Some improvments of the S-adic Conjecture

Transcription

1 Some improvments of the S-adic Conjecture Julien Leroy Laboratoire Amiénois de Mathématiques Fondamentales et Appliquées CNRS-UMR 6140, Université de Picardie Jules Verne, 33 rue Saint Leu, Amiens Cedex 01, France September 9, 2010

2 Sturmian sequences Recall that Sturmian sequences are binary sequences w satisfying p(n) = n+1 for all n.

3 Sturmian sequences Recall that Sturmian sequences are binary sequences w satisfying p(n) = n+1 for all n. Define the 4 substitutions: { a a L a : b ab { a ba L b : b b { a a R a : b ba { a ab R b : b b Then w = L n 1 a R n 2 a L n 3 b Rn 4 b Ln 5 a R n 6 a (aaa ) with some particular conditions on (n k ) k.

4 S-adicity w A N is S-adic if there are

5 S-adicity w A N is S-adic if there are S = finite set of non-erasing morphisms σ.

6 S-adicity w A N is S-adic if there are S = finite set of non-erasing morphisms σ. a sequence (σ n : A n+1 A n) n in S N ;

7 S-adicity w A N is S-adic if there are S = finite set of non-erasing morphisms σ. a sequence (σ n : A n+1 A n) n in S N ; a sequence (a n A n ) n of letters such that:

8 S-adicity w A N is S-adic if there are S = finite set of non-erasing morphisms σ. a sequence (σ n : A n+1 A n) n in S N ; a sequence (a n A n ) n of letters such that: w = lim n + σ 0σ 1 σ n (a n+1 a n+1 a n+1 ) with lim n + σ 0 σ 1 σ n (a n+1 ) = +

9 S-adicity Example ϕ : { a ab b a µ : { a ab b ba w = lim n + ϕµϕ2 µ 2 ϕ n µ n (aaa )

10 S-adicity Example ϕ : { a ab b a µ : { a ab b ba w = lim n + ϕµϕ2 µ 2 ϕ n µ n (aaa ) ϕ(a) ab ϕµ(a) aba ϕµϕ 2 (a) abaaababa ϕµϕ 2 µ 2 (a) abaaababababbbababbabbbabababaaababa (ϕµϕ 2 µ 2 ϕ n µ n (aaa )) n converges in A N.

11 Some other S-adic sequences Many classical examples are S-adic:

12 Some other S-adic sequences Many classical examples are S-adic: any ultimately periodic sequence is morphic;

13 Some other S-adic sequences Many classical examples are S-adic: any ultimately periodic sequence is morphic; any automatic sequence is the image under a letter-to-letter morphism of a fixed point of uniform substitution;

14 Some other S-adic sequences Many classical examples are S-adic: any ultimately periodic sequence is morphic; any automatic sequence is the image under a letter-to-letter morphism of a fixed point of uniform substitution; any episturmian sequence is S-adic with S = {L a, R a a A} and L a { a a b ab b a R a { a a b ba b a

15 Some other S-adic sequences Many classical examples are S-adic: any ultimately periodic sequence is morphic; any automatic sequence is the image under a letter-to-letter morphism of a fixed point of uniform substitution; any episturmian sequence is S-adic with S = {L a, R a a A} and L a { a a b ab b a linearly recurrent sequences;. R a { a a b ba b a

16 S-adic conjecture All examples previously cited have an at most linear complexity (p(n) Cn) Question: Is there any relation between "S-adic" and "low complexity"?

17 S-adic conjecture All examples previously cited have an at most linear complexity (p(n) Cn) Question: Is there any relation between "S-adic" and "low complexity"? There is a condition C such that w has an at most linear complexity if and only if it is a S-adic sequence satisfying the condition C.

18 S-adic conjecture Example of condition: Theorem (Durand) w is linearly recurrent if and only if it is a primitive and proper S-adic sequence.

19 S-adic conjecture Example of condition: Theorem (Durand) w is linearly recurrent if and only if it is a primitive and proper S-adic sequence. Primitive S-adic: s 0 such that r, b A r, c A r+s0 +1: b occurs in σ r σ r+1 σ r+s0 (c).

20 S-adic conjecture Example of condition: Theorem (Durand) w is linearly recurrent if and only if it is a primitive and proper S-adic sequence. Primitive S-adic: s 0 such that r, b A r, c A r+s0 +1: b occurs in σ r σ r+1 σ r+s0 (c). Proper S-adic: σ S, b, c A such that σ(a) ba c a A.

21 A sufficient condition Proposition (Durand) If w is S-adic satisfying max a,b then p(n) Kn for all n. σ 0 σ 1 σ n (a) σ 0 σ 1 σ n (b) C,

22 A sufficient condition Proposition (Durand) If w is S-adic satisfying max a,b then p(n) Kn for all n. σ 0 σ 1 σ n (a) σ 0 σ 1 σ n (b) C, Corollary (Durand) Primitive S-adic sequences and S-adic sequences with S containing only uniform morphisms have an at most linear complexity.

23 Ferenczi s work Theorem (Ferenczi 1996) Let w A N be aperiodic, uniformly recurrent and such that p w (n) Cn. There exist a finite number of non-erasing (σ(a) ε) morphisms σ i, 1 i c, over an alphabet D = {0,..., d 1}; an application α from D to A and an infinite sequence (i n ) n N {1,..., c} N such that inf 0 r d 1 σ i0 σ i1 σ in (r) tends to infinity with n and any factor of w is a factor of ασ i0 σ i1 σ in (0) for some n.

24 Some improvments Theorem With the same conditions:

25 Some improvments Theorem With the same conditions: w is S-adic;

26 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n;

27 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n; n N, a, c : σ n (c) / A naa naa n;

28 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n; n N, a, c : σ n (c) / A naa naa n; n N, a, b, c, d : (σ(c),σ n (d)) / A n aa n ba n A n ba n aa n ;

29 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n; n N, a, c : σ n (c) / A naa naa n; n N, a, b, c, d : (σ(c),σ n (d)) / A n aa n ba n A n ba n aa n ; n, a, b, c : σ n (a) A n ba+ n σ n (c) / A n b.

30 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n; n N, a, c : σ n (c) / A naa naa n; n N, a, b, c, d : (σ(c),σ n (d)) / A n aa n ba n A n ba n aa n ; n, a, b, c : σ n (a) A n ba+ n σ n (c) / A n b. r 0, s > r such that all a A r occur in all σ r σ r+1 σ s (c);

31 Some improvments Theorem With the same conditions: w is S-adic; inf a An+1 σ 0 σ 1 σ n (a) tends to infinity with n; n N, a, c : σ n (c) / A naa naa n; n N, a, b, c, d : (σ(c),σ n (d)) / A n aa n ba n A n ba n aa n ; n, a, b, c : σ n (a) A n ba+ n σ n (c) / A n b. r 0, s > r such that all a A r occur in all σ r σ r+1 σ s (c); all morphisms in S can be decomposed as a product of morphisms in T = {G} {E ij i, j A } {M i i A } with G(0) = 10 and G(i) = i for all letters i 0; Eij exchange i and j and fix the other letters; Mi maps i to 0 and fix the other letters;

32 Another question: which complexities are reachable? For substitutive sequences, p(n) can have only 5 asymptotic behaviors (Pansiot): O(1), O(n), O(n log n), O(n log log n) and O(n 2 ) depending on the growth of the letters.

33 Another question: which complexities are reachable? For substitutive sequences, p(n) can have only 5 asymptotic behaviors (Pansiot): O(1), O(n), O(n log n), O(n log log n) and O(n 2 ) depending on the growth of the letters. Quid for S-adic sequences?

34 Any sequence is S-adic Cassaigne s construction: Let w = w 0 w 1 w 2 w 3 A N and l / A. { l la a A: σ a : b b b l { l w 0 τ : a a a A Then w = τσ w1 σ w2 (lll ).

35 Growth of images The growth of letters seems to be important to have a "low" complexity:

36 Growth of images The growth of letters seems to be important to have a "low" complexity: 1. Pansiot: maximal complexity O(n 2 ) can be reached only if there are some non-growing letters ( σ n (a) C for all n);

37 Growth of images The growth of letters seems to be important to have a "low" complexity: 1. Pansiot: maximal complexity O(n 2 ) can be reached only if there are some non-growing letters ( σ n (a) C for all n); 2. Cassaigne: any high complexity and many non-growing letters.

38 Growth of images Question: Can we reach any complexity with S-adic sequences such that inf a σ 0 σ 1 σ n (a) tends to infinity?

39 Growth of images Question: Can we reach any complexity with S-adic sequences such that inf a σ 0 σ 1 σ n (a) tends to infinity? Theorem Let w be a S-adic sequence such that σ(a) 2 for all σ S and all letters a. Then p w (n) Cn log n.

40 Ideas of the proof of Ferenczi s result Definition The Rauzy graph of order n of a sequence w is the directed graph G n = (V(n), E(n)) with V(n) = L(w) A n and u b a v ub = av L(w).

41 Ideas of the proof of Ferenczi s result Definition The Rauzy graph of order n of a sequence w is the directed graph G n = (V(n), E(n)) with V(n) = L(w) A n and u b a v ub = av L(w). Let w be a Sturmian word. Since p(n+1) p(n) = 1 for all n, there is only one right special factor and one left special factor of each length.

42 Ideas of the proof of Ferenczi s result Rauzy graph G n (w): 1 left and 1 right special factor: L and R L R

43 Ideas of the proof of Ferenczi s result Rauzy graph G n (w): w being recurrent: L R

44 Ideas of the proof of Ferenczi s result Rauzy graph G n (w): Forbidden: L R

45 Ideas of the proof of Ferenczi s result Definition A n-segment is a path in G n (w) starting in a right special factor, ending in a right special factor and passing only through non-right special factors. In Sturmian case: L A a R B b

46 Ideas of the proof of Ferenczi s result Definition A n-segment is a path in G n (w) starting in a right special factor, ending in a right special factor and passing only through non-right special factors. In Sturmian case: L A a R B b

47 Ideas of the proof of Ferenczi s result Facts about n-segments (in Sturmian case):

48 Ideas of the proof of Ferenczi s result Facts about n-segments (in Sturmian case): the label of any n-segment is a factor of w;

49 Ideas of the proof of Ferenczi s result Facts about n-segments (in Sturmian case): the label of any n-segment is a factor of w; the length of a n-segment is ultimately increasing;

50 Ideas of the proof of Ferenczi s result Facts about n-segments (in Sturmian case): the label of any n-segment is a factor of w; the length of a n-segment is ultimately increasing; any factor u of a given length will be factor of the label of a n-segment for n large enough;

51 Ideas of the proof of Ferenczi s result Facts about n-segments (in Sturmian case): the label of any n-segment is a factor of w; the length of a n-segment is ultimately increasing; any factor u of a given length will be factor of the label of a n-segment for n large enough; We want to study the behavior of the n-segments as n increases.

52 Ideas of the proof of Ferenczi s result First case: L R: G n L R a b

53 Ideas of the proof of Ferenczi s result First case: L R: G n α β L x y R a b

54 Ideas of the proof of Ferenczi s result First case: L R: G n α β L x y R a b

55 Ideas of the proof of Ferenczi s result First case: L R: G n+1 αl βl Lx yr Ra Rb

56 Ideas of the proof of Ferenczi s result First case: L R: G n+1 αl βl Lx yr Ra Rb

57 Ideas of the proof of Ferenczi s result First case: L R: G n+1 αl βl Lx yr Ra Rb - The (n+1)-segments have the same labels of the n-segments.

58 Ideas of the proof of Ferenczi s result First case: L R: G n+1 αl βl Lx yr Ra Rb - The (n+1)-segments have the same labels of the n-segments. - The length of the path from L to R deacreased by 1.

59 Ideas of the proof of Ferenczi s result Second case: L = R: G n A α β B a b B

60 Ideas of the proof of Ferenczi s result Second case: L = R: G n+1 αb Ba βb Bb

61 Ideas of the proof of Ferenczi s result Second case: L = R: G n+1 A αb Ba βb Bb B

62 Ideas of the proof of Ferenczi s result Second case: L = R: G n+1 A αb Ba βb Bb B

63 Ideas of the proof of Ferenczi s result Second case: L = R: G n+1 A αb Ba βb Bb B

64 Ideas of the proof of Ferenczi s result Second case: L = R: A = A G n+1 A αb Ba βb Bb B

65 Ideas of the proof of Ferenczi s result Second case: L = R: G n+1 A αb Ba βb Bb B = BA B

66 Ideas of the proof of Ferenczi s result This provides a morphism R A : { A A B BA

67 Ideas of the proof of Ferenczi s result This provides a morphism R A : { A A B BA When the new right special factor is βb, we get { A AB R B : B B

68 Ideas of the proof of Ferenczi s result This provides a morphism R A : { A A B BA When the new right special factor is βb, we get { A AB R B : B B Any factor of w is a factor of for some k. R n 1 A Rn 2 B Rn 3 A Rn k 1 A R n k B (A)

69 Thank you