Recall that if is a word of length , where are letters and , then the word is called the reversal of . We have that is a palindrome if . The palindromic length of the word is equal to the minimal number such that and are nonempty palindromes, where . In 2013, Frid, Puzynina, Zamboni conjectured that [FrPuZa]:
If is an infinite word and is an integer such that for every factor of then is ultimately periodic.
Conjecture 1.1 has been solved for some classes of words. Most notably it is known that if is an infinite -power free word for some positive integer , then the conjecture holds for [FrPuZa]. Also, for example, the conjecture has been confirmed for Sturmian words [FRID2018202]. However, in general, the conjecture remains open.
In [10.1007/978-3-319-66396-8_19], another version of Conjecture 1.1 was presented:
Every non-ultimately periodic infinite word has prefixes of arbitrarily high palindromic length.
There are quite many papers dealing with palindromic length. In [AMBROZ201974], the authors study the palindromic length of factors of fixed points of primitive morphisms. In [10.1007/978-3-030-62536-8_14], the palindromic length of factors with many periodic palindromes is investigated.
In addition, algorithms for computing the palindromic length were researched [borozdin_et_al:LIPIcs:2017:7338], [FICI201441], [RuSh15]. In [RuSh15], the authors present a linear time online algorithm for computing the palindromic length.
Given an infinite word , let be the palindromic length of the prefix of with the length . In 2019, Frid conjectured that [10.1007/978-3-030-24886-4_17]:
If an infinite word is -power free for some positive integer , then .
Let denote the set of all positive integers and let denote the set of all real numbers. It is quite easy to see that if is a function with and for all (so that the inverse function exists), then there is an infinite non-ultimately periodic word , such that . To see this, consider the alphabet . Let . It is straightforward to prove that . It means that the palindromic length of prefixes can grow arbitrarily slow. Note that for every , we have that is a factor of . Thus there is no such that is -power free.
Based on this observation and Conjecture 1.3, we could say that the power factors in allow us to restrict the growth rate of palindromic length of prefixes . In the current paper we investigate the relation between the palindromic length and the presence of power factors in an infinite words. More specifically, we reduce the powers of a given recurrent factor in an infinite word in such a way that the palindromic length does not “significantly” change. The basic idea of our powers reduction is as follows: For a given factor of an infinite word , we replace every factor with by a factor , where is a function such that for some constant .
One of the motivations for this investigation is this idea: Given an infinite non-ultimately periodic word with , if we could construct an infinite -power free word with from by reducing the power factors, we could prove Conjecture 1.1.
Given a finite or infinite word , let denote the set of all finite factors of and let . Given an infinite word and a finite nonempty word , let
The main result of the current article is the following theorem.
If is an infinite non-ultimately periodic word, , , is primitive, is recurrent in , , and then there is an infinite non-ultimately periodic word such that , , and .
Let be an infinite non-ultimately periodic word with .
In [BucMichGreedy2018], it was shown that has an infinite suffix such that has infinitely many periodic prefixes. It follows that has only finitely many non-recurrent factors. This result justifies that we consider that is recurrent.
In [FrPuZa], it was shown that does not satisfy the so-called -condition. It follows that the set is infinite for every . Hence an iterative application of Theorem 1.4 to construct a -power free non-ultimately periodic word would require an infinite number of iterations. In consequence the constructed word would have an unbounded palindromic length of its factors.
Let denote a finite alphabet. Given , let denote the set of all finite words of length over the alphabet , let denote the empty word, let , and let . Let denote the set of all infinite words over the alphabet ; i.e. .
A word is called ultimately periodic, if there are such that . If there are no such then is called non-ultimately periodic.
Given a word , let denote the set of all finite factors of including the empty word. If is finite then .
Given a word , let and denote the set of all prefixes and suffixes of , respectively. We have that .
Given a word , let denote the set of all finite prefixes of including the empty word.
Let denote the set of all nonnegative integers. Let denote the set of all rational numbers bigger or equal to .
Let . A word is a -power of the word if , where and . We write that .
A word is called primitive if there are no and such that .
Let be the set of all palindromes.
Given and , we say that is recurrent in if . It means that has infinitely many occurrences in .
Let and , where and . We denote by and the factors and , respectively, where .
3 Non-ultimately periodic words
Given , let . Fix with . Given , let
The condition guarantees the power is not a prefix of . This is just for our conveniance, when definining the factorization of in next sections. Since non-ultimately periodic, there are infinitely many suffixes of without the prefix . We apply this obervation in the proof of Theorem 1.4.
Let be non-ultimately periodic and let .
We call an element a -run or simply a run. The term “run” has been used also in [FrPuZa]. Our definition of a run is slightly different. Our definition guarantees that runs do not overlap with each other, as shown in the next lemma. In addition, note that if is a -run, it does not imply that .
Suppose that . We have that
If then .
If then .
If then .
If then .
Suppose that .
Suppose that . Then since . This contradicts that . Thus .
Suppose that and . Let , let , and let . Then and are periodic words with the period and is such that , , and , since . To clarify the inequality the border case is depicted in Table 1. It follows that is periodic with the period and consequently . This is a contradiction to that , since implies that
and implies that
We conclude that .
Table 1: Case for and .
The case is analogous. Also it is straightforward to see that if and only if . This completes the proof. ∎
To simplify the presentation of our result we introduce a function. Given and with , let
such that .
If then , , and thus .
The next proposition shows that a mirror image of a run in a palindrome is also a run on condition that the border of a palindrome is sufficiently far from the border of the run.
If , , , , , and then .
From , , and , it follows that
Since we have that
From (1) it follows that
Consequently . This completes the proof. ∎
4 Factorization and the reduced word
We define an order on the set as follows: if and only if , where . Lemma 3.2 implies that is totally ordered and since is a positive integer for every , we have also that is well ordered. It means that the function is well defined for every .
Let be defined as follows:
and for every we define that
We define a special factorization of the word , that will allow us to construct an infinite power free word based on with “reduced” powers of .
Let be defined as follows:
and are such that ,
, , and .
It is clear that exists and is uniquely determined. Obviously we have that
We define a set of functions, that we will use to “reduce” the exponents of factors . Fix with . Let
The next lemma, less formally said, shows that the reduction by preserves the reverse relation.
If , , , , and , then .
Let . Then and , where and . Let .
We have that . Let be such that . Then and . Since , we have that and . Because , it follows that if then . This completes the proof. ∎
Given , let
We call the word the reduced word of . The basic property of the reduced word is that powers of in are bounded by .
If and then .
Realize that for all and if , then
The lemma follows. ∎
For our convenience when working with the reduced word we introduce some more functions. Given and , let
In addition we define and .
We define a set to be the set of positions that are covered by runs; formally let Let and let be a function defined as follows:
Given , let and be such that and .
We define that .
We will need the following simple lemma.
For every we have that .
For the the lemma follows from the definition of . For , the lemma from Lemma 3.2.
This ends the proof. ∎
The next lemma says that the function forms a “natural” bijection between -runs of and -runs of the reduced word. Realize that is the end position of the -th run of . Then we have that and consequently is defined. We show that is the end position of the -th run in the reduced word.
If , , and then .
We have that and that .
The definition of implies that for every we have that , and there is such that . Let be such that and . Lemma 4.6 implies that such exist, because .
We have that . Since , it follows easily that . Since and , it follows also that and . Since and , this implies that and consequently that and .
Suppose that there is . From the definition of we have that
Lemma 3.2 implies that there is such that
and consequently . This is a contradiction, since by definition the factors do not contain -runs.
It follows that . This completes the proof. ∎
From the proof of Proposition 4.7 it follows also that is a bijection between positions of and that are not covered by -runs.
If and then is a bijection.
For our main result we need that is non-ultimately periodic. We show we can select in such a way that this requirement is satisfied.
There is such that is non-ultimately periodic.
Clearly . Let and let . If is non-ultimately periodic, then let and we are done. Suppose that is ultimately periodic. Let be such that for every .
We have that . Proposition 4.7 implies that , where .
Let . If is such that then obviously
since and is ultimately periodic. It follows that there are such that
Since is non-ultimately periodic, it follows that there is such that for every we have that and the sequence is non-ultimately periodic.
It is straightforward to verify that there exists such that the sequence
is non-ultimately periodic. In consequence is also non-ultimately periodic.
This completes the proof. ∎
Let us fix such that is non-ultimately periodic. Theorem 4.9 asserts that such exists.
We will need the following elementary Lemma concerning the palindromic length. We omit the proof.
If then .
Given , let
By means of the function we map the positions of into the positions of the reduced word . Thus by mapping two positions , we get a map of factors on the factors of . The next proposition shows an upper bound on the palindromic length of the factors of as a function of the number of runs in the preimage of these factors of .
If , , , , and then