We answer a question of Harju: For every n ≥ 3 there is a square-free ternary word of length n with a square-free self-shuffle.http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i1p

Currie, James D.

Saari, Kalle

The Electronic Journal of Combinatorics

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Square-free words with square-free self-shufflesJames D. Currie & Kalle SaariDepartment of Mathematics and StatisticsUniversity of Winnipeg515 Portage AvenueWinnipeg, MB R3B 2E9, Canadaj.currie@uwinnipeg.ca, kasaar2@gmail.comSubmitted: Jul 27, 2013; Accepted: Jan 5, 2014; Published: Jan 12, 2014AbstractWe answer a question of Harju: For every n > 3 there is a square-free ternaryword of length n with a square-free self-shuffle.1 IntroductionShuffles of words are natural objects of study in combinatorics on words, and a varietyof interesting problems have been posed. (See [5], for example.) Recently, self-shuffles ofwords have been studied. (See, for example [7, 8] which independently show that it isNP-complete to decide whether a finite word can be written as a self-shuffle.) If a wordw is factored asw = Πai = Πbi,where ai, bi 6= ε, then we callΠ(aibi)a self-shuffle of w. For example, letting w = 01101001, a1 = 011, a2 = 01, a3 = 001,b1 = 01, b2 = 1010, a3 = 01, we get the self-shuffle of w0110101101000101.(Here the bi have been underlined for ease of reading.) The notion of a self-shuffle equallyapplies to infinite words, and in [3] it is shown that the Fibonacci word has a self-shufflewhich is equal to the Fibonacci word; similarly, it is shown that the Thue-Morse word isequal to one of its self-shuffles.The recent note of Harju [4] poses this problem:Problem 1.1. For every n > 3 is there a square-free word of length n with a square-freeself-shuffle?the electronic journal of combinatorics 21(1) (2014), #P1.9 1In this paper we answer this question in the affirmative; in fact the desired square-freewords can be found over a ternary alphabet. In what follows, we freely use the usualnotions of combinatorics on words. A standard reference is [6].2 Long finite square-free words with square-free self-shufflesConsider a square-free word u ∈ {0, 1, 2}∗ such that neither of 010 and 212 is a factor ofu, and u is of the formu = 0120w0Πmi=1(awi)2012 (1)where m ∈ {0, 1, 2, 3}, the wi ∈ {0, 1, 2}∗ and a = 2021020. We will show later that suchwords u of length n exist for all large enough n ≡ 3 (mod 4).Let b = 2021201020. Let ū be the wordū = 0120w0Πmi=1(bwi)2012. (2)The longest prefix of b not containing 212 is 2021, which is also a prefix of a. The longestsuffix of b not containing 010 is 1020, which is also a suffix of a. It follows that any factorof ū not containing 010 or 212 is itself a factor of u.Now consider the self-shuffle w of ū given byw = ū2−1020−1ū = 0120w0Πmi=1(bwi)20102120w0Πmi=1(bwi)2012. (3)The prefix of w of length |ū| − 1 is a prefix of ū, while the prefix of w of length |ū| hassuffix 010. The suffix of w of length |ū| − 1 is a suffix of ū, while the suffix of w of length|ū| has prefix 212. It follows that the only factors of w not containing either 010 or 212must themselves be factors either of ū or of 1021; by the previous paragraph, they arefactors of u or of 1021, and in particular are square-free. At this point we will mentionthat many arguments can be shortened by noting that the definitions of a, b, u, ū andw are invariant under the operation combining reversal with the substitution k → 2 − kon each letter. Particular words u and ū need not be invariant under this operation, butthey are sent to words of the same form.Lemma 2.1. Consider a square-free word u of the form (1) and let ū and w be defined asin (2) and (3). Fix j, 0 6 j 6 m, and let a word U be obtained from ū by replacing somej occurrences of b by a. Let W be obtained from w by making the analogous replacements.Thus W = U2−1020−1U . Then U and W are square-free. In particular, words ū and ware square-free.Proof. Suppose not. Consider a word U obtained from ū such that one of U and Wcontains a square, and such that m− j is as small as possible.We deal first with the case where m − j = 0. In this case, U = u is automaticallysquare-free, and any factor of W = w not containing 010 or 212 is square-free. Let yy bea factor of W = w, y 6= ε. Thus, one of 010 or 212 is a factor of yy.the electronic journal of combinatorics 21(1) (2014), #P1.9 2If |y|010 > 1 then |yy|010 > 2|y|010 > 2; however, |yy|010 6 |w|010 = 1. It follows that infact |y|010 = 0. Similarly, |y|212 = 0. Now if 010212 is a factor of yy, then depending onhow 010212 is distributed between the two copies of y, at least one of 010 and 212 mustbe a factor of y. This is impossible, so that 010212 is not a factor of yy. It follows that yymust be a factor of one of 0120w0Πmi=1(awi)201021 and 102120w0Πmi=1(awi)2012. (Theseare, respectively, the longest prefix and the longest suffix of w not containing 010212.)Suppose then that yy is a factor of 0120w0Πmi=1(awi)201021. (The other case is similar.)Then 212 is not a factor of yy, forcing 010 to be a factor of yy. However, 010 must notbe a factor of y, so that, depending on how 010 is split between copies of y, we canwrite y = p0 = 10s or y = p01 = 0s, where s must be a prefix of 21, p a suffix of0120w0Πmi=1(awi)2. However, y = p0 = 10s is impossible; if s 6= ε, then the word on theright-hand side of this equation ends in 1 or 2, while the left-hand word ends in 0; if s = ε,p = 1, which is not a suffix of 0120w0Πmi=1(awi)2. Again, y = p01 = 0s forces s = 21,since the left-hand word ends in 1; however p01 doesn’t end in 21.This shows that m − j = 0 is impossible. We now have m − j > 0, so that multiplecopies of 010 and 212 appear in W . It will be useful to work out the distances betweenoccurrences of 010, that is, the minimum value of |010v| such that 010v010 is a factorof W . From the definition of W , any word 010v such that 010v010 is a factor of Wis at least as long as a word of the form 01020wi20212, 01020wm2 or 0102120w020212.From the definition of u, factor 020wm2012 of awm2012 is square-free, and doesn’t contain010 or 212. This implies that wm has prefix 1 and suffix 0. However, wm 6= 10 or elseawm would contain 0wm, which starts with 010. In particular |wm| > 3, and |010v| >|01020| + 3 + |2| = 9, and |v| > 6. From (3) we see that this argument also guaranteesthat any factor 010v010 of U will also have |v| > 6.Suppose yy is a square inW or in U , y 6= ε. Suppose now that |y|010 > 0. Note that 010occurs in W or U in one of only two possible contexts: either 2021201020 or 020102120.Observing the 3 characters to the left of an occurrence of 010 is enough to identify thiscontext. If the 3-character string to the left is 212, then the context is 2021201020; ifthe 3-character string is not 212, then the context is 020102120 (since wm ends in 0.)Similarly, examining the three characters to the right of an occurrence of 010 establishesits local context. Let us write y = p010s. Then 010sp010 is a factor of W or U and|sp| > 6, so that at least one of |p|, |s| > 3. This establishes the local context of a certainoccurrence of 010 in both copies of y, and these contexts must be the same. Since thelocal context 20102120 only occurs exactly once inW , and never in U , both local contextsof 010 in y are as a factor of b. Similarly, if |y|212 > 0, then 212 appears in a local contextcoming from b. In fact, this argument shows that |yy|010212 = 0; if |yy|010212 = 1, then atleast half of the occurrence of 010212 lies inside one copy of y, so that an occurrence of010 or of 212 in y comes from 20102120, which is impossible. Therefore, if yy is a factorof W , we conclude that yy is a factor of one of U2−1 and 0−1U , the longest prefix andsuffix, respectively, of W not containing a 010 or 212 coming from 20102120; however,this prefix and suffix are themselves factors of W , so that we see that yy must be a factorof U .We have shown that any occurrences of 010 in y arise as factors of b. Write b′ = 20212,the electronic journal of combinatorics 21(1) (2014), #P1.9 3b′′ = 20, so that b = b′010b′′. We are thus saying that any occurrence of 010 in y is preceded(in W) by b′ and followed by b′′. Suppose |y|010 > 1. Write y = p010s. Suppose |y|b = 0.Then either |p| < |b′| or |s| < |b′′|. If |p| < |b′|, write W = xyyz. Then b′ must be a suffixof both xp and yp. Let σ be the common suffix of x and y such that σp = b′. Replacingy by σyσ−1, we have a square yy in W such that |y|b = 1. The case where |s| < |b′′|is similar; in either case, if |y|010 > 0, then adjusting yy cyclically if necessary, we canassume that |y|b > 0. Now, replacing b’s in y (and hence in U) by a’s yields a square ina word of the form of U , with the same m, but larger j. This contradicts the minimalityof m− j.From now on, we can assume that |y|010, |y|212 = 0 and yy is a factor of U . If|yy|212010 > 0, then depending on how 212010 is split between the copies of y, at least oneof |y|010 and |y|212 is non-zero. We conclude that |yy|212010 = 0. By the same argumentas earlier, any factors of U not containing 010 or 212 are square-free. It follows that atleast one of |yy|010 and |yy|212 is non-zero. Without loss of generality (up to reversal and2-complementation) suppose that |yy|010 > 0. Since |y|010 = 0, we must be able to writey = p0 = 10s or y = p01 = 0s where p is a suffix of 12 (since |y|212 = 0.) If y = p0 = 10s,each of p = 12, 2, ε is seen to be impossible. If y = p01 = 0s, then p begins with 0, whichis also impossible.We conclude that W and U , and hence w and ū, cannot contain a non-empty squareyy.As promised, we now show that words of the form u = 0120w0Πmi=1(awi)2102 of lengthn exist for all large enough n ≡ 3 (mod 4).The Thue-Morse word is the sequence t = µω(0) where µ(0) = 01, µ(1) = 10. Word tis well-known to be overlap-free. From the definition of t it is clear that t ∈ {01, 10}∗. Onoccasion it is useful to add ‘bar lines’ to a factor of t indicating the parsing of t in termsof 01 and 10. These bar lines always split any occurrence of 00 or 11; viz, 0|0 or 1|1, not|00| or |11|. It is proved in [1, Lemma 4] that t contains a factor of the form 10x01 ofevery length greater than or equal to 6.Consider the word s obtained from the Thue-Morse word by counting 1’s betweensubsequent 0’s. Thus if we writet = Π01si ,thens = Πsi.It is well-known that s is square-free. It is also well-known and easily verified that neitherof 010 and 212 is a factor of s.Lemma 2.2. Word s contains a factor of the form 0120x2012 of every length n ≡ 3 (mod4), n > 23.Proof. A factor of s of the form z = 0120x2012 corresponds to a factorv = 00101100y0110010110the electronic journal of combinatorics 21(1) (2014), #P1.9 4of t. For clarity, add ‘bar lines’ to v:v = 0|01|01|10|0y01|10|01|01|10.The number of 0’s in v is one more than the length of z, giving |z| = |v|0−1 = (|v|−1)/2.s contains a factor of form z of length k⇒ t contains a factor of form v of length 2k + 1⇒ t contains a factor of form 10|01|0y′0|10|01 of length k + 1⇒ k is odd and t contains a factor of form 10|0y′′1|10 of length (k + 1)/2⇒ (k + 3)/2 is even and t contains a factor of form 10ŷ01 of length (k + 1)/4The result follows.The words z of the last lemma begin and end in the form desired for u. We will nowshow when z is long enough, word a = 2021020 is a factor of z at least 5 times. Althoughthe first and last occurrences of a may overlap with the prefix 0120 or suffix 2012 of z,there will be at least three other occurrences of a in z, so that for any m ∈ {0, 1, 2, 3} wecan write z in the formz = 0120w0Πmi=1(awi)2012,as desired.Lemma 2.3. Suppose that 02102v02102 is a factor of s, but that 02102 is not a factor of2102v0210. Then |02102v02102| 6 41.Proof. A factor 02102 of s corresponds to a factor 0|01|10|10|01|10| of t. Such fac-tors of t occur precisely in the context 01|10|01|10|10|01|10|01 = µ2(0011). A factor02102v01202 of s such that 02102 is not a factor of 2102v0210 corresponds to a factor(011)−1µ2(0011u0011)(01)−1 of t which does not contain 0011 as an internal factor. Wordt is concatenated from µ4(0) = 0110100110010110 and µ4(1) = 1001011001101001, andeach of these contains a factor 0011. In addition, concatenating suffix 0 and prefix 011of µ4(0) produces a factor 0011; so does concatenating suffix 001 and prefix 1 of µ4(1).We therefore see that the longest factor 0011u0011 of t with no internal 0011 is the word00110010110|10010110011, of length 22.We have determined that 0210v0210 corresponds to a factorz = (011)−1µ2(0011u0011)(01)−1of t where |0011u0011| 6 22. Because s is obtained from t by counting 0’s and z beginsand ends with 0,|02102v02102| = |z|0 − 1.the electronic journal of combinatorics 21(1) (2014), #P1.9 5Every second letter of µ2(0011u0011) is a 0, so that|z|0 = |µ2(0011u0011)|0 − |011|0 − |01|0= |µ2(0011u0011)|/2− 2= 2|0011u0011| − 26 2(22)− 2= 42.We conclude that |02102v02102| 6 41.Corollary 2.4. Any factor of s of length 40 contains 02102 as a factor.Corollary 2.5. Any factor of s of length 42 contains a = 2021020 as a factor.Proof. The word 02102 cannot be preceded by 1 or 0 in s; It follows that 02102 can onlybe preceded by 2 in s. Similarly, 02102 is only followed by 0. Any length 42 factor v ofs contains 02102. Extending v before and after by one character then forces a to be afactor.Corollary 2.6. Any factor z of s of the form 0120x2012 of length at least 134 can bewritten in the formz = 0120w0Πmi=1(awi)2012.Proof. Since 134 = |0120|+3(42)+|2012|, the result follows by the previous Corollary.Theorem 2.7. For every n > 143 there is a square-free word u ∈ {0, 1, 2}∗ of length nwhich permits a square-free self-shuffle.Proof. We note that |b| − |a| = 3. Given n > 143, let m be least such that n − 3m ≡ 3(mod 4). We have |n− 3m| > 143− 3(3) = 134. By Lemma 2.2 there is a factor u of s ofthe form u = 0120x2012, |z| = n− 3m. By Lemma 2.6, word u has the formu = 0120w0Πmi=1(awi)2012.Lettingū = 0120w0Πmi=1(awi)2012gives a word ū of length n, and by Lemma 2.1, both ū and the self-shufflew = ū2−1020−1ūof ū are square-free.the electronic journal of combinatorics 21(1) (2014), #P1.9 63 Short square-free words with square-free self-shufflesIt is well-known that s is the fixed point of 2 7→ 210, 1 7→ 20, 0 7→ 1.Lemma 3.1. For every n with 3 6 n 6 200, there exists a ternary square-free word witha self-shuffle that is also square-free.Proof. The following claims can be checked computationally1.For each n with 29 6 n 6 200, s has a factor w of length |w| = n such that the shufflep1p2s1s2 is square-free, where w = p1s1 = p2s2. Furthermore, the lengths of s1 and p2 canbe restricted to satisfy 1 6 |s2|, |p1| 6 3.For each n with 3 6 n 6 28 except for n = 10, there exist a ternary square-free word wwith a square-free self-shuffle p1p2s1s2 as above. The difference with the above is that wecannot always take w to be a factor of s and the lengths of s1 and p2 cannot be restrictedas much.Finally, for n = 10, one can take the square-free word w = 0102120102, which has thefollowing square-free self-shuffle:01020121020102120102.Combining this with the result of the previous section solves Harju’s problem:Theorem 3.2. For every n > 3, there exists a ternary square-free word of length n havinga square-free self-shuffle.References[1] A. Aberkane & J. D. Currie, There exist binary circular 5/2+ power free words ofevery length, Elec. J. Comb. 11 (2004), R10.[2] F.-J. Brandenburg, Uniformly growing k-th power-free homomorphisms, Theoret.Comput. Sci. 23 (1983), 69–82.[3] Émilie Charlier, Teturo Kamae, Svetlana Puzynina, & Luca Q. Zamboni. Self-shuffling words, arXiv: 1302.3844 (2013)[4] T. Harju, A note on square-free shuffles of words, LNCS, WORDS 2013. To appear.[5] D. Henshall, N. Rampersad, & J. Shallit, Shuffling and unshuffling, Bull. EATCS,107 (2012), 131–142.[6] M. Lothaire, Combinatorics on Words, Encyclopedia of Mathematics and its Appli-cations 17, Addison-Wesley, Reading, 1983.[7] S. Buss, M. Soltys, Unshuffling a square is NP-hard, arXiv: 1211.7161 (2013).[8] R. Rizzi and S. Vialette. On recognizing words that are squares for the shuffle product,CSR 2013, LNCS 7913 (2013), 235–245.1An IPython notebook showing these computations can be found in http://users.utu.fi/kasaar/square-free_shuffles.ipynbthe electronic journal of combinatorics 21(1) (2014), #P1.9 7

Square-free Words with Square-free Self-shuffles

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Square-free Words with Square-free Self-shuffles

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