Circular critical exponents for Thue-Morse factors
CCircular critical exponents for Thue-Morse factors
Jeffrey Shallit and Ramin ZarifiSchool of Computer ScienceUniversity of WaterlooWaterloo, ON N2L 3G1Canada [email protected]@edu.uwaterloo.ca
August 9, 2018
Abstract
We prove various results about the largest exponent of a repetition in a factor ofthe Thue-Morse word, when that factor is considered as a circular word. Our resultsconfirm and generalize previous results of Fitzpatrick and Aberkane & Currie.
Consider the English word amalgam ; it has a factor ama of period 2 and length 3, so wecan consider ama to be a power. However, if we think of amalgam as a “circular word” or“necklace”, where the word “wraps around”, then it has the factor amama of period 2 andlength 5. We say that amalgam has a circular critical exponent of .The famous Thue-Morse infinite word t = t t t · · · = · · · has been studied extensively since its introduction by Thue in 1912 [10, 4]. In particular,Thue proved that the largest repetitions in t are 2-powers (also called “squares”).It was only fairly recently, however, that the repetitive properties of its factors, consideredas circular words , have been studied. Fitzpatrick [6] showed that, for all n ≥
1, there is alength- n factor of t with circular critical exponent <
3. Aberkane and Currie [1] conjecturedthat for every n ≥
1, some length- n factor of t has circular critical exponent ≤ , and, usinga case analysis, they later proved this conjecture [2].In this paper we show how to obtain the Aberkane-Currie result, and much more, usingan approach based on first-order logic and the Walnut prover, written by Hamoon Mousavi. A factor is a contiguous block lying inside another word. a r X i v : . [ c s . F L ] A ug Basics
The i ’th letter of a word w is written w [ i ]. The notation w [ i..j ] represents the word w [ i ] w [ i + 1] · · · w [ j ] . If i ≥ j + 1, then w [ i..j ] = (cid:15) , the empty word.An infinite (resp., nonempty finite) word w has a period p ≥ w [ i ] = w [ i + p ] for all i ≥ i with 0 ≤ i < | w | − p ). For finite words of length n , we restrict our attentionto periods that are ≤ n . A word can have multiple periods; for example, the English word alfalfa has periods 3 ,
6, and 7. The smallest period is called the period and is denoted p ( w ). The exponent of a finite word w is defined to be exp( w ) = | w | /p ( w ); it measures thelargest amount of (fractional) repetition of a word. The period of alfalfa is 3, and it haslength 7; hence its exponent is .A word is called a square if its exponent is 2. If its exponent is greater than 2, it is calledan overlap . Thus, for example, the English word murmur is a square and the French word entente is an overlap.The critical exponent of a word w is the supremum, over all finite nonempty factors x of w , of exp( x ); it is denoted ce( w ). For example, Mississippi has critical exponent 7 / ississi .We can also define this notion for “circular words” (aka “necklaces”). We say two words x, y are conjugate if one is a cyclic shift of the other; alternatively, if there exist (possiblyempty) words u, v such that x = uv and y = vu . For example, the English words listen and enlist are conjugates.We let conj( w ) denote the set of all cyclic shifts of w :conj( w ) = { yx : ∃ x, y such that w = xy } . For example, the conjugates of ate are { ate , tea , eat } .Here is the most fundamental definition of our paper: Definition 1.
The circular critical exponent of a word w , denoted by cce( w ), is the supre-mum of exp( x ) over all finite nonempty factors x of all conjugates of w .Note that cce( w ) can be as much as twice as large as ce( w ). See [9] for more about thisnotion for infinite words. The Thue-Morse word t has many equivalent definitions [3], but for us it will be sufficientto describe it as the fixed point, starting with , of the morphism µ mapping → and → .A basic fact about the binary alphabet is that every word of length ≥ roposition 2. Let x be a nonempty factor of the Thue-Morse word. Then ce( x ) ∈ { , , } .Furthermore, ce( x ) = 2 if | x | ≥ . In this paper, we prove the analogue of Proposition 2 for the circular critical exponent.Here the statement is more complicated and the analysis more difficult.
Walnut
Our main software tool is the
Walnut prover, written by Hamoon Mousavi [8]. This Javaprogram deals with deterministic finite automata with output (DFAO’s) and k -automaticsequences ( a n ) n ≥ . A k -DFAO is a finite-state machine M = ( Q, Σ k , δ, q , ∆ , τ ), where Q isa finite nonempty set of states, Σ k = { , , . . . , k − } is the input alphabet, δ : Q × Σ k → Q is the transition function (which is extended to Q × Σ ∗ k in the obvious way), q is the initialstate, ∆ is the output alphabet, and τ : Q → ∆ is the output mapping. DFAO’s are anobvious generalization of ordinary DFA’s. A sequence ( a n ) n ≥ is said to be computed by the k -DFAO M if τ ( δ ( q , ( n ) k )) = a n , where ( n ) k denotes the base- k representation of n . (Unlessotherwise stated, we assume that all automata read the base- k representation of n from leftto right, starting with the most significant digit.) If a sequence ( a n ) n ≥ is computed by a k -DFAO, it is said to be k -automatic . Walnut can evaluate the truth of a first-order statement S involving indexing of k -automatic sequences, logical connectives, and quantifiers ∃ and ∀ . If there are free vari-ables, it produces an automaton accepting the base- k representation of the values of the freevariables for which S evaluates to true. One minor technical point is that the automata itproduces give the correct answer, even when the input is prefixed by any number of leadingzeroes.The syntax of Walnut statements is more or less self-explanatory. The interested readercan enter the
Walnut commands we give and directly reproduce our results.All computations in this paper, unless otherwise indicated, were performed on an AppleMacBook Pro with 16 GB of memory, running macOS High Sierra, version 10.13.3. All thecode we discuss is available for download at https://cs.uwaterloo.ca/~shallit/papers.html . For the Thue-Morse word, the computations all run in a matter of seconds.
There is a notion of minimality for DFAO’s that exactly parallels the notion of ordinaryDFA’s. We say that two states p, q of a k -DFAO are distinguishable if there exists a string x ∈ Σ ∗ k such that τ ( δ ( p, x )) (cid:54) = τ ( δ ( q, x )). Then the analogue of the Myhill-Nerode theoremfor DFAO’s is the following, which is easily proved: Proposition 3.
There is a unique minimal k -DFAO equivalent to any given k -DFAO. Fur-thermore, a k -DFAO M is minimal iff (a) every state of M is reachable from the start stateand (b) every pair of distinct states is distinguishable. We observe that the automata that
Walnut computes are guaranteed to be minimal.We will need the following lemma. 3 emma 4.
Let M = ( Q , Σ k , δ , q , ∆ , τ ) and M = ( Q , Σ k , δ , q , ∆ , τ ) be two minimalDFAO’s. Let M = ( Q, Σ k × Σ k , δ, q , ∆ × ∆ , τ ) be the cross product automaton defined by • Q = Q × Q ; • δ ([ p, q ] , a ) = [ δ ( p, a ) , δ ( q, a )] ; • q = [ q , q ] ; • τ ([ p, q ]) = [ τ ( p ) , τ ( q )] .Then every pair of distinct states of M is distinguishable.Proof. Let [ p, q ] and [ p (cid:48) , q (cid:48) ] be two distinct states of M . Without loss of generality, assume p (cid:54) = p (cid:48) . Then, since M is minimal, we know that p and p (cid:48) are distinguishable, so there exists x such that τ ( δ ( p, x )) (cid:54) = τ ( δ ( p (cid:48) , x )). Then τ ( δ ([ p, q ] , x )) = [ τ ( δ ( p, x )) , τ ( δ ( q, x ))] (cid:54) =[ τ ( δ ( p (cid:48) , x ) , τ ( δ ( q (cid:48) , x ))] = τ ( δ ([ p (cid:48) , q (cid:48) ] , x )). So [ p, q ] and [ p (cid:48) , q (cid:48) ] are distinguishable by x . Corollary 5.
Let M and M be minimal k -DFAO’s. Form their cross product automaton,and remove all states unreachable from the start state. The result is minimal. Corollary 5 gives a way to form the minimal cross product automaton, but in practice wecan do something even more efficient: namely, using a breadth-first approach, we can startfrom the start state [ q , q ] and incrementally add only those states reachable from it. We start by developing a useful first-order logical formula with free variables i, m, n, p, s . Wewant it to assert thatin the circular word given by the length- n word startingat position s in the Thue-Morse word, there is a factor (1) w of length m and (not necessarily least) period p ≥ i .In order to do this, we will conceptually repeat the word x = t [ s..s + n −
1] twice,as depicted below, where the black vertical line separates the two copies. The factor w isindicated in grey; it may or may not straddle the boundary between the two copies.4 s n + s n -1 s n +2 -1... i m + s ix second copy of x first copy of pmp Figure 1: Factor of a circular word of length n Here indices should be interpreted as “wrapping around”; the index s + n + j is the sameas s + j for 0 ≤ j < n . Then the assertion that w has period p potentially corresponds tothree different ranges of j : • Both j and j + p lie in the first copy of x , so we compare t [ j ] to t [ j + p ] for all j inthis range: i ≤ j < min( s + n − p, i + m − p ). • j lies in the first copy of x , but j + p lies in the second copy, so we compare t [ j ] to t [ j + p − n ] for all j in this range: max( i, s + n − p ) ≤ j < min( s + n, i + m − p ). • Both j and j + p lie in the second copy of x , so we compare t [ j − n ] to t [ j + p − n ] forall j in this range: max( i, s + n ) ≤ j < i + m − p .Putting this all together, we get the following logical formula that asserts the truth ofstatement (1):crep( i, m, n, p, s ) :=( ∀ j (( j ≥ i ) ∧ ( j < s + n − p ) ∧ ( j < i + m − p )) = ⇒ t [ j ] = t [ j + p ]) ∧ ( ∀ j (( j ≥ i ) ∧ ( j < s + n ) ∧ ( j ≥ s + n − p ) ∧ ( j < i + m − p )) = ⇒ t [ j ] = t [ j + p − n ]) ∧ ( ∀ j (( j ≥ i ) ∧ ( j ≥ s + n ) ∧ ( j < i + m − p )) = ⇒ t [ j − n ] = t [ j + p − n ])The translation into Walnut is as follows: def crep "(Aj ((j>=i)&(j+p T[j]=T[j+p]) &(Aj ((j>=i)&(j=s+n)&(j+p T[j]=T[(j+p)-n]) &(Aj ((j>=i)&(j>=s+n)&(j+p T[j-n]=T[(j+p)-n])":
The resulting automaton implementing crep( i, m, n, p, s ) has 1423 states. Note that ourformula does not impose conditions such as p ≥ p ≤ n or m ≤ n , which are required forcrep to make sense. These conditions (or stronger ones that imply them) must be includedin any predicate that makes use of crep. Neither does the predicate assert that the givenfactor’s smallest period is p ; just that p is one of the possible periods.5 Prefixes
In this section we prove the following theorem:
Theorem 6.
Every nonempty prefix of the Thue-Morse word has circular critical exponentin S := { , , , , , , } . Furthermore, we will precisely characterize the n for which the circular critical exponentis each member of S .We start by creating a first-order formula asserting that the length- n prefix, consideredas a circular word, has some factor of length m and (not necessarily least) period p , satisfying m/p = a/b :prefgeab( n ) := ∃ i, m, p ( p ≥ ∧ ( m ≤ n ) ∧ ( i < n ) ∧ ( bm ≥ ap ) ∧ crep( i, m, n, p, . Note that the condition p ≤ n need not be included explicitly, as it is implied by theconjunction of m ≤ n and bm ≥ ap .Next, we create a formula asserting that the length- n prefix, considered as a circularword, has a factor with exponent > a/b :prefgtab( n ) := ∃ i, m, p ( p ≥ ∧ ( m ≤ n ) ∧ ( i < n ) ∧ ( bm > ap ) ∧ crep( i, m, n, p, . Finally, we create a formula asserting that the length- n prefix has some factor of exponentexactly a/b : prefeqab( n ) := prefgeab( n ) ∧ ¬ prefgtab( n ) . No single
Walnut command can be the direct translation of the formulas above, as thereis no way to take arbitrary integer parameters a, b as input and perform multiplication bythem. Nevertheless, since there are only finitely many possibilities, we can translate theabove logical statements to finitely many individual
Walnut commands for each exponent a/b . For example, for 7 / def prefge73 "E i,m,p (p>=1) & (m<=n) & (i
We can now prove Theorem 6 by executing the
Walnut command eval testpref "An (n>=1) => ($prefeq11(n) | $prefeq21(n) | $prefeq73(n) |$prefeq52(n) | $prefeq135(n) | $prefeq83(n) | $prefeq31(n))": (where | represents OR), and verifying that Walnut returns true .Table 4 gives information about the state sizes of the automata for the exponents in S := { , , , , , , } .In fact, even more is true. We can create a single n outputsthe circular critical exponent of the prefix of length n of t . We do this by computing the6utomaton for each of the possible exponents, forming the cross product automaton, andproducing the appropriate output.number of number of number of a/b states for states for states for first few n accepted by prefeqabprefgeab prefgtab prefeqab1/1 2 4 3 1, 22/1 4 4 4 3 , , , , , , , , , , , . . . , , , , , , , . . . , , , , , , , , , , . . . , , , , , , . . . , , , , , , , , , , . . . , , , , , , , , , , . . . Table 1: State sizes for repetition of prefixes
Theorem 7.
There is a -DFAO of 29 states that, on input ( n ) , returns the circular criticalexponent of the prefix of length n of t .Proof. We cannot compute this automaton directly in
Walnut in its current version, but itcan be computed easily from the individual automata
Walnut computes for each exponentin S = { , , , , , , } .Now we can finish the (computational) proof of Theorem 7. We start with the automaton prefeq11 discussed above. Next, for each of the remaining exponents a/b , we iterativelyform the cross product of the current automaton with the automaton prefeqab , and removeunreachable states. After all exponents are handled, this gives the 29-state automatondepicted in Figure 2. 7
00 1 011
00 1 1
01 01 01 Figure 2: Automaton for prefixes of Thue-Morse
Remark . We tested the automaton in Figure 2 by explicitly calculating cce for the first500 prefixes of t and comparing the results. They agreed in every case. Instead of just prefixes, we can carry out the calculations of the previous section for allfactors. The goal is to prove the following result.
Theorem 9.
Every factor of the Thue-Morse word has circular critical exponent lying inthe finite set U := { , , , , , , , , , , , } .Proof. We can mimic the previous analysis. A length- n factor t [ s..s + n −
1] can be specifiedby the pair ( n, s ).We first make the assertion that the factor specified by ( n, s ), considered as a circularword, has a factor of length m that has a period p with m/p = a/b :facgeab( n ) = ∃ i, m, p ( p ≥ ∧ ( m ≤ n ) ∧ ( i ≥ s ) ∧ ( i < s + n ) ∧ ( bm ≥ ap ) ∧ crep( i, m, n, p, s ) . Next, we make the assertion that t [ s..s + n − > a/b :facgtab( n ) = ∃ i, m, p ( p ≥ ∧ ( m ≤ n ) ∧ ( i ≥ s ) ∧ ( i < s + n ) ∧ ( bm > ap ) ∧ crep( i, m, n, p, s ) . t [ s..s + n − a/b and no larger:faceqab( n ) = facgeab( n ) ∧ ¬ facgtab( n ) . number of number of number of first occurrence a/b states for states for states for ( n, s ) of factorfacgeab facgtab faceqab with cce = a/b eval testfac "An (n>=1) => (As ($faceq11(n,s) | $faceq21(n,s) |$faceq73(n,s) | $faceq177(n,s) | $faceq52(n,s) | $faceq135(n,s) |$faceq83(n,s) | $faceq31(n,s) | $faceq103(n,s) | $faceq72(n,s) |$faceq113(n,s) | $faceq41(n,s)))": and Walnut evaluates it to be true. Furthermore, it is easy to check that each possibilityoccurs at least once, as given in the table.
Theorem 10.
There is a 204-state -DFAO that, on input ( n, s ) in base , outputs cce( t [ s..s + n − . Proof.
As before, we use the product construction to combine the automata for faceqab( n, s )for all twelve possibilities for a/b . The automaton is too large to display here, but it isavailable at https://cs.uwaterloo.ca/~shallit/papers.html . Remark . We tested the correctness of our automaton by comparing its result to the resultof thousands of randomly-chosen factors of varying lengths of t . It passed all tests.9 .1 Smallest circular critical exponents for each length For every length n , we can consider the least circular critical exponent over all factors t [ s..s + n −
1] of length n . Definelcce( n ) = min x a factor of t | x | = n cce( x ) . Theorem 12.
For all n ≥ we have lcce( n ) ∈ T where T := { , , , , } .Proof. First, we create a first-order logic statement asserting that there exists some length- n factor whose circular exponent equals a/b :facab( n ) := ∃ s faceqab( n, s ) . Next, we create a statement asserting that a/b is the least circular critical exponent forwords of length n ; in other words, that there exists some length- n factor whose circular criti-cal exponent equals a/b , and furthermore every length- n factor has circular critical exponent ≥ a/b : facsmallab( n ) := facab( n ) ∧ ( ∀ s facgeab( n, s )) . Finally, we just assert that for every n ≥
1, at least one of the five alternatives holds: eval smallfactest "An (n >=1) => ($facsmall11(n) | $facsmall21(n) |$facsmall73(n) | $facsmall177(n) | $facsmall52(n))":Walnut evaluates this to be true.The sizes of the automata occurring in the proof are summarized below.number of number of a/b states for states for first few n accepted by facsmallabfacab facsmallab1/1 3 3 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Theorem 13.
There is a 25-state -DFAO, that on input ( n ) computes the least circularcritical exponent over all factor of t of length n . It is given in Figure 3. Figure 3: Automaton computing least possible cce of a factor, for each length
Proof.
Combine, using the cross product construction, the five automata facsmallab for a/b ∈ { / , / , / , / , / } as before. The output of each state is depicted in the centerof the corresponding circle. We can also consider the greatest circular critical exponent over all factors t [ s..s + n −
1] oflength n . Define gcce( n ) = max x a factor of t | x | = n cce( x ) . Theorem 14.
For all n ≥ we have gcce( n ) ∈ V where V := { , , , , } .Proof. We definefaclargeab( n ) = ( ∃ s faceqab( n, s )) ∧ ( ∀ s ¬ facgtab( n, s )) . The number of states, and the first few n that match the category, are given in Table 4.We then verify the claim by writing eval largefactest "An (n>=1) => ($faclarge11(n) | $faclarge21(n) |$faclarge31(n) | $faclarge72(n) | $faclarge41(n))": which returns true . 11umber of a/b states for first few n matching the casefaclargeab1/1 2 12/1 3 2,33/1 8 4,5,9,13,15,17,21,25,29,33,37,41,45,49,53,57,61, . . .7/2 7 7,11,19,23,27,31,35,39,43,47,51,55,59,63, . . .4/1 5 6,8,10,12,14,16,18,20,22,24,26,28,30,32, . . .Table 4: State sizes for faclargeab Theorem 15.
There is a 9-state -DFAO, that on input ( n ) , returns the greatest circularcritical exponent over all length- n factors of t .Proof. We follow the same approach as before, using the cross product construction tocombine the automata faclargeab for a/b ∈ { , , , , } . The result is depicted in Figure 4. Figure 4: Automaton computing greatest possible cce of a factor, for each length12 .3 Sets of circular critical exponents
We can get even more! Define the set of all possible circular critical exponents of factors oflength n as follows:ace( n ) = { cce( x ) : | x | = n ≥ x is a factor of t } . Theorem 16.
The range of ace( n ) consists of exactly distinct sets as enumerated inTable 5.Proof. This follows immediately from our proof of the next result.
Theorem 17.
There is a 49-state -DFAO that, on input n written in base , outputs ace( n ) .Proof. We use the same cross product automaton technique as before. This time, we use theautomata facab for each a/b ∈ U . The result is depicted in Figure 5. The outputs associatedwith each state are encoded as 12-bit numbers, one for each of the 12 possible exponents inincreasing order, with least significant bit corresponding to exponent 4. Square states are“transient” and circular states are “recurrent”. Evidently one could (in principle) perform the same sort of analysis for many other famousinfinite words. We carried this out for the regular paperfolding word p = 00100110001101100010 · · · (see, for example, [7, 5]), and the results are summarized below. We omit the details, butthe Walnut code proving these results is available at https://cs.uwaterloo.ca/~shallit/papers.html . The computations were nontrivial.
Walnut was invoked using the Linuxcommand java -Xmx16000M -d64 Main.prover on a 4 CPU AMD Opteron 6380 SE with 256GB RAM. The analogue of crep for p has 4226states and took 9 minutes to compute. The largest intermediate automaton had 822,161states. Theorem 18. (a) Every nonempty prefix of p has circular critical exponent lying in { , , , , , , , } .(b) Every nonempty factor of p has circular critical exponent lying in { , , , , , , , , , , , , } .(c) The least circular critical exponent of p , over all factors of length n , lies in { , , , , , , } . d) The greatest circular critical exponent of p , over all factors of length n , lies in { , , , , , } .(e) There are exactly 16 distinct possible sets of circular critical exponents for factors oflength n ≥ of p . In principle, we could also treat the Rudin-Shapiro sequence. For example, one might beable to prove the following.
Conjecture 19.
Every nonempty factor of the Rudin-Shapiro sequence has a circular criticalexponent lying in { , , , , , , , , , , , , , , , , } . However, so far we have not been able to complete the computations with
Walnut (it runsout of space).For some infinite words, the sets under consideration will be infinite, and hence anotherkind of analysis will be needed. As an example, consider the infinite word 210201 · · · thatis a fixed point of 2 → →
20, 0 →
1. It is well-known that this word is squarefree,but contains factors with exponent arbitrarily close to 2. In this case there will be no finiteanalogue of our Proposition 2 and Theorem 6. The same case occurs for the Fibonacci word(the fixed point of 0 →
01 and 1 → The first author acknowledges, with thanks, conversations with Daniel Goˇc on the subjectin 2013. 14et of circular encoding first few n forcritical exponents S in automaton ace( n ) = S { } { }{ , } { }{ } { }{ , } { }{ , } { }{ , } { }{ , , } { }{ , , } { , , , , , , , , , , . . . }{ , , } { , }{ , , , } { , , , , , . . . }{ , , } { }{ , , , } { }{ , , , } { , , , , , . . . }{ , , , } { , , , . . . }{ , , , , } { , , , , , , , . . . }{ , , , } { , , . . . }{ , , , } { }{ , , , , } { , , , , . . . }{ , , , } { , , . . . }{ , , , , } { , , , , , , , , , , , , , . . . }{ , , , , , } { , , , , . . . }{ , , , , } { , , , , , , , , , , , , . . . }{ , , , , } { , , , , , , , , , , , , . . . }{ , , , , , , } { , , , , , . . . }{ , , , , } { , , , , . . . }{ , , , , , , } { , , , . . . }{ , , , , , } { , , , , . . . }{ , , , , , , , } { , , , , , , , , , , . . . }{ , , , , , , } { , , , , , , . . . }{ , , , , , , } { , , , , , , . . . }{ , , , , , , } { , , . . . } Table 5: Sets of circular critical exponents for lengths n / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /
010 1 / / / / / / /
010 10 1 010 10 1 / /
01 0 F i g u r e : A u t o m a t o n c o m pu t i n g s e t s o f c i r c u l a r c r i t i c a l e x p o n e n t s f o r f a c t o r s o f l e n g t h n eferences [1] A. Aberkane and J. D. Currie. There exist binary circular 5 / + power free wordsof every length. Electronic J. Combinatorics (1) (2004), [2] A. Aberkane and J. D. Currie. The Thue-Morse word contains circular 5 / + power freewords of every length. Theoret. Comput. Sci. (2005), 573–581.[3] J.-P. Allouche and J. O. Shallit. The ubiquitous Prouhet-Thue-Morse sequence. InC. Ding, T. Helleseth, and H. Niederreiter, editors,
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