DOI:10.1051/ita:2007064 www.rairo-ita.org
ON SOME PROBLEMS RELATED TO PALINDROME CLOSURE
∗Michelangelo Bucci
1, Aldo de Luca
1, Alessandro De Luca
1and Luca Q. Zamboni
2Abstract. In this paper, we solve some open problems related to (pseudo)palindrome closure operators and to the infinite words gen- erated by their iteration, that is, standard episturmian and pseudo- standard words. We show that ifϑis an involutory antimorphism of A∗, then the right and left ϑ-palindromic closures of any factor of a ϑ-standard word are also factors of some ϑ-standard word. We also introduce the class of pseudostandard words with “seed”, obtained by iterated pseudopalindrome closure starting from a nonempty word. We show that pseudostandard words with seed are morphic images of stan- dard episturmian words. Moreover, we prove that for any given pseu- dostandard word s with seed, all sufficiently long left special factors ofsare prefixes of it.
Mathematics Subject Classification.68R15.
Introduction
Sturmian words are a classical subject of combinatorics on words (see for in- stance [3]); by definition, they are infinite words havingn+ 1 factors (i.e., blocks of consecutive symbols) of each lengthn. Sturmian words enjoy many interesting characterizations and have a wide range of applications, from discrete geometry to crystallography.
Palindrome closure operators, introduced in [5], have had an important role in the study of Sturmian words. Ifwis a word, itsright (resp.left)palindrome closure
Keywords and phrases.Palindromes, palindrome closures, Sturmian and episturmian words, involutory antimorphisms, pseudopalindromes, pseudostandard words.
∗ The work for this paper has been supported by the Italian Ministry of Education under Project COFIN 2005 – Automi e Linguaggi Formali: aspetti matematici e applicativi.
1 Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Universit`a degli Studi di Napoli Federico II. Via Cintia, Monte S. Angelo, I-80126 Napoli, Italy;[email protected];
[email protected]; [email protected]
2Department of Mathematics, PO Box 311430, University of North Texas. Denton TX, USA;
Article published by EDP Sciences c EDP Sciences 2008
M. BUCCIET AL.
w(+) (resp. w(−)) is the shortest palindrome having w as a prefix (resp. suffix).
Standard Sturmian words can be constructed byiterated palindrome closure, that is, by the following procedure. Start from the empty word, and successively add a letter from{a, b}and apply the right palindrome closure operator. In this way one generates a sequence of palindromes, each one being a prefix of the next one, so that a limit is naturally defined. If both a and b are used infinitely many times during such process, the infinite word obtained as a limit is aperiodic, and is exactly a standard Sturmian word. For any Sturmian word, there exists a (unique) standard Sturmian word having the same factors.
In recent years, many extensions of Sturmian words have appeared. In particu- lar, in [7]episturmian words were introduced (see also [10]). They can be defined as words having the same set of factors of astandard episturmian word, which is just a word obtained by iterated palindrome closure over an arbitrary alphabet (and without the aperiodicity condition).
A further generalization was introduced in [6], by substituting palindrome clo- sure with pseudopalindrome closure. Apseudopalindrome is a fixed point of some involutory antimorphismϑof a free monoidA∗. Thus ordinary palindromes are a special case of pseudopalindromes where the antimorphism is simply the reversal operator. We speak of ϑ-palindromes when a particular antimorphism ϑis cho- sen. It is then natural to considerϑ-palindrome closure operators, and to look at words obtained by iteratedϑ-palindrome closure, called ϑ-standard (or generally pseudostandard) words.
In this paper, we discuss some properties related to (pseudo)palindrome closure, episturmian and ϑ-standard words. In [6] it was proven that both palindromic closuresw(+)andw(−)of a factorwof a Sturmian word are themselves factors of Sturmian words. In Section2, this property is proved for episturmian words.
In Section 3, the closure property is extended to factors of ϑ-standard words too. We also show that aϑ-standard word having both closures as factors always exists. Moreover, we prove that every left special factor of aϑ-standard word t, whose length is at least 3, is a prefix of t. Recall that a factor uof a (finite or infinite) wordw over an alphabetA is left (resp. right) special if there exist at least two distinct lettersa, b∈A such that bothauandbu (resp.uaandub) are factors ofw.
In the last section we introduce the class ofϑ-standard words with seed. They are infinite words obtained by iterated ϑ-palindrome closure, starting from an arbitrary wordu0 (called seed) instead of the empty word. We show that every ϑ-standard word with seed is a morphic image of a standard episturmian word.
More precisely, if Δ = xx1x2. . . xn. . . is the infinite sequence of letters which directs the construction of aϑ-standard wordtwith a seed, thent=φx(s), where φxis a morphism depending onϑandu0, andsis the standard episturmian word directed by Δ =x1x2. . . xn. . .
Finally, we show that every sufficiently long left special factor of aϑ-standard word with seed is a prefix of it, and give an upper bound for the minimal length from which this occurs, in terms of the length of the rightϑ-palindrome closure of u0x. This proves a conjecture posed in [6].
1. Preliminaries
LetAbe a finite alphabet andA∗thefree monoidgenerated byA. The elements ofAare usually calledletters and those ofA∗ words. The identity element ofA∗ is calledempty word and denoted by ε. We set A+=A∗\ {ε}. A wordw∈A+ can be written uniquely as a sequence of letters w = a1a2. . . an, with ai ∈ A, i= 1, . . . , n. The integer nis called the length of w and is denoted by|w|. The length ofεis conventionally 0.
Letw∈A∗. A wordv is a factor of wif there exist words rand ssuch that w=rvs. Ifw=vsfor somes (resp.w=rv for some wordr), then v is called a prefix (resp. asuffix) ofw. A word which is both a prefix and a suffix ofwis called aborder ofw. We shall denote respectively by Fact(w), Pref(w), and Suff(w) the sets of all factors, prefixes, and suffixes of the wordw.
ForX⊆A∗ andu∈A∗,u−1X andXu−1denote respectively the sets {w∈A∗|uw∩X =∅} and{w∈A∗|wu∩X =∅}.
When X is a singleton {x} and u−1X = ∅ (resp. Xu−1 = ∅), the unique word w∈u−1{x}(resp.w∈ {x}u−1) is denoted byu−1x(resp.xu−1).
Ifw=a1. . . an ∈A∗, ai∈A, i= 1, . . . , n, the reversal, ormirror image, ofw is the word
˜
w=an. . . a1.
One sets ˜ε = ε. A word is called palindrome if it is equal to its reversal. Any border of a palindrome is trivially a palindrome. We shall denote byPAL(A), or simplyPAL, the set of all palindromes on the alphabetA.
An infinite word (from left-to-right) x over the alphabet A is any map x : N+−→AwhereN+ is the set of positive integers. We can representxas
x=x1x2. . . xn. . . ,
where for anyi >0,xi=x(i)∈A. A (finite)factor ofxis either the empty word or any sequenceu=xi. . . xj withi≤j,i.e., any block of consecutive letters ofx.
Ifi= 1, thenuis aprefix ofx. We shall denote by Fact(x) and Pref(x) the sets of finite factors and prefixes ofxrespectively. The set of all infinite words overA is denoted byAω. Moreover, we setA∞=A∗∪Aω.
The product between a finite wordwand an infinite onexis naturally defined as the infinite wordwx. Anoccurrenceof the wordvinw∈A∞is any pair (r, s), withr∈A∗ ands∈A∞, such thatw=rvs. Ifw∈A∗ and a∈A,|w|a denotes the number of distinct occurrences ofain w.
Ifx∈Aandvx (resp. xv) is a factor of w∈A∞, thenvx(resp. xv) is called a right (resp. left)extension of v in w. We recall that a factorv of a (finite or infinite) wordwis calledright special if it has at least two distinct right extensions in w,i.e., there exist at least two distinct lettersa, b∈Asuch that both vaand vbare factors ofw. Left special factors are defined analogously. A factor of wis calledbispecial if it is both right and left special.
M. BUCCIET AL.
We call a factorw of a words∈ A∞ a first return tou ifw contains exactly two distinct occurrences ofu, one as a prefix and the other as a suffix,i.e.,
w=uλ=μu withλ, μ∈A+ andw /∈A+uA+.
We observe that in such a case,wu−1=μ is usually called areturn word over u ins(see [8]).
A words ∈ Aω is said to be closed under reversal if for anyu∈ Fact(s) one has ˜u∈Fact(s). In this case, a factoruof sis right special if and only if ˜uis a left special factor ofs.
A word w∈ Aω is called episturmian if it is closed under reversal and it has at most one right (or equivalently, left) special factor of each length. We recall (see [7]) that every episturmian word isuniformly recurrent, i.e., every factor of an episturmian word occurs infinitely often, with bounded gaps.
An episturmian wordw is calledstandard if every left special factor ofwis a prefix of it. We denote byEp(A), or simplyEp, the set of all episturmian words overA, and bySEpthe set of standard ones.
Proposition 1.1(cf.[7]). For every episturmian wordw, there exists a standard episturmian words such thatFact(s) = Fact(w).
Thus Fact(Ep) = Fact(SEp). The elements of Fact(Ep) are called finite epis- turmian words.
Given a wordw∈A∗, we denote byw(+)itsright palindrome closure, i.e., the shortest palindrome havingw as a prefix. Similarly, w(−) is the left palindrome closure of w. For instance, if w = abacbca, then w(+) =abacbcaba and w(−) = acbcabacbca.
For anyw∈A∗, one hasw(−)= ˜w(+). Moreover, ifQis the longest palindromic suffix ofwandw=sQ, thenw(+)=sQ˜s.
Letψ:A∗→A∗ be defined byψ(ε) =εandψ(va) = (ψ(v)a)(+)for anya∈A andv ∈A∗. For any u, v∈A∗, one hasψ(uv)∈ψ(u)A∗∩A∗ψ(u). The map ψ can then be naturally extended toAωby setting, for any infinite wordx,
ψ(x) = lim
n→∞ψ(wn), where{wn}= Pref(x)∩An for alln≥0.
Proposition 1.2(cf.[7]). Let s∈Aω. The following conditions are equivalent:
(1) sis a standard episturmian word,
(2) for any prefixuofs,u(+) is also a prefix ofs, (3) there existsx∈Aω such thats=ψ(x).
Given a standard episturmian word s, the (unique) infinite word x such that s =ψ(x) is called directive word of s and is denoted by Δ(s), or simply by Δ.
From the preceding proposition, one can easily derive (cf. [7]) that the set of palindromic prefixes of a standard episturmian wordscoincides with
{ψ(u)|u∈Pref(Δ(s))}.
A standard episturmian wordsover the alphabetAis called a (standard)Arnoux- Rauzy word if every symbol ofAoccurs infinitely often in the associated directive word Δ(s). We will denote by AR(A), or simply AR, the set of Arnoux-Rauzy words over A. In the case of a binary alphabet, an AR-word is usually called standard Sturmian word.
Example 1.3. LetA={a, b} andx= (ab)ω. One has that f =ψ(x) =abaababaabaababa . . .
is the famousFibonacci word, a standard Sturmian word. On an alphabet with three lettersA={a, b, c}, if we takex= (abc)ω as a directive word, then
τ =ψ(x) =abacabaabacababacabaabac . . .
is a standard Arnoux-Rauzy word, often calledTribonacci word. The word s = cabaabacababacabaab . . . such thatabas=τis an example of an episturmian word which is not standard, asais a left special factor ofsbut not a prefix of it.
The periodic words= (abac)ωis standard episturmian, but not Arnoux-Rauzy.
Its directive word is Δ(s) =abcω.
The following proposition can be easily proved using well-known results on episturmian words (see [7]).
Proposition 1.4. Let s be a standard episturmian word. Any bispecial factor of sis a palindromic prefix of s. Ifsis not periodic, the converse holds too.
Proposition 1.5. Fact(Ep) = Fact(AR).
Proof. Let u ∈ Fact(Ep) = Fact(SEp). Hence there exists s ∈ SEp such that u∈Fact(s). Now let bes=ψ(Δ) where Δ =t1t2. . . tn. . ., withti∈Afori≥1.
Therefore there exists a palindromic prefix p of s such that u ∈ Fact(p). Now p=ψ(t1. . . ti) for somei. We can consider Δ =t1. . . tit witht ∈Aω such that any letter ofA occurs infinitely many times in t. Hence s = ψ(Δ)∈ AR and contains p as a factor, so that u ∈ Fact(s). Therefore, Fact(Ep) ⊆ Fact(AR).
Since the inverse inclusion is trivial, the result follows.
The following proposition collects two properties of standard episturmian words (cf. Lems. 1 and 4 in [7]) which will be useful in the sequel.
Proposition 1.6(cf. [7]). Let s be a standard episturmian word. The following hold:
(1) Any prefixpof s has a palindromic suffix which has a unique occurrence inp.
(2) The first letter ofsoccurs in every factor of shaving length 2.
Clearly, ifpis a prefix of a standard episturmian word, then the palindromic suffix ofpwhich has a unique occurrence in pis the longest palindromic suffix ofp.
M. BUCCIET AL.
2. A closure property
We want to show that ifw∈Fact(Ep), then also its right and left palindrome closures belong to Fact(Ep); since episturmian words are closed under reversal, andw(−) = ˜w(+), it suffices to prove only the right palindrome closure case. We have the following:
Proposition 2.1. Let ube a non-palindromic finite episturmian word; let Q be the longest palindromic suffix ofuand writeu=saQwherea∈Aands∈A∗ (s possibly empty). Thenua=saQais a finite episturmian word.
Before proving the proposition we need some lemmas. The first lemma was proved in [1], Theorem 1.1. We report here a different and simpler proof.
Lemma 2.2. Let wbe an episturmian word andP ∈PAL∩Fact(w). Then every first return toP in wis a palindrome.
Proof. By Proposition 1.1, we may always suppose that w is a standard epis- turmian word. Let u ∈ Fact(w) be a first return to the palindrome P, i.e., u = P λ = ρP, λ, ρ ∈ A∗, and the only two occurrences of P in u are as a prefix and as a suffix ofu. If |P|>|ρ|, then the prefix P ofuoverlaps with the suffixP in uand this implies, as is easily to verify, thatuis a palindrome. Then let us suppose thatu=P vP withv∈A∗.
Now we consider the first occurrence ofuor of ˜uinw. Without loss of generality, we may suppose that w = αuw and that ˜u does not occur in the prefix of w having length|αu| −1. LetQbe the palindromic suffix ofαuof maximal length.
If |Q|>|u|, then we have that ˜uoccurs in αu beforeu, which is absurd. Then suppose|Q| ≤ |u|. If|u|>|Q|>|P|, then one contradicts the hypothesis thatu is a first return toP. If|Q|=|P|, thenQ=P has more than one occurrence in αu, which is absurd in view of Proposition1.6. The only remaining possibility is
Q=u,i.e.,uis a palindrome.
The following lemma is well-known. We report here a proof for the sake of completeness.
Lemma 2.3. Let w∈ARandsbe the unique right special factor of lengthn. If B1, . . . , Bm, . . .are the bispecial factors of w ordered by increasing length, then s is a suffix of anyBm such that|s| ≤ |Bm| and, for anyx∈A,sx∈Fact(w).
Proof. Sincewis not periodic, by Proposition1.4 the bispecial factorsBi, i >0, are its palindromic prefixes. Moreover, ift =t1t2. . . tn. . . ∈Aω is the directive word ofw, thenBi+1= (Biti)(+)for anyi >0. Sincesis a right special factor of w, ˜sis left special and thus a prefix ofw. Therefore,sis a suffix of any palindromic prefixBmofwsuch that|s| ≤ |Bm|. Asw∈AR, any letterx∈Aoccurs infinitely often int; hence there exists k≥m such thatx=tk, so thatBkxis a factor of w. SinceBmis a suffix ofBk, it followssx∈Fact(w).
Lemma 2.4. Let w andw be Arnoux-Rauzy words on the alphabet A. If wand w have the same right special factor of lengthn, then they share the same factors up to lengthn+ 1.
Proof. Trivial ifn = 0. By induction, suppose we have proved the assertion for the integern−1 ≥0. LetQ be the common right special factor ofw andw of lengthn. If we writeQ=aQ, witha∈A, thenQ is the only right special factor of lengthn−1 of both w andw. Hence w and w have the same factors up to lengthn.
By symmetry, it suffices to prove that any factorvofw, of length|v|=n+ 1, is also a factor ofw. Letv=vb,b∈A. Suppose first thatv=Q. By Lemma2.3, each right extensionQx, withx∈A, is a factor of bothw andw; in particular, vis a factor of w.
Now assume thatv=Q. Letv =cvwith c∈A, and suppose thatv=Q. One has then c = a. In this case, since v = cvb and Qb = avb are different factors ofw, one has thatvbis left special inw. Since|vb|=n, one derives that vb = ˜Q is a left special factor of w too, so that v=cvb is a factor of w as a consequence of Lemma2.3.
Ifv=Q, thenvb is the unique right extension ofv in w. As |vb|=n, it is also a factor of w, and no other letterx is such that vx∈ Fact(w). Hence v=cvb is the only right extension inw of the factorcv=Q.
We can now proceed to prove Proposition2.1.
Proof of Proposition2.1. We first observe thatucontains a single occurrence ofQ.
Indeed, ifucontained other occurrences ofQ, by Lemma2.2the suffix ofubegin- ning with the penultimate occurrence would be a palindromic suffix ofustrictly longer thanQ, contradicting the hypothesis of maximality of the length ofQ.
By Proposition1.5there exists an Arnoux-Rauzy wordwsuch thatu∈Fact(w).
We can assume thatua /∈Fact(w) (otherwise uais in Fact(AR) as required); so there existb∈Asuch thatb=aandub∈Fact(w). ThusaQb∈Fact(w); sinceQ is a palindrome andw∈AR, alsobQa∈Fact(w) andQis a bispecial factor ofw.
Then it follows that every left special factor ofwlonger thatQmust containQas a prefix, and since there is only a single occurrence ofQinu,Qitself is the longest suffix ofuwhich is left special in w. Thus every occurrence of aQin wmust be
“preceded” bys,i.e., ifw=λaQμ, thenw=λsaQμ, withλ=λs. In particular aQais not a factor ofw, for otherwiseuawould be in Fact(w), contradicting our assumption.
Set Δ(w) =t1t2. . . LetB1=ε, B2, . . . be the sequence of all bispecial factors of w, ordered by increasing length,i.e.,|Bi|<|Bi+1|for alli >0. By Proposition1.4, they are the palindromic prefixes of w as w is not periodic. Moreover, for each i >0 we haveBi+1 = (Biti)(+), so thatBitiis left special andtiBiis right special.
SinceQis a bispecial factor ofw, one hasQ=Bmfor somem >1. Let|Q|= n−1 forn≥2. We then have thattmQis right special inwand, from Lemma2.3, tmQx ∈Fact(w) for all x∈A. It is clear thattm =asince aQa /∈Fact(w) and tmQa ∈ Fact(w), then we have that aQb and tmQb are distinct factors of w, thusQb is left special andbQ is the unique right special factor ofw of lengthn.
Sotm=b.
Letw be any Arnoux-Rauzy sequence overA, whose directive word Δ(w) = t1t2. . . satisfiesti=ti for 0< i≤m−1 and tm=a. SinceQis the unique right
M. BUCCIET AL.
special factor ofwandw of lengthn−1, from Lemma2.4, we obtain thatwand w have the same factors of lengthkfor eachk≤n. However, they differ on some factors of lengthn+ 1. Indeed, from the definition of w, we have thataQis its unique right special factor of lengthn, so that by Lemma 2.3, for all x∈ A we have thataQx∈Fact(w). ThereforeaQa∈Fact(w)\Fact(w).
Now let us prove that, as inw, each occurrence ofaQ in w is preceded by s.
Letp∈A∗ be such that|p|=|s| andpaQ∈Fact(w). Let thenS be the largest common suffix ofpaQ andsaQand Q its prefix of lengthn−1. ClearlyQ=Q since there is only one occurrence ofQinsaQ. If we assume thatS =paQ, then there existx, y ∈A such thatx=y, xS ∈Suff (saQ) and yS ∈Suff(paQ); then xQ and yQ are both factors of wandw since these latter words have the same factors of length n. Thus Q is a left special factor of w and w, and that is a contradiction, since the only left special factor of lengthn−1 inwand inw isQ.
Thusp=sand so every occurrence ofaQinw is preceded bys.
SinceaQais a factor ofw, it follows that saQa=uais a factor ofw. Hence
uais in Fact(AR) as required.
From the preceding proposition one derives the following theorem, announced without proof in [6].
Theorem 2.5. If w is a finite episturmian word, then so is each of w(+) and w(−).
Proof. Trivial if w ∈ PAL. Let then w = a1. . . anQ, where ai ∈ A for i = 1, . . . , n andQis the longest palindromic suffix ofw. By Proposition2.1,wan = a1. . . anQan is a finite episturmian word; since its longest palindromic suffix is anQan, also wanan−1 is episturmian. In this way, by applying Proposition 2.1 exactlyntimes, one eventually obtains that
a1a2. . . anQan. . . a2a1=w(+)
is episturmian. Sincew(−)= ˜w(+), the assertion follows.
Corollary 2.6. Let a∈A andu∈A∗. If au is a finite episturmian word, then so isau(+).
Proof. If au is not a palindrome, then by Theorem 2.5, (au)(+) =au(+)a is an episturmian word and therefore so is au(+). Let us then suppose that au is a palindrome.
By Theorem 2.5 one has u(+) ∈ Fact(s) for a suitable s ∈ AR. Since s is recurrent there exist lettersx, y∈Asuch that
xu(+)y∈Fact(s).
If x = y, then, since s is closed under reversal, one has also yu(+)x ∈ Fact(s).
Hence u(+) is bispecial, so that it follows au(+) ∈Fact(s). Let us now consider the casex=y. Ifx=a, then the assertion is trivially verified.
Suppose then x = a. As au is a palindrome, we can write u = ua with u ∈PAL. Hence,
x(ua)(+)x∈Fact(s).
Since (ua)(+) begins with ua and ends with au, one has that xua and aux are factors of s, so that u is bispecial and then a palindromic prefix of s by Proposition1.4.
Let Δ(s) =t1t2. . . tn. . . be the directive word of s. There exists an integerk such thatu =ψ(t1t2. . . tk). We consider anyAR words whose directive word Δ(s) has the prefix t1t2. . . tka. Thusua=uis a prefix of s. This implies, by Propositions1.2and1.4, thatu(+)is a bispecial prefix ofs. From this one derives
au(+)∈Fact(s).
3. Pseudostandard words
Aninvolutory antimorphism of the free monoidA∗ is a mapϑ:A∗→A∗such thatϑ(uv) =ϑ(v)ϑ(u) for anyu, v∈A∗, and ϑ◦ϑ= id. The reversal operator
R:w∈A∗→w˜∈A∗
is the basic example of involutory antimorphism ofA∗. Any involutory antimor- phism is the compositionϑ=τ◦R=R◦τ whereτ is an involutory permutation of the alphabetA. Thus it makes sense to callϑ-palindromes the fixed points of an involutory antimorphismϑ. We shall denote byPALϑthe set ofϑ-palindromes overA. One can then define theϑ-palindrome closure operators: w⊕ϑ (resp.wϑ) denotes the shortestϑ-palindrome havingwas a prefix (resp. suffix).
Some properties and results relatingϑ-palindrome closure operators with peri- odicity and conjugacy are in [6]. Further interesting combinatorial properties of ϑ-palindromes, motivated by problems of molecular biology, have been recently studied in [11].
In the following, we shall fix an involutory antimorphismϑofA∗, and use the notation ¯w for ϑ(w). We shall also drop the subscript ϑfrom the ϑ-palindrome closure operator⊕ϑ when no confusion arises. Ifw=sQ=P t, whereQ(resp.P) is the longestϑ-palindromic suffix (resp. prefix) ofw, then
w⊕ =sQ¯s and w= ¯tP t (see [6]). Moreover, from the definition it follows
w = ¯w⊕ for anyw∈A∗.
For example, whenA={a, b},ϑ=E◦RwhereEis the interchange morphism defined by E(a) = b and E(b) = a, one has (aabab)⊕ =aababb and (aabab) = ababbaabab.
The following lemma will be useful in the sequel.
M. BUCCIET AL.
Lemma 3.1. For any u∈PALϑ\ {ε} and a∈ A, (ua)⊕ is a first return to u, i.e., if(ua)⊕=λuρwith λ, ρ∈A∗, then either λ=εorρ=ε.
Proof. By contradiction, letλ, ρ∈A+ be such that
(ua)⊕=λuρ. (1)
Clearly|λ|+|u|+|ρ|=|(ua)⊕| ≤2|u|+ 2, which implies|λ| ≤ |u|+ 2−|ρ| ≤ |u|+ 1.
Let us show that actually one has|λ| ≤ |u|. Indeed, ifλ=uathen from (1) one derives|(ua)⊕|= 2|u|+ 2; this implies thata /∈PALϑand (ua)⊕=ua¯au=uauρ, so thatuρ= ¯au. It follows that for somek >0,u= ¯ak ∈/ PALϑ, a contradiction.
Let thenv, w∈A∗ be such thatu=λvand (ua)⊕ =uw= ¯wu, whenceλuρ= uw=λvw. Thus uρ=vw, so that v is also a prefix of uand therefore a border of u. Since u is a ϑ-palindrome, v is a ϑ-palindrome too, so that u=λv = vλ.¯ Therefore
(ua)⊕ =λuρ=λv¯λρ.
Thus λvλ¯ is a ϑ-palindrome beginning with uaand strictly shorter than (ua)⊕,
which is a contradiction.
We can naturally define a mapψϑ:A∗→A∗byψϑ(ε) =εand ψϑ(ua) = (ψϑ(u)a)⊕
for u ∈ A∗, a ∈ A. For any u, v ∈ A∗ one has ψϑ(uv) ∈ ψϑ(u)A∗∩A∗ψϑ(u), so that, as done for the iterated palindrome closure, the domain of ψϑ can be extended to infinite words too. More precisely, ifx∈Aω, then
ψϑ(x) = lim
n→∞ψϑ(wn),
where {wn} = Pref(x)∩An for all n ≥ 0. The word x is called the directive word ofψϑ(x) and is denoted by Δ(ψϑ(x)). The images of infinite words over A byψϑ have been calledϑ-standard words in [6]. If ϑ =R, then ψR = ψ, where ψis the iterated palindrome closure operator introduced in Section1, so that an R-standard word is a standard episturmian word. A ϑ-standard word, without specifying the antimorphismϑ, has been calledpseudostandard word.
Example 3.2. LetA={a, b} andϑ=E◦R, so that ¯a=b. Forx= (ab)ω, we haveψϑ(a) =ab,ψϑ(ab) =abbaab, and
s=ψϑ(x) =abbaababbaabbaab . . .
The wordsis theϑ-standard word havingxas its directive word.
The following theorem, proven in [6], shows that any ϑ-standard word is a morphic image of the standard episturmian word having the same directive word.
Theorem 3.3. For any w ∈ A∞, one has ψϑ(w) = μϑ(ψ(w)), where μϑ is the injective morphism defined asμϑ(a) =a⊕ for any letter a∈A.
For instance, one easily verifies that the wordsof Example3.2is equal toμ(f), wheref is the Fibonacci word and μ=μϑ is the Thue-Morse morphism defined asμ(a) =ab,μ(b) =ba.
A new proof of Theorem 3.3 will be given in Section 4, as a consequence of a more general result. Some general properties of ϑ-standard words have been considered in [6]. In particular, we recall that
Proposition 3.4. Let s=ψϑ(x)be aϑ-standard word. The following hold:
(1) wis a prefix ofs if and only ifw⊕ is a prefix ofs,
(2) the set of allϑ-palindromic prefixes ofs is given byψϑ(Pref(x)), (3) sis closed underϑ, i.e., ifw∈Fact(s), then w¯∈Fact(s).
Moreover, the following holds:
Proposition 3.5. If s is a ϑ-standard word over A and two letters of A occur infinitely often in Δ(s), then any prefix of sis a left special factor of s.
Proof. A prefixpofsis also a prefix of anyϑ-palindromic prefixB ofssuch that
|p| ≤ |B|. SinceBis a suffix of anyϑ-palindromic prefix ofswhose length is at least
|B|, and there exist two distinct letters (sayaandb) which occur infinitely often in Δ(s), by Proposition3.4one derivesBa, Bb∈Fact(s). Therefore, as ¯p∈Suff(B), we have ¯pa,pb¯ ∈Fact(s),i.e., ¯pis right special. Since by Proposition3.4sis closed underϑ, one has ¯ap,¯bp∈Fact(s); as ¯a= ¯b,pis left special.
For the converse of the previous proposition, we observe that aϑ-standard word scan have left special factors which are not prefixes ofs. For instance, consider theϑ-standard wordsin Example3.2. As one easily verifies,bandbaare two left special factors ofs, which are not prefixes.
However, we will show that if a left special factorwof aϑ-standard wordsis not a prefix of s, then|w| ≤ 2. For a proof of this we need a couple of lemmas.
We denote byA=A\PALϑthe set of letters ofA that are notϑ-palindromic.
Lemma 3.6. The following holds:
Aμϑ(A∗)∩μϑ(A∗) =μϑ(A∗)A∩μϑ(A∗) =∅.
Proof. It is sufficient to observe that any word inμϑ(A∗) has an even number of
occurrences of letters inA.
Lemma 3.7. Let b, c ∈ A, and let f = ¯bμϑ(u) and g =μϑ(v)c be factors of a ϑ-standard wordt=μϑ(s), withs∈SEp. Then:
(1) Ifbu, vc∈Fact(s)and|f|>1, thenf =g.
(2) Ifu∈Fact(s) and|f|>3, thenbu∈Fact(s).
Proof. (1) Since |f|>1, one hasu=ε. By contradiction, if f =g, one has also v=ε, so that, from the definition ofμϑ, ¯bbis a prefix ofμϑ(v). Thenb¯bis a prefix ofμϑ(u), and so on; therefore, f = ¯b(b¯b)k = (¯bb)k¯b for k=|u|=|v| ≥1. Hence c= ¯b, u=bk, andv= ¯bk. Ask≥1, by Proposition1.6,bu=bk+1 andvc= ¯bk+1 cannot be both factors of the episturmian words, a contradiction.
M. BUCCIET AL.
(2) Since|f|>3, one derives|u|>1. By contradiction, supposebu /∈Fact(s).
By the preceding lemma and by Theorem3.3, one derivesf =μϑ(v)c for some suitablev ∈ A∗ and c ∈A such that vc ∈Fact(s). As done before, one then obtainsf = (¯bb)k¯bso thatbk,¯bk∈Fact(s), which is absurd by Proposition1.6, as
k≥2.
Theorem 3.8. Letwbe a left special factor of aϑ-standard wordt=μϑ(s), with s∈SEp. If |w| ≥3, then wis a prefix oft.
Proof. By Theorem3.3,wcan be written in one of the following ways:
(1) w=μϑ(u), withu∈Fact(s),
(2) w= ¯bμϑ(u), withbu∈Fact(s) andb∈A, (3) w=μϑ(u)c, withuc∈Fact(s) andc∈A, (4) w= ¯bμϑ(u)c, withbuc∈Fact(s) andb, c∈A.
In case 1, letxw, yw∈Fact(t) withx=yletters ofA. Ifxisϑ-palindromic, then clearly one must havexu∈Fact(s). Ifx∈A, then by the preceding lemma one has ¯xu∈Fact(s), as|xw|>3. Since the same holds fory,uis a left special factor of the episturmian words, and therefore a prefix of it. Thusw=μϑ(u) is a prefix oft.
Cases 2 and 4 are absurd; indeed, by the preceding lemma one derives that every occurrence ofwis preceded byb.
Finally, in case 3, by the preceding lemma one derives that every occurrence of wis followed by ¯c. Henceμϑ(uc) is a left special factor oftand one can apply the same argument as in case 1 to show that it is a prefix oft.
An infinite word t is a ϑ-word if there exists a ϑ-standard word s such that Fact(t) = Fact(s). AnR-word is an episturmian word.
Proposition 2.1 and Theorem 2.5 can be extended to the class of ϑ-words, showing that if w is a factor of a ϑ-word, then w⊕ and w are also factors of ϑ-words. A proof can be obtained as a consequence of Theorems2.5and3.3and of Corollary2.6. However, we need the following lemma (cf.[6]):
Lemma 3.9. Let u∈A∗ andx∈A∪ {ε}. Then (μϑ(u)x)⊕=μϑ
(ux)(+)
.
Theorem 3.10. Let w be a factor of a ϑ-standard word. Then each of w⊕ and w is a factor of a ϑ-standard word.
Proof. We shall supposew /∈PALϑ, otherwise the result is trivial. Sincew= ¯w⊕, it suffices to prove the result forw⊕. Let A =A\PALϑ as above. From Theo- rem3.3, one derives thatwcan be written in one of the following ways:
(1) w=μϑ(u)x, with x∈A∪ {ε}andux∈Fact(Ep), (2) w= ¯aμϑ(u)b, witha, b∈A andaub∈Fact(Ep), (3) w= ¯aμϑ(u), witha∈A andau∈Fact(Ep).
In the first case, by Theorem 2.5 there exists a standard episturmian word s =ψ(Δ) such that (ux)(+) ∈ Fact(s). Thus, by Lemma 3.9 and Theorem 3.3, w⊕=μϑ
(ux)(+)
is a factor of theϑ-standard wordψϑ(Δ) =μϑ(s).
In the second case, by using Lemma3.9, one has:
w⊕= ¯a(μϑ(u)b)⊕a= ¯aμϑ
(ub)(+)
a∈Fact
μϑ
a(ub)(+)a
.
Moreover,aubis not a palindrome, since otherwise one would derive, for instance using Lemma 3.9, that w = ¯aμϑ(u)b is a ϑ-palindrome, which contradicts our assumption. Thus (aub)(+)=a(ub)(+)aand the result is a consequence of Theo- rem3.3.
In the third case, sincewis not aϑ-palindrome, by Lemma3.9one obtains w⊕ = ¯aμϑ(u)⊕a∈Fact
μϑ(au(+)a)
.
Ifu=ak for somek≥0, thenau(+)a=ak+2 ∈Fact(Ep); otherwiseau(+) is not a palindrome andau(+)a = (au(+))(+), so that au(+)a is episturmian by Corol- lary2.6and Theorem2.5. Once again, the assertion follows from Theorem3.3.
Corollary 3.11. Let w be a factor of a ϑ-standard word. Then there exists a ϑ-standard word having bothw⊕ and w as factors.
Proof. Trivial if w ∈ PALϑ. Let then w = P bt = saQ, where P (resp. Q) is the longestϑ-palindromic prefix (resp. suffix) of w, anda, b ∈ A. Thusw¯a and
¯bw, being respectively factors of w⊕ = saQ¯a¯s and w = ¯t¯bP bt, are factors of ϑ-standard words by Theorem3.10.
Supposew¯a /∈ PALϑ. Then (w¯a) = aw¯a, so that w¯a is a factor of some ϑ-standard word, by Theorem3.10. Consider the word
(w¯a)⊕ = (¯t¯bP bt¯a)⊕= (¯t¯bsaQ¯a)⊕,
and callQthe longestϑ-palindromic suffix ofw¯a; thenQ=aQ¯a. Indeed, since aQ¯ais aϑ-palindrome, one has|Q| ≥ |aQ¯a|; but|aQ¯a|<|Q| ≤ |saQ¯a|is absurd, forQwould not be the longestϑ-palindromic suffix ofw, and|Q|>|saQ¯a|cannot happen, for otherwise there would exist aϑ-palindromic proper suffix ofwhaving was a suffix, contradicting the definition ofw. Thus
(wa)¯ ⊕= ¯t¯bsaQ¯a¯sbt= ¯t¯bP bt¯a¯sbt
is a factor of someϑ-standard word, again by Theorem3.10, and it contains both w⊕ andw as factors.
Ifw¯a ∈PALϑ but ¯bw /∈ PALϑ, one can prove by a symmetric argument that (¯bw⊕) is a factor of someϑ-standard word having bothw⊕ and w as factors.
Let thenw¯a,¯bw∈PALϑ, so that
w⊕=w¯a=aw¯ andw= ¯bw= ¯wb. (2)
M. BUCCIET AL.
If w is a single letter, one derives w =a =b, so that w⊕ = a¯a and w = ¯aa.
Therefore w⊕ and w are factors of any ϑ-standard word whose directive word begins with a2. Let us then suppose |w| > 1. From (2) it follows w = aRbfor someR∈A∗ such thataR= ¯R¯a=P andRb= ¯bR¯=Q. Moreover,
w=aRb=a¯bR¯= ¯R¯ab, (3) showing that ¯Ris a border ofw. Therefore one has eitherw= (a¯b)korw= (a¯b)ka, for some k > 0. In the first case, from (3) one derives a = ¯a and b = ¯b, so that anyϑ-standard word whose directive word begins with abk+1 contains both w⊕ = (ab)kaand w =b(ab)k as factors. In the latter case, by (3) one obtains a=b, so that anyϑ-standard word whose directive word begins withak+1contains
bothw⊕= (a¯a)k andw= (¯aa)k as factors.
Remark 3.12. For a finite episturmian wordw, the proof of the preceding result can be simplified by using Theorem2.5 and Corollary 2.6. Indeed, if w is not a palindrome, we can write w =P bt =saQ, whereP and Q are respectively the longest palindromic prefix and suffix ofw, and a, b∈ A. By Theorem2.5, w(+) and w(−) are finite episturmian words; moreover bw is a factor of w(−), so that by Corollary2.6,bw(+)is a finite episturmian word. By Theorem2.5,
bw(+)(−) is a finite episturmian word, which has also w(−) as a factor, as one can prove similarly as in the proof of Corollary3.11.
In the case of Sturmian words, results analogous to Theorem3.10 and Corol- lary3.11were proven in [6] with a different and simpler technique based on the structure of finite Sturmian words.
Example 3.13. Letτ be the Tribonacci word
τ=ψ((abc)ω) =abacabaabacababacabaabacabac . . .
Ifw=bac∈Fact(τ), one has thatw(+)=bacabandw(−)=cabacare factors ofτ.
However, in the case of the factorv=abacabab, one hasv(+)=abacababacaba∈ Fact(τ), whereasv(−)=babacababis not a factor ofτ, since otherwisevwould be a left special factor ofτ, which is a contradiction asv /∈Pref(τ). Nevertheless, both v(+) and v(−) are factors of any episturmian word whose directive word begins withabcbb. Indeed, v =P b whereP =abacabais the longest palindromic prefix ofv, and
bv(+) (−)
=abacababacababacaba=ψ(abcbb).
4. Words generated by nonempty seeds
We now consider a generalization of the construction ofϑ-standard words. De- fine the map ˆψϑ : A∗ → A∗ by setting ˆψϑ(ε) = u0 with u0 a fixed word of A∗ calledseed, and
ψˆϑ(ua) =
ψˆϑ(u)a ⊕ foru∈A∗and a∈A.
As usual, we can extend this definition to infinite wordst∈Aω by:
ψˆϑ(t) = lim
n→∞ψˆϑ(wn),
where{wn}= Pref(t)∩An for alln≥0. The wordt is called thedirective word of ˆψϑ(t), and denoted by Δ( ˆψϑ(t)). When the seedu0is empty, one has ˆψϑ=ψϑ so that one obtainsϑ-standard words. If u0 = ε, then any word ˆψϑ(t) is called ϑ-standard with seed.
Example 4.1. LetA={a, b, c},ϑbe the involutory antimorphism exchanginga andband fixingc,u0=acbbc, andw=abc. Then
ψˆϑ(w) =
ψˆϑ(ab)c ⊕
=
ψˆϑ(a)b ⊕
c
⊕
=
(acbbca)⊕b⊕ c
⊕
=
(acbbcaacbb)⊕c⊕
=acbbcaacbbcaacbcacbbcaacbbcaacb.
Let t =xt1t2. . ., withx ∈ A and ti ∈ A for i ≥1. We remark that the set of ϑ-palindromic prefixes of the wordw= ˆψϑ(t) is
(PALϑ∩Pref(u0))∪ {un|n≥1}, whereu1= (u0x)⊕ andui+1= (uiti)⊕ fori≥1.
Define the endomorphismφx ofA∗ by setting φx(a) = ˆψϑ(xa) ˆψϑ(x)−1
for any lettera∈A. From the definition, one has thatφx depends onϑand u0; moreover,φx(a) ends with ¯afor alla∈A, so that any word of the setX=φx(A) is uniquely determined by its last letter. ThusX is a suffix code, and φx is an injective morphism.
Example 4.2. Let A, ϑ, and u0 be defined as in Example 4.1, and let x= a.
Then
φa(a) = ψˆϑ(aa) ˆψϑ(a)−1=acbbcaacb,
φa(b) = ψˆϑ(ab) ˆψϑ(a)−1=acbbca, (4) φa(c) = ψˆϑ(ac) ˆψϑ(a)−1=acbbcaacbc.
To simplify the notation, in the following we shall often omit in the proofs the subscriptxfrom φx, when no confusion arises.
Theorem 4.3. Fixx∈Aandu0∈A∗. Letψˆϑ andφx be defined as above. Then for anyw∈A∗, the following holds:
ψˆϑ(xw) =φx(ψ(w)) ˆψϑ(x).
M. BUCCIET AL.
Proof. In the following we shall often use the property that ifγis an endomorphism ofA∗ andv is a suffix ofu∈A∗, then γ(uv−1) =γ(u)γ(v)−1.
We will prove the theorem by induction on|w|. It is trivial that forw=ε the claim is true sinceψ(ε) =ε=φ(ε). Suppose that for all the words shorter than w, the statement holds. For|w|>0, we setw=vy withy∈A.
First we consider the case|v|y= 0. We can then writev=v1yv2with|v2|y= 0, so that
ψˆϑ(xv) = ˆψϑ(xv1yv2) = ˆψϑ(xv1)yλ= ¯λ¯yψˆϑ(xv1),
for a suitable λ ∈ A∗. Note that ˆψϑ(xv1) is the largest ϑ-palindromic prefix (resp. suffix) followed (resp. preceded) byy (resp. ¯y) in ˆψϑ(xv). Therefore,
ψˆϑ(xvy) = ¯λ¯yψˆϑ(xv1)yλ= ˆψϑ(xv) ˆψϑ(xv1)−1ψˆϑ(xv). (5) By a similar argument one has:
ψ(vy) =ψ(v)ψ(v1)−1ψ(v). (6) By induction we have:
ψˆϑ(xv) =φ(ψ(v)) ˆψϑ(x), ψˆϑ(xv1) =φ(ψ(v1)) ˆψϑ(x).
Replacing in (5), and by (6), we obtain
ψˆϑ(xvy) = φ(ψ(v))φ(ψ(v1))−1φ(ψ(v)) ˆψϑ(x)
= φ(ψ(v)ψ(v1)−1ψ(v)) ˆψϑ(x)
= φ(ψ(vy)) ˆψϑ(x), which was our aim.
Now suppose that|v|y = 0 andPALϑ∩Pref(u0x)y−1=∅. Letαybe the longest word inPALϑ∩Pref(u0x)y−1, that is the longestϑ-palindromic prefix ofu0xwhich is followed byy. Since|v|y = 0, one derives that the longestϑ-palindromic suffix of ˆψϑ(xv)y is ¯yαyy, whence
ψˆϑ(xvy) =
ψˆϑ(xv)y ⊕
= ˆψϑ(xv)α−1y ψˆϑ(xv). (7) By induction, this implies
ψˆϑ(xvy) =φ(ψ(v)) ˆψϑ(x)α−1y φ(ψ(v)) ˆψϑ(x). (8) By using (7) forv=ε, one has ˆψϑ(xy) = ˆψϑ(x)α−1y ψˆϑ(x), and
φ(y) = ˆψϑ(xy) ψˆϑ(x)
−1
= ˆψϑ(x)α−1y .
