Journal of Discrete MathematicsVolume 2013 (2013), Article ID 219291, 8 pageshttp://dx.doi.org/10.1155/2013/219291

Research Article

Equivalence of Right Infinite Words

Liga Kulesa

University of Latvia, 8 Zellu street, Riga LV-1002, Latvia

Received 31 October 2012; Accepted 18 February 2013

Academic Editor: Zhan Zhou

Copyright © 2013 Liga Kulesa. 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.

Closure properties of some classes of right infinite words have been studied extensively; we are interested in the general algebraic structure of right infinite words. We investigate preorder of morphism invariant classes and show that it is not a semilattice.

1. Introduction

Right infinite words are ubiquitous in mathematics and theoretical computer science, therefore closure properties of some classes of right infinite words have been studied extensively. We consider simple function called morphism, which transforms right infinite word into a right infinite word.

There are several kinds of properties that are typically studied about a class of words (a language), being finite or infinite. These include categorization, the study of the relationships between different languages; representation, different ways of describing the words belonging to a language; closure properties, the invariance of the language when certain transformations are applied; for example, Cobham [1] proved that the class of automatic sequences is closed under uniform transductions, that is, under the transducers where every transition outputs a string of the same length. In 1994, Dekking presented proof that finite state transduction of morphic sequence is still morphic, even if the transducer is nonuniform [2].

In [3] Belovs introduced partial ordering on right infinite words: if and only if there exists a Mealy machine that transforms to . He proved that this ordering is an upper semilattice, but it is not a lower semilattice. Similarly we can introduce partial ordering or right infinite words by transducers.

We consider a poset of morphism invariant classes, which is simpler structure, therefore from an algebraic point of view it is surprising that it is neither an upper semilattice, nor a lower semilattice. Recently, Muchnik et al. [4] proved that transducer can be expressed as a composition of morphism and Mealy machine. Thus the ordering introduced by transducers is not an upper semilattice, nor a lower semilattice. In this sense the ordering defined by Mealy machine crucially differs.

2. Preliminaries

Let be a finite nonempty set, and let be a free monoid generated by . The identity element of is called the empty word and is denoted by . A function is called morphism if and . The morphism is uniquely defined by its values on the elements.

The morphism is called nonerasing if for all . It is called uniform if for all . It is called literal if for all . The morphism can also be applied to an infinite word as follows: It is possible to define morphism with more than one argument in the same way. Let and be a function from to , then A set with defined binary relation on it is called a partially ordered set (further in text—a poset) or just an order, if it fulfills the following three axioms: If a relation fulfills only the reflexivity and transitivity axioms then such an algebraic structure is called a preorder.

An element of is called the largest element if and only if for all . An element is called maximal if and only if for all .

For each two elements and of their upper bound is such an element where and . The supremum is the smallest upper bound; that is, such element that not only , but also for any other element , the expression holds. For each two elements and of their lower bound is such an element where and . The infimum is such element that not only , but also for any other element holds .

A poset whose every two elements have a supremum is called an upper semilattice. If every two elements have infimum, it is called a lower semilattice.

For more in depth coverage of the previous topics see, for example, [5, 6].

3. Semilattice

In this section we will prove that algebraic structure is not a semilattice, where(1); (2) is a relation, which holds if and only if there exists a morphism such that ;(3) if and only if

One can check that is a poset. By the semilattice definition every two elements must have a supremum and an infimum. We show that there are two elements in , which do not have a supremum, and that there are two elements in , which do not have an infimum, therefore, is not a semilattice.

Let denote a word, such that Let morphisms and be defined : Morphism maps the word as follows: Morphism maps the word as follows: Now we will define new word from and : Mappings define morphisms From these definitions we obtain For technical reasons we will write instead of .

Lemma 1. For all factor appears in words , ,  and once at most.

Proof. Obvious.

Next we consider following diagram: (11)

Lemma 2. If , then .

Proof. Let us consider the first position in word ; suppose that , then(1)if and (12) then , thus ;(2), since word begins with ;(3)if , then (13)(3.1). Indeed, otherwise will generate ;(3.2). Indeed, otherwise will generate ;(3.3), because and will generate , a contradiction with Lemma 1;(3.4), since word begins with ;(4)if , then (14)(4.1). Indeed, otherwise will generate ;(4.2), since ;(4.3);(4.4), where , since ;(5), where , since ;(6), since .From (4.3) it follows that .

Lemma 3. If , then .

Proof. Let us consider the first position in word ; suppose that , then(1)if and (15) then , thus ;(2), since word begins with ;(3)if , then (16)(3.1). Indeed, otherwise will generate ;(3.2). Indeed, otherwise will generate ;(3.3), because then and will generate , a contradiction with Lemma 1;(3.4), since word begins with ;(4)if , then (17)(4.1). Indeed, otherwise will generate ;(4.2). Indeed, otherwise will generate ;(4.3), because and will generate , a contradiction with Lemma 1; (4.4), since word begins with ;(5)if , then (18)(5.1). Indeed, otherwise will generate ;(5.2), since ;(5.3), since ;(5.4);(5.5), where , since ;(6), where , since ;(7), since .From (5.4) it follows that .

Lemma 4. If , then .

Proof. Let us consider the first position in word ; suppose that , then (1) if and , then    (1.1)if , then (19) Next we consider position in word ; suppose that , then    (1.1.1), otherwise we obtain periodic word, but word is not periodic;    (1.1.2), since word begins with ;    (1.1.3), otherwise we obtain , but it is a contradiction with construction;    (1.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (1.2), since word begins with ;    (1.3)if , then (20) Next we consider position in word ; suppose that , then    (1.3.1)if , then (21) Next we consider position in word ; suppose that , then    (1.3.1.1), otherwise we obtain periodic word ;    (1.3.1.2), since word begins with ;    (1.3.1.3), otherwise we obtain periodic word ;    (1.3.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (1.3.2), since word begins with ;    (1.3.3), otherwise we obtain , but it is a contradiction with construction;    (1.3.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (1.4), where , since and will generate , but from construction such situation occurs only in beginning of the word;    (1.5), since, for example, and will generate , but it is a contradiction with Lemma 1; (2) , since word begins with ; (3) if and , then    (3.1)if , then (22) Next we consider position in word ; suppose that , then    (3.1.1)if , then (23) Next we consider position in word ; suppose that , then    (3.1.1.1), otherwise we obtain periodic word ;    (3.1.1.2), since word begins with ;    (3.1.1.3), otherwise we obtain periodic word ;    (3.1.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.1.2), since word begins with ;    (3.1.3), otherwise we obtain , but it is a contradiction with construction;    (3.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.2), since word begins with ;    (3.3)if , then (24) Next we consider position in word ; suppose that , then    (3.3.1)if , then (25)  Next we consider position in word ; suppose that , then    (3.3.1.1), otherwise we obtain periodic word , but word is not periodic;    (3.3.1.2), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.3.1.3), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.3.1.4), otherwise we obtain periodic word , but word is not periodic;    (3.3.1.5), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.3.2)if , then (26) Next we consider position in word ; suppose that , then    (3.3.2.1), otherwise we obtain word ;    (3.3.2.2);    (3.3.2.3), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.3.2.4), since, for example, and will generate , but it is a contradiction with Lemma 1;    (3.3.2.5), otherwise we obtain word ;    (3.3.2.6), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.3)if , then (27) Next we consider position in word ; suppose that , then(3.3.3.1), otherwise we obtain word ;(3.3.3.2), otherwise we obtain word ;(3.3.3.3), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.3.4), otherwise we obtain word ;(3.3.3.5), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.4)if , then (28) Next we consider position in word ; suppose that , then(3.3.4.1), otherwise we obtain word ;(3.3.4.2), otherwise we obtain word ;(3.3.4.3), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.4.4), otherwise we obtain word ;(3.3.4.5), where , otherwise we obtain word ;(3.3.4.6), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.5), where , otherwise we obtain word ;(3.3.6), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.7), since word begins with ;(3.4), where since word begins with ; (3.5), since, for example, and will generate , but it is a contradiction with Lemma 1. (4) , where since word begins with . (5) , since, for example, and will generate , but it is a contradiction with Lemma 1.From (3.3.2.2) it follows that .

Lemma 5. If , then .

Proof. Let us consider the first position in word ; suppose that , then (1) if and , then(1.1)if , then (29) Next we consider position in word ; suppose that , then(1.1.1), otherwise we obtain periodic word, but word is not periodic;(1.1.2), since word begins with ;(1.1.3), otherwise we obtain , but it is a contradiction with construction;(1.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;(1.2), since word begins with ;(1.3), otherwise we obtain , but it is a contradiction with construction;(1.4), since, for example, and will generate , but it is a contradiction with Lemma 1; (2) , since word begins with ; (3) if , then(3.1)if , then (30) Next we consider position in word ; suppose that , then(3.1.1)if , then (31) Next we consider position in word ; suppose that , then(3.1.1.1), otherwise we obtain periodic word, but word is not periodic;(3.1.1.2), otherwise we obtain word ;(3.1.1.3), since word begins with ;(3.1.2)if , then (32)  Next we consider position in word ; suppose that , then(3.1.2.1), otherwise we obtain word ;(3.1.2.2), otherwise we obtain word ;(3.1.2.3), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.1.2.4), since word begins with ;(3.1.3)if , then (33) Next we consider position in word ; suppose that , then(3.1.3.1), otherwise we obtain word ;(3.1.3.2), otherwise we obtain word ;(3.1.3.3), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.1.3.4), since word begins with ;(3.1.4), where , since word begins with ;(3.1.5), since word begins with ;(3.2)if , then (34) Next we consider position in word ; suppose that , then(3.2.1)if , then (35) Next we consider position in word ; suppose that , then(3.2.1.1), otherwise we obtain word ; (3.2.1.2), otherwise we obtain word ; (3.2.1.3), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.2.1.4), since word begins with ;(3.2.2)if , then (36) Next we consider position in word ; suppose that , then(3.2.2.1), otherwise we obtain word ;(3.2.2.2), since word begins with .(3.2.2.3);(3.2.2.4), where , otherwise we obtain word ;(3.2.2.5), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.2.3), where , otherwise we obtain word ;(3.2.4), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.2.5), since word begins with ;(3.3)if , then (37) Next we consider position in word ; suppose that , then(3.3.1)if , then (38) Next we consider position in word ; suppose that , then(3.3.1.1), otherwise we obtain word ;(3.3.1.2), since word begins with ;(3.3.1.3), otherwise we obtain word ;(3.3.1.4), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.3.2), since word begins with ;(3.3.3), otherwise we obtain , but it is a contradiction with construction;(3.3.4), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.4), since, for example, and will generate , but it is a contradiction with Lemma 1;(3.5), since word begins with ; (4) , where , otherwise we obtain word ; (5) , since, for example, and will generate , but it is a contradiction with Lemma 1;From (3.2.2.3) it follows that .

Theorem 6. If , , , and are morphisms, then the diagram is not commutative. (39)

Proof. We have shown that the compositions of morphisms , , , and coincide, thus, for example, if maps word letter by letter, then does the same.Next, we consider morphisms , , , and .Denote that We denote by the word which is obtained by mapping first 8 letters of word with morphism as follows: Since and map letter by letter, then and will map the word letter by letter from Lemmas 4 and 5 as follows: Next, we denote . Since maps letter by letter, then by Lemma 2   will map the word letter by letter, thus therefore Since maps word letter by letter, then will map word letter by letter from Lemma 3 as follows: then which leads to a contradiction, and the diagram is not commutative. Hence two elements in such an algebraic structure do not have a supremum, since , but there does not exist such that and . Also two elements in such an algebraic structure do not have an infimum, since and , but there does not exist such that ,   and . Thus algebraic structure is not a semilattice.

Acknowledgment

This work has been supported by the European Social Fund within the Project no. 2009/0223/1DP/1.1.1.2.0/09/APIA/VIAA/008.

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