# Transition Property for α-Power Free Languages with α≥ 2 and k≥ 3 Letters

In 1985, Restivo and Salemi presented a list of five problems concerning power free languages. Problem 4 states: Given α-power-free words u and v, decide whether there is a transition from u to v. Problem 5 states: Given α-power-free words u and v, find a transition word w, if it exists. Let Σ_k denote an alphabet with k letters. Let L_k,α denote the α-power free language over the alphabet Σ_k, where α is a rational number or a rational "number with +". If α is a "number with +" then suppose k≥ 3 and α≥ 2. If α is "only" a number then suppose k=3 and α>2 or k>3 and α≥ 2. We show that: If u∈ L_k,α is a right extendable word in L_k,α and v∈ L_k,α is a left extendable word in L_k,α then there is a (transition) word w such that uwv∈ L_k,α. We also show a construction of the word w.

## Authors

• 6 publications
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## 1 Introduction

The power free words are one of the major themes in the area of combinatorics on words. An -power of a word is the word such that and is a prefix of , where is a rational number . For example and . We say that a finite or infinite word is -power free if has no factors that are -powers for and we say that a finite or infinite word is -power free if has no factors that are -powers for , where are rational numbers. In the following, when we write “-power free” then means a number or a “number with ”. The power free words, also called repetitions free words, include well known square free (-power free), overlap free (-power free), and cube free words (-power free). Two surveys on the topic of power free words can be found in [Rampersad_Narad2007] and [10.1007/978-3-642-22321-1_3].

One of the questions being researched is the construction of infinite power free words. We define the repetition threshold to be the infimum of all rational numbers such that there exists an infinite -power-free word over an alphabet with letters. Dejean’s conjecture states that , , , and for each [DEJEAN197290]. Dejean’s conjecture has been proved by the work of several authors [CARPI2007137, Rampersad2011_dejean, DEJEAN197290, OLLAGNIER1992187, PANSIOT1984297, Rao2011_dejean].

It is easy to see that -power free words form a factorial language [10.1007/978-3-642-22321-1_3]; it means that all factors of a -power free word are also -power free words. Then Dejean’s conjecture implies that there are infinitely many finite -power free words over , where .

In [10.1007/978-3-642-82456-2_20], Restivo and Salemi presented a list of five problems that deal with the question of extendability of power free words. In the current paper we investigate Problem and Problem :

• Problem : Given -power-free words and , decide whether there is a transition word , such that is -power free.

• Problem : Given -power-free words and , find a transition word , if it exists.

A recent survey on the progress of solving all the five problems can be found in [10.1007/978-3-030-19955-5_27]; in particular, the problems and are solved for some overlap free (-power free) binary words. In addition, in [10.1007/978-3-030-19955-5_27] the authors prove that: For every pair of cube free words (-power free) over an alphabet with letters, if can be infinitely extended to the right and can be infinitely extended to the left respecting the cube-freeness property, then there exists a “transition” word over the same alphabet such that is cube free.

In 2009, a conjecture related to Problems and Problem of Restivo and Salemi appeared in [10.1007/978-3-642-02737-6_38]: Let be a power-free language, , where is the set of extendable words of . Then for some word . In 2018, this conjecture was presented also in [SHALLIT201996] in a slightly different form.

Let denote the set of natural numbers and let denote the set of rational numbers.

###### Definition 1.1.

Let

 Υ={(k,α)∣k∈N and α∈Q and k=3 and α>2}∪{(k,α)∣k∈N and α∈Q and k>3 and α≥2}∪{(k,α+)∣k∈N and α∈Q and k≥3 and α≥2}.
###### Remark 1.2.

The definition of says that: If and is a “number with ” then and . If and is “just” a number then and or and .

Let be a language. A finite word is called left extendable (resp., right extendable) in if for every there is a word with such that (resp., ).

In the current article we improve the results addressing Problems and Problem of Restivo and Salemi from [10.1007/978-3-030-19955-5_27] as follows. Let denote an alphabet with letters. Let denote the -power free language over the alphabet . We show that if , is a right extendable word in , and is a left extendable word in then there is a word such that . We also show a construction of the word .

We sketch briefly our construction of a “transition” word. Let be a right extendable -power free word and let be a left extendable -power free word over with letters. Let be an -power free right infinite word having as a prefix and let be an -power free left infinite word having as a suffix. Let be a letter that is recurrent in both and . We show that we may suppose that and have a common recurrent letter. Let be a right infinite -power free word over . Let be a left infinite -power free word such that the set of factors of is a subset of the set of recurrent factors of . We show that such exists. We identify a prefix of such that is a prefix of and is an -power free right infinite word. Analogously we identify a suffix of such that is a suffix of and is an -power free left infinite word. Moreover our construction guarantees that is a prefix of and is a suffix of . Then we find a prefix of such that is a suffix of and such that both and are “sufficiently long”. Then we show that is an -power free word having as a prefix and as a suffix.

The very basic idea of our proof is that if are -power free words and is a letter such that is not a factor of both and , then clearly is -power free on condition that . Just note that there cannot be a factor in which is an -power and contains , because has only one occurrence in . Our constructed words , , and have “long” factors which does not contain a letter . This will allow us to apply a similar approach to show that the constructed words does not contain square factor such that contain the letter .

Another key observation is that if , , and is an -power free word over then there is also an -power free word over . This is an implication of Dejean’s conjecture. Less formally said, if are -power free words over an alphabet with letters, then we construct a “transition” word over an alphabet with letters such that is -power free.

Dejean’s conjecture imposes also the limit to possible improvement of our construction. The construction cannot be used for , where , because an infinite (or “sufficiently long”) word over an alphabet with letters is not -power free. Also for and our technique fails. On the other hand, based on our research, it seems that our technique, with some adjustments, could be applied also for and .

## 2 Preliminaries

Recall that denotes an alphabet with letters. Let denote the set of all finite words over including the empty word , let denote the set of all right infinite words over , and let denote the set of all left infinite words over . Let . We call an infinite word.

Let denote the number of occurrences of the nonempty factor in the word . If and , then we call a recurrent factor in .

Let denote the set of all finite factors of a finite or infinite word . The set contains the empty word and if is finite then also . Let denote the set of all recurrent nonempty factors of .

Let denote the set of all prefixes of and let denote the set of all suffixes of . We define that and if is finite then also .

We have that . Let denote the set of all infinite -power free words over . Obviously . In addition we define and ; it means the sets of right infinite and left infinite -power free words.

## 3 Power Free Languages

Let and are -power free words. The first lemma says that is -power free if there are no word and no nonempty prefix of such that is a suffix of and is longer than .

###### Lemma 3.1.

Suppose , , and . Let

 Π={(r,¯v)∣r∈Σ∗k∖{ϵ} and ¯v∈Prf(v)∖{ϵ} and rr∈Suf(u¯v) and |rr|>|¯v|}.

If then .

###### Proof.

Suppose that is not -power free. Since is -power free, then there is and such that , and . It means that there is such that for some ; or if is a “number with ”. Because , see Definition 1.1 of , this implies that . If follows that . We proved that implies that . The lemma follows. ∎

The following technical set of -tuples will simplify our propositions.

###### Definition 3.2.

Given , we define that if

1. ,

2. ,

3. ,

4. ,

5. ,

6. ,

7. , where is such that , and

8. .

###### Remark 3.3.

Less formally said, the -tuple is in if is -power free word over , is a right infinite -power free word over , has no occurrence of (thus is a word over ), is a prefix of , has only one occurrence in , where is a letter such that is a prefix of , and the number of occurrences of in is bigger than the number of occurrences of in , where are finite words and is a letter.

The next proposition shows that if is from the set then is a right infinite -power free word, where is from the set .

If and then .

###### Proof.

Lemma 3.1 implies that it suffices to show that there are no with and no such that and . Recall that is an -power free word, hence we consider . We distinguish following cases.

• If : Since it follows that and hence . It is clear that if and only if . Since and thus , this is a contradiction.

• If and : Let be such that . Since we have that and . Since we have that . In consequence . But Property 7 of Definition 3.2 states that . Since , this is a contradiction.

• If and and : Let and be such that , , and ; see Figure below.

It follows that and . It is easy to see that . For we have that and . Because it follows that . This is a contradiction to Property 8 of Definition 3.2.

• If and and : Let such that and ; see Figure below.

It follows that

 occur(w12,x)=occur(w13,x)+occur(w2,x)+occur(xu,x).

This is a contradiction to Property 8 of Definition 3.2.

We proved that for each prefix we have that . The proposition follows. ∎

We prove that if then there is a right infinite -power free word over . In the introduction we showed that this observation could be deduced from Dejean’s conjecture. Here additionally, to be able to address Problem from the list of Restivo and Salemi, we present in the proof also examples of such words.

###### Lemma 3.5.

If then the set is not empty.

###### Proof.

If then . It is well known that the Thue Morse word is a right infinite -power free word over an alphabet with letters [SHALLIT201996]. It follows that the Thue Morse word is -power free for each .

If then . It is well known that there are -power free infinite words over an alphabet with letters [SHALLIT201996]. Suppose . An example is the fixed point of the morphism defined by , , and [SHALLIT201996]. If an infinite word is -power free then obviously is -power free and -power free for each .

This completes the proof. ∎

We define the sets of extendable words.

###### Definition 3.6.

Let . We define

 lext(L)={w∈L∣w is % left extendable in L}

and

 rext(L)={w∈L∣w is % right extendable in L}.

If then let be the set of all left infinite words such that and . Analogously if then let be the set of all right infinite words such that and .

###### Remark 3.7.

König’s Infinity Lemma implies that and are nonempty if is left extendable in and is right extendable in [Koenig1926], [Rampersad_Narad2007], because (resp., ) implies that there are infinitely many finite words in having as a suffix (resp., as a prefix).

The next proposition proves that if , is a right extendable -power free word, is a right infinite -power free word having the letter as a recurrent factor and having as a prefix, and is a right infinite -power free word over , then there are finite words such that the -tuple is in the set and is a prefix of .

###### Proposition 3.8.

If , , , , , and then there are finite words such that and .

###### Proof.

Let be the set of recurrent factors of that are also prefix of the word . Based on the size of the set we construct the words and we show that . The properties 1, 2, 3, 4, 5, and 6 of Definition 3.2 are easy to verify. Hence we explicitely prove only properties 7 and 8 and that .

• If is an infinite set. It follows that . Let be such that and let be such that and . Let .

Property 7 of Definition 3.2 follows from . Property 8 of Definition 3.2 is obvious, because is the empty word. Since we have that .

• If is a finite set. Let be the set of prefixes of that are recurrent in . Since is recurrent in we have that and thus is not empty. Let be such that is the longest element in . Let be the shortest prefix of such that if is a non-recurrent prefix of in then . Such obviously exists, because is a finite set and non-recurrent factors have only a finite number of occurrences. Let be the shortest factor of such that , , and . Since is recurrent in it is clear such exists.

We show that Property 7 of Definition 3.2 holds. Let be such that . Suppose that . It would imply that is recurrent in , since all occurrences of non-recurrent words from are in . But we defined to be the longest recurrent word . Hence it is contradiction to our assumption that .

Property 8 of Definition 3.2 and are obvious from the construction of .

This completes the proof. ∎

We define the reversal of a finite or infinite word as follows: If and , where and , then . If and , where and , then . Analogously if and , where and , then .

Proposition 3.4 allows to construct a right infinite -power free word with a given prefix. The next simple corollary shows that in the same way we construct a left infinite -power free word with a given suffix.

###### Corollary 3.9.

If , , , , , and then there are finite words such that , , and .

###### Proof.

Let . Realize that if and only if . Then the corollary follows from Proposition 3.4 and Proposition 3.8. ∎

Given and a right infinite word , let be the set of all left infinite words such that . It means that all factors of are recurrent factors of . We show that the set is not empty.

If and then .

###### Proof.

Since is an infinite word, the set of recurrent factors of is not empty. Let be a recurrent nonempty factor of ; may be a letter. Obviously there is such that is also recurrent in . This implies that the set is infinite. The lemma follows from König’s Infinity Lemma [Koenig1926], [Rampersad_Narad2007]. ∎

The next lemma shows that if is a right extendable -power free word then for each letter there is a right infinite -power free word such that is recurrent in and is a prefix of .

###### Lemma 3.11.

If , , and then there is such that .

###### Proof.

Let . If then we are done. Suppose that . Let . Clearly . Proposition 3.8 implies that there is such that . The proof of Lemma 3.5 implies that we can choose in such a way that is recurrent in . Then and . This completes the proof. ∎

The next proposition shows that if is left extendable and is right extendable then there are finite words , a letter , a right infinite word , and a left infinite word such that are infinite -power free words, has no occurrence of , every factor of is a recurrent factor in , is a prefix of , and is a suffix of .

###### Proposition 3.12.

If , , and then there are , , , and such that , , , , , and .

###### Proof.

Let and be such that . Lemma 3.11 implies that such exist. Let . It means that the letter is recurrent in both and .

Let be a right infinite -power free word over . Lemma 3.5 asserts that such exists. Let ; Lemma 3.10 shows that . It is easy to see that , because and .

Proposition 3.8 and Corollary 3.9 imply that there are such that

• ,

• ,

• , and

• ; it follows that .

Proposition 3.4 implies that . It follows that . Let and . This completes the proof. ∎

The main theorem of the article shows that if is a right extendable -power free word and is a left extendable -power free word then there is a word such that is -power free. The proof of the theorem shows also a construction of the word .

###### Theorem 3.13.

If , , and then there is such that .

###### Proof.

Let be as in Proposition 3.12. Let be the shortest suffix such that . Let be the shortest prefix such that and ; such exists, because is a recurrent factor of ; see Proposition 3.12. We show that .

We have that , since and Proposition 3.12 states that . Lemma 3.1 implies that it suffices to show that there are no and no such that and . We distinguish following cases.

• If then , because and . This is a contradiction, since and ; see Proposition 3.12.

• If then , otherwise cannot be a suffix of . Because we have that . Since it follows that there is are words such that , and . It follows that and because we have that . Since we get that ; this is a contradiction.

We conclude that there are no word and a no prefix such that with . Hence . Due to the construction of and we have that and . This completes the proof. ∎

## Acknowledgments

The author acknowledges support by the Czech Science Foundation grant GAČR 13-03538S and by the Grant Agency of the Czech Technical University in Prague, grant No. SGS14/205/OHK4/3T/14.