DOI:10.1051/ita:2007042 www.rairo-ita.org
ON CRITICAL EXPONENTS IN FIXED POINTS OF k -UNIFORM BINARY MORPHISMS
Dalia Krieger
1Abstract. Let w be an infinite fixed point of a binary k-uniform morphism f, and let E(w) be the critical exponent of w. We give necessary and sufficient conditions forE(w) to be bounded, and an explicit formula to compute it when it is. In particular, we show that E(w) is always rational. We also sketch an extension of our method to non-uniform morphisms over general alphabets.
Mathematics Subject Classification.68R15.
1. Introduction
Letwbe a right-infinite word over a finite alphabet Σ. Thecritical exponent of w, denoted byE(w), is the supremum of the set of exponentsr∈Q≥1, such that w contains anr-power (see Sect.2 for the definition of fractional powers). Given an infinite wordw, a natural question is to determine its critical exponent.
The first critical exponent to be computed was probably that of the Thue- Morse word,t[2,22]. This word, defined as the fixed point beginning with 0 of the Thue-Morse morphismμ(0) = 01,μ(1) = 10, was proved by Thue in 1912 to be overlap-free, that is, to contain no subword of the formaxaxa, where a∈ {0,1}
and x ∈ {0,1}∗. In other words, t is r-power-free for all r > 2; and since it contains 2-powers (squares), by our definitionE(t) = 2. Another famous word for which the critical exponent has been computed is the Fibonacci wordf, defined as the fixed point of the Fibonacci morphismf(0) = 01,f(1) = 0. In 1992, Mignosi and Pirillo [19] showed thatE(f) = 2 +ϕ, whereϕ = (1 +√
5)/2 is the golden mean. This gives an example of an irrational critical exponent. Another result we mention here is the critical exponent of the ternary Arshon sequence, generated
Keywords and phrases.Critical exponent, binaryk-uniform morphism.
1School of Computer Science, University of Waterloo, Waterloo, ON N2L 3G1, Canada;
Article published by EDP Sciences c EDP Sciences 2007
In a more general setting, critical exponents have been studied mainly with relation to Sturmian words (for the definition, properties and structure of Sturmian words, see e.g. [16] Chap. 2). In 1989, Mignosi [18] proved that for a Sturmian words,E(s)<∞if and only if the continued fraction expansion of the slope ofs has bounded partial quotients; an alternative proof was given in 1999 by Berstel [3]. In 2000, Vandeth [23] gave an explicit formula forE(s), wheresis a Sturmian word which is a fixed point of a morphism, in terms of the continued fraction expansion of the slope ofs. In particular,E(s) is algebraic quadratic. Alternative proofs for the results of Mignosi and Vandeth, with some generalizations, were given in 2000 by Carpi and de Luca [5], and in 2001 by Justin and Pirillo [11].
Carpi and de Luca also showed that 2 +ϕ is the minimal critical exponent for any Sturmian word. In 2002, Damanik and Lenz [7] gave a formula for critical exponents of general Sturmian words, again in terms of the continued fraction expansion of the slope. An alternative proof for this result was given in 2003 by Cao and Wen [4].
In this work we consider infinite words generated by iterating a morphism, also known as pure morphic sequences or D0L-words. For such words, most of the research has focused on deciding whether a given word has bounded critical exponent; see [6,8,13,20,21]. We study words generated by binaryk-uniform mor- phisms. Letf be a binaryk-uniform morphism, and letw be a fixed point off. After some preliminary definitions and results, given in Section2, we analyze in Section 3 the structure of powers occurring inw. We show that when E(w) is bounded, sufficiently large powers must have a power block divisible byk, and must be produced by a simple iterative process, which we describe. Based on this anal- ysis, we give necessary and sufficient conditions forE(w) to be bounded, and an explicit formula to compute it when it is. In particular, we show that ifE(w)<∞ then it is rational. We also show that, given a rational number 0< r <1, we can construct a binaryk-uniform morphismf suchE(fω(0)) =n+rfor some positive integern.
The results presented in this paper where generalized in [15] to non-uniform morphisms over an arbitrary finite alphabet. The main result, sketched in Sec- tion4, is that if f is uniform, then E(w) is rational when bounded; and iff is non-erasing, thenE(w) lies in the field extensionQ[λ1, . . . , λ], where λ1, . . . , λ
are the eigenvalues of the incidence matrix of f. In particular, it is algebraic of degree at most|Σ|. In [15] we also gave (under certain conditions) an algorithm for computingE(w).
Though the results of [15] are much more general, two points distinguish the results of this paper: first, we give a simple formula for computingE(w), based on only 4 iterations off; for the general case, there are no such formula and bound.
Secondly, the proof is based on very elementary tools, while the general results relies on heavy combinatorial and algebraic machinery.
This paper is an extended and corrected version of [14].
2. Preliminaries
We useZ≥r (and similarlyQ≥r,R≥r) to denote the integers (similarly rational or real numbers) greater than or equal tor.
Let Σ be a finite alphabet. We use the notation Σ∗, Σ+ and Σω to denote the sets of finite words, non-empty finite words, and right-infinite words over Σ, respectively. We use ε to denote the empty word. Infinite words are usually denoted by bold letters. For a finite wordw∈Σ∗,|w|is the length ofw, and|w|a
is the number of occurrences inwof the lettera∈Σ. For both finite and infinite words, wi is the letter at position i, starting from zero; e.g., w = w0w1· · ·wn, w=w0w1w2· · ·, wherewi∈Σ. A wordu∈Σ∗is asubwordof a wordw∈Σ∗∪Σω ifw=xuy for some wordsx∈Σ∗ andy∈Σ∗∪Σω. Ifx=ε(resp. y=ε) thenu is aprefix (resp. asuffix) ofw. A prefix (resp. suffix)uofw isproper ifu=w.
The set of subwords of a wordwis denoted by S(w).
Let w be an infinite word. Anoccurrence of a subword within w is a triple (z, i, j), wherez ∈S(w), 0 ≤i≤j, andwi· · ·wj =z. In other words,z occurs inw at positionsi, . . . , j. For convenience, we usually omit the indices, and refer to an occurrence (z, i, j) asz =wi· · ·wj. The set of all occurrences of subwords withinw is denoted byOC(w).
We use the notation to denote the relation of subword orsuboccurrence: if xand y are words, x finite and y finite or infinite, then x y stands for “xis a subword ofy”; if (x, i, j) and (y, i, j) are occurrences within a word w, then xy means thati ≤iand j≥j.
Letz=a0· · ·an−1 ∈Σ+, ai∈Σ. A positive integer q≤ |z|is aperiod of zif ai+q =aifori= 0,· · ·, n−1−q. An infinite wordz=a0a1· · · ∈Σωhas a period q∈Z≥1 ifai+q=ai for alli≥0; in this case,zisperiodic, and we writez=xω, wherex=a0· · ·aq−1. Ifzhas a periodic suffix, we say it isultimately periodic.
Afractional power is a word of the formz=xny, wheren∈Z≥1,x∈Σ+, and yis a proper prefix ofx. Equivalently,zhas a|x|-period and|y|=|z| mod|x|. If
|z|=pand|x|=q, we say that zis ap/q-power, orz =xp/q. Sinceqstands for both the fraction’s denominator and the period, we use non-reduced fractions to denote fractional powers: for example, 10101 is a 52-power (as well as a 54-power), while 1010101010 is a 104-power (as well as a 102-power). The wordxis referred to as thepower block.
Letαbe a real number. We say that a wordw(finite or infinite) isα-power-free if no subword of it is anr-power for any rationalr≥α; otherwise,wcontains an α-power. Thecritical exponent of an infinite wordwis defined by
E(w) = sup{r∈Q≥1:wcontains an r-power}. (1) By this definition, w containsα-powers for all 1 ≤α < E(w), but no α-powers forα > E(w); it may or may not containE(w)-powers.
A morphismf : Σ∗ →Γ∗ is k-uniform if|f(a)|=k for all a∈Σ, where k is a positive integer. A morphism f : Σ∗ →Σ∗ is prolongable on a letter a∈ Σ if f(a) =ax for some x∈Σ+, and furthermorefn(x) =εfor all n ≥0. If this is
fω(a) = lim
n→∞fn(a) =axf(x)f2(x)f3(x)· · ·
Moreover, for a uniform morphismf = Id, w is a fixed point of f if and only if w=fω(a) for somea∈Σ on whichf is prolongable. Such fixed points are called pure morphic sequences or D0L words. In this work we consider powers in fixed points of uniform morphisms defined over a binary alphabet Σ ={0,1}, therefore we assume thatf is prolongable on 0.
We now give three key definitions.
Definition 1. Letz∈Σ+be ap/q-power. We say thatzisreducible if it contains a p/q-power, such that p/q > p/q, or p/q =p/q and q < q. If p/q > p/q thenzisstrictly reducible.
Example 1. The 84-power 1011 1011 is strictly reducible, since it contains the
31-power 111. The 63-power 101 101 is reducible since it contains the 21-power 11.
The word 1111 is strictly reducible as a 42-power and irreducible as a 41-power.
Definition 2. Letw∈Σω, and letz=wi· · ·wj∈OC(w) be ap/q-power. We say thatz is left stretchable (resp. right stretchable) ifwi−1· · ·wj (resp. wi· · ·wj+1) is a (p+ 1)/q-power. If z is neither left nor right stretchable we say it is an unstretchable power.
Example 2. Let f : {0,1}∗ → {0,1}∗ be the morphism defined by f(0) = 01 and f(1) = 00, and letw = fω(0) = 0100 01 01 01 0 0· · ·. Then the 62-power w4· · ·w9= 010101 is right-stretchable to the 72-powerw4· · ·w10= 0101010.
SinceE(w) is an upper bound, it is enough to consider irreducible, unstretchable powers when computing it. Therefore, over a binary alphabet, we can assume that p/q≥2: since any binary word of length 4 or more contains a square, ap/q-power over{0,1}with 1< p/q <2 is always reducible, save for the 32-power 101 (010).
Definition 3. Let f be a binary k-uniform morphism. The left stretch of f, denoted byσf, is the longest wordσ∈ {0,1}∗satisfyingf(0) =xσandf(1) =yσ for some x, y ∈ {0,1}∗. Similarly, the right stretch of f, denoted by ρf, is the longest wordρ∈ {0,1}∗ satisfyingf(0) =ρx, f(1) =ρy for some x, y ∈ {0,1}∗. Thestretch size off is the combined length λf =|ρf|+|σf|.
Example 3. The morphism defined by f(0) = 0110 and f(1) = 1010 satisfies σf = 10,ρf =ε, and λf = 2.
We can now state our main theorem:
Theorem 1. Let f be a binary k-uniform morphism prolongable on 0, and let w=fω(0). Then:
(1) E(w) =∞if and only if at least one of the following holds:
(a) f(0) =f(1);
(b) f(0) = 0k; (c) f(1) = 1k;
(d) k= 2m+ 1,f(0) = (01)m0,f(1) = (10)m1.
(2) Suppose E(w) < ∞. Let E be the set of exponents r = p/q, such that q < kandf4(0) contains anr-power. Then
E(w) = max
p/q∈E
p(k−1) +λf
q(k−1)
·
In particular, E(w), when bounded, is always rational. The bound E(w) is attained if and only if λf = 0.
Here is an example of an application of Theorem1:
Example 4. The Thue-Morse word is overlap-free.
Proof. The Thue-Morse morphismμsatisfiesλμ= 0; and since the largest power inμ4(0) is a square, we get thatE(μω(0)) = 2, and the bound is attained.
The three following theorems are fundamental results in the area of combina- torics on words. They constitute the main tools we use in this paper. Theorem2 can be found in [16], Theorem 8.1.4; Theorems3and4can be found in [1], Theo- rems 1.5.2 and 1.5.3. In this setting, Σ is any finite alphabet.
Theorem 2 (Fine and Wilf [9]). Let w be a word having periods p and q, with p≤q, and suppose that |w| ≥p+q−gcd(p, q). Thenw also has periodgcd(p, q).
Theorem 3 (Lyndon and Sch¨utzenberger [17]). Let y∈Σ∗ andx, z∈Σ+. Then xy=yzif and only if there existu, v∈Σ∗ and an integere≥0such thatx=uv, z=vu, andy= (uv)eu.
Theorem 4(Lyndon and Sch¨utzenberger [17]). Letx, y∈Σ+. Then the following three conditions are equivalent:
(1) xy=yx;
(2) There exist integersi, j >0 such thatxi=yj;
(3) There existz∈Σ+ and integers k, >0 such thatx=zk andy=z. We end this section by stating a theorem that will be useful later; the proof can be found in,e.g., [1] Theorem 10.9.5. A wordw ∈Σω is recurrent if every finite subword of w occurs infinitely often. It is uniformly recurrent if for each finite subwordxofw there exists an integerm, such that every subword ofwof length mcontains x. A morphism h: Σ∗ →Σ∗ is primitive if there exists an integern such thatboccurs inhn(a) for all lettersa, b∈Σ.
Theorem 5. Let h: Σ∗→Σ∗ be a primitive morphism, prolongable on a. Then hω(a)is uniformly recurrent.
Figure 1. Applyingf towi−1· · ·wj+1.
3. Power structure
For the rest of this section, Σ ={0,1}.
Lemma 6. Let f be a binary k-uniform morphism prolongable on 0, and let w=fω(0). Let σ,ρandλ be the left stretch, right stretch, and stretch size off, respectively. Supposez=wi· · ·wj ∈OC(w)is ap/q-power. Then
E(w)≥p(k−1) +λ q(k−1) ·
Proof. If f(0) = f(1) or f(0) = 0k or f(1) = 1k, then it is easy to see that E(w) =∞. Otherwise,f is primitive, thusw is recurrent by Theorem5, and we can assume thati >0. Letp=nq+r, wheren, r∈Nandr < q, and letz=xny, where x=a0· · ·aq−1 and y =a0· · ·ar−1. Letf(wi−1) =uσ and f(wj+1) =ρv for someu, v∈Σ∗. Applyingf to wi−1· · ·wj+1, we get a subword ofw which is a fractional power with periodkq, as illustrated in Figure1.
Sinceσ is a common suffix of f(0) and f(1), it is a suffix off(aq−1) as well;
similarly,ρis a prefix off(ar). Therefore, we can stretch thekq-period off(z) by σto the left andρto the right. We get thatz =σf(z)ρis a (kp+λ)/kq-power.
The process of applyingf and stretching the resulting power can be repeated infinitely. Successive applications of f give a sequence of powers, {pm/qm}m≥0, which satisfyp0 =p,q0=q, and form >0,pm=kpm−1+λ, andqm=kqm−1. Letπ:OC(w)×Q→OC(w)×Qbe the map defined by
π
z,p q
=
σf(z)ρ,kp+λ kq
· (2)
We use π(z) and π(p/q) to denote the first and second component, respectively.
Iteratingπonp/q, we get
πm p
q
=kmp+λm−1 i=0 ki
kmq = kmp+λkkm−1−1 kmq
−−−−−→
m→ ∞ p(k−1) +λ q(k−1) · Our goal is to show that theπ map defined in (2) is what generates E(w), and that it is enough to apply it to powers that appear inf4(0). Though the details are a bit tedious, the proof idea is very simple:
(1) every p/q-power inw that satisfies q≡0 (modk) is an image under the πmap;
(2) everyp/q-power inw that satisfiesq≡0 (modk) and q > kis reducible to ap/q-power that satisfiesq ≤k;
(3) all the distinctp/q-powers inw that satisfyq < kmust appear inf4(0).
We start with a few lemmas which describe power behavior in a more general setting, namely, in an infinite word v = h(u), where h is a k-uniform binary morphism andu∈Σω is an arbitrary infinite word.
Definition 4. Lethbe a binaryk-uniform morphism, and letv=h(u) for some u∈ Σω. We refer to the decomposition of v into images of has decomposition intok-blocks. Let α=vi· · ·vj ∈OC(v). Theouter closure and inner closure of α, denoted by ˆα, ˇα, respectively, are defined as follows:
ˆ
α=vˆı· · ·vˆj, ˆı= i
k
k, jˆ= j+ 1 k
k−1;
ˇ
α=vˇı· · ·vˇj, ˇı= i k
k, jˇ= j+ 1
k
k−1.
Thus ˆα∈OC(v) consists of the minimal number ofk-blocks that containα; sim- ilarly, ˇα∈OC(v) consists of the maximal number ofk-blocks that are contained in α. By this definition, both ˆα and ˇα have inverse images under h, denoted by h−1( ˆα) andh−1( ˇα), respectively. Note that ˇαmay be empty.
Lemma 7. Let h be an injective binary k-uniform morphism, let v = h(u) for some u ∈ Σω, and let α = vi· · ·vj ∈ OC(v) be an unstretchable p/q-power.
Supposeq≡0 (modk). Then αis an image under theπ map defined in (2).
Proof. Letq=mk,m≥1. By definition, ˇı−i, j−jˇ≤k−1. Therefore,
|α|ˇ =|α| −(ˇı−i)−(j−ˇj)≥ |α| −(2k−2).
Since|α| =p andp/q ≥2, we get that|αˇ| ≥2q−(2k−2) = (2m−2)k+ 2. If m ≥2, this implies that |αˇ| > q; if m = 1, this implies that |αˇ| ≥2, and since
|ˇα| ≡0 (modk), necessarily|ˇα| ≥k=q. Thus |ˇα| ≥q, and since ˇαα, we get that ˇαhas aq-period. It follows that ˇαis ap/q-power, wherep =nkfor some n≥1. Letα =h−1( ˇα). Sincehis injective, necessarilyα is ann/m-power.
Now applyhto α. By the proof of Lemma 6, the mk-period ofh(α) can be stretched by at least|σ| to the left and |ρ| to the right, to create a power with exponent (kn+λ)/km= (p+λ)/q. On the other hand,h(α) = ˇα, and by inner closure definition, theq-period of ˇαcan be stretched by at least ˇı−i to the left and at leastj−ˇjto the right to create thep/q-powerα. But, since by assumption αis unstretchable, it cannot be stretched more than this, i.e., it can be stretched by exactly ˇı−i to the left and j−jˇto the right. Therefore, σ = wi· · ·wˇı−1, ρ=wˇj+1· · ·wj, and
α=σαρˇ =σh(α)ρ=π(α).
Lemma 8. Lethbe a binaryk-uniform morphism, letv=h(u)for someu∈Σω, and letα∈OC(v)be ap/q-power withp/q≥2. LetQbe the power block. Suppose
Figure 2. yz=zt,y=uj,|z|=i|u|.
q >2kandq≡0 (modk). Then either h(0) =h(1), orαis reducible to ap/q- power, which satisfiesq≤k. If the second case holds, we get one of the following:
(1) Q=uc for some integer c≥4 andu∈Σ+ satisfying |u|< k;
(2) p/q <5/2.
Proof. We start with three propositions that will be useful for the proof. All are extensions of the theorems of Lyndon and Sch¨utzenberger (Thms.3,4) to systems of word equations.
Proposition 9. Let x, y, z, t∈Σ+, and suppose the following equalities hold:
(1) xy=yx;
(2) yz=zt(equivalently, tz=zy);
(3) |z|=|x|.
Then x=z andy=t.
Proof. By Theorem 4, xy = yx if and only if there exists u ∈ Σ+ and integers i, j >0 such thatx=uiandy=uj. Therefore, it is enough to prove the following:
ujz=ztand|z|=i|u|implyz=ui andt=uj. We consider two cases:
i < j: in this case,|z|=|x|<|y|, and sinceyz=zt,zis a prefix (equivalently suffix) ofy. Sincey=uj and|z|=i|u|, necessarilyz=ui=x. This gives usuj+i=uit, which impliest=uj=y.
i≥j: here we use induction on |z|. If|z| = 1, then either |z|<|y|, which impliesi < j, or |y|= 1, which implies x=y =z =t ∈Σ. Let |z|>1.
Since|z| ≥ |y|, we get thaty =uj is a prefix (suffix) ofz. Let z =ujz (z=zuj). Thenujz=zt(zuj =tz) and|z|=|ui| − |uj|= (i−j)|u|.
Therefore, by the induction hypothesis, z = ui−j, and so z = ui and
t=uj.
Figure2illustrates Proposition9.
Proposition 10. Let x, y, z, t∈Σ+, and suppose the following equalities hold:
(1) xy=yz;
(2) yx=zt(equivalently, tx=zy).
Then x=z andy=t.
Proof. We prove this proposition foryx=zt. The proof fortx=zyis identical.
By Theorem3,xy=yzif and only if there existu, v∈Σ∗and an integere≥0 such thatx=uv,z=vu, andy= (uv)eu. Ife >0, theny hasuvas a prefix, and
soyx=ztimpliesuv=vu,i.e., x=z. The conditions translate toxy=yx=xt, thusy=t.
Supposee= 0. Then yx =zttranslates to uuv=vut. By Theorem3, there exist α, β ∈ Σ∗ and a c ≥ 0 such that uu = αβ, ut = βα, and v = (αβ)cα.
Therefore,uuv= (αβ)c+1α= (uu)c+1α=u2c+2α, whereαis a prefix ofuu. We get thatuuv has a|u|period, and thereforevut has a|u|period. Since |t| =|u|
(as implied by|uuv|=|vut|), we get thatt=u=y.
The equality yx = zt now translates to uuv = vuu. By Theorem 4, there exists w ∈ Σ+ and i, j > 0 such that uu = wi and v = wj. If i > 1, then uu has both periods |u| and |w| and |uu| > |u|+|w| −1, so by Theorem 2, it has a g = gcd(|u|,|w|) period; thus there exists w ∈ Σ+ such that u =w|u|/g, w=w|w|/g, andv=wj|w|/g. If i= 1, thenuu=w, andv=u2j. In any case, u andv are integral powers of the same word, which meansuv=vu, andx=z.
Proposition 11. Let x, y, z, t∈Σ+, and suppose the following equalities hold:
(1) xy=yz;
(2) zt=tx;
Then there exists u ∈ Σ+, v ∈ Σ∗, and integers i ≥ 1, j, m ≥ 0, such that x= (uv)i, z= (vu)i,y = (uv)ju,t= (vu)mv. If in addition |y|=|t|, then either v=ε andm=j+ 1 (y=t=um), or|u|=|v| andm=j.
Proof. From the first equation, we get by Theorem3 that x = rs, z =sr, and y = (rs)er for some r, s ∈ Σ∗, and an integer e ≥ 0. Plugging into the second equation, we getsrt=trs, thereforesrtr=trsr, and so by Theorem 4, sr=wi andtr=wi for someu∈Σ+ and i, i ≥1. This implies s=wav, r=uwb, and t=wmv, wherew=vu, 0≤ |v|< w,i=a+b+ 1, andi=m+b+ 1. Altogether we get:
x = rs=u(vu)b(vu)av= (uv)a+b+1= (uv)i; z = sr= (vu)i;
y = (rs)er= (uv)ieu(vu)b= (uv)ie+bu;
t = (vu)mv.
Ifv=ε, the lemma assertion holds forj =ie+b; ifv=ε, it holds forj=ie+b+1.
Now suppose that|y| =|t|. If v =ε, necessarilyy =t=um. Otherwise, the equation|(uv)ju|=|(vu)mv|impliesj=mand|u|=|v|. We now go back to the proof of Lemma 8. Let α=Qp/q ∈OC(v) be a p/q- power, wherep/q ≥ 2 and q > 2k, q ≡ 0 (mod k). Since v is an image under a k-uniform morphism, it can be decomposed into k-blocks, which are images of either 0 or 1. Assume the decomposition of α into k-blocks starts from its first character; we will show at the end of the proof that this assumption causes no loss of generality. Since q ≡0 (modk), the last k-block of the first q-block extends into the secondq-block; since the first and second q-blocks are identical, we get overlaps ofk-blocks in the secondq-block. An example is given in Figure3.
Figure 3. Overlaps ofk-blocks. The bold rectangles represent the power blocks, the light grey ones stand forh(0), the dark grey ones forh(1).
The bold rectangles denote the power’s q-blocks; the light grey and dark grey rectangles stand forh(0), h(1), respectively; the top line ofhrectangles stands for thek-decomposition ofα; and the bottom line shows the repetition of theq-block.
The fact thatq >2k implies that there are at least 5k-blocks involved in the overlap. We shall now analyze the different overlap cases. For case notation, we order thek-blocks by their starting index (the numbers 1-5 in Fig.3), and denote the case by the resulting 5-letter binary word; in the Figure3case, it is 01001. By symmetry arguments, it is enough to consider words that start with 0, therefore we need to consider 16 overlap combinations. Figure4shows these overlaps. Each overlap induces a partition on thek-blocks involved, denoted by dashed lines. We mark the k-block parts by the letters x, y, z, tin the following manner: we start by marking the leftmost part byx, and then mark byxall the parts we know are identical to it. We then mark the leftmost unmarked part byy, and so on. Since thek-decomposition starts from the first letter ofα, we have|x|=q mod k,i.e.,
|x|>0.
We begin with combinations that implyh(0) =h(1) straightforwardly.
00010, 00100, 01000, 01110: h(0) =xy=yx=h(1).
00011: h(0) =xy=yx, h(1) =yz=zt. By Proposition9,x=z andy=t,i.e., h(0) =h(1).
01001: h(0) =xy=yz,h(1) =yx=zt. By Proposition10,x=zandy=t.
00111: h(0) = xy =yz, h(1) =zt =tz. By Proposition9 (set x↔ t, y ↔ z), x=z andy=t.
01101: h(0) =xy =tz, h(1) =yz=zt. By Proposition10 (set x→t, y →x, z→y,t→z),x=z andy=t.
For the rest of the combinations, we need to consider possible continuations of theq-block. As mentioned above,q =mk+|x| for some m ≥2. Ifm ≥3, the q-block continues with anotherk-block on the bottom row; the top row continues with the k-decomposition regardless of m. Let Kt be the next k-block of thek- decomposition, and letKb be the nextk-block in the bottom row ifm≥3. Note that by the first block,Qmust end with anx.
Figure 4. Possible overlaps ofk-blocks.
00000: h(0) =xy =yx, therefore, by Theorem 4, x=ui and y =uj for some u∈Σ+,i, j∈Z≥1. Ifm= 2, orQis continued withh(0) blocks all the way, then Q= (yx)mx=um(i+j)+i=uc,c≥5; we get that thep/q-power contains at least a 5p/q-power,i.e., it is reducible withq =|u|< k. Otherwise, ifm >2 and the continuation ofk-blocks is not strictly byh(0) blocks, then at some point anh(1) block is introduced, and the behavior is similar to one of 00001, 00010, or 00011.
The latter two implyh(0) =h(1), as was shown above; the first one is discussed next.
00001: h(0) =xy=yx, h(1) =xt. By Theorem 4, x=ui and y =uj for some u∈Σ+ andi, j∈Z≥1. Thus|t|=|y|=j|u|.
Suppose m ≥ 3. If Kb = h(0), we get t = y, and so h(1) = xy = h(0); if Kb =h(1), we get h(1) =xt=tz, where |t|=|y|, therefore by Proposition9(set x↔yandz↔t)y=tandx=z, and againh(1) =xy=h(0). Therefore we can assume thatm = 2. Thus Q=yxyxx=u3i+2j =uc, c ≥5, i.e., α is reducible withq=|u|< k.
00101: h(0) = xy=yz, h(1) = zy. If we switch the roles ofh(0) and h(1), we get a mirror image of case 01011, which is proved below to be reducible.
Figure 5. Overlap 00110,Q= (h(0)h(1))(m−1)/2h(0)x.
00110: h(0) = xy =yz, h(1) =zt =tx. By Proposition11, either x=z =ui and y = t = uj, or there exist u, v ∈ Σ+ and integers i ≥ 1, j ≥ 0, such that
|u| = |v|, x= (uv)i, z = (vu)i, y = (uv)ju, and t = (vu)jv. In the first case, h(0) =h(1) =u, where=i+j; in the second,h(0) = (uv)uandh(1) = (vu)v, thush(0)h(1) = (uv)2+1.
Supposem≥3. IfKb=h(1), we gety=t, and by Proposition10,h(0) =h(1).
IfKb=h(0) andKt=h(0), we getx=z, and againh(0) =h(1). The only way to continue theq-block without forcingh(0) =h(1) is to haveh(0)h(1)h(0)h(1). . . in the bottom row andyh(1)h(0)h(1)h(0). . .in the top row. Thus we can assume thatQhas the form (h(0)h(1))m/2x(meven) or (h(0)h(1))(m−1)/2h(0)x(modd);
herem≥2.
Ifm is odd, then eitherxis a prefix of t or t is a prefix ofx(Fig. 5); since x begins with u, tbegins with v and|u|=|v|, we getu=v andh(0) =h(1). Ifm is even, thenQ= (h(0)h(1))m/2x= ((uv)2+1)m/2(uv)i = (uv)c,c ≥4, thusαis reducible withq=|uv|=|x|/i < k.
01010: h(0) = xy, h(1) = yx. Suppose m ≥ 3. The 4 possible continuations for KbKt are 00, 01, 11, 10. The first two, when combined with the last 3 k- blocks (010) yield the combinations 01000, 01001, which were shown above to implyh(0) =h(1). The 11 continuation yields the combination 01011, which will be dealt with next. The 10 continuation yields the original configuration again.
Therefore, we can assume that theq-block continues withh(1), and Q=h(1)mx.
Supposep/q≥5/2. The first two blocks have the form:
QQ= (yx)mx(yx)mx= (yx)m(xy)mxx=h(1)mh(0)mxx.
We get that eitherxxis a prefix of ak-block, orvice versa; and since thekblocks arexyandyx, eitherxis a prefix ofy ory is a prefix ofx. Ifx=y, then h(0) = h(1). If x=yy1, thenh(0) =yy1y, h(1) =yyy1, and xx=yy1yy1 =h(0)y1; we get thaty1 begins at the beginning of ak-block, thereforey1 is a prefix ofy ory is a prefix ofy1. Ify=xx1, thenh(0) =xxx1,h(1) =xx1x, and the continuation of xx implies again that xis a prefix ofx1 or vice versa. By induction, we can continue to split the k-block into shorter and shorter substrings, until finally we must get equality. We conclude that forp/q≥5/2,h(0) =h(1).
Ifp/q <5/2 then obviouslyαis reducible, since it contains the cubeh(0)3.
Figure 6. Overlap 01011,Q=h(1)h(1)x.
01100: h(0) = xy = tx, h(1) =yz = zt. This case is symmetric to the 00110 case. Use Proposition11to show that eitherh(0) =h(1) orαis reducible.
01011: h(0) = xy, h(1) = yx = xt. Suppose m ≥ 3. Similar arguments to the ones used in previous cases show that any continuation of theq-block implies h(0) =h(1). Therefore, we can assume m= 2,Q=h(1)h(1)x. This means that eitherxis a prefix oft ortis a prefix ofx(Fig.6).
By Theorem 3, there exist u, v ∈ Σ∗ and e ≥ 0 such that y = uv, t = vu, x= (uv)eu. If|x|>|y|thene >0, and sincetis a prefix ofxwe getuv=vu=wj for some j ≥1, which implies h(0) =h(1). If |x|<|y|, then x=u, i.e., it is a prefix ofy. Since theq-block ends withxxand starts withy,αcontains the cube xxx.
Supposep/q≥5/2. Sincexis a prefix of bothh(0),h(1), the thirdq-block will give us the equality tx= xy, which together with xt =yx implies h(0) = h(1).
Thush(0)=h(1) impliesp/q <5/2, andαis reducible withq =|x|< k.
01111: h(0) = xy, h(1) = yz = zy. By Theorem 4 z = ui, y = uj, therefore
|x|=|z|=i|u|.
Suppose m ≥ 3. If Kt = h(0), then for Kb = h(1) we get x = z, and so h(0) = h(1); for Kb = h(0) we get case 00011 with 0 and 1 flipped, and again h(0) = h(1). Therefore we can assume Kt = h(1). If Kb = h(0) we then get h(0) = yz = h(1). The only continuation that does not immediately force h(0) = h(1) is Kb =Kt =h(1). Thus we can assumeQ =h(1)mx=um(i+j)x, m≥2. But then, like in the 00110 case,xis a prefix of y or a prefix of yz, and since|x|=i|u|, we getx=ui=z andh(0) =h(1).
To finish the proof of the lemma, we need to justify our assumption that thek- decomposition starts from the first letter ofα. Recall that α=vi· · ·vj∈OC(v).
Suppose i ≡ 0 (mod k). Let ˇα = vˇı· · ·vjˇ be the inner closure of α, and let β=vˇı· · ·vj. Letc= ˇı−i. Thenβ ∈OC(v) is a (p−c)/q-power,βα, and the k-decomposition of β starts from the first letter. Also, by the definition of ˇαwe havec≤k−1, and sop/q−(p−c)/q≤(k−1)/q <12.
LetQbe the power block ofβ. Supposep/q≥5/2. Ifp/q≥3, then (p−c)/q≥ 5/2; by the analysis above, eitherh(0) =h(1), or Q =ud for some u∈Σ∗ and d≥4. If 5/2≤p/q <3, then (p−c)/q≥2, and by the analysis above β contains at least a cube. In both cases, α is reducible. We can therefore assume that p/q <5/2.
Letp= 2q+r, where 0 ≤r < q, and letq =mk+s, where 0≤s < k. By the theorem’s conditions,m≥2 ands≥1. Ifr≥c then β is at least a square,
Figure 7. n≥3.
Figure 8. Overlaps ofk-blocks,q= 2k+s,p= 2q+r.
and by the analysis aboveαis reducible to a cube. We can therefore assume that r < c. We can also assume that r >0, since a square of length more than 4 is always reducible.
Suppose thatm≥3. Then
p−q−c=q+r−c≥3k+ 2−(k−1) = 2k+ 3.
Therefore, the bottom row ofk-blocks contains more than 2 blocks, and altogether we have at least 5 blocks involved in the overlap. By the analysis above,β (and therefore α) is reducible to a cube. We can therefore assume that m = 2, i.e., q= 2k+s, andp−q−c= 2k+s+r−c. If s+r≥c then again we get thatβ is reducible to a cube, and so we can assume that 2≤s+r < c. Sincec < k we also get that 2k+s+r−c > k+s+r. To summarize the setting, we have:
• 2< p/q <5/2;
• q= 2k+s, p= 2q+r;
• 1≤r, s;r+s < c≤k−1;
• k+s+r < p−q−c <2k.
Again we use overlap analysis. This time we have 4 combinations to consider (Fig.8). We will use the following notation:
• K1- the white k-block in Figure8;
• K2- thek-block precedingβ.
Since|x|=s < c, the |x|letters to the right ofβ belong to αand must equalx.
ThereforeK2=wxfor some w∈Σ+.
000:h(0) =xy=yx. By Theorem4,xyx=ufor some≥3, andαis reducible.
001: h(0) =xy=yz,h(1) =zt. By Theorem3,x=uv,z=vu, andy= (uv)eu, for someu, v∈Σ∗ ande ≥0. If e >0, then xyz contains the cube (uv)3. Sup- posee = 0. If K1 = h(0), then αcontains the cube h(0)3, thus we can assume K1 =h(1). Sincex is a suffix ofK2, we get that α contains the suboccurrence xh(0)h(1) =xxyzt=uvuvuvut, which again contains the cube (uv)3.
010: h(0) = xy, h(1) = yx. Here, we do not show that α is reducible per se;
rather, we show thatv contains a cube with power block of length at mostk.
Consider possible choices forK1 andK2. IfK1=h(0), we get the cube h(0)3. AssumeK1=h(1). IfK2=h(1), we get the cubeh(1)3. Otherwise, ifK2=h(0), we get thath(0) =wx=xy, and so w=uv,x= (uv)eu,y =vu. Ife >0, then wxy(which must occur in v, being a prefix ofh(0)h(1)) contains the cube (uv)3; otherwise,h(1)h(0) =yxxy=vuuuvu, containing the cube u3.
011: h(0) =xy, h(1) =yz=zt. By Theorem3,y=uv,t=vu, andz= (uv)eu, for someu, v∈Σ∗ ande≥0. Ife >0, thenyztcontains the cube (uv)3. Suppose z=u. IfK2 =h(1), then bothh(1) =yz =wx, and since|z|=|x| we get that x=z =u. Thush(0) = uuv, h(1) =uvu, and h(1)h(0) contains the cubeuuu.
If K2 =h(0), thenh(0) = wx= xy, therefore w =ab, x= (ab)ea, and y =ba for somea, b∈Σ∗ ande≥0. Since|x|=|u|<|uv|=|y|, necessarilye= 0, and x=a. We get:
w = ab, x = a, y = ba;
y = uv, z = u, t = vu;
|a| = |u|, |b| = |v|.
Assume |b| ≤ |u|. Then y = ba =uv implies that u =bd and a =dv for some d∈Σ∗, and so
h(0)h(1) =xy zt=aba uvu=dvbdv bdvbd,
containing the cube (dvb)3. Now assume|b|>|u|. Then b=ud and v =dafor somed∈Σ∗, and
h(0)h(1) =auda udau= (aud)(6|u|+2|d|)/(2|u|+|d|).
If |d| ≤ 2|u|, then (6|u|+ 2|d|)/(2|u|+|d|) ≥5/2, making αreducible. Suppose
|d|>2|u|. Sincet=dau, andp−q−c > k+s+r, eitheraorumust be a prefix of d, depending on whether K1 =h(0) or K1 = h(1) (recall that |a| =|u| =s, thus the firstsletters of tare still part of β). If K1=h(1), thenuis a prefix of d, andh(1)h(1) contains the cubeuuu. IfK1 =h(0), then ais a prefix ofd, and h(1)h(0) contains the subwordauaua, which is a 5s/2s-power. In both cases,αis
reducible.
sufficiently long to imply easy constraints. Things are a little more subtle when dealing with short powers, especially when k < q < 2k and p/q = 2. While powers withq > 2k are always strictly reducible, and in most cases to a power with a much bigger exponent, this is not necessarily the case for small powers.
Consider the following example of a 10-uniform morphism:
h(0) = 0110010110 h(1) = 1001011010
h(101) = 1001011010 0110010110 1001011010.
The prefix of length 24 ofh(101) is a square with block size 12, which contains no overlaps; thus it is not strictly reducible, though it does contain squares of block size smaller thank. This example demonstrates that we need to be more careful when analyzing powers withk < q <2k.
Definition 5. Lethbe a binary k-uniform morphism, and letv =v0v1v2· · · = h(u) for someu∈Σω. Let 0< i < k. Thek-partition of thei-shift ofv, denoted byTi,k(v) is the partition ofvivi+1vi+2· · · intok-blocks.
Letα=vi· · ·vj ∈OC(v) be a p/q-power, and suppose i≡0 (mod k). Con- siderαas an occurrence of Ti,k(v). In general,Ti,k(v) can have up to 4 different k-blocks. However, if there are only two composing α, we can do the overlap analysis ofα using the blocks of Ti,k(v). Clearly, if α is reducible with respect to Ti,k(v) then it is also reducible with respect to v. If the k-blocks ofTi,k(v) are proved to be equal it does not necessarily imply thath(0) =h(1); however, it would still imply thatαis reducible.
Lemma 12. Let hbe a binary k-uniform morphism, let v=h(u) for some u∈ Σω, and let α = vi· · ·vj ∈ OC(v) be a p/q-power with p/q ≥ 2. Let Q be the power block. Supposek < q <2k. Then either h(0) =h(1), or αis reducible to a p/q-power, which satisfiesq≤k.
Proof. Letq=k+s, 1≤s≤k−1, and letp=nq+r=nk+ns+r, 0≤r≤q−1.
Letβ=vˇı· · ·vj, c= ˇı−i, and letQ be the power block ofβ. Ifp−q−c≥2k, then there are at least 5k-blocks involved in the overlap ofβ. Using the same analysis applied in the previous lemma, we can show that either h(0) =h(1), or Q = u for some ≥ 4 and u ∈ Σ+ satisfying |u| ≤ k. The two cases where this assertion does not hold are 01010 and 01011, but these cases cannot hold for k < q <2k(see Fig.9). Sinceβis a (p−c)/q-power andc/q≤(k−1)/(k+ 1)<1, we get thatαis also reducible.
Supposep−q−c= (n−1)k+ (n−1)s+r−c <2k. Then we get:
(n−1)k+n≤(n−1)k+ (n−1)s+r <2k+c≤3k−1.
Therefore,n≤(4k−2)/(k+ 1)<4.
Figure 9. Overlaps ofk-blocks,k < q <2k. Thek-block marked by ‘*’ cannot be anh(1) block, or the period will be violated.
Figure 10. Overlaps ofk-blocks,k < q <2k,n= 3.
Supposen= 3 and p−q−c= 2k+ 2s+r−c <2k. Then 2s+r < c≤k−1, i.e.,s < k/2. Also,p−q−c≥k+ 2s+r+ 1, and so there are 3k-blocks involved in the overlap. The cases are illustrated in Figure10. We denote byKthek-block preceding β (the white block in Fig. 10). Note that in all cases |x| < |y|, since
|x|=s < k/2, and|x|+|y|=k.
000: h(0) =xy = yx. By Theorem 4, h(0)h(0) = u for some ≥ 4, and αis reducible.
001: h(0) = xy =yz, h(1) = zt. Since |x| < |y|, we get that x=uv, z = vu, and y = (uv)eu for some u, v ∈ Σ∗ and e ≥ 1. If e > 2, then xyz contains the 4-power (uv)4, makingαreducible. Ife= 1, then|t|=|y|=|uvu|=s+|u|<2s, and so all of the h(1) block is included in β, implying that t = y. Consider K: since |x| = s < c, the last |x| letters of K belong to α, and must equal x.
Since bothh(0) andh(1) end with y, and|y|>|x|, necessarily xis a suffix ofy.
Therefore, the wordyxyzcontains the wordxxyz= (uv)4u. Again,αis reducible.
010: h(0) =xy,h(1) =yx. Also, the last block on the second row implies thatx is a prefix ofy. Letybe the suffix ofyof lengths. ConsiderK: as in the previous case,xis a suffix ofK. IfK=h(0) =xy, theny =x. If K=h(1) =yx, then, sincec >2s, the first power block ofβ must end withyx, and againy =x. We get thatxis both a prefix and a suffix ofy, and so yxxy containsx4, makingα reducible.
Figure 11. Overlaps ofk-blocks,k < q <2k,n= 2.
011: h(0) =xy,h(1) =yz =zt. Since|z|<|y|, we get thaty =uv,z =u, and t = vu for some u, v ∈ Σ∗. Suppose that K =h(1). Then x is a suffix of h(1) thereforex= z =u. Also, the last block on the second row implies thatx is a prefix oft=vu. Therefore,zxyz=uuuvucontainsu4.
Now assume thatK=h(0). Thenαh(0)h(1)h(0)h(1), and contains at most two differentk-blocks when considered as a subword ofTi,k(v): the block resulting from h(0)h(1), and the block resulting from h(1)h(0). Using theTi,k(v) decom- position, we get 5 blocks involved in the overlap ofα, making it reducible.
Finally, suppose thatn = 2. The cases are illustrated in Figure 11. We denote by K1 the k-block that continues the k-decomposition on the first row, and by K2 the k-block preceding it. Figure 11 depicts powers that start at an index i≡0 (modk), but we also consider the case whereαbegins at some indexi, and p−c−q≥k.
000: αcontainsh(0)3, and must be reducible.
001: h(0) =xy=yz, and so x=uv, z=vu, andy= (uv)eufor someu, v∈Σ∗ ande≥0. If|x|<|y|(as illustrated in Fig.10), thene≥1, andxyz contains the cube (uv)3. Otherwise,|x|<|y|andh(1) =zy(as illustrated in Fig.11). We get thath(0) =uvu,h(1) =vuu, and h(1)h(0) contains the cubeu3.
Ifαstarts at some indexi≡0 (modk), considerK2: if K2=h(0), we get the cubeh(0)3. IfK2=h(1), assumep−c−q≥k,i.e., the whole of thezpart of the h(1) block is included inα. Thenh(1) =vutfor somet, and by comparingK2 to x, we get thatt=v. Thereforeh(0) =uvu,h(1) =vuv, andh(0)h(1) = (uv)3. 010: h(0) =xy,h(1) =yx. Also, the lastk-block in the second row implies that eitherxis a prefix ofyorvice versa. Ifx=yafor somea∈Σ∗, thenh(0) =yay, h(1) = yya, andh(0)h(1) contains the cube y3; if y =xb for some b∈ Σ∗, then h(0) = xxb, h(1) = xbx, and h(1)h(0) contains the cube x3. In any case, α is reducible.
Ifαstarts at some indexi≡0 (mod k) andp−c−q≥k, we still geth(0) =xy andh(1) =yx. IfK2=h(0), thenyis a suffix ofxorvice versa, and again we get a cube. IfK2=h(1), then eitherαis the square (xyx)2 (in which case it must be reducible), or again eitherxis a prefix (suffix) ofy orvice versa.
