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Patrice Séébold
To cite this version:
Patrice Séébold. Length-k-overlap-free Binary Infinite Words. Fundamenta Informaticae, Polskie Towarzystwo Matematyczne, 2012, 116, pp.251-263. �10.3233/FI-2012-682�. �lirmm-00803975�
IOS Press
Length-k-overlap-free binary infinite words
Patrice S´e´ebold∗
Univ. Montpellier 3 Paul Val´ery, Montpellier, F-34199 LIRMM, CNRS, F-34392
Patrice.Seebold@lirmm.fr
Abstract. We study length-k-overlap-free binary infinite words, i.e., binary infinite words which can contain only overlapsxyxyxwith|x| ≤k−1. We prove that no such word can be generated by a morphism, except ifk= 1. On the other hand, for everyk≥2, there exist length-k-overlap-free binary infinite words which are not length-(k−1)-overlap-free. As an application, we prove that, for every non-negative integern, there exist infinitely many length-k-overlap-free binary infinite partial words withnholes.
Keywords: Repetition-freeness, length-k-overlaps, partial words, Thue-Morse word, infinite words Mathematical Subject Classification:68R15
1. Introduction
Repetitions, i.e., consecutive occurrences of a given factor within a word, and repetition-freeness have been fundamental research subjects in combinatorics on words since the seminal papers of Thue [13, 14]
in the beginning of the 20th century (see also [2]). In particular, Thue and Morse independently showed the existence of an overlap-free binary infinite word (the Thue-Morse word [8, 14]), i.e., an infinite word using only two different letters and which does not contain any factorxyxyxwithxa non-empty word.
However overlap-freeness is a very restrictive condition, in the binary case, because every binary word of length at least 5 contains a square, and an overlap is just a square plus one single letter. So it
∗Dpt Math´ematiques et Informatique Appliqu´ees, Univ. Montpellier 3, Route de Mende, 34199 Montpellier Cedex 5, France;
Laboratoire d’Informatique, de Robotique et de Micro-´electronique de Montpellier, UMR 5506, CNRS, 161 rue Ada, 34392 Montpellier, France
is a natural question to know if the fact that words contain a restricted kind of overlaps gives interesting families of words (similar questions are dealt with, e.g., in [12], [9]).
In the present paper we study the case where x must be of length at leastk. Words which do not contain any factorxyxyxwith|x| ≥kare calledlength-k-overlap-free binary infinite words. Note that ymay be empty, thus a length-k-overlap-free binary infinite word is also length-k-cube-free.
The paper is organized as follows. After general definitions and notations given in Section 2, the notion of length-k-overlap-freeness is introduced in Section 3 where it is proved that no length-k-overlap- free binary infinite word can be generated by a morphism, except ifk= 1. In Section 4 we introduce the concept of0-limited square property (a word has this property if the squares it contains have a particular form) to prove that, for every integerk, there exist length-k-overlap-free binary infinite words that are not length-(k−1)-overlap-free. In Section 5 we consider the particular case of length-k-overlap-free words which do not contain cubes of some letters. Section 6 is then dedicated to an application to partial words1.
2. Preliminaries
Generalities on combinatorics on words can be found, e.g., in [7].
LetAbe a finite alphabet. The elements ofAare calledletters. A wordw=a1a2· · ·anof lengthn over the alphabetAis a mappingw:{1,2, . . . , n} → Asuch thatw(i) =ai. The length of a wordw is denoted by|w|, andεis the empty word of length zero. For a wordwand a lettera,|w|adenotes the number of occurrences of the letterain the wordw. By a (right) infinite wordw=a1a2a3· · · we mean a mappingwfrom the positive integersN+to the alphabetAsuch thatw(i) =ai. The set of all finite words is denoted byA∗, infinite words are denoted byAω andA+ = A∗\ {ε}. A finite wordv is a factorofw∈A∗∪Aω ifw=xvy, wherex∈A∗andy∈A∗∪Aω. Wordsxvandvyare respectively calleda prefixanda suffixofw.
A morphism onA∗ is a mappingf:A∗ →A∗satisfyingf(xy) =f(x)f(y)for allx, y∈A∗. The morphismf iserasingif there existsa∈Asuch thatf(a) =ε. Note thatf is completely defined by the valuesf(a)for every letteraonA. We can also apply morphisms to infinite words: ifw=c0c1c2. . .is an infinite word then we definef(w) =f(c0)f(c1)f(c2). . .
A morphism is called prolongable on a letteraif f(a) = aw for some word w ∈ A+ such that fn(w)6=εfor all integersn≥1. This implies thatfn(a)is a prefix offn+1(a)for all integersn≥0 andais agrowing letter forf, that is,|fn(a)|<|fn+1(a)|for everyn∈N. Consequently, the sequence (fn(a))n≥0converges to the unique infinite word generated byf from the lettera,
fω(a) := lim
n→∞fn(a) =awf(w)f2(w)· · ·,
which is a fixed point off.
A morphismf: A∗ →A∗ generates an infinite wordwfrom the lettera∈ Aif there existsp ∈N such that the morphismfpis prolongable ona. We say that the morphismf generates an infinite word if it generates an infinite word from at least one letter.
1A preliminary version of this paper was presented at JORCAD’08 [11]. Some results about the casek = 2appeared in [6].
But in these two papers, the0-limited square property was replaced by the restricted square property, a much more restrictive condition.
Akthpowerof a wordu 6= εis the worduk prefix of lengthk· |u|ofuω, where uω denotes the infinite catenation of the wordu, andkis a rational number such thatk· |u|is an integer. A wordwis calledk-freeif there does not exist a wordxsuch thatxkis a factor ofw. Ifk= 2ork = 3, then we talk aboutsquare-freeorcube-freewords, respectively. Anoverlapis a word of the formxyxyxwhere x ∈ A+ andy ∈ A∗. A word is calledoverlap-free if it does not contain overlaps. Therefore, it can contain squares but it cannot contain any longer repetitions such as overlaps or cubes. For example, over the alphabet{a, b}the wordabbabaais overlap-free but it contains squaresbb,aa, andbaba. It is easy to verify that there does not exist a square-free infinite word over a binary alphabet, but as we recall in the next section there exist overlap-free binary infinite words.
Throughout this paper, we shall refer to the alphabetsA={a, b}, andB={0,1,2}.
3. Length-k-overlap-free binary words
In [14], Thue introduced the morphism
µ: A∗ → A∗ a 7→ ab b 7→ ba
The Thue-Morse word is the overlap-free binary infinite word t:= lim
n→∞µn(a) =abbabaabbaababbaba· · ·
generated by µfrom the letter a (see, e.g., [1] for other definitions and properties, see also [2] for a translation of the contribution of Thue to the combinatorics on words). Another overlap-free binary infinite word ist0, the word generated by the morphismµfrom the letterb. Note that the wordt0can be obtained from the wordtby exchanging all thea’s andb’s.
We generalize the notion of overlap with the following definition.
Definition 3.1. A length-k-overlap is a word of the form xyxyx where x and y are two words with
|x|=k.A word islength-k-overlap-free2if it does not contain length-k-overlaps.
For example, the wordbaabaabis not overlap-free (takex =b,y =aa) but it is length-2-overlap- free while the wordbaabaabais not (takex=ba,y=a).
It is important to note that a length-k-overlap-free word can contain a3k−1 for each lettera. More generally, a length-k-overlap-free word can contain`-powers u` where the value of `, which can be greater thank, depends on the wordu. For example, the wordu = ababababab, which is a5-power, is length-3-overlap-free. In contrast, the wordv = abaabaaba, which is only a 3-power and whose length is smaller than|u|, is a length-3-overlap thus not being length-3-overlap-free! This peculiarity is one reason for restricting the definition, for example to the case of cube-free words. However, such a
2While it is not exactly the same, this notion of length-k-overlap-freeness resembles that ofk-bounded overlaps introduced by Thue in [14]. Note also that length-k-overlap-freeness is inappropriately calledk-overlap-freeness in [6] and [11].
restriction seems to be very drastic and it is generally enough to avoid the powers of letters. In Section 5, we study the change in our results when restricting to the case of words withouta3and we will see that the results are not very different.
By definition, it is evident that every length-k-overlap-free word is also length-k0-overlap-free for k0 ≥ k. Note that a word is length-1-overlap-free if and only if it is overlap-free. So, an overlap-free infinite word is a length-k-overlap-free infinite word for every positive integerk.
It is a well-known important property that tand t0 are the only overlap-free binary infinite words which are generated by morphisms (see, e.g., [5], [10]). Since length-k-overlap-freeness does not imply length-`-overlap-freeness for ` < k, length-k-overlap-freeness is weaker than overlap-freeness when k≥2. Therefore, we might suppose that there exist binary infinite words, generated by morphisms, that are length-k-overlap-free for somek ≥ 2 but that are not overlap-free (therefore different fromt and t0). In fact, rather surprisingly, for that property, length-k-overlap-freeness does not give more than only length-1-overlap-freeness.
Theorem 3.2. Letk∈N+and letwbe a length-k-overlap-free binary infinite word. Thenwis generated by a morphism if and only ifw=torw=t0.
Proof:
The only if part is obvious sincet=µω(a)andt0 =µω(b)are length-k-overlap-free for every positive integerk.
Conversely, as we have already seen, if an overlap-free (k = 1) binary infinite wordwis generated by a morphism thenw= torw =t0.Thus it remains to prove that an infinite word which contains an overlap but is length-k-overlap-free for some integerk≥2cannot be generated by a morphism.
Assume, contrary to what we want to prove, that an infinite wordwoverAwhich contains an overlap but is length-k-overlap-free for some integer k ≥ 2, is generated by a morphismf. Then there exists a positive integer r such that fr is prolongable on the first letter ofw. Without loss of generality we may assume that this first letter is the lettera. In particular,wbegins with(fr)n(a)for everyn ∈ N. Moreover, by definitionais a growing letter forfr, which implies that there exists a positive integerN0 such that|frN0(a)| ≥k.
Iff(b) = εor if|f(a)|b = 0(which meansf(a) =ap,p≥2, becauseais a growing letter forf), thenwis the periodic word(f(a))ωwhich contains arbitrarily large powers off(a).
Iff(b) =bp,p ≥2, or iffr(a)ends withbandf(b) =b(which implies thatfrn(a)ends withbn for every integern), thenwcontains arbitrarily large powers ofb.
Iff(a)ends withaand begins withabpa,p≥1, andf(b) =b, thenw, which begins with(fr)2(a), contains a factorauauawithu=bp.
Ifwcontains an overlapauauaas a factor, thenfrN0(auaua)is also a factor ofw.
The only remaining case iswcontains an overlapbububas a factor andf(b) =xayfor some words x, y∈A∗. ThenfrN0(bubub)contains the factorfrN0(a)frN0(yux)frN0(a)frN0(yux)frN0(a).
Consequently, since frN0(w) = w (because w = (fr)ω(a)), in all cases w contains a length-k- overlap, which contradicts with the hypothesis.
Remark. Although the case of alphabets with more than two letters is out of the scope of the present paper, one can notice that Theorem 3.2 is no more true if we consider larger alphabets. Indeed, over a
3-letter alphabet it is possible, for every integerk ≥2, to find length-k-overlap-free words that are not length-(k−1)-overlap-free and that are generated by morphisms.
For example, let us consider the morphism
µc: (A∪ {c})∗ → (A∪ {c})∗
a 7→ ac3(k−1)b
b 7→ ba
c 7→ c
This morphism is obtained from the morphism µby adding c3(k−1) at the middle ofµ(a). Since µis an overlap-free morphism, the only overlaps in the wordµωc(a)are powers of the letterc. Therefore the wordµωc(a)is length-k-overlap-free but not length-(k−1)-overlap-free.
4. The0-limited square property
We have seen with Theorem 3.2 that the Thue-Morse wordstandt0 are the only length-k-overlap-free binary infinite words generated by morphisms, whatever be the value ofk. So it is natural to ask about the existence of length-k-overlap-free binary infinite words withk ≥2which are not length-`-overlap-free for` < k. The answer is given in the present section where a family of such words is characterized.
Before this, we have to recall some works of Thue.
In order to prove the existence of infinite cube-free words over a two-letter alphabet from square-free words over three letters, Thue used in [13] the mapping
δ : B∗ → A∗ 0 7→ a 1 7→ ab 2 7→ abb Six years later he proved the following result.
Proposition 4.1. [14] [2] Letu∈Aω andv∈Bω be such thatδ(v) =u.The worduis overlap-free if and only if the wordvis square-free and does not contain010nor212as a factor.
Thue also remarked that if the wordδ(w)is not overlap-free for a square-free wordw(thus containing 010or212) then every overlapxyxyxinδ(w)is such thatx is a single letter. Therefore, it suffices to prove the existence of a square-free ternary infinite word containing either010or212overBto obtain a length-2-overlap-free binary infinite word that is not overlap-free. Here again such a word is found in [14].
Letτ be the morphism
τ : B∗ → B∗ 0 7→ 01201 1 7→ 020121 2 7→ 0212021
Proposition 4.2. [14] The wordτω(0)is square-free and it contains212as a factor.
Now, to prove the existence, for every integerk≥2, of length-k-overlap-free binary infinite words that are not length-(k−1)-overlap-free, we generalize Thue’s idea with the following notion.
Definition 4.3. An infinite wordvoverBhas the0-limited square propertyif
• the wordvdoes not contain00as a factor,
• whenevervcontains a non-empty squarerras a factor, then, inv, the factorrris preceded (if it is not a prefix ofv) and followed by the letter 0.
Note that if a wordv ∈ Bω has the0-limited square property thenv is overlap-free and ifvcontains a non-empty squarerras a factor, the wordrdoes not begin nor end with the letter0.
The following corollary is straightforward from Proposition 4.2 because each square-free word ob- viously has the0-limited square property.
Corollary 4.4. The wordτω(0)has the0-limited square property.
Now, letk, pbe two integers withk≥2and1≤p≤k−1.We associate to(k, p)the mapping δk,p: B∗ → A∗
0 7→ ak−p 1 7→ ak−pbp 2 7→ ak−pbp+1
Of course,δ2,1=δthus our affirming that this is a generalization of Thue’s idea.
Theorem 4.5. Letu∈Aωandv∈Bωbe such thatδk,p(v) =u.If the wordvhas the0-limited square property then the worduis length-k-overlap-free.
Proof:
Suppose thatuis not length-k-overlap-free. Sinceu=δk,p(v), the following cases are possible:
• ucontains a factorakxakxak
If|x|b = 0, or ifx =zx0 with|z| ≥k−2p+ 1and|z|b = 0thenucontainsak−pak−pawhich means thatvcontains00.
Henceforth,ucontains a factoranakamx0akamx0akwithn+m+k= 2(k−p), i.e.,k−p=n+
m+p, andx0begins with the letterb. Therefore,ucontains a factoran+m+pak−px0ap+mak−px0ap+map+n, which implies thatvcontains a factor0yywhereyis such thatδk,p(y) = ak−px0ap+m(in partic-
ular,y6=ε). But in this case,ynecessarily ends with0becausep+m≥p≥1. Therefore, either yyis followed by the letter0implying thatvcontains00as a factor, oryyis not followed by the letter0.
• ucontains a factoranbp+1amxanbp+1amxanbp+1amwithn+m =k−p−1(this includes the case whereucontainsbkxbkxbkwhenp=k−1)
In this case,vcontains a factor2y2y2withδk,p(y2) =amxanbp+1.
• ucontains a factoranbpamxanbpamxanbpamwithn+m=k−p Here, two cases are possible.
1. m6= 0
Thenucontains a factorbpamxanbpamxanbpak−p, which implies thatv contains a square yy, preceded by1or2, withδk,p(y) =amxanbp.
2. m= 0(thenn=k−p)
Thenucontains a factorak−pbpxak−pbpxak−pbp, which implies thatvcontains a squareyy, followed by1or2, withδk,p(y) =ak−pbpx.
• ucontains a factorbnak−pbmxbnak−pbmxbnak−pbmwithn+m=p Here again, two cases are possible.
1. m6= 0
Then u contains a factor ak−pbmxbnak−pbmxbnak−pbp, which implies that v contains a squareyy, followed by1or2, withδk,p(y) =ak−pbmxbn.
2. m= 0(thenn=p)
Thenucontains a factorbpak−pxbpak−pxbpak−p, which implies thatvcontains a squareyy, preceded by1or2, withδk,p(y) =ak−pxbp.
In all the casesvhas not the0-limited square property.
The conditions given in Definition 4.3 are not sufficient to guarantee that the word v has the 0- limited square property whenu =δk,p(v)is length-k-overlap-free. For example, the wordv = 0τω(0) contains only one square, the factor00whichvbegins with. But, since the wordτω(0)has the0-limited square property, the wordδk,p(τω(0))is length-k-overlap-free which implies thatδk,p(v)is also length-k- overlap-free (otherwise,δk,p(v)begins with a length-k-overlap whose prefix isak−pak−pak−p, implying thatδk,p(τω(0)) contains an occurrence of this factorak−pak−pak−p from whichτω(0)contains00, a contradiction). However, it is possible to obtain an equivalence by giving conditions on the words u andv.
Corollary 4.6. Letu, an infinite word over A, which does not contain the factora2(k−p)+1, andv, an infinite word over B which does not begin with a square, be such that δk,p(v) = u. The word u is length-k-overlap-free if and only if the wordvhas the0-limited square property.
Proof:
Letuandvbe as in the statement. It is of course equivalent thatudoes not contain the factora2(k−p)+1 andvdoes not contain the factor00, thus our assuming that00is not a factor ofv.
From Theorem 4.5, it suffices to prove the necessary condition.
Letrrbe a factor ofvwithr6=ε.According to the hypothesis,rris not at the beginning ofvwhich means that inv,rris preceded (and followed) by at least one letter.
• If r begins with the letter 0 then, since 00 is not a factor of v, r does not end with 0. Thus δk,p(r) =ak−psbp.For the same reason,rris preceded by the letter1or by the letter2, soδk,p(rr)
is preceded bybp.Whatever be the letter followingrr,δk,p(rr)is followed byak−p.Consequently, ucontains the factorbpδk,p(rr)ak−p =bpak−psbpak−psbpak−p, a length-k-overlap. This implies thatuis not length-k-overlap-free.
• Ifr ends with the letter0then, sincev does not contain00as a factor, r does not begin with0, which implies thatδk,p(r)begins withak−pbp. Moreover, inv, the factorrris followed either by1 or by2. Thenucontains the factorδk,p(rr)ak−pbp =ak−pbpsak−pbpsak−pbp, a length-k-overlap.
This implies thatuis not length-k-overlap-free.
• Now if r begins with 1 or 2, and rr is not followed by 0 then δk,p(r) begins with ak−pbp and δk,p(rr)is followed byak−pbp, which means thatuis not length-k-overlap-free.
• Finally, ifrends with1or2, andrris not preceded by0thenδk,p(r)begins withak−pand ends withbp, andδk,p(rr)is preceded bybp.This implies that, sinceδk,p(rr)is followed byak−p, uis not length-k-overlap-free.
Consequently, ifuis length-k-overlap-free thenvhas the0-limited square property (remark that here the worduis also[2(k−p) + 1]-free).
Theorem 4.5 gives the first part of the answer to the question we asked at the beginning of this section by showing the existence of length-k-overlap-free binary infinite words for every integerk≥2.
It remains to prove that some wordsusatisfying Theorem 4.5 can effectively be constructed containing length-(k−1)-overlaps. This is done by using again Thue’s morphismτ.
Proposition 4.7. For every integer k ≥ 2, the word δk,k−1(τω(0)) is length-k-overlap-free but not length-(k−1)-overlap-free.
Proof:
Since from Corollary 4.4 the wordτω(0)has the0-limited square property, the wordδk,k−1(τω(0))is length-k-overlap-free from Theorem 4.5.
Now, we know from Proposition 4.2 thatτω(0)contains212as a factor. Therefore,δk,k−1(τω(0)) contains the factorδk,k−1(212) =abkabk−1abk, which implies that the length-(k−1)-overlapbk−1abk−1abk−1 is a factor ofδk,k−1(τω(0)).
5. Strongly length-k-overlap-free binary words
A length-k-overlap-free binary infinite word must contain occurrences ofa2 or b2 (or both). For if it were not the case the word would be (ab)ω or (ba)ω which obviously contains length-k-overlaps for everyk∈N.
As mentioned after Definition 3.1, the particular case of length-k-overlap-free binary infinite words withoutx3 for some letterx∈Ais of interest. We define such words as follows.
Definition 5.1. A word overAisx-strongly length-k-overlap-freeif it is length-k-overlap-free and if it does not containx3, wherexis a letter. A word is strongly length-k-overlap-freeif it isx-strongly length-k-overlap-free for every letterx∈A.
For example, the word a5 is length-2-overlap-free; it is b-strongly length-2-overlap-free, but it is not strongly length-2-overlap-free because it containsa3thus being nota-strongly length-2-overlap-free.
Notice that there exists effectively strongly length-k-overlap-free words that are not length-(k−1)- overlap-free: for example, from Proposition 4.2, the word δ(τω(0)) is strongly length-2-overlap-free without being overlap-free.
Since every strongly length-k-overlap-free binary infinite word is length-k-overlap-free and since the Thue-Morse wordstandt0are cube-free, Theorem 3.2 remains true in the present case.
Theorem 5.2. Letk∈N+and letwbe a strongly length-k-overlap-free binary infinite word. Thenwis generated by a morphism if and only ifw=torw=t0.
It is obvious thatµ(u)does not contain neithera3norb3, whatever be the value ofu. This implies that if µ(u) is a length-k-overlap-free binary infinite word then it is indeed strongly length-k-overlap-free.
Let us recall the two lemmas used by Thue to prove that the Thue-Morse wordtis overlap-free.
Lemma 5.3. LetΣ ={ab, ba}. Ifu∈Σ∗thenaua6∈Σ∗ andbub6∈Σ∗.
Lemma 5.4. A wordu∈A∗∪Aωis overlap-free if and only if the wordµ(u)is overlap-free.
The following result is an extension of Lemma 5.4.
Proposition 5.5. Letw∈A∗∪Aω and letk∈N+. The wordwis length-k-overlap-free if and only if the wordµ(w)is strongly length-(2k−1)-overlap-free.
Proof:
Ifk= 1, the equivalence is true from Lemma 5.4, thus our assuming thatk≥2.
If the wordwis not length-k-overlap-free then it contains a factorXY XY X with|X| = k. This implies that µ(w), which contains the factor µ(X)µ(Y)µ(X)µ(Y)µ(X) with |µ(X)| = 2k, is not length-(2k−1)-overlap-free.
Conversely, if the wordµ(w)is not length-(2k−1)-overlap-free then it contains a factorXxY XxY Xx whereX, Y ∈A∗,|X|= 2k−2, andx∈A.
If|Y|is even thenY 6=ε. For if notµ(w)would containXxXxXxwhich implies that bothXand xXxare inΣ∗, a contradiction with Lemma 5.3. So, letZ ∈A∗andy, z ∈ Abe such thatY =Zyz.
Then XxY XxY Xx = XxZyzXxZyzXx which implies that both X, xZy, yz, zXx, and Z are elements ofΣ∗.
From Lemma 5.3,X ∈Σ∗andzXx∈Σ∗implyx6=z, andZ ∈Σ∗andxZy ∈Σ∗ implyx6=y.
Therefore,y=zwhich contradicts withyz∈Σ∗.
Consequently, |Y| is odd so |XxY XxY Xx| is odd, and two cases are possible depending on whether, inµ(w), the factorXxY XxY Xxappears at an even index or at an odd index.
• µ(w) =µ(w1)XxY XxY Xxyµ(w2)for a lettery.
In this case, by definition ofµ, the letteryis also the first letter ofY. This implies thatXxY XxY Xxy= µ(ZY0ZY0Z)withµ(Z) =Xxy. Since|Xxy|= 2k,|Z|=kand the wordwis not length-k- overlap-free.
• µ(w) =µ(w1)yXxY XxY Xxµ(w2)for a lettery.
In this case, since k ≥ 2 one has X 6= ε, so let X = zX0, z ∈ A, X0 ∈ A+. Then yXxY XxY Xx = yzX0xY zX0xY zX0x, and by definition of µ, the letter y is also the last letter ofY. This implies thatyzX0xY zX0xY zX0x=µ(ZY0ZY0Z)withµ(Z) =yzX0x. Since
|yzX0x|= 2k,|Z|=kand the wordwis not length-k-overlap-free.
Now, we consider the mapping δk,k−1 (k ≥ 2) already used above. Since δk,k−1 is defined by δk,k−1(0) = a,δk,k−1(1) = abk−1,δk,k−1(2) = abk, it is straightforward that ifu ∈A∗∪Aω is such thatu=δk,k−1(v)for somev∈B∗∪Bωthenucontainsa3if and only ifvcontains00. Consequently, from Theorem 4.5, ifu ∈ Aω andv ∈ Bω are such thatu = δk,k−1(v)then uisa-strongly length-k- overlap-free whenevervhas the0-limited square property. We have seen above that ifk= 2, i.e., in the case of Thue’s original mappingδ, the worduis strongly length-2-overlap-free.
Now we notice that, in the case ofδk,k−1, the statement of Corollary 4.6 can be simplified because 2(k−p) + 1 = 3whenp=k−1.
Corollary 5.6. Letu∈Aω, andv, an infinite word overBwhich does not begin with a square, be such thatδk,k−1(v) = u.The word u is a-strongly length-k-overlap-free if and only if the word v has the 0-limited square property.
In this section we have seen that we obtain results similar to those given in Section 4 when adding the condition that words overAdo not contain cubes of some single letter, in particular when using the mappingδk,k−1. In the next section we give another interesting use of this mapping.
6. Length-k-overlap-free binary partial words
Apartial worduof lengthnover an alphabetAis a partial functionu:{1,2, . . . , n} →A. This means that in some positions the worducontainsholes, i.e., “do not know”-letters. The holes are represented by , a symbol that does not belong to A. Classical words (called full words) are only partial words without holes. Partial words were first introduced by Berstel and Boasson [3] (see the survey [4], and the references therein).
Similarly to finite words, we define infinite partial words to be partial functions fromN+toA. We denote byA∗andAω the sets of finite and infinite partial words, respectively.
A partial wordu∈A∗is afactorof a partial wordv∈A∗∪Aω if there exist wordsx, u0∈A∗ and y∈A∗∪Aω such thatv=xu0ywithu0(i) =u(i)whenever neitheru(i)noru0(i)is a hole.Prefixes and suffixes are defined in the same way.
For example, letu=abbbaa. The length ofuis|u|= 8, anducontains two holes in positions 3 and 7. Letv =aabbbaabbaa. The wordvcontains the worduas a factor in positions 3 and 8. The worduis a suffix of the wordv.
Note that a partial word is a factor of all the (full) words of the same length in which eachis replaced by any letter ofA.We call these (full) words thecompletionsof the partial word. In the previous example, ifA={a, b}, the partial worduhas four completions:ababbaaa,ababbaba,abbbbaaa, andabbbbaba.
