$ is the number of quotients of $L$; it is equal to the state complexity of $L$, which is the number of states in a minimal deterministic finite automaton accepting $L$. An atom of $L$ is an equivalence class of the relation in which two words are equivalent if for each quotient, they either are both in the quotient or both not in it; hence it is a nonempty intersection of complemented and uncomplemented quotients of $L$. A right (respectively, left and twosided) ideal is a language $L$ over an alphabet $\Sigma$ that satisfies $L=L\Sigma^*$ (respectively, $L=\Sigma^*L$ and $L=\Sigma^*L\Sigma^*$). We compute the maximal number of atoms and the maximal quotient complexities of atoms of right, left and twosided regular ideals.  [Show abstract] [Hide abstract]
ABSTRACT: We solve an open problem concerning syntactic complexity: We prove that the cardinality of the syntactic semigroup of a suffixfree language with $n$ left quotients (that is, with state complexity $n$) is at most $(n1)^{n2}+n2$ for $n\ge 7$. Since this bound is known to be reachable, this settles the problem. We also reduce the alphabet of the witness languages reaching this bound to five letters instead of $n+2$, and show that it cannot be any smaller. Finally, we prove that the transition semigroup of a minimal deterministic automaton accepting such a witness language is unique for each $n\ge 7$.  [Show abstract] [Hide abstract]
ABSTRACT: We solve two open problems concerning syntactic complexity: We prove that the cardinality of the syntactic semigroup of a left ideal or a suffixclosed language with $n$ left quotients (that is, with state complexity $n$) is at most $n^{n1}+n1$, and that of a twosided ideal or a factorclosed language is at most $n^{n2}+(n2)2^{n2}+1$. Since these bounds are known to be reachable, this settles the problems.  [Show abstract] [Hide abstract]
ABSTRACT: The quotient complexity of a regular language L, which is the same as its state complexity, is the number of left quotients of L. An atom of a nonempty regular language L with n quotients is a nonempty intersection of the n quotients, which can be uncomplemented or complemented. An NFA is atomic if the right language of every state is a union of atoms. We characterize all reduced atomic NFAs of a given language, i.e., those NFAs that have no equivalent states. We prove that, for any language L with quotient complexity n, the quotient complexity of any atom of L with r complemented quotients has an upper bound of 2n − 1 if r = 0 or r = n; for 1 ≤ r ≤ n − 1 the bound is For each n ≥ 2, we exhibit a language with 2n atoms which meet these bounds. 
Article: Theory of Átomata
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ABSTRACT: We show that every regular language defines a unique nondeterministic finite automaton (NFA), which we call “átomaton”, whose states are the “atoms” of the language, that is, nonempty intersections of complemented or uncomplemented left quotients of the language. We describe methods of constructing the átomaton, and prove that it is isomorphic to the reverse automaton of the minimal deterministic finite automaton (DFA) of the reverse language. We study “atomic” NFAs in which the right language of every state is a union of atoms. We generalize Brzozowski's doublereversal method for minimizing a deterministic finite automaton (DFA), showing that the result of applying the subset construction to an NFA is a minimal DFA if and only if the reverse of the NFA is atomic. We prove that Sengoku's claim that his method always finds a minimal NFA is false. 
Article: Large Aperiodic Semigroups
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ABSTRACT: The syntactic complexity of a regular language is the size of its syntactic semigroup. This semigroup is isomorphic to the transition semigroup of the minimal deterministic finite automaton accepting the language, that is, to the semigroup generated by transformations induced by nonempty words on the set of states of the automaton. In this paper we search for the largest syntactic semigroup of a starfree language having $n$ left quotients; equivalently, we look for the largest transition semigroup of an aperiodic finite automaton with $n$ states. We introduce two new aperiodic transition semigroups. The first is generated by transformations that change only one state; we call such transformations and resulting semigroups unitary. In particular, we study complete unitary semigroups which have a special structure, and we show that each maximal unitary semigroup is complete. For $n \ge 4$ there exists a complete unitary semigroup that is larger than any aperiodic semigroup known to date. We then present even larger aperiodic semigroups, generated by transformations that map a nonempty subset of states to a single state; we call such transformations and semigroups semiconstant. In particular, we examine semiconstant tree semigroups which have a structure based on full binary trees. The semiconstant tree semigroups are at present the best candidates for largest aperiodic semigroups. We also prove that $2^n1$ is an upper bound on the state complexity of reversal of starfree languages, and resolve an open problem about a special case of state complexity of concatenation of starfree languages.  [Show abstract] [Hide abstract]
ABSTRACT: A right ideal is a language L over an alphabet A that satisfies L = LA*. We show that there exists a stream (sequence) (R_n : n \ge 3) of regular right ideal languages, where R_n has n left quotients and is most complex under the following measures of complexity: the state complexities of the left quotients, the number of atoms (intersections of complemented and uncomplemented left quotients), the state complexities of the atoms, the size of the syntactic semigroup, the state complexities of the operations of reversal, star, and product, and the state complexities of all binary boolean operations. In that sense, this stream of right ideals is a universal witness.  [Show abstract] [Hide abstract]
ABSTRACT: The quotient complexity of a regular language L is the number of left quotients of L, which is the same as the state complexity of L. Suppose that L and L′ are binary regular languages with quotient complexities m and n, and that the subgroups of permutations in the transition semigroups of the minimal deterministic automata accepting L and L′ are the symmetric groups S m and S n of degrees m and n, respectively. Denote by ∘ any binary boolean operation that is not a constant and not a function of one argument only. For m,n ≥ 2 with \((m,n)\not \in \{(2,2),(3,4),(4,3),(4,4)\}\) we prove that the quotient complexity of L ∘ L′ is mn if and only either (a) \(m\not= n\) or (b) m = n and the bases (ordered pairs of generators) of S m and S n are not conjugate. For (m,n) ∈ {(2,2),(3,4),(4,3),(4,4)} we give examples to show that this need not hold. In proving these results we generalize the notion of uniform minimality to direct products of automata. We also establish a nontrivial connection between complexity of boolean operations and group theory. 
Article: Maximally Atomic Languages
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ABSTRACT: We explore the relationship between the transition semigroup of the minimal deterministic finite automaton (DFA) of a regular language, and the state/quotient complexities of the language's atoms. The atoms of a language are nonempty intersections of complemented and uncomplemented quotients of the language. Tight upper bounds on the number of atoms of a language and on the quotient complexities of atoms are known. We introduce a new class of regular languages called the \emph{maximally atomic languages}  languages with the maximal number of atoms, for which every atom has the maximal possible quotient complexity. We establish several relationships between transition semigroups and atoms; in particular, we give necessary and sufficient conditions a transition semigroup must meet for the associated language to be maximally atomic. Our methods of proof give new insights into the structures of minimal DFAs of atoms. 
Article: Maximal Syntactic Complexity of Regular Languages Implies Maximal Quotient Complexities of Atoms
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ABSTRACT: We relate two measures of complexity of regular languages. The first is syntactic complexity, that is, the cardinality of the syntactic semigroup of the language. That semigroup is isomorphic to the semigroup of transformations of states induced by nonempty words in the minimal deterministic finite automaton accepting the language. If the language has n left quotients (its minimal automaton has n states), then its syntactic complexity is at most n^n and this bound is tight. The second measure consists of the quotient (state) complexities of the atoms of the language, where atoms are nonempty intersections of complemented and uncomplemented quotients. A regular language has at most 2^n atoms and this bound is tight. The maximal quotient complexity of any atom with r complemented quotients is 2^n1, if r=0 or r=n, and 1+\sum_{k=1}^{r} \sum_{h=k+1}^{k+nr} \binom{h}{n} \binom{k}{h}, otherwise. We prove that if a language has maximal syntactic complexity, then it has 2^n atoms and each atom has maximal quotient complexity, but the converse is false.  [Show abstract] [Hide abstract]
ABSTRACT: We examine the NFA minimization problem in terms of atomic NFA's, that is, NFA's in which the right language of every state is a union of atoms, where the atoms of a regular language are nonempty intersections of complemented and uncomplemented left quotients of the language. We characterize all reduced atomic NFA's of a given language, that is, those NFA's that have no equivalent states. Using atomic NFA's, we formalize Sengoku's approach to NFA minimization and prove that his method fails to find all minimal NFA's. We also formulate the KamedaWeiner NFA minimization in terms of quotients and atoms.  [Show abstract] [Hide abstract]
ABSTRACT: The syntactic complexity of a regular language is the cardinality of its syntactic semigroup. The syntactic complexity of a subclass of the class of regular languages is the maximal syntactic complexity of languages in that class, taken as a function of the state complexity n of these languages. We study the syntactic complexity of R and Jtrivial regular languages, and prove that n! and floor of [e(n1)!] are tight upper bounds for these languages, respectively. We also prove that 2^{n1} is the tight upper bound on the state complexity of reversal of Jtrivial regular languages. 
Conference Paper: In Search of Most Complex Regular Languages
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ABSTRACT: Regular languages that are most complex under common complexity measures are studied. In particular, certain ternary languages Un(a,b,c), n ≥ 3, over the alphabet {a,b,c} are examined. It is proved that the state complexity bounds that hold for arbitrary regular languages are also met by the languages Un(a,b,c) for union, intersection, difference, symmetric difference, product (concatenation) and star. Maximal bounds are also met by Un(a,b,c) for the number of atoms, the quotient complexity of atoms, the size of the syntactic semigroup, reversal, and 22 combined operations, 5 of which require slightly modified versions. The language Un(a,b,c,d) is an extension of Un(a,b,c), obtained by adding an identity input to the minimal DFA of Un(a,b,c). The witness Un(a,b,c,d) and its modified versions work for 14 more combined operations. Thus Un(a,b,c) and Un(a,b,c,d) appear to be universal witnesses for alphabets of size 3 and 4, respectively. 
Article: Universal Witnesses for State Complexity of Boolean Operations and Concatenation Combined with Star
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ABSTRACT: We study the state complexity of boolean operations and product (concatenation, catenation) combined with star. We derive tight upper bounds for the symmetric differences and differences of two languages, one or both of which are starred, and for the product of two starred languages. We prove that the previously discovered bounds for the union and the intersection of languages with one or two starred arguments, for the product of two languages one of which is starred, and for the star of the product of two languages can all be met by the recently introduced universal witnesses and their variants.  [Show abstract] [Hide abstract]
ABSTRACT: We study the state complexity of boolean operations, concatenation and star with one or two of the argument languages reversed. We derive tight upper bounds for the symmetric differences and differences of such languages. We prove that the previously discovered bounds for union, intersection, concatenation and star of such languages can all be met by the recently introduced universal witnesses and their variants.  [Show abstract] [Hide abstract]
ABSTRACT: We study the syntactic complexity of finite/cofinite, definite and reverse definite languages. The syntactic complexity of a class of languages is defined as the maximal size of syntactic semigroups of languages from the class, taken as a function of the state complexity n of the languages. We prove that (n1)! is a tight upper bound for finite/cofinite languages and that it can be reached only if the alphabet size is greater than or equal to (n1)!(n2)!. We prove that the bound is also (n1)! for reverse definite languages, but the minimal alphabet size is (n1)!2(n2)!. We show that \lfloor e\cdot (n1)!\rfloor is a lower bound on the syntactic complexity of definite languages, and conjecture that this is also an upper bound, and that the alphabet size required to meet this bound is \floor{e \cdot (n1)!}  \floor{e \cdot (n2)!}. We prove the conjecture for n\le 4.
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19762015

University of Waterloo
 David R. Cheriton School of Computer Science
Ватерлоо, Ontario, Canada
