Uniformly growing k-th power-free homomorphisms

Theoretical Computer Science 23 (1983) 69-82 North-Holland 69 zyxwvutsrqpon Publishing Company UNIFORMLY GROWING k-TH POWER-FREE HOMOMORPHISMS Franz-Josef BRANDENBURG zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Imtirrtt fiir hfonnarik, Uthwsitiit Botw , 53(10 Bontt. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLK Fed. Rep. Gertvat~ y Communicated by A. Salomaa Received July 198 1 Revised February 1982 Abstract. A string is called k th power-free, if ir ,iocs not have x’ as a nonempty zyxwvutsrqponmlkjihgfed s&string . Fo r all nonnegative rational numhcrs k. k th power-free strings and k th po we r- fre e ho m o m o rphism s are investigated and the shortest uniformly gr<>winp square-free tk = 2) a nd c ub e - fre e tk 7 3) homomorphisms mapping into least alphabets with three and two.letters are intro duc e d. it is shown that there exist exponentially many square-free and c ub e - fre e string s o f e a c h le ng th o \ c r the se alphabets. Sharpening the kth power-freeness to the repetitive threshold RT(~I 1of II letter RTI~I rth po we r- fre e alphahct>. wc provide arguments for the nonexlstence o f va rio us ho m o m o rphism s. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA 1. introduction Since the work of Thue [9, IO] at the beginning of this century there have been man investigations on the construction of strings without repetitions. The simplest sue; strings are the square-free and the cube-free strings, which do not have s’ and .v3 as a nonempty substring. Curiously the English words square-free and repetitive each have a repetition and they are examples of non square-free strings, j3 = such as the constants e=2.718281828..., mathematical 1.7320 5080 8 . . . , and 1~= 3.1415 9265 3589 7932 3846 2643 3 . . . (see [7]). The existence of square-free strings of arbitrary length over three letter alp ‘Iabets and of cube-free strings over two letter alphabets has originally been disk lvered by Thue [9, IO]. This is in fact surprising and a remarkable combinatorial pr. >perty of strings, Square-free strings have been applied m various situations, e.g., in UI. ending chess, in group theory, and in fh,rmal language theory. Details can bc found I 1 [8] and the references given there. In this paper we generalize the notion of a power of a string to rational powers and strict rational pol#ers. Square-free and cube-free strings now are special cases with k = 2 and k = 3, and strongly cube-free or overlap-free strings are weakly 2nd power-free strings. This generalization is’a real simplification over existing notions. Furthermore, it allows to define partial repetitions of strings and repetitive thresholds of alphahers. ‘:304-397S/83,‘0000-0000/$03.00 @ 1983 North-Holland 70 F.-J. Brmufenburg Particular emphasis is laid upon the sets of square-free and cube-free strings over least alphabets. It is easy to see that there are no square-free striilgs of length four over a two letter alphabet and no cube-free strings of length four over a single letter alphabet. However, there are infinite such strings over alphabets with three resp. two letters. Here we improve these results and show that for every positive integer n there are exponentially many square-free strings of length n over a three letter alphabet and exponentially many cube-free strings of length n over a two letter alphabet. Hence, the sets of square-free and cube-free strings are either trivial or exponentially dense. For the proofs of these results we use square-free and cube-free homomorphisms from an arbitrary alphabet into the sets of strings over a three and a two letter alphabet, respectively. We introduce the shortest uniformly growing square-free homomorphisms from alphabets up to six letters into the set of strings over a three letter alphabet and the shortest cube-free homomorphisms from alphabets up to four letters into the set of strings over a two letter alphabet. Much longer and not uniformly growing homomorphisms of this kind have been introduced in [I]. We also sharpen conditions on uniformly growing homomorphisms to be square-free to I he very optimum. This improves results by Bean er (71.[l], Berstel [2] and Thue [9. lo]. i.?nally, the repetitive threshold RT(rz) of II letter alphabets I,, is detined. RTcrl i is the least zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA k such that there are infinitely m:eny weakly kth power-free strings ov’:r S,,. This continues the work of Dejean [3]. We establish certain properties of weakly RT(n)th power-free homomorphisms. In particular, we show the nonexistcncc of nontrivial such homomorphisms from E:, , into .Yz. and from I’: mto -T”-Pl1 if RTI/I 1c ;. These results imply that new technique< are necessary to determine the unknown va!ues of the repetitive threshold RT(u ) for tt ‘~3, and they show that our proof techniques for establishing lower bounds on the numhcrs of k th power-fret: strings fail here. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA 2. k th po we r- fre e string s kth power-free homomorphisms 71 For example, aba is the $th power of zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIH ab, and abab is the strict $th power of ab. If u’ is a nonempty string and a is the first letter of cv, then wa is the strict first power of w. In the following we shall see that the use of rational and strict rational powers of strings both generalizes and simplifies existing notions. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQP A (finite or infinite) string IV is kth power- free ( weakly kth power- free 1, if )Vdoes not have the (strict) k th pob er of a non-empty string as a substring, i.e., IV# UX~V,where k is a nonnegative rational number. In accordance with the commonly used terminology, second dnd third power-free strings are called squrrrefree and cube- free, respectively. Let FREEZ(<k 1,FE.EEr(<k) and FREEIdenote the ws of‘kth pmler- free strings, weakly kth power- free strings and exactly ktlt power- free strings over the alphabet S, respectively, where FREEx(=k I= FREEY(sk) -FREEI( When it is appropriate we replace the subscript ,’ by its cardinality. Thus FREE,(<2), e.g., denotes the set of square-free or second power-free strings over any fixed Definition. three letter alphabet. The languages FREEX ( < k b and FREE2 (s k ) consists of all strings over 2. which have at most 1~th powers, where m < k and nc s k, respectively. Every string w in FREEI( = k I has a k th power, i.e., 11’= l(xkt’, but IV does not have an snth power with IV > k. Obviously, every k th power-free string is weakly k th power-free, and every weakly k th power-free string is IPIth power-free, if k < m. Thus FR13EL (<k 1G F’REE\(Irk)cFREE~(‘m). It is easy to see that FREE1( s k ) contains only strings up to length [k + 11, and that FREEz(Z) contains seT*en strings up to length three. To the contrary, Thue [9] has discovered the existence of infinite square-free strings over three letter alphabets and of infinite cube-free strings over two letter alphabets. One of the airs pursued in this paper is to count square-free strings over a three letter alphabet and cube-free strings over a two letter alphabet, i.e., to establish lower and upper bounds on the numbers of such strings cjf length II for every IZ. A trivial upper bound stems from the number of all strings, which grows exponentially in the cardinality of the alphabet. For our proofs of exponential lower bounds we make use of k th power-free homomorphisms, which are the most useful tools in the theory of k th power-free strings. A Ilo,?lnt~lorp/ri.sr,lh is a mapping between free monoids S* and d” with h(.~y ) = h is iengrh mifurrn. if jh (n )I = I/z(h !I If (.c )/I (y ) for every X, y E -t-* . A homomorphism for every (1, 11E 1. h is grolt*iljg, if II(CJ) f A for every a E 1 and jh (a)/ - b 1 for some (1 E Z;, and II is uni~bwd~ growhg, if tl is length uniform and growing, i.c., jh (a 11= t > 1 for e:‘ery a E E. A homomorphism is compatible with the product of strings, but it is not compatible with r,ltional powers of strings. Hence, it may occur that h (w )’ Z lz(1%~~ 1. AS ;ti: example consider k =:, h(u) =a. Atb)=bcd, s =ub and y = bn. Then h(s”)= 72 F.-J. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Brandenburg zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJ and k(yJk = h(x)k zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA =abcdk =ahcdab, h(y k)=h(bab)=bcdabcd, hcda k = bcdabc. Thus the repetitive power of a string may increase or decrease under a homomorphism. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA h(aba)=abcda, Definition. A homomorphism h is (rvenkl~) kth power- frw, if h(FREEl;(<k FREE,(<k) Ih(FREE,(~k))EFREE~~~k)). )) 5 Every (weakly) kth power-free homomorphism map:; (weakly) kth power-free strings into such strings. It tnay however occur that a (weakly) kth power-free homomorphism is not (weakly) mth power-free, where III> k or wt-Ck. Examples are easy to find due to the incompatibility mentionad above. However, (weakly) k th power-free homomorphisms can be composed without harm. Theorem 1. It III nrtd hZ arc (wcakfy rlrpircompo.sitiotl121 - h 2. ) ktlt power- free hnntor,torphisr,ts, thm so is This property provides us with a powerful and elegant tool for defining infinite (weakly) k th power-free string:, and we (as others before) maku use thereof. In particular, if it is growing and , z (a ) = as for a letter a and a nonempty string s. then the limit of the sequence h”(a) = iz (/I” !(a )) is an infinite (weakly)k th powerfree string, if h is such a homomorphism. Furthermore, Thue [IO], Bean clt 01. [l] and Bcrstel [2] have established conditions which guarantee that a homomorphism is k th power-free. These conditions say that it is sutlicient to check the k th power-freeness of a homomorphism on all kth power-free strings of length k + I :tnd tither a certain substring property of the homomorphic images of the letters ~scc [ I, 1(I]) or, for square-free homomorphisms. the square-freeness on all squarcfree strings of Icngth 2 +2p(h 1. where ~(11) = max{h((z jjo E Z}/min{h(cl )jn E 11 I st’c 12 ] 1. l’hcse conditions arc too restrictive for uniformly growing square-free hl>!nomorphisms and optimal ones arc established by Theorem 2. homcmorphsims w’ does not necessarily have this form. Note that the lengthuniformity of 11will not be used elsewhere. If Ix’1= jy’], then s’ = y’> II (14)= II (c ), u = c and y =2. Now a = zyxwvutsrqponmlkjihgfedcbaZY b or b = c, since (I f b # c implies It (abc) = ss’y_t-‘yz’. Hence H’= bubrtc or w = aubnb, and w has a square. Let lx’/ # jy’l and assume Ix’/ > Iy’l. The case lx’1< Iy’l is similar. Then x’ = y’x” with x”# A, and c = v’&” with d E Ir. Ih (o’)l< Ix”/ and Ih(u’d)l zz Ix”]. If h is length uniform, then C’ = A, which simplifies this analysis a bit. Now zyxwvutsrqponmlkjihgfedcbaZYXWVUTS h(d) = SC?’ and h (n ) = xy ‘h (o’)S, where ISI = Ix”1- IIt (c’)l > 0. Thus It (ad) has a square, which implies that n = d. Hence, h(a) begins and ends with S and 2 - ISI < ]h(a)j, i.e., 11(a) = S&3 with p f A ; otherwise, h(n) has a square by results in [5,6]. Furthermore, p is a prefix of h (K/P). Thus h (nub) = &36py has a square, contradicting the minimality of h t w 1. Hence, h is sqlrare-free. It is easy tc? see that the homomorphism g with ~(a I= ah, g:(b) = cb and g(c) = cd is square-free on al! square-free strings of length one or two, but g is not square-free on uhc. Thus the bound three is optimal. 9 q For non uniformly growing homomorphisms consider the homomorphism g from Example I .6 in [ 11. which maps a, b, c, d. e to ad, b, cdbadce, cdabdce, dad/m, respectively. g is square-free for all square-free strings of length three. However, ,q(ahtr 1 = adhadcdhacicc has the square (&a&)‘. Here, M’= abac factors into a~ with ~:(a I= AX and ,4(c) = ys’h(u)yz. Thus g(c) forces the square and prevents the fxtorization of w into nltbm with the properties as in Theorem 2 Note that the substring property of Thue and Bean et al. [lo, l] is too restrictive. This property guargntees the square-freeness of every homomorphism. It requires that h(a 1 is not a substring of I?(h) for letters a, b with a f 6, which means that the homomorphic image of each letter cap uniquely be determined in a string. However. the homomorphism h with Ir (n ) = (I and I?(b 1 = hat, e.g., is k th power-free for every k p :, and h violates the substring property from above, as does the ‘Thue’ homomorphism, which is of interest for its own right. 2 + p +,r with O- r r-. I. Suppose that Ir,(rre) has the kth power of a 2 . Pt’ “I ntlnctnptv string as a substring and is of minimal lrrlgth. Then Il,clr*) =CL with C, tl c ((1. ‘1)v (A }. If IL’\is odd, then there exists no x such that Al(s) =TCCW, whcrc either t* = (1= A or C, t’ E {rc,h} and LJis the first letter of c. Hence, II:) is Eden. Then c .= A by the minimality of h I(~iq) and w =x7- “‘y, where x = It 1 1(1: 1 and \’ = A, ’ IL\‘), if 1~~1is even, and J’ -=h , ’ (r’ci), if IU’/ is odd, Then y = .xr and M’= _” ‘y has a k th power, which is a strict k th power, if h 1(~t*) has a strict kth power. ti -Ice, It, is 1weakly) k th power-free. 3 Proof. I ct k I- F J. Brmdenburg 74 We close this section with examlrles of two particular weakly kth power-free homomorphisms, which are optimal in a sense made clear below. zyxwvutsrqponmlkjihgfedcbaZYXWV Example 1. Let Ill(a) = ab and hi(b) = ba. hl is called the ‘Thue’ homomorphism, who has studied the string obtained by iterating hl. See [&-lo]. In particular, II 1 is weakly square-free and defines an infinite weakly square-free string over {a, b} by iteration. Notice that there are no growing square-free homomorphisms mapping into the set of strings over two letter alphabets, since this set is finite. Example 2. Let II?(a ) = nhcachcdwbacbcncba, h,(b) = ocabacabcncbacabacb, hz(c ) = ~~abchabcabacbabcbac. ?he homomorphism hz is due to Dijean 131. She has shown that It2 is weakly ath power-free, and that ith power-free string; over three letter alphabets are of length at most 38. Thus there exists an infinite (or infinitely many) weakly $th power-free string over a three letter alphabet, but there exist no growing $th power-free homomorphisms over three letter alphabets. Under various perspectives hz is an cxtcnsion of h, to three letter alphabets; details are discussed in Section 4. 3. Density of square-free and cube-free sets of strings Our first investigation and experience with square-free strings was the attempt of computing all initial square-free strings over {N,6, c} and listing these strings as the paths of a tree with root A. See [IO]. This representation immediately asks for bounds on the width of the tree so obtained, which equals the number of square-free strings of length tz. kth pow+free homomorphisms 75 Each of these homomorphisms is square-free on all square-free strings of length three, and thus square-free by Theorem 2. If .Z has more than six letters, then define copies 1~:’ of hh from {u~~__~, . . . , ah,}* into (Uji-2, a3i-1, <Lji}*, identifying a and aai-2, 6 and a3i-1, and c and U3i under hf’. Suppose that 2 has 3 - 2’ letters and let r = 2’-‘. For homomorphisms gl, g2 from ET into AT with ZI nXz = Al nA2 = 0 define the zyxwvutsrqponmlkjihgfedcbaZYXWVUTSR parallel compasitior: gl zyxwvutsrqponmlkji x g2 from (EILJ.&)* into (AluAd* by (glxg2)(a)=gl(a), if aEEl, and (glxg&z)= ~~(a), if a ~2~. Obviously, gl x g2 is a square-free homomorphism, if gl and ~2 are square-free. Hence, hk” x - * * x hi” is a square-free homomorphism from (of depth p) Ia,, -. -, Us,}* into (~41,. . . , a3,}*. Now the repeated composition (11 x 11:‘) is a square-free homomorphism from Z* into 116 ~...c(h;“x**~ {ci,, cll,%)*. 0: Remark. It should be noted that the homomorphisms from Theorem 4 are the shortest uniformly growing square-free homomorphisms from alphabets up to six letters into {a, b, c}“. Furthermore, for n < 3, every uniformly growing square-free homomorphism from {a,, . . . , a,,)* into {a, 6, ci* equals lzl, hz, h.~ or Iti up to a renaming of the letters and a reversal of the strings, where 11;(a 1) = uhcuchcubuc, Ir g (n 2) = ubcbucubncb, and It 4 (u3) = ubcbacbcuch. The fact that these homomorphisms are the shortest square-free homomorphisms of their kind was checked by a PL/l computer program, which was run on the IBM 370-165 of the RHRZ Bonn, and consumed 30 min CPU time. The program first generated all square-free strings of length n over {a, h, c}. Then it exhaustively searched all k-suples of these strings, which are compatible with each other according to the conditions of Theorem 2. These strings can be used as the homomorphic images of k letters. The existence of square-free homomorphisms, which properly reduce arbitrary alphabets to three letter alphabets, can be used to show that the set of square-free strings over a three (and more) letter alphabet is exponentially dense. 11,.07) denote the number of stlmgs of For a language L and II z-0 let zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHG length II in L. /I,_ is called the dcmity frrnctiort of L. Definition. Proof. There are 1172 square-free strings of length 24 over {a, h, c} beginning with ub. Hence, each square-free string uv with IL’\ = 2 has at most 1172 square-free extensions IWCYwith ICY/ = 22. Since there are 6 square-free strings of length two, l-zFlr;RL’t:J,._L) (n)-G6 * CT ‘, where c2 = 1172 I’*’ < 1.38. To establish the lower bound 76 F.-J. Bratldenburg !et IV be a square-free string of length I > 0 over (a, 6, c), which exists by Example 1. Define i finite substitution by r(u ) = {a, a’}, T(b) = (6, 6') and T(C) = {c, c’}. Then each stringx E T(W) is a square-free string over {a, a’, b, b’, c, c’} and hh(x) E {a, b, c}* yr, Y~E~{w} with yr # ~2, is a square-free string of length 221. I17fu)(I) = 2’. Forevery zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLK hh(yl) and h6(y2) have different prefixes of lengrh m with 22(1- 1)s~ ~2221. This follows from the fact that hg is injective, which is a consequence of the squarefreeness. Hence, for every n > 2 there exist at least 6 * 2’ square-free strings of length n ‘:n {a, b, c}*, where I = [n/22]. Thus Z7FREE3(-_r2, (n) 2 6 - 2”‘22 2 6 - 1.032”. IJ Forn=l,2,..., 24 that actual numbers of square-free strings over (n, b, c) are as follows: 3, 6, 12, 18, 30, 42, 60, 78, 108, 144; 204, 264, 342, 456, 618, 798, 1044, 1392, 1830, 2388; 3180,4146, 5418, 7032. This sequence suggests that the density function of the set of square-free strings over a three letter alphabet grows at least as 1.3”, which means that our upper bound is better than our lower bound. Important is that both bounds are exponential. Notice that the iteration of a single square-free homomorphism defines only sparse languages of density O(rz - log 12) as it has been shown in [4]. For cube-free strings over two letter alphabets we proceed in a similar way, improving again a result by Bean et al. Cl]. Proof. I-et l = (~1, , u7, . . . , a,,} and dcfinc homomorphisms h ,. . . . ,/I~ by the table Ohvlously, /I, is cuhc-free, and II: i!*cub:-free by Theorem 3. The homomorphism It i is cube-free on all cube-free strings of length four. but 1~~does not satisfy the -\uhstring property’ from [ 11, which requir!:s that 11(trh) = rtlr (c )c implies that II = A and u = I., or I’ -=A and b L c for all letters n. h. C. Here the homomorphic images r)f the lcttcrs arc of the form .uJ*,1’s and sz, and I’S is a substring of s\‘sz. However, catch of the strings I, !‘, 2 of Icngth three is unique. and the prefix (rtrhA of hJ(~r?) \,crvcs as a separation marker so th;)t its occurrence in a string h ;(N-) uniquely determines the occurrence of CI?ii] W. kth zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA power-free homomorphisms 77 a simple case analysis as in Theorem 2 shows cy = p or y = p, such that w has a cube. Conversely, if zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA aabb does not occur in U, then a2 OCCUB at most as the first or the last letter of w. By the uniqueness of X, y and z, M’= alua3ua3ua3, w = alcalualuaz or M’= LW with t’ E {a 1, a3}*. In each case, u’ has a cube. Hence, hj is cube-free. The homomorphism h4 is cube-free on all cube-free strings of length four. But h4 does not satisfy the “substring property”, since h,(a,) occurs as a substring of h4(ala4). However, aabaa, ababa and aababb serve as separation markers and uniquely identify h4(al), hd(aZ), and h4(a4), respectively. Now an analysis as above for h3 shows that h4 is a cube-free homomorphism. If the alphabet contains more than four letters, then repeated compositions of 4 define a cube-free (extensions of) the homomorphism h 4 as. in Theorem homomorphism from Z* into {a, b}*. El zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFE Remark. It should be noted that the homomorphisms ErBm ‘rheorem shortest uniformly growing cube-free homomorphisms frcm alphabets letters into two letter alphabets. Furthermore, the homomorphisms h I, 11: and It.; are unique up to a renaming of the letters or a reversal of where II G(a 1) = aabbab, II[1(a:) = abbaab, h [t((13) = habaa!,, and h I:(a, ltI;((1;1)= baabah, 6 arc the up to four h,, and h3, the strings, ) = nabhnb, hli’(a.~) = babaah. Using the fact that there are 1251 cube-free strings of length 18 in {a, b}” which begin with LJ and the homomorphism h, from Theorem 6 we obtain that the set of cube-free strings over two letter alphabets is exponentially dense. 4. Repetitive thresholds From the aforesaid we know that there exist infinitely many square-free strings over a three letter alphabet and infinitely many cube-free strings over a fwo letter alphabet. However, if the size of the alphabets is reduced by one or if the repetitive power is decreased e.g., to ;, then the sets of k th power-free strings are finite. Thus there is the problem of determining the least repetitive power k such that tar every .Z, FREE-(<k) is finite, and FREEL(sk) is infinite, and of If letter alphabet estabL!.ing lower bounds on the density of FREEI-(ck zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQP j. We cannot solve these problems here, since our techniques from above fail. repetitive threshold RT(rt) is the rational number Definition. For every n 2 1 ,the zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA is finite and FREEY(<k) is k: such that for every n letter alphabet X FREEr(<k) infinite. Since FREEZ(<RT(n)) is finite for every n letter alphabet 2 there is an upper bound on the length of the strings in FREEm(<RT(n)). Hence the set of exactly RT(n )th power-free strings FREEX (= RT(n )) is infinite. The notion of a repetitive threshold is due to DCjean [3]. Our definition is based on powers of strings, whereas Dejean has considered the relationship between the length of some strings 11 and u such that UVN is the kth power of uu for some !c. For alphabets up to three letters the repetitive thresholds are known. Theorem 8. RTr 1) = 4x, RT(2) = 2, RT(3)=:, I c’:RTf n + 1) -: RT( II ) zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDC for ewry positiw irtregcr t1. Proof. FREE,(-.k) contains [X:+ 11 elements, which implies that RT( 1) = m. FREEi,,.,,; 1~2) = {A, n, h, cth, ha, N’I~, hnb}, whereas FREEj,,,hi(s2) is infinite, since it contains the set of prefixes of the infinite sequence generated by the ‘Thuc’ homomorphism from Example 1. Finally, FREE,(<:) contains 3196 strings of length up to 3X as shown by Dcjean [3], and her homomorphism from Example 2 is weakly ith power-free and defines an infinite weakly zth power-free string so that FREE,(- 1) is infinite. Finally, it is obvious that the sequence of repetitive thresholds dccrcases, when the alphabets grow, and that RT(tz i ) 1. 3 Remark. For II -~4 the repetitive thresholds are unknown. There is good evidence that RTr4) = i and RT(tr )=: rz/(n - 1) for tf zS, as proposed by Dejean [3]. FREE,I,:) is finite and :onsists of 236 345 strings of length up to 121, and every string of length II I 7 over a II letter alphabet has n tt/(tl - I)th power, so that FRF:E.,(,:u/I~ -- 1 II is finite. Hence, one condition of a rcpctitive threshold is satisticd hy thcsc values. However, we do not know whether they satisfy the second condrtitrn, too, since infinite or infinitely many weakly k th power-free strings have rult vet heen found in these casts. The following results show that new techniques iirc ncedcd to determine the rcprtirive thresholds. kth power-free h (w ) E FREEd( hornomorphisnts zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONM 79 =2). Consider n 2 3. Since RT(n zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLK j s i by Theorem 8, that h(w)&FREE,(=RT(nj) for some w EFREE~(=RT(~)), i.e., ha FREEA(<RT(n)).Then zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Ih(y)j>t * Ih(x)I,which implies that /h(a)\ f Ih(b #,(y) < t - #.(x), and #d(y) > t * #d(y) for some letters n, b, c, d ~2, where #,(y) Suppose denotes the number of occurrences of c in y. Consider w’ = U’X’Y’X’UE FREEr( =RT(n)), which is obtained from w = uxyxv by a renaming of the letters is maximal among all strings so obtained, Then t . Ih (x’)l- ih( > such that Ih( t. I/t(x)\-Ih(y which implies ~(w’)&FREE~(GRT(~~)), and It is not weakly RT(fz )th power-free. III From the proof of Theorem 9 we obtain that every weakly RT(,t)th power-free homomorphism is either length uniform or the letters are uniformly distributed in each exactly RT(11)th power-free string ZIX~XL’,such that # <,(y )/ # ,,(s ) = (2 - RT(n ))/(RT(n) - 1) for all letters n and xyx E FREEY( =RT(n 1). zyxwvutsrqponmlkjihgfedcbaZYX Theorem 10. If RT(4) = z nrtti RT(n) = rt/(n - 1) for tz 3 5, therl ecery zyxwvutsrqponmlkjihgfed ttwzkly RT(rl )th power-free homomorphism over 11ietler alphabets Z and J is lerzgth uttiforrn for eccrp II 2 3. Proof. Assume t:;p contrary. Let n, b ~2 with /h(a)\= max{(h(c)((c ES:), and (h 01 )I = min{jh (c)I jc E Z;). Thus )h (a )I > /h (611. Now abcbabc E FREEx( =:I, but h(abcbabc)& FREEJ( =iJ, abcdbacbdcabcd EFREEx( =:), but h(abc(ibacbdcabcd 1 - 111, but and for nzS, ala?... a,, lal EFREE\-(=n,/(n &FREE,!=;), - 1)) with ~1~=N and a2=b. h(n,n,. . . a,, la,i@FREEJ(=r~/(rl C! Another lowing: restriction on weakly RT(rz)th power-free homomorphisms is the fol- Theorem 11. Let h be a weakly RT(n )th power-free homomorphism ouer tt letter alphnbets Z arld A. Zfh(a) = CII and h(b) = du(h(a) = UC, h(b) -ud) with a, b, c, d E v the/r(I f h impiies c # d. Thus the begirming and the end of tfle holnomorphic ‘L, images of different letters must be d;fjerent. Proof. Assume the contrary and let II (a) = cu and h(b) = CL’with a f 6. For )I = 2, h (a, b ) = CLICIKC is not weakly 2nd power-free, and thus contradicts the assumption. Consider rt 3 3 and RT(n ) s i. Since FREEr( = RT(n )) is infinite and invariant under a renaming of the letters there exists xyxb E FREEz(=RT(n)) with /y/ = f * 1x1, xyz E FREEI-(=RT(n)), htxyz) E where t = (2 - RT(n))/(kT(n ) - 1). Now Then h (xyx-h ) FREE,,(=RT(n)), and y f bz, i.e., y = az by renaming. g FREE~(~RT(H )), which contradicts the assumption that h is weakly RT(n )th power-free. 0 F.-J. Brandenburg 80 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA As a consequence we obtain: zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA Theorem 12. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA There exists nu weakly RT(rl)th power- free homomorphism from X* nz > n. irttoA*, if 2 is an m letter alphaber, d is an n letteralphabet, and zyxwvutsrqponmlkjihgfedcbaZYX Theorem 12 shows that the technique employed in Theorem 5 and in Theorem 7 cannot be used to show that there exist exponentially many weakly RT(n)th power-free strings of each length. In fact, the density function of the set of weakly square-free strings over two letter alphabets is not strictly increasing (it makes some zig-zags; see IZ= 24 - 27) and seems to grow slower than 2’1”‘,as the tirst 95 values of the density function may suggest. 2, 4, 6, 10, 14, 120, 136, 148, 228, 248, 272, 380, 382; 376, 500, 522, 525, 724, 740, 724, 91%. 960, 976, 20, 24, 30, 36, 164, 152., 154, 284, 296, 300; 1382, 356. 374, 540, 548, 568, 724; 716, 748, 1008, 1000. 44; 48, 60, 60, 148, 162, 176, 296, 320, 332, 392, 410, 432, 560, 592; 592, 788, 824, 816, 72, 82, 88, 96, 190; 196, 210, 356, 356, 376, 458, 464, 486; 620, 660, 688, 856, 868, 880, 112, 216, 400, 476, 688, 868, 120; 224, 416, 498, 722, 912; If the repetitive threshold is less than :, which is likely to be true for alphabets tvith at Icast four letters, the situation is even worse Proof. Let II be a homomorphism from ,V” into _I* and assume II 24, since RT(3) = :. Ry Theorem 1 1, II ((1 ! = CY(n )y(o )@ (a) for each letter a, where cy and /3 are renamings (permutations) of j. Assume that 11((I 1 = (I 1u3u with (I ,, 11,ES. From Theorem 11 we obtain @(/I I f (1, for all h E 1 with h # (7 and II (C‘, = cw, for some (. 5 I. Thus h (cn 1 = cog Iu_a and 1’1is not weakly RT(ri jth power-free. 5 t+y Theorem 13 it is irnpossibk to define an infinite weakly RT(/z)th power-free using weakly R~‘(II )th power-fret 1~otnomc~rphisms, if RT(n ) <I 1. Since RT(r. ) is a\hum:tJ to hc less than J for !I ---3, this tool fails to work for determining the vnlucs of the repctitivc thresholds. Nevertheless, an infinite weakly RT\H )th powcrfree string rn:ty btb definable by the iteration of ;I hrmomorphism, which is weakly RT!H )th power-free only on its secJuence of strings of the form h’(\t*) for some :r.r;iom 11’and all i ‘2 1, i.e., a weakly RT(n jth power-fret DOL. system may exist IX*C [4j). In the cast of four lotter alphabets and RT(rl )-c i we disprove this ;ls+umy,tion for uniformly grciwing homomorphisms. So there is no hope to deterUI~I~Cthe L.i1Itlcof RT1-I) caciiy. string krh zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDC power- free homomorphisms zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONM 81 Lemma. Let E = {a, 6, c, d) and w E FREE\-( =k ) with k <$. (i) If the ietter u does not occur in w , then jw 1d 4. zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGF (ii) zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA If the string ab does not occur in w as II substring, then 1w 1s 22. zyxwvutsrqponmlkjihgfedcbaZY Proof. The longest string satisfying (i) is, e.g. bcdb, which proves (i). All strings in FREEI( which do not contain ab as a substring and begin with ac are shown in Fig. 1. By the symmetry of c and d, a similar set of strings is obtained for the prefix ad. All strings so obtained are no longer than 22, so that (i) implies /w 1s 26 for all strings satisfying (ii). In fact if w satisfies (ii) and begins with 6, c, or ci, then ]w] 5 22. q b-cAa ‘d-a-c-b-d-c-a--d-b(c a-c<: -a ‘c-d-a b a-c’ a-d--‘b -c-b”’ ‘d-a ‘c-a ‘d a -cl \ d/ \ a a-d-c’ b / \ .a ~b_d_a-c-d-b-c-a-d-c-~-a-c-d/ a ‘b /a a-c-d,b c-a-d-c-b/ a Ld-a-c-d-b-3-d-c<b Fig. 1 Notice that the particular lemma, whereas substrings avoidable used here is more notion of avoiding accr and and Sat2 10-j. strings a and ah are avoided in the strings of the above such as b or hu or hc may occur. Thus the notion of restrictive than the one in [l]. It coincides with Thue’s i~-b in square-free strings over {n, h, c}. See [9, Satz 4 Theorem 14. If RT(4) <: 1, t1retr there emts power-free hornomorphisrn RT(4)tit power-free h 01; a four for all i 2 HO un(formly letter alphabet growing such that weakly RT(4)tlz h’lirf) is wuk/?* 1 md Some axiom w . Proof. Let Z = {a, b, c, d} and consider i 2 1 such that /h’(n)f ~23. By the previous Lemma, h’(a) contains all substrings pq with p, q E,T. Hence h(Pq 1E FREEr (<‘T(4)) for all letters p, q with p f q. Since FREEz( CRT(~)) is finite there is an upper bound K on the length of its strings. Thus for all i >K/Ih(nll, h’(a) is inFREE\-(=RT(4)), andif ]h’(a)j>K -t2, then h’(a) =nxy xu with u, u #A, and?syx is the exact RT(4)th power of ~1. Since h is growing appropriate t’s exist. 82 F.- J. Brandenburg If h is weakly RT(4)th power-free on its images h’(w), then u and y end with different letters and the last letters of the homomorphic images of these letters are different. Otherwise, IIX~XI!resp. h(uxyxu) is not weakly RT (4)th power-free. By symmetry L‘ and y begin with different letters and the homomorphic images of these letters begin with different letters. We now try to ftx the homomorphism h, and for our convenience we assume y = zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA ay ‘, u = bv’, h(a) = ua and h(6) = b@. Since h(cu), h (cb), h(du) and h (db) are weakly RT (4)th power-free we obtain that ir (c) = yc and h(d) = Sd or h (c) = yd and h(d) = SC. Simply assume the first case. Then we can determine the secoAd letters of the homomorphic images and obtain /l(a)=abcu’, h(b)=b@‘, h(c)=y’dc and h(d)=S’cd, where cy’, /3’. y’, S’#h. Continuing in this way h(a) may begin with abc or abd, but h (ub) forces Ih(u)) a 4. However, there is no possibility left for the forth letter of h(u) such that both h (cu ) and h (da i are weakly RT (4)th power-free. Hence, h does not exist. •i Concluding we have not been able to determine the values of RT(n) for 1134, since new techniques are necessary for that purpose. Further open problems are the decidability of the (weakly) kth power-freeness of homomorphisms for nonintegral rationals k, and optimal conditions for a kth power-freeness check of homomorphisms. References