There are k-uniform cubefree binary morphisms for all k >= 0

THERE ARE k -UNIF ORM CUBEFREE BINAR Y MORPHISMS F OR ALL k ≥ 0 JAMES CURRIE AND NARAD RAMPERSAD Abstract. A w ord is c ub efree if it co n tains no non-empty subw ord of the for m xxx . A morphism h : Σ ∗ → Σ ∗ is k -uniform if h ( a ) has length k for all a ∈ Σ. A mo rphism is cube fr ee if it maps cub efree words to cub efree words. W e show tha t for a ll k ≥ 0 there exists a k -uniform cubefr e e binary morphism. 1. Introduction A squar e is a non- empt y w ord of the form xx , and a cub e is a non-empt y w ord of the form xxx . An ov erlap is a w ord of the form axaxa , whe re a is a letter a nd x is a w ord (p ossibly empty ). A word is squar efr e e (resp. cub e fr e e , overlap-fr e e ) if none o f its factors are squares (resp. cu b es, ov erlaps). The construction of infinite squarefree, cub efree, and o v erlap-free w o r ds is typically done by iterating a suitable morphism. Unifo rm morphisms ha v e particularly nice prop erties. In this note w e sho w tha t for all k ≥ 0 there exists a k -uniform cub efree binary morphism. Let Σ ∗ denote the set of a ll finite w ords ov er the alphab et Σ. A morphism h : Σ ∗ → Σ ∗ is k - uniform if h ( a ) has length k for all a ∈ Σ; it is uniform if it is k -uniform for some k . A morphism h : Σ ∗ → Σ ∗ is squar efr e e (resp. cub efr e e , overlap-fr e e ) if h ( w ) is squar efr e e (resp. cub efr e e , overlap-fr e e ) wh enev er w ∈ Σ ∗ is squar efr e e (r esp. cub efr e e , overlap-fr e e ). Squarefree, cub efree, and ov erlap-free morphisms hav e b een studied extens iv ely [3, 4, 5, 6, 7, 8, 9, 10, 11, 12]. W e denote the T hue–Morse morphism b y θ : θ (0) = 01 θ (1) = 10 . The T hue–Morse w or d is the infinite fixed p oint of θ : t = θ ω (0) = 011010 0110010110 · · · It is w ell-known that the Thue–Morse w ord is ov erlap-free [1 4]. Moreov er, the Th ue–Morse morphism is b oth ov erlap-free and cub efree (see [5, 13] for ev en stronger resu lts). Berstel and S ´ e ´ eb old [4] g av e a remark able c haracterization of o verlap-free binary morphisms: namely , that a binary morphism h is ov erlap-free if and only if h (01101001 ) is o v erlap-free. F urther- more, they sho w ed that if h is an o v erlap-free binary morphism then h is a p o we r of θ (o r its complemen t). Th us an y ov erlap-free binary morphism is k - uniform where k is a p o w er of 2. Date : Aug ust 10, 2018. 2000 Mathematics Subje ct Classific ation. 68R15. The first author is supp orted b y an NSERC Disco very Gran t. The second author is supp o rted by a n NSER C Postdoctor al F ellowship. 1 It is natural to inquire if cub efree binary morphisms exhibit similar b eha viour. In this case the answ er is no , as we are able to construct uniform binary mor phisms o f ev ery length. F or further bac kground material concerning com binatorics on w ords we refer the reader to [2]. 2. Main resul t The main result of this note is tha t for all k ≥ 0 there exists a k -uniform cub efree bina r y morphism. W e b egin with some preliminary lemmas. Lemma 1. L et k ≥ 4 b e an inte ger. Then the Thue–Morse wor d t c ontains two distinct wor ds of length k of the form 0 y 0 and two distinct w or ds of length k of the f orm 0 z 1 . Pr o of. F or k = 4 , 5 , 6 the following table gives the required pairs of sub w ords. k = 4 (0010 , 0100) (0101 , 0011) k = 5 (00110 , 01100) (01101 , 01001) k = 6 (001100 , 011010) (001011 , 0 1 0011) Supp ose then that k > 6. If k is eve n, let k = 2 r ; otherwise, let k = 2 r − 1. Supp ose inductiv ely that t con tains t wo distinct w ords 0 y 0 and 0 y ′ 0 of length r and t w o distinct w or ds 0 z 1 and 0 z ′ 1of length r . If k is ev en then the w or ds 01 θ ( y )01, 01 θ ( y ′ )01, and 01 θ ( z ) 1 0, 01 θ ( z ′ )10 are the desired w or ds of length k . If k is o dd then the w o rds 01 θ ( y )0, 01 θ ( y ′ )0, and 01 θ ( z )1 , 01 θ ( z ′ )1 are the desired w ords of length k .  The pro of of the follo wing lemma essen tially follo ws that of [1, Lemma 4]. Lemma 2. L e t k ≥ 7 b e an inte ger. Then t c ontains two distinct subw or ds of leng th k of the form 01 x 01 and two distinct subwo r ds of length k of the fo rm 0 1 x 10 . Pr o of. W e o nly g ive the details for 01 x 01, the pro of for 01 x 10 being analogous. If k is ev en, let k = 2 r . W e ha v e r = k / 2 ≥ 4, so that t contains distinct words u = 0 v 0 a nd u ′ = 0 v ′ 0 of length r by Lemma 1 . The words θ ( u ) = 01 θ ( v )01 and θ ( u ′ ) = 01 θ ( v ′ )01 are therefore w ords of the required fo rm of length k . If k is o dd and k ≥ 23, w e can write k as 8 r − 9, 8 r − 7, 8 r − 5 or 8 r − 3 for some r ≥ 4. Let u = 0 v 0 and u ′ = 0 v ′ 0 b e distinct words of length r in t . The word θ 3 ( u ) = 01101001 θ 3 ( v )01101001 con t ains words 01 x 01 o f lengths 8 r − 9 (including the first and second underlined 01 ’s) and 8 r − 3 (including the first and third underlined 01’s.) Similarly , the word θ 3 ( u ′ ) = 0110100 1 θ 3 ( v ′ )0110100 1 con t ains w ords 01 x ′ 01 of lengths 8 r − 9 and 8 r − 3. Moreo v er, since v 6 = v ′ , these w ords are distinct f rom the corresp onding sub w o rds of θ 3 ( u ). Let z = 0 v 1 and z ′ = 0 v ′ 1 b e distinct words of length r in t . The word θ 3 ( z ) = 01 101 001 θ 3 ( v )10010110 2 con t ains words 01 x 01 o f lengths 8 r − 7 (including the first and second underlined 01 ’s) and 8 r − 5 (including the first and third underlined 01’s.) Similarly , the word θ 3 ( z ′ ) = 0110100 1 θ 3 ( v ′ )1001011 0 con t ains w ords 01 x ′ 01 of lengths 8 r − 7 and 8 r − 5. Moreo v er, since v 6 = v ′ , these w ords are distinct f rom the corresp onding sub w o rds of θ 3 ( z ). F or k o dd, 7 ≤ k ≤ 21 , the following table giv es the required pair s of sub w ords. k = 7 0100101 0101101 k = 9 01001100 1 01100110 1 k = 11 0100110 0101 01100101 101 k = 13 0100101 101001 01101001 01101 k = 15 0110010 11001101 01001100 1011001 k = 17 0100101 1001101001 01101001011 0 01101 k = 19 0100101 10011010010 1 010110100110 0101101 k = 21 0110100 11001011001 1 01 01100110 100110010110 1  Lemma 3. L et k ≥ 9 b e a n inte ger. T hen ther e exist two distinct cub efr e e wor ds of length k of the f orm 00 x 11 . Pr o of. F or 9 ≤ k ≤ 14, the follow ing ta ble giv es the required pairs of sub w ords. k = 9 00100101 1 00101001 1 k = 10 0010011 011 0010110011 k = 11 0010011 0011 00101001 011 k = 12 0010010 10011 001001011011 k = 13 0010010 110011 00100110 01011 k = 14 0010010 1001011 00100101 101011 Supp ose k ≥ 15. If k is ev en, let k = 2 r − 2; otherwise, let k = 2 r + 1. Note that r ≥ 7, so by Lemma 2, there ar e distinct sub w ords 01 x 10 and 01 x ′ 10 of t o f length r . If k is ev en, then the complemen ts o f the w ords 0 − 1 θ (01 x 10)1 − 1 = 110 θ ( x )100 and 0 − 1 θ (01 x ′ 10)1 − 1 = 110 θ ( x ′ )100 ar e cub efree words of the desired fo rm of length k . If k is o dd, then let u = 11 θ (01 x 10)1 − 1 = 110 110 θ ( x )100 and u ′ = 11 θ (01 x ′ 10)1 − 1 = 110110 θ ( x ′ )100. W e claim that u and u ′ are cub efr ee. Supp ose to the contrary t ha t u con t ains a cub e. Since 0110 θ ( x )100 is o v erlap- free, any suc h cub e w ould ha v e to start with either the first or second 1, but in either case, b y insp ection the p eriod of the cub e is at least 3, whic h forces an ov erlap in 0110 θ ( x )100, a con tradiction. Similarly , u ′ is cub efree. T aking the complemen ts of u and u ′ giv es cub efree w ords of the desired fo rm of length k .  Theorem 4. L et k ≥ 5 b e an inte ger. L et w 0 , w 1 ∈ 00 { 0 , 1 } k 11 b e distinct cub e-fr e e wo r ds. The morphism φ : { 0 , 1 } ∗ → { 0 , 1 } ∗ given b y φ ( i ) = θ ( w i )(010) − 1 is cub e-fr e e. 3 Pr o of. The exis tence of w 0 and w 1 is gua r a n teed b y Lemma 3. Supp ose that v ∈ { 0 , 1 } ∗ is cub e-free, but φ ( v ) con tains a cub e xxx . Let p = | x | . F or i = 0 , 1, since neither 000 nor 111 is a factor of w i , w o rd φ ( i ) cannot ha v e 10101 as a fa ctor; for the same reason, w ord φ ( i ) has prefix 01011 and suffix 1 1 . Th us 10101 o ccurs a s a factor in φ ( v ) exactly at the b oundaries b et wee n images of letters of v . It follo ws that the indices o f an y o ccurences of 10101 in φ ( v ) differ b y m ultiples of | φ (0) | . Again, since 10101 a lwa ys o ccurs in φ ( v ) in the con t ext 1101 0 11, no prop er extension of 10 101 in v has p erio d 1, 2, 3 or 4. Since w i and θ are cub e-free, for i ∈ { 0 , 1 } , the word φ ( i )010 = θ ( w i ) is cub e-free. It follo ws that xxx spans the bo r der b et w een φ ( i ) and φ ( j ) for some i , j ∈ { 0 , 1 } a nd in fa ct xxx con tains factor 10101. Since 10101 is cube-fr ee, xxx is a prop er extension of 10101, and th us has p erio d at least 5 . Note that an y f a ctor u of xxx with | u | ≤ p o ccurs tw ice in xxx with indices differing by p . In particular, since | 10101 | = 5 ≤ p , tw o o ccurrences of 10 101 in xxx hav e indices differing b y p . W e conclude that p is a m ultiple of | φ (0) | . W rite x = aφ ( u ) b where u ∈ { 0 , 1 } ∗ , | ab | = | φ (0) | . W e ha v e xxx = aφ ( u ) baφ ( u ) baφ ( u ) b , and ba = φ ( i 0 ) for some i 0 ∈ { 0 , 1 } . How ev er, since w 1 6 = w 2 , w e also ha v e φ (0) 6 = φ (1) so that either • at most one of φ (0 ), φ (1) has b as a prefix OR • at most one of φ (0 ), φ (1) has a as a suffix. Supp ose that at mos t one o f φ ( 0), φ (1) has b as a prefix. (The other case is similar.) Without loss of g enerality , sa y that φ (0) has b as a prefix. It follo ws that φ ( v ) con tains φ ( u 0 u 0 u 0), and v con tains u 0 u 0 u 0 a s a factor. This is a con t r a diction.  Corollary 5. F or every inte ger k ≥ 0 , ther e exists a k -unifo rm cub e fr e e binary morph ism. Pr o of. If k is o dd and k ≥ 15, then Theorem 4 give s a cub efree morphism of length k . F or k ∈ { 3 , 5 , 7 , 11 , 13 } , the morphisms giv en in the table b elo w are cub efree. φ 3 0 → 001 1 → 011 φ 5 0 → 01001 1 → 10110 φ 7 0 → 001001 1 1 → 001101 1 φ 11 0 → 001010 0 1011 1 → 001 01001101 φ 13 0 → 001001 0 110011 1 → 001 0 011001011 The cubefreeness of these morphisms can b e established b y a criterion of Ker¨ ane n [8], whic h states that to confirm that a uniform binary morphism is cubefree, it suffices to c hec k that the images of all words of length at most 4 are cub efree. F or k = 1, the iden tit y morphism is certainly cubefree, and for k = 9, clearly w e ma y tak e φ 2 3 . This establishes the result for all o dd k . If k = 0, the morphism that maps ev ery word to the empt y w ord is trivially cub efree. If k is p ositiv e, ev en and not a p o w er o f 2, then k = 2 a (2 r + 1) for some p ositiv e a, r . If φ is a (2 r + 1)-uniform cub efree morphism, then the morphism θ a ◦ φ is a k -uniform cub efree morphism. Similarly , if k = 2 a , then θ a is a k -uniform cub efree morphism. This completes the pro of.  Branden burg [5] gav e an example of an 11-unifo rm squarefree ternary morphism and stated further that there are no smaller uniform squarefree ternary morphisms (excluding 0-uniform and 1- uniform morphisms). W e therefore conclude b y asking: 4 Do there exist k - uniform squarefree ternary morphisms for a ll k ≥ 11 ? Reference s [1] A. Ab erk ane, J. C ur rie, “Ther e exist bina ry circula r 5 / 2 + power free words of e v ery length”, Ele ctr on. J. Combinatorics 11 (20 04), #R10 . [2] J.-P . Allouche, J. 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Dep ar tment o f Ma thema tics a nd St a tistics, University of Winnipeg, 515 Por t ag e A venue, Winnipeg, Manitoba R3B 2E9 (Canada) E-mail addr ess : { j.currie ,n.rampersad } @uwinnipeg.ca 5