Binary Morphisms to Ultimately Periodic Words

Binary Morphisms to Ultimately P erio dic W ords Brendan Lucier ∗ Septem b er 13, 2004 Abstract This pap e r classifies morphisms from { 0 , 1 } that map to ultimately p eriodic words. I n particular, if a morphism h maps an infinite non-ultimately p eriod ic word to an u ltimately period ic word th en it must b e true that h (0) commutes with h (1). 1 In tro d uction In th is short note w e p resent a rough solution to an op en problem in t h e stu dy of com binatorics on w ords, due to Jean-Paul Allouc he [4]. W e omit a full discussion of the sub ject matter in this man uscript; there are numerous excellent texts th at p ro v ide bac kground o n the sub ject o f combinatori cs on w ords, and sp ecifically the study of morphisms [1, 2]. The problem of interest is the follo wing: Problem 1.1. L et w b e an infinite wor d over { 0 , 1 } that is not ul timately p erio dic, and let h b e a morphism. Supp ose h ( w ) is ul timately p erio di c. Pr ove (or dispr ove) that h (0) c ommutes with h (1) . This problem has applications to con t inued fraction expansions [3]. In this p ap er we shall prov e that, indeed, h (0) must comm u t e with h (1). A few short notes on notatio n : given a string x , w e s h all use x ω to denote the infi nite rep etition of x (i.e. the word xxx · · · ). Also, we shall write x [ i ] to mean the chara cter of x a t index i , and x [ i, j ] to mean the substring of x consisting of those characters at indexes i to j , inclusive. F or examp le, if x is the binary word 0100110 then x [2 , 5] = 10 01. 2 Main Result Before proving th e main result, we requ ire a simple prop osition regarding in jectiv e morphisms. Proposition 2.1. Supp ose h is a morphism f r om { 0 , 1 } such that h (0) and h (1) do not c ommute. Then for al l a, b ∈ { 0 , 1 } ∗ , a 6 = b = ⇒ h ( a ) 6 = h ( b ) . Pr o of . Assume that h maps to an alphab et that do es not contain 0 or 1. D enote x = h (0) and y = h (1). First note that h (01) 6 = h (10) implies that x 6 = y . Also, if w e had k := | x | = | y | , then given h ( a ) w e could u n iquely determine a simply by m atching the letters of h ( a ) to the letters of x and y , k at a time. Being able to uniq uely determine a from h ( a ) wo u ld imply th at a 6 = b = ⇒ h ( a ) 6 = h ( b ), as req uired. So w e can assume | x | 6 = | y | . Let n = max {| x | , | y |} . Assume without loss of generalit y t hat | y | > | x | , so we hav e n = | y | . W e now proceed by ind u ction on n. If n = 1 then w e must hav e x = ǫ . But then we w ould hav e xy = y = y x , a contradiction. If n = 2 then | x | = 1 and | y | = 2. But since x and y do not commute, we cann ot hav e y = xx . Let c b e the fi rst letter th at app ears in y that is not x . Now, given a string h ( a ) , we can uniquely determine the va lue of a as follo ws. Scan the word h ( a ) from l eft to right. Every time we fin d a c , w e map it p lu s the preceeding character ( if y = xc ) or the follo wing character (otherwise) to 1. Once th at is done, map all of the remaining x c h aracters to 0. This pro cess generates th e string a in a unique wa y . W e conclude that if a 6 = b then we m u st hav e h ( a ) 6 = h ( b ). ∗ Sc ho ol of Compute r Science, Universit y of W aterlo o. Email: b lucier@alumni.u waterloo.ca . Supported in pa r t b y NSER C. 1 This concludes th e base cases. So s u pp ose n o w that max {| x | , | y |} = n > 2. Sup p ose also f or contradictio n that there exist binary w ords a, b such that a 6 = b but h ( a ) = h ( b ). Since h ( a ) = h ( b ) and neither x n or y is ǫ it cannot b e the case that either a or b is a prefix of t he other. There must therefore b e some minimal index i ≤ min {| a | , | b |} such th at a [ i ] 6 = b [ i ]. But then if w e let z = h ( a [1 , i − 1]) we ha ve that both zx and zy are prefixes of h ( a ). W e conclude that x is a prefix of y . Sa y y = xy ′ , with | y ′ | < | y | . Let f : { 0 , 1 } ∗ → { 0 , 1 } ∗ b e the morphism f (0) = 0 , f (1) = 0 1. Let h be the morphism on { 0 , 1 } ∗ giv en by h (0) = x , h (1) = y ′ . Note that h = g ◦ f . By our base case, a 6 = b = ⇒ f ( a ) 6 = f ( b ). Also, max {| x | , | y ′ |} < | y | = n , so by ind uction we must now hav e h ( a ) = g ( f ( a )) 6 = g ( f ( b )) = h ( b ) (1) as required. W e are now ready to prov e the main result. Theorem 2.2. Supp ose w is an infinite wor d ov er { 0 , 1 } that is not ultimately p erio di c, an d let h b e a morphism. I f h ( w ) is ultimately p erio di c then h (0) c omm utes with h (1) . Pr o of . Supp ose for contradiction that h (0) d o es not commute with h (1). Since h ( w ) is ultimately p eri- odic, we can write h ( w ) = y z ω (2) for finite strings y and z . Note that every prefix of w must map to a prefix of y z ω , so in p articular there must be infin itely many prefixes of w that map t o a string of the form y z ∗ z [1 , k ] for any k ≤ | z | . But there are only finitely many p ossible v alues for k . There must t herefore b e some prefix z 1 of z such that infin itely many prefix es of w map to strings of the form y z ∗ z 1 for any t ≥ 0. Now say z = z 1 z 2 . Say th at x, xa 1 , xa 1 a 2 , . . . are the infinitely many prefixes of w discussed ab ov e, where each a i is chosen to minimize | a i | . Then we h av e h ( a i ) = z 2 z p i z 1 ∀ i ≥ 1 (3) where p i ≥ 0 for all i ≥ 1. Supp ose first that all p i are eq ual. Then since h (0) doesn’t commute with h (1), Proposition 2.1 tells us that since all h ( a i ) are equal, all a i must b e equal. But then w = xa 1 a 1 a 1 · · · is ultimately p erio dic, a contradiction. Supp ose instead n ot all p i are equal, so there is some i such that p i 6 = p i +1 . Then we hav e h ( a i a i +1 ) = z 2 z p i z 1 z 2 z p i +1 z 1 = z 2 z p i +1 z 1 z 2 z p i z 1 = h ( a i +1 a i ) (4) So by Prop osition 2.1, a i a i +1 = a i +1 a i . The second theorem of Lyndon a n d Sc hutzenberger now tells us that t here e x ists so me b such that a i = b k , a i +1 = b l . Since we assumed p i 6 = p i +1 w e must ha ve k 6 = l . If k < l then | a i | < | a i +1 | and a i +1 has a i as a strict prefix. But a i is of the form z 2 z ∗ z 1 , so this contra d icts the assumed minimalit y of | a i +1 | . If k > l then an identical argument contradicts the minimality of | a i | . References [1] J.-P . A llouc h e, J. O. Sh allit, Automatic Se quenc es: The ory, Appli c ations, Gener ali zations , Cam- bridge Universit y Press, 2003. [2] M. Lothaire, Combinatorics on Wor ds , vol. 17 of Encyclop edia of Mathematics and its Ap p lications, Addison-W esley , Reading, Massac husetts, 1983. [3] D. R o y , O n the Continue d F r action Exp ansion of a Class of Numb ers , 2008; arXiv:math/0409233 v2. [4] J. O. Sh allit, p ersonal communication, August 2004. 2