On Collapsing Prefix Normal Words

TL;DR

This paper explores properties of prefix normal words, focusing on prefix normal palindromes and collapsing words, providing characterizations and connections to help address open problems in the field.

Contribution

It introduces characterizations of prefix normal palindromes and collapsing words, extending the understanding of prefix normal words and proposing ways to decompose related open problems.

Findings

Characterizations of prefix normal palindromes

Characterizations of collapsing words

Connection between palindromes and collapsing words

Abstract

Prefix normal words are binary words in which each prefix has at least the same number of $\so$s as any factor of the same length. Firstly introduced by Fici and Lipt\'ak in 2011, the problem of determining the index of the prefix equivalence relation is still open. In this paper, we investigate two aspects of the problem, namely prefix normal palindromes and so-called collapsing words (extending the notion of critical words). We prove characterizations for both the palindromes and the collapsing words and show their connection. Based on this, we show that still open problems regarding prefix normal words can be split into certain subproblems.

Tables

Table 1. Table 1: Prefix normal palindromes (pnPals).

word length	prefix normal palindromes	# prefix normal words
$1$	$\mathrm{𝟢}$, $\mathrm{𝟣}$	$2$
$2$	${\mathrm{𝟢}}^2$, ${\mathrm{𝟣}}^2$	$3$
$3$	${\mathrm{𝟢}}^3$, $\mathrm{𝟣𝟢𝟣},{\mathrm{𝟣}}^3$	$5$
$4$	${\mathrm{𝟢}}^4$, $\mathrm{𝟣𝟢𝟢𝟣},{\mathrm{𝟣}}^4$	$8$
$5$	${\mathrm{𝟢}}^5$, $\mathrm{𝟣𝟢𝟢𝟢𝟣},\mathrm{𝟣𝟢𝟣𝟢𝟣},\mathrm{𝟣𝟣𝟢𝟣𝟣},{\mathrm{𝟣}}^5$	$14$
$6$	${\mathrm{𝟢}}^6$, $\mathrm{𝟣𝟢𝟢𝟢𝟢𝟣},\mathrm{𝟣𝟣𝟢𝟢𝟣𝟣},{\mathrm{𝟣}}^6$	$23$
$7$	${\mathrm{𝟢}}^7$, ${\mathrm{𝟣𝟢}}^2{\mathrm{𝟣𝟢}}^2\mathrm{𝟣},{\mathrm{𝟣𝟢}}^5\mathrm{𝟣},\mathrm{𝟣𝟢𝟣𝟢𝟣𝟢𝟣},{\mathrm{𝟣}}^2{\mathrm{𝟢𝟣𝟢𝟣}}^2,{\mathrm{𝟣}}^2{\mathrm{𝟢}}^3{\mathrm{𝟣}}^2,{\mathrm{𝟣}}^3{\mathrm{𝟢𝟣}}^3,{\mathrm{𝟣}}^7$	$41$
$8$	${\mathrm{𝟢}}^8$, ${\mathrm{𝟣𝟢}}^6\mathrm{𝟣},{\mathrm{𝟣𝟢𝟣𝟢}}^2\mathrm{𝟣𝟢𝟣},{\mathrm{𝟣}}^2{\mathrm{𝟢}}^4{\mathrm{𝟣}}^2,{\mathrm{𝟣}}^2{\mathrm{𝟢𝟣}}^2{\mathrm{𝟢𝟣}}^2,{\mathrm{𝟣}}^3{\mathrm{𝟢}}^2{\mathrm{𝟣}}^3,{\mathrm{𝟣}}^8$	$70$

Table 2. Table 2: Number of pnPals. [ 15 ] (A308465)

$i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	$13$	$14$	$15$
#	$2$	$2$	$3$	$3$	$5$	$4$	$8$	$7$	$12$	$11$	$21$	$18$	$36$	$31$	$57$

Equations

$$p_w(n)=p_w(n-i)+s_w(i)\text{mbox}and{s}_w(n)=p_w(i)+s_w(n-i).$$

$$f_v(i)=f_w(i)=p_w(i-1)=p_v(i-1)=p_v(i)-1\le f_v(i)-1<f_v(i).$$

$$s_{w^R}=p_w=f_w=f_{w^R}$$

$$p_{w^R}(i)=p_w^R(i)-p_w^R(i+1)-p_{w^R}(i-1).$$

$$s_w(i)\ge \{\begin{array}{ll}{s}_w(n)-s_w(n-i+1)& \text{mbox}if{s}_w(n-i+1)=s_w(n-i),\\ s_w(n)-s_w(n-i+1)+1& \text{mbox}otherwise.\end{array}$$

$$s_w(i)=|{\mathrm{Suff}}_i(w){|}_1\ge |{\mathrm{Pref}}_i(w){|}_1=r=s+r-s=s_w(n)-s_w(n-i+1).$$

$$s_w(i)$$

$$=s_w(n)-s_w(n-i+1)+1.$$

$$f_w(k)=\{\begin{array}{ll}1& \text{if k=1,}\\ f_v(k-1)+1& \text{if 1<k≤∣w∣−1,}\\ f_w(|v|+1)+1& \text{if k=∣w∣.}\end{array}$$

$$f_v(k-1)+1=|{\mathrm{Pref}}_{k-1}(v){|}_1+1=|{\mathrm{Pref}}_k(w){|}_1=p_w(k).$$

$$f_w(|v|+1)+1$$

$$=f_w(|w|).$$

$${\overline{p}}_w^R(n-i)$$

$$={\overline{p}}_w(i)-\left(p_w(i)-p_w(i-1)\right)$$

$$={\overline{p}}_w^R(n-i+1)+(-s_w(i)+s_w(i-1))$$

$$=s_w(n-i+1)-w[i]$$

$$=s_w(n-i)+w[n-i+1]-w[i]$$

$$=s_w(n-i).$$

$${\overline{f}}_v(n-i+1)$$

$$={\overline{f}}_v^R(i-1)+x$$

$$={\overline{f}}_v(n-i+2)+x$$

$$={\overline{f}}_v(n-i+1)-\left(f_v(n-i+1)-f_v(n-i)\right)+x.$$

$$f_{1w}(j)=s_{1w}(j)=s_w(j)\ne f_w(j).$$

$$f_v(i)=s_v(i)=s_{1v}(i)\le f_{1v}(i)=f_{1w}(i)=s_{1w}(i)=s_w(i)=f_w(i).$$

$$\sigma (v)$$

$$<s_w(j)+\sum _{i\in [n]\backslash \{j\}}{s}_w(j)=\sum _{i\in [n]}{s}_w(j)=\sum _{i\in [n]}{f}_w(j)$$

$$=\sigma (w).\text{squareforqed}$$

$$s_{1v}(n-i)=s_v(n-i)=s_w(n-i)+1=f_w(n-i)+1=f_{1w}(n-i)+1=f_{1v}(n-i)+1.$$

$$f_w(n-i+1)=s_w(n-i+1)=|{\mathrm{Suff}}_{n-i+1}(w){|}_1=|w{|}_1-|{\mathrm{Pref}}_{i-1}(w){|}_1.$$

$$|{\mathrm{Pref}}_{n-i+1}(1w){|}_1=|{\mathrm{Pref}}_{n-i}(w){|}_1+1=|w{|}_1-|{\mathrm{Suff}}_i(w){|}_1+1=|w{|}_1-s_w(i)+1.$$

$$s-f_{w^{\prime }}(n-i+1)$$

$$<|{\mathrm{Suff}}_i(w^{\prime }){|}_1\le f_{w^{\prime }}(i).$$

$$f_{1w^{\prime }}(i)=f_{w^{\prime }}(i)+1=f_w(i)-1+1=f_w(i)=f_{1w}(i).$$

$$f_{1w}$$

$$=f_{1u},$$

$$1v>1w[n-\mathrm{1..1}]=w^{R}1=(1w{)}^R$$

$$(f_{w^{\prime }}(i)=f_w(i)$$

Full text

1 Introduction

Two words are called abelian equivalent if the amount of each letter is identical in both words, e.g. rotor and torro are abelian equivalent albeit banana and ananas are not. Abelian equivalence has been studied with various generalisations and specifications such as abelian-complexity, $k$-abelian equivalence, avoidability of ($k$-)abelian powers and much more (cf. e.g., [6, 10, 13, 11, 17, 22, 24, 23] ). The number of occurrences of each letter is captured in the Parikh vector (also known as Parikh image or Parikh mapping) ([21]): given a lexicographical order on the alphabet, the i$${}^{\text{th}} component of this vector is the amount of the i$${}^{\text{th}} letter of the alphabet in a given word. Parikh vectors have been studied in [12, 16, 19] and are generalised to Parikh matrices for saving more information about the word than just the amount of letters (cf. eg., [20, 25]).

A recent generalisation of abelian equivalence, for words over the binary alphabet $\{\mathsf{0},\mathsf{1}\}$, is prefix normal equivalence (pn-equivalence) [14]. Two binary words are pn-equivalent if their maximal numbers of $\mathsf{1}$s in any factor of length $n$ are equal for all $n\in\mathbb{N}$. Burcsi et al. [5] showed that this relation is indeed an equivalence relation and moreover that each class contains exactly one uniquely determined representative - called a prefix normal word. A word $w$ is said to be prefix normal if the prefix of $w$ of any length has at least the number of $\mathsf{1}$s as any of $w$’s factors of the same length. For instance, the word $\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{0}\mathsf{1}$ is prefix normal but $\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}$ is not, witnessed by the fact that $\mathsf{1}\mathsf{1}$ is a factor but not a prefix. Both words are pn-equivalent. In addition to being representatives of the pne-classes, prefix normal words are also of interest since they are connected to Lyndon words, in the sense that every prefix normal word is a pre-necklace [14]. Furthermore, as shown in [14], the indexed jumbled pattern matching problem (see e.g. [2, 4, 18]) is connected to prefix normal forms: if the prefix normal forms are given, the indexed jumbled pattern matching problem can be solved in linear time $\mathcal{O}(n)$ of the word length $n$. The best known algorithm for this problem has a run-time of $\mathcal{O}(n^{1.864})$(see [7]). Consequently there is also an interest in prefix normal forms from an algorithmic point of view. An algorithm for the computation of all prefix normal words of length $n$ in run-time $\mathcal{O}(n)$ per word is given in [8]. Balister and Gerke [1] showed that the number of prefix normal words of length $n$ is $2^{n-\Theta(\log^{2}(n))}$ and the class of a given prefix normal word contains at most $2^{n-O(\sqrt{n\log(n)})}$ elements. A closed formula for the number of prefix normal words is still unknown. In “OEIS” [15] the number of prefix normal words of length $n$ (A194850), a list of binary prefix normal words (A238109), and the maximum size of a class of binary words of length $n$ having the same prefix normal form (A238110), can be found. An extension to infinite words is presented in [9].

Our contribution. In this work we investigate two conspicuities mentioned in [14, 3]: palindromes and extension-critical words. Generalising the result of [3] we prove that prefix normal palindromes (pnPal) play a special role since they are not pn-equivalent to any other word. Since not all palindromes are prefix normal, as witnessed by $\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}$, determining the number of pnPals is an (unsolved) sub-problem. We show that solving this sub-problem brings us closer to determining the index, i.e. number of equivalence classes w.r.t. a given word length, of the pn-equivalence relation. Moreover we give a characterisation based on the maximum-ones function for pnPals. The notion of extension-critical words is based on an iterative approach: compute the prefix normal words of length $n+1$ based on the prefix normal words of length $n$. A prefix normal word $w$ is called extension-critical if $w\mathsf{1}$ is not prefix normal. For instance, the word $\mathsf{1}\mathsf{0}\mathsf{1}$ is prefix normal but $\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}$ is not and thus $\mathsf{1}\mathsf{0}\mathsf{1}$ is called extension-critical. This means that all non-extension-critical words contribute to the class of prefix normal words of the next word-length. We investigate the set of extension-critical words by introducing an equivalence relation collapse, grouping all extensional-critical words that are pn-equivalent w.r.t. length $n+1$. Finally we prove that (prefix normal) palindromes and the collapsing relation (extensional-critical words) are related. In contrast to [14] we work with suffix-normal words (least representatives) instead of prefix-normal words. It follows from Lemma 1 that both notions lead to the same results.

Structure of the paper. In Section 2, the basic definitions and notions are presented. In Section 3, we present the results on pnPals. Finally, in Section 4, the iterative approach based on collapsing words is shown. This includes a lower bound and an upper bound for the number of prefix normal words, based on pnPals and the collapsing relation. Due to space restrictions all proofs are in the appendix.

2 Preliminaries

Let $\mathbb{N}$ denote the set of natural numbers starting with $1$, and let $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$. Define $[n]=\{1,\dots,n\}$, for $n\in\mathbb{N}$, and set $[n]_{0}=[n]\cup\{0\}$.

An alphabet is a finite set $\Sigma$, the set of all finite words over $\Sigma$ is denoted by $\Sigma^{\ast}$, and the empty word by $\varepsilon$. Let $\Sigma^{+}=\Sigma^{\ast}\backslash\{\varepsilon\}$ be the free semigroup for the free monoid $\Sigma^{\ast}$. Let $w[i]$ denote the i$${}^{\text{th}} letter of $w\in\Sigma^{\ast}$ that is $w=\varepsilon$ or $w=w[1]\dots w[n]$. The length of a word $w=w[1]\dots w[n]$ is denoted by $|w|$ and let $|\varepsilon|=0$. Set $w[i..j]=w[i]\dots w[j]$ for $1\leq i\leq j\leq|w|$. Set $\Sigma^{n}=\{w\in\Sigma^{\ast}|\,|w|=n\}$ for all $n\in\mathbb{N}_{0}$. The number of occurrences of a letter $\mathsf{x}\in\Sigma$ in $w\in\Sigma^{\ast}$ is denoted by $|w|_{\mathsf{x}}$. For a given word $w\in\Sigma^{n}$ the reversal of $w$ is defined by $w^{R}=w[n]\dots w[1]$. A word $u\in\Sigma^{\ast}$ is a factor of $w\in\Sigma^{\ast}$ if $w=xuy$ holds for some words $x,y\in\Sigma^{\ast}$. If $x=\varepsilon$ then $u$ is called a prefix of $w$ and a suffix if $y=\varepsilon$. Let $\operatorname{Fact}(w),\operatorname{Pref}(w),\operatorname{Suff}(w)$ denote the sets of all factors, prefixes, and suffixes respectively. Define $\operatorname{Fact}_{k}(w)=\operatorname{Fact}(w)\cap\Sigma^{k}$ and $\operatorname{Pref}_{k}(w)$, $\operatorname{Suff}_{k}(w)$ are defined accordingly. Notice that $|\operatorname{Pref}_{k}(w)|=|\operatorname{Suff}_{k}(w)|=1$ for all $k\leq|w|$. The powers of $w\in\Sigma^{\ast}$ are recursively defined by $w^{0}=\varepsilon$, $w^{n}=ww^{n-1}$ for $n\in\mathbb{N}$.

Following [14], we only consider binary alphabets, namely $\Sigma=\{\mathsf{0},\mathsf{1}\}$ with the fixed lexicographic order induced by $\mathsf{0}<\mathsf{1}$ on $\Sigma$. In analogy to binary numbers we call a word $w\in\Sigma^{n}$ odd if $w[n]=\mathsf{1}$ and even otherwise.

For a function $f:[n]\rightarrow\Delta$ for $n\in\mathbb{N}_{0}$ and an arbitrary alphabet $\Delta$ the concatenation of the images defines a finite word $\operatorname{\mathsf{serialise}}(f)=f(1)f(2)\dots f(n)\in\Delta^{\ast}$. Since $\operatorname{\mathsf{serialise}}$ is bijective, we will identify $\operatorname{\mathsf{serialise}}(f)$ with $f$ and use in both cases $f$ (as long as it is clear from the context). This definition allows us to access $f$’s reversed function $g:[n]\rightarrow\Delta;k\mapsto f(n-k+1)$ easily by $f^{R}$.

Definition 1

The maximum-ones functions is defined for a word $w\in\Sigma^{\ast}$ by $f_{w}:[|w|]_{0}\rightarrow[|w|]_{0};\,k\mapsto\max\left\{\,|{v}|_{\mathsf{1}}\mid v\in\operatorname{Fact}_{k}(w)\right\},$ giving for each $k\in[|w|]_{0}$ the maximal number of $\mathsf{1}$s occuring in a factor of length $k$. Likewise the prefix-ones and suffix-ones functions are defined by $p_{w}:[|w|]_{0}\rightarrow[|w|]_{0};k\mapsto|\operatorname{Pref}_{k}(w)|_{\mathsf{1}}$ and $s_{w}:[|w|]_{0}\rightarrow[|w|]_{0};k\mapsto|\operatorname{Suff}_{k}(w)|_{\mathsf{1}}$.

Definition 2

Two words $u,v\in\Sigma^{n}$ are called prefix-normal equivalent (pn-equivalent, $u\equiv_{n}v$) if $f_{u}=f_{v}$ holds and $v$’s equivalence class is denoted by $[v]_{\equiv}=\{u\in\Sigma^{n}|\,u\equiv_{n}v\}$. A word $w\in\Sigma^{\ast}$ is called prefix (suffix) normal iff $f_{w}=p_{w}$ ($f_{w}=s_{w}$ resp.) holds. Let $\sigma(w)=\sum_{i\in[n]}f_{w}(i)$ denote the maximal-one sum of a $w\in\Sigma^{n}$.

Remark 1

Notice that $s_{w}=p_{w^{R}}$, $f_{w}=f_{w^{R}}$, $p_{w}(i),s_{w}(i)\leq f_{w}(i)$ for all $i\in\mathbb{N}_{0}$. By $p_{w^{R}}=s_{w}$ and $f_{w}=f_{w^{R}}$ follows immediately that a word $w\in\Sigma^{\ast}$ is prefix normal iff its reversal is suffix normal.

Fici and Lipták [14] showed that for each word $w\in\Sigma^{\ast}$ there exists exactly one $w^{\prime}\in[w]_{\equiv}$ that is prefix normal - the prefix normal form of $w$. We introduce the concept of least representative, which is the lexicographically smallest element of a class and thus also unique. As mentioned in [5] palindromes play a special role. Immediately by $w=w^{R}$ for $w\in\Sigma^{\ast}$, we have $p_{w}=s_{w}$, i.e. palindromes are the only words that can be prefix and suffix normal. Recall that not all palindromes are prefix normal witnessed by $\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}$.

Definition 3

A palindrome is called prefix normal palindrome (pnPal) if it is prefix normal. Let $\operatorname{NPal}(n)$ denote the set of all prefix normal palindromes of length $n\in\mathbb{N}$ and set $\operatorname{npal}(n)=|\operatorname{NPal}(n)|$. Let $\operatorname{Pal}(n)$ be the set of all palindromes of length $n\in\mathbb{N}$.

3 Properties of the Least-Representatives

Before we present specific properties of the least representatives (LR) for a given word length, we mention some useful properties of the maximum-ones, prefix-ones, and suffix-ones functions (for the basic properties we refer to [14, 5] and the references therein). Since we are investigating only words of a specific length, we fix $n\in\mathbb{N}_{0}$.

Beyond the relation $p_{w}=s_{w^{R}}$ the mappings $p_{w}$ and $s_{w}$ are determinable from each other. Counting the $\mathsf{1}$s in a suffix of length $i$ and adding the $\mathsf{1}$s in the corresponding prefix of length $(n-i)$ of a word $w$, gives the overall amount of $\mathsf{1}$s of $w$, namely

[TABLE]

For suffix (resp. prefix) normal words this leads to $p_{w}(i)=f_{w}(n)-f_{w}(n-i)$ resp. $s_{w}(i)=f_{w}(n)-f_{w}(n-i)$ witnessing the fact $p_{w}=s_{w}$ for palindromes (since both equation hold). Before we show that indeed pnPals form a singleton class w.r.t. $\equiv_{n}$, we need the relation between the lexicographical order and prefix and suffix normality.

Lemma 1

The prefix normal form of a class is the lexicographically largest element in the class and the suffix-normal of a class is a LR.

Proof

Let $w\in\Sigma^{n}$ be the prefix normal form of the class $[w]_{\equiv}$. Suppose there existed $v\in[w]_{\equiv}$ with $v>w$. Let $i\in[n]$ be the smallest index with $v[i]\neq w[i]$. Since we are only considering binary alphabets we get $v[i]=1$ and $w[i]=0$. By the prefix normality of $w$ we have $f_{w}(i)=p_{w}(i)=p_{w}(i-1)$ but on the other hand, $v\in[w]_{\equiv}$ and the minimality of $i$ implies

[TABLE]

This contradiction shows that the prefix normal form of a class is the lexicographically largest element. The reverse $w^{R}$ is thus the lexicographically smallest element of the class which is by definition the $LR$. By Remark 1 follows

[TABLE]

and hence the suffix normality of the $LR$. ∎

Lemma 1 implies that a word being prefix and suffix normal forms a singleton class w.r.t. $\equiv_{n}$. As mentioned $p_{w}=s_{w}$ only holds for palindromes.

Proposition 1

For a word $w\in\Sigma^{n}$ it holds that $|[w]|_{\equiv}=1$ iff $w\in\operatorname{NPal}(n)$.

Proof

Already in [5], the authors proved that $|[w]|_{\equiv}=1$ implies $w\in\operatorname{NPal}(n)$ for $w\in\Sigma^{n}$. The other direction follow immediately from Lemma 1: if $w$ is a prefix normal palindrome it is by definition prefix normal and by $w=w^{R}$, $w$ is the lexicographically largest and smallest element of the class. This implies that the class is a singleton.∎

The general part of this section is concluded by a somewhat artificial equation which is nevertheless useful for pnPals : by $s_{w}(i)=p_{w}^{R}(i)-p_{w}^{R}(i+1)+s_{w}(i-1)$ with $p_{w}^{R}(n+1)=0$ for $i\in[n]$ and $s_{w}=p_{w^{R}}$ we get

[TABLE]

The rest of the section will cover properties of the LRs of a class.

Remark 2

For completeness, we mention that $\mathsf{0}^{n}$ is the only even LR w.r.t. $\equiv_{n}$ and the only pnPal starting with $\mathsf{0}$. Moreover, $\mathsf{1}^{n}$ is the largest LR. As we show later in the paper $\mathsf{0}^{n}$ and $\mathsf{1}^{n}$ are of minor interest in the recursive process due to their speciality.

The following lemma is an extension of [5, Lemma 1] for the suffix-one function by relating the prefix and the suffix of the word $s_{w}$ for a least representative. Intuitively the suffix normality implies that the $\mathsf{1}$s are more at the end of the word $w$ rather than at the beginning: consider for instance $s_{w}=1123345$ for $w\in\Sigma^{7}$. The associated word $w$ cannot be suffix normal since the suffix of length two has only one $\mathsf{1}$ ($s_{w}(2)=1$) but by $s_{w}(5)=3$, $s_{w}(6)=4$, and $s_{w}(7)=5$ we get that within two letters two $\mathsf{1}$s are present and consequently $f_{w}(2)\geq 2$. Thus, a word $w$ is only least representative if the amount of $\mathsf{1}$s at the end of $s_{w}$ does not exceed the amount of $\mathsf{1}$s at the beginning of $s_{w}$.

Lemma 2

Let $w\in\Sigma^{n}$ be a LR. Then we have

[TABLE]

Proof

Since $w\in\Sigma^{n}$ is least representative we have $f_{w}(i)=s_{w}(i)=|\operatorname{Suff}_{i}(w)|_{\mathsf{1}}$ and $|\operatorname{Suff}_{i}(w)|_{\mathsf{1}}\geq|\operatorname{Pref}_{i}(w)|_{\mathsf{1}}$ for all $i\in[n]$. Let $i\in[n]$, $|\operatorname{Suff}_{n-i}(w)|_{\mathsf{1}}=s$, and $|\operatorname{Pref}_{i}(w)|_{\mathsf{1}}=r$. This implies $s_{w}(n-i)=s$ and $s_{w}(n)=s+r$. If $s_{w}(n-i+1)=s_{w}(n-i)$ then $w[i]=\mathsf{0}$ and $s_{w}(n-i+1)=s$. This implies

[TABLE]

If $s_{w}(n-i+1)\neq s_{w}(n-i)$ then $w[i]=\mathsf{1}$, $s_{w}(n-i+1)=s+1$ and

[TABLE]

The remaining part of this section presents results for prefix normal palindromes. Notice that for $w\in\operatorname{NPal}(n)$ with $w=\mathsf{x}v\mathsf{x}$ with $\mathsf{x}\in\Sigma$, $v$ is not necessarily a pnPal; consider for instance $w=\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{0}\mathsf{1}$ with $\mathsf{0}\mathsf{1}\mathsf{0}\in\operatorname{Pal}(3)\backslash\operatorname{NPal}(3)$. The following lemma shows a result for prefix normal palindromes which is folklore for palindromes substituting $f_{w}$ by $p_{w}$ or $s_{w}$.

Lemma 3

For $w\in\operatorname{NPal}(n)\backslash\{\mathsf{0}^{n}\}$, $v\in\operatorname{Pal}(n)$ with $w=\mathsf{1}v\mathsf{1}$ we have

[TABLE]

Proof

For $k=1$ we have $f_{w}(1)=|\operatorname{Pref}_{1}(w)|_{1}=1$ since $w\neq\mathsf{0}^{n}$. For $k\in[|w|-1]_{>1}$ we have

[TABLE]

Finally we have

[TABLE]

In the following we give a characterisation of when a palindrome $w$ is prefix normal depending on its maximum-ones function $f_{w}$ and a derived function $\overline{f_{w}}$. In particular we observe that $f_{w}=\overline{f_{w}}^{R}$ if and only if $w$ is a prefix normal palindrome. Intuitively $\overline{f_{w}}$ captures the progress of $f_{w}$ in reverse order. This is an intriguing result because it shows that properties regarding prefix and suffix normality can be observed when $f_{w},s_{w},p_{w}$ are considered in their serialised representation.

Definition 4

For $w\in\Sigma^{n}$ define $\overline{f}_{w}:[n]\rightarrow[n]$ by $\overline{f}_{w}(k)=\overline{f}_{w}(k-1)-(f_{w}(k-1)-f_{w}(k-2))$ with the extension $f_{w}(-1)=f_{w}(0)=0$ of $f$ and $\overline{f}_{w}(0)=f_{w}(n)$. Define $\overline{p}_{w}$ and $\overline{s}_{w}$ analogously.

Example 1

Consider the pnPal $w=\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}$ with $f_{w}=12234$. Then $\overline{f}_{w}$ is $43221$ and we have $f_{w}=\overline{f}_{w}^{R}$. On the other hand for $v=\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\in\operatorname{Pal}(6)\backslash\operatorname{NPal}(6)$ we have $p_{v}=112334$ and $f_{v}=122334$ and $\overline{f}_{v}=432211$ and thus $\overline{f}_{v}^{R}\neq f_{v}$.

The following lemma shows a connection between the reversed prefix-ones function and the suffix-ones function that holds for all palindromes.

Lemma 4

For $w\in\mathrm{Pal}(n)$ we have $s_{w}\equiv\overline{p}_{w}^{R}$.

Proof

Let $w\in\mathrm{Pal}(n)$. We get $\overline{p}_{w}^{R}(n)=p_{w}^{R}(1)=|w|_{1}=s_{w}(n)$. Now let $i\in[n]_{0}$. Assume $s_{w}(n-i+1)=\overline{p}_{w}^{R}(n-i+1)$ holds. We have by induction

[TABLE]

By Lemma 4 we get $p_{w}\equiv\overline{p}_{w}^{R}$ since $p_{w}\equiv s_{w}$ for a palindrome $w$. As advocated earlier, our main theorem of this part (Theorem 3.1) gives a characterisation of pnPals. The theorem allows us to decide if a word is a pnPal by only looking at the maximum-ones-function, thus a comparison of all factors is not required.

Theorem 3.1

Let $w\in\Sigma^{n}\setminus\Set{0^{n}}$. Then $w$ is a pnPal if and only if $f_{w}=\overline{f}^{\mathsf{R}}_{w}$.

Proof

Let $w\in\Sigma^{n}$. By definition of $\operatorname{NPal}(n)$, $w$ is prefix normal and a palindrome, i.e. $s_{w}=f_{w}$. By Lemma 4 and Definition 4 we get $f_{w}=s_{w}=\overline{p}_{w}^{R}=\overline{f}_{w}^{R}$. This proves $\Rightarrow$. Let $w\in\Sigma^{n}\setminus\Set{0^{n}}$ such that $f_{w}=\overline{f}^{\mathsf{R}}_{w}$. If $w=\varepsilon$, then obviously $w\in\operatorname{NPal}(n)$ holds. Otherwise, if $w\neq\varepsilon$, there exists a least representative $v\in\Sigma^{n}$ with $w\in[v]_{\equiv}$. This implies $f_{v}=f_{w}=\overline{f}_{w}^{R}=\overline{f}_{v}^{R}$, therefore the assumption also holds for $v$. Firstly, we will prove that $v$ is a palindrome. Let $x\in\Set{\mathsf{0},\mathsf{1}}$ and $i\in[n]$. Thus we have $f_{v}(i-1)+x=f_{v}(i)$. Since $v$ is least representative this implies $v[i]=x$. By the assumption we get $\overline{f}_{v}^{R}(i-1)+x=\overline{f}_{v}^{R}(i)$ and applying the definition of the reversal and $\overline{f}_{v}$ leads to

[TABLE]

Hence we get $f_{v}(n-i+1)=f_{v}(n-i)+x$, i.e. $v[n-i+1]=x$. Thus $v[n-i+1]=v[i]$ and therefore $v$ is a palindrome. As proven in [5], prefix normal (and thus suffix normal) palindromes are not pn-equivalent to any different word. Consequently $v=w$ and $w\in\operatorname{NPal}(n)$. ∎

Table 2 presents the amount of pnPals up to length $30$. These results support the conjecture in [5] that there is a different behaviour for even and odd length of the word.

4 Recursive Construction of Prefix Normal Classes

In this section we investigate how to generate LRs of length $n+1$ using the the LRs of length $n$. This is similar to the work of Fici and Lipták [14] except they investigated appending a letter to prefix normal words while we explore the behaviour on prepending letters to LRs. Consider the words $v=\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}$ and $w=\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}$, both being (different) LRs of length $4$. Prepending a $\mathsf{1}$ to them leads to $\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}$ and $\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}$ which are pn-equivalent. We say that $v$ and $w$ collapse and denote it by $v\leftrightarrow w$. Hence for determining the index of $\equiv_{n}$ based on the least representatives of length $n-1$, only the least representative of one class matters.

Definition 5

Two words $w,v\in\Sigma^{n}$ collapse if $\mathsf{1}w\equiv_{n+1}\mathsf{1}v$ holds. This is denoted by $w\leftrightarrow v$.

Prepending a $\mathsf{1}$ to a non LR will never lead to a LR. Therefore It is sufficient to only look at LRs. Since collapsing is an equivalence relation, denote the equivalence class w.r.t. $\leftrightarrow$ of a word $w\in\Sigma^{\ast}$ by $[w]_{\leftrightarrow}$. Next, we present some general results regarding the connections between the LRs of lengths $n$ and $n+1$. As mentioned in Remark 2, $\mathsf{0}^{n}$ and $\mathsf{1}^{n}$ are for all $n\in\mathbb{N}$ LRs. This implies that they do not have to be considered in the recursive process.

Remark 3

By [14] a word $w\mathsf{0}\in\Sigma^{n+1}$ is prefix-normal if $w$ is prefix-normal. Consequently we know that if a word $w\in\Sigma^{n}$ is suffix normal, $\mathsf{0}w$ is suffix normal as well. This leads in accordance to the naïve upper bound of $2^{n}+1$ to a naïve lower bound of $|\Sigma^{n}/\equiv_{n}|$ for $|\Sigma^{n+1}/\equiv_{n+1}|$.

Remark 4

The maximum-ones functions for $w\in\Sigma^{\ast}$ and $\mathsf{0}w$ are equal on all $i\in[|w|]$ and $f_{\mathsf{0}w}(|w|+1)=f_{w}(|w|)$ since the factor determining the maximal number of $\mathsf{1}$’s is independent of the leading $\mathsf{0}$. Prepending $\mathsf{1}$ to a word $w$ may result in a difference between $f_{w}$ and $f_{\mathsf{1}w}$, but notice that since only one $\mathsf{1}$ is prepended, we always have $f_{\mathsf{1}w}(i)\in\{f_{w}(i),f_{w}(i)+1\}$ for all $i\in[n]$. In both cases we have $s_{w}(i)=s_{\mathsf{x}w}(i)$ for $\mathsf{x}\in\{\mathsf{0},\mathsf{1}\}$ and $i\in[|w|]$ and $s_{\mathsf{0}w}(n+1)=s_{w}(n)$ as well as $s_{\mathsf{1}w}(n+1)=s_{w}(n)+1$.

Firstly we improve the naïve upper bound to $2|\Sigma^{n}/\equiv_{n}|$ by proving that only LRs in $\Sigma^{n}$ can become LRs in $\Sigma^{n+1}$ by prepending $\mathsf{1}$ or $\mathsf{0}$.

Proposition 2

Let $w\in\Sigma^{n}$ not be LR. Neither $\mathsf{0}w$ nor $\mathsf{1}w$ are LRs in $\Sigma^{n+1}$.

Proof

Suppose firstly $\mathsf{0}w$ is a least representative, i.e. $f_{\mathsf{0}w}(i)=s_{\mathsf{0}w}(i)$ for all $i\in[n+1]$. By $s_{\mathsf{0}w}(i)=s_{w}(i)$ and $f_{\mathsf{0}w}(i)=f_{w}(i)$ for $i\in[n]$, we have $s_{w}(i)=f_{w}(i)$ and thus $w$ would be a least representative. Suppose that secondly $\mathsf{1}w$ is a least representative. Since $w$ is not a least representative there exists a $j\in[|w|]$ with $s_{w}(j)\neq f_{w}(j)$. Choose $j$ minimal. Since $\mathsf{1}w$ is a least representative, we get

[TABLE]

By Remark 4 we have $f_{\mathsf{1}w}(j)=f_{w}(j)+1$ $s_{w}(j)\leq f_{w}(j)$ implies $f_{w}(j)\geq s_{w}(j)=f_{\mathsf{1}w}(j)=f_{w}(j)+1$ - a contradiction. ∎

By Proposition 1 prefix (and thus suffix) normal palindromes form a singleton class. This implies immediately that a word $w\in\Sigma^{n}$ such that $\mathsf{1}w$ is a prefix normal palindrome, does not collapse with any other $v\in\Sigma^{n}\backslash\{w\}$. The next lemma shows that even prepending once a $\mathsf{1}$ and once a $\mathsf{0}$ to different words leads only to equivalent words in one case.

Lemma 5

Let $w,v\in\Sigma^{n}$ be different LRs. Then $\mathsf{0}w\equiv_{n}\mathsf{1}v$ if and only if $v=\mathsf{0}^{n}$ and $w=\mathsf{0}^{n-1}\mathsf{1}$.

Proof

The equivalence of $\mathsf{0}\mathsf{0}^{n-1}\mathsf{1}=\mathsf{0}^{n}\mathsf{1}$ and $\mathsf{1}\mathsf{0}^{n}$ is immediate. This proves the $\Leftarrow$-direction. For the other direction assume $\mathsf{0}w\equiv_{n}\mathsf{1}v$. By definition we get $f_{\mathsf{0}w}(i)=f_{\mathsf{1}v}(i)$ for all $i\in[n+1]$ and moreover by Remark 4 $f_{w}(i)=f_{\mathsf{1}v}(i)$ for all $i\in[n]$. By $s_{w}(1)=f_{w}(1)=f_{\mathsf{1}v}(1)=1$ we get $w[|w|]=\mathsf{1}$ and by $s_{w}(n)=f_{\mathsf{1}v}(n)=|w|_{\mathsf{1}}$ there exists $u\in\operatorname{Fact}_{n}(\mathsf{1}v)$ with $|u|_{\mathsf{1}}=|w|_{\mathsf{1}}$. The equivalence of $\mathsf{0}w$ and $\mathsf{1}v$ implies $|w|_{\mathsf{1}}=|v|_{\mathsf{1}}+1$ and thus $u$ has to be a prefix of $\mathsf{1}v$. Hence $u[2..n]$ is a prefix of $v$ of length $n-1$ with $|w|_{\mathsf{1}}-1$ $\mathsf{1}$s. Since this is the overall amount of $\mathsf{1}$ in $v$, $v[n]=0$ follows and implies immediately $v=\mathsf{0}^{n}$. By $s_{w}(1)=f_{\mathsf{1}v}(1)=1$ follows $w[n]=\mathsf{1}$ and the claim follows with $|\mathsf{0}w|_{\mathsf{1}}=|\mathsf{1}v|_{\mathsf{1}}$.∎

By Lemma 5 and Remark 3 it suffices to investigate the collapsing relation on prepanding $\mathsf{1}$s. The following proposition characterises the LR $\mathsf{1}w$ among the elements $\mathsf{1}v\in[\mathsf{1}w]_{\equiv}$ for all LRs $v\in\Sigma^{n}$ with $w\leftrightarrow v$ for $w\in\Sigma^{n}$.

Proposition 3

Let $w\in\Sigma^{n}$ be a LR. Then $\mathsf{1}w\in\Sigma^{n+1}$ is a LR if and only if $f_{1w}(i)=f_{w}(i)$ holds for $i\in[n]$ and $f_{1w}(n+1)=f_{w}(n)+1$.

Proof

Let $w\in\Sigma^{n}$ be a least representative. Consider firstly $\mathsf{1}w\in\Sigma^{n+1}$ to be least representatives as well. Since $w$ is a least representative we have $s_{w}(i)=f_{w}(i)$ for all $i\in[n]$. By Remark 4 follows $s_{w}(i)=s_{\mathsf{1}w}(i)$ for all $i\in[n]$ and with $\mathsf{1}w$ being a least representative we get $f_{w}(i)=f_{\mathsf{1}w}(i)$ for all $i\in[n]$. By the same arguments we get $f_{\mathsf{1}w}(n+1)=s_{\mathsf{1}w}(n+1)=s_{w}(n)+1=f_{w}(n)+1$. Similarly we get for the second direction $f_{\mathsf{1}w}(i)=f_{w}(i)=s_{w}(i)=s_{\mathsf{1}w}(i)$ for all $i\in[n]$ and $f_{\mathsf{1}w}(n+1)=f_{w}(n)+1=s_{w}(n)+1=s_{\mathsf{1}w}(n+1)$.∎

Corollary 1

Let $w\in\operatorname{NPal}(n)$. Then $f_{w\mathsf{1}}(i)=f_{w}(i)$ for $i\in[n]$ and $f_{w\mathsf{1}}(n+1)=f_{w}(n)+1$. Moreover $s_{w\mathsf{1}}(i)=s_{w}(i)$ for $i\in[n]$ and $s_{w\mathsf{1}}(n+1)=s_{w}(n)+1$.

Proof

Since $w$ is a prefix normal palindrome, we have $(\mathsf{1}w)^{R}=w^{R}\mathsf{1}=w\mathsf{1}$. This implies $f_{w}(i)=f_{\mathsf{1}w}(i)=f_{(\mathsf{1}w)^{R}}(i)=f_{w\mathsf{1}}(i)$ for all $i\in[n]$ and $f_{w\mathsf{1}}(n+1)=f_{w}(n)+1$. If $s_{w\mathsf{1}}(i)=s_{w}(i)+1$ then $s_{w\mathsf{1}}(i)=s_{w}(i)+1=f_{w}(i)+1=f_{w\mathsf{1}}(i)+1$ would contradict $s_{w\mathsf{1}}(i)\leq f_{w\mathsf{1}}(i)$ for some $i\in[n+1]$. This proves the claim for the suffix-ones function.∎

This characterization is unfortunately not convenient for determining either the number of LRs of length $n+1$ from the ones from length $n$ or the collapsing LRs of length $n$. For a given word $w$, the maximum-ones function $f_{w}$ has to be determined, $f_{w}$ to be extended by $f_{w}(n)+1$, and finally the associated word - under the assumption $f_{\mathsf{1}w}\equiv s_{\mathsf{1}w}$ has to be checked for being suffix normal. For instance, given $w=\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{0}\mathsf{1}$ leads to $f_{w}=11223$, and is extended to $f_{\mathsf{1}w}=112234$. This would correspond to $\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{0}\mathsf{1}$ which is not suffix normal and thus $w$ is not extendable to a new LR. The following two lemmata reduce the amount of LRs that needs to be checked for extensibility.

Lemma 6

Let $w\in\Sigma^{n}$ be a LR such that $\mathsf{1}w$ is a LR as well. Then for all LRs $v\in\Sigma^{n}\backslash\{w\}$ collapsing with $w$, $f_{v}(i)\leq f_{w}(i)$ holds for all $i\in[n]$, i.e. all other LRs have a smaller maximal-one sum.

Proof

Let $v\in\Sigma^{n}\backslash\{w\}$ a least representative with $w\leftrightarrow_{n}v$. By the property of being least representatives, the definition of the maximum-ones and suffix-ones functions follows for all $i\in[n]$

[TABLE]

By $v\neq w$ there exists at least one $j\in[n]$ with $s_{w}(j)>s_{v}(j)$. This implies

[TABLE]

Corollary 2

If $w,v\in\Sigma^{n}$ and $\mathsf{1}w\in\Sigma^{n+1}$ are LRs with $w\leftrightarrow v$ and $v\neq w$ then $w\leq v$.

Proof

By $v\neq w$ exists an $i\in[n]$ minimal with $w[i]\neq v[i]$. Suppose $w[i]=1$ and $v[i]=0$. By $f_{\mathsf{1}v}\equiv f_{\mathsf{1}w}$ we get $|w[i+1..n]|_{\mathsf{1}}+1=|v[i+1..n]|_{\mathsf{1}}$. Thus

[TABLE]

This contradicts $s_{\mathsf{1}v}(n-i)\leq f_{\mathsf{1}v}(n-i)$.∎

Remark 5

By Corollary 2 the lexicographically smallest LR $w$ among the collapsing leads to the LR of $[\mathsf{1}w]$. Thus if $w$ is a LR not collapsing with any lexicographically smaller word then $\mathsf{1}w$ is LR.

Before we present the theorem characterizing exactly the collapsing words for a given word $w$, we show a symmetry-property of the LRs which are not extendable to LRs, i.e. a property of words which collapse.

Lemma 7

Let $w\in\Sigma^{n}$ be a LR. Then $f_{1w}(i)\neq f_{w}(i)$ for some $i\in[n]$ iff $f_{1w}(n-i+1)\neq f_{w}(n-i+1)$.

Proof

Since $w$ is least representative, we have

[TABLE]

From $f_{w}(i)\neq f_{\mathsf{1}w}(i)$ follows $f_{\mathsf{1}w}(i)=f_{w}(i)+1=s_{w}(i)+1$. Thus $\mathsf{1}w$ has a factor of length $i$ with $s_{w}(i)+1$ $\mathsf{1}$s. The suffix normality of $w$ implies that this factor needs to be the prefix of $\mathsf{1}w$ of length $i$, i.e. $|\operatorname{Pref}_{i-1}(w)|_{\mathsf{1}}=s_{w}(i)$. Thus we get $f_{w}(n-i+1)=|w|_{\mathsf{1}}-s_{w}(i)$. On the other hand we have

[TABLE]

Consequently $f_{\mathsf{1}w}(n-i+1)\geq|w|_{1}-s_{w}(i)+1>f_{w}(n-i+1)$. The second direction follows immediately with $j:=n-i+1$ and $f_{\mathsf{1}w}(j)\neq f_{w}(j)$. ∎

By [5, Lemma 10] a word $w\mathsf{1}$ is prefix normal if and only if $|\operatorname{Suff}_{k}(w)|_{\mathsf{1}}<|\operatorname{Pref}_{k+1}(w)|_{\mathsf{1}}$ for all $k\in\mathbb{N}$. The following theorem extends this result for determining the collapsing words $w^{\prime}$ for a given word $w$.

Theorem 4.1

Let $w\in\Sigma^{n}$ be a LR and $w^{\prime}\in\Sigma^{n}\backslash\{w\}$ with $|w|_{1}=|w^{\prime}|_{1}=s\in\mathbb{N}$. Let moreover $v\not\leftrightarrow w$ for all $v\in\Sigma^{\ast}$ with $v\leq w$. Then $w\leftrightarrow w^{\prime}$ iff

$f_{w^{\prime}}(i)\in\{f_{w}(i),f_{w}(i)-1\}$* for all $i\in[n]$,* 2. 2.

$f_{w^{\prime}}(i)=f_{w}(i)$* implies $f_{1w^{\prime}}(i)=f_{w}(i)$,* 3. 3.

$f_{w^{\prime}}(i)\geq\begin{cases}f_{w^{\prime}}(n)-f_{w^{\prime}}(n-i+1)&\mbox{if }f_{w^{\prime}}(n-i+1)=f_{w^{\prime}}(n-i),\\ f_{w^{\prime}}(n)-f_{w^{\prime}}(n-i+1)+1&\mbox{otherwise}.\end{cases}$**

Proof

Notice that $|w|_{1}=|w^{\prime}|_{1}=s\in\mathbb{N}$ implies immediately $f_{w}(1)=f_{w^{\prime}}(1)=1$ and $f_{w^{\prime}}(n)=f_{w}(n)=s$. Moreover for all $i\in[n]$ by Lemma LABEL:f_w(k+1) we have $f_{w^{\prime}}(i)\in\{f_{w^{\prime}}(i-1),f_{w^{\prime}}(i-1)+1\}$ and $f_{w}(i)\in\{f_{w}(i-1),f_{w}(i-1)+1\}$ and by Lemma 7 we get $f_{w^{\prime}}(i)\neq f_{1w^{\prime}}(i)$ iff $f_{w^{\prime}}(n-i+1)\neq f_{1w^{\prime}}(n-i+1)$.

Firstly consider the $\Leftarrow$-direction, i.e. let $w^{\prime}\in\Sigma^{n}$ with $|w|_{1}=s$ and the properties 1, 2, and 3. We have to prove $w^{\prime}\leftrightarrow w$, hence we have to prove $f_{\mathsf{0}w}(i)=f_{\mathsf{0}w^{\prime}}(i)$ for all $i\in[n]$. Since $w$ does not collapse with any lexicographically smaller $v\in\Sigma^{n}$, $\mathsf{0}w$ is a least representative by Remark 5. From Proposition 3 follows $f_{w}(i)=f_{1w}(i)$ for all $i\in[n]$. Obviously we have $f_{\mathsf{1}w}(1)=1=f_{\mathsf{1}w^{\prime}}(1)$ and hence the claim holds for $i=1$. By $f_{w^{\prime}}(n)=s$ we get $f_{1w^{\prime}}(n)\in\{s,s+1\}$. If $f_{1w^{\prime}}(n)$ were $s+1$ then by $f_{1w^{\prime}}(n)\neq f_{w^{\prime}}(n)$ and consequently by Lemma 7 we would have $1=f_{1w^{\prime}}(1)\neq f_{w^{\prime}}(1)=1$. Hence the claim holds for $i=n$. Let $i\in[n-1]_{>1}$. The claim holds by property 1 and Proposition 3 if $f_{w^{\prime}}(i)=f_{w}(i)$. Hence, assume $f_{w^{\prime}}(i)\neq f_{w}(i)$ for an $i\in[n-1]_{>1}$, i.e. $f_{w^{\prime}}(i)=f_{w}(i)-1$ by property 1. By Remark 4 we have $f_{\mathsf{1}w^{\prime}}(i)\in\{f_{w^{\prime}}(i),f_{w^{\prime}}(i)+1\}$.

case 1: $f_{w^{\prime}}(i)=f_{1w^{\prime}}(i)$

If $w^{\prime}$’s prefix of length $i-1$ had more (or equal) $\mathsf{1}$s than the suffix of length $i$, then the prefix of $\mathsf{1}w^{\prime}$ of length $i$ would have strictly more $\mathsf{1}$s than the suffix of length $i$. This contradicts $f_{w^{\prime}}(i)={1w^{\prime}}(i)$ and thus we have $|\operatorname{Pref}_{i-1}(w^{\prime})|_{\mathsf{1}}<|\operatorname{Suff}_{i}(w^{\prime})|_{\mathsf{1}}$. By $f_{w^{\prime}}(n-i+1)\geq|\operatorname{Suff}_{n-i+1}(w^{\prime})|_{\mathsf{1}}$ and $|\operatorname{Suff}_{n-i+1}(w^{\prime})|_{\mathsf{1}}+\|\operatorname{Pref}_{i-1}(w^{\prime})|_{\mathsf{1}}=s$ we get

[TABLE]

This is a contradiction to property 3.

case 2: $f_{w^{\prime}}(i)+1=f_{1w^{\prime}}(i)$

In this case we get immediately

[TABLE]

Thus $f_{1w^{\prime}}(i)=f_{w}(i)$ for all $i\in[n]$ which means that $f_{1w^{\prime}}$ and $f_{1w}$ are identical, i.e. $w$ and $w^{\prime}$ collapse.

For the $\Rightarrow$-direction, assume $w\leftrightarrow w^{\prime}$, i.e. $f_{1w^{\prime}}=f_{1w}$. Proposition 2 implies that $w^{\prime}$ can be assumed as a least representative since $\mathsf{1}w$ is a least representative and by $w$ and $w^{\prime}$ collapsing, $\mathsf{1}w^{\prime}$ is one as well. By Proposition 3 we have $f_{1w}(i)=f_{w}(i)$ for all $i\in[n]$ and thus $f_{1w^{\prime}}(i)=f_{w}(i)$ which proves (2). Since $f_{w}(i)=f_{1w^{\prime}}(i)\in\{f_{w^{\prime}}(i),f_{w^{\prime}}(i)+1\}$ for all $i\in[n]$ we get property 1. Since $w^{\prime}$ is a least representative, Lemma 2 implies property 3.∎

Theorem 4.1 allows us to construct the equivalence classes w.r.t. the least representatives of the previous length but more tests than necessary have to be performed: Consider, for instance $w=\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{1}$ which is a smallest LR of length $17$ not collapsing with any lexicographically smaller LR. For $w$ we have $f_{w}=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 5\cdot 6\cdot 7\cdot 8\cdot 8\cdot 8\cdot 9\cdot 10\cdot 10\cdot 11\cdot 12\cdot 13$ where the dots just act as separators between letters. Thus we know for any $w^{\prime}$ collapsing with $w$, that $f_{w^{\prime}}(1)=1$ and $f_{w^{\prime}}(17)=13$. The constraints $f_{w^{\prime}}(2)\in\{f_{w^{\prime}}(2),f_{w^{\prime}}(2)+1\}$ and $f_{w^{\prime}}(2)\leq f_{w}(2)$ implies $f_{w^{\prime}}(2)\in\{1,2\}$. First the check that $f_{w^{\prime}}(10)=4$ is impossible excludes $f_{w^{\prime}}(2)=1$. Since no collapsing word can have a factor of length $2$ with only one $\mathsf{1}$, a band in which the possible values range can be defined by the unique greatest collapsing word $w^{\prime}$. It is not surprising that this word is connected with the prefix normal form. The following two lemmata define the band in which the possible collapsing words $f_{w}$ are.

Lemma 8

Let $w\in\Sigma^{n}\backslash\{\mathsf{0}^{n}\}$ be a LR with $v\not\leftrightarrow w$ for all $v\in\Sigma^{n}$ with $v\leq w$. Set $u:=(\mathsf{1}w[1..n-1])^{R}$. Then $w\leftrightarrow u$ and for all LRs $v\in\Sigma^{n}\backslash\{u\}$ with $v\leftrightarrow w$ and all $i\in[n]$ $f_{v}(i)\geq f_{u}(i)$, thus $\sigma(u)=\sum_{i\in[n]}f_{u}(i)\leq\sum_{i\in[n]}f_{v}(i)=\sigma(v)$.

Proof

Set $u=(1w[1..n-1])^{R}$. Then by $w$ odd follows

[TABLE]

i.e. $w\leftrightarrow u$. Since $w$ does not collapse with any lexicographically smaller word, $\mathsf{1}w$ is a least representative by Remark 5. By Remark 1 $(\mathsf{1}w)^{R}\in[\mathsf{1}w]_{\equiv}$ and $w^{R}\mathsf{1}$ is lexicographically the largest element in the class. If there existed a $v\in[w]_{\leftrightarrow}$ with $v>w[n-1..1]\mathsf{1}$, then

[TABLE]

would hold which contradicts the maximality of $(\mathsf{1}w)^{R}$.∎

Notice that $w^{\prime}=(\mathsf{1}w[1..n-1])^{R}$ is not necessarily a LR in $\Sigma^{n}/\equiv_{n}$ witnessed by the word of the last example. For $w$ we get $u=\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{1}$ with $f_{u}(8)=f_{w}(8)$ and $f_{u}(10)=7\neq 8=f_{w}(10)$ violating the symmetry property given in Lemma 7. The following lemma alters $w^{\prime}$ into a LR which represents still the lower limit of the band.

Lemma 9

Let $w\in\Sigma^{n}$ be a LR such that $\mathsf{1}w$ is also a LR. Let $w^{\prime}\in\Sigma^{n}$ with $w\leftrightarrow w^{\prime}$, and $I$ the set of all $i\in[\lfloor\frac{n}{2}\rfloor]$ with

[TABLE]

and $f_{w}(j)=f_{w^{\prime}}(j)$ for all $j\in[n]\backslash I$. Then $\hat{w}$ defined such that $f_{\hat{w}}(j)=f_{w^{\prime}}(j)$ for all $j\in[n]\backslash I$ and $f_{\hat{w}}(n-i+1)=f_{w^{\prime}}(n-i+1)+1$ ($f_{\hat{w}}(i)=f_{\hat{w}}(i)+1$ resp.) for all $i\in I$ holds, collapses with $w$.

Proof

Let $k\in[n]$. Since $\mathsf{1}w$ is least representative, we have $f_{\mathsf{1}w}(k)\geq f_{\mathsf{1}\hat{w}}(k)$. If $k\not\in I$ we get

[TABLE]

and thus $f_{\mathsf{1}w}(k)=f_{\mathsf{1}\hat{w}}(k)$. If $k\in I$ we get in the first case

[TABLE]

and thus $f_{\mathsf{1}w}(n-k+1)=f_{\mathsf{1}\hat{w}}(n-k+1)$. This second case holds analogously.∎

Remark 6

Lemma 9 applied to $(\mathsf{1}w[1..n-1])^{R}$ gives the lower limit of the band. Let $\hat{w}$ denote the output of this application for a given $w\in\Sigma^{n}$ according to Lemma 9.

Continuing with the example, we firstly determine $\hat{w}$ for $w=\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{1}\mathsf{1}$. We get with $u=w[n-1..1]1$ Since for all collapsing $w^{\prime}\in\Sigma^{n}$ we have $f_{\hat{w}}(i)\leq f_{w^{\prime}}(i)\leq f_{w}(i)$, $w^{\prime}$ is determined for $i\in[17]\backslash\{5,9,13\}$. Since the value for $5$ de-

termines the one for $13$ there are only two possibilities, namely $f_{w^{\prime}}(5)=5$ and $f_{w^{\prime}}(9)=7$ and $f_{w^{\prime}}(5)=4$ and $f_{w^{\prime}}(9)=8$. Notice that the words $w^{\prime}$ corresponding to the generated words $f_{w^{\prime}}$ are not necessarily LRs of the shorter length as witnessed by the one with $f_{w^{\prime}}(5)=5$ and $f_{w^{\prime}}(9)=7$. In this example this leads to at most three words being not only in the class but also in the list of former representatives. Thus we are able to produce an upper bound for the cardinality of the class. Notice that in any case we only have to test the first half of $w^{\prime}$’s positions by Lemma 7. This leads to the following definition.

Definition 6

Let $h_{d}:\Sigma^{\ast}\times\Sigma^{\ast}\rightarrow\mathbb{N}_{0}$ be the Hamming-distance. The palindromic distance $p_{d}:\Sigma^{\ast}\rightarrow\mathbb{N}_{0}$ is defined by $p_{d}(w)=h_{d}(w[1..\lfloor\frac{n}{2}\rfloor],(w[\lceil\frac{n}{2}\rceil+1..|w|])^{R})$. Define the palindromic prefix length $p_{\ell}:\Sigma^{\ast}\rightarrow\mathbb{N}_{0}$ by $p_{\ell}(w)=\max\left\{\,k\in[|w|]\,|\,\exists u\in\operatorname{Pref}_{k}(w):\,p_{d}(u)=0\,\right\}$.

The palindromic distance gives the minimal number of positions in which a bit has to be flipped for obtaining a palindrome. Thus, $p_{d}(w)=0$ for all palindromes $w$, and, for instance, $p_{d}(\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{0}\mathsf{1})=2$ since the first half of $w$ and the reverse of the second half mismatch in two positions. The palindromic prefix length determines the length of $w$’s longest prefix being a palindrome. For instance $p_{\ell}(\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1})=2$ and $p_{\ell}(\mathsf{0}\mathsf{1}\mathsf{1}\mathsf{0}\mathsf{1})=4$. Since a LR $w$ determines the upper limit of the band and $w[n-1..1]\mathsf{1}$ the lower limit, the palindromic distance of $ww[n-1..1]\mathsf{1}$ is in relation to the positions of $f_{w}$ in which collapsing words may differ from $w$.

Theorem 4.2

If $w\in\Sigma^{n}$ and $\mathsf{1}w$ are both LRs then $|[w]_{\leftrightarrow}|\leq 2^{\lceil\frac{p_{d}(ww[n-1..1]\mathsf{1}}{2}\rceil}$.

Proof

By Lemma 8 $w[n-1..1]\mathsf{1}$ determines the lower bound of the band for collapsing words. Let $s_{1},\dots,s_{\ell}\in[n]$ with $s_{i}<s_{i+1}$ for $i\in[\ell],\ell\in[n]$ be the positions with $w[s_{i}]\neq(w[n-1..1]\mathsf{1})[s_{i}]$. Thus for all odd $i\in[\ell-1]$, $f_{w}$ and $f_{w[n-1..1]\mathsf{1}}$ are different between $s_{i}$ and $s_{i+1}-1$, since a different bit leads to a different number of $\mathsf{1}$s. By the same argument, $f_{w}$ and $f_{w[n-1..1]\mathsf{1}}$ are identical between $s_{i}$ and $s_{i+1}-1$ for all even $i\in[\ell-1]$. This implies that only the differences in odd positions lead to differences values of the corresponding maximum-ones function. Since each difference in the maximum-ones functions can be altered independently for obtaining a potential collapsing word, the number of collapsing words is exponential in half the palindromic distance. ∎

For an algorithmic approach to determine the LRs of length $n$, we want to point out that the search for collapsing words can also be reduced using the palindromic prefix length. Let $w_{1},\dots,w_{m}$ be the LRs of length $n-1$. For each $w$ we keep track of $|w|-p_{\ell}(w)$. For each $w_{i}$ we check firstly if $|w_{i}|-p_{\ell}(w_{i})=1$ since in this case the prepended $\mathsf{1}$ leads to a palindrome. Only if this is not the case, $[w_{i}]_{\leftrightarrow}$ needs to be determined. All collapsing words computed within the band of $w_{i}$ and $\hat{w_{i}}$ are deleted in $\{w_{i+1},\dots,w_{m}\}$.

In the remaining part of the section we investigate the set $\operatorname{NPal}(n)$ w.r.t. $\operatorname{NPal}(\ell)$ for $\ell<n$. This leads to a second calculation for an upper bound and a refinement for determining the LRs of $\Sigma^{n}/\equiv_{n}$ faster.

Lemma 10

If $w\in\operatorname{NPal}(n)\backslash\{\mathsf{1}^{n}\}$ then $\mathsf{1}w$ is not a LR but $w\mathsf{1}$ is a LR.

Proof

It suffices to prove that $w\mathsf{1}$ is a least representative. Then $(w\mathsf{1})^{R}=\mathsf{1}w$ is prefix normal and since $w\mathsf{1}$ is not a palindrome, $\mathsf{1}w$ is not a least representative. By Corollary 1 follows immediately that $w\mathsf{1}$ is a least representative.∎

Remark 7

By Lemma 10 follows that all words $w\in\operatorname{NPal}(n)$ collapse with a smaller LR. Thus, for all $n\in\mathbb{N}$, an upper bound for $|\Sigma^{n+1}/\equiv_{n+1}|$ is given by $2|\Sigma^{n}/\equiv_{n}|-\operatorname{npal}(n)$.

For a closed recursive calculation of the upper bound in Remark 7, the exact number $\operatorname{npal}(n)$ is needed. Unfortunately we are not able to determine $\operatorname{npal}(n)$ for arbitrary $n\in\mathbb{N}$. The following results show relations between prefix normal palindromes of different lengths. For instance, if $w\in\operatorname{NPal}(n)$ then $1w1$ is a prefix normal palindrome as well. The importance of the the pnPals is witnessed by the following estimation.

Theorem 4.3

For all $n\in\mathbb{N}_{\geq 2}$ and $\ell=|\Sigma^{n}/\equiv_{n}|$ we have

[TABLE]

Proof

By Corollary 1, $w\mathsf{1}$ is a least representative. From Lemma 8 follows that the band for possible collapsing words is given by $(\mathsf{1}(w\mathsf{1})[1..n])^{R}=(\mathsf{1}w)^{R}=w^{R}1=w1$. If the band is empty, $\mathsf{1}w\mathsf{1}$ is the single element in $[\mathsf{1}w\mathsf{1}]$ and consequently a palindrome [14]. This implies the lower bound. For the upper bound firstly all $\mathsf{0}w$, for $w$ being a least representative in $\Sigma^{n}/\equiv_{n}$ have to be counted and secondly all prefix normal palindromes. All other elements collapse with at least one different element.∎

The following results only consider pnPals that are different from $\mathsf{0}^{n}$ and $\mathsf{1}^{n}$. Notice for these special palindromes that $\mathsf{0}^{n}\mathsf{0}^{n}$, $\mathsf{1}^{n}\mathsf{1}^{n}$, $\mathsf{1}^{n}\mathsf{1}\mathsf{1}^{n}$, $\mathsf{0}^{n}\mathsf{0}\mathsf{0}^{n}$, $\mathsf{1}\mathsf{1}^{n}\mathsf{1}^{n}\mathsf{1}$, $\mathsf{1}\mathsf{0}^{n}\mathsf{0}^{n}\mathsf{1}\in\operatorname{NPal}(k)$ for an appropriate $k\in\mathbb{N}$ but $\mathsf{0}^{n}\mathsf{1}\mathsf{0}^{n}\not\in\operatorname{NPal}(2n+1)$.

Lemma 11

If $w\in\operatorname{NPal}(n)\backslash\{\mathsf{1}^{n},\mathsf{0}^{n}\}$ then neither $ww$ nor $w\mathsf{1}w$ are prefix normal palindromes.

Proof

Let $k\in[n-1]$ be minimal with $f_{w}(k)=f_{w}(k+1)$ (exists by $w\not\in\{\mathsf{1}^{n},\mathsf{0}^{n}\}$). Thus we have $f_{w}(k+1)=k$. Since $w\in\operatorname{NPal}(n)$ we have $|\operatorname{Suff}_{k}(w)|_{\mathsf{1}}=|\operatorname{Pref}_{k}(w)|_{\mathsf{1}}=k$ and $|\operatorname{Suff}_{k+1}(w)|_{\mathsf{1}}=|\operatorname{Pref}_{k+1}(w)|_{\mathsf{1}}=k$. This implies $f_{ww}(k+1)=k+1$ and hence $f_{ww}(k+1)\neq s_{ww}(k+1)$. The proof of $w\mathsf{1}w\not\in\operatorname{NPal}(2n+1)$ is similar to the proof of Lemma 11: in the middle of $w\mathsf{1}w$ is a larger block of $\mathsf{1}$’s than at the end.∎

Lemma 12

Let $w\in\operatorname{NPal}(n)\backslash\{\mathsf{0}^{n}\}$ with $n\in\mathbb{N}_{\geq 3}$. If $w\mathsf{0}w$ is also a prefix normal palindrome then $w=\mathsf{1}^{k}$ or $w=\mathsf{1}^{k}\mathsf{0}\mathsf{1}u\mathsf{1}\mathsf{0}\mathsf{1}^{k}$ for some $u\in\Sigma^{\ast}$ and $k\in\mathbb{N}$.

Proof

Consider $w\in\operatorname{NPal}(n)\backslash\{\mathsf{0}^{n}\}$. By $w\neq\mathsf{0}^{n}$ follows $w=\mathsf{1}u\mathsf{1}$ for an $u\in\operatorname{Pal}(n)$. If $w\neq\mathsf{1}^{k}$, there exists $k$ minimal with $w[k]=\mathsf{0}$. Suppose $w=\mathsf{1}^{k}\mathsf{0}^{\ell}u\mathsf{0}^{\ell}\mathsf{1}^{k}$ for some $k\in\mathbb{N}$ and $\ell\in\mathbb{N}_{>1}$. Then $p_{w\mathsf{0}w}(k+2)=k$ but $|w\mathsf{0}w[n-k,n+2]|_{\mathsf{1}}=k+1$. This is a contradiction to $w\mathsf{0}w\in\operatorname{NPal}(2n+1)$. This implies $w=\mathsf{1}^{k}\mathsf{0}\mathsf{1}u\mathsf{1}\mathsf{0}\mathsf{1}^{k}$. ∎

A characterisation for $w\mathsf{1}w$ being a pnPal is more complicated. By $w\in\operatorname{NPal}(n)$ follows that a block of $\mathsf{1}$s contains at most the number of $\mathsf{1}$s of the previous block. But if such a block contains strictly less $\mathsf{1}$s the number of $\mathsf{0}$s in between can increase by the same amount the number of $\mathsf{1}$s decreased.

Lemma 13

Let $w\in\operatorname{NPal}(n)\backslash\{\mathsf{1}^{n},\mathsf{0}^{n}\}$. If $\mathsf{1}ww\mathsf{1}$ is also a prefix normal palindrome then $\mathsf{1}\mathsf{0}\in\operatorname{Pref}(w)$.

Proof

Let $\mathsf{1}ww\mathsf{1}\in\operatorname{NPal}(2n+2)$. Since $w\neq\mathsf{0}^{n}$, there exists $u\in\Sigma^{\ast}$ with $w=\mathsf{1}u\mathsf{1}$. Since $w\neq\mathsf{1}^{n}$ there exists a minimal $k\in\mathbb{N}$ with $u[k]=\mathsf{0}$. If $k>1$, then $\mathsf{1}ww\mathsf{1}=\mathsf{1}^{k}\mathsf{0}\mathsf{1}^{2k}\mathsf{0}\mathsf{1}^{k}$ or $\mathsf{1}ww\mathsf{1}=\mathsf{1}^{k}\mathsf{0}v\mathsf{0}\mathsf{1}^{2k}\mathsf{0}v\mathsf{0}\mathsf{1}^{k}$. In both cases a contradiction to $\mathsf{1}ww\mathsf{1}\in\operatorname{NPal}(2n+2)$. ∎

Lemma 11, 12, and 13 indicate that a characterization of prefix normal palindromes based on smaller ones is hard to determine.

5 Conclusion

Based on the work in [14], we investigated prefix normal palindromes in Section 3 and gave a characterisation based on the maximum-ones function. At the end of Section 4 results for a recursive approach to determine prefix normal palindromes are given. These results show that easy connections between prefix normal palindromes of different lengths cannot be expected. By introducing the collapsing relation we were able to partition the set of extension-critical words introduced in [14]. This leads to a characterization of collapsing words which can be extended to an algorithm determining the corresponding equivalence classes. Moreover we have shown that palindromes and the collapsing classes are related.

The concrete values for prefix normal palindromes and the index of the collapsing relation remain an open problem as well as the cardinality of the equivalence classes w.r.t. the collapsing relation. Further investigations of the prefix normal palindromes and the collapsing classes lead directly to the index of the prefix equivalence.

Acknowledgments. We would like to thank Florin Manea for helpful discussions and advice.