International audienceFor all α ≥ RT (3) (where RT (3) = 7/4 is the repetition threshold for the 3-letter alphabet), there exists an infinite word over 3 letters whose critical exponent is α

Vaslet, Elise

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Critical exponents of words over 3 lettersElise VasletInstitut de Mathe´matiques de Luminy, Universite´ Aix-Marseille IIMarseille, Francevaslet@iml.univ-mrs.frSubmitted: Oct 5, 2010; Accepted: May 25, 2011; Published: Jun 6, 2011Mathematics Subject Classification: 68R15AbstractFor all α ≥ RT (3) (where RT (3) = 7/4 is the repetition threshold for the 3-letteralphabet), there exists an infinite word over 3 letters whose critical exponent is α.1 IntroductionLet A be a finite alphabet. Any finite word v over A, v 6= ǫ, can be factorized as v = pke,where:- k ≥ 1- e is a prefix of p- |p| is minimalWe then say that v has period p, excess e, and exponent E(v) = |v|/|p|. For example,the English word church has period chur, excess ch, and exponent 3/2, while the Frenchword entente has period ent, excess e, and exponent 7/3. A (finite or infinite) word issaid to be α-free (resp. α+-free) if none of its subwords has exponent β, with β ≥ α (resp.β > α).The critical exponent of an infinite word w over A is defined asEc(w) = sup{E(v) ∈ Q, v subword of w}.For example, the binary word abab2 · · ·abn · · · has critical exponent +∞. The Thue-Morseword, fixed point of the morphism 0 7→ 01, 1 7→ 10, has critical exponent 2 ([10] and [1]).The Fibonacci word, fixed point of the morphism 0 7→ 01, 1 7→ 0, has critical exponent2 + φ, where φ is the golden number [8].The problem to determine if, for a given real number α > 1, there is an infinite wordwα with critical exponent α, has been solved by Krieger and Shallit [7]. The number ofthe electronic journal of combinatorics 18 (2011), #P125 1letters they use to construct the word wα grows very fast as α tends to 1, and they leftopen the construction of wα over an alphabet with a fixed size.Let k be a natural number, and let Ak be the k-letter alphabet. The repetitionthreshold on k letters is the real number (see [6] and [3] for more details)RT (k) = inf{Ec(w), w ∈ Aωk}.Dejean [6] conjectured thatRT (k) =2 if k = 27/4 if k = 37/5 if k = 4k/(k − 1) if k > 4and this conjecture is now proved (see [9] and [5]).It is clear that if α < RT (k), no word over Ak has critical exponent α. Currie andRampersad [4] proved the following result for a binary alphabet:For each α ≥ 2 = RT (2), there is an infinite binary word with critical exponent α.And they conjectured:Let k ≥ 2. For each α ≥ RT (k), there is an infinite word over k letters with criticalexponent α.We will prove that this is true for k = 3. Let A3 = {a, b, c} be the 3-letter alphabet.Dejean [6] proved that RT (3) = 7/4. We immediately remark that if α ≥ 2, the result ofCurrie and Rampersad gives us a binary word, and then also a word over A3, with criticalexponent α. That’s why we only have to consider the case 7/4 ≤ α < 2. To demonstrateher result, Dejean considered the morphism µ (which we will call Dejean’s morphism)defined as follows:µ :a 7→ abc acb cab c bac bca cbab 7→ π(µ(a)) = bca bac abc a cba cab acbc 7→ π2(µ(a)) = cab cba bca b acb abc bacwhere π is the permutation (a b c), and proved that its fixed point µ∞(a) has criticalexponent 7/4.2 Exponents and Dejean’s morphismDejean [6] noticed the existence of specific subwords in µ(a), µ(b), and µ(c), which canbe used to desubstitute µ. She called them characteristic factors:the electronic journal of combinatorics 18 (2011), #P125 2Proposition 1. The words f1 = abcacbc, f2 = cabcbac and f3 = cbcacba only appear inµ(A∗3) respectively as prefix, central factor, and suffix, of µ(a).Similarly, π(f1), π(f2), π(f3) only appear as prefix, central factor, and suffix of µ(b), andπ2(f1), π2(f2), π2(f3) as prefix, central factor, and suffix of µ(c).Definition 1. The words f1, f2, and f3 (resp. π(f1), π(f2), π(f3) ; resp. π2(f1), π2(f2),π2(f3)) are called characteristic factors of µ(a) (resp. µ(b); resp. µ(c)).We use these characteristic factors to prove the following desubstitution results for µ.Proposition 2. Let w be a word over A3, and u a subword of µ(w). If |u| ≥ 18, then,there exists a unique x ∈ A3, a unique y ∈ A3, and there exist some unique s, v, p ∈ A∗3,such that u = sµ(v)p, where s 6= ǫ is a suffix of µ(x), and p 6= ǫ is a prefix of µ(y).Proof. As |u| ≥ 18, u has a characteristic factor as a subword. The result is then clearsince µ is a 19-uniform morphism, and since ∀x ∈ A3, µ(x) begins and ends with x.Theorem 1. Let w ∈ A∗3, and let u be a subword of µ(w). Assume u has period p, excesse and exponent 7/4 < |u|/|p| < 2. Then, w has a subword v of length |v| ≥ ⌈|u|/19⌉, withperiod q such that |q| = ⌈|p|/19⌉, and with exponent E(v) ≥ |u|/|p|.Proof. There are two cases: either |e| ≥ 18, or |e| < 18.Suppose first that |e| ≥ 18. Then, we also have |u| ≥ 18. Then, by Proposition 2,- There exist some unique xu, yu ∈ A3, and some unique mu, su, pu ∈ A∗3, su 6= ǫ suffixof µ(xu), pu prefix of µ(yu), such that u = suµ(mu)pu,- There exist some unique xe, ye ∈ A3, and some unique me, se, pe ∈ A∗3, se 6= ǫ suffixof µ(xe), pe prefix of µ(ye), such that e = seµ(me)pe.Let:f = xemeyev = xumuyuAs e is a suffix of u, we have: pu = pe, yu = ye, and µ(xe)µ(me) is a suffix ofµ(xu)µ(mu), thus xeme is a suffix of xumu. Moreover, e is a prefix of u, so su = se,xe = xu, µ(me)µ(ye) is a prefix of µ(mu)µ(yu), and meye is a prefix of muyu (see thefollowing figure).Therefore, f is a prefix and a suffix of v, and v has excess f (|f | is maximal, otherwise|e| would not be maximal in u). Let us denote its period by q, i.e. v = qf .It is clear that|v| ≥ ⌈|u|19⌉.the electronic journal of combinatorics 18 (2011), #P125 3u =µ(xu)µ(mu)µ(mu) µ(yu)su puµ(me)µ(me)µ(me)µ(me)pe pese seµ(xe) µ(xe) µ(ye)µ(ye)eeMoreover, q has length|q| = |v| − |f | = |mu| − |me|=|µ(mu)| − |µ(me)|19=(|u| − |su| − |pu|)− (|e| − |se| − |pe|)19=|u| − |e|19(because su = se and pu = pe)=|p|19Finally, v has exponentE(v) =|v||q|≥ ⌈|u|19⌉/|p|19≥|u|19.19|p|= E(u)Thus, v is the word we were looking for.Suppose now that |e| < 18. Then, as E(u) > 7/4, |u| ≤ 24 + 18 = 42. As µ is 19-uniform, u is a subword of a word µ(x), where x ∈ A∗3, |x| ≤ 4. Moreover, E(u) > 7/4, sox is not a subword of µ∞(a). Then, by looking at the µ(x) obtained if x is not a subwordof µ∞(a), we can again reduce the set of possible x:x ∈ {aa, bb, cc, aaa, bbb, ccc, aaaa, bbbb, cccc, abab, acac, baba, bcbc, caca, cbcb}since in the other cases, µ(x) has no subword u such that |u| ≤ 42 and E(u) > 7/4. Then,we have:- if u is a subword of µ(x), with x ∈ {aa, bb, cc}, consider v = x. v has exponent 2,and period q of length |q| = 1. We can also remark that u necessarily has a periodof length 19 = 19.|q|, and has exponent E(u) ≤ 2 = E(v). Therefore, v is the wordwe were looking for.the electronic journal of combinatorics 18 (2011), #P125 4- if u is a subword of µ(x), with x ∈ {aaa, bbb, ccc}, consider v = x. v has exponent 3,and period q of length |q| = 1. As u necessarily has a period of length 19 = 19.|q|,and has exponent E(u) ≤ 3 = E(v), v is the word we were looking for.- if u is a subword of µ(x), with x ∈ {aaaa, bbbb, cccc}, consider v = x. v has exponent4, and period q of length |q| = 1. As u necessarily has a period of length 19 = 19.|q|,and has exponent E(u) ≤ 4 = E(v), v is the word we were looking for.- finally, if u is a subword of µ(x), with x ∈ {abab, acac, baba, bcbc, caca, cbcb}, considerv = x. v has exponent 2, and period q of length q = 2. As u necessarily has a periodof length 2.19 = 19.|q|, and has exponent E(u) ≤ 2 = E(v), v is the word we werelooking for.3 Construction of an infinite α-free word over A3In the following, we will use the operator, denoted by δ, that removes the first letter of aword: for example, δ(0110) = 110.Lemma 1. Let L be the set Fact(µ(A∗3)) of all subwords of the words in µ(A∗3). Letα ∈]7/4, 2[ and v ∈ A∗3, such that:- abcbabcv ∈ L- abcbabcv is α-free.Suppose that babcbabcv = xuy, where u has exponent E(u) ≥ α. Then, x = ǫ, andu = babcbabc.Proof. By hypothesis, abcbabcv is α-free. Since E(u) ≥ α, u is necessarily a prefix ofbabcbabcv. Moreover, babcbab has exponent 7/4 < α ≤ E(u). Therefore, u = babcbabcv′,where v′ is a prefix of v. Suppose that v′ 6= ǫ.By hypothesis, abcbabcv′ ∈ L. Moreover, the subword abcbabc only appears in L as asubword of µ(c). So v′ = abacbabcbac....The excess of u is at most babcbab. Indeed, otherwise, the word babcbabc, whose exponent is2, is a subword of abcbabcv, which is impossible since abcbabcv is α-free. Then, u has excesse, with |e| ≤ 7, and so has period p with |p| ≤ 9. So u is a subword of babcbabcabacbabc. Bylooking at the factors of this word, we deduce that the only possibility is u = babcbabc.Lemma 2. Let α ∈]7/4, 2[ be given. Let s, t be natural numbers such that µs(b) =xabcbabcy, with |x| = t. Let β = 2 − t4.19s. Suppose that 7/4 < β < α, and thatabcbabcv ∈ L is α-free. Consider the word w = δtµs(babcbabcv). Then, we have:1. w has a prefix with exponent β.the electronic journal of combinatorics 18 (2011), #P125 52. If abcbabcv has a subword with exponent γ and period p, then, w has a subword withexponent γ and a period of length 19s|p|.3. w is α-free.Proof. 1. µs(babcbabc) has exponent 2 and period µs(babc). We have |µs(babc)| = 4.19s,and |µs(babcbabc)| = 8.19s, so the prefix δtµs(babcbabc) of w has exponent|w||µs(babc)|=|µs(babcbabc)| − t|µs(babc)|= β2. Let u be a subword of abcbabcv, with exponent γ and period p. Then µs(u) is asubword of µs(abcbabcv), with exponent γ and period 19s|p|. Moreover, µs(abcbabcv)is a suffix of δtµs(babcbabcv), since t = |x| ≤ |µs(b)|. So µs(u) is a subword of w,with the required properties.3. Suppose that w has a subword u, with exponent κ ≥ α and period p. Then, byiteration of Theorem 2, babcbabcv has a subword u′ with exponent κ′ ≥ κ and qsuch that |q| = |p|19s. By Lemma 1, as κ′ ≥ α, we deduce that κ′ = 2 and thatu′ = babcbabc. Then q = babc, and |p| = |q|.19s = 4.19s.Moreover, u is not a subword of µs(abcbabcv), otherwise abcbabcv has a subword withexponent ≥ α. u is not a subword of δtµs(babcbabc) either, otherwise, we would have|u| ≤ |δtµs(babcbabc)| = 8.19s − t, and so:E(u) =|u||p|≤8.19s − t4.19s= β < α,which is absurd since E(u) ≥ α.Therefore, u has a prefix z such that z = z1µs(abcbabc)z2, where z1 6= ǫ is a suffixof µs(b), and z2 6= ǫ is a prefix of µs(a) (since we remarked that the first letter ofv is a a). z being a prefix of u, z has a period of length 4.19s. Then, z has periodz1µs(abc)z′1, with z′1 such that µs(b) = z′1z1. So we have:z = z1µs(abc)z′1z1µs(abc)z2.We deduce that either z2 is a prefix of z′1, or z′1 is a prefix of z2. Yet neither ispossible. Indeed, z2 begins with the letter a, and z′1 begins with the letter b.Finally, w is α-free.4 A word over A3 with critical exponent α ≥ RT (3)Definition 2. A real number β < α is said to be obtainable if β can be written asβ = 2− t4.19s, where the natural numbers s and t verify:- s ≥ 3the electronic journal of combinatorics 18 (2011), #P125 6- the word δt(µs(b)) begins with abcbabc.We note that for any given s ≥ 3, it is possible to choose t such that- 7/4 < β = 2− t4.19s< α- |α− β| ≤ 1924.19sIndeed, µ2(a), µ2(b), and µ2(c) have length 192, and each have abcbabc as a subword.Therefore, choosing a large enough s, we can always find some obtainable real numbersβ, arbitrarily close to α.Theorem 2. Let α ≥ RT (3) = 7/4. Then, there is an infinite word over A3 with criticalexponent α.Proof. If α = 7/4, we already know that µ∞(a) has critical exponent 7/4. If α ≥ 2, bythe theorem for k = 2, we can find a word over A∗3 with critical exponent α. Now, letα ∈]7/4, 2[.Let (βi)i∈N be an increasing sequence of obtainable numbers, converging to α. For eachi, we write βi as:βi = 2−ti4.19siwhere si and ti are such that:- si ≥ 3- δtiµsi(b) begins with abcbabc.For all words v ∈ L, let Φi(v) denote the word δtiµsi(bv), and consider the followingsequence:v1 = Φ1(abcbabc) = δt1µs1(babcbabc)v2 = Φ1Φ2(abcbabc) = δt1µs1(bδt2µs2(babcbabc))v3 = Φ1Φ2Φ3(abcbabc)...vn = Φ1Φ2Φ3...Φn(abcbabc)...By iteration of Lemma 2, as abcbabc ∈ L is α-free, we deduce that each vi is α-free.Moreover, once again by Lemma 2, each vi has a subword with exponent βj, j = 1, 2, ..., i.Finally, consider the word w = limn→∞ ∈ Aω3 (it is possible to take this limit since eachvi is a prefix of vi+1). w then has critical exponent α: it is α-free, yet has subwords withexponents βi converging to α.the electronic journal of combinatorics 18 (2011), #P125 7The conjecture proposed by Currie and Rampersad in [4] is then true for alphabets of size2 and 3. It still have to be proved for alphabets of size ≥ 4. For that, another methodmust be found, because of Brandenburg’s result in [2] : if k ≥ 4, there is no RT (k)-freemorphism, i.e., no morphism which maps, as Thue-Morse morphism for k = 2 or Dejean’smorphism for k = 3, every RT (k)-free word to an RT (k)-free word.References[1] J. Berstel. Axel Thue’s papers on repetitions in words: a translation. Publicationsdu LaCIM, 20, 1995.[2] F. J. Brandenburg. Uniformly growing k-th powerfree homomorphisms. Theoret.Comput. Sci., 23:69–82, 1983.[3] A. Carpi. On Dejean’s conjecture over large alphabets. Theoret. Comput. Sci.,385:137–151, 2007.[4] J. D. Currie and N. Rampersad. For each α > 2 there is an infinite binary word withcritical exponent α. Electron. J. Combin., 15:N34, 2008.[5] J. D. Currie and N. Rampersad. A proof of Dejean’s conjecture. Preprint,http://arxiv.org/abs/0905.1129, 2009.[6] F. Dejean. Sur un the´ore`me de Thue. J. Comb. Theory A, 13:90–99, 1972.[7] D. Krieger and J. Shallit. Every real number greater than 1 is a critical exponent.Theoret. Comput. Sci., 381:177–182, 2007.[8] F. Mignosi and G. Pirillo. Repetitions in the Fibonacci infinite word. RAIRO Inform.Theor. Appl., 26:199–204, 1992.[9] M. Rao. Last cases of Dejean’s conjecture. WORDS 2009,http://www.labri.fr/perso/rao/publi/dejean.ps, 2009.[10] A. Thue. U¨ber die gegenseitige Lage gleicher Teile gewisser Zeichenreihen. NorskeVid. Selsk. Skr. Mat. Nat. Kl., 10:1–67, 1912.the electronic journal of combinatorics 18 (2011), #P125 8

Critical Exponents of Words over 3 Letters

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Critical Exponents of Words over 3 Letters

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