Fife's Theorem for (7/3)-Powers
PP. Ambroˇz, ˇS. Holub and Z. Mas´akov´a (Eds.):8th International Conference WORDS 2011EPTCS 63, 2011, pp. 189–198, doi:10.4204/EPTCS.63.25
Fife’s Theorem for -Powers Narad Rampersad
Department of Mathematics, University of Li`ege, Grande Traverse, 12 (Bat. B37), 4000 Li`ege, Belgium [email protected]
Jeffrey Shallit
School of Computer Science, University of Waterloo, Waterloo, ON N2L 3G1, Canada [email protected]
Arseny Shur
Department of Algebra and Discrete Mathematics, Ural State University, Ekaterinburg, Russia [email protected]
We prove a Fife-like characterization of the infinite binary -power-free words, by giving a finiteautomaton of 15 states that encodes all such words. As a consequence, we characterize all suchwords that are 2-automatic. An overlap is a word of the form axaxa , where a is a single letter and x is a (possibly empty) word. In1980, Earl Fife [8] proved a theorem characterizing the infinite binary overlap-free words as encodings ofpaths in a finite automaton. Berstel [4] later simplified the exposition, and both Carpi [6] and Cassaigne[7] gave an analogous analysis for the case of finite words.In a previous paper [13], the second author gave a new approach to Fife’s theorem, based on thefactorization theorem of Restivo and Salemi [12] for overlap-free words. In this paper, we extend thisanalysis by applying it to the case of -power-free words.Given a rational number pq >
1, we define a word w to be a pq -power if w can be written in the form x n x ′ where n = ⌊ p / q ⌋ , x ′ is a (possibly empty) prefix of x , and | w | / | x | = p / q . The word x is called a period of w , and p / q is an exponent of w . If p / q is the largest exponent of w , we write exp ( w ) = p / q .We also say that w is | x | -periodic . For example, the word alfalfa is a -power, and the correspondingperiod is alf . Sometimes, as is routine in the literature, we also refer to | x | as the period; the contextshould make it clear which is meant.A word, whether finite or infinite, is b -power-free if it contains no factor w that is an a -power for a ≥ b . A word is b + -power-free if it contains no factor w that is an a -power for a > b . Thus, theconcepts of “overlap-free” and “2 + -power-free” coincide. Let S be a finite alphabet. We let S ∗ denote the set of all finite words over S and S w denote the set of all(right-) infinite words over S . We say y is a factor of a word w if there exist words x , z such that w = xyz .If x is a finite word, then x w represents the infinite word xxx · · · .90 Fife’s Theorem for 7 / S = { , } . The most famous infinite binary overlap-free word is t , the Thue-Morse word, defined as the fixed point, starting with 0, of the Thue-Morse morphism m , which maps 0to 01 and 1 to 10. We have t = t t t · · · = · · · . The morphism m has a second fixed point, t , which is obtained from t by applying the complementationcoding defined by 0 = = F / denote the set of (right-) infinite binary -power-free words. We point out that thesewords are of particular interest, because is the largest exponent a such that there are only polynomially-many a -power-free words of length n [9]. The exponent plays a special role in combinatorics on words,as testified to by the many papers mentioning this exponent (e.g., [10, 14, 9, 11, 1, 5]).We now state a factorization theorem for infinite -power-free words: Theorem 1.
Let x ∈ F / , and let P = { p , p , p , p , p } , where p = e , p = , p = , p = , andp = . Then there exists y ∈ F / and p ∈ P such that x = p m ( y ) . Furthermore, this factorization isunique, and p is uniquely determined by inspecting the first letters of x .Proof. The first two claims follow immediately from the version for finite words, as given in [9]. Thelast claim follows from exhaustive enumeration of cases.We can now iterate this factorization theorem to get
Corollary 2.
Every infinite -power-free word x can be written uniquely in the form x = p i m ( p i m ( p i m ( · · · ))) (1) with i j ∈ { , , , , } for j ≥ , subject to the understanding that if there exists c such that i j = forj ≥ c, then we also need to specify whether the “tail” of the expansion represents m w ( ) = t or m w ( ) = t .Furthermore, every truncated expansionp i m ( p i m ( p i m ( · · · p i n − m ( p i n ) · · · ))) is a prefix of x , with the understanding that if i n = , then we need to replace p i n with either (if the“tail” represents t ) or (if the “tail” represents t ).Proof. The form (1) is unique, since each p i is uniquely determined by the first 5 characters of theassociated word.Thus, we can associate each infinite binary -power-free word x with the essentially unique infinitesequence of indices i : = ( i j ) j ≥ coding elements in P , as specified by (1). If i ends in 0 w , then we needan additional element (either 1 or 3) to disambiguate between t and t as the “tail”. In our notation, weseparate this additional element with a semicolon so that, for example, the string 000 · · · ; 1 represents t and 000 · · · ; 3 represents t .Of course, not every possible sequence of ( i j ) j ≥ of indices corresponds to an infinite -power-free word. For example, every infinite word coded by an infinite sequence beginning 400 · · · has a -power.Our goal is to characterize precisely, using a finite automaton, those infinite sequences corre-sponding to -power-free words.Next, we recall some connections between the morphism m and the powers over the binary alphabet.Below x is an arbitrary finite or right-infinite word. Lemma 3.
If the word m ( x ) has a prefix zz, then the word x has the prefix m − ( z ) m − ( z ) . .Rampersad, J.Shallit,and A.Shur 191 Proof.
Follows immediately from [3, Lemma 1.7.2].
Lemma 4. (1) For any real b > , we have exp ( x ) = b iff exp ( m ( x )) = b .(2) For any real b ≥ + , the word x is b -power-free iff m ( x ) is b -power-free.Proof. For (1), see [14, Prop. 1.1]. For (2), see [14, Prop. 1.2] or [9, Thm. 5].
Lemma 5.
Let p be a positive integer. If the longest p-periodic prefix of the word m ( x ) has the exponent b ≥ , then the longest ( p / ) -periodic prefix of x also has the exponent b .Proof. Let zzz ′ (where | z | = p and z ′ is a possibly empty prefix of z w ) be the longest p -periodic prefix of m ( x ) . Lemma 3 implies that p is even. If | z ′ | is odd, let a be the last letter of z ′ . The next letter b in m ( x ) is fixed by the definition of m : b = a . By the definition of period, another a occurs p symbols to the leftof the last letter of z ′ . Since p is even, this a also fixes the next letter b . Hence the prefix zzz ′ b of m ( x ) is p -periodic, contradicting the definition of zzz ′ . Thus | z ′ | is even. Therefore x begins with the b -power m − ( z ) m − ( z ) m − ( z ′ ) of period p / x has a ( p / ) -periodic prefix y of exponent a > b , then by Lemma 4 (1), the p -periodic prefix m ( y ) of m ( x ) also has the exponent a , contradicting the hypotheses of the lemma. For each finite word w ∈ { , , , , } ∗ , w = i i · · · i r , and an infinite word x ∈ { , } w , we define C w ( x ) = p i m ( p i m ( p i m ( · · · x · · · ))) and F w = { x ∈ S w : C w ( x ) ∈ F / } . Note that F w ⊆ F / for any w in view of Lemma 4 (2). Lemma 6.
The sets F w satisfy the equalities listed in Fig. 1. In particular, there are only 15 differentnonempty sets F w ; they areF e , F , F , F , F , F , F , F , F , F , F , F , F , F , F . Proof.
Due to symmetry, it is enough to prove only the 30 equalities from the upper half of Fig. 1 andthe equality F = F e . We first prove the emptiness of 15 sets from the upper half of Fig. 1.Four sets: F , F , F , and F , consist of words that start 000.Eight sets consist of words that contain the factor 0 m ( ) = F , F , F , F , F , and F ), its m -image ( F ), or the complement of its m -image ( F ).Two sets: F and F , consist of words that start 00 m ( ) = F have the form 00 m ( x ) ; each of these words starts either 000 or 0010010.Each of the 16 remaining equalities has the form F w = F w . We prove them by showing that foran arbitrary x ∈ F / , the words u = C w ( x ) and u = C w ( x ) are either both -power-free or bothnot. In most cases, some suffix of u coincides with the image of u under some power of m . Then byLemma 4 (2) the word u can be -power-free only if u is -power-free. In these cases, it suffices tostudy u assuming that u is -power-free.When we refer to a “forbidden” power in what follows, we mean a power of exponent ≥ . F = F e : By Lemma 4 (2), u = m ( x ) is -power-free iff u = x is -power-free.92 Fife’s Theorem for 7 / F e F = F e F F F = F e F F = F F F = ∅ F = ∅ F = ∅ F F = F F = ∅ F = F F = F F = F F = F F = ∅ F F = F F = ∅ F = ∅ F = ∅ F = ∅ F = ∅ F = ∅ F F = F F = F F = F F = F F = ∅ F = ∅ F = F F = ∅ F = ∅ F = F F = F F F F = F e F F = F F F = ∅ F = ∅ F = ∅ F F = F F = ∅ F = F F = F F = F F = F F = ∅ F F = F F = ∅ F = ∅ F = ∅ F = ∅ F = ∅ F = ∅ F F = F F = F F = F F = F F = ∅ F = ∅ F = F F = ∅ F = ∅ F = F F = F Figure 1:
Equationsbetweenlanguages F w . .Rampersad, J.Shallit,and A.Shur 193 F = F e : The word u = m ( m ( x )) contains a m -image of u = x . If x is -power-free, then so is m ( x ) . Hence, if u has a forbidden power, then this power must be a prefix of u .Now let b < / x and q be the smallest period of aprefix of exponent b in x . Write b = p / q . Then the word m ( x ) has a prefix of exponent b and of period4 q by Lemma 4 (1), but no prefixes of a bigger exponent or of the same exponent and a smaller period byLemma 5. Hence u has no prefixes of exponent greater than ( p + ) / ( q ) . Since p and q are integers,we obtain the required inequality: pq < = ⇒ p < q = ⇒ p + < q = ⇒ p + q < . F = F : The word u = m ( m ( x )) contains a m -image of u = m ( x ) . Suppose that u is -power-free. Then it starts 0010011. Since the factor 001001 cannot occur in a m -image, we note that ( ⋆ ) the word 00 m ( x ) has only two prefixes of exponent 2 (00 and 001001) and no prefixes of biggerexponents.By Lemma 5, the word m ( u ) has only two prefixes of exponent 2 ( m ( ) and m ( ) ) and noprefixes of bigger exponents. Thus, the word u = m ( u ) is obviously -power-free. F = F : We have u = m ( m ( x )) = u . Suppose the word u is -power-free; then it starts010011. A forbidden power in u , if any, occurs at the beginning and hence contains 0010011. But00100 does not occur later in this word, so no such forbidden power exists. F = F : The word u = m ( m ( m ( x ))) = m ( x ) is a suffix of the m -image of u = m ( x ) .Hence, if u is -power-free, then by Lemma 4 (2) u is -power-free as well.For the other direction, assume u is -power-free and then m ( x ) is -power-free. So, if u containssome power yyy ′ with | y ′ | ≥ | y | /
3, then this power must be a prefix of u . Put y = z and y ′ = z ′ . Theword m ( x ) starts z z z ′ . Hence x starts m − ( z ) m − ( z ) by Lemma 3. So we conclude that | z | = | y | isan even number. Now let | y | = q and p = | yyy ′ | so that p / q ≥ /
3. Thus the word 1 u = m ( u ) startswith a ( p / q ) -power of period 4 q . Since u is -power-free, we have ( p − ) / q < / p ≥ q and 3 p − / < q . Since p and q are integers this means3 p = q and hence q is divisible by 3. On the other hand from above q is even. So q is divisible by 6.Now | y ′ | = | y | / | y ′ | is even. But then z ′ is odd, and begins at an even position in a m -image, so thecharacter following z ′ is fixed and must be the same character as in the corresponding position of z , say a . Thus z z z ′ a is a ( / ) -power occurring in m ( x ) , a contradiction. F = F : The word u = m ( m ( m ( x ))) = m ( x ) contains a m -image of u = m ( m ( x )) = m ( x ) . Suppose u is -power-free but to the contrary u = m ( u ) has a forbidden power. ByLemma 4 (2), this power must be a prefix of u . Note that this power can be extended to the left by 1(not by 0, because a m -image cannot contain 000). Hence the word 1 u = m ( x ) starts with a forbid-den power. This induces a forbidden power at the beginning of 1 x ; this power has a period q and someexponent p / q ≥ /
3. Then u has a prefix of exponent ( p − ) / q ≥ /
3. On the other hand the word m ( u ) is -power-free, whence ( p − ) / q < /
3. So we get the system of inequalities 3 p − / ≥ q ,3 p − / < q . This system has no integer solutions, a contradiction. F = F : We have u = m ( m ( m ( x ))) = m ( m ( x )) = m ( u ) . In view of ( ⋆ ) , one caneasily check that if u is -power-free, then so is u . F = F : We have u = m ( m ( m ( x ))) = m ( u ) . Suppose u is -power-free. Then m ( x ) starts 01101001. Assume to the contrary that u has a forbidden power. By Lemma 4 (2), this power94 Fife’s Theorem for 7 / u . Again, this power can be extended to the left by 1, not by 0. Hence the word1 u = m ( m ( x )) starts with a forbidden power, thus inducing a forbidden power at the beginning of u = m ( x ) = · · · . The word u has only two squares as prefixes (11 and 110110, cf. ( ⋆ ) ). Hence u has the prefix 11011010 and no forbidden factors except for the ( / ) -power prefix 1101101. Therefore,the word u has no prefixes of exponent ≥ / F = F : We have u = m ( m ( m ( x ))) = m ( u ) . Suppose u is -power-free. Then the word u = m ( m ( x )) is -power-free by the equality F = F , which is symmetric to F = F provedabove. But u is a suffix of m ( u ) , whence the result. F = F : We have u = m ( m ( m ( x ))) = m ( u ) . Suppose u is -power-free. Using theobservation symmetric to ( ⋆ ) , we check by inspection that u contains no forbidden power. F = F : Neither one of the words u = m ( m ( m ( m ( x )))) = m ( x ) , u = m ( m ( m ( x ))) = m ( x ) contains an image of the other. The proofs for both directions are essentially the same, so wegive only one of them. Let u be -power-free; then the words m ( x ) , x , and m ( x ) are -power-free aswell, and x starts 0. A simple inspection of short prefixes of u shows that if this word is not -power-free, then some b -power with b ≥ m ( x ) . By Lemma 5, the word x also starts with a b -power. The argument below will be repeated, with small variations, for several identities. ( ∗ ) Consider a prefix yyy ′ of x which is the longest prefix of x with period | y | . Then | y ′ | < | y | / m ( x ) having period 8 | y | is m ( yyy ′ ) . If some word z m ( yyy ′ ) also has period 8 | y | , then z should be a suffix of a m -image of some word. Sincethe word 010 is not such a suffix, then 10 m ( yyy ′ ) is the longest possible ( | y | ) -periodic wordcontained in u . Let us estimate its exponent. Since | z | and | y ′ | are integers, we have8 | y ′ | < | y | / = ⇒ | y ′ | < | y | = ⇒ | y ′ | + < | y | = ⇒ | y ′ | + < | y | / , whence exp ( m ( yyy ′ )) < /
3. Since we have chosen an arbitrary prefix yyy ′ of x , we concludethat the word u is -power-free. F = F : The word u = m ( m ( m ( m ( x )))) = m ( m ( x )) contains a m -image of u = m ( x ) .Suppose u is -power-free. It suffices to check that the prefix 010 of u does not complete any prefix of m ( u ) to a forbidden power. For short prefixes, this can be checked directly, while long prefixes that canbe completed in this way should have exponents ≥
2. By Lemma 5, a prefix of period p and exponent b ≥ m ( u ) corresponds to the prefix of the word u having the exponent b and the period p /
8. So, we repeat the argument ( ∗ ) replacing x with u to obtain that u is -power-free. F = F : We have u = m ( m ( m ( m ( x )))) = m ( m ( x )) = m ( u ) . Suppose u is -power-free. By ( ⋆ ) and Lemma 5, among the prefixes of m ( u ) there are only two squares, m ( ) and m ( ) , and no words of bigger exponent. By direct inspection, u is ( / ) -free. F = F : Neither one of the words u = m ( m ( m ( m ( x )))) = m ( x ) and u = m ( m ( m ( x ))) = m ( x ) contains an image of the other. If the word u is assumed to be -power-free, then the proof repeatsthe proof of the identity F = F , up to renaming all 0’s to 1’s and vice versa. Let u be -power-free. The words m ( x ) and x are also -power-free, and x begins with 1, assuring that there are no shortforbidden powers in the beginning of u . Concerning long forbidden powers, we consider, similar to ( ∗ ) ,a prefix yyy ′ of x which is the longest prefix of x with period | y | . The longest possible ( | y | ) -periodic.Rampersad, J.Shallit,and A.Shur 195word contained in u is 01001 m ( yyy ′ ) , because 001001 is not a suffix of a m -image. As in ( ∗ ) , weobtain 16 | y ′ | + < | y | /
3, implying exp ( m ( yyy ′ )) < /
3. Hence the word u is -power-free. F = F : The word u = m ( m ( m ( m ( x )))) = m ( m ( x )) contains a m -image of u = m ( m ( x )) = m ( x ) . Again, if the word u is -power-free, then so is m ( u ) , and it suffices to checkthat the prefix 00 of u does not complete any prefix of m ( u ) to a forbidden power. Similar to ( ∗ ) ,consider a prefix yyy ′ of u which is the longest prefix of u with period | y | . The longest possible ( | y | ) -periodic word contained in u is 0 m ( yyy ′ ) , because 00 is not a suffix of a m -image. As in ( ∗ ) , we seethat 4 | y ′ | + < | y | /
3, implying exp ( m ( yyy ′ )) < /
3, and conclude that the word u is -power-free. F = F : The word u = m ( m ( m ( m ( x )))) = m ( m ( x )) contains a m -image of u = m ( x ) . Suppose u is -power-free. Using ( ⋆ ) and Lemma 5, we conclude that among the prefixes of m ( u ) there are only two squares, m ( ) and m ( ) , and no words of bigger exponent. By directinspection, u is ( / ) -free. e
031 001300 24241303 310131 0 3 440 1 2213 31 0 2 1200 4 340
Figure 2: Automaton coding infinite binary -power-free wordsFrom Lemma 6 and the results above, we get Theorem 7.
Every infinite binary -power-free word x is encoded by an infinite path, starting in F e ,through the automaton in Figure 2.Every infinite path through the automaton not ending in w codes a unique infinite binary -power-free word x . If a path i ends in w and this suffix corresponds to a cycle on state F e then x is coded byeither i ; 1 or i ; 3 . If a path i ends in w and this suffix corresponds to a cycle on F , then x is coded by i ; 3 . If a path i ends in w and this suffix corresponds to a cycle on F , then x is coded by i ; 1 .Remark . Blondel, Cassaigne, and Jungers [5] obtained a similar result, and even more general ones,for finite words. The main advantage to our construction is its simplicity.96 Fife’s Theorem for 7 / Corollary 9.
Each of the 15 sets F e , F , F , F , F , F , F , F , F , F , F , F , F , F , F isuncountable.Proof. It suffices to provide uncountably many distinct paths from each state to itself. By symmetry, itsuffices to prove this for all the states labeled e or below in Figure 2. These are as follows: • e : ( + ) w • ( + ) w • ( + ) w • ( + ) w • ( + ) w • ( + ) w • ( + ) w • ( + ) w . Corollary 10.
For all words w ∈ { , , , , } ∗ , either F w is empty or uncountable. -power-free word Theorem 11.
The lexicographically least infinite binary -power-free word is t .Proof. By tracing through the possible paths through the automaton we easily find that 2030 w ; 1 is thecode for the lexicographically least sequence. Remark . This result does not seem to follow directly from [2] as one referee suggested. -power-free words As a consequence of Theorem 7, we can give a complete description of the infinite binary -power-free words that are 2-automatic [3]. Recall that an infinite word ( a n ) n ≥ is k -automatic if there existsa deterministic finite automaton with output that, on input n expressed in base k , produces an outputassociated with the state last visited that is equal to a n . Alternatively, ( a n ) n ≥ is k -automatic if its k -kernel { ( a k i n + j ) n ≥ : i ≥ ≤ j < k i } consists of finitely many distinct sequences. Theorem 13.
An infinite binary -power-free word is -automatic if and only if its code is both specifiedby the DFA given above in Figure 2, and is ultimately periodic. First, we need a lemma:
Lemma 14.
An infinite binary word x = a a a · · · is -automatic if and only if m ( x ) is -automatic.Proof. Proved in [13]..Rampersad, J.Shallit,and A.Shur 197Now we can prove Theorem 13.
Proof.
Suppose the code of x is ultimately periodic. Then we can write its code as yz w for some finitewords y and z . Since the class of 2-automatic sequences is closed under appending a finite prefix [3,Corollary 6.8.5], by Lemma 14, it suffices to show that the word coded by z w is 2-automatic.The word z w codes a -power-free word w satisfying w = t j ( w ) , where t is a finite word and j = m k .Hence, by iteration, we get that w = t j ( t ) j ( t ) · · · . It is now easy to see that the 2-kernel of w iscontained in S : = { u m i ( v ) m i + k ( v ) m i + k ( v ) · · · : | u | ≤ | t | and v ∈ { t , t } and 1 ≤ i ≤ k } , which is a finite set.On the other hand, suppose the code for x is not ultimately periodic. Then we show that the 2-kernelis infinite. Now it is easy to see that if the code for x is a y for some letter a ∈ { , , } then one of thesequences in the 2-kernel (obtained by taking either the odd- or even-indexed terms) is either coded by y or its complement is coded by y . On the other hand, if the code for x is a y with a ∈ { , } , then y beginswith 0 , , or 3, say y = b z . It follows that the subsequences obtained by taking the terms congruent to0 , , , or 3 (mod 4) is coded by z , or its complement is coded by z . Since the code for x is not ultimatelyperiodic, there are infinitely many distinct sequences in the orbit of the code for x , under the shift. By theinfinite pigeonhole principle, infinitely many correspond to a sequence in the 2-kernel, or its complement.Hence x is not 2-automatic. We are grateful to the referees for their helpful suggestions.
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