Inverse Lyndon words and Inverse Lyndon factorizations of words

TL;DR

This paper introduces inverse Lyndon factorizations, a new string decomposition method that extends Lyndon factorizations, with a unique, efficiently computable canonical form that preserves key sorting properties.

Contribution

It defines inverse Lyndon words, proves the existence of a unique canonical inverse Lyndon factorization (ICFL), and demonstrates its computational efficiency and sorting compatibility.

Findings

ICFL(w) can be computed in linear time

ICFL(w) is uniquely determined for each word w

ICFL(w) preserves suffix sorting properties

Abstract

Motivated by applications to string processing, we introduce variants of the Lyndon factorization called inverse Lyndon factorizations. Their factors, named inverse Lyndon words, are in a class that strictly contains anti-Lyndon words, that is Lyndon words with respect to the inverse lexicographic order. The Lyndon factorization of a nonempty word w is unique but w may have several inverse Lyndon factorizations. We prove that any nonempty word w admits a canonical inverse Lyndon factorization, named ICFL(w), that maintains the main properties of the Lyndon factorization of w: it can be computed in linear time, it is uniquely determined, it preserves a compatibility property for sorting suffixes. In particular, the compatibility property of ICFL(w) is a consequence of another result: any factor in ICFL(w) is a concatenation of consecutive factors of the Lyndon factorization of w with respect to the inverse lexicographic order.

Equations

$$xz$$

$$x$$

$$w=xr=ry,$$

$$w=x^{\prime }{r}^{\prime }=r^{\prime }{y}^{\prime },$$

$$x$$

$$r^{\prime }=\{\begin{array}{ll}u& \text{mbox}ifu\ne 1\\ v& \text{mbox}ifu=1\end{array}$$

$$x^{\prime }=\{\begin{array}{ll}x(uv{)}^n=(uv{)}^{n+1}& \text{mbox}ifu\ne 1\\ v^n& \text{mbox}ifu=1\end{array}{y}^{\prime }=\{\begin{array}{ll}(vu{)}^{n}y=(vu{)}^{n+1}& \text{mbox}ifu\ne 1\\ v^n& \text{mbox}ifu=1\end{array}$$

$$w$$

$$w={\ell }_1^{n_1}\cdots {\ell }_r^{n_r}$$

$$\forall a,b\in \Sigma b{<}_{in}a\iff a<b$$

$$su{f}_u(i)\prec su{f}_u(j)⟺suf(i)\prec suf(j).$$

$${\mathcal{L}}_{loc}(bbcbc,w)$$

$${\mathcal{L}}_{glob}(w,w)$$

$$s$$

$$p$$

$$w=(daba)(dabdab)(dadac)=(dabadab)(dabda)(dac).$$

$$daba\ll dabdab\ll dadac,dabadab\ll dabda\ll dac.$$

$$(dab)(dadacd)(db)(dc)=(dabda)(dac)(ddbdc)$$

$${\mathrm{Pref}}_{bre}(w)=\{(p,\overline{p})|p\text{mbox}isaninverseLyndonwordwhichisanonemptyproperprefixofw\}.$$

$$b=b^{\prime },p\overline{p}=q\overline{q}=rasrb=r^{\prime }{a}^{\prime }{s}^{\prime }{r}^{\prime }b,a<b,a^{\prime }<b.$$

$$rurb$$

$$a{u}^{\prime }=zt,u^{\prime \prime }=tz,r=t(zt{)}^n.$$

$$\mathrm{ICFL}(w)=\{\begin{array}{ll}(p,\mathrm{ICFL}(v))& \text{mbox}if\overline{p}=rb{\le }_p{m}_1^{\prime }\\ (p{m}_1^{\prime },m_2^{\prime },\dots ,m_k^{\prime })& \text{mbox}if{m}_1^{\prime }{\le }_{p}r\end{array}$$

$$\mathrm{ICFL}(w)=\{\begin{array}{ll}(p,\mathrm{ICFL}(v))& \text{mbox}ifrb{\le }_p{m}_1^{\prime }\\ (p{m}_1^{\prime },m_2^{\prime },\dots ,m_{k^{\prime }}^{\prime })& \text{mbox}if{m}_1^{\prime }{\le }_{p}r\end{array}$$

$$w[t]=a<b=w[k]\le w[h_k].$$

$$w[t]=a<b=w[n]\le w[i].$$

$$f(i)=\max \{k|k<i\text{mbox}andw[\mathrm{1..}k]\text{mbox}isasuffixofw[\mathrm{1..}i]\}.$$

$$f^{(0)}(i)$$

$$f^{(\ell )}(i)$$

$$f^{\ast }(i)$$

$$f^{(\ell )}(i)=f(f^{(\ell -1)}(i))=f^{(\ell -1)}(f(i)).$$

Full text

1 Introduction

Lyndon words were introduced in [25], as standard lexicographic sequences, and then used in the context of the free groups in [6]. A Lyndon word is a word which is strictly smaller than each of its proper cyclic shifts for the lexicographical ordering. A famous theorem concerning Lyndon words asserts that any nonempty word factorizes uniquely as a nonincreasing product of Lyndon words, called its Lyndon factorization. This theorem, that can be recovered from results in [6], provides an example of a factorization of a free monoid, as defined in [32] (see also [4, 23]). Moreover, there are several results which give relations between Lyndon words, codes and combinatorics of words [3].

The Lyndon factorization has recently revealed to be a useful tool also in string processing algorithms [2, 29] with strong potentialities that have not been completely explored and understood. This is due also to the fact that it can be efficiently computed. Linear-time algorithms for computing this factorization can be found in [11, 12] whereas an $\mathcal{O}(\lg{n})$-time parallel algorithm has been proposed in [1, 10]. A connection between the Lyndon factorization and the Lempel-Ziv (LZ) factorization has been given in [18], where it is shown that in general the size of the LZ factorization is larger than the size of the Lyndon factorization, and in any case the size of the Lyndon factorization cannot be larger than a factor of 2 with respect to the size of LZ.

Relations between Lyndon words and the Burrows-Wheeler Transform (BWT) have been discovered first in [9, 26] and, more recently, in [21]. Variants of BWT proposed in the previous papers are based on combinatorial results proved in [14] (see [30] for further details and [13] for more recent related results).

Lyndon words are lexicographically smaller than all its proper nonempty suffixes. This explains why the Lyndon factorization has become of particular interest also in suffix sorting problems. The suffix array (SA) of a word $w$ is the lexicographically ordered list of the starting positions of the suffixes of $w$. The connection between Lyndon factorizations and suffix arrays has been pointed out in [17], where the authors show a method to construct the Lyndon factorization of a text from its SA. Conversely, the computation of the SA of a text from its Lyndon factorization has been proposed in [5] and then explored in [27, 28].

The algorithm proposed in [27, 28] is based on the following interesting combinatorial result, proved in the same papers: if $u$ is a concatenation of consecutive Lyndon factors of $w=xuy$, then the position of a nonempty suffix $u_{i}$ in the ordered list of suffixes of $u$ (called local suffixes) is the same position of the nonempty suffix $u_{i}y$ in the ordered list of the suffixes of $w$ (called global suffixes). In turn, this result suggests a divide and conquer strategy for the sorting of the suffixes of a word $w=w_{1}w_{2}$: we order the nonempty suffixes of $w_{1}$ and the nonempty suffixes of $w_{2}$ independently (or in parallel) and then we merge the resulting lists (see Section 2.5 for further details).

However, in order to have a practical interest, the divide and conquer approach proposed in [28] would require two main ingredients: an efficient algorithm to perform the merging of two sorted lists, which is still to be improved, and a strategy to manage the size of the factors in a Lyndon factorization. Indeed, we may have extreme cases of very short or very long factors (a word $a^{k}$ is factorized into factors of size $1$, while we may have Lyndon words of huge size). On the other hand LZ may produce factorizations of very large sizes which are not useful for a good compression (as in the case of genomic sequences).

In this paper we face the following main question, raised by the above discussion: can we define a factorization of $w$, which maintains some useful properties of the Lyndon factorization but that allows us to manage the size of the factors? We first introduce the notion of an inverse Lyndon word, that is a word greater than any of its proper nonempty suffixes (Section 4). We compare this notion with that of Lyndon words with respect to the inverse lexicographic order (or anti-Lyndon words [15]) and with the notion of strict sesquipowers of Lyndon words (see [11]). We show that the set of the inverse Lyndon words is equal to the set of the strict sesquipowers of anti-Lyndon words and, consequently, it strictly contains the class of the anti-Lyndon words. Then we give the definition of an inverse Lyndon factorizations of a word, whose factors are inverse Lyndon words (Section 4). The Lyndon factorization of a nonempty word $w$ is unique but $w$ may have several inverse Lyndon factorizations. As a main result, we define a canonical inverse Lyndon factorization of a nonempty word $w$, denoted by $\operatorname{ICFL}(w)$. We prove that $\operatorname{ICFL}(w)$ can be still computed in linear time and it is uniquely determined. Moreover, if $w$ is a Lyndon word different from a letter, then $\operatorname{ICFL}(w)$ has at least two factors and a converse holds for an inverse Lyndon word and its Lyndon factorization. Finally, we prove that $\operatorname{ICFL}(w)$ belongs to a special class of inverse Lyndon factorizations, called groupings (see Section 7), and then each of its factors is a concatenation of consecutive factors of the Lyndon factorization of $w$ with respect to the inverse lexicographic order. Hence the compatibility property proved in [27] applies also to $\operatorname{ICFL}(w)$, with respect to the inverse lexicographic order.

In order to answer the above question, we propose to combine the Lyndon factorization of a word $w$ with $\operatorname{ICFL}(w)$. We test our proposal by running an experimental analysis over two biological datasets 111http://www.di.unisa.it/professori/zizza/EXP/experiments.pdf. Experiments confirm that we obtain a factorization of intermediate size between that of LZ and that of the Lyndon factorization.

The paper is organized as follows. In Section 2, we gathered the basic definitions and known results we need. Inverse Lyndon words are discussed in Section 3. Inverse Lyndon factorizations and $\operatorname{ICFL}(w)$ are presented in Section 4. More precisely, for the construction of $\operatorname{ICFL}(w)$ we need a special prefix of $w$, defined in Section 4.2, whereas we give the recursive definition of the factorization in Section 4.4. A linear-time algorithm for computing $\operatorname{ICFL}(w)$ is presented in Section 6. This algorithm uses two subroutines described in Section 5. We introduce groupings in Section 7 and we prove that $\operatorname{ICFL}(w)$ falls in this class of factorizations in Section 7.3.

2 Preliminaries

For the material in this section see [4, 7, 23, 24, 31].

2.1 Words

Let $\Sigma^{*}$ be the free monoid generated by a finite alphabet $\Sigma$ and let $\Sigma^{+}=\Sigma^{*}\setminus 1$, where $1$ is the empty word. For a set $X$, $\operatorname{Card}(X)$ denotes the cardinality of $X$. For a word $w\in\Sigma^{*}$, we denote by $|w|$ its length. A word $x\in\Sigma^{*}$ is a factor of $w\in\Sigma^{*}$ if there are $u_{1},u_{2}\in\Sigma^{*}$ such that $w=u_{1}xu_{2}$. If $u_{1}=1$ (resp. $u_{2}=1$), then $x$ is a prefix (resp. suffix) of $w$. A factor (resp. prefix, suffix) $x$ of $w$ is proper if $x\not=w$. Two words $x,y$ are incomparable for the prefix order, and we write $x\Join y$, if neither $x$ is a prefix of $y$ nor $y$ is a prefix of $x$. Otherwise, $x,y$ are comparable for the prefix order. We write $x\leq_{p}y$ if $x$ is a prefix of $y$ and $x\geq_{p}y$ if $y$ is a prefix of $x$. The following result, named overlapping-suffix lemma in [8], is a direct consequence of the definitions.

Lemma 2.1

Let $x,y,w\in\Sigma^{+}$ such that $x$ and $y$ are both prefixes (resp. suffixes) of $w$. If $|x|<|y|$, then $x$ is a proper prefix (resp. proper suffix) of $y$. If $|x|=|y|$, then $x=y$. Conversely, let $x$ be a prefix (resp. a suffix) of $y$ and let $y$ be a prefix (resp. a suffix) of $w$. If $|x|<|y|$, then $x$ is a proper prefix (resp. a proper suffix) of $w$.

We recall that two words $x,y$ are called conjugate if there exist words $u,v$ such that $x=uv,y=vu$. The conjugacy relation is an equivalence relation. A conjugacy class (or necklace) is a class of this equivalence relation. The following is Proposition 1.3.4 in [22].

Proposition 2.1

Two words $x,y\in\Sigma^{+}$ are conjugate if and only if there exists $z\in\Sigma^{*}$ such that

[TABLE]

More precisely, equality (2.1) holds if and only if there exist $u,v\in\Sigma^{*}$ such that

[TABLE]

A sesquipower of a word $x$ is a word $w=x^{n}p$ where $p$ is a proper prefix of $x$ and $n\geq 1$. A nonempty word $w$ is unbordered if no proper nonempty prefix of $w$ is a suffix of $w$. Otherwise, $w$ is bordered. A nonempty word $w$ is primitive if $w=x^{k}$ implies $k=1$. An unbordered word is primitive. Let $r,w$ be nonempty words over $\Sigma$. We say that two occurrences of $r$ as a factor of $w$ overlap if $w=xrz=x^{\prime}rz^{\prime}$ with $|x^{\prime}|<|x|<|x^{\prime}r|$. Therefore $r$ is bordered. The following lemma will be used in Section 5.

Lemma 2.2

Let $x,y,w,r\in\Sigma^{+}$ be such that

[TABLE]

with $|x|<|r|$, i.e., $r$ occurs twice in $w$ and these two occurrences of $r$ in $w$ overlap. Then there exists $r^{\prime}\in\Sigma^{+}$ such that

[TABLE]

with $|r^{\prime}|<|x^{\prime}|$, and $y^{\prime},y$ start with the same letter.

Proof :

Let $x,y,r\in\Sigma^{+}$ be as in the statement. By Proposition 2.1, there are $u,v\in\Sigma^{*}$ and $n\in\mathbb{N}$ such that

[TABLE]

Set

[TABLE]

and

[TABLE]

By using Eqs. (2.3)-(2.5), we can easily see that $x^{\prime}r^{\prime}=r^{\prime}y^{\prime}=ry=xr$. Moreover, since $|x|<|r|$, if $u\not=1$, then $n\geq 1$ and if $u=1$, then $n\geq 2$. Hence, in both cases, $|r^{\prime}|<|x^{\prime}|$. Finally, $y$ is a prefix of $y^{\prime}$, therefore they start with the same letter.

2.2 Lexicographic order and Lyndon words

Definition 2.1

Let $(\Sigma,<)$ be a totally ordered alphabet. The lexicographic (or alphabetic order) $\prec$ on $(\Sigma^{*},<)$ is defined by setting $x\prec y$ if

• •

$x$* is a proper prefix of $y$, or*

• •

$x=ras$, $y=rbt$, $a<b$, for $a,b\in\Sigma$ and $r,s,t\in\Sigma^{*}$.

For two nonempty words $x,y$, we write $x\ll y$ if $x\prec y$ and $x$ is not a proper prefix of $y$ [2]. We also write $y\succ x$ if $x\prec y$. Basic properties of the lexicographic order are recalled below.

Lemma 2.3

For $x,y\in\Sigma^{*}$, the following properties hold.

• (1)

$x\prec y$* if and only if $zx\prec zy$, for every word $z$.*

• (2)

If $x\ll y$, then $xu\ll yv$ for all words $u,v$.

• (3)

If $x\prec y\prec xz$ for a word $z$, then $y=xy^{\prime}$ for some word $y^{\prime}$ such that $y^{\prime}\prec z$.

Definition 2.2

A Lyndon word $w\in\Sigma^{+}$ is a word which is primitive and the smallest one in its conjugacy class for the lexicographic order.

Example 2.1

Let $\Sigma=\{a,b\}$ with $a<b$. The words $a$, $b$, $aaab$, $abbb$, $aabab$ and $aababaabb$ are Lyndon words. On the contrary, $aba$ and $abaab$ are not Lyndon words. Indeed, $aab\prec aba$ and $aabab\prec abaab$.**

Lyndon words are also called prime words and their prefixes are also called preprime words in [19]. Interesting properties of Lyndon words are recalled below.

Proposition 2.2

Each Lyndon word $w$ is unbordered.

Proposition 2.3

A word $w\in\Sigma^{+}$ is a Lyndon word if and only if $w\prec s$, for each nonempty proper suffix $s$ of $w$.

2.3 The Lyndon factorization

A family $(X_{i})_{i\in I}$ of subsets of $\Sigma^{+}$, indexed by a totally ordered set $I$, is a factorization of the free monoid $\Sigma^{*}$ if each word $w\in\Sigma^{*}$ has a unique factorization $w=x_{1}\cdots x_{n}$, with $n\geq 0$, $x_{i}\in X_{j_{i}}$ and $j_{1}\geq j_{2}\ldots\geq j_{n}$ [4]. A factorization $(X_{i})_{i\in I}$ is called complete if each $X_{i}$ is reduced to a singleton $x_{i}$ [4]. Let $L=L_{(\Sigma^{*},<)}$ be the set of Lyndon words, totally ordered by the relation $\prec$ on $(\Sigma^{*},<)$. The following theorem, shows that the family $(\ell)_{\ell\in L}$ is a complete factorization of $\Sigma^{*}$.

Theorem 2.1

Any word $w\in\Sigma^{+}$ can be written in a unique way as a nonincreasing product $w=\ell_{1}\ell_{2}\cdots\ell_{h}$ of Lyndon words, i.e., in the form

[TABLE]

The sequence $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$ in Eq. (2.6) is called the Lyndon decomposition (or Lyndon factorization) of $w$. It is denoted by $\operatorname{CFL}(w)$ because Theorem 2.1 is usually credited to Chen, Fox and Lyndon [6]. Uniqueness of the above factorization is a consequence of the following result, proved in [11].

Lemma 2.4

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$. Then the following properties hold:

• (i)

$\ell_{h}$* is the nonempty suffix of $w$ which is the smallest with respect to the lexicographic order.*

• (ii)

$\ell_{h}$* is the longest suffix of $w$ which is a Lyndon word.*

• (iii)

$\ell_{1}$* is the longest prefix of $w$ which is a Lyndon word.*

Therefore, given $w\in\Sigma^{+}$, if $\ell_{1}$ is its longest prefix which is a Lyndon word and $w=\ell_{1}w^{\prime}$, then $\operatorname{CFL}(w)=(\ell_{1},\operatorname{CFL}(w^{\prime}))$. As a consequence of Theorem 2.1, for any word $w$ there is a factorization

[TABLE]

where $r>0$, $n_{1},\ldots,n_{r}\geq 1$, and $\ell_{1}\succ\ldots\succ\ell_{r}$ are Lyndon words, also named Lyndon factors of $w$. There is a linear time algorithm to compute the pair $(\ell_{1},n_{1})$ and thus, by iteration, the Lyndon factorization of $w$. It is due to Fredricksen and Maiorana [12] and it is also reported in [24]. It can also be used to compute the Lyndon word in the conjugacy class of a primitive word in linear time [24]. Linear time algorithms may also be found in [11] and in the more recent paper [16].

2.4 Inverse lexicographic order and anti-Lyndon words

We also need the following well-known definition.

Definition 2.3

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $<_{in}$ be the inverse of $<$, defined by

[TABLE]

The inverse lexicographic or inverse alphabetic order, denoted $\prec_{in}$, on $(\Sigma^{*},<)$ is the lexicographic order on $(\Sigma^{*},<_{in})$.

From now on, $L_{in}=L_{(\Sigma^{*},<_{in})}$ denotes the set of the Lyndon words on $\Sigma^{*}$ with respect to the inverse lexicographic order. A word $w\in L_{in}$ will be named an anti-Lyndon word.

2.5 Sorting suffixes of a text

In [27, 28], the authors found interesting relations between the sorting of the suffixes of a word $w$ and that of its factors. Let $w,x,u,y\in\Sigma^{*}$, and let $u$ be a nonempty factor of $w=xuy$. Let $first(u)$ and $last(u)$ denote the position of the first and the last symbol of $u$ in $w$, respectively. If $w=a_{1}\cdots a_{n}$, $a_{i}\in\Sigma$, $1\leq i\leq j\leq n$, then we also set $w[i,j]=a_{i}\cdots a_{j}$. A local suffix of $w$ is a suffix of a factor of $w$, specifically $suf_{u}(i)=w[i,last(u)]$ denotes the local suffix of $w$ at the position $i$ with respect to $u$, $i\geq first(u)$. The corresponding global suffix $suf_{u}(i)y$ of $w$ at the position $i$ is denoted by $suf_{w}(i)=w[i,last(w)]$ (or simply $suf(i)$ when it is understood). We say that $suf_{u}(i)y$ is associated with $suf_{u}(i)$.

Definition 2.4

[27, 28]** Let $w\in\Sigma^{+}$ and let $u$ be a factor of $w$. We say that the sorting of the nonempty local suffixes of $w$ with respect to $u$ is compatible with the sorting of the corresponding nonempty global suffixes of $w$ if for all $i,j$ with $first(u)\leq i<j\leq last(u)$,

[TABLE]

Let $u=\ell_{r}\cdots\ell_{s}$ be a concatenation of consecutive Lyndon factors of $w$. Let $\mathcal{L}_{\mathit{l}oc}(u,w)=(s_{1},\ldots,s_{t})$ be the ordered list of the nonempty local suffixes of $w$ with respect to $u$ and let $\mathcal{L}_{\mathit{g}lob}(u,w)=(s^{\prime}_{1},\ldots,s^{\prime}_{t})$ be the ordered list of the corresponding nonempty global suffixes of $w$. We name it the global list associated with $\mathcal{L}_{\mathit{l}oc}(u,w)$. The following result proved in [27, 28] shows that each $s^{\prime}_{i}$ in $\mathcal{L}_{\mathit{g}lob}(u,w)$ is associated with $s_{i}$.

Theorem 2.2

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$ be its Lyndon factorization. Then, for any $r,s$, $1\leq r\leq s\leq h$, the sorting of the nonempty local suffixes of $w$ with respect to $u=\ell_{r}\cdots\ell_{s}$ is compatible with the sorting of the corresponding nonempty global suffixes of $w$.

Remark 2.1

The compatibility property stated in Theorem 2.2 does not hold for the empty suffix of a word. Indeed, let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$. Let $w=bbcbcacad$, thus $\operatorname{CFL}(w)=(bbc,bc,acad)$. Consider the local suffixes $1,bbc$ of $w$ with respect to $u=bbc$ and the corresponding global suffixes $bcacad,w$. We have $1\prec bbc$ but $w\prec bcacad$.**

If $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are two sorted lists of elements taken from any totally ordered set, then the result of the operation $\operatorname{merge}(\mathcal{L}_{1},\mathcal{L}_{2})$ is a single sorted list containing the elements of $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$. Theorem 2.2 could be considered in a merge sort algorithm for the sorting of the suffixes of $w$, as suggested in [27, 28]. Starting with the list $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$, it could operate as follows.

Divide the sequence into two subsequences $(\ell_{1},\ldots,\ell_{r})$, $(\ell_{r+1},\ldots,\ell_{h})$, where $r=\lceil k/2\rceil$

• -

Let $\mathcal{L}_{1}$ be the list of the nonempty suffixes of $u=\ell_{1}\cdots\ell_{r}$, let $\mathcal{L}_{2}$ be the list of the nonempty suffixes of $y=\ell_{r+1}\cdots\ell_{h}$. Sort the two subsequences $\mathcal{L}_{1},\mathcal{L}_{2}$ recursively using merge sort, thus obtaining $\mathcal{L}_{\mathit{l}oc}(u,w),\mathcal{L}_{\mathit{l}oc}(y,w)$

• -

Merge the two subsequences $\mathcal{L}_{\mathit{g}lob}(u,w),\mathcal{L}_{\mathit{g}lob}(y,w)$ to produce $\mathcal{L}_{\mathit{g}lob}(w,w)$.

Notice that in the third step we change $\mathcal{L}_{\mathit{l}oc}(u,w),\mathcal{L}_{\mathit{l}oc}(y,w)$ into $\mathcal{L}_{\mathit{g}lob}(u,w),\mathcal{L}_{\mathit{g}lob}(y,w)$. Indeed, as pointed out in Example 2.2, two problems arise if we considered the local lists in the merge step. First, there could exist two local suffixes $s,s^{\prime}$, with corresponding global suffixes $sz,s^{\prime}z^{\prime}$, such that $s\prec s^{\prime}$ but $s^{\prime}z^{\prime}\prec sz$. Furthermore, an element $s$ could occur twice in $\operatorname{merge}(\mathcal{L}_{\mathit{l}oc}(u,w),\mathcal{L}_{\mathit{l}oc}(y,w))$. In this case, when we produce the global list $\mathcal{L}_{\mathit{g}lob}(w,w)$, we have to change the second occurrence of $s$ into $sy$.

Example 2.2

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$. Let $w=bbcbcacad$, thus $\operatorname{CFL}(w)=(bbc,bc,acad)$. Then, $\mathcal{L}_{\mathit{l}oc}(bbc,w)=(bbc,bc,c)$, $\mathcal{L}_{\mathit{l}oc}(bc,w)=(bc,c)$, and $\mathcal{L}_{\mathit{l}oc}(acad,w)=(acad,ad,cad,d)$. Consider the local suffix $c$ in $\mathcal{L}_{\mathit{l}oc}(bbc,w)$, the local suffix $cad$ in $\mathcal{L}_{\mathit{l}oc}(acad,w)$ and the corresponding associated global suffixes $cbcacad$ and $cad$. We see that $c\prec cad$ but $cbcacad\succ cad$. Let $u=bbcbc$, $y=acad$, then

[TABLE]

If in the third step we merged $\mathcal{L}_{\mathit{l}oc}(bbc,u),\mathcal{L}_{\mathit{l}oc}(bc,u)$, we would obtain $(bbc,bc,bc,c,c)$. Then, in order to obtain $\mathcal{L}_{\mathit{l}oc}(bbcbc,w)$, we have to change $bbc$ into $bbcbc$, the second occurrence of $bc$ into $bcbc$ and the second occurrence of $c$ into $cbc$.**

3 Inverse Lyndon words

As mentioned in Section 1, the Lyndon factorization of a word may generate very long or too short factors, thus becoming unsatisfactory with respect to a parallel strategy. We face this problem in Section 4, where we introduce another factorization which maintains the main properties of the Lyndon factorization but that allows us to overcome the glitch. This factorization is based on the notion of inverse Lyndon words, given below.

Definition 3.1

A word $w\in\Sigma^{+}$ is an inverse Lyndon word if $s\prec w$, for each nonempty proper suffix $s$ of $w$.

Example 3.1

The words $a$, $b$, $bbba$, $baaab$, $bbaba$ and $bbababbaa$ are inverse Lyndon words on $\{a,b\}$, with $a<b$. On the contrary, $aaba$ is not an inverse Lyndon word since $aaba\prec ba$. Analogously, $aabba\prec ba$ and thus $aabba$ is not an inverse Lyndon word.**

In Section 7.1 we will see that the set of inverse Lyndon words properly contains the set of Lyndon words with respect to the inverse lexicographic order (or anti-Lyndon words). Then, in Section 7.4 we will prove that the set of the inverse Lyndon words is equal to the set of the strict sesquipowers of anti-Lyndon words. Some useful properties of the inverse Lyndon words are proved below. The following is a direct consequence of Definitions 2.1, 3.1.

Lemma 3.1

If $w\in\Sigma^{+}$ is not an inverse Lyndon word, then there exists a nonempty proper suffix $s$ of $w$ such that $w\ll s$.

Next lemma shows that the set of the inverse Lyndon words (with the empty word) is a prefix-closed set, that is, it contains the prefixes of its elements.

Lemma 3.2

Any nonempty prefix of an inverse Lyndon word is an inverse Lyndon word.

Proof :

Let $w\in\Sigma^{+}$ be an inverse Lyndon word. By contradiction, assume that there is a proper nonempty prefix $p$ of $w=ps$ which is not an inverse Lyndon word. By Lemma 3.1, there is a proper nonempty suffix $v$ of $p$ such that $p\ll v$. Hence, by item (2) in Lemma 2.3, $w=ps\ll vs$. Thus $w$ is smaller than its proper nonempty suffix $vs$, in contradiction with the hypothesis that $w$ is an inverse Lyndon word.

Lemmas 3.3 and 3.4 are needed for the definition of our new factorization of a word.

Lemma 3.3

If $w\in\Sigma^{+}$ is not an inverse Lyndon word, then there exists a nonempty proper prefix $p$ of $w=ps$ such that $p\ll s$.

Proof :

Let $w$ be as in the statement. By Lemma 3.1, there exists a shortest nonempty proper suffix $s$ of $w=ps$ such that $w\ll s$. Hence, $p$ is nonempty. Moreover there are words $r,t,q\in\Sigma^{*}$ and letters $a,b\in\Sigma$, with $a<b$, such that

[TABLE]

We show that $p\ll s$. If $|p|\geq|ra|$, then $p\ll s$ and the proof is ended. Thus assume $0<|p|\leq|r|$. By Eq. (3.1), there exists a prefix $t^{\prime}$ of $t$ such that $pr=rat^{\prime}$. Therefore, by Proposition 2.1, there are $u,v\in\Sigma^{*}$ such that

[TABLE]

If $u=1$, then $p$ starts with $a$ and so does $w$. Thus $w\ll bq$ and, by the hypothesis on $s$, we have $s=bq$. Hence $r=1$, a contradiction. The same argument applies if $v=1$. Thus assume $u\not=1$, $v\not=1$. Set $r=u(vu)^{k}$, $k\geq 0$. We also have $k>0$, since, otherwise, by Eq. (3.2), $p=rv$, with $v\not=1$, and consequently $|r|<|p|\leq|r|$, a contradiction. Finally, by Eqs. (3.1) and (3.2), $w=ps=uvs=(uv)^{k+1}ubq$ and $v$ starts with $a$. Consequently, $w\ll ubq$, with $ubq$ shorter than $s$, in contradiction with the hypothesis on the length of $s$.

Corollary 3.1

A word $w\in\Sigma^{+}$ is not an inverse Lyndon word if and only if there are words $r,u,t\in\Sigma^{*}$ and letters $a,b\in\Sigma$, with $a<b$ such that $w=raurbt$.

Proof :

Let $r,u,t\in\Sigma^{*}$ and $a,b\in\Sigma$, with $a<b$. By Definition 2.1 $w=raurbt\ll rbt$, hence $w$ is not an inverse Lyndon word. Conversely, assume that $w\in\Sigma^{+}$ is not an inverse Lyndon word. By Lemma 3.3, there exists a nonempty proper prefix $p$ of $w=ps$ such that $p\ll s$. By Definition 2.1, there are $r,u,t\in\Sigma^{*}$, $a,b\in\Sigma$, with $a<b$, such that $p=rau$, $s=rbt$, and thus $w=ps=raurbt$.

The following lemma shows that there exists $p$ satisfying Lemma 3.3 and which is in addition an inverse Lyndon word.

Lemma 3.4

If $w\in\Sigma^{+}$ is not an inverse Lyndon word, then there exists a nonempty proper prefix $p$ of $w=ps$ such that $p$ is an inverse Lyndon word and $p\ll s$.

Proof :

Let $w\in\Sigma^{+}$ a word which is not an inverse Lyndon word. The proof is by induction on $|w|$. The shortest nonempty word which is not an inverse Lyndon word has the form $w=ab$, with $a,b$ letters such that $a<b$. In this case the nonempty proper prefix $a$ of $w$ is an inverse Lyndon word and $a\ll b=s$. Now assume $|w|>2$. By Lemma 3.3, there exists a nonempty proper prefix $p$ of $w=ps$ such that $p\ll s$. If $p$ is an inverse Lyndon word, then the proof is ended. Otherwise, by induction hypothesis there exists a nonempty proper prefix $p^{\prime}$ of $p=p^{\prime}s^{\prime}$ such that $p^{\prime}$ is an inverse Lyndon word and $p^{\prime}\ll s^{\prime}$. Hence, by item (2) in Lemma 2.3, $p^{\prime}\ll s^{\prime}s$. Thus, $p^{\prime}$ is a nonempty proper prefix of $w=p^{\prime}s^{\prime}s$ such that $p^{\prime}$ is an inverse Lyndon word and $p^{\prime}\ll s^{\prime}s$.

4 Variants of the Lyndon factorization

4.1 Inverse Lyndon factorizations

We give below the notion of an inverse Lyndon factorization.

Definition 4.1

Let $w\in\Sigma^{+}$. A sequence $(m_{1},\ldots,m_{k})$ of words over $\Sigma$ is an inverse Lyndon factorization of $w$ if it satisfies the following conditions

• (1)

$w=m_{1}\cdots m_{k}$,

• (2)

for any $j\in\{1,\ldots,k\}$, the word $m_{j}$ is an inverse Lyndon word,

• (3)

$m_{1}\ll m_{2}\ll\ldots\ll m_{k}$.

Example 4.1 shows that a word may have different inverse Lyndon factorizations even with a different number of factors.

Example 4.1

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, let $w=dabadabdabdadac\in\Sigma^{+}$. The two sequences $(daba,dabdab,dadac)$, $(dabadab,dabda,dac)$ are both inverse Lyndon factorizations of $w$. Indeed,

[TABLE]

Moreover, $daba,dabdab,dadac,dabadab,dabda,dac$ are all inverse Lyndon words. Furthermore,

[TABLE]

As another example, consider the following two factorizations of $dabdadacddbdc$

[TABLE]

It is easy to see that the two sequences $(dab,dadacd,db,dc)$, $(dabda,dac,ddbdc)$ are both inverse Lyndon factorizations of $dabdadacddbdc$. The first factorization has four factors whereas the second one has three factors.**

In Section 2.3 we have given the definition of a complete factorization $(x_{i})_{i\in I}$ of the free monoid $\Sigma^{*}$. By a result of Schützenberger, if $(x_{i})_{i\in I}$ is a complete factorization of $\Sigma^{*}$, then the set $X=\{x_{i}~{}|~{}i\in I\}$ is a set of representatives of the primitive conjugacy classes (see [4, Corollary 8.1.7]). In particular, any $x_{i}$ is a primitive word. The fact that a word $w$ may have several different inverse Lyndon factorizations is a consequence of this result since an inverse Lyndon word is not necessarily primitive (take $baba$, with $a<b$, for instance).

However, we focus on a special canonical inverse Lyndon factorization, denoted by $\operatorname{ICFL}(w)$, which maintains three important features of $\operatorname{CFL}(w)$: it is uniquely determined (Proposition 4.2), it can be computed in linear time (Section 6) and it maintains the compatibility property of the nonempty suffixes with respect to the inverse lexicographic order (Theorem 7.2). We give the definition of $\operatorname{ICFL}(w)$ in Section 4.4. It is based on the definition of the bounded right extension of a prefix of a word, defined in Section 4.2.

4.2 The bounded right extension

The bounded right extension, abbreviated bre, of a prefix of a word $w$, defined below, allows us to define the first factor in the inverse Lyndon factorization $\operatorname{ICFL}(w)$.

Definition 4.2

Let $w\in\Sigma^{+}$, let $p$ be an inverse Lyndon word which is a nonempty proper prefix of $w=pv$. The bounded right extension $\overline{p}_{w}$ of $p$ (relatively to $w$), denoted by $\overline{p}$ when it is understood, is a nonempty prefix of $v$ such that:

• (1)

$\overline{p}$* is an inverse Lyndon word,*

• (2)

$pz^{\prime}$* is an inverse Lyndon word, for each proper nonempty prefix $z^{\prime}$ of $\overline{p}$,*

• (3)

$p\overline{p}$* is not an inverse Lyndon word,*

• (4)

$p\ll\overline{p}$.

Moreover, we set

[TABLE]

Notice that, given a word $w$, a nonempty proper prefix $p$ of $w=pv$ can have at most one bounded right extension $\overline{p}_{w}$. Indeed, two different prefixes $p_{1},p_{2}$ of $v$ are comparable for the prefix order, say $p_{1}\leq_{p}p_{2}$. If $p_{2}=\overline{p}_{w}$, then $pp_{1}$ is an inverse Lyndon word and $p_{1}$ cannot be a bounded right extension of $p$. Analogously, if $p_{1}=\overline{p}_{w}$, then $p_{2}$ cannot be a bounded right extension of $p$ because $p_{2}$ does not satisfy condition (2) in Definition 4.2. Furthermore, the bounded right extension $\overline{p}_{w}$ of $p$ might not exist. For instance, let $\Sigma=\{a,b,c\}$ with $a<b<c$. For the prefix $ba$ of $baababc$, $\overline{ba}$ does not exist since any nonempty prefix $p^{\prime}$ of $ababc$ starts with $a$, thus $p^{\prime}\ll ba$. On the contrary, for the prefix $baa$ of $baababc$, we have $bab=\overline{baa}$. As another example, for the prefix $bab$ of $babc$ we have $c=\overline{bab}$ but for the prefix $ba$ of the same word $babc$, $\overline{ba}$ does not exist. Moreover, it is clear that if $w$ is a letter, then $\operatorname{Pref}_{bre}(w)$ is empty. More generally, if $w$ is an inverse Lyndon word, then $\operatorname{Pref}_{bre}(w)$ is empty. A more precise result will be proved below. We will see that the set $\operatorname{Pref}_{bre}(w)$ is either empty or it is a singleton. In other words, given a word $w$, either there is no prefix of $w$ which has a bounded right extension or this prefix is unique. This result will be proved through Lemmas 4.3, 4.4 and Proposition 4.1. Lemmas 4.1, 4.2 below show interesting properties of $\operatorname{Pref}_{bre}(w)$.

Lemma 4.1

Let $w\in\Sigma^{+}$ and let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. Then, there are $r,s,t\in\Sigma^{*}$, $a,b\in\Sigma$, with $a<b$, such that $p=ras$ and $\overline{p}=rb$.

Proof :

Let $w\in\Sigma^{+}$ and let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. By Definition 4.2, $p\ll\overline{p}$, hence $p=ras$, $\overline{p}=rbt$, with $r,s,t\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Moreover $t=1$, otherwise for any proper prefix $t^{\prime}$ of $t$, the word $z^{\prime}=rbt^{\prime}$ would be a proper prefix of $\overline{p}$ such that $p\ll z^{\prime}$, thus $pz^{\prime}\ll z^{\prime}$ and $pz^{\prime}$ would not be an inverse Lyndon word, in contradiction with Definition 4.2.

Recall that, if $w\in\Sigma^{+}$ is not an inverse Lyndon word, then there exists a nonempty proper prefix $p$ of $w=ps$ such that $p\ll s$ (Lemma 3.3).

Lemma 4.2

Let $w$ be a nonempty word which is not an inverse Lyndon word but all its proper nonempty prefixes are inverse Lyndon words. If $p$ is the longest nonempty proper prefix of $w=ps$ such that $p\ll s$ for the corresponding suffix $s$, then $(p,s)\in\operatorname{Pref}_{bre}(w)$. In other words, if $(p,s)\in\mathcal{G}=\{(p^{\prime},s^{\prime})~{}|~{}p^{\prime}s^{\prime}=w,~{}p^{\prime}\ll s^{\prime}\}$ and $|p|\geq|p^{\prime}|$, for any $(p^{\prime},s^{\prime})\in\mathcal{G}$, then $(p,s)\in\operatorname{Pref}_{bre}(w)$.

Proof :

Let $w,p,s$ be words as in the statement. If $s$ is an inverse Lyndon word, then conditions (1)-(4) in Definition 4.2 are satisfied with $s=\overline{p}$ and we have done. Assume that $s$ is not an inverse Lyndon word and set $p=rax$, $s=rby$, where $r,x,y\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. We notice that $y=1$, otherwise $prb=raxrb$ would be a nonempty proper prefix of $w=ps=prby$ which is not an inverse Lyndon word, since $prb=raxrb\prec rb$, in contradiction with the hypotheses. Thus $r\not=1$ (otherwise, $s=b$ would be an inverse Lyndon word). Moreover, by Lemma 3.3, there exists a nonempty proper prefix $q$ of $s=rb=qt$ such that $q\ll t$. Notice that $q$ is a prefix of $r=qq^{\prime}$ and $w=pqt$. Hence, by item (2) in Lemma 2.3 and $q\ll t$, we get $pq=qq^{\prime}axq\ll t$, with $pq$ longer than $p$, a contradiction.

Lemma 4.3

Let $w\in\Sigma^{+}$. For any nonempty prefix $x$ of $w$, we have $\operatorname{Pref}_{bre}(x)\subseteq\operatorname{Pref}_{bre}(w)$.

Proof :

Let $w\in\Sigma^{+}$ and let $x$ be a proper nonempty prefix of $w=xv$. If $\operatorname{Pref}_{bre}(x)=\emptyset$, then the proof is ended. Otherwise, let $(p,\overline{p}_{x})\in\operatorname{Pref}_{bre}(x)$. By Definition 4.2, $p$ is a nonempty prefix of $x=pv^{\prime}$, hence of $w=pv^{\prime}v$, which is an inverse Lyndon word. Moreover, $\overline{p}_{x}$ is a nonempty prefix of $v^{\prime}$, thus of $v^{\prime}v$, such that conditions (1)-(4) holds. Therefore, $(p,\overline{p}_{x})=(p,\overline{p}_{w})\in\operatorname{Pref}_{bre}(w)$.

Lemma 4.4

A word $w\in\Sigma^{+}$ is not an inverse Lyndon word if and only if $\operatorname{Pref}_{bre}(w)$ is nonempty.

Proof :

Let $w\in\Sigma^{+}$ be a word such that $\operatorname{Pref}_{bre}(w)\not=\emptyset$ and let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. By Definition 4.2, the nonempty prefix $p\overline{p}$ of $w$ is not an inverse Lyndon word, thus $w$ is not an inverse Lyndon word by Lemma 3.2.

Conversely, let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. We prove that $\operatorname{Pref}_{bre}(w)$ is nonempty by induction on $|w|$. The shortest nonempty word which is not an inverse Lyndon word has the form $w=ab$, with $a,b$ letters such that $a<b$ and $\operatorname{Pref}_{bre}(ab)=\{(a,b)\}$.

Assume $|w|>2$ and let $p^{\prime}$ be the shortest nonempty prefix of $w$ which is not an inverse Lyndon word. Hence $|p^{\prime}|\geq 2$. If $|p^{\prime}|<|w|$, then $\emptyset\not=\operatorname{Pref}_{bre}(p^{\prime})\subseteq\operatorname{Pref}_{bre}(w)$ (induction hypothesis and Lemma 4.3). Otherwise, $p^{\prime}=w$ and all nonempty proper prefixes of $w$ are inverse Lyndon words. Thus, by Lemma 3.3, there exists a nonempty proper prefix $p$ of $w=ps$ such that $p\ll s$. Choose it of maximal length. By Lemma 4.2, $(p,s)\in\operatorname{Pref}_{bre}(w)$.

Proposition 4.1

If $w\in\Sigma^{+}$ is not an inverse Lyndon word, then $\operatorname{Card}(\operatorname{Pref}_{bre}(w))=1$.

Proof :

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. By Lemma 4.4, the set $\operatorname{Pref}_{bre}(w)$ is nonempty. By contradiction, let $(p,\overline{p}),(q,\overline{q})\in\operatorname{Pref}_{bre}(w)$, with $p\not=q$. Since $p\overline{p}$ and $q\overline{q}$ are both prefixes of $w$, they are comparable for the prefix order and one of the following three cases holds.

• (1)

$p\overline{p}$ is a proper prefix of $q\overline{q}$,

• (2)

$q\overline{q}$ is a proper prefix of $p\overline{p}$,

• (3)

$p\overline{p}=q\overline{q}$.

In case (1), either $p\overline{p}$ is a prefix of $q$ or $p\overline{p}=qz^{\prime}$, for a proper prefix $z^{\prime}$ of $\overline{q}$. Since $p\overline{p}$ is not an inverse Lyndon word, both cases are impossible (the first contradicts Lemma 3.2, the second Definition 4.2). We may exclude case (2) by a similar reasoning, thus assume that $p\overline{p}=q\overline{q}$, with $q$ being a nonempty proper prefix of $p$ (the same argument applies if $p$ is a proper prefix of $q$). By Lemma 4.1, there are $r,s,r^{\prime},s^{\prime}\in\Sigma^{*}$, $a,b,a^{\prime},b^{\prime}\in\Sigma$, with $a<b$, $a^{\prime}<b^{\prime}$, such that $p=ras$, $\overline{p}=rb$, $q=r^{\prime}a^{\prime}s^{\prime}$, $\overline{q}=r^{\prime}b^{\prime}$. Hence

[TABLE]

Since $q$ is a nonempty proper prefix of $p$, we have $|q|<|p|$ which yields $|rb|=|\overline{p}|<|\overline{q}|=|r^{\prime}b|$, i.e., $|r^{\prime}|>|r|$. By Eq. (4.1), the word $ra$ is a prefix of $r^{\prime}$, i.e., there is a word $x$ such that $r^{\prime}=rax$. Furthermore, by Eq. (4.1) again, since $q$ is a nonempty proper prefix of $p$, $\overline{p}=rb$ is a nonempty proper suffix of $\overline{q}$, i.e., there is a nonempty word $y$ such that $\overline{q}=yrb$. In conclusion, $\overline{q}=r^{\prime}b=raxb=yrb$, thus $\overline{q}\ll rb$, that is, $\overline{q}$ is not an inverse Lyndon word, a contradiction.

4.3 Some technical lemmas

In this section we prove some technical results which allows us to compute the pair $(p,s)$ in $\operatorname{Pref}_{bre}(w)$, for a word $w$ which is not an inverse Lyndon word. They will be used in Section 5. In detail, Lemma 4.5 is a more precise reformulation of Lemma 3.3. Lemma 4.6 is preliminary to Lemma 4.7 which in turn characterizes $(p,s)$ through two simple conditions.

Lemma 4.5

Let $w\in\Sigma^{+}$. If $w$ has no prefix with the form $raurb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, with $a<b$, then $w$ is an inverse Lyndon word.

Proof :

Let $w\in\Sigma^{+}$ be as in the statement. If $w$ were not an inverse Lyndon word, then, by Corollary 3.1, there would be $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, with $a<b$ such that $w=raurbt$, a contradiction since $raurb$ would be a prefix of $w$.

Lemma 4.6

Let $w,u,v\in\Sigma^{+}$, $r\in\Sigma^{*}$, and $a,b\in\Sigma$ be such that

• (1)

$w=rurv$,

• (2)

$rur$* is an inverse Lyndon word,*

• (3)

$u$* starts with $a$, $v$ starts with $b$, with $a<b$,*

• (4)

$r$* is the shortest prefix of $w$ such that conditions (1)-(3) hold.*

Then $w$ is not an inverse Lyndon word and $(ru,rb)\in\operatorname{Pref}_{bre}(w)$.

Proof :

Let $w,u,v\in\Sigma^{+}$, $r\in\Sigma^{*}$, and $a,b\in\Sigma$ be as in the statement. By item (2) in Lemma 2.3, $ru\ll rb$ implies $rurb\ll rb$. Therefore, $rurb$ is not an inverse Lyndon word, hence, by Lemma 3.2 $w$ is not an inverse Lyndon word too. Moreover, by (2), each proper nonempty prefix of $rurb$ is an inverse Lyndon word. Thus, by Lemmas 4.2, 4.3, if $p$ is the longest nonempty proper prefix of $rurb=ps$ such that $p\ll s$ for the corresponding suffix $s$, then $(p,s)\in\operatorname{Pref}_{bre}(rurb)\subseteq\operatorname{Pref}_{bre}(w)$. In particular, $s$ is an inverse Lyndon word (Definition 4.2).

If $(ru,rb)\not=(p,s)$, then $|p|>|ru|$. Moreover, since $p\ll s$, there are $r^{\prime},x,y\in\Sigma^{*}$, $c,d\in\Sigma$, with $c<d$, such that $x$ starts with $c$ and $p=r^{\prime}x$, $s=r^{\prime}dy$. Since any prefix of $rur$ is an inverse Lyndon word and $r^{\prime}xr^{\prime}d$ is not an inverse Lyndon word, the word $rur$ is a proper prefix of $r^{\prime}xr^{\prime}d$, thus, by $rurb=ps=r^{\prime}xr^{\prime}dy$, we have $dy=b=d$ which implies

[TABLE]

Looking at Eq. (4.2), if $|r|=|r^{\prime}|$, we would have $r=r^{\prime}$, hence $p=ru$, a contradiction. Consequently, since $r$ is the shortest prefix of $w=rurv=r^{\prime}xr^{\prime}v$ such that conditions (1)-(3) hold, and $r^{\prime}$ satisfies conditions (1)-(3), we have $|r^{\prime}|>|r|$. Since $u$ starts with $a$, by Eq. (4.2) again, we get $r^{\prime}=rau^{\prime}=u^{\prime\prime}r$. Thus, by Proposition 2.1 there are $z,t\in\Sigma^{*}$, $n\geq 0$, such that

[TABLE]

Of course $zt\not=1$. If $z\not=1$, $z$ starts with $a$ and $s=r^{\prime}b=t(zt)^{n}ztb\ll tb$, a contradiction since $s$ is an inverse Lyndon word. The same argument applies if $z=1$. In this case, $t$ starts with $a$ and $s=r^{\prime}b=t(zt)^{n}ztb\ll b$, a contradiction since $s$ is an inverse Lyndon word.

Lemma 4.7

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. Let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$.

• (1)

$p\overline{p}$* is the shortest nonempty prefix of $w$ which is not an inverse Lyndon word.*

• (2)

$p=rau$* and $\overline{p}=rb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$ and $r$ is the shortest prefix of $p\overline{p}$ such that $p\overline{p}=raurb$, with $a<b$.*

Proof :

Let $w,p,\overline{p}$ be as in the statement.

(1) Let $x$ be the shortest nonempty prefix of $w$ which is not an inverse Lyndon word. Since $x$ and $p\overline{p}$ are both prefixes of $w$, they are comparable for the prefix order. Since $x$ is not an inverse Lyndon word whereas any proper nonempty prefix of $p\overline{p}$ is an inverse Lyndon word (Definition 4.2), the word $p\overline{p}$ is a prefix of $x$, hence $|x|\geq|p\overline{p}|$. Since $x$ is the shortest nonempty prefix of $w$ which is not an inverse Lyndon word, we have $|x|\leq|p\overline{p}|$, hence $|x|=|p\overline{p}|$. Therefore, since $x$ and $p\overline{p}$ are comparable for the prefix order, $x=p\overline{p}$.

(2) By Lemma 4.1, there exists a prefix $r$ of $p\overline{p}$ such that $p\overline{p}=raurb$, with $a<b$. The proper nonempty prefix $raur$ of $p\overline{p}$ is an inverse Lyndon word (Definition 4.2). Choose $r$ of minimal length. Thus, by Lemma 4.6, $(rau,rb)\in\operatorname{Pref}_{bre}(p\overline{p})\subseteq\operatorname{Pref}_{bre}(w)$.

Lemma 4.8

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. The shortest nonempty prefix of $w$ which is not an inverse Lyndon word is the shortest nonempty prefix $x$ of $w$ such that $x=raurb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$.

Proof :

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word, let $x^{\prime}$ be the shortest nonempty prefix of $w$ which is not an inverse Lyndon word, and let $x$ be the shortest nonempty prefix of $w$ such that $x=raurb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. By Lemmas 4.1 and 4.7, there are $r^{\prime},u^{\prime}\in\Sigma^{*}$, $c,d\in\Sigma$, $c<d$, such that $x^{\prime}=r^{\prime}cu^{\prime}r^{\prime}d$. The words $x,x^{\prime}$ are both prefixes of $w$, thus they are comparable for the prefix order. If they were different, by the hypothesis on $|x|$, $x$ would be a proper nonempty prefix of $x^{\prime}$. This fact contradicts the hypothesis on $|x^{\prime}|$ because $x$ is not an inverse Lyndon word (Corollary 3.1).

Example 4.2

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, let $w=cbabcbad\in\Sigma^{+}$. The word $w$ is not an inverse Lyndon word but any nonempty proper prefix of $w$ is an inverse Lyndon word. By Lemma 4.7, item (1), we have $w=p\overline{p}$, with $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. We have $cbab\ll cbad$ but $cba$ is not the shortest prefix $r$ of $w=p\overline{p}$ satisfying item (2) in lemma 4.7. Notice that $cbad$ is not an inverse Lyndon word. Since $cbabcba\ll d$, the shortest prefix $r$ of $w=p\overline{p}$ satisfying item (2) in lemma 4.7 is the empty word. Consequently, $(p,\overline{p})=(cbabcba,d)$.**

Example 4.3

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, let $v=dabdabdadac$. We can check that $dabdabdad$ is the shortest nonempty prefix of $v$ which is not an inverse Lyndon word, hence $dabdabdad=p\overline{p}$, with $(p,\overline{p})\in\operatorname{Pref}_{bre}(v)$. The shortest prefix $r$ of $p\overline{p}$ satisfying item (2) in lemma 4.7 is $da$, thus $(p,\overline{p})=(dabdab,dad)\in\operatorname{Pref}_{bre}(v)$. As another example, consider $w=dabadabdabdadac$. We can check that $dabadabd$ is the shortest nonempty prefix of $w$ which is not an inverse Lyndon word, hence $dabadabd=q\overline{q}$, with $(q,\overline{q})\in\operatorname{Pref}_{bre}(w)$. The shortest prefix $r$ of $q\overline{q}$ satisfying item (2) in lemma 4.7 is $dab$, thus $(q,\overline{q})=(daba,dabd)\in\operatorname{Pref}_{bre}(w)$.**

Example 4.4

Let $\Sigma=\{a,b,c\}$ with $a<b<c$, let $v=cbabacbac$. We can check that $v=cbabacbac$ is the shortest nonempty prefix of $v$ which is not an inverse Lyndon word, hence $v=cbabacbac=p\overline{p}$, with $(p,\overline{p})\in\operatorname{Pref}_{bre}(v)$. The shortest prefix $r$ of $p\overline{p}$ satisfying item (2) in lemma 4.7 is $cba$, thus $(p,\overline{p})=(cbaba,cbac)\in\operatorname{Pref}_{bre}(v)$. As another example, consider $w=cbabacaacbabacbac$. We can check that $cbabacaacbabacb$ is the shortest nonempty prefix of $w$ which is not an inverse Lyndon word, hence $cbabacaacbabacb=q\overline{q}$, with $(q,\overline{q})\in\operatorname{Pref}_{bre}(w)$. The shortest prefix $r$ of $q\overline{q}$ satisfying item (2) in lemma 4.7 is $cbabac$, thus $(q,\overline{q})=(cbabacaa,cbabacb)\in\operatorname{Pref}_{bre}(w)$.**

4.4 A canonical inverse Lyndon factorization: $\operatorname{ICFL}(w)$

We give below the recursive definition of the canonical inverse Lyndon factorization $\operatorname{ICFL}(w)$.

Definition 4.3

Let $w\in\Sigma^{+}$.

(Basis Step)* If $w$ is an inverse Lyndon word, then $\operatorname{ICFL}(w)=(w)$.*

(Recursive Step)* If $w$ is not an inverse Lyndon word, let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$ and let $v\in\Sigma^{*}$ such that $w=pv$. Let $\operatorname{ICFL}(v)=(m^{\prime}_{1},\ldots,m^{\prime}_{k})$ and let $r\in\Sigma^{*}$ and $a,b\in\Sigma$ such that $p=rax$, $\overline{p}=rb$ with $a<b$.*

[TABLE]

Remark 4.1

With the same notations as in the recursive step of Definition 4.3, we notice that $rb$ and $m^{\prime}_{1}$ are both prefixes of the same word $v$. Thus they are comparable for the prefix order, that is, either $rb\leq_{p}m^{\prime}_{1}$ or $m^{\prime}_{1}\leq_{p}r$.**

Example 4.5

Let $\Sigma=\{a,b,c\}$ with $a<b<c$, let $v=cbabacbac$. Let us compute $\operatorname{ICFL}(v)$. As showed in Example 4.4, we have $(p,\bar{p})=(cbaba,cbac)\in\operatorname{Pref}_{bre}(v)$. Therefore, $cbaba,cbac$ are both inverse Lyndon words and we have $\operatorname{ICFL}(cbaba)=(cbaba)$, $\operatorname{ICFL}(\bar{p})=\operatorname{ICFL}(cbac)=(m^{\prime})=(cbac)$. Since $\bar{p}=cbac\leq_{p}m^{\prime}$, we are in the first case of Definition 4.3, hence $\operatorname{ICFL}(v)=\operatorname{ICFL}(cbabacbac)=(p,\operatorname{ICFL}(\bar{p}))=(cbaba,cbac)=(m^{\prime}_{1},m^{\prime}_{2})$. Now, let $w=cbabacaacbabacbac$ and let us compute $\operatorname{ICFL}(w)$. In Example 4.4 we showed that $(q,\bar{q})=(cbabacaa,cbabacb)\in\operatorname{Pref}_{bre}(w)$. Thus, $w=qv$, where $v=cbabacbac$ is the above considered word and $\bar{q}=r^{\prime}b$, where $r^{\prime}=cbabac$. Since $m^{\prime}_{1}=cbaba\leq_{p}cbabac$, we are in the second case of Definition 4.3, therefore $\operatorname{ICFL}(w)=(qm^{\prime}_{1},m^{\prime}_{2})=(cbabacaacbaba,cbac)$.**

Example 4.6

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, let $v=dabdabdadac$. Let us compute $\operatorname{ICFL}(v)$. As showed in Example 4.3, we have $(p,\bar{p})=(dabdab,dad)\in\operatorname{Pref}_{bre}(v)$. Hence, $v=pv^{\prime}$, where $v^{\prime}=dadac=\overline{p}ac$. On the other hand, we can easily see that $dadac$ is an inverse Lyndon word, hence $\operatorname{ICFL}(dadac)=(dadac)$. Since $\overline{p}=dad\leq_{p}dadac$, by Definition 4.3 (first case), $\operatorname{ICFL}(v)=(dabdab,dadac)$. Now, let us compute $\operatorname{ICFL}(w)$, where $w=dabadabdabdadac$. In Example 4.4 we showed that $(q,\bar{q})=(daba,dabd)\in\operatorname{Pref}_{bre}(w)$. Thus, $w=qv$, where $v$ is the above considered word. Since $\overline{q}=dabd\leq_{p}dabdab$, by Definition 4.3 (first case), $\operatorname{ICFL}(w)=(daba,dabdab,dadac)$. In Example 4.1 we showed two different inverse Lyndon factorizations of $w^{\prime}=dabdadacddbdc$. Both are different from $\operatorname{ICFL}(w^{\prime})=(dab,dadac,ddbdc)$.**

As proved below, $\operatorname{ICFL}(w)$ is uniquely determined.

Proposition 4.2

For any word $w\in\Sigma^{+}$, there is a unique sequence $(m_{1},\ldots,m_{k})$ of words over $\Sigma$ such that $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$.

Proof :

The proof is by induction on $|w|$. If $w$ is a letter, then the statement is proved since $w$ is an inverse Lyndon word and $\operatorname{ICFL}(w)=(w)$ (Definition 4.3). Thus, assume $|w|>1$. If $w$ is an inverse Lyndon word we have done since, by Definition 4.3, $\operatorname{ICFL}(w)=(w)$. Otherwise, $w$ is not an inverse Lyndon word and there is a unique pair $(p,\overline{p})$ in $\operatorname{Pref}_{bre}(w)$ (Proposition 4.1). Let $v\in\Sigma^{+}$ such that $w=pv$. Since $|v|<|w|$, by induction hypothesis there is a unique sequence $(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$ of words such that $\operatorname{ICFL}(v)=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$. Let $r\in\Sigma^{*}$ and $a,b\in\Sigma$ such that $p=rax$, $\overline{p}=rb$ with $a<b$. By the recursive step of Definition 4.3, we have

[TABLE]

In both cases, by the above arguments, the sequence $\operatorname{ICFL}(w)$ is uniquely determined.

As a main result, we now prove that $\operatorname{ICFL}(w)$ is an inverse Lyndon factorization of $w$.

Lemma 4.9

For any $w\in\Sigma^{+}$, the sequence $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$ is an inverse Lyndon factorization of $w$, that is, $w=m_{1}\cdots m_{k}$, $m_{1}\ll\ldots\ll m_{k}$ and each $m_{i}$ is an inverse Lyndon word.

Proof :

The proof is by induction on $|w|$. If $w$ is a letter, then the statement is proved since $w$ is an inverse Lyndon word and $\operatorname{ICFL}(w)=(w)$ (Definition 4.3). Thus, assume $|w|>1$. If $k=1$ we have done since, by Definition 4.3, $w$ is an inverse Lyndon word and $\operatorname{ICFL}(w)=(w)$. Otherwise, $w$ is not an inverse Lyndon word. Let $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$ and let $v\in\Sigma^{+}$ such that $w=pv$. Let $\operatorname{ICFL}(v)=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$ and let $r,x\in\Sigma^{*}$ and $a,b\in\Sigma$ such that $p=rax$, $\overline{p}=rb$ with $a<b$. By the recursive step of Definition 4.3, we have

[TABLE]

Moreover, since $|v|<|w|$, by induction hypothesis, $v=m^{\prime}_{1}\cdots m^{\prime}_{k^{\prime}}$, $m^{\prime}_{1}\ll\ldots\ll m^{\prime}_{k^{\prime}}$ and each $m^{\prime}_{i}$ is an inverse Lyndon word. Therefore, $w=pv=pm^{\prime}_{1}\cdots m^{\prime}_{k^{\prime}}$.

If $rb\leq_{p}m^{\prime}_{1}$, then there is $z\in\Sigma^{*}$ such that $m^{\prime}_{1}=rbz$ and $\operatorname{ICFL}(w)=(p,m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$. By Definition 4.2, $p$ is an inverse Lyndon word and so are all the words in $\operatorname{ICFL}(w)$. Furthermore, $p\ll m^{\prime}_{1}\ll\ldots\ll m^{\prime}_{k^{\prime}}$ and the proof is ended. Otherwise, there is $z\in\Sigma^{*}$ such that $r=m^{\prime}_{1}z$, hence $p=rax=m^{\prime}_{1}zax$ and $\operatorname{ICFL}(w)=(pm^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$. The word $pm^{\prime}_{1}$ is a proper prefix of $p\overline{p}$, thus, by Definition 4.2, $pm^{\prime}_{1}$ is an inverse Lyndon word and so are all words in $\operatorname{ICFL}(w)$. Finally, by $m^{\prime}_{1}\ll m^{\prime}_{2}\ldots\ll m^{\prime}_{k^{\prime}}$ we have $pm^{\prime}_{1}=m^{\prime}_{1}zaxm^{\prime}_{1}\ll m^{\prime}_{2}\ldots\ll m^{\prime}_{k^{\prime}}$ (Lemma 2.3).

We end this section with the following result showing that inverse Lyndon factorizations of Lyndon words are not trivial.

Proposition 4.3

Any word $w\in\Sigma^{+}$ has an inverse Lyndon factorization $(m_{1},\ldots,m_{k})$. Moreover, if $w$ is a Lyndon word which is not a letter, then $k>1$.

Proof :

The first part of the statement follows by Proposition 4.2 and Lemma 4.9. Assume that $w$ is a Lyndon word which is not a letter and let $(m_{1},\ldots,m_{k})$ be one of its inverse Lyndon factorizations. By Proposition 2.3, there is a proper nonempty suffix $v$ of $w$ such that $w\prec v$. Hence, $w$ is not an inverse Lyndon word, which yields $w\not=m_{1}$ and consequently $k>1$.

Of course a converse of Proposition 4.3 can also be stated.

Proposition 4.4

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$. If $w$ is an inverse Lyndon word which is not a letter, then $h>1$.

Proof :

Let $w$ be an inverse Lyndon word which is not a letter and let $\operatorname{CFL}(w)=(\ell_{1},\ldots,\ell_{h})$. By Definition 3.1, there is a proper nonempty suffix $v$ of $w$ such that $v\prec w$. Hence, by Proposition 2.3, $w$ is not a Lyndon word, which yields $w\not=\ell_{1}$ and consequently $h>1$.

5 An algorithm for finding the bounded right extension

In section 6 we will give a linear time recursive algorithm, called Compute-ICFL, that computes $\operatorname{ICFL}(w)$, for a given nonempty word $w$. By Definition 4.3, we know that the computation of $\operatorname{ICFL}(w)$, when $w$ is not an inverse Lyndon word, is based on that of the pair $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. In this section we give algorithms to compute the above pair $(p,\overline{p})$. By Lemmas 4.7, 4.8, we are faced with the problems of

• (1)

stating whether $w$ is an inverse Lyndon word;

• (2)

if not, finding the shortest prefix $x$ of $w$ such that $x=raurb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Therefore, $x=p\bar{p}$.

• (3)

Finding the shortest $r$ such that $x=raurb$, with $r,u,a,b$ as in (2). Thus $p=rau$, $\bar{p}=rb$.

The first two tasks are carried out by algorithm Find-prefix, the third is performed by algorithm Find-bre. Algorithm Find-prefix is very similar to Duval’s algorithm in [11] which computes the longest Lyndon prefix of a given string $w$. This is not surprising because Find-prefix computes the longest prefix $raur$ of $w$ which is an inverse Lyndon word. Algorithm Compute-ICFL uses algorithm Find-bre as a subroutine. In turn, the latter uses the output of algorithm Find-prefix. Algorithm Find-bre also calls a procedure to compute the well known failure function (named the prefix function in [8]) of the Knuth-Morris-Pratt matching algorithm [20].

Firstly, we give a high-level description of algorithm Find-prefix followed by its pseudocode (Section 5.1) and we prove its correctness through some loop invariants (Section 5.2). Next, in Section 5.3, we recall the definition of the failure function and we prove some results concerning this function which will be useful later. Finally, we give a high-level description of algorithm Find-bre followed by its pseudocode (Section 5.4) and we end the section with the proof of its correctness through some loop invariants (Section 5.5).

5.1 Description of Find-prefix

Let us describe the high-level structure of algorithm Find-prefix. As already said, given $w$, the algorithm looks for the shortest prefix $raurb$ of $w$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$ and $raur$ is an inverse Lyndon word. This is equivalent to find, if it exists, a nonempty word $r$ which is a prefix of $w$ and a prefix of a suffix of $w$, and these two occurrences of $r$ are followed by $a,b$ as above.

As usual, the word is represented as an array $w[1..n]$ containing the sequence of the letters in $w$, with $n=|w|$. The algorithm uses two indices $i,j$ to scan the word $w$. Initially, these indices denote the position of the first letter of a candidate common prefix $r$.

A while-loop is used to compare the two letters $w[i]$ and $w[j]$. While $j$ is incremented at each iteration, $i$ is incremented only when $w[i]=w[j]$. Notice that the algorithm does not test if eventually the two occurrences of $r$ overlap. Lemma 2.2 allows us to avoid this test.

If $w[i]>w[j]$, then the algorithm resets $i$ to the first position of $w$ and examines a new candidate common prefix $r$, whose position of the first letter in the second occurrence of $r$ is indicated by the value of $j$ at this time.

The loop condition is false when $w[i]<w[j]$ or when $j$ denotes the last letter of $w$. In the first case, our search has been successful and the algorithm returns $raurb$. If $j$ denotes the last letter of $w$ (and $w[i]\geq w[j]$), then $w$ has no prefix $raurb$, with $r,u,a,b$ as above, and the algorithm returns w\$$, where $$is a letter such that$$\not\in\Sigma. In this case, by Lemma [4.5](#S4.Thmlemma5), we know that w$ is an inverse Lyndon word.

In conclusion, Find-prefix allows us to state whether $w$ is an inverse Lyndon word or not and, in the latter case, it finds the prefix $x$ of $w$ such that $x=p\bar{p}$, with $(p,\bar{p})\in\operatorname{Pref}_{bre}(w)$. Algorithm 1 describes the procedure Find-prefix. It is understood that the empty array, namely $w[j+1..n]$ with $j=n$, represents the empty word.

We notice that we can reach line (11) in two different cases: either when $x=w$ or when $w$ is an inverse Lyndon word. We distinguish these two cases: in the former case the output is $(w,1)$ and in the latter case the output is (w\,1)$ (see example below).

Example 5.1

Let $\Sigma=\{a,b,c\}$ with $a<b<c$. Algorithm Find-prefix$(bac)$ returns $(bac,1)$. Indeed, $|w|=3$ and $w[2]<w[1]$ in the first iteration of the while-loop. Then, $i=1$ (line $(7)$), $j=3$ (line $(10)$), and we reach line $(11)$ with $j=3=|w|$. Since $w[j]=w[3]=c>w[1]=w[i]$, the algorithm returns $(w[1..j],w[j+1..|w|])=(bac,1)$.

Consider now the inverse Lyndon word $w=bab$. Algorithm Find-prefix$(bab)$ returns (bab\,1)$. Indeed,$|w|=3$and, as before,$w[2]<w[1]$in the first iteration of the while-loop. Then,$i=1$(line$(7)$),$j=3$(line$(10)$), and we reach line$(11)$with$j=3=|w|$. Since$w[j]=w[3]=b=w[1]=w[i]$, the algorithm returns$(bab$,1)$. The same argument applies when$w=baa$.**

Example 5.2

Let $\Sigma=\{a,b\}$ with $a<b$. Algorithm Find-prefix$(bbabbabbb)$ outputs $(bbabbabbb,1)$. **

5.2 Correctness of Find-prefix

In this section we prove that Find-prefix does what it is claimed to do. To begin, notice that each time around the while-loop of lines (5) to (10), $j$ is increased by 1 at line (10). Therefore, when $j$ becomes equal to $|w|$, if we do not break out of the while-loop earlier, the loop condition $j<|w|$ will be false and the loop will terminate. In order to prove that Find-prefix does what it is intended to do, we define the following loop-invariant statement, where we use $k$ to stand for one of the values that the variable $j$ assumes and $h_{k}$ for the corresponding variable $i$, as we go around the loop. Finally $h^{\prime}_{k}$ depends on $h_{k}$ and $k$.

$S(k)$: If we reach the loop test “$j<|w|$ and $w[j]\leq w[i]$” with the variable $i$ having the value $h_{k}$ and the variable $j$ having the value $k$, then

• (a)

$1\leq h_{k}<k$. Set $h^{\prime}_{k}=k-h_{k}+1$. We have $1<h^{\prime}_{k}\leq|w|$ and the following conditions holds

• (a1)

$w[1..h_{k}-1]=w[h^{\prime}_{k}..k-1]$, that is, $w[1..h_{k}-1]$ is a proper prefix of $w[1..k-1]$ and $w[1..h_{k}-1]$ is also a suffix of $w[1..k-1]$.

• (a2)

For any $t^{\prime}$, $1<t^{\prime}<h^{\prime}_{k}$, $w[t^{\prime}..k-1]$ is not a prefix of $w[1..k-1]$.

• (b)

$w[1..k-1]$ has no prefix with the form $raurb$, $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Therefore, by Lemma 4.5, $w[1..k-1]$ is an inverse Lyndon word.

Remark 5.1

Notice that $w[t^{\prime}..k-1]$ is nonempty, for any $t^{\prime}$, $1<t^{\prime}<h^{\prime}_{k}$. Indeed, $h^{\prime}_{k}=k-h_{k}+1\leq k$. Thus, since $t^{\prime}\leq h^{\prime}_{k}-1\leq k-1$, we have $|w[t^{\prime}..k-1]|\geq|w[h^{\prime}_{k}-1..k-1]|\geq|w[k-1..k-1]|=|w[k-1]|=1$.**

Loosely speaking, at the beginning of each iteration of the loop of lines (5)-(10), indices $i,j$ store the end of a candidate common prefix $r$ (item (a1)) and we do not yet find a prefix with the required form in the examined part of the array (items (a2), (b)). Note that the examined part may be a single letter and $r$ could be the empty word.

We need to show that the above loop invariant is true prior to the first iteration and that each iteration of the loop maintains the invariant. This will be done in Proposition 5.1, where we prove $S(k)$ by (complete) induction on $k$. Then, we show that the invariant provides a useful property to prove correctness when the loop terminates in Proposition 5.2. Even if $S(k)$ clearly holds for $k=0,1$, we prove it for $k\geq 2$.

Proposition 5.1

For any $k\geq 2$, $S(k)$ is true.

Proof :

(Basis) Let us prove that $S(2)$ is true. We reach the test with $j$ having value $2$ only when we enter the loop from the outside. Prior to the loop, lines $(3)$ and $(4)$ set $i$ to 1 and $j$ to 2. Therefore $h_{k}=1<k=2$ and $h^{\prime}_{k}=k-h_{k}+1=2$, hence $1<h^{\prime}_{k}\leq|w|$. Part (a1) of $S(2)$ is true. Indeed $w[1..1-1]=w[2..2-1]$ holds since there are no elements in both the descriptions. Part (a2) of $S(2)$ is also true since there are no integers $t^{\prime}$ such that $1<t^{\prime}<2$. Similarly, since $w[1..2-1]=w[1]$, part (b) is true.

(Induction) We suppose that $S(k)$ is true and prove that $S(k+1)$ is true. We may assume $k<|w|$ and $w[k]\leq w[h_{k}]$ (otherwise we break out of the while-loop when $j$ has the value $k$ or earlier and $S(k+1)$ is clearly true, since it is a conditional expression with a false antecedent). Moreover, by using the inductive hypothesis (part (a)), $1\leq h_{k}<k$, $w[1..h_{k}-1]=w[h^{\prime}_{k}..k-1]$, where $h^{\prime}_{k}=k-h_{k}+1>1$ and $w[t^{\prime}..k-1]$ is not a prefix of $w[1..k-1]$, $1<t^{\prime}<h^{\prime}_{k}$.

To prove $S(k+1)$, we consider what happens when we execute the body of the while-loop with $j$ having the value $k$ and $i$ having the value $h_{k}$. Moreover, to prove part (a) of $S(k+1)$, we distinguish two cases:

• (1)

$w[k]<w[h_{k}]$,

• (2)

$w[k]=w[h_{k}]$.

(Case (1)) If $w[k]<w[h_{k}]$, then line $(7)$ set $i$ to 1. Thus, when we reach the loop test with $j$ having the value $k+1$, the variable $i$ has the value $h_{k+1}=1$. Consequently, $h^{\prime}_{k+1}=k+1-h_{k+1}+1=k+1$ which implies $1<h^{\prime}_{k+1}=k+1\leq|w|$ and $w[1..h_{k+1}-1]=w[1..0]=w[k+1..k]$. Therefore, part (a1) is proved in this case. Let us prove (a2). Let $t^{\prime}$ be any integer such that $1<t^{\prime}<h^{\prime}_{k+1}=k+1$. If $w[t^{\prime}..k]$ were a prefix of $w[1..k]$, then there would exist $t$ such that $w[1..t]=w[t^{\prime}..k]$, with $t\geq 1$ (see Remark 5.1). Moreover, $w[t^{\prime}..k-1]$ would be a prefix of $w[1..k-1]$ with $t^{\prime}>1$ thus, by inductive hypothesis (part (a2)), we would have $t^{\prime}\geq h^{\prime}_{k}$. Consequently, $|w[1..t]|=|w[t^{\prime}..k]|\leq|w[h^{\prime}_{k}..k]|=|w[h^{\prime}_{k}..k-1]|+1=|w[1..h_{k}-1]|+1$ which implies $t\leq h_{k}$. Notice that $w[t]=w[k]<w[h_{k}]$, hence $t\leq h_{k}-1$.

Since $t\leq h_{k}-1$, the word $w[1..t-1]=w[t^{\prime}..k-1]$ is a proper prefix of $w[1..h_{k}-1]$ and, since $t^{\prime}\geq h^{\prime}_{k}$, the same word $w[1..t-1]=w[t^{\prime}..k-1]$ is a suffix of $w[h^{\prime}_{k}..k-1]=w[1..h_{k}-1]$. Notice that each time around the while-loop, $j$ is increased by $1$ at line $(10)$ and $2\leq h_{k}+1\leq k$, by inductive hypothesis (part (a)). Therefore, when we reach the loop test with $j$ having the value $h_{k}+1$, the word $w[1..h_{k}]$ has a prefix with the form $raurb$, $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Indeed, $w[1..h_{k}]=raurb$, where $a=w[t]<w[h_{k}]=b$ and $r=w[1..t-1]$ if $t=1$ or if the two occurrences of $w[1..t-1]$ (as a prefix and a suffix) in $w[1..h_{k}-1]$ do not overlap, otherwise it is its prefix given by Lemma 2.2 applied to $w[1..t-1]$. Hence part (b) of $S(h_{k}+1)$ is not true, a contradiction.

(Case (2)) Assume that $w[k]=w[h_{k}]$. Thus, by $w[1..h_{k}-1]=w[h^{\prime}_{k}..k-1]$, we also have $w[1..h_{k}]=w[h^{\prime}_{k}..k]$. Therefore, when we reach the loop test with $j$ having the value $k+1$, the variable $i$ has value $h_{k+1}=h_{k}+1$. Consequently, $h^{\prime}_{k+1}=k+1-h_{k+1}+1=k-h_{k}+1=h^{\prime}_{k}$ and $w[1..h_{k+1}-1]=w[1..h_{k}]=w[h^{\prime}_{k}..k]=w[h^{\prime}_{k+1}..k]$, with $1<h^{\prime}_{k+1}=h^{\prime}_{k}\leq|w|$. Thus part (a1) is proved in case (2). Furthermore, by inductive hypothesis (part (a2)), for any $t^{\prime}$, $1<t^{\prime}<h^{\prime}_{k}=h^{\prime}_{k+1}$, the word $w[t^{\prime}..k-1]$ is not a prefix of $w[1..k-1]$, hence $w[t^{\prime}..k]$ is not a prefix of $w[1..k]$ and part (a2) holds.

To prove part (b) of $S(k+1)$, assume on the contrary that $w[1..k]$ has a prefix with the form $raurb$, $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Thus, by using the inductive hypothesis (part (b)), we would have $w[1..k]=raurb$, that is $w[k]=b$. Moreover, there would exist $t,t^{\prime}$, with $1\leq t=|r|+1<k$ and $1<t^{\prime}\leq k-1$, such that $w[t]=a$ and $r=w[1..t-1]=w[t^{\prime}..k-1]$. Hence,

[TABLE]

Since $t<k$, the word $w[1..t-1]=w[t^{\prime}..k-1]$ is a prefix of $w[1..k-1]$. Therefore, by using the inductive hypothesis, part (a2), we have $t^{\prime}\geq h^{\prime}_{k}$. Now we apply the same argument as above. By $t^{\prime}\geq h^{\prime}_{k}$, we have $|w[1..t-1]|=|w[t^{\prime}..k-1]|\leq|w[h^{\prime}_{k}..k-1]|=|w[1..h_{k}-1]|$ which implies $t-1\leq h_{k}-1$. Moreover $t-1<h_{k}-1$, by Eq. (5.1). In turn, since $t-1<h_{k}-1$, the word $r=w[1..t-1]=w[t^{\prime}..k-1]$ is a proper prefix of $w[1..h_{k}-1]$ and, since $t^{\prime}\geq h^{\prime}_{k}$, the same word $w[1..t-1]=w[t^{\prime}..k-1]$ is a suffix of $w[h^{\prime}_{k}..k-1]=w[1..h_{k}-1]$. As already noticed, each time around the while-loop, $j$ is increased by $1$ at line $(10)$ and $1\leq t<h_{k}<k$ (see above and the inductive hypothesis, part (a)). Therefore, when we reach the loop test with $j$ having the value $h_{k}+1$, we have $w[1..h_{k}]=r^{\prime}au^{\prime}r^{\prime}b^{\prime}$, where $u^{\prime}\in\Sigma^{*}$, $b^{\prime}\in\Sigma$, $a=w[t]<w[h_{k}]=b^{\prime}$ (Eq. (5.1)) and $r^{\prime}=r$ if $r=1$ or if the two occurrences of $r$ (as a prefix and a suffix) in $w[1..h_{k}-1]$ do not overlap, otherwise it is its prefix $r^{\prime}$ given by Lemma 2.2 applied to $r$. Hence part (b) of $S(h_{k}+1)$ is not true, a contradiction.

Of course, when we reach the loop test with $j$ having the value $k+1$, the variable $i$ has either value $h_{k+1}=1$ or $h_{k+1}=h_{k}+1$. We always have $k+1>1$ and, if $1\leq h_{k}<k$, then $1\leq h_{k}+1<k+1$. Thus, we have proved the inductive step.

Proposition 5.2

Algorithm Find-prefix allows us to state whether $w$ is an inverse Lyndon word or not and, in the latter case, it finds the shortest prefix $x$ of $w$ such that $x=raurb$ where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. More precisely, if $w$ is an inverse Lyndon word, then Find-prefix outputs (w\,1)$; if$w$is not an inverse Lyndon word, then **Find-prefix** outputs$(p\bar{p},y)$, where$w=p\bar{p}y$and$(p,\bar{p})\in\operatorname{Pref}_{bre}(w)$.

Proof :

Let $n=|w|$. We have already shown that the while-loop will terminate. Evidently it terminates in two cases: when $w[i]<w[j]$, $j\leq n$, or when $j=n$ and $w[i]\geq w[n]$. Moreover, when we reach line $(11)$, statement $S(j)$ must hold. In particular, $w[1..i-1]=w[h^{\prime}_{j}..j-1]$, where $1<h^{\prime}_{j}=j-i+1\leq n$, $1\leq i<j$.

If $w[i]<w[j]$, the word $w[1..j]$ is a prefix of $w=w[1..n]$ which has the form $raurb$ where $r,u\in\Sigma^{*}$, $a=w[i]<w[j]=b$ and $r=w[1..i-1]=w[h^{\prime}_{j}..j-1]$, if $r=1$ or if the two occurrences of $r$ (as a prefix and a suffix) in $w[1..j-1]$ do not overlap, otherwise it is its prefix $r^{\prime}$ given by Lemma 2.2 applied to $r$. By Corollary 3.1, $w[1..j]=raurb$ is not an inverse Lyndon word. On the contrary, by part (b) of $S(j)$, $raur=w[1..j-1]$ is an inverse Lyndon word. Thus, $raurb$ is the shortest prefix of $w$ which is not an inverse Lyndon word, hence, by Lemma 4.7, $raurb=w[1..j]=p\bar{p}$, where $(p,\bar{p})\in\operatorname{Pref}_{bre}(w)$. Finally, Find-prefix outputs $(w[1..j],w[j+1..|w|])=(p\bar{p},y)$, where $w=p\bar{p}y$ (line (14)).

If $j=n$ and $w[i]\geq w[n]$, then we prove below that $w$ has no prefix with the form $raurb$, $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. As a consequence, $w$ is an inverse Lyndon word (Lemma 4.5) and Find-prefix outputs (w\,1)$(line (13)). If$w$had a prefix with the form$raurb$,$r,u\in\Sigma^{*}$,$a,b\in\Sigma$,$a<b$, then, by part (b) of$S(j)=S(n)$, we would have$w=raurb$, that is$w[n]=b$. Moreover, there would exist$t,t^{\prime}$, with$1\leq t=|r|+1<n$,$1<t^{\prime}\leq n$, such that$r=w[1..t-1]=w[t^{\prime}..n-1]$would be a prefix and a suffix of$w[1..n-1]$ and

[TABLE]

By using part (a2) of $S(n)$, we would have $t^{\prime}\geq h^{\prime}_{n}$, hence $|r|=|w[1..t-1]|=|w[t^{\prime}..n-1]|\leq|w[h^{\prime}_{n}..n-1]|=|w[1..i-1]|$, which implies $t-1\leq i-1$ and finally $t-1<i-1$, by Eq. (5.2). In turn, since $t-1<i-1$, the word $r=w[1..t-1]=w[t^{\prime}..n-1]$ would be a proper prefix of $w[1..i-1]$ and, since $t^{\prime}\geq h^{\prime}_{n}$, the same word $r=w[1..t-1]=w[t^{\prime}..n-1]$ would be a suffix of $w[h^{\prime}_{n}..n-1]=w[1..i-1]$. In conclusion, $w[1..i]=r^{\prime}aur^{\prime}b^{\prime}$, where $b^{\prime}\in\Sigma$, $a=w[t]<b=w[n]\leq w[i]=b^{\prime}$, and $r^{\prime}=r$ if $r=1$ or if the two occurrences of $r$ (as a prefix and a suffix) in $w[1..i-1]$ do not overlap, otherwise it is its prefix $r^{\prime}$ given by Lemma 2.2 applied to $r$. Hence part (b) of $S(i+1)$ would be false, with $i+1\leq j=n$, a contradiction.

5.3 The failure function

We recall that, given a word, represented by the array $w[1..n]$, the failure function for $w$, is the function $f:\{1,\dots,n\}\rightarrow\{0,\ldots,n-1\}$ such that

[TABLE]

That is, $f(i)$ is the length of the longest prefix of $w[1..n]$ which is a proper suffix of $w[1..i]$. Thus, by Lemma 2.1, $f(i)$ is the length of the longest proper prefix of $w[1..i]$ which is a suffix of $w[1..i]$. It is known that by iterating the failure function $f$, we can enumerate all the prefixes $w[1..k]$ that are suffixes of a given prefix $w[1..i]$ (Lemma 5.2). A slightly more precise version of this result is given in Lemma 5.3.

Definition 5.1

For a positive integer $i$, we set

[TABLE]

where it is understood that the sequence in $f^{*}(i)$ stops when $f^{(m)}(i)=0$ is reached.

Lemma 5.1

For any positive integer $i$ and $\ell\geq 1$

[TABLE]

Consequently, $\operatorname{Card}(f^{*}(f(i)))=\operatorname{Card}(f^{*}(i))-1$.

Proof :

The proof is by induction on $\ell\geq 1$. Clearly $f^{(1)}(i)=f(f^{(0)}(i))=f(i)=f^{(0)}(f(i))$. Let $\ell\geq 2$. Then, by using induction hypothesis,

[TABLE]

and the proof is complete.

The following result is proved in [8, Lemma 32.5] (Prefix-function iteration lemma).

Lemma 5.2

Let $w$ be a word of length $n$ with failure function $f$. For any $i$, $1\leq i\leq n$, we have $f^{*}(i)=\{k~{}|~{}k<i\mbox{ and }w[1..k]\mbox{ is a suffix of }w[1..i]\}$.

Lemma 5.3

Let $w$ be a word of length $n$ with failure function $f$. For any $i$, $1\leq i\leq n$, $f^{*}(i)$ lists the sequence of the lengths of all the words that are prefixes and suffixes of $w[1..i]$ in decreasing order. That is,

• (1)

a word $x$ is a prefix and a suffix of $w[1..i]$ if and only if there is $k$, $0\leq k\leq\operatorname{Card}(f^{*}(i))-1$, such that $x=w[1..f^{(k)}(i)]$.

• (2)

If $k<k^{\prime}$, then $|w[1..f^{(k^{\prime})}(i)]|<|w[1..f^{(k)}(i)]|$ and $w[1..f^{(k^{\prime})}(i)]$ is a prefix and a suffix of $w[1..f^{(k)}(i)]$.

Proof :

(1) Let $x$ be a prefix and a suffix of $w[1..i]$. We prove that there is $k$, $0\leq k\leq\operatorname{Card}(f^{*}(i))-1$, such that $x=w[1..f^{(k)}(i)]$ by induction on $i$. If $i=1$, then $x=w[1]$ or $x=1$. Since $f^{*}(i)=\{1,0\}$ the statement is proved for $i=1$. Assume $i>1$. If $x=w[1..i]=w[1..f^{(0)}(i)]$, then we have done. Otherwise, by the definition of the failure function, $x$ is a prefix and a suffix of $w[1..f(i)]$. Of course $f(i)<i$, hence, by induction hypothesis, there is $k$, $0\leq k\leq\operatorname{Card}(f^{*}(f(i)))-1$, such that $x=w[1..f^{(k)}(f(i))]$. Thus, by Lemma 5.1, there is $k+1$, $0\leq k+1\leq\operatorname{Card}(f^{*}(i))-1$, such that $x=w[1..f^{(k+1)}(i)]$. Conversely, by Lemma 5.2, if there is $k$, $0\leq k\leq\operatorname{Card}(f^{*}(i))-1$, such that $x=w[1..f^{(k)}(i)]$, then $x$ is a prefix and a suffix of $w[1..i]$.

(2) By the definition of the failure function, one has that the sequence $f^{*}(i)$ is strictly decreasing, that is, if $k<k^{\prime}$, with $0\leq k,k^{\prime}\leq\operatorname{Card}(f^{*}(i))-1$, then $f^{(k^{\prime})}(i)<f^{(k)}(i)$ and consequently $|w[1..f^{(k^{\prime})}(i)]|<|w[1..f^{(k)}(i)]|$. Therefore $w[1..f^{(k^{\prime})}(i)]$ is a prefix and a suffix of $w[1..f^{(k)}(i)]$ by Lemma 2.1.

The longest proper prefix of $w[1..i]$ which is a suffix of $w[1..i]$ is also called a border of $w[1..i]$ [24]. It is known that there is an algorithm that outputs the array $\mathcal{F}$ of $n=|w|$ integers such that $\mathcal{F}[i]=f(i)$ is the length of the border of $w[1..i]$ in time $\mathcal{O}(n)$ (see [8] for a description of this procedure, called Compute-Prefix-Function or [24], where it is called Border). Moreover, by Lemma 2.1, the array $\mathcal{F}$ determines the set $f^{*}(n)\setminus\{n\}=\{f(n),f^{(2)}(n),\ldots,f^{(m)}(n)\}$. Indeed, $\mathcal{F}[n]=f(n)$ and, if $f^{k}(n)=\mathcal{F}[i]=f(i)=j$, then $f^{k+1}(n)=f(f^{k}(n))=f(j)=\mathcal{F}[j]$.

Example 5.3

Let us consider again $w=bbabbabbb$ as in Example 5.2. We have that $\mathcal{F}=[0,1,0,1,2,3,4,5,2]$. **

5.4 Description of Find-bre

In this section we present Find-bre, which applies to the output $(x,y)$ of Find-prefix$(w)$ when $w$ is not an inverse Lyndon word. In this case, $x=p\overline{p}$, where $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. As already said, the task of Find-bre$(x,y)$ is to find the shortest $r$ such that $x=raurb$, where $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$. Hence, by Lemma 4.7, we have $p=rau$ and $\bar{p}=rb$. Therefore, Find-bre$(x,y)$ computes the prefix $p$ and its bounded right extension $\overline{p}$ and outputs the quadruple $(p,\overline{p},y,|r|)$. Find-bre uses the array $\mathcal{F}$ computed by Border$(raur)$.

Example 5.4

Let $\Sigma=\{a,b\}$ with $a<b$, let $w=bbabbabbb$. In Example 5.2 we noticed that Find-prefix$(bbabbabbb)$ outputs $(bbabbabbb,1)$. We can check that Find-bre$(bbabbabbb,1)$ returns $(bbabba,bbb,1,2)$.

5.5 Correctness of Find-bre

In this section we prove that Find-bre does what it is claimed to do. To begin, notice that each time around the while-loop of lines (3) to (6), $i$ decreases since $f(i)<i$. Thus when $i$ becomes zero, the loop condition $i>0$ will be false and the loop will terminate. Then, consider the following loop-invariant statement.

S(t): If we reach the loop test “$i>0$” with the variable $i$ having the value $h$ and the variable $\operatorname{LAST}$ having the value $k$, after $t$ iterations of the while-loop, then

• •

$h=f^{t}(n)$.

• •

$k\geq h$. Precisely, if $w[h+1]<w[n+1]$, then $k=h$ else $k>h$.

Proposition 5.3

For any $t\geq 0$, $S(t)$ is true.

Proof :

(Basis) Let us prove that $S(t)$ is true for $t=0$. We reach the test after [math] iterations of the while-loop only when we enter the loop from the outside. Prior to the loop, lines (1) and (2) set $i$ to $n=f^{0}(n)$ and $\operatorname{LAST}$ to $n+1$. Hence, we reach the test after [math] iterations of the while-loop with $h=f^{0}(n)$ and $k>h$. Thus, since $w[h+1]=w[n+1]$, clearly $S(0)$ is true.

(Induction) We suppose that $S(t)$ is true and prove that $S(t+1)$ is true. Therefore, after $t$ iterations of the while-loop, we reach the loop test “$i>0$”, with the variable $i$ having the value $h=f^{t}(n)$. We may assume $h>0$ (otherwise we break out of the while-loop after $t$ iterations or earlier and $S(t+1)$ is clearly true, since it is a conditional expression with a false antecedent). Let us consider what happens when we run the $(t+1)$th iteration of the while-loop and we execute the body of the while-loop with $i$ having the value $h$ and $\operatorname{LAST}$ having the value $k$.

The variable $i$ assumes value $f(h)=f^{t+1}(n)$ on line (6). Moreover, if $w[f(h)+1]=w[f^{t+1}(n)+1]<w[n+1]$, then $\operatorname{LAST}$ assumes value $f^{t+1}(n)$ (lines (4)-(5)). Otherwise $\operatorname{LAST}$ remains unchanged, hence $k\geq h=f^{t}(n)>f^{t+1}(n)$ (induction hypothesis). In both cases, $S(t+1)$ is true.

Proposition 5.4

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. Algorithm Find-bre, applied to the output $(x,y)$ of Find-prefix$(w)$, outputs the quadruple $(p,\overline{p},y,|r|)$, where $w=xy=p\overline{p}y$, $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$, $p=rau$, $\overline{p}=rb$.

Proof :

Let $w\in\Sigma^{+}$ be a word which is not an inverse Lyndon word. By Proposition 5.2, if $(x,y)$ is the output of Find-prefix$(w)$, then $x=p\overline{p}$, where $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$, and $w=p\overline{p}y$. Let $z$ be the prefix of $x$ such that $|z|=|x|-1$. By Lemma 4.7, it suffices to prove that Find-bre outputs $(rau,rb,y,|r|)$, where $r$ is the shortest prefix and suffix of $z$ such that $zb=x=p\overline{p}=raurb$, with $r,u\in\Sigma^{*}$, $a,b\in\Sigma$, $a<b$.

Of course after $m=\operatorname{Card}(f^{*}(n))-1$ iterations, the while-loop of lines (3) to (6) terminates. Recall also that the sequence $f^{*}(n)$ is strictly decreasing (Lemma 5.3). Let $r$ be the shortest prefix and suffix of $z$ such that $z=raur$ and $w[|r|+1]=a<b=w[n+1]$. By Lemma 5.3, there is $q\geq 0$ such that $|r|=f^{q}(n)$. Moreover, at line (6), the integer $t$ such that $i=f^{t}(n)$ is incremented by 1 each time around the loop. Therefore, after $q$ iterations of the while-loop we have $i=|r|=f^{q}(n)$. By Proposition 5.3, $S(q)$ is true. Thus, $i=|r|=f^{q}(n)=\operatorname{LAST}$ since $w[|r|+1]=w[i+1]=a<b=w[n+1]$. This value of $\operatorname{LAST}$ remains unchanged until we break out of the while-loop since, otherwise, for $s>q$, we would have $w[f^{s}(n)+1]=c<w[n+1]$ and there would exist a shorter word $r^{\prime}=w[1..f^{s}(n)]$ such that $z=r^{\prime}u^{\prime}r^{\prime}$, where $u^{\prime}$ starts with $c$, a contradiction. Finally, Find-bre outputs $(w[1..n-\operatorname{LAST}],w[n-\operatorname{LAST}+1..n]b,y,\operatorname{LAST})$. Now $w[n-\operatorname{LAST}+1..n]$ is the suffix of $z$ of length $\operatorname{LAST}=|r|$, hence $w[n-\operatorname{LAST}+1..n]=r$. Of course $w[1..n-\operatorname{LAST}]=rau$.

6 Computing $\operatorname{ICFL}$ in linear time

In this section we give a linear time algorithm, called Compute-ICFL to compute $\operatorname{ICFL}(w)$. For the sake of simplicity we present a recursive version of Compute-ICFL. In this case the correctness of the algorithm easily follows from the definition of $\operatorname{ICFL}$. The output of Compute-ICFL is represented as a list denoted by $list$.

Let us describe the high-level structure of algorithm Compute-ICFL$(w)$. The algorithm firstly calls Find-prefix$(w)$ (line (1)), that, in view of Proposition 5.2, allows us to state whether $w$ is an inverse Lyndon word or not.

If $w$ is an inverse Lyndon word, then Find-prefix$(w)$ returns (x,y)=(w\,1)$and **Compute-ICFL** stops and returns$(w)(lines (2)-(3)), according to Definition [4.3](#S4.Thmdefinition3). Ifw$is not an inverse Lyndon word, then **Find-prefix$(w)$** returns the pair$(x,y)$such that$w=xy=p\overline{p}y$,$(p,\overline{p})\in\operatorname{Pref}{bre}(w)$, and **Compute-ICFL** calls **Find-bre$(x,y)$** (line (4)). In turn, **Find-bre$(x,y)$** returns a quadruple$(x{1},x_{2},y,\operatorname{LAST})$, where$x_{1}=p$,$x_{2}=\overline{p}$and$\operatorname{LAST}=|r|$. Next, **Compute-ICFL** recursively calls itself on$x_{2}y$(line (5)) and returns$list=\operatorname{ICFL}(x_{2}y)=\operatorname{ICFL}(\bar{p}y)$. Let$z=m^{\prime}{1}$be the first element of$list(line (6)). According to Definition [4.3](#S4.Thmdefinition3), we have to test whetherx{2}=\overline{p}=rb\leq_{p}m^{\prime}{1}=z$, that is if$|z|>|r|=\operatorname{LAST}$. This is done on line (7). If$|z|>|r|=\operatorname{LAST}$, then we add$x{1}=p$at the first position of$list$(line (8)), otherwise we replace$z=m^{\prime}{1}$in$list$with$x{1}z=pm^{\prime}_{1}$(line (10)). In both cases, **Compute-ICFL** returns$list=\operatorname{ICFL}(w)$.

It is worth of noting that there is no preprocessing of $w$ for computing the failure function used by Find-bre$(x,y)$. Each call to Find-bre$(x,y)$ calls Border$(x^{\prime})$, where $(x,y)$ is the output of Find-prefix$(w)$ and $x^{\prime}$ is the prefix of $x$ of length $|x|-1$.

Example 6.1

Let $\Sigma=\{a,b,c\}$ with $a<b<c$, let $w=cbabacaacbabacbac$, already considered in Example 4.5. Suppose that we call Compute-ICFL on $w$ and on the empty list $list$. The three tables below illustrate the sequence of calls made to Compute-ICFL, Find-prefix and Find-bre if we read the first column downward. For instance, since Find-prefix$(cbac)$ returns (cbac\,1)$, **Compute-ICFL$(cbac)$** returns$(cbac)$without invoking itself again and the recursion stops. **Compute-ICFL$(cbabacbac)$** calls **Find-prefix** on$cbabacbac$, **Find-bre** on$(cbabacbac,1)$and then **Compute-ICFL$(cbac)$** which returns$(cbac)$. Since$|z|=|cbac|=4>\operatorname{LAST}=3$, **Compute-ICFL$(cbabacbac)$** returns$(cbaba,cbac)$. Finally, **Compute-ICFL$(w)$** calls **Find-prefix** on$w$, **Find-bre** on$(cbabacaacbabacb,ac)$and then **Compute-ICFL$(cbabacbac)$** which returns$(cbaba,cbac)$. Since$|z|=|cbaba|=5\leq\operatorname{LAST}=6$, **Compute-ICFL$(w)$** replaces$cbaba$with the concatenation of$cbabacaa$and$cbaba$and returns$list=(cbabacaacbaba,cbac)=\operatorname{ICFL}(w)$.

CALL RETURN

Compute-ICFL$(cbabacaacbabacbac)$ $(cbabacaacbaba,cbac)$

Compute-ICFL$(cbabacbac)$ $(cbaba,cbac)$

Compute-ICFL$(cbac)$ $(cbac)$

CALL RETURN

Find-prefix$(cbabacaacbabacbac)$ $(cbabacaacbabacb,ac)$

Find-prefix$(cbabacbac)$ $(cbabacbac,1)$

Find-prefix$(cbac)$ (cbac\,1)$

CALL RETURN

Find-bre$(cbabacaacbabacb,ac)$ $(cbabacaa,cbabacb,ac,6)$

Find-bre$(cbabacbac,1)$ $(cbaba,cbac,1,3)$

6.1 Performance of Compute-ICFL

Let us compute the running time $T(n)$ of Compute-ICFL when $w$ has length $n$. We can check that the running time of Find-prefix$(w)$ is $\mathcal{O}(|x|)$, when the output of the procedure is $(x,y)$. Then, Compute-ICFL calls the procedure Find-bre$(x,y)$, where $x=p\overline{p}$ and $(p,\overline{p})\in\operatorname{Pref}_{bre}(w)$. By Lemma 4.1, we know that $|\overline{p}|\leq|p|$, and so the running time of Find-bre is $\mathcal{O}(|p|+|\overline{p}|)=\mathcal{O}(|p|)$. Let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$ and let $n_{j}$ be the length of $m_{j}$, $1\leq j\leq k$. The recurrence for $T(n)$ is defined as $T(n)=T(n-n_{1})+\mathcal{O}(n_{1})$, where $T(n_{k})$ is $\mathcal{O}(n_{k})$, because there is no recursive call in this case. It is easy to see that the solution to this recurrence is $T(n)=\sum_{j=1}^{k}\mathcal{O}(n_{j})=\mathcal{O}(n)$, since $w=m_{1}\cdots m_{k}$.

7 Groupings

Let $(\Sigma,<)$ be a totally ordered alphabet. As we know, for any word $w\in\Sigma^{+}$, there are three sequences of words associated with $w$: the Lyndon factorization of $w$ with respect to the order $\prec$, denoted $\operatorname{CFL}(w)$, the Lyndon factorization of $w$ with respect to the inverse lexicographic order $\prec_{in}$, denoted $\operatorname{CFL}_{in}(w)$ and the inverse Lyndon factorization $\operatorname{ICFL}(w)$ of $w$. In this section, we compare $\operatorname{CFL}_{in}(w)$ and $\operatorname{ICFL}(w)$.

We begin by proving some relations between inverse Lyndon words and anti-Lyndon words (Section 7.1). Then, we point out some relations between $\operatorname{ICFL}(w)$ and $\operatorname{CFL}_{in}(w)$. Precisely, starting with $\operatorname{CFL}_{in}(w)$, we define a family of inverse Lyndon factorizations of $w$, called groupings of $\operatorname{CFL}_{in}(w)$ (Section 7.2). We prove that $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$ in Section 7.3. Then, we prove that the set of the inverse Lyndon words is equal to the set of the strict sesquipowers of anti-Lyndon words (Section 7.4). Finally, we prove that Theorem 2.2 may be generalized to groupings when we refer to the sorting with respect to the inverse lexicographic order (Section 7.5).

7.1 Inverse Lyndon words and anti-Lyndon words

Let $(\Sigma,<)$ be a totally ordered alphabet, let $<_{in}$ be the inverse of $<$ and let $\prec_{in}$ be the inverse lexicographic order on $(\Sigma,<)$. The following proposition justifies the adopted terminology.

Proposition 7.1

Let $(\Sigma,<)$ be a totally ordered alphabet. For all $x,y\in\Sigma^{*}$ such that $x\Join y$,

[TABLE]

Moreover, in this case $x\ll y$.

Proof :

Let $x,y\in\Sigma^{*}$ such that $x\Join y$. Assume $y\prec_{in}x$. Thus, by Definition 2.3, there are $a,b\in\Sigma$, with $b<_{in}a$, and $r,s,t\in\Sigma^{*}$ such that $y=rbs$, $x=rat$. Hence $a<b$ and $x\prec y$, by Definition 2.1. Conversely, if $x\prec y$ we have $y\prec_{in}x$ by a similar argument. The second part of the statement follows by Definition 2.1.

Lemma 7.1 is a dual version of item (2) in Lemma 2.3.

Lemma 7.1

If $y\prec_{in}x$ and $x\Join y$, then $yu\prec_{in}xv$ for all words $u,v$.

Proof :

If $y\prec_{in}x$ and $x\Join y$, then $x\ll y$, by Proposition 7.1. Then, by item (2) in Lemma 2.3, $xv\ll yu$ for all words $u,v$. Since $xv\Join yu$, the conclusion follows by Proposition 7.1.

The following proposition characterizes the set $L_{in}=L_{(\Sigma^{*},<_{in})}$ of the anti-Lyndon words on $\Sigma^{*}$.

Proposition 7.2

A word $w\in\Sigma^{+}$ is in $L_{in}$ if and only if $w$ is primitive and $w\succ vu$, for each $u,v\in\Sigma^{+}$ such that $w=uv$.

Proof :

By Definition 2.2, if $w\in L_{in}$, then $w$ is nonempty and primitive. Moreover, if $w=uv$, with $u,v\not=1$, then $w\prec_{in}vu$. Since $uv\Join vu$, by Proposition 7.1 one has $vu\prec w$, i.e., $w\succ vu$. A similar argument shows that if $w$ is a primitive nonempty word and $w\succ vu$, for each $u,v\in\Sigma^{+}$ such that $w=uv$, then $w\in L_{in}$.

We state below a slightly modified dual version of Proposition 2.3. It shows that anti-Lyndon words are inverse Lyndon words.

Proposition 7.3

A word $w\in\Sigma^{+}$ is in $L_{in}$ if and only if $w$ is unbordered and $w\succ v$, for each proper nonempty suffix $v$.

Proof :

Let $w\in L_{in}=L_{(\Sigma^{*},<_{in})}$. Therefore, $w$ is nonempty and unbordered (Definition 2.2, Proposition 2.2). Moreover, if $w=uv$, with $u,v\not=1$, then $w\prec_{in}v$, by Proposition 2.3. In addition, $w\Join v$ since $w$ is unbordered and $|v|<|w|$. Consequently, $w\succ v$ (Proposition 7.1).

Conversely, let $w=uv$ a nonempty unbordered word such that $w\succ v$, for each proper nonempty suffix $v$. Thus, $v\Join w$, hence $w\prec_{in}v$ (Proposition 7.1). By Proposition 2.3, the word $w$ is in $L_{in}$.

Of course there are inverse Lyndon words which are not anti-Lyndon words. For instance consider $\Sigma=\{a,b\}$, with $a<b$. The word $bab$ is an inverse Lyndon words but it is not unbordered, thus it is not an anti-Lyndon word.

The following result give more precise relations between words in $L_{in}$ and their proper nonempty suffixes.

Proposition 7.4

If $v$ is a proper nonempty suffix of $w\in L_{in}$, then $v\ll w$.

Proof :

Let $v$ be a proper nonempty suffix of $w\in L_{in}$. By Proposition 7.3, we have $v\prec w$. Moreover, since $w$ is unbordered, we have $v\Join w$, hence $v\ll w$.

Some compositional properties of the inverse Lyndon words are proved below.

Proposition 7.5

For any $w\in L_{in}$ and $h\geq 1$, the word $w^{h}$ is an inverse Lyndon word on $(\Sigma^{*},<)$.

Proof :

Let $w\in L_{in}$ and let $vw^{i}$ be a proper nonempty suffix of $w^{h}$, $h\geq 1$. If $v=1$, then $0<i<h$, hence $w^{i}$ is a nonempty proper prefix of $w^{h}$. By Definition 2.1, $w^{h}\succ w^{i}$. Otherwise, $v$ is a proper nonempty suffix of $w$. Thus, by Proposition 7.4, $v\ll w$ and by item (2) in Lemma 2.3, $vw^{i}\ll w^{h}$, i.e., $w^{h}\succ vw^{i}$.

Proposition 7.6

Let $\ell_{1},\ell_{2}\in L_{in}$. If $\ell_{2}$ is a proper prefix of $\ell_{1}$, then $\ell_{1}\ell_{2}$ is an inverse Lyndon word.

Proof :

Let $\ell_{1},\ell_{2}$ be as in the statement. Set $\ell_{1}\ell_{2}=\ell_{2}x\ell_{2}=w$. If $s$ is a nonempty proper suffix of $w$, then one of the following cases holds

• (1)

$s$ is a nonempty proper suffix of $\ell_{2}$

• (2)

$s=\ell_{2}$

• (3)

$s=s^{\prime}\ell_{2}$, where $s^{\prime}$ is a nonempty proper suffix of $\ell_{1}$.

Assume that case (1) holds. Thus $s\ll\ell_{2}$, by Proposition 7.4. Consequently, $s\ll\ell_{2}(x\ell_{2})=w$, by item (2) in Lemma 2.3. Of course, $w=\ell_{2}x\ell_{2}\succ\ell_{2}$ (case (2)), therefore assume that case (3) holds, i.e., $s=s^{\prime}\ell_{2}$, where $s^{\prime}$ is a nonempty proper suffix of $\ell_{1}$. Arguing as before, $s^{\prime}\ll\ell_{1}$ (Proposition 7.4), hence $s^{\prime}\ell_{2}\ll\ell_{1}\ell_{2}=w$ (item (2) in Lemma 2.3).

Example 7.1

Let $\ell_{1},\ldots,\ell_{h}$ be words in $L_{in}$ which form a non-increasing chain $\ell_{1}\geq_{p}\ldots\geq_{p}\ell_{h}$ for the prefix order, i.e., $\ell_{i}$ is a prefix of $\ell_{i-1}$, $1<i\leq h$. The word $\ell_{1}\cdots\ell_{h}$ is not necessarily an inverse Lyndon word. As an example, let $\Sigma=\{a,b\}$ with $a<b$. The sequence $baa,ba,b$ is such that $baa\geq_{p}ba\geq_{p}b$. The word $baabab$ is not an inverse Lyndon word since $baabab\ll bab$.**

7.2 A family of inverse Lyndon factorizations of words

Groupings of $\operatorname{CFL}_{in}(w)$ are special inverse Lyndon factorizations. They are constructed in a very natural way. We first give some needed definitions and results.

Definition 7.1

Let $w\in\Sigma^{+}$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$ and let $1\leq r<s\leq h$. We say that $\ell_{r},\ell_{r+1}\ldots,\ell_{s}$ is a non-increasing maximal chain for the prefix order in $\operatorname{CFL}_{in}(w)$, abbreviated $\mathcal{PMC}$, if $\ell_{r}\geq_{p}\ell_{r+1}\ldots\geq_{p}\ell_{s}$. Moreover, if $r>1$, then $\ell_{r-1}\not\geq_{p}\ell_{r}$, if $s<h$, then $\ell_{s}\not\geq_{p}\ell_{s+1}$. Two $\mathcal{PMC}$ $\mathcal{C}_{1}=\ell_{r},\ell_{r+1}\ldots,\ell_{s}$, $\mathcal{C}_{2}=\ell_{r^{\prime}},\ell_{r^{\prime}+1}\ldots,\ell_{s^{\prime}}$ are consecutive if $r^{\prime}=s+1$ (or $r=s^{\prime}+1$).

Lemma 7.2

Let $x,y$ be nonempty words such that $x\succeq_{in}y$. Then either $x\geq_{p}y$ or $x\ll y$.

Proof :

Let $x,y$ be nonempty words such that $x\succeq_{in}y$. Therefore $x$ is not a proper prefix of $y$. If $y$ is not a prefix of $x$, then $x\Join y$ and $y\prec_{in}x$. Hence, by Proposition 7.1, we have $x\ll y$.

The following is a direct consequence of Lemma 7.2.

Proposition 7.7

Let $w\in\Sigma^{+}$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Then

[TABLE]

where any $\mathcal{C}_{j}$, $1\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. Moreover, $\mathcal{C}_{j}$, $\mathcal{C}_{j+1}$ are consecutive and $\ell\ll\ell^{\prime}$, where $\ell$ is the last word in $\mathcal{C}_{j}$ and $\ell^{\prime}$ is the first word in $\mathcal{C}_{j+1}$, for $1\leq j\leq t-1$.

Proof :

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$, i.e.,

[TABLE]

By Lemma 7.2, each symbol $\succeq_{in}$ in Eq. (7.1) may be replaced either by $\geq_{p}$ or by $\ll$. Therefore, the conclusion follows.

The definition of a grouping of $\operatorname{CFL}_{in}(w)$ is given below in two steps. We first define the grouping of a $\mathcal{PMC}$. Then a grouping of $\operatorname{CFL}_{in}(w)$ is obtained by changing each $\mathcal{PMC}$ with one of its groupings.

Definition 7.2

Let $\ell_{1},\ldots,\ell_{h}$ be words in $L_{in}$ such that $\ell_{i}$ is a prefix of $\ell_{i-1}$, $1<i\leq h$. We say that $(m_{1},\ldots,m_{k})$ is a grouping of $(\ell_{1},\ldots,\ell_{h})$ if the following conditions are satisfied.

• (1)

$m_{j}$* is an inverse Lyndon word which is a product of consecutive $\ell_{q}$, $1\leq j\leq k$.*

• (2)

$\ell_{1}\cdots\ell_{h}=m_{1}\cdots m_{k}$,

• (3)

$m_{1}\ll\ldots\ll m_{k}$.

We now extend Definition 7.2 to $\operatorname{CFL}_{in}(w)$.

Definition 7.3

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. We say that $(m_{1},\ldots,m_{k})$ is a grouping of $\operatorname{CFL}_{in}(w)$ if it can be obtained by replacing any $\mathcal{PMC}$ $\mathcal{C}$ in $\operatorname{CFL}_{in}(w)$ by a grouping of $\mathcal{C}$.

Proposition 7.8 shows that groupings of $\operatorname{CFL}_{in}(w)$ are inverse Lyndon factorizations of $w$. However, as Example 7.2 shows, there are inverse Lyndon factorizations which are not groupings.

Example 7.2

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, and let $w=dabadabdabdadac$. Thus, $d<_{in}c<_{in}b<_{in}a$ and $\operatorname{CFL}_{in}(w)=(daba,dab,dab,dadac)$. The two sequences $(daba,dabdab,dadac)$, $(dabadab,dabda,dac)$ are both inverse Lyndon factorizations of $w$ and moreover $\operatorname{ICFL}(w)=(daba,dabdab,dadac)$ (Examples 4.1, 4.6). However, $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$, whereas $(dabadab,dabda,dac)$ is not a grouping of $\operatorname{CFL}_{in}(w)$. In Example 4.1 we also considered the following two inverse Lyndon factorizations of $z=dabdadacddbdc$

[TABLE]

Both of them are not groupings since $\operatorname{CFL}_{in}(z)=(dab,dadac,ddbdc)$. Notice that $\operatorname{CFL}_{in}(z)=\operatorname{ICFL}(z)$ (see Corollary 7.1).**

Proposition 7.8

Let $w\in\Sigma^{+}$. If $(m_{1},\ldots,m_{k})$ is a grouping of $\operatorname{CFL}_{in}(w)$, then $(m_{1},\ldots,m_{k})$ is an inverse Lyndon factorization of $w$.

Proof :

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Therefore, by Proposition 7.7, we have

[TABLE]

where any $\mathcal{C}_{j}$, $1\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. Moreover, $\mathcal{C}_{j}$, $\mathcal{C}_{j+1}$ are consecutive and $\ell\ll\ell^{\prime}$, where $\ell$ is the last word in $\mathcal{C}_{j}$ and $\ell^{\prime}$ is the first word in $\mathcal{C}_{j+1}$, for $1\leq j\leq t-1$.

Now let $(m_{1},\ldots,m_{k})$ be a grouping of $\operatorname{CFL}_{in}(w)$. Any $m_{j}$ is an inverse Lyndon word since it is an element of a grouping of a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. Then, let $\mathcal{S}_{j}$ be the product of the words in $\mathcal{C}_{j}$, $1\leq j\leq t$. It is clear that

[TABLE]

Finally, set $\mathcal{C}_{j}=(\ell_{i},\ldots,\ell_{g})$ and $\mathcal{C}_{j+1}=(\ell_{g+1},\ldots,\ell_{f})$, for $1\leq j\leq t-1$. Thus $\ell_{g}\ll\ell_{g+1}$. It suffices to show that if $(m_{r},\ldots,m_{s})$ is the grouping of $\mathcal{C}_{j}$ that replaces $\mathcal{C}_{j}$ and $(m_{s+1},\ldots,m_{v})$ is the grouping of $\mathcal{C}_{j+1}$ that replaces $\mathcal{C}_{j+1}$, then $m_{s}\ll m_{s+1}$. But this is clear since $\ell_{g}$ is a suffix of $m_{s}$, hence it is also a prefix of $m_{s}$ ($\mathcal{C}_{j}$ is a $\mathcal{PMC}$) and $\ell_{g+1}$ is a prefix of $m_{s+1}$. Therefore, by item (2) in Lemma 2.3, $\ell_{g}\ll\ell_{g+1}$ implies $m_{s}\ll m_{s+1}$.

7.3 $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$

Let us outline how we prove below that $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$. Let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$ and let $\ell_{1},\ldots,\ell_{q}$ be a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$, $1\leq q\leq k$. The proof will be divided into four steps.

• (1)

We prove that $m_{1}$ cannot be a proper prefix of $\ell_{1}$ (Proposition 7.10).

• (2)

We prove that if $m_{1}$ is a prefix of $\ell_{1}\cdots\ell_{q}$, then $m_{1}=\ell_{1}\cdots\ell_{q^{\prime}}$, for some $q^{\prime}$, $1\leq q^{\prime}\leq q$ (Proposition 7.11).

• (3)

We prove that $\ell_{1}\cdots\ell_{q}$ cannot be a proper prefix of $m_{1}$ (Proposition 7.12).

• (4)

We complete the proof by induction on $|w|$ (Proposition 7.13).

We say that a sequence of nonempty words $(m_{1},\ldots,m_{k})$ is a factorization of $w$ if $w=m_{1}\cdots m_{k}$. It is worth of noting that steps (1) and (2) are proved under the more general hypothesis that $(m_{1},\ldots,m_{k})$ is a factorization of $w$ such that $m_{1}\ll m_{2}$ and, for step (2), where $m_{1}$ is an inverse Lyndon word. Proposition 7.9 and Corollary 7.1 deal with two extremal cases where there is only one grouping of $\operatorname{CFL}_{in}(w)$, namely $\operatorname{ICFL}(w)$.

Proposition 7.9

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. If $w$ is an inverse Lyndon word, then either $w$ is unbordered or $\ell_{1},\ldots,\ell_{h}$ is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. In both cases $\operatorname{ICFL}(w)=(w)$ is the unique grouping of $\operatorname{CFL}_{in}(w)$.

Proof :

Let $w\in\Sigma^{+}$ be an inverse Lyndon word and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. We know that $\operatorname{ICFL}(w)=(w)$ (Definition 4.3). By Proposition 7.3, if $w$ is unbordered, then $w$ is an anti-Lyndon word. Thus, by item (iii) in Lemma 2.4, $\operatorname{CFL}_{in}(w)=(w)$ and of course this is the unique grouping of $\operatorname{CFL}_{in}(w)$.

Otherwise, $w$ is bordered and, again by Proposition 7.3, $h>1$. By contradiction assume that $\ell_{1},\ldots,\ell_{h}$ is not a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. By Lemma 7.2, there would be a smallest $q$, $1\leq q\leq h-1$ such that

[TABLE]

Hence, since $\ell_{q}$ is a prefix of $w$, by item (2) in Lemma 2.3, we would have $w\ll\ell_{q+1}\cdots\ell_{h}$, which is a contradiction since $w$ is an inverse Lyndon word and $\ell_{q+1}\cdots\ell_{h}$ is a proper nonempty suffix of $w$. Thus, $\ell_{1},\ldots,\ell_{h}$ is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$ and $\operatorname{ICFL}(w)=(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$. By contradiction, assume that $(m_{1},\dots,m_{k})$, $k\geq 2$, is another grouping of $\operatorname{CFL}_{in}(w)$. Therefore, $m_{1}\ll m_{2}$ and by item (2) in Lemma 2.3, we would have $w\ll m_{2}\cdots m_{k}$, which is a contradiction since $w$ is an inverse Lyndon word and $m_{2}\cdots m_{k}$ is a proper nonempty suffix of $w$.

Proposition 7.10

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. For any factorization $(m_{1},\ldots,m_{k})$ of $w$, with $k>1$, if $m_{1}\ll m_{2}$, then $m_{1}$ cannot be a proper prefix of $\ell_{1}$.

Proof :

Let $(m_{1},\ldots,m_{k})$ be a factorization of $w$ such that $m_{1}\ll m_{2}$. By contradiction assume that $m_{1}$ is a proper prefix of $\ell_{1}$. Therefore, there are two nonempty words $x,y$ such that $m_{1}=x$, $\ell_{1}=xy$, and $v\in\Sigma^{*}$ such that $m_{2}\cdots m_{k}=yv$. On one hand, by Proposition 7.4 we have $y\ll\ell_{1}$. On the other hand, $x=m_{1}\ll m_{2}$, thus $x=m_{1}\ll m_{2}\cdots m_{k}=yv$, by item (2) in Lemma 2.3. Consequently, $x=ras$, $yv=rbt$, with $a<b$. Hence either $y$ is a prefix of $r$, which is a contradiction since $\ell_{1}=xy=rasy$ is unbordered (Proposition 7.3), or $rb$ is a prefix of $y$, which is once again a contradiction since we would have $x\ll y$ and consequently, by item (2) in Lemma 2.3, $\ell_{1}\ll y$.

Proposition 7.11

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Let $(m_{1},\ldots,m_{k})$ be a factorization of $w$ such that $k>1$, $m_{1}$ is an inverse Lyndon word and $m_{1}\ll m_{2}$. Let $\ell_{1},\ldots,\ell_{q}$ be a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$, $1\leq q\leq h$. If $m_{1}$ is a prefix of $\ell_{1}\cdots\ell_{q}$, then $m_{1}=\ell_{1}\cdots\ell_{q^{\prime}}$, for some $q^{\prime}$, $1\leq q^{\prime}\leq q$.

Proof :

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Let $(m_{1},\ldots,m_{k})$ be a factorization of $w$ such that $k>1$, $m_{1}$ is an inverse Lyndon word and $m_{1}\ll m_{2}$. Let $\ell_{1},\ldots,\ell_{q}$ be a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$, $1\leq q\leq h$.

By contradiction, assume that there are two nonempty words $x,y$ such that $m_{1}=\ell_{1}\cdots\ell_{j-1}x$, $xy=\ell_{j}$, $1\leq j\leq q$ (where it is understood that $m_{1}=x$ when $j=1$). By Proposition 7.10, we have $j>1$. Moreover there is $v\in\Sigma^{*}$ such that $m_{2}\cdots m_{k}=yv$.

Since $m_{1}\ll m_{2}$, we have $m_{1}=\ell_{1}\cdots\ell_{j-1}x\ll m_{2}\cdots m_{k}=yv$. Hence, there are words $r,s,t\in\Sigma^{*}$ and letters $a,b\in\Sigma$, with $a<b$ such that $m_{1}=ras$, $m_{2}\cdots m_{k}=yv=rbt$. If $|r|<|y|$, then $rb$ is a prefix of $y$, therefore it is a factor of $\ell_{j}$. In turn, $\ell_{j}$ is a prefix of $\ell_{j-1}$, thus there is $\gamma\in\Sigma^{*}$ such that $rb\gamma$ is a proper nonempty suffix of $m_{1}$. Then, $m_{1}=ras\ll rb\gamma$, a contradiction, since $m_{1}$ is an inverse Lyndon word. Hence $|r|\geq|y|$, i.e., $y$ is a prefix of $r$ and consequently it is a prefix of $m_{1}$. The word $\ell_{j}=xy$ is also a prefix of $m_{1}$, thus $y$ and $\ell_{j}=xy$ are comparable for the prefix order. Therefore $y$ is both a nonempty proper prefix and a suffix of $\ell_{j}$, i.e., $\ell_{j}=xy$ is bordered, a contradiction since $\ell_{j}\in L_{in}$ (see Proposition 7.3).

Proposition 7.12

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$ and let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$. Let $\ell_{1},\ldots,\ell_{q}$ be a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$, $1\leq q\leq h$. Then $m_{1}=\ell_{1}\cdots\ell_{q^{\prime}}$, for some $q^{\prime}$, $1\leq q^{\prime}\leq q$.

Proof :

Let $w\in\Sigma^{+}$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$, and let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$. We prove the statement by induction on $|w|$. If $|w|=1$, then $w$ is an inverse Lyndon word and we have done, by Proposition 7.9. Hence assume $|w|>1$. If $w$ is an inverse Lyndon word, then the proof is ended, once again by Proposition 7.9. Therefore, assume that $w$ is not an inverse Lyndon word.

Let $\ell_{1},\ldots,\ell_{q}$ be a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$, $1\leq q\leq h$. Since $m_{1}$ and $\ell_{1}\cdots\ell_{q}$ are both prefixes of $w$, $m_{1}$ and $\ell_{1}\cdots\ell_{q}$ are comparable for the prefix order. By contradiction, assume that $m_{1}$ violates the statement. Therefore, by Propositions 7.10 and 7.11, $m_{1}$ is not a prefix of $\ell_{1}\cdots\ell_{q}$. Hence, $\ell_{1}\cdots\ell_{q}$ is a proper prefix of $m_{1}$, i.e., there are two words $x,y$ and $j$, with $q<j\leq h$, such that $m_{1}=\ell_{1}\cdots\ell_{j-1}x$, $xy=\ell_{j}$. Moreover, $\ell_{q}\ll\ell_{q+1}$ and, if $j-1=q$, then $x\not=1$. We must have $j-1=q$ (thus $x\not=1$ also) and $y\not=1$. Indeed, otherwise $j-1>q$ or $x=\ell_{j}$, $j\geq q+1$. In both cases, $\ell_{q+1}\cdots\ell_{j-1}x$ would be a proper nonempty suffix of $m_{1}$ and $\ell_{q}$ is a prefix of $\ell_{1}$, hence $\ell_{q}$ is a prefix of $m_{1}$. By item (2) in Lemma 2.3 applied to $\ell_{q}\ll\ell_{q+1}$, we would have $m_{1}\ll\ell_{q+1}\cdots\ell_{j-1}x$, a contradiction since $m_{1}$ is an inverse Lyndon word. In conclusion, there are words $x,y,y^{\prime}$, with $x\not=1$ and $y\not=1$, such that

[TABLE]

Let $(p,\bar{p})\in\operatorname{Pref}_{bre}(w)$. Let $v\in\Sigma^{+}$ be such that $w=pv$ and let $\operatorname{ICFL}(v)=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$. By Definition 4.3, one of the following two cases holds

• (1)

$m_{1}=p$

• (2)

$m_{1}=pm^{\prime}_{1}$.

In both cases, $p$ is a prefix of $\ell_{1}\cdots\ell_{q}x$, therefore $|p|\leq|\ell_{1}\cdots\ell_{q}x|$. We claim that we also have $|p|>|\ell_{1}\cdots\ell_{q}|$. By Eq. (7.2), this is clearly true in case (1), hence assume $m_{1}=pm^{\prime}_{1}$. Since $p\ll\bar{p}$, by item (2) in Lemma 2.3, we have $p\ll v$. Thus, Propositions 7.10 and 7.11 apply to the factorization $(p,v)$ of $w$. Therefore, if $|p|\leq|\ell_{1}\cdots\ell_{q}|$, then $p=\ell_{1}\cdots\ell_{j}$, with $j\leq q$. Hence, $v=\ell_{j+1}\cdots\ell_{h}$ and, by Theorem 2.1, $\operatorname{CFL}_{in}(v)=(\ell_{j+1},\ldots,\ell_{h})$. On the other hand, by Eq. (7.2), we would have $m^{\prime}_{1}=\ell_{j+1}\cdots\ell_{q}x$, $\ell_{q+1}=xy$, $\ell_{q}\ll\ell_{q+1}$, and $x\not=1$, in contradiction with induction hypothesis applied to $v$. In conclusion, in both cases (1) and (2), we have $|\ell_{1}\cdots\ell_{q}|<|p|\leq|\ell_{1}\cdots\ell_{q}x|$, thus there are words $x_{1},x_{2}$ such that

[TABLE]

Then, by Definitions 4.2 and 4.3, there are words $r,s,t\in\Sigma^{*}$ and letters $a,b\in\Sigma$, with $a<b$ such that

[TABLE]

Since $\ell_{q}\ll\ell_{q+1}$, there are words $z,f,g\in\Sigma^{*}$ and letters $c,d\in\Sigma$, with $c<d$ such that

[TABLE]

Observe that $x$ and $zd$ are both prefixes of $\ell_{q+1}$, hence they are comparable for the prefix order. If $zd$ would be a prefix of $x$, then for a word $\gamma$, $zd\gamma$ would be a proper nonempty suffix of $m_{1}$ (see Eq. (7.2)) and $\ell_{q}$ would be a prefix of $\ell_{1}$, thus of $m_{1}$ such that $\ell_{q}\ll zd\gamma$ (see Eq. (7.5)). By item (2) in Lemma 2.3 applied to $\ell_{q}\ll zd\gamma$, we would have $m_{1}\ll zd\gamma$, a contradiction since $m_{1}$ is an inverse Lyndon word. Therefore, $x$ is a proper prefix of $zd$, i.e., there is $z^{\prime}\in\Sigma^{*}$ such that

[TABLE]

Eqs. (7.5) and (7.6) yield $xy=zdg=xz^{\prime}dg$, therefore $z^{\prime}$ is a prefix of $y$ and thus $x_{2}z^{\prime}$ is a prefix of $x_{2}m_{2}\cdots m_{k}$ (see Eq. (7.2)). On the other hand $rb$ is also a prefix of $x_{2}m_{2}\cdots m_{k}$ (see Eq. (7.4)). Hence, $rb$ and $x_{2}z^{\prime}$ are comparable for the prefix order and one of the following two cases is satisfied

• (i)

$rb$ is a prefix of $x_{2}z^{\prime}$,

• (ii)

$x_{2}z^{\prime}$ is a prefix of $r$.

Assume that case (i) holds. In this case, since $x_{2}z^{\prime}$ is a suffix of $z$ (Eq. (7.6)) and $z$ is a prefix of $\ell_{q}$ (Eq. (7.5)), the word $rb$ would be a factor of $\ell_{q}$. Thus, by Eq. (7.2), there would be $\gamma\in\Sigma^{*}$ such that $rb\gamma$ would be a proper nonempty suffix of $m_{1}$. Since $ras=p$ (Eq. (7.4)), the word $ras$ would be a prefix of $m_{1}$ and we would have $m_{1}\ll rb\gamma$, a contradiction since $m_{1}$ is an inverse Lyndon word. Case (ii) also leads to a contradiction. Indeed, set $p^{\prime}=\ell_{1}\cdots\ell_{q}$. Then, by Eqs. (7.3) and (7.6), $px_{2}z^{\prime}=p^{\prime}x_{1}x_{2}z^{\prime}=p^{\prime}z$. If $x_{2}z^{\prime}$ is a prefix of $r$, then $px_{2}z^{\prime}=p^{\prime}z$ is a prefix of $pr$, thus $px_{2}z^{\prime}=p^{\prime}z$ is a proper prefix of $p\bar{p}$ (Eq. (7.4)), hence $px_{2}z^{\prime}=p^{\prime}z$ is an inverse Lyndon word, as do all nonempty prefixes of $p^{\prime}z$, by Definition 4.2. Moreover, by Eq. (7.5), $zc$ is a prefix of $\ell_{q}$, thus of $p^{\prime}$. Therefore, we have $p^{\prime}=zcf^{\prime}$, for a word $f^{\prime}$, and $p^{\prime}zd$ is not an inverse Lyndon word since $p^{\prime}zd\ll zd$. On the contrary, $zd$ is an inverse Lyndon word, because it is a prefix of $\ell_{q+1}$ (Eq. 7.5). Finally, $p^{\prime}=zcf^{\prime}\ll zd$. By Definition 4.2, the pair $(p^{\prime},zd)\in\operatorname{Pref}_{bre}(w)$. Since $x_{1}\not=1$, we have $p^{\prime}\not=p=p^{\prime}x_{1}$, in contradiction with Proposition 4.1.

Proposition 7.13

Let $(\Sigma,<)$ be a totally ordered alphabet. For any $w\in\Sigma^{+}$, $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$.

Proof :

Let $w\in\Sigma^{+}$, let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$, and let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$. The proof is by induction on $|w|$. If $|w|=1$, then $w$ is an inverse Lyndon word and we have done, by Proposition 7.9. Hence assume $|w|>1$.

If $w$ is an inverse Lyndon word, once again by Proposition 7.9, $\operatorname{ICFL}(w)=(w)=(\ell_{1}\cdots\ell_{h})$ is a grouping of $\operatorname{CFL}_{in}(w)$. Therefore, assume that $w$ is not an inverse Lyndon word. By Propositions 7.10-7.12, there is $j$, $1\leq j\leq h$ such that $m_{1}=\ell_{1}\cdots\ell_{j}$, where $\ell_{1}\geq_{p}\ldots\geq_{p}\ell_{j}$.

Let $(p,\bar{p})\in\operatorname{Pref}(w)_{bre}$. Let $\operatorname{ICFL}(v)=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$, where $v\in\Sigma^{+}$ is such that $w=pv$. By induction hypothesis, $(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$ is a grouping of $\operatorname{CFL}_{in}(v)$. By Definition 4.3, one of the following two cases holds

• (1)

$m_{1}=p$, $(m_{2},\ldots,m_{k})=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$, i.e., $k=k^{\prime}+1$, $m_{j}=m^{\prime}_{j-1}$, $2\leq j\leq k$.

• (2)

$m_{1}=pm^{\prime}_{1}$, $(m_{2},\ldots,m_{k})=(m^{\prime}_{2},\ldots,m^{\prime}_{k^{\prime}})$, i.e., $k=k^{\prime}$, $m_{j}=m^{\prime}_{j}$, $2\leq j\leq k$.

(Case (1)). In this case, since $p=m_{1}=\ell_{1}\cdots\ell_{j}$, we have $v=\ell_{j+1}\cdots\ell_{h}$, where any $\ell_{g}$ is in $L_{in}$ and $\ell_{j+1}\succeq_{in}\ldots\succeq_{in}\ell_{h}$. By Theorem 2.1, $\operatorname{CFL}_{in}(v)=(\ell_{j+1},\ldots,\ell_{h})$. Consequently, if $(\ell_{j+1},\ldots,\ell_{h})=(\mathcal{C}_{1},\ldots,\mathcal{C}_{t})$, where any $\mathcal{C}_{j}$, $1\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(v)$, then $(\ell_{1},\ldots,\ell_{h})=(\mathcal{C},\mathcal{C}_{1},\ldots,\mathcal{C}_{t})$ where $\mathcal{C}=\ell_{1},\ldots,\ell_{j}$, any $\mathcal{C}_{j}$, $2\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$ and either $\mathcal{C},\mathcal{C}_{1}$ represents two $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$ or it represents a single $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. In both cases, since $m_{1}=\ell_{1}\cdots\ell_{j}$ and given that $(m_{2},\ldots,m_{k})=(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$ is a grouping of $\operatorname{CFL}_{in}(v)$, we conclude that $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$.

(Case (2)). Set $\operatorname{CFL}_{in}(v)=(\ell^{\prime}_{1},\ldots,\ell^{\prime}_{h^{\prime}})$, and let $j^{\prime}$ be such that $m^{\prime}_{1}=\ell^{\prime}_{1}\cdots\ell^{\prime}_{j^{\prime}}$, $\ell^{\prime}_{1}\geq_{p}\ldots\geq_{p}\ell^{\prime}_{j^{\prime}}$. Therefore we have

[TABLE]

Since $(pm^{\prime}_{1})=(\ell_{1}\cdots\ell_{j})=(p\ell^{\prime}_{1}\cdots\ell^{\prime}_{j^{\prime}})$, Eq. (7.7) implies

[TABLE]

All the words $\ell_{g},\ell^{\prime}_{g^{\prime}}$ are in $L_{in}$ and $\ell_{j+1}\succeq_{in}\ldots\succeq_{in}\ell_{h}$, $\ell^{\prime}_{j^{\prime}+1}\succeq_{in}\ldots\succeq_{in}\ell^{\prime}_{h^{\prime}}$. Hence, by Theorem 2.1, applied to $v^{\prime}=m_{2}\cdots m_{k}=m^{\prime}_{2}\cdots m^{\prime}_{k^{\prime}}$, Eq. (7.8) implies $(\ell_{j+1},\ldots,\ell_{h})=(\ell^{\prime}_{j^{\prime}+1},\ldots,\ell^{\prime}_{h^{\prime}})$ and $\operatorname{CFL}_{in}(v^{\prime})=(\ell_{j+1},\ldots,\ell_{h})$. Now, $(m^{\prime}_{1},\ldots,m^{\prime}_{k^{\prime}})$ is a grouping of $\operatorname{CFL}_{in}(v)$, where $v=m^{\prime}_{1}v^{\prime}$ and $m^{\prime}_{1}=\ell^{\prime}_{1}\cdots\ell^{\prime}_{j^{\prime}}$. Thus, $(m_{2},\ldots,m_{h})=(m^{\prime}_{2},\ldots,m^{\prime}_{k^{\prime}})$ is a grouping of $(\ell_{j+1},\ldots,\ell_{h})=(\ell^{\prime}_{j^{\prime}+1},\ldots,\ell^{\prime}_{h^{\prime}})=\operatorname{CFL}_{in}(v^{\prime})$. The rest of the proof runs as in Case (1). Namely, if $(\ell_{j+1},\ldots,\ell_{h})=(\mathcal{C}_{1},\ldots,\mathcal{C}_{t})$, where any $\mathcal{C}_{j}$, $1\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(v^{\prime})$, then $(\ell_{1},\ldots,\ell_{h})=(\mathcal{C},\mathcal{C}_{1},\ldots,\mathcal{C}_{t})$ where $\mathcal{C}=\ell_{1},\ldots,\ell_{j}$, any $\mathcal{C}_{j}$, $2\leq j\leq t$, is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$ and either $\mathcal{C},\mathcal{C}_{1}$ represents two $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$ or it represents a single $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. In both cases, since $m_{1}=\ell_{1}\cdots\ell_{j}$ and given that $(m_{2},\ldots,m_{k})=(m^{\prime}_{2},\ldots,m^{\prime}_{k^{\prime}})$ is a grouping of $\operatorname{CFL}_{in}(v^{\prime})$, we conclude that $\operatorname{ICFL}(w)$ is a grouping of $\operatorname{CFL}_{in}(w)$.

Corollary 7.1

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$, with $h>1$. If $\ell_{1}\ll\ldots\ll\ell_{h}$, then $\operatorname{ICFL}(w)=\operatorname{CFL}_{in}(w)$ and this is the unique grouping of $\operatorname{CFL}_{in}(w)$.

Proof :

Let $w\in\Sigma^{+}$ and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$, with $h>1$. If $\ell_{1}\ll\ldots\ll\ell_{h}$, then $\operatorname{CFL}_{in}(w)$ is the unique grouping of $\operatorname{CFL}_{in}(w)$ (Definition 7.3). Thus, by Proposition 7.13, $\operatorname{ICFL}(w)=\operatorname{CFL}_{in}(w)$.

The following example shows that $\operatorname{ICFL}(w)$ is in general different from $\operatorname{CFL}_{in}(w)$.

Example 7.3

Let $\Sigma=\{a,b\}$ with $a<b$, let $w=bab\in\Sigma^{+}$. Therefore, $b<_{in}a$ and $\operatorname{CFL}_{in}(w)=(ba,b)$. Since $bab$ is an inverse Lyndon word, we have $\operatorname{ICFL}(w)=(bab)$.**

The following example shows that there are words $w$ such that $\operatorname{CFL}_{in}(w)$ has more than one grouping, thus there are groupings of $\operatorname{CFL}_{in}(w)$ different from $\operatorname{ICFL}(w)$.

Example 7.4

Let $\Sigma=\{a,b,c,d\}$ with $a<b<c<d$, let $w=dabadabdabdabdadac\in\Sigma^{+}$. Therefore, $d<_{in}c<_{in}b<_{in}a$. Moreover, $\operatorname{CFL}_{in}(w)=(daba,dab,dab,dab,dadac)$. The two sequences $(dabadab,(dab)^{2},dadac)$, $(daba,(dab)^{3},dadac)$ are both groupings of $\operatorname{CFL}_{in}(w)$. Let us compute $\operatorname{ICFL}(w)$. The shortest prefix of $w$ which is not an inverse Lyndon word is $dabadabd$ and $(daba,dabd)\in\operatorname{Pref}_{bre}(w)$ by Lemma 4.7. Set $w=dabay$, we have to compute $\operatorname{ICFL}(y)$. The shortest prefix of $y=(dab)^{3}dadac$ which is not an inverse Lyndon word is $(dab)^{3}dad$ and $((dab)^{3},dad)\in\operatorname{Pref}_{bre}(y)$ by Lemma 4.7. On the other hand, since $dadac$ is an inverse Lyndon word, we have $\operatorname{ICFL}(dadac)=(dadac)$ and thus $\operatorname{ICFL}(y)=((dab)^{3},dadac)$ (see Definition 4.3). Finally, by Definition 4.3, $\operatorname{ICFL}(w)=(daba,(dab)^{3},dadac)$. **

We end the section with Proposition 7.14. It shows that nonempty words which are both proper prefix and suffix of an inverse Lyndon word have a special form.

Proposition 7.14

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ be an inverse Lyndon word and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. For any proper nonempty suffix $u$ of $w$ such that $u$ is also a prefix of $w$, there is $j_{u}$ such that $u=\ell_{j_{u}}\cdots\ell_{h}$, $2\leq j_{u}\leq h$.

Proof :

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ be a bordered inverse Lyndon word and let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Let $u\in\Sigma^{+}$ be such that $u$ is a proper nonempty suffix of $w$ and $u$ is also a prefix of $w$. By contradiction, assume that $u=s\ell_{j_{u}}\cdots\ell_{h}$, $2\leq j_{u}\leq h$, where $s$ is a nonempty proper suffix of $\ell_{j_{u}-1}$. Since $w$ is a bordered inverse Lyndon word, by Proposition 7.9, $\ell_{1},\ldots,\ell_{h}$ is a $\mathcal{PMC}$ in $\operatorname{CFL}_{in}(w)$. Hence $\ell_{j_{u}-1}$ is a prefix of $w$. It follows that $\ell_{j_{u}-1}$ and $s$ are comparable for the prefix order and consequently $\ell_{j_{u}-1}$ is bordered, in contradiction with $\ell_{j_{u}-1}\in L_{in}$.

7.4 Strict sesquipowers of anti-Lyndon words

For a word $w\in\Sigma^{*}$ and a letter $a\in\Sigma$, we denote by $|w|_{a}$ the number of occurrences of $a$ in $w$. Then, the set

[TABLE]

is the set of all letters occurring at least once in $w$.

Let $\mathcal{P}_{aL}=\{w\in\Sigma^{+}~{}|~{}w\Sigma^{*}\cap L_{in}\not=\emptyset\}$ be the set of the nonempty prefixes of anti-Lyndon words, let $a$ be the minimum letter in $\Sigma$ and let $P^{\prime}=\mathcal{P}_{aL}\cup\{a^{m}~{}|~{}m\geq 2\}$. Let $\mathcal{SP}$ be the set of strict sesquipowers of the words in $L_{in}$, that is

[TABLE]

The following result has been proved in [11].

Proposition 7.15

[TABLE]

The aim of this section is to prove that $\mathcal{SP}$ is equal to the class of the inverse Lyndon words.

Proposition 7.16

For any $a\in\Sigma$ and $m\geq 1$, the word $a^{m}$ is an inverse Lyndon word.

Proof :

By definition, any letter $a\in\Sigma$ is in $L_{in}$. Therefore, by Proposition 7.5, the word $a^{m}$ is an inverse Lyndon word, for $m\geq 1$.

Proposition 7.17

Let $(\Sigma,<)$ be a totally ordered alphabet. Let $w\in\Sigma^{+}$ be an inverse Lyndon word. If $w$ starts with the minimum letter in $\Sigma$, say $a$, then $w=a^{m}$, with $m\geq 1$.

Proof :

Let $(\Sigma,<)$ be a totally ordered alphabet and let $a$ be the minimum letter in $\Sigma$. Let $w,w^{\prime}$ be words such that $w=aw^{\prime}$ is an inverse Lyndon word. By contradiction, let $w_{1},w_{2}$ be words such that $w=aw_{1}bw_{2}$, with $b\not=a$. Hence $b>a$ and $w\ll bw_{2}$, in contradiction with the hypothesis.

As another step in our proof, we show that any inverse Lyndon word can be extended as a word which is still an inverse Lyndon word.

Proposition 7.18

Let $(\Sigma,<)$ be a totally ordered alphabet and let $a$ be the minimum letter in $\Sigma$. For any inverse Lyndon word $w$ and for any nonnegative integer $n$, the word $wa^{n}$ is an inverse Lyndon word.

Proof :

Let $(\Sigma,<)$ be a totally ordered alphabet and let $a$ be the minimum letter in $\Sigma$. Let $w$ be an inverse Lyndon word.

If $w$ starts with $a$, then $w=a^{m}$, with $m\geq 1$, by Proposition 7.17. Thus, by Proposition 7.16, the word $wa^{n}=a^{m+n}$ is still an inverse Lyndon word.

Now, let $b\in\Sigma$, $w^{\prime}\in\Sigma^{*}$ be such that $w=bw^{\prime}$, with $b>a$. We first prove that $wa$ is an inverse Lyndon word, that is, $wa\succ s$, for any proper nonempty suffix of $wa$. Any proper nonempty suffix $s$ of $wa$ has the form $s=s^{\prime}a$, where $s^{\prime}$ is a proper suffix of $w$. If $s^{\prime}=1$, then $a=s\ll wa=bw^{\prime}a$ and we have done. If $s^{\prime}\not=1$, then $w\succ s^{\prime}$ because $w$ is an inverse Lyndon word. In addition, if $s^{\prime}\ll w$, then $s=s^{\prime}a\ll wa$, by item (2) in Lemma 2.3. Otherwise, $s^{\prime}$ is both a prefix and a proper nonempty suffix of $w$, hence $w=s^{\prime}y$ for a nonempty word $y$. By contradiction, assume $s^{\prime}\prec wa=s^{\prime}ya\prec s^{\prime}a$. By item (3) in Lemma 2.3, $ya\prec a$ which implies $ya\ll a$, a contradiction since $a$ is the minimum letter in $\Sigma$. The proof is completed by using induction. Indeed, if $wa^{n}$ is an inverse Lyndon word which starts with $b$, the above argument shows that $wa^{n+1}$ is still an inverse Lyndon word which starts with $b$.

We prove below that any inverse Lyndon word can be extended as a word which is still an inverse Lyndon word and which is additionally unbordered. Proposition 7.19 shows that this can be easily obtained if we do not limit the size of the alphabet whereas Proposition 7.20 states the result for a fixed alphabet. Let $(\Sigma_{1},<_{1}),(\Sigma_{2},<_{2})$ be totally ordered alphabets such that $\Sigma_{1}\subseteq\Sigma_{2}$. We say that $<_{2}$ is an extension of $<_{1}$ if the following holds

[TABLE]

If $<_{2}$ is an extension of $<_{1}$, then inverse Lyndon words in $\Sigma_{1}^{+}$ are also inverse Lyndon words in $\Sigma_{2}^{+}$.

Proposition 7.19

Let $(\Sigma_{1},<_{1})$ be a totally ordered alphabet. For any inverse Lyndon word $w\in\Sigma_{1}^{+}$, there is a totally ordered alphabet $(\Sigma_{2},<_{2})$, where $\Sigma_{1}\subseteq\Sigma_{2}$ and $<_{2}$ is an extension of $<_{1}$, such that $w\in\mathcal{SP}$.

Proof :

Let $(\Sigma_{1},<_{1})$ be a totally ordered alphabet and let $w\in\Sigma_{1}^{+}$ be an inverse Lyndon word. Let $\alpha$ be a letter such that $\alpha\not\in\Sigma_{1}$ and set $\Sigma_{2}=\Sigma_{1}\cup\{\alpha\}$. Let $(\Sigma_{2},<_{2})$ be the totally ordered alphabet, where $<_{2}$ is the extension of $<_{1}$ defined as follows

[TABLE]

By Proposition 7.18, $w\alpha$ is an inverse Lundon word. Moreover, $w\alpha$ is unbordered, thus $w\alpha\in L_{in}$ (Proposition 7.3) and consequently $w\in\mathcal{P}_{aL}\subseteq\mathcal{SP}$.

Proposition 7.20

Let $(\Sigma,<)$ be a totally ordered alphabet and let $w\in\Sigma^{+}$. If $w$ is an inverse Lyndon word, then $w\in\mathcal{SP}$.

Proof :

Let $(\Sigma,<)$ be a totally ordered alphabet and let $w\in\Sigma^{+}$ be an inverse Lyndon word. Let $a$ be the minimum letter in $\Sigma$. If $w$ starts with $a$, then $w=a^{m}$, with $m\geq 1$ (Proposition 7.17), hence $w\in\mathcal{SP}$. Otherwise, let $b\in\Sigma$, $w^{\prime}\in\Sigma^{*}$ be such that $w=bw^{\prime}$, with $b>a$. If $w=bw^{\prime}$ is unbordered, then $w$ is an anti-Lyndon word (Proposition 7.3) and consequently $w\in\mathcal{SP}$. Otherwise, $b\in\operatorname{alph}(w^{\prime})$. Let $x$ be the word that precedes the last occurrence of $b$ in $w^{\prime}$, that is, let $x,y\in\Sigma^{*}$ be such that $w^{\prime}=xby$ and $b\not\in\operatorname{alph}(y)$. Let $n>|x|+1$ and let $z=wa^{n}=bxbya^{n}$. By Proposition 7.18, the word $z$ is an inverse Lyndon word. We claim that $z$ is unbordered, thus $z=wa^{n}\in L_{in}$ (Proposition 7.3) and consequently $w\in\mathcal{P}_{aL}\subseteq\mathcal{SP}$. On the contrary, let $u$ be a proper prefix of $z=wa^{n}=bxbya^{n}$ which is also a suffix of $z$. On one hand, $|u|_{b}<|z|_{b}$, because $u$ is a proper suffix of $z$. On the other hand, $u$ starts with $b$ since it is a prefix of $z$. Thus $|u|\geq|bya^{n}|>|x|+2=|bxb|$, because $b\not\in\operatorname{alph}(ya^{n})$. Finally, since $u$ is a prefix of $z$ such that $|u|>|bxb|$, it follows that $|u|_{b}=|z|_{b}$. This contradiction ends the proof.

Proposition 7.21

A word $w$ is an inverse Lyndon word if and only if $w\in\mathcal{SP}$.

Proof :

By Proposition 7.20, if $w$ is an inverse Lyndon word, then $w\in\mathcal{SP}$. Conversely, let $w\in\mathcal{SP}$. If $w=a^{m}$, $m\geq 1$, where $a$ is (the minimum letter) in $\Sigma$, then $w$ is an inverse Lyndon word by Proposition 7.16. Otherwise, $w\in\mathcal{P}_{aL}$ is a prefix of an anti-Lyndon word, hence $w$ is an inverse Lyndon word by Lemma 3.2.

7.5 Sorting suffixes in $\operatorname{ICFL}(w)$

In this section we use the same notation and terminology as in Section 2.5. We prove that the same compatibility property proved in [28] holds for the sorting of the nonempty suffixes of a word $w$ with respect to $\prec_{in}$ if we replace $\operatorname{CFL}(w)$ with $\operatorname{ICFL}(w)$.

Theorem 7.1

Let $w$ be a word and let $(m_{1},\ldots,m_{k})$ be a grouping of $\operatorname{CFL}_{in}(w)$. Then, for any $r,s$, $1\leq r\leq s\leq k$, the sorting with respect to $\prec_{in}$ of the nonempty local suffixes of $w$ with respect to $u=m_{r}\cdots m_{s}$ is compatible with the sorting with respect to $\prec_{in}$ of the corresponding nonempty global suffixes of $w$.

Proof :

Let $w$ and $(m_{1},\ldots,m_{k})$ be as in the statement. Let $\operatorname{CFL}_{in}(w)=(\ell_{1},\ldots,\ell_{h})$. Let $u=m_{r}\cdots m_{s}$, with $1\leq r\leq s\leq k$. By Definitions 7.2, 7.3, any $m_{j}$ is a concatenation of consecutive $\ell_{q}$. Hence $u$ is also a concatenation of consecutive $\ell_{q}$. By Theorem 2.2, for all $i,j$ with $first(u)\leq i<j\leq last(u)$, we have

[TABLE]

The following corollary is a direct consequence of Proposition 7.13 and Theorem 7.1.

Corollary 7.2

Let $w$ be a word and let $\operatorname{ICFL}(w)=(m_{1},\ldots,m_{k})$. Then, for any $r,s$, $1\leq r\leq s\leq k$, the sorting with respect to $\prec_{in}$ of the nonempty local suffixes of $w$ with respect to $u=m_{r}\cdots m_{s}$ is compatible with the sorting with respect to $\prec_{in}$ of the corresponding nonempty global suffixes of $w$.

On the contrary, we give below a counterexample showing that the compatibility property of local and global nonempty suffixes does not hold in general for inverse Lyndon factorizations with respect to $\prec_{in}$ (and with respect to $\prec$).

Example 7.5

Let $(\Sigma,<)$ be as in Example 4.1 and let $w=daddbadc\in\Sigma^{+}$. Therefore, $d<_{in}c<_{in}b<_{in}a$. Consider the inverse Lyndon factorization $(dad,dba,dc)$ of $w$, with $dad\ll dba\ll dc$ and the factor $u=daddba$. Consider the local suffixes $a,addba$ of $u$ and the corresponding global suffixes $adc$ and $addbadc$. We have that $addbadc\prec_{in}adc$ while $a\prec_{in}addba$. Consequently, in general, the compatibility property does not hold for inverse Lyndon factorizations with respect to $\prec_{in}$. It does not hold also with respect to $\prec$ and even for $\operatorname{ICFL}$. Indeed, let $w=dabadabdabdabdadac\in\Sigma^{+}$. We know that $\operatorname{ICFL}(w)=(daba,(dab)^{3},dadac)$ (see Example 7.4). For the local suffixes $dab,dabdab$ of $(dab)^{3}$ we have $dab\prec dabdab$ but for the corresponding global suffixes $dabdadac,dabdabdadac$ we have $dabdabdadac\prec dabdadac$.**