Table of Contents
Journal of Discrete Mathematics
Volume 2014, Article ID 891760, 6 pages
http://dx.doi.org/10.1155/2014/891760
Research Article

Modularity in the Semilattice of ω-Words

Department of Mathematics, University of Latvia, 19 Rainis Boulevard, Riga 1586, Latvia

Received 31 August 2013; Accepted 6 December 2013; Published 18 May 2014

Academic Editor: Teturo Kamae

Copyright © 2014 Jānis Buls and Edmunds Cers. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

A partial ordering of ω-words can be introduced with regard to whether an ω-word can be transformed into another by a Mealy machine. It is known that the poset of ω-words that is introduced by this ordering is a join-semilattice. The width of this join-semilattice has the power of continuum while the depth is at least . We have created a technique for proving that power-characteristic ω-words are incomparable. We use this technique to show that this join-semilattice is not modular.

1. Introduction

Infinite words (-words) provide a fertile ground of research and many classes of -words are known. While closure properties of some classes of -words have been studied extensively (see, e.g., [13]), we are interested in the general algebraic structure of -words. Mealy machines are a simple model of a word transforming automaton with the beneficial property of always transforming an -word into an -word. A partial ordering is defined on -words by the existence of a Mealy machine transforming one word into another (we write if such a machine exists). When both and are true, we say that   and    are machine equivalent. A class of -words is called machine invariant if contains all possible transformations of its elements.

Buls [4] has shown that machine invariant classes of -words form a completely distributive lattice. Belovs [5] showed that the poset of machine equivalent classes of -words is a join-semilattice and that the width of this join-semilattice has the power of continuum while the depth is at least . We show in this paper that this join-semilattice is not modular.

2. Preliminaries

Let be a finite, nonempty set and the free monoid generated by . Call an alphabet, its elements letters, and the elements of    finite words. The identity element of is called the empty word and is denoted by .

Let be a word over a finite and non-empty alphabet . We denote the length of by . Similarly, the cardinality of is denoted by . The concatenation of the words is denoted by or simply . Define and for all   . We say that (resp., ) is a prefix (resp., suffix) of . Denote by and the respective sets of prefixes and suffixes of . Let denote the set of integers : Call a total map an (indexed) -word of the alphabet . For any define and write The set of all -words over is denoted by . We say that is a prefix of , if there exists an integer such that . An -word is called ultimately periodic if there exist integers and such that for all . In this case is called preperiod and a period of . An ultimately periodic -word with a preperiod of zero is called periodic. We say a finite word is (ultimately) periodic if it is a prefix of some (ultimately) periodic -word. We recall the important periodicity theorem due to Fine and Wilf [6].

Theorem 1. Let be a word having periods and and denote by the greatest common divisor of and . If , then has also the period .

Corollary 2. Let and be words having periods, respectively, and . If , then has the period .

This is almost folklore in combinatorics on words. Nevertheless, for the sake of completeness, we will give the proof of this corollary.

Proof. Since , then (Theorem 1) has the period .
(i) At first we will prove by induction on that has the period . Let . If , then for some letter. Since the period of is then .
Notice that divides . Since the period of is , then . Hence, .
(ii) If , then can be represented as concatenation , where . By assumption the period of is . Now from (i) follows that the period of is too. We have completed the inductive step.
Now we shall prove by induction on that has the period .
(iii) If , then for some letter. Since the period of is , then . Notice Hence, .
(iv) If , then can be represented as concatenation , where . By assumption the period of is . Now from (iii) follows that the period of is too. We have completed the inductive step.

A 3-sorted algebra is called an initial Mealy machine if , , are finite, nonempty sets, called the set of states, the input alphabet, and the output alphabet, respectively, is called the initial state, is a total function called the transition function, and is a total surjective function called the output function. We write or even if there is no danger of confusion. The mappings and are extended to by defining for all , . Henceforth, we shall omit parentheses if there is no danger of confusion. So, for example, we will write instead of . Let . If for some Mealy machine : for all   , we say that   transforms   to   and write or . We write if there exists such that ; otherwise, we write . We write if and and say that and are machine equivalent.

Given the integers , let denote the least common multiple of . Given a number , denote by the greatest integer less than or equal to and by the least integer greater than or equal to .

3. Machine Transformations of Power-Characteristic -Words

Definition 3. We will call the -word the characteristic word of the power if For example, is the characteristic word of the squares.

Convention. Henceforth, we assume that and it is a natural number.

More generally, let be any total increasing function; then Let be a Mealy machine, where Applying the pigeonhole principle, we can state that for every there is a least integer such that for some . The integer is called the index of , and is called the period of . We can visualize this as the diagrams (see Figure 1).

891760.fig.001
Figure 1: Indices and periods.

Claim 1. If is the index and is the period of , then

Claim 2. If and , where then Let be an integer polynomial; that is, . The following theorem is known from elementary number theory.

Theorem 4. If , then .

If we take , then we can express where . Hence, .

Corollary 5. If , then .

Let Then Let . We define a sequence where .

Corollary 6. The sequence is ultimately periodic.

Proof. Let . There exists such that Now consider the sequence . Since , then—by the pigeonhole principle—there must exist two equal states Hence The rest follows by induction.

The following lemma is very easy, but it turns out to be useful.

Lemma 7. Let be a Mealy machine. If and , then there exists such that

Proof. (i) If , then , and we can choose , .
(ii) Let and states, where Since , then—by the pigeonhole principle—there must exist two equal states; namely, there exist and , , such that Putting then and .

Proposition 8. If , and then is ultimately periodic.

Proof . Since and , there exist Mealy machines such that and .
(i) First, we express with and look at the sequence , where We have shown (Corollary 6) that the sequence is ultimately periodic. Assume its period is and the preperiod .
(ii) By assumption (see (23)) we can choose integers and such that and, moreover, and . Now, is a word of the form , and thus (by Lemma 7) must be ultimately periodic with both its period and preperiod not greater than . We denote this preperiod by and the least period by . Since then is periodic with the period . Notice that and that the sequence of states is also periodic. Therefore where . So we have two periodic words and . Hence by Corollary 2 is periodic with period . Therefore we can conclude that is periodic with period for all ; besides, is the least period for all .
(iii) Let and . Then where We have shown in (ii) that is periodic. Therefore there is a such that where and . Since divides , then . But then for some number and . It follows from (33) that and therefore .
(iv) Finally, we can select integers , such that and Now we repeat the proof from (ii). So we can conclude that there is the least period of the word . A period of is too. Hence (Theorem 1) . Denote . Since (from formula (33)) there is an integer such that . As it was shown in (iii) we can choose such that But then , which means that is periodic with the period . Now suppose that , where and . Then there exists such that . From (see formula (33)) , we can conclude that It follows from what we have shown in (iii) that there are such that with and . But then and This means that is periodic with period . Now, by induction, we have is periodic with the period for any . Hence, is ultimately periodic.

4. Modularity in the Semilattice of -Words

Our main object of investigation is the machine poset of infinite words. In order to avoid some set-theoretical problems, we make some assumptions. Let us take the set . We shall assume that the states of the involved Mealy machines as well as their input and output alphabets all are from the set . If another input or output alphabet is used, we assume that there exists a bijection and that this bijection is applied to the input or output word, respectively.

We suppose that the reader is familiar with the basic notions of ordered sets [7]. If is used as an algebraic relation on , then the algebraic structure defines a preorder [5], while the quotient set becomes the ordered set . It has been shown that this poset is a join-semilattice [5], where the join .

Definition 9. A join-semilattice is distributive when A join-semilattice is modular when

Theorem 10. The join-semilattice is not modular.

Proof. We start by showing that , where By definition, . Define the Mealy machine by We illustrate this by the diagram in Figure 2. It follows straightforwardly from the construction that . Now suppose that there exists such that and . But then too. Notice that for Hence, by Proposition 8 is ultimately periodic. But if so, then . However this is a contradiction because then .

891760.fig.002
Figure 2: .

Corollary 11 (see [8]). The join-semilattice is not distributive.

We recall that every distributive join-semilattice is modular.

5. Conclusion

We agree that it is not sufficient to state that is very complicated as it was done in Wikipedia writing review about Turing degrees [9]. Nevertheless it is clear that is not modular and that there is no classical algebraic notion that describes such semilattice axiomatically at this moment. So we have challenge to extract axioms for semilattice .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The useful suggestions of two referees are gratefully acknowledged.

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