Unary Pattern Avoidance in Partial Words Dense with Holes

Transcription

1 Unary Pattern Avoidance in Partial Words Dense with Holes F Blanchet-Sadri 1 Kevin Black 2 Andrew Zemke 3 1 University of North Carolina at Greensboro 2 Harvey Mudd College 3 Rochester Institute of Technology LATA 2011 This material is based upon work supported by the National Science Foundation under Grant No DMS The Department of Defense is also gratefully acknowledged

2 Outline 1 Avoidable patterns in words 2 Avoidable patterns in partial words 3 Hole sparsity 4 Non-trivial minimum hole sparsity for unary patterns 5 Minimum hole sparsity for unary patterns 6 Classification of all unary patterns with respect to hole sparsity

3 1 Avoidable patterns in words A pattern is a word over an alphabet of variables, which are denoted by α, β, γ, etc The Thue-Morse word w meets α 2 (or α 2 occurs in w): abbabaabbaababbabaababbaabbabaab But w avoids α 3 (we say that α 3 is 2-avoidable or is avoidable over the 2-letter alphabet)

4 Deciding the avoidability of a pattern Bean et al and Zimin proved that it is decidable whether a pattern p is avoidable; in fact, if p is over m variables, then p is avoidable if and only if w m avoids p, where w m is recursively defined by w 1 = 1 and w m = w m 1 mw m 1, m > 1 For example, w 2 = 121 does not avoid αβα but avoids ααβα, implying that αβα is unavoidable but ααβα is avoidable However, the complexity of deciding avoidability has remained open D R Bean, A Ehrenfeucht and G McNulty, Avoidable patterns in strings of symbols, Pacific Journal of Mathematics 85 (1979) A I Zimin, Blocking sets of terms, Mathematics of the USSR-Sbornik 47 (1984)

5 Computing the avoidability index of a pattern The problem Is it decidable, given a pattern p and an integer k, whether p is k-avoidable (or avoidable over a k-letter alphabet)? has also remained open An alternative is the problem of classifying all the patterns over a fixed number of variables m, that is, to find the smallest k such that p is k-avoidable, called the avoidability index of p, where p is such pattern For m = 1: α is unavoidable; α 2 is 2-unavoidable but 3-avoidable; α n, n 3, is 2-avoidable

6 2 Avoidable patterns in partial words A partial word is a sequence of symbols over a finite alphabet that may have some undefined positions, called holes and denoted by s, that match every letter of the alphabet (we also say that is compatible with each letter of the alphabet) a bca b is a partial word with two holes over {a, b, c} A full word is a partial word without holes aabcabb is a full word over {a, b, c} How dense with holes can a partial word defined over a fixed alphabet be, while still avoiding a given pattern?

7 2 Avoidable patterns in partial words A partial word is a sequence of symbols over a finite alphabet that may have some undefined positions, called holes and denoted by s, that match every letter of the alphabet (we also say that is compatible with each letter of the alphabet) a bca b is a partial word with two holes over {a, b, c} A full word is a partial word without holes aabcabb is a full word over {a, b, c} How dense with holes can a partial word defined over a fixed alphabet be, while still avoiding a given pattern?

8 Compatibility and containment Let u, v be partial words over A of equal length u and v are compatible, u v, if u(i) = v(i) whenever u(i), v(i) A a b a a b a b a a a a a u is contained in v, u v, if u(i) = v(i) whenever u(i) A a b a a b a a b a a a a a

9 Occurrence of a pattern in a partial word The pattern p = α 0 α n 1, where each α i is a variable, occurs in a partial word w, or w meets p, if there is a factor u 0 u n 1 in w, where u i, u j are non-empty and compatible whenever α i = α j ; otherwise, w avoids p or w is p-free An occurrence u 0 u n 1 of p is non-trivial if u i for all i = 0,, n 1 Otherwise, the occurrence is called trivial We call w non-trivially p-free if w contains no non-trivial occurrences of p αββα occurs in ab b a bba, while babbbaaab avoids αββα (the underlined occurrence of αββα is trivial)

10 Pattern (un)avoidability A pattern p is k-avoidable if there are infinitely many partial words over A with h holes, for any integer h > 0, that avoid p, where A is any alphabet of size k Note that if there is a partial word over A with infinitely many holes that avoids p, then p is obviously k-avoidable If, for some integer h 0, p occurs in every long enough partial word over A with h holes, then p is k-unavoidable A pattern p which is k-avoidable for some k is simply called avoidable, and a pattern which is k-unavoidable for every k is called unavoidable The avoidability index of p is the smallest integer k such that p is k-avoidable, or is if p is unavoidable

11 Division and equivalence of patterns If a pattern p occurs in a pattern q, then p divides q, denoted by p q αα αβα but αα αβαβ When both p q and q p hold, p and q are equivalent αα and ββ are equivalent

12 Classification of all binary patterns with respect to non-trivial partial word avoidability With respect to non-trivial avoidability in partial words, the avoidability index of a binary pattern is the same as in the full word case, that is, binary patterns fall into three categories: 1 The binary patterns ε, α, αβ, αβα, and their complements, are unavoidable (or have avoidability index ) 2 The binary patterns αα, ααβ, ααβα, ααββ, αβαβ, αββα, ααβαα, ααβαβ, their reverses, and complements, have avoidability index 3 3 All other binary patterns, and in particular all binary patterns of length six or more, have avoidability index 2 F Blanchet-Sadri, R Mercaş, S Simmons and E Weissenstein, Avoidable binary patterns in partial words, Acta Informatica 48 (2011) 25 41

13 Classification of all binary patterns with respect to partial word avoidability For partial words, binary patterns fall into four categories: 1 The binary patterns ε, α, αα, ααβ, ααβα, ααβαα, αβ, αβα, and their complements, are unavoidable (or have avoidability index ) 2 The binary patterns ααβαβ, ααββ, αβαβ and αβαβα, their reverses, and complements, have avoidability index 3 3 The binary pattern αββα has avoidability index 3 or 4 4 All other binary patterns, and in particular all binary patterns of length six or more, have avoidability index 2 F Blanchet-Sadri, R Mercaş, S Simmons and E Weissenstein, Avoidable binary patterns in partial words, Acta Informatica 48 (2011) 25 41

14 3 Hole sparsity Results for pattern avoidance in partial words have thus far been obtained by hole insertions in selected positions of words For example, ϕ ω (a) avoids αβαβα, where ϕ is the Thue-Morse morphism such that ϕ(a) = ab and ϕ(b) = ba There are infinitely many non-overlapping occurrences of ϕ 7 (a) starting at an even position in ϕ ω (a) and for each such occurrence, put a hole at position 47: abbabaabbaababbabaababbaabbabaabbaababbaabb abaa abbabaabbaababbabaababbaabbabaababbaba abbaababbaabbabaabbaababbabaababbaabbabaab The resulting partial word, which non-trivially avoids αβαβα, has more than 120 letters between any two consecutive holes F Blanchet-Sadri, R Mercaş, S Simmons and E Weissenstein, Avoidable binary patterns in partial words, Acta Informatica 48 (2011) 25 41

15 A measure of the frequency of holes in a partial word The hole sparsity of a partial word w is the smallest positive integer λ such that every factor of w of length λ contains at least one hole In this case, we call w λ-sparse ab b ac bc a is 3-sparse For a fixed pattern p and fixed alphabet A of size k, we want to determine the smallest λ so that an infinite λ-sparse word over A avoids p

16 (Non-trivial) minimum hole sparsity The non-trivial minimum hole sparsity for a pattern p over an alphabet of size k, χ k (p), is the smallest positive integer λ such that there exists an infinite λ-sparse word w over a k-letter alphabet that avoids all non-trivial occurrences of p If no such integer exists, then χ k (p) = The minimum hole sparsity for a pattern p over an alphabet of size k, χ k (p), is the smallest positive integer λ such that there exists an infinite λ-sparse word w over a k-letter alphabet that avoids all occurrences of p (including trivial occurrences) If no such integer exists, then χ k (p) =

17 A technical lemma For patterns p, q and integer k > 0, the following hold: 1 χ k (p) 2; 2 If k < k, then χ k (p) χ k (p); 3 If p q, then χ k (p) χ k (q); 4 If p is k-unavoidable, then χ k (p) = Statements 1 4 hold for the χ k function as well Proof We prove the lemma for the χ k function Statement 4 is trivial, while for Statement 1, we point out that if an infinite word is 1-sparse, then it consists only of holes, and therefore meets every pattern For Statement 2, if k < k, then a λ-sparse word avoiding p over the k-letter alphabet will also avoid p over the k -letter alphabet We prove Statement 3 similarly by noting that if p q then a word that avoids p also avoids q

18 Adaptation of the backtracking algorithm Algorithm 1 backtrack( w, p, λ, A ): Given as input partial word w, pattern p, sparsity λ, and alphabet A of size k, the algorithm outputs, if p is k-avoidable over λ-sparse partial words, the list of partial words over A with prefix w and hole sparsity λ or smaller that avoid p; otherwise, the algorithm never terminates 1: if w meets p then 2: return 3: run backtrack( w, p, λ, A ) 4: if among the last λ 1 symbols of w there is at least one then 5: for all a A do 6: run backtrack( wa, p, λ, A ) 7: print w 8: return J Cassaigne, Motifs évitables et régularités dans les mots, PhD Thesis, Université Paris VI (1994)

19 4 Non-trivial minimum hole sparsity for unary patterns We will fill the table with the values of χ k (α n ), k, n 1 k \ α n α α 2 α 3 α 4 α

20 k \ α n α α 2 α 3 α 4 α α is unavoidable over a k-letter alphabet

21 k \ α n α α 2 α 3 α 4 α α n is unavoidable over the unary alphabet

22 k \ α n α α 2 α 3 α 4 α Thue showed that α 2 is 2-unavoidable For all k 1, χ k (α 2 ) > 3 This is due to the fact that a b, a b c, ab are compatible with the non-trivial squares, (ab) 2, (acb) 2 and (ba) 2

23 k \ α n α α 2 α 3 α 4 α Thue showed that α 2 is 2-unavoidable For all k 1, χ k (α 2 ) > 3 This is due to the fact that a b, a b c, ab are compatible with the non-trivial squares, (ab) 2, (acb) 2 and (ba) 2

24 Lemma For all k 4, χ k (α 2 ) = 4 k \ α n α α 2 α 3 α 4 α

25 Proof (sketch) Let A = {a, b, c, d} and ρ : A A be defined by ρ(a) = ad, ρ(b) = bc, ρ(c) = ab, ρ(d) = ba, and σ : A A by σ(a) = dca, σ(b) = bca, σ(c) = dba, σ(d) = bda We show that the 4-sparse word σ(ρ ω (a)) avoids non-trivial squares First, we show that ρ ω (a) is square-free Second, we assume, to the contrary, that σ(ρ ω (a)) contains a non-trivial square uv Furthermore, every fourth symbol of σ(ρ ω (a)) is an a Third, we show that u 4, and u is divisible by 4 It follows that u v implies u = v, hence, we may refer to uv as u 2

26 Proof (continued) Finally, we look at the position of the first hole in u If u(3) =, then u starts with one of the images of σ, and since u is divisible by 4, it follows that for some x we have u 2 = σ(xx), which is a contradiction with the fact that ρ ω (a) is square-free If u(0) = (resp, u(1) = ), we consider the square u[1 u )u (resp, u[2 u )ua ), which leads to a contradiction as above If u(2) = and u starts with c, say, then u 2 = ca u zca u z, where u = u and z {b, d} In this case, suppose u 2 is preceded by d In order to avoid a contradiction similar to the previous cases, we assume z = b Since the preimage of dca under σ is a, and, in ρ ω (a), a is always followed by b or d, it follows that the first letter of u is b However, by similar reasoning the first letter in u is d, a contradiction since the latter d corresponds to the b in the first copy of u

27 Lemma The equality χ 3 (α 2 ) = 7 holds k \ α n α α 2 α 3 α 4 α

28 Proof (sketch) Let A = {a, b, c, d}, B = {a, b, c}, ρ : A A be defined by ρ(a) = ad, ρ(b) = bc, ρ(c) = ab, ρ(d) = ba, and π : A B by π(a) = abcbac b acbabc, π(b) = bacabc a bcabac, π(c) = abcacb, and π(d) = bacbca For the sake of contradiction, we suppose that Π = π(ρ ω (a)) contains a non-trivial square uv where u v Every length two factor of ρ ω (a) has an image under π of length at least 23 Moreover, all squares of length at most 24 are contained in the image of a factor of length three of ρ ω (a): aba, abc, adb, bab, bad, bca, cab, cad, dba and dbc Their images under π being square-free, uv > 25 Looking at all 54 length nine factors of Π obtained from the above ten factors, we get that no two are compatible Hence, all compatible factors of Π of length nine or greater must be equal Since u = v > 12, we must have u = v Contradictions follow based on the position of the first hole in u

29 Lemma The equality χ 2 (α 3 ) = 3 holds k \ α n α α 2 α 3 α 4 α

30 Proof (sketch) Let A = {a, b, c}, B = {a, b}, δ : A A be defined by δ(a) = ab, δ(b) = bc and δ(c) = ab, and υ : A B by υ(a) = aa, υ(b) = ab and υ(c) = bb Replacing each c in δ with a yields the Thue-Morse morphism It is well-known that the Thue-Morse word avoids α 3 and αβαβα Thus, δ ω (a) also avoids these patterns Suppose towards a contradiction, that there exists a non-trivial cube u 1 u 2 u 3 in υ(δ ω (a)), where u 1, u 2, u 3 u We first look at the possible starting letters of u 1, u 2 and u 3 in order to show that u 0 mod 3, and u = u 1 = u 2 = u 3 Contradictions follow based on the position of the first hole in u

31 Lemma For all k, n 3, χ k (α n ) = 2 k \ α n α α 2 α 3 α 4 α

32 Proof (sketch) Let A = {a, b, c}, ϕ : A A be defined by ϕ(a) = a b a c, ϕ(b) = a b, ϕ(c) = a c, and ϕ( ) = Suppose towards a contradiction that ϕ ω (a) has a non-trivial third power u 1 u 2 u 3 with u 1, u 2, u 3 u Consider u is even, in which case u = u 1 = u 2 = u 3, a appears in ϕ ω (a) exactly once every four symbols, and each u i contains an a Thus, u 0 mod 4 Assume without loss of generality that u 3 begins with an a Break u 3 into factors s = a b and t = a c The cubes in ϕ ω (a) correspond to cubes in ϕ ω (s), where ϕ : {s, t} {s, t} is given by ϕ (s) = sts and ϕ (t) = stt We reach the desired contradiction using a result of Richomme and Wlazinski which implies that ϕ avoids cubes if and only if ϕ (ssttststtsttsstsststsstt) is cube-free G Richomme and F Wlazinski, Some results on k-power-free morphisms, Theoretical Computer Science 273 (2002)

33 Lemma For all n 4, χ 2 (α n ) = 2 k \ α n α α 2 α 3 α 4 α

34 Proof (sketch) Let A = {a, b} and define µ : A A by µ(a) = a b, µ(b) = a a, and µ( ) = For the sake of contradiction, assume that µ ω (a) has a non-trivial 4th power u 1 u 2 u 3 u 4 such that u 1, u 2, u 3, u 4 u u is even: Since the holes align, u = u 1 = u 2 = u 3 = u 4 Assume without loss of generality that u 4 begins with a letter Break u 4 into factors s = a and t = b The 4th powers in µ ω (a) correspond to 4th powers in µ ω (s), where µ : {s, t} {s, t} is given by µ (s) = st and µ (t) = ss We then show that µ ω (s) is 4th power free, a contradiction u is odd: Note that u 1 u 3 implies u 1 = u 3 and u 2 u 4 implies u 2 = u 4, and a b occurs at least every 8 symbols Since u 4 8, there is at least one b in u 4 Assume without loss of generality that b occurs in u 1 Then, there is a corresponding b in u 3, and between the two b s there are 2 u 1 symbols We then reach a contradiction

35 5 Minimum hole sparsity for unary patterns We will fill the table with the values of χ k (αn ), k, n 1 k \ α n α α 2 α 3 α 4 α 5 α 6 α

36 k \ α n α α 2 α 3 α 4 α 5 α 6 α

37 k \ α n α α 2 α 3 α 4 α 5 α 6 α factors of the form a or a appear in all infinite partial words having holes

38 Looking at trivial cubes, any 2-sparse word meets the pattern α 3, since any letter a is preceded and followed by a hole Thus, for all k 1, we have the lower bound χ k (α3 ) 3 For k 3, we show that the bound is tight Lemma For all k 3, χ k (α3 ) = 3 k \ α n α α 2 α 3 α 4 α 5 α 6 α

39 Proof (sketch) Let A = {a, b} and B = {a, b, c}, and define the morphisms ξ : A A by ξ(a) = ab and ξ(b) = ba, and τ : B B by τ(a) = ab and τ(b) = ac We can show that τ(ξ ω (a)) avoids cubes τ(ξ ω (a)) = τ(abbabaabbaababbabaababbaabbaba ) = ab ac ac ab ac ab ab ac ac ab

40 Lemma The equality χ 2 (α3 ) = 7 holds k \ α n α α 2 α 3 α 4 α 5 α 6 α Let A = {a, b}, ξ : A A the Thue-Morse morphism defined by ξ(a) = ab and ξ(b) = ba, and define ζ : A A by ζ(a) = babaab abbaba and ζ(b) = baabba The word ζ(ξ ω (a)) avoids cubes

41 Lemma The equality χ 2 (α3 ) = 7 holds k \ α n α α 2 α 3 α 4 α 5 α 6 α Let A = {a, b}, ξ : A A the Thue-Morse morphism defined by ξ(a) = ab and ξ(b) = ba, and define ζ : A A by ζ(a) = babaab abbaba and ζ(b) = baabba The word ζ(ξ ω (a)) avoids cubes

42 Lemma The equality χ 2 (α4 ) = 3 holds k \ α n α α 2 α 3 α 4 α 5 α 6 α Let A = {a, b}, ξ : A A the Thue-Morse morphism defined by ξ(a) = ab and ξ(b) = ba, and define κ : A A by κ(a) = ab ab ba ba and κ(b) = ab ab ab ba ba The word κ(ξ ω (a)) avoids 4th powers

43 Lemma The equality χ 2 (α4 ) = 3 holds k \ α n α α 2 α 3 α 4 α 5 α 6 α Let A = {a, b}, ξ : A A the Thue-Morse morphism defined by ξ(a) = ab and ξ(b) = ba, and define κ : A A by κ(a) = ab ab ba ba and κ(b) = ab ab ab ba ba The word κ(ξ ω (a)) avoids 4th powers

44 Lemma For all k 3, χ k (α4 ) = 2 k \ α n α α 2 α 3 α 4 α 5 α 6 α

45 Proof (sketch) Let A = {a, b} and define µ : A A as µ(a) = a b, µ(b) = a a, and µ( ) = We showed earlier that µ ω (a) is free of non-trivial 4th powers a b a a a b a b a b a a a b a a Let B = {a, b, c} and define µ : B B as µ (a) = a b, µ (b) = a c, µ (c) = a b, and µ ( ) = Note that µ ω (a) is simply a copy of µ ω (a) in which some a s have been replaced with c s a b a c a b a b a b a c a b a c Since a non-trivial 4th power in µ ω (a) implies the presence of a non-trivial 4th power in µ ω (a), which is impossible, we have that µ ω (a) is non-trivially 4th power free It is easy to check that µ ω (a) avoids trivial 4th powers as well, since no two consecutive letters of µ ω (a) are the same

46 Lemma The equality χ 2 (α5 ) = 3 holds k \ α n α α 2 α 3 α 4 α 5 α 6 α Proof The upper bound χ 2 (α5 ) 3 comes from the equality χ 2 (α4 ) = 3 along with the technical lemma, while the backtracking algorithm provides the lower bound χ 2 (α5 ) 3

47 Lemma For all n 6 and k 2, χ k (αn ) = 2, and for all n 5 and k 3, χ k (αn ) = 2 k \ α n α α 2 α 3 α 4 α 5 α 6 α For the first statement, the Thue-Morse word with a hole inserted between every two letters is 6th power free a b b a b a a b b a a b a b b a

48 Lemma For all n 6 and k 2, χ k (αn ) = 2, and for all n 5 and k 3, χ k (αn ) = 2 k \ α n α α 2 α 3 α 4 α 5 α 6 α For the first statement, the Thue-Morse word with a hole inserted between every two letters is 6th power free a b b a b a a b b a a b a b b a

49 6 Classification of all unary patterns with respect to hole sparsity Theorem The values of χ k (α n ) and χ k (αn ) for all k, n 1 are as follows: k \ α n α α 2 α 3 α 4 α α 2 α 3 α 4 α 5 α

50 Acknowledgement We thank Robert Mercaş for very valuable help in the writing of this paper

51 wwwuncgedu/cmp/research/patterns