Construction Of A Rich Word Containing Given Two Factors

TL;DR

The paper addresses the open problem of determining whether two given rich words can be factors of a larger rich word, providing an explicit bound on the length of such a word for decision purposes.

Contribution

It establishes a constructive bound on the length of a rich word containing two given rich factors, enabling a finite check for their coexistence within a larger rich word.

Findings

Provides a bound on the length of a rich word containing two given rich factors.

Shows it is sufficient to check all rich words up to a certain length to decide factor inclusion.

Addresses an open problem in the combinatorics of rich words.

Abstract

A finite word $w$ with $\vert w\vert=n$ contains at most $n+1$ distinct palindromic factors. If the bound $n+1$ is attained, the word $w$ is called \emph{rich}. Let $\Factor(w)$ be the set of factors of the word $w$. It is known that there are pairs of rich words that cannot be factors of a common rich word. However it is an open question how to decide for a given pair of rich words $u,v$ if there is a rich word $w$ such that $\{u,v\}\subseteq \Factor(w)$. We present a response to this open question:\\ If $w_1, w_2,w$ are rich words, $m=\max{\{\vert w_1\vert,\vert w_2\vert\}}$, and $\{w_1,w_2\}\subseteq \Factor(w)$ then there exists also a rich word $\bar w$ such that $\{w_1,w_2\}\subseteq \Factor(\bar w)$ and $\vert \bar w\vert\leq m2^{k(m)+2}$, where $k(m)=(q+1)m^2(4q^{10}m)^{\log_2{m}}$ and $q$ is the size of the alphabet. Hence it is enough to check all rich words of length equal or lower to $m2^{k(m)+2}$ in order to decide if there is a rich word containing factors $w_1,w_2$.

Equations

$$|F_p(w)\cap A^n|\le (q+1)n(4q^{10}n{)}^{{\log }_{2}n}\text{mbox}.$$

$$|T(w)|\le (q+1)n^2(4q^{10}n{)}^{{\log }_{2}n}\text{mbox}.$$

Full text

If $w_{1},w_{2},w$ are rich words, $m=\max{\{|w_{1}|,|w_{2}|\}}$, and $\{w_{1},w_{2}\}\subseteq\operatorname{F}(w)$ then there exists also a rich word $\bar{w}$ such that $\{w_{1},w_{2}\}\subseteq\operatorname{F}(\bar{w})$ and $|\bar{w}|\leq m2^{k(m)+2}$, where $k(m)=(q+1)m^{2}(4q^{10}m)^{\log_{2}{m}}$ and $q$ is the size of the alphabet. Hence it is enough to check all rich words of length equal or lower to $m2^{k(m)+2}$ in order to decide if there is a rich word containing factors $w_{1},w_{2}$.

1 Introduction

In the last years there have appeared several articles dealing with rich words; see, for instance, [1], [2], [3], [5]. Recall that a palindrome is a word that reads the same forwards and backwards, for example “noon” and “level”. Rich words are those words that contain the maximal number of palindromic factors. It is known that a word of length $n$ can contain at most $n+1$ palindromic factors including the empty word. The notion of a rich word has been extended also to infinite words. An infinite word is called rich if its every finite factor is rich [4], [3].

Let $\operatorname{lps}(w)$ and $\operatorname{lpp}(w)$ denote the longest palindromic suffix and the longest palindromic prefix of a word $w$, respectively. The authors of [1] showed the following property of rich words:

Proposition 1.1**.**

If $r,t$ are two factors of a rich word $w$ such that $\operatorname{lps}(r)=\operatorname{lps}(t)$ and $\operatorname{lpp}(r)=\operatorname{lpp}(t)$, then $r=t$.

Two related open questions can be found:

• •

In [5]: Is the condition in Proposition 1.1 sufficient for joining two rich words $u$ and $v$ into factors of a same rich word?

• •

In [3]: We do not know how to decide whether two rich words $u$ and $v$ are factors of a common rich word $w$.

In the current article we present a response to the question from [3] in the following form: We prove that if $w_{1},w_{2},w$ are rich words, $m=\max{\{|w_{1}|,|w_{2}|\}}$, and $\{w_{1},w_{2}\}\subseteq\operatorname{F}(w)$ then there exists a rich word $\bar{w}$ such that $\{w_{1},w_{2}\}\subseteq\operatorname{F}(\bar{w})$ and $|\bar{w}|\leq m2^{k(m)+2}$, where $k(m)=(q+1)m^{2}(4q^{10}m)^{\log_{2}{m}}$ and $q$ is the size of the alphabet. Thus it is enough to check all rich words of length equal or lower to $m2^{k(m)+2}$ in order to decide if there is a rich word containing factors $w_{1},w_{2}$.

We describe the basic ideas of the proof. If $w$ is a rich word, then let $a$ be a letter such that $\operatorname{lps}(wa)=a\operatorname{lpps}(w)a$, where $\operatorname{lpps}$ denotes the longest proper palindromic suffix. It is known and easy to show that $wa$ is a rich word [5]. Thus every rich word $w$ can be richly extended to a word $wa$. We will call $wa$ a standard extension of $w$. If there is a letter $b$ such that $a\not=b$ and $wb$ is also a rich word, then we call the longest palindromic suffix of $wb$ a flexed palindrome; the explication of the terminology is that $wb$ is not a standard extension of $w$, hence $wb$ is “flexed” from the standard extension.

We define a set $\Gamma$ of pairs of rich words $(w,r)$, where $r$ is a flexed palindrome of $w$, the longest palindromic prefix of $w$ does not contain the factor $r$, and $|r|\geq|\bar{r}|$ for each flexed palindrome $\bar{r}$ of $w$. If $(w,r)\in\Gamma$, $w_{1}$ is the prefix of $w$ with $|w_{1}|=|r|-1$ and $w_{2}$ is the suffix of $w$ with $|w_{2}|=|r|-1$ then we construct a rich word $\bar{w}$ possessing the following properties:

• •

The word $w_{1}$ is a prefix of $\bar{w}$.

• •

The word $w_{2}$ is a suffix of $\bar{w}$.

• •

The number of occurrences of $r$ in $\bar{w}$ is strictly smaller than the number of occurrences of $r$ in $w$.

• •

The set of flexed palindromes of $\bar{w}$ is a subset of the set of flexed palindromes of $w$.

Iterative applying of this construction will allow us for a given rich word $w$ with a prefix $w_{1}$ and a suffix $w_{2}$ to construct a rich word $t$ containing factors $w_{1},w_{2}$ and having no flexed palindrome longer than $m$, where $m=\max\{|w_{1}|,|w_{2}|\}$.

Another important, but simple, observation is that if $w$ is a rich word with prefix $u$ such that the number of flexed palindromes in $w$ is less than $k$ and $u$ has exactly one occurrence in $w$ then there is an upper bound for the length of $w$. We show this upper bound as a function of $k$ and consequently we derive an upper bound for the length of $t$.

2 Preliminaries

Let $\operatorname{A}$ be a finite alphabet with $q=|\operatorname{A}|$. The elements of $\operatorname{A}$ will be called letters.

Let $\epsilon$ denote the empty word.

Let $\operatorname{A}^{*}$ be the set of all finite words over $\operatorname{A}$ including the empty word, let $\operatorname{A}^{n}\subset\operatorname{A}^{*}$ be the set of all words of length $n$, and let $\operatorname{A}^{+}=\operatorname{A}^{*}\setminus\{\epsilon\}$.

Let $\operatorname{R}\subset\operatorname{A}^{*}$ denote the set of all rich words and let $\operatorname{R}^{+}=\operatorname{R}\cap\operatorname{A}^{+}$.

Let $\operatorname{F}(w)\subset\operatorname{A}^{*}$ denote the set of all factors of $w\in\operatorname{A}^{*}$; we state explicitly that $\epsilon,w\in\operatorname{F}(w)$.

Let $\operatorname{F}(S)=\bigcup_{v\in S}\operatorname{F}(v)$, where $S\subseteq\operatorname{A}^{*}$.

Let $\operatorname{F}_{p}(w)\subseteq\operatorname{F}(w)$ be set of all palindromic factors of $w\in\operatorname{A}^{*}$.

Let $\operatorname{F}(w,r)=\{u\mid u\in\operatorname{F}(w)\mbox{ and }r\not\in\operatorname{F}(u)\}\subseteq\operatorname{F}(w)$, where $w,r\in\operatorname{A}^{*}$. The set $F(w,r)$ contains factors of $w$ avoiding the factor $r$.

Let $\operatorname{F}_{p}(w,r)=\operatorname{F}_{p}(w)\cap\operatorname{F}(w,r)$.

Let $\operatorname{Prf}(w)$ and $\operatorname{Suf}(w)$ be the set of all prefixes and all suffixes of $w\in A^{*}$ respectively; we define that $\{\epsilon,w\}\subseteq\operatorname{Prf}(w)\cap\operatorname{Suf}(w)$.

Let $w^{R}$ denote the reversal of $w\in A^{*}$; formally if $w=w_{1}w_{2}\dots w_{k}$ then $w^{R}=w_{k}\dots w_{2}w_{1}$, where $w_{i}\in\operatorname{A}$ and $i\in\{1,2,\dots,k\}$. In addition we define that $\epsilon^{R}=\epsilon$.

Let $\operatorname{lps}(w)$ and $\operatorname{lpp}(w)$ denote the longest palindromic suffix and the longest palindromic prefix of $w\in\operatorname{A}^{*}$ respectively. We define that $\operatorname{lps}(\epsilon)=\operatorname{lpp}(\epsilon)=\epsilon$.

Let $\operatorname{lpps}(w)$ and $\operatorname{lppp}(w)$ denote the longest proper palindromic suffix and the longest proper palindromic prefix of $w\in\operatorname{A}^{*}$ respectively, where $|w|\geq 2$.

Let $\operatorname{trim}(w)=v$, where $v,w\in\operatorname{A}^{*}$, $x,y\in\operatorname{A}$, $w=xvy$, and $|w|\geq 2$.

Let $\operatorname{rtrim}(w)=v$, where $v,w\in\operatorname{A}^{*}$, $y\in\operatorname{A}$, $w=vy$, and $|w|\geq 1$.

Let $\operatorname{ltrim}(w)=v$ , where $v,w\in\operatorname{A}^{*}$, $x\in\operatorname{A}$, $w=xv$, and $|w|\geq 1$.

Example 2.1**.**

• •

$\operatorname{A}=\{1,2,3,4,5\}$.

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$w=124135$.

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$\operatorname{trim}(w)=2413$.

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$\operatorname{ltrim}(w)=24135$.

• •

$\operatorname{rtrim}(w)=12413$.

Let $\operatorname{pc}(w)$ be the palindromic closure of $w\in\operatorname{A}^{*}$; formally $\operatorname{pc}(w)=uvu^{R}$, where $w=uv$ and $v=\operatorname{lps}(w)$.

Let $\operatorname{MinLenWord}(U)$ and $\operatorname{MaxLenWord}(U)$ be the shortest and the longest word from the set $U$ respectively, where either $U\subseteq\operatorname{Prf}(w)$ or $U\subseteq\operatorname{Suf}(w)$ for some $w\in A^{*}$. If $U=\emptyset$ then we define $\operatorname{MinLenWord}(U)=\epsilon$ and $\operatorname{MaxLenWord}(U)=\epsilon$.

Let $\operatorname{lcp}(w_{1},w_{2})$ be the longest common prefix of words $w_{1},w_{2}\in\operatorname{A}^{*}$; formally $\operatorname{lcp}(w_{1},w_{2})=\operatorname{MaxLenWord}(\operatorname{Prf}(w_{1})\cap\operatorname{Prf}(w_{2}))$.

Let $\operatorname{lcs}(w_{1},w_{2})$ be the longest common suffix of words $w_{1},w_{2}\in\operatorname{A}^{*}$; formally $\operatorname{lcs}(w_{1},w_{2})=\operatorname{MaxLenWord}(\operatorname{Suf}(w_{1})\cap\operatorname{Suf}(w_{2}))$.

Let $\operatorname{occur}(u,v)$ be the number of occurrences of $v$ in $u$, where $u,v\in\operatorname{A}^{*}$ and $|v|>0$; formally $\operatorname{occur}(u,v)=|\{w\mid w\in\operatorname{Suf}(u)\mbox{ and }v\in\operatorname{Prf}(w)\}|$.

Recall the notion of a complete return [2]: Given a word $w$ and factors $r,u\in\operatorname{F}(w)$, we call the factor $r$ a complete return to $u$ in $w$ if $r$ contains exactly two occurrences of $u$, one as a prefix and one as a suffix.

We list some known properties of rich words that we use in our article. All of them can be found, for instance, in [2].

Proposition 2.2**.**

If $w,u\in R^{+}$ and $u\in\operatorname{F}_{p}(w)$ then all complete returns to $u$ in $w$ are palindromes.

Proposition 2.3**.**

If $w\in\operatorname{R}$ and $p\in\operatorname{F}(w)$ then $p,p^{R}\in\operatorname{R}$.

Proposition 2.4**.**

A word $w$ is rich if and only if every prefix $p\in\operatorname{Prf}(w)$ has a unioccurrent palindromic suffix.

3 Standard Extensions and Flexed Palindromes

We start with a formal definition of a standard extension and a flexed palindrome introduced at the beginning of the article.

Definition 3.1**.**

Let $j\geq 0$ be a nonnegative integer, $w\in\operatorname{R}$, and $|w|\geq 2$. We define $\operatorname{StdExt}(w,j)$ as follows:

• •

$\operatorname{StdExt}(w,0)=w$.

• •

$\operatorname{StdExt}(w,1)=wa$* such that $\operatorname{lps}(wa)=a\operatorname{lpps}(w)a$ and $a\in\operatorname{A}$.*

• •

$\operatorname{StdExt}(w,j)=\operatorname{StdExt}(\operatorname{StdExt}(w,j-1),1)$, where $j>1$.

*Let $\operatorname{StdExt}(w)=\{\operatorname{StdExt}(w,j)\mid j\geq 0\}$. If $p\in\operatorname{StdExt}(w)$ then we call $p$ a standard extension of $w$.

Let $\operatorname{T}(w)=\{\operatorname{lps}(ub)\mid ub\in\operatorname{Prf}(w)\mbox{ and }b\in\operatorname{A}\mbox{ and }ub\not=\operatorname{StdExt}(u,1)\}$. If $r\in\operatorname{T}(w)$ then we call $r$ a flexed palindrome of $w$.*

For a given rich word $w\in\operatorname{R}$ having a flexed palindrome $r$ we define a standard palindromic replacement of $r$ to be the longest palindromic suffix of a standard extension of a prefix $p$ of $w$ such that $\operatorname{lps}(px)=r$, where $px$ is a prefix of $w$ and $x\in\operatorname{A}$. The idea is that we can “replace” $r$ with the standard palindromic replacement.

Definition 3.2**.**

*Let $\operatorname{stdPalRep}(w,r)=\operatorname{lps}(\operatorname{StdExt}(h,1))$, where $w,r\in\operatorname{R}$, $r\in\operatorname{T}(w)$, $hx\in\operatorname{Prf}(w)$, $x\in\operatorname{A}$, and $\operatorname{lps}(hx)=r$.

We call $\operatorname{stdPalRep}(w,r)$ a standard palindromic replacement of $r$ in $w$.*

Example 3.3**.**

• •

$\operatorname{A}=\{0,1\}$.

• •

$w=110101100110011$.

• •

$001100\in\operatorname{T}(w)$.

• •

$\operatorname{lps}(1101011001100)=001100$.

• •

$\operatorname{StdExt}(110101100110,1)=1101011001101$.

• •

$\operatorname{stdPalRep}(w,001100)=\operatorname{lps}(1101011001101)=1011001101$.

We show that the length of a flexed palindrome $r$ is less than the length of the standard palindromic replacement $\operatorname{stdPalRep}(w,r)$.

Lemma 3.4**.**

If $ux,uy\in\operatorname{R}$, $x,y\in\operatorname{A}$, $x\not=y$, and $ux=\operatorname{StdExt}(u,1)$ then $|\operatorname{lps}(ux)|>|\operatorname{lps}(uy)|$.

Proof.

Let $yty=\operatorname{lps}(uy)$. From the definition of a standard extension we have $\operatorname{lps}(ux)=xvx$, where $v=\operatorname{lpps}(u)$ and hence $t\in\operatorname{Suf}(v)$. Since $y\not=x$ we have also $yt\in\operatorname{Suf}(v)$. The lemma follows. ∎

An obvious corollary is that a flexed palindrome of $w$ is not a prefix of $w$.

Corollary 3.5**.**

If $w,r\in\operatorname{R}$ and $r\in\operatorname{T}(w)$ then $r\not\in\operatorname{Prf}(w)$.

In [5] the standard extension has been used to prove that each rich word $w$ can be extended “richly”; this means that there is $a\in A$ such that $wa$ is rich.

Lemma 3.6**.**

If $w\in\operatorname{R}$ and $|w|\geq 2$ then $\operatorname{StdExt}(w)\subset R$.

Proof.

Obviously it is enough to prove that $\operatorname{StdExt}(w,1)\in\operatorname{R}$, since for every $t\in\operatorname{StdExt}(w)\setminus\{w\}$ there is a rich word $\bar{t}$ such that $t=\operatorname{StdExt}(\bar{t},1)$.

Let $xpx=\operatorname{lps}(\operatorname{StdExt}(w,1))$, where $x\in\operatorname{A}$. Proposition 2.4 implies that we need to prove that $xpx$ is unioccurrent in $\operatorname{StdExt}(w,1)$. Realize that $p$ is unioccurrent in $w$, hence $xpx$ is unioccurrent in $\operatorname{StdExt}(w,1)$. ∎

To simplify the proofs of the paper we introduce a function $\operatorname{MaxStdExt}(u,v)$ to be the longest prefix $z$ of $u$ such that $z$ is also a standard extension of $v$:

Definition 3.7**.**

Let $\operatorname{MaxStdExt}(u,v)=\operatorname{MaxLenWord}(\{\operatorname{StdExt}(v)\cap\operatorname{Prf}(u)\})$. We call $\operatorname{MaxStdExt}(u,v)$ a maximal standard extension of $v$ in $u$.

The next lemma shows that if a rich word contains factors $ypx$ and $ypy$, where $p$ is a palindrome, $p$ is not a prefix of $w$, $x,y$ are distinct letters, and $ypx$ “occurs” before $ypy$ in $w$ then $ypy$ is a flexed palindrome.

Lemma 3.8**.**

If $w,v,p\in\operatorname{R}$, $v\in\operatorname{Prf}(w)$, $p\not\in\operatorname{Prf}(w)$, $x,y\in\operatorname{A}$, $x\not=y$, $ypx\in\operatorname{Suf}(v)$, $ypy\not\in\operatorname{F}(v)$, and $ypy\in\operatorname{F}(w)$ then $ypy\in\operatorname{T}(w)$.

Proof.

Let $\bar{v}$ be such that $\bar{v}y\in\operatorname{Prf}(w)$, $ypy\in\operatorname{Suf}(\bar{v}y)$, and $\operatorname{occur}(\bar{v}y,ypy)=1$. Let $u=\operatorname{lps}(\bar{v})$. Because $p\not\in\operatorname{Prf}(w)$ it follows that $u=\operatorname{lpps}(\bar{v})=\operatorname{lps}(\bar{v})$ and thus there is $z\in\operatorname{A}$ such that $zu\in\operatorname{Suf}(\bar{v})$. Obviously $v\in\operatorname{Prf}(\bar{v})$ and hence $\operatorname{occur}(\bar{v},p)>1$. Proposition 2.2 implies that $\operatorname{occur}(u,p)>1$, since the complete return to $p$ which is a suffix of $\bar{v}$ must a suffix of $u$. It follows that $yp\in\operatorname{Suf}(u)$ and Lemma 3.4 implies that $ypy\in\operatorname{T}(w)$. ∎

4 Removing flexed points

We define formally the set $\Gamma$ mentioned in the introduction. An element $(w,r)$ of the set $\Gamma$ represents a rich word $w$ for which we are able to construct a new rich word $\bar{w}$ such that $\bar{w}$ does not contain the flexed palindrome $r$, but $\bar{w}$ have certain common prefixes and suffixes with $w$. We define that $r$ is one of the longest flexed palindromes of $w$ and that $r$ is not a factor of the longest palindromic prefix of $w$. In addition we require that $|r|>2$ so that the standard extension of $\operatorname{rtrim}(r)$ would be defined.

Definition 4.1**.**

Let $\Gamma$ be a set defined as follows: $(w,r)\in\Gamma$ if

$w,r\in\operatorname{R}$* and* 2. 2.

$|r|>2$. 3. 3.

$r\in\operatorname{T}(w)$* and* 4. 4.

$r\not\in\operatorname{F}(\operatorname{lpp}(w))$* and* 5. 5.

$|r|\geq|\bar{r}|$* for each $\bar{r}\in\operatorname{T}(w)$.*

Given $(w,r)\in\Gamma$, we need to express $w$ as a concatenation of its factors having some special properties. For this reason we define a function $\operatorname{parse}(w,r)$:

Definition 4.2**.**

If $(w,r)\in\Gamma$ then let $\operatorname{parse}(w,r)=(v,z,t)$, where

• •

$v,z,t\in\operatorname{R}$* and*

• •

$vzt=w$* and*

• •

$r\in\operatorname{Suf}(v)$* and*

• •

$\operatorname{occur}(w,r)=\operatorname{occur}(v,r)$* and*

• •

$vz=\operatorname{MaxStdExt}(vzt,v)$.

Remark 4.3**.**

The prefix $v$ is the shortest prefix of $w$ that contains all occurrences of $r$. The prefix $vz$ is the maximal standard extension of $v$ in $w$, and $t$ is such that $vzt=w$. It is easy to see that $v,z,t$ exist and are uniquely determined for $(w,r)\in\Gamma$.

For an element $(w,r)\in\Gamma$ we define a function $\operatorname{rdcPrf}(w,r)$ (a reduced prefix), which is a prefix of the palindromic closure of some prefix of $w$. In Theorem 4.12 we show that the concatenation of $\operatorname{rdcPrf}(w,r)$ and $t$ is a rich word having a strictly smaller number of occurrences of $r$ than in $w$, where $(v,z,t)=\operatorname{parse}(w,r)$.

Definition 4.4**.**

*If $w,r\in\Gamma$ and $(v,z,t)=\operatorname{parse}(w,r)$ then let $\operatorname{rdcPrf}(w,r)$ be defined as follows:

It follows from Property 4 of Definition 4.1 that there is $h\in\operatorname{Prf}(w)$ such that $w=hz^{R}\operatorname{lps}(v)zt$. Note that $\operatorname{lps}(v)\not=v$ since $r\in\operatorname{T}(w)$ and thus $r\not\in\operatorname{Prf}(w)$, see Corollary 3.5. It is clear that $r\in\operatorname{Prf}(\operatorname{lps}(v))\cap\operatorname{Suf}(\operatorname{lps}(v))$. This implies that $hz^{R}r\in\operatorname{Prf}(w)$. We distinguish two cases:*

• •

$r\in\operatorname{F}(hz^{R}\operatorname{rtrim}(r))$:

Let $g$ be the complete return to $r$ such that $g\in\operatorname{Suf}(hz^{R}r)$. Clearly $rz\in\operatorname{Prf}(g)$ and $z^{R}r\in\operatorname{Suf}(g)$, since $r\not\in\operatorname{F}(\operatorname{ltrim}(r)z)$; recall $r\in\operatorname{Suf}(v)$ and $\operatorname{occur}(v,r)=\operatorname{occur}(vzt,r)$. Let $\bar{g}$ be such that $\bar{g}g=hz^{R}r$.

We define $\operatorname{rdcPrf}(w,r)=\bar{g}rz$. Note that $\operatorname{rdcPrf}(w,r)\in\operatorname{Prf}(w)$.

• •

$r\not\in\operatorname{F}(hz^{R}\operatorname{rtrim}(r))$:

Let $\bar{u}=\operatorname{stdPalRep}(hz^{R}r,r)$. Clearly $\operatorname{lps}(hz^{R}r)=r$ and $\bar{u}\not=r$. Because $z^{R}\operatorname{rtrim}(r)\in\operatorname{Suf}(hz^{R}\operatorname{rtrim}(r))$, then obviously $U\not=\emptyset$ and $r\not\in\operatorname{F}(U)$, where $U=\{u\mid u\in\operatorname{Prf}(\operatorname{pc}(hz^{R}\operatorname{rtrim}(r)))\mbox{ and }\operatorname{ltrim}(r)z\in\operatorname{Suf}(u)\}$. We define $\operatorname{rdcPrf}(w,r)=\operatorname{MinLenWord}(U)$. Note that $r\not\in\operatorname{F}(\operatorname{rdcPrf}(w,r))$.

We call $\operatorname{rdcPrf}(w,r)$ a reduced prefix of $w$ by $r$.

Remark 4.5**.**

Note in Definition 4.4 in the second case where $r\not\in\operatorname{F}(hz^{R}\operatorname{rtrim}(r))$ that it may happen that $\operatorname{rdcPrf}(w,r)$ is not a prefix of $w$. However it is a prefix of a palindromic closure of $hz^{R}\operatorname{rtrim}(r)$, hence the number of flexed palindromes remains the same; formally $|\operatorname{T}(hz^{R}\operatorname{rtrim}(r)))|=|\operatorname{T}(\operatorname{rdcPrf}(w,r))|$. Realize that $\operatorname{pc}(t)\in\operatorname{StdExt}(t)$ for each $t\in\operatorname{R}$ and $|t|\geq 2$.

To clarify the definition of the reduced prefix $\operatorname{rdcPrf}(w,r)$ we present below two examples representing those two cases in the definition.

Example 4.6**.**

• •

$\operatorname{A}=\{1,2,3,4,5,6,7,8,9\}$.

• •

$w=123999322399932442399932255223993$.

• •

$r=999$.

• •

$v=1239993223999324423999$.

• •

$z=322$.

• •

$t=55223993$.

• •

$\operatorname{lps}(v)=999324423999$.

• •

$h=1239993$.

• •

$w=hz^{R}\operatorname{lps}(v)zt$.

• •

$g=9993223999\in\operatorname{Suf}(hz^{R}r)=\operatorname{Suf}(1239993223999)$.

• •

$\bar{g}=123$.

• •

$\operatorname{rdcPrf}(w,r)=123999322$.

Example 4.7**.**

• •

$\operatorname{A}=\{1,2,3,4,5,6,7,8,9\}$.

• •

$w=123999599932239949$.

• •

$r=999$.

• •

$v=1239995999$.

• •

$z=32$.

• •

$t=239949$.

• •

$\operatorname{lps}(v)=9995999$.

• •

$h=1$.

• •

$w=hz^{R}\operatorname{lps}(v)zt$.

• •

$\operatorname{StdExt}(hz^{R}\operatorname{rtrim}(r),1)=\operatorname{StdExt}(12399,1)=123993$.

• •

$\bar{u}=\operatorname{stdPalRep}(123999,999)=3993$.

• •

$\operatorname{pc}(12399)=12399321$.

• •

$U=\{1239932\}$.

• •

$\operatorname{rdcPrf}(w,r)=1239932$.

We know that the reduced prefix $\operatorname{rdcPrf}(w,r)$ may not be a prefix of $w$, however we show that the longest common prefix of $\operatorname{rdcPrf}(w,r)$ and $w$ is longer than $|r|-1$.

Lemma 4.8**.**

If $(w,r)\in\Gamma$ and $u=\operatorname{rdcPrf}(w,r)$ then $|\operatorname{lcp}(u,w)|\geq|r|-1$.

Proof.

In Definition 4.4 in the first case where $r\not\in\operatorname{F}(hz^{R}\operatorname{rtrim}(r))$ and $u\in\operatorname{Prf}(w)$, it is clear that $r\in\operatorname{F}(u)$ and thus $|u|\geq|r|$. Hence we need to verify only the second case, where $r\not\in\operatorname{F}(hz^{R}\operatorname{rtrim}(r))$. Either $hz^{R}\operatorname{rtrim}(r)\in\operatorname{Prf}(u)$ or $u\in\operatorname{Prf}(hz^{R}\operatorname{rtrim}(r))$. Since $\operatorname{ltrim}(r)z^{R}\in\operatorname{Suf}(u)$ the lemma follows. ∎

Using the reduced prefix we can now define the word $\operatorname{rdcWrd}(w,r)$ (a reduced word):

Definition 4.9**.**

Let $\operatorname{rdcWrd}(w,r)=\operatorname{rdcPrf}(w,r)t$, where $(v,z,t)=\operatorname{parse}(w,r)$ and $(w,r)\in\Gamma$. We call $\operatorname{rdcWrd}(w,r)$ a reduced word of $w$ by $r$.

We show that the longest common suffix of the reduced word $\operatorname{rdcWrd}(w,r)$ and $w$ is longer than $|r|-1$.

Lemma 4.10**.**

If $(w,r)\in\Gamma$ then $\operatorname{lcs}(\operatorname{rdcWrd}(w,r),w)|\geq|r|-1$.

Proof.

Given $(w,r)\in\Gamma$ and $(v,z,t)=\operatorname{parse}(w,r)$. From Definition 4.4 of the reduced prefix, it is obvious that $\operatorname{ltrim}(r)z\in\operatorname{Suf}(\operatorname{rdcPrf}(w,r))$ and consequently $\operatorname{ltrim}(r)zt\in\operatorname{Suf}(\operatorname{rdcWrd}(w,r))$. Recall Definition 4.2 of the function $\operatorname{parse}(w,r)$. Since $w=vzt$ and $r\in\operatorname{Suf}(v)$ it follows that $\operatorname{ltrim}(r)zt\in\operatorname{Suf}(w)$. This implies that $\operatorname{ltrim}(r)zt$ is a common suffix of $w$ and $\operatorname{rdcWrd}(w,r)$. Because $|\operatorname{ltrim}(r)|=|r|-1$ the lemma follows. ∎

As already mentioned the reduced prefix $\operatorname{rdcPrf}(w,r)$ is not necessarily a prefix of $w$. In such a case $\operatorname{rdcPrf}(w,r)$ contains palindromic factors that are not factors of the longest common prefix $\operatorname{lcp}(w,\operatorname{rdcPrf}(w,r))$. We show that none of these palindromes is a factor of $w$. This will be important when proving richness of the word $\operatorname{rdcWrd}(w,r)$.

Proposition 4.11**.**

If $(w,r)\in\Gamma$, $u=\operatorname{rdcPrf}(w,r)$, $\bar{u}=\operatorname{stdPalRep}(w,r)$, and $g=\operatorname{lcp}(w,u)$, then $\operatorname{F}_{p}(u,\bar{u})\subseteq\operatorname{F}_{p}(g)$ and $\bar{u}\not\in\operatorname{F}_{p}(w)$.

Proof.

From the properties of the palindromic closure it is easy to see that $\operatorname{F}_{p}(\operatorname{pc}(f),\operatorname{lps}(f))\subseteq\operatorname{F}_{p}(f)$ for each $f\in\operatorname{R}$. It means that every palindromic factor of $f$ that is not a factor of $\operatorname{pc}(f)$ contains the factor $\operatorname{lps}(f)$. It follows that $\operatorname{F}_{p}(u,\bar{u})\subseteq\operatorname{F}_{p}(g)$.

We show that $\operatorname{occur}(w,\bar{u})=0$. Let $\bar{u}=xtx$ and $r=ypy$, where $x,y\in\operatorname{A}$. Obviously $x\not=y$, $py\in\operatorname{Prf}(t)$, and $yp\in\operatorname{Suf}(p)$. Thus $xty\in\operatorname{F}(w)$. Lemma 3.8 implies that $\bar{u}\in\operatorname{F}(w)$ if and only if $\bar{u}\in\operatorname{T}(w)$. In addition Lemma 3.4 implies that $|\bar{u}|>|r|$. This is a contradiction to Property 5 of Definition 4.1. Hence $\bar{u}\not\in\operatorname{F}_{p}(w)$. This completes the proof. ∎

The main theorem of the paper states the the reduced word $\operatorname{rdcWrd}(w,r)$ is rich, where $(w,r)\in\Gamma$. In addition the theorem asserts that the set of flexed palindromes of $\operatorname{rdcWrd}(w,r)$ is a subset of the set of flexed palindromes of the word $w$, the number of occurrences of $r$ is strictly smaller in $\operatorname{rdcWrd}(w,r)$ than in $w$, and the longest common prefix and suffix of $\operatorname{rdcWrd}(w,r)$ and $w$ are longer than $|r|-1$.

Theorem 4.12**.**

If $(w,r)\in\Gamma$ then

• •

$\operatorname{rdcWrd}(w,r)\in\operatorname{R}$* and*

• •

$\operatorname{T}(\operatorname{rdcWrd}(w,r))\subseteq\operatorname{T}(w)$* and*

• •

$\operatorname{occur}(\operatorname{rdcWrd}(w,r),r)<\operatorname{occur}(w,r)$* and*

• •

$|\operatorname{lcp}(\operatorname{rdcWrd}(w,r),w)|\geq|r|-1$* and*

• •

$|\operatorname{lcs}(\operatorname{rdcWrd}(w,r),w)|\geq|r|-1$.

Proof.

Recall that $\operatorname{rdcWrd}(w,r)=ut$, where $(v,z,t)=\operatorname{parse}(w,r)$ and $u=\operatorname{rdcPrf}(w,r)$. Suppose that $up\in\operatorname{R}$, $\operatorname{T}(up)\subseteq\operatorname{T}(vzp)$, where $p\in\operatorname{Prf}(\operatorname{rtrim}(t))$. In addition suppose that if $|p|\geq 1$ then $\operatorname{lps}(up)=\operatorname{lps}(vzp)$ and $r\not\in\operatorname{F}(\operatorname{lps}(vzp))$. The assumptions obviously hold for $p=\epsilon$.

Let $x\in\operatorname{A}$ be such that $px\in\operatorname{Prf}(t)$. We show that the assumptions hold also for $upx$.

Proposition 2.4 implies that if $f\in\operatorname{R}$ and $y\in\operatorname{A}$ then $fy\in\operatorname{R}$ if and only if $fy$ has a unioccurrent palindromic suffix. Using this property we prove the theorem. We distinguish two cases:

• •

If $\operatorname{lps}(vzpx)\in\operatorname{T}(w)$ then Property 5 of Definition 4.1 implies that $\operatorname{lps}(vzpx)\in\operatorname{Suf}(\operatorname{ltrim}(r)zpx)$. From Definition 4.4 we know that $\operatorname{ltrim}(r)z\in\operatorname{Suf}(u)$. This implies that $\operatorname{lps}(vzpx)\in\operatorname{Suf}(upx)$ and $r\not\in\operatorname{F}(\operatorname{lps}(vzpx))$.

Proposition 4.11 implies that $\operatorname{lps}(vzpx)$ is unioccurrent in $upx$. In consequence $\operatorname{lps}(upx)=\operatorname{lps}(vzpx)$ and $upx\in\operatorname{R}$. Because $\operatorname{lps}(vzpx)\in\operatorname{T}(w)$ and $\operatorname{T}(up)\subseteq\operatorname{T}(vzp)$ we conclude that $\operatorname{T}(upx)\subseteq\operatorname{T}(vzpx)$. We do not need to prove that $\operatorname{lps}(upx)\in\operatorname{T}(upx)$, although it would not be difficult.

• •

If $\operatorname{lps}(vzpx)\not\in\operatorname{T}(w)$, then $|p|\geq 1$, because $vz=\operatorname{MaxStdExt}(vzt,v)$. Realize that $\operatorname{lps}(vzy)\in\operatorname{T}(w)$, where $y\in\operatorname{A}$ and $vzy\in\operatorname{Prf}(vzt)$.

Hence according to our assumptions we have $\operatorname{lps}(up)=\operatorname{lps}(vzp)$. Obviously $\operatorname{lps}(vzpx)=x\operatorname{lpps}(vzp)x$ and $r\not\in\operatorname{lpps}(vzp)$.

Suppose that $r\in\operatorname{F}(\operatorname{lps}(vzpx))$. Then $r\in\operatorname{Prf}(\operatorname{lps}(vzpx))\cap\operatorname{Suf}(\operatorname{lps}(vzpx))$. This is a contradiction since $\operatorname{occur}(v,r)=\operatorname{occur}(w,r)$, see Definition 4.2. This implies that $r\not\in\operatorname{lps}(vzpx)$. It follows that $|\operatorname{lps}(vzpx)|<|rzpx|$ and that $\operatorname{lps}(vzpx)\in\operatorname{Suf}(upx)$. In consequence $\operatorname{lps}(upx)=x\operatorname{lpps}(up)x$. Thus $upx$ is a standard extension of $up$. We conclude that $\operatorname{lps}(upx)=\operatorname{lps}(vzpx)$, $\operatorname{lps}(upx)\not\in\operatorname{T}(upx)$, $\operatorname{T}(upx)\subseteq\operatorname{T}(vzpx)$, and $upx\in\operatorname{R}$.

So we have $upx\in\operatorname{R}$ and $\operatorname{T}(upx)\subseteq\operatorname{T}(vzpx)$ for each $px\in\operatorname{Prf}(t)$. The fact that $\operatorname{occur}(ut,r)<\operatorname{occur}(w,r)$ follows simply from the construction of $u=\operatorname{rdcPrf}(w,r)$, see Definition 4.4.

Lemma 4.8 and Lemma 4.10 imply that $|\operatorname{lcp}(\operatorname{rdcWrd}(w,r),w)|\geq|r|-1$ and $|\operatorname{lcs}(\operatorname{rdcWrd}(w,r),w)|\geq|r|-1$. This completes the proof. ∎

Two more examples illuminate the construction of $\operatorname{rdcWrd}(w,r)$.

Example 4.13**.**

• •

$\operatorname{A}=\{1,2,3,4,5,6,7,8\}$.

• •

$w=vzt=12145656547745656545656547874$.

• •

$r=656$.

• •

$v=12145656547745656545656$.

• •

$z=547$.

• •

$t=874$.

• •

$\operatorname{lps}(v)=656545656$.

• •

$u=\operatorname{rdcPrf}(w,r)=12145656547$.

• •

$\operatorname{rdcWrd}(w,r)=ut=12145656547874$.

Example 4.14**.**

• $\operatorname{A}=\{1,2,3,4,5,6,7,8\}$.
• •

$w=vzt=12145656547874$.

• •

$r=656$.

• •

$v=12145656$.

• •

$z=54$.

• •

$t=7874$.

• •

$\operatorname{lps}(v)=656$.

• •

$u=\operatorname{rdcPrf}(w,r)=12145654$.

• •

$\operatorname{rdcWrd}(w,r)=ut=121456547874$.

If a rich word $w$ has a factor $u$, then the palindromic closure of $w$ is rich and contains the factor $u^{R}$. Hence for us when constructing a rich word containing given factors, it does not matter if $w$ contains $u$ or $u^{R}$. We introduce the notion of a reverse-unioccurrent factor.

Definition 4.15**.**

If $|\{u,u^{R}\}\cap\operatorname{F}(w)|=1$ then we say that a word $u$ is reverse-unioccurrent in $w$, where $w,u\in\operatorname{R}$.

We introduce a function $\operatorname{ruo}(w,u,v)$ (a reverse unioccurrence of $u,v$ in $w$) which returns a factor of $w$ such that $u,v$ are reverse unioccurrent. In addition we require that $u$ or $u^{R}$ is a prefix and $v$ or $v^{R}$ is a suffix of $\operatorname{ruo}(w,u,v)$.

Definition 4.16**.**

If $w_{1},w_{2},w\in\operatorname{R}$, $w_{1}\in\operatorname{Prf}(w)$ and $w_{2}\in\operatorname{Suf}(w)$, then let $\operatorname{M}(w,w_{1},w_{2})\subset\operatorname{F}(w)$ such that $t\in\operatorname{M}(w,w_{1},w_{2})$ if:

• •

$t\in\operatorname{F}(w)$* and*

• •

$w_{1},w_{2}$* are reverse-unioccurrent in $t$ and*

• •

$\{w_{1},w_{1}^{R}\}\cap\operatorname{Prf}(t)\not=\emptyset$* and*

• •

$\{w_{2},w_{2}^{R}\}\cap\operatorname{Suf}(t)\not=\emptyset$.

Let the set $\operatorname{M}(w,w_{1},w_{2})$ be ordered and let $\operatorname{ruo}(w,w_{1},w_{2})$ be the first element of $\operatorname{M}(w,w_{1},w_{2})$.

Remark 4.17**.**

It is not difficult to see that the function $\operatorname{ruo}(r,w_{1},w_{2})$ is well defined and the set $\operatorname{M}(w,w_{1},w_{2})$ is nonempty.

We define maximal flexed palindrome of a rich word $w$, which is a flexed palindrome $r$ of $w$, such $(w,r)\in\Gamma$ and $|r|>n$, where $n$ is a positive integer.

Definition 4.18**.**

Let $\operatorname{H}(w,n)=\{r\mid(w,r)\in\Gamma\mbox{ and }|r|>n\}$, let the set $\operatorname{H}(w,n)$ be ordered and let $\operatorname{maxFlxPal}(w,n)$ be the first element of $\operatorname{H}(w,n)$. If $\operatorname{H}(w,n)=\emptyset$ then we define $\operatorname{maxFlxPal}(w,n)=\epsilon$. We call $\operatorname{maxFlxPal}(w,n)$ a maximal flexed palindrome of $w$.

Remark 4.19**.**

The name “maximal flexed palindrome” comes from the properties of $\Gamma$. Recall that for a pair $(w,r)$ to be in the set $\Gamma$, it is necessary that $r$ is one of the longest flexed palindromes of $w$.

Next we define the function $\operatorname{elmWrd}(w,w_{1},w_{2})$ (eliminated word) that constructs a rich word from $w$ by “eliminating all” flexed palindromes longer than $m=\max\{|w_{1}|,|w_{2}|\}$ and keeping the prefix $w_{1}$ and the suffix $w_{2}$ of $w$.

Definition 4.20**.**

If $w,w_{1},w_{2}\in\operatorname{R}$, $m=\max\{|w_{1}|,|w_{2}|\}$, $w_{1}\in\operatorname{Prf}(w)$, and $w_{2}\in\operatorname{Suf}(w)$, then let $\operatorname{elmWrd}(w,w_{1},w_{2})$ be the result of the following procedure:

01 INPUT: w,m,w_1,w_2; 02 res: = ruo(w,w_1,w_2); 03 r := maxFlxPal(res,m); 04 WHILE r is nonempty word 05 DO 06 res := rdcWrd(res,r); 07 res := ruo(res,w_1,w_2); 08 r := maxFlxPal(res,m); 09 END-DO; 10 RETURN res; *

The call of the function $\operatorname{ruo}$ on the lines $02$ and $07$ guarantees that $w_{1},w_{2}$ are reverse-unioccurrent in the word $res$ and that $\{w_{1},w_{1}^{R}\}\cap\operatorname{Prf}(res)\not=\emptyset$ and $\{w_{2},w_{2}^{R}\}\cap\operatorname{Suf}(res)\not=\emptyset$. Realize that it is not guaranteed that $w_{1},w_{2}$ are reverse-unioccurrent in $\operatorname{rdcWrd}(res,r)$, even if $w_{1},w_{2}$ are reverse-unioccurrent in $res$.

Clearly, the facts that $\bar{t}$ is reverse unioccurrent in a rich word $t$ and $\bar{t}\in\operatorname{Prf}(t)$ imply that $\operatorname{lppp}(t)\in\operatorname{Prf}(\bar{t})$. Thus if $r$ is a flexed palindrome of $t$ longer than the prefix $\bar{t}$, then $r$ is not a factor of $\operatorname{lppp}(t)$ and hence $r$ satisfies Property 4 of Definition 4.1. In consequence the word $\operatorname{elmWrd}(w,w_{1},w_{2})$ contains no flexed palindrome longer than $m$.

The call of the function $\operatorname{rdcWrd}(res,r)$ on the line $06$ makes obviously sense, since if $\operatorname{maxFlxPal}(w,m)\not=\epsilon$ then $(w,\operatorname{maxFlxPal}(w,m))\in\Gamma$.

In addition, because $|r|>\max\{|w_{1},w_{2}\}$, Theorem 4.12 asserts that $\{w_{1},w_{1}^{R}\}\cap\operatorname{Prf}(\operatorname{rdcWrd}(res,r))\not=\emptyset$ and $\{w_{2},w_{2}^{R}\}\cap\operatorname{Prf}(\operatorname{rdcWrd}(res,r))\not=\emptyset$; consequently $\{w_{1},w_{1}^{R}\}\cap\operatorname{Prf}(res)\not=\emptyset$ and $\{w_{2},w_{2}^{R}\}\cap\operatorname{Suf}(res)\not=\emptyset$ on the line $06$.

Moreover Theorem 4.12 implies that the procedure finishes after a finite number of iterations, because $\operatorname{occur}(\operatorname{rdcWrd}(w,r),r)<\operatorname{occur}(w,r)$ and $\operatorname{T}(\operatorname{rdcWrd}(w,r))\subseteq\operatorname{T}(w)$. The number of iterations is bounded by the number $\sum_{r\in\operatorname{T}(w)}\operatorname{occur}(w,r)$. Note that several occurrences of $r$ may be “eliminated” in one iteration. Hence we proved the following lemma:

Lemma 4.21**.**

If $(w,r)\in\Gamma$, $w_{1}\in\operatorname{Prf}(w)$, $w_{2}\in\operatorname{Prf}(w)$, $m=\max\{|w_{1}|,|w_{2}|\}$, and $t=\operatorname{elmWrd}(w,w_{1},w_{2})$ then

• •

$t\in\operatorname{R}$* and*

• •

$\{w_{1},w_{1}^{R}\}\cap\operatorname{Prf}(t)\not=\emptyset$* and*

• •

$\{w_{2},w_{2}^{R}\}\cap\operatorname{Suf}(t)\not=\emptyset$* and*

• •

for each $r\in\operatorname{T}(t)$ we have $|r|\leq m$.

5 Words with limited number of flexed points

What is the maximal length of a word $u$ such that $w$ is reverse-unioccurrent in $u$, $w$ is a prefix of $u$, and $u$ has a given maximal number of flexed palindromes? The proposition below answers this question.

Proposition 5.1**.**

If $u,w\in\operatorname{R}^{+}$, $w\in\operatorname{Prf}(u)$, $|\operatorname{T}(u)\setminus\operatorname{T}(w)|\leq k$, $|w|\leq m$, and $w$ is reverse-unioccurrent in $u$ then $|u|\leq m2^{k+1}$.

Proof.

Let $\bar{u}=\operatorname{StdExt}(u,1)$; then obviously $|\operatorname{pc}(\bar{u})|<2|\bar{u}|$, $\operatorname{pc}(\bar{u})\in\operatorname{StdExt}(u)$, and $w$ is not reverse-unioccurrent in $\operatorname{pc}(\bar{u})$, since $w^{R}\in\operatorname{Suf}(\operatorname{pc}(\bar{u}))$.

It follows that if $v_{1},v_{2}\in\operatorname{Prf}(\bar{u})$ such that $v_{1}$ is reverse unioccurrent in $\bar{u}$, $v_{1}\in\operatorname{Prf}(v_{2})$, $|\operatorname{T}(v_{2})\setminus\operatorname{T}(v_{1})|=1$, and $\operatorname{lps}(v_{2})\in\operatorname{T}(v_{2})$ then $|\operatorname{ltrim}(v_{2})|<2|v_{1}|$, since $\operatorname{ltrim}(v_{2})\in\operatorname{StdExt}(v_{1})$. This implies that $|v_{2}|\leq 2|v_{1}|$. The proposition follows. ∎

Remark 5.2**.**

The proof asserts that if $v_{1},v_{2}$ are two prefixes of a word $u$ such that the longest palindromic suffix of $v_{2}$ is the only flexed palindrome in $v_{2}$ which is not a factor of $v_{1}$, then $v_{2}$ is at most twice longer than $v_{1}$ on condition that $v_{1}^{R}$ is not a factor of $\operatorname{ltrim}(v_{2})$. Less formally it means that the length of a word can grow at most twice before next flexed palindrome appears. Note that for $k=1$ we have $|u|\leq 2m$, which makes sense, since the palindromic closure of a nonpalindromic word $w$ is at most twice longer than $w$ and $w$ is not reverse-unioccurrent in $\operatorname{pc}(w)$; realize that $w^{R}\in\operatorname{Suf}(\operatorname{pc}(w))$.

In [4] the author showed an upper bound for the number of palindromic factors of given length in a rich word:

Proposition 5.3** ([4],Corollary 2.23).**

If $w\in\operatorname{R}$ and $n>0$ then

[TABLE]

Proposition 5.3 implies an upper bound for the number of flexed palindromes:

Lemma 5.4**.**

If $w\in\operatorname{R}$, $n>0$, and $\operatorname{T}(w)\cap\operatorname{A}^{j}=\emptyset$ for each $j>n$ then

[TABLE]

Proof.

Just realize that $\sum_{j=1}^{n}(q+1)j(4q^{10}j)^{\log_{2}{j}}\leq(q+1)n^{2}(4q^{10}n)^{\log_{2}{n}}$. ∎

From Lemma 4.21, Lemma 5.4 and Proposition 5.1 we obtain the result of the article:

Corollary 5.5**.**

If $w,w_{1},w_{2}$ are rich words, $w_{1},w_{2}\in\operatorname{F}(w)$, $m=\max{\{|w_{1}|,|w_{2}|\}}$ then there exists also a rich word $\bar{w}$ such that $w_{1},w_{2}\in\operatorname{F}(\bar{w})$ and $|\bar{w}|\leq m2^{k(m)+2}$, where $k(m)=(q+1)m^{2}(4q^{10}m)^{\log_{2}{m}}$.

Proof.

Let $t\in\operatorname{F}(\operatorname{pc}(w))$ such that $w_{1}\in\operatorname{Prf}(t)$ and $w_{2}\in\operatorname{Suf}(t)$. Obviously such $t$ exists. Consider the word $g=\operatorname{elmWrd}(t,w_{1},w_{2})$. Let $k(m)=(q+1)m^{2}(4q^{10}m)^{\log_{2}{m}}$. Lemma 5.4 and Proposition 5.1 imply that $|g|\geq m2^{k(m)+1}$. Lemma 4.21 implies that $g\in\operatorname{R}$, $\{w_{1},w_{1}^{R}\}\cap\operatorname{F}(g)\not=\emptyset$, and $\{w_{2},w_{2}^{R}\}\cap\operatorname{F}(g)\not=\emptyset$. Let $\bar{w}=\operatorname{pc}(g)$. It follows that $w_{1},w_{2}\in\operatorname{F}(\bar{w})$. Because $|\operatorname{pc}(g)|\leq 2|g|$, the corollary follows. ∎

Acknowledgments

The author wishes to thank to Štěpán Starosta for his useful comments. 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.