Palindromic Length of Words with Many Periodic Palindromes

The palindromic length PL(v) of a finite word v is the minimal number of palindromes whose concatenation is equal to v. In 2013, Frid, Puzynina, and Zamboni conjectured that: If w is an infinite word and k is an integer such that PL(u)≤ k for every factor u of w then w is ultimately periodic. Suppose that w is an infinite word and k is an integer such PL(u)≤ k for every factor u of w. Let Ω(w,k) be the set of all factors u of w that have more than √(k^-1| u|) palindromic prefixes. We show that Ω(w,k) is an infinite set and we show that for each positive integer j there are palindromes a,b and a word u∈Ω(w,k) such that (ab)^j is a factor of u and b is nonempty. Note that (ab)^j is a periodic word and (ab)^ia is a palindrome for each i≤ j. These results justify the following question: What is the palindromic length of a concatenation of a suffix of b and a periodic word (ab)^j with "many" periodic palindromes? It is known that |PL(uv)-PL(u)|≤PL(v), where u and v are nonempty words. The main result of our article shows that if a,b are palindromes, b is nonempty, u is a nonempty suffix of b, | ab| is the minimal period of aba, and j is a positive integer with j≥3PL(u) then PL(u(ab)^j)-PL(u)≥ 0.

Authors

• 6 publications
• Palindromic Length and Reduction of Powers

Given a nonempty finite word v, let PL(v) be the palindromic length of v...
03/26/2021 ∙ by Josef Rukavicka, et al. ∙ 0

• Prefix palindromic length of the Thue-Morse word

The prefix palindromic length PPL_u(n) of an infinite word u is the mini...
06/22/2019 ∙ by Anna E. Frid, et al. ∙ 0

• Computing the k-binomial complexity of the Thue--Morse word

Two words are k-binomially equivalent whenever they share the same subwo...
12/18/2018 ∙ by Marie Lejeune, et al. ∙ 0

• Undecidability of MSO+"ultimately periodic"

We prove that MSO on ω-words becomes undecidable if allowing to quantify...
07/23/2018 ∙ by Mikołaj Bojańczyk, et al. ∙ 0

• Faster Recovery of Approximate Periods over Edit Distance

The approximate period recovery problem asks to compute all approximate ...
07/27/2018 ∙ by Tomasz Kociumaka, et al. ∙ 0

• Characteristic Parameters and Special Trapezoidal Words

Following earlier work by Aldo de Luca and others, we study trapezoidal ...
06/04/2019 ∙ by Alma D'Aniello, et al. ∙ 0

• A Unique Extension of Rich Words

A word w is called rich if it contains | w|+1 palindromic factors, inclu...
10/02/2019 ∙ by Josef Rukavicka, et al. ∙ 0

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1 Introduction

In 2013, Frid, Puzynina, and Zamboni introduced a palindromic length of a finite word [FrPuZa]. Recall that the word of length is called a palindrome if , where are letters and . The palindromic length of the word is defined as the minimal number such that and are palindromes, where ; note that the palindromes are not necessarily distinct. Let denote the empty word. We define that .

In general, the factorization of a finite word into the minimal number of palindromes is not unique; for example and the word can be factorized in two ways: .

The authors of [FrPuZa] conjectured that:

Conjecture 1.1.

If is an infinite word and is an integer such that for every factor of then is ultimately periodic.

So far, Conjecture 1.1 remains open. We call an infinite word that satisfies the condition from Conjecture 1.1 a word with a bounded palindromic length. Note that there are infinite periodic words that do not have a bounded palindromic length; for example . Hence the converse of Conjecture 1.1 does not hold.

In [FrPuZa] the conjecture was proved for infinite words that are -power free for some positive integer . It follows that if is an infinite word with a bounded palindromic length, then for each positive integer there is a nonempty factor such that is a factor of .

In [10.1007/978-3-319-66396-8_19], another variation of Conjecture 1.1 was considered:

Conjecture 1.2.

Every aperiodic (not ultimately periodic) infinite word has prefixes of arbitrarily high palindromic length.

In [10.1007/978-3-319-66396-8_19], the author proved that Conjecture 1.1 and Conjecture 1.2 are equivalent. More precisely, it was proved that if every prefix of an infinite word is a concatenation of at most palindromes then every factor of is a concatenation of at most palindromes. It follows that Conjecture 1.2 remains also open.

In [FRID2018202] Conjecture 1.1 and Conjecture 1.2 have been proved for all Sturmian words. The properties of the palindromic length of Sturmian words have been investigated also in [10.1007/978-3-030-24886-4_18]. In [AMBROZ201974], the authors study the palindromic length of factors of fixed points of primitive morphisms. In [10.1007/978-3-030-24886-4_17], the lower bounds for the palindromic length of prefixes of infinite words can be found.

In [BucMichGreedy2018], a left and right greedy palindromic length have been introduced as a variant to the palindromic length. It is shown that if the left (or right) greedy palindromic lengths of prefixes of an infinite word is bounded, then is ultimately periodic.

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.

In the current paper we investigate infinite words with a bounded palindromic length. Let be a positive integer, let be an infinite word such that for every factor of , and let be the set of all factors of that have more than palindromic prefixes. We show that is an infinite set and we show that for each positive integer there are palindromes and a word such that is a factor of and is nonempty. Note that is a periodic word and is a palindrome for each . In this sense we can consider that has infinitely many periodic palindromes with an arbitrarily high exponent .

The existence of infinitely many periodic palindromes in is not surprising. It can be deduced also from the result in [FrPuZa], which says, as mentioned above, that if is an infinite word with a bounded palindromic length, then for each positive integer there is a nonempty factor such that is a factor of .

These results justify the following question: What is the palindromic length of a concatenation of a suffix of and a periodic word with “many” periodic palindromes?

It is known that if are nonempty words then [10.1007/978-3-319-66396-8_19]. Less formally said, it means that by concatenating a word to a word the change of the palindromic length is at most equal to the palindromic length of . The main result of our article shows that if are palindromes, is nonempty, is a nonempty suffix of , is the minimal period of , and is a positive integer with then .

The results of our article should shed some light on infinite words for which Conjecture 1.1 and Conjecture 1.2 remain open.

2 Preliminaries

Let denote the set of all positive integers, let denote the set of all nonnegative integers, let denote the set of all real numbers, and let denote the set of all positive real numbers.

Let denote a finite alphabet with letters. Let denote the set of all finite nonempty words over the alphabet and let ; recall that denotes the empty word. Let denote the set of all right infinite words.

Let and let , where and . We denote by the factor of starting at position and ending at position , where and

Let , where and . We denote by the factor of starting at position and ending at position , where and .

We call the word a factor of the word if there are words and such that . Given a word , we denote by the set of all factors of . It follows that and if then also .

We call the word a prefix of the word if there is such that . Given a word , we denote by the set of all prefixes of . It follows that and if then also .

We call the word a suffix of the word if there is such that . Given a word , we denote by the set of all suffixes of . It follows that .

Let , where and . Let denote the reversal of the word ; it means . In addition we define that the reversal of the empty word is the empty word; formally .

Realize that is a palindrome if and only if . Let denote the set of all palindromes over the alphabet . We define that . Let be the set of all nonempty palindromes.

Given , let be the set of all palindromic prefixes of .

Given , let denote the set of all -tuples of palindromes whose concatenation is equal to and ; formally

 MPF(w)={(t1,t2,…,tk)∣k=PL(w) and t1t2…tk=w and t1,t2,…,tk∈Pal+}.

We call a -tuple a minimal palindromic factorization of .

Let denote the set of all rational numbers. We say that the word is a periodic word, if there are , , and such that , , and ; note that is uniquely determined by . We write and the period of is equal to . For example and .

Given , let

 Period(w)={(r,α)∣rα=w and r∈Prf(w)∖{ϵ} and α∈Q%andα>1}.

The set contains all couples such that . Let

 MinPer(w)=min{|r|∣(r,α)∈Period(w)}∈N.

The positive integer is the minimal period of the word . The word has a period if there is a couple such that .

We will deal a lot with periodic palindromes. The two following known lemmas will be useful for us.

Lemma 2.1.

(see [10.1007/978-3-662-46078-8_24, Lemma 1]) Suppose is a period of a nonempty palindrome ; then there are palindromes and such that , , and .

Lemma 2.2.

(see [10.1007/978-3-662-46078-8_24, Lemma 2]) Suppose is a palindrome and is its proper suffix-palindrome or prefix-palindrome; then the number is a period of .

3 Periodic palindromic factors

We start the section with a definition of a set of real non-decreasing functions that diverge as tends towards the infinity.

Let denote the set of all functions such that

• ,

• , and

• .

Let , let , let , and let

 Ω(w,k)={t∈Fac(w)∣|PalPrf(t)|≥τ(|t|,k)}.

The definition says that the set contains a factor of if the number of palindromic prefixes of is bigger than or equal to .

The next proposition asserts that if is an infinite word with a bounded palindromic length, then the set of factors that have more than palindromic prefixes is infinite, where is the length of the factor in question and for each factor of .

If , and then .

Proof.

Suppose that and let

 K=max{|PalPrf(t)|∣t∈Ω(w,k)}.

Less formally said, the value is the maximal value from the set of numbers of palindromic prefixes of factors of that have more than palindromic prefixes. Clearly , because of the assumption .

Let be the shortest prefix of such that . Since , it is clear that such prefix exists.

To get a contradiction suppose that for some . Since and thus , it follows that and consequently . It is a contradiction, because . Hence we have that

 |PalPrf(t)|<τ(|p|,k) for each t∈Fac(p). (1)

Let and let

 Θ(n,j)={(v1,v2,…,vj)∣vi∈Pal+ and i∈{1,2,…,j} and |v1v2…vj|≤n and v1v2…vj∈Prf(w)}.

The set contains -tuples of nonempty palindromes whose concatenation is of length lower than or equal to and also the concatenation is a prefix of .

Thus from (1) we get that

 |Θ(|p|,j)|<(τ(|p|,k))j. (2)

The equation (2) follows from the fact that each factor of has at most palindromic prefixes. In consequence there are at most of -tuples of palindromes.

Let . Since we have from (2) that

 |¯Θ(|p|,k)|≤k|Θ(|p|,k)|

The inequality (3) says that the number of prefixes of having the form , where and is lower than the length of . But has nonempty prefixes. It is a contradiction. Since we conclude that is an infinite set. ∎

Remark 3.2.

In the proof of Proposition 3.1, we used the idea that the number of prefixes of a word of length that are a concatenation of at most palindromes is lower than . This idea was used also in Theorem in [FrPuZa].

We show that if is an infinite set of words such that the number of nonempty palindromic prefixes of grows more than as tends towards infinity then for each positive integer there are palindromes and a word such that is a prefix of and is nonempty. Realize that is a palindrome for each . This means that contains infinitely many words that have a periodic palindromic prefix of arbitrarily high exponent .

Proposition 3.3.

If , , , , and for each then for each there are palindromes , and a word such that .

Proof.

Given , let be the lengths of all palindromic prefixes of such that (a letter is a palindrome) and , where and . The integer is the number of nonempty palindromic prefixes of . Let . It is clear that

 μ(t,i+1)=μ(t,i)μ(t,i+1)μ(t,i). (4)

From (4) we have that

 μ(t,ht)μ(t,ht−1)μ(t,ht−1)μ(t,ht−2)μ(t,ht−2)μ(t,ht−3)⋯μ(t,2)μ(t,1)=μ(t,ht)≤|t|. (5)

Suppose that there is such that and for each and for each we have that . It follows from (5) that

 αht−1≤ht≤|t|. (6)

Let . Then . Since we get that

 αht−1|t|≥αϕ(|t|)−1|t|=αϕ(|t|)−1αcln|t|=αϕ(|t|)−1−cln|t|. (7)

Because the equation (7) implies that there is such that for each with we have that

 αht−1|t|≥αϕ(|t|)−1−cln|t|>1. (8)

From (6) and (8) we have that and , which is a contradiction. We conclude there is no such . In consequence, for each with there is and such that .

Let , let

 γ≤1j+1∈R+, (9)

let , and be such that . Let . Let be such that and . Then is a periodic palindrome with a period ; see Lemma 2.2. Lemma 2.1 implies that there are and such that for some . From Lemma 2.1 we have also that is the period of . Thus

 |ab|=μ(t,i)(δ−1)≤μ(t,i)(γ−1). (10)

From (9) and (10) it follows that

 |ab|≤μ(t,i)(γ−1)≤μ(t,i)1j. (11)

Note that and . Since we get that . From (11) we have that

 j≤μ(t,i)|ab|≤k.

Thus for arbitrary we found such that and . The proposition follows. ∎

A corollary of Proposition 3.1 and Proposition 3.3 says that if is an infinite word with a bounded palindromic length then for each positive integer there are palindromes such that is a factor of and is a nonempty word.

Corollary 3.4.

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

Proof.

Just take . Obviously . Then Proposition 3.3 implies the corollary. ∎

4 Palindromic length of concatenation

In this section we present some known results about the palindromic length of concatenation of two words.

The first lemma shows the very basic property of the palindromic length that the palindromic length of concatenation of two words and is lower than or equal to the sum of palindromic length of and .

If then .

Proof.

If or then obviously . Hence suppose that . Let and . Let and . Then is a factorization of into palindromes. Consequently . This completes the proof. ∎

An another basic property of the palindromic length says that if

 (t1,t2,…,tk)∈MPF(w)

is a minimal palindromic factorization of the word then the palindromic length of the factor is equal to for each and .

Lemma 4.2.

If , , and then for each with we have that .

Proof.

Since the word is concatenated of palindromes it is clear that . Suppose that . It would follow from Lemma 4.1 that

 PL(t1t2…tk)≤PL(t1t2…ti−1)+PL(titi+1d…tj)+PL(tj+1tj+2…tk)≤i−1+m+k−j

This is contradiction, since . The lemma follows. ∎

The following result has been proved in [10.1007/978-3-319-66396-8_19]. It says that if are words then the palindromic length of is the maximal absolute difference of palindromic lengths of and ; i.e. .

Lemma 4.3.

(see [10.1007/978-3-319-66396-8_19, Lemma ]) If then

• and

• .

The immediate corollary of Lemma 4.3 is that if is a word and is a palindrome then the absolute difference of palindromic lengths of and is at most .

If and then .

Proof.

It is enough to consider in Lemma 4.3 to be a palindrome. Thus we have if or if . The corollary follows. ∎

The next simple Corollary of Lemma 4.3 says that if are words such that is a palindrome then the absolute difference in palindromic lengths of and is at most .

If and then .

Proof.

If then , because clearly . Suppose that . It follows that , since . Without loss of generality suppose that . Let be such that . Then . Thus . Corollary 4.4 implies that . The corollary follows. ∎

5 Concatenation of periodic palindromes

To simplify the notation of the next two lemmas and the theorem we define an auxiliary set . Let be the set of all -tuples such that

• ,

• ,

• ,

• ,

• , and

• .

Remark 5.1.

The set contains all -tuples such that is a nonempty palindrome, is a palindrome (possibly empty), is a nonempty suffix of , is the minimal period of the word , and is a positive integer such that . It follows that , since is nonempty and thus .

Lemma 5.2.

If , , and then .

Proof.

Let , let with , and let be such that . Let be such that . Note that and thus , where .

Let and be such that , , and . Obviously such and exist. Let and . It is easy to see that .

We distinguish:

• If then for some and for such that .

• If then for some and for such that .

• If then for some and for such that .

In all three cases one can see that . It is easy to see that if then for each with . The lemma follows. ∎

Remark 5.3.

Note in the previous proof that with the condition it would be possible that . In the cases and we would have . That is why the condition necessary is. For this reason in the definition of we state that .

The next lemma shows that if , is the palindromic length of , and is a minimal palindromic factorization of then there is such that is a palindrome having the factor in the “center” of ; formally for some positive integer and for some proper suffix of .

Lemma 5.4.

If , , , and

 (t1,t2,…,tk)∈MPF(w)

then there are , , and such that .

Proof.

Suppose that for each . It follows that

 |t1t2…tk|<3k|vd|.

Since and it is a contradiction. It follows that there is such that . Lemma 5.2 asserts that . Then clearly there are and such that , , and .

To get a contradiction suppose that . Without loss of generality suppose that . It follows that . Obviously . Thus we have two palindromes and . Lemma 2.2 implies that is periodic with a period

 δ=|p1d(vd)γpR1|−|p1d(vd)γp2|=|p1|−|p2|.

Clearly . This is a contradiction to the condition , see Definition of . We conclude that . The lemma follows. ∎

The main theorem of the article says that if are palindromes, is nonempty, is a nonempty suffix of , , is the minimal period of , and is a positive integer such that then the palindromic length of the word is bigger than or equals to the palindromic length of .

Theorem 5.5.

If , , and then .

Proof.

Let . Lemma 5.4 asserts that there are , , and such that .

Let and be such that . Realize that and . Note that or can be the empty word; then or respectively. Lemma 4.2 implies that

 PL(w)=PL(t1t2…tj−1)+PL(tj)+PL(tj+1tj+2…tk)=PL(a)+PL(tj)+PL(b). (12)

We distinguish three distinct cases.

1. : This case is depicted in Table 1. Let be such that . Let be such that . It follows that and .

Then we have that for some . Hence . In consequence and

 PL(b)≥PL(u2)−1, (13)

since and ; see Corollary 4.5.

Lemma 4.1 implies that

 PL(a)+PL(u2)≥PL(u). (14)

From (12), (13), and (14) we have that

 PL(w)=PL(a)+PL(tj)+PL(b)≥PL(a)+1+PL(u2)−1≥PL(u).