Complexité palindromique des codages de rotations et conjectures

Transcription

1 Complexité palindromique des codages de rotations et conjectures Sébastien Labbé Laboratoire d Informatique, de Robotique et de Microélectronique de Montpellier Université Montpellier 2 Laboratoire de Combinatoire et d Informatique Mathématique Université du Québec à Montréal Séminaire Ernest Dynamique, Arithmétique, Combinatoire Institut de Mathématique de Luminy Marseille, 20 avril 2010 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 1 / 25

2 Plan 1 Introduction Complexité palindromique Mots pleins 2 Codages de rotations sur deux intervalles sont pleins travail commun avec Blondin Massé, Brlek et Vuillon 3 Conclusion Défaut palindromique Les 4 classes de complexité palindromique Conjecture de Hof, Knill et Simon Conjectures Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 2 / 25

3 Palindromes G D D D G G G tenet ressasser kisik E S E E S E reconocer Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 3 / 25

4 The Fibonacci word We define f 1 = b, f 0 = a and, for n 1, f n = f n 1 f n 2. Therefore, we have f 0 = a f 1 = ab f 2 = aba f 3 = abaab f 4 = abaababa f 5 = abaababaabaab.. The infinite word f is called the Fibonacci word. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 4 / 25

5 Palindromes in the Fibonacci word w = a Palindromes a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

6 Palindromes in the Fibonacci word w = a b Palindromes a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

7 Palindromes in the Fibonacci word w = a b a Palindromes a b a b a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

8 Palindromes in the Fibonacci word w = a b a a Palindromes a b a b a a a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

9 Palindromes in the Fibonacci word w = a b a a b Palindromes a b a b a a a b a a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

10 Palindromes in the Fibonacci word w = a b a a b a Palindromes a b a b a a a b a a b a b a a b a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

11 Palindromes in the Fibonacci word w = a b a a b a b Palindromes a b a b a a a b a a b a b a a b a b a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

12 Palindromes in the Fibonacci word w = a b a a b a b a Palindromes a b a b a a a b a a b a b a a b a b a b a b a b a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

13 Palindromes in the Fibonacci word w = a b a a b a b a a Palindromes a b a b a a a b a a b a b a a b a b a b a b a b a a a b a b a a... Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 5 / 25

14 Number of distinct palindrome factors w = p q Assume that the first occurrence of some distinct palindromes p and q ends at the same position. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 6 / 25

15 Number of distinct palindrome factors w = q = q p q Assume that the first occurrence of some distinct palindromes p and q ends at the same position. Then q has a previous occurence in w which is a contradiction. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 6 / 25

16 Number of distinct palindrome factors w = q = q p q Assume that the first occurrence of some distinct palindromes p and q ends at the same position. Then q has a previous occurence in w which is a contradiction. Theorem (Droubay, Justin and Pirillo, 2001) Let w be a finite word. Then Pal(w) w + 1. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 6 / 25

17 More on the palindrome complexity Theorem (Droubay, Justin and Pirillo, 2001) Sturmian words w realize the upper bound, i.e. Pal(w) = w + 1. Definition (Brlek, Hamel, Nivat, Reutenauer, 2004) A word w is called full if Pal(w) = w + 1. Since 2008, some authors say rich instead of full. Personally, I prefer to use the word full because it suggests that a limit is reached. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 7 / 25

18 More on the palindrome complexity Theorem (Brlek, Hamel, Nivat, Reutenauer, 2004) A infinite periodic word w is full if and only if w = (uv) ω with both u and v are palindromes and (uv) 1+e is full where e = u v 3 / uv. Theorem (Bucci, De Luca, Glen, Zamboni, 2008) An infinite word w whose set of factors is closed under reversal is full if and only if for all n. Pal n (w) + Pal n+1 (w) = Fact n+1 (w) Fact n (w) + 2 What about words that are not full? Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 8 / 25

19 The Thue-Morse word We define t 0 = a and, for n 1, t n = t n 1 t n 1. so that t 0 = a t 1 = ab t 2 = abba t 3 = abbabaab t 4 = abbabaabbaababba t 5 = abbabaabbaababbabaababbaabbabaab.. The infinite word t is called the Thue-Morse word. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 9 / 25

20 Palindromes in the Thue-Morse w = a Palindromes a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

21 Palindromes in the Thue-Morse w = a b Palindromes a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

22 Palindromes in the Thue-Morse w = a b b Palindromes a b b b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

23 Palindromes in the Thue-Morse w = a b b a Palindromes a b b b a b b a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

24 Palindromes in the Thue-Morse w = a b b a b Palindromes a b b b a b b a b a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

25 Palindromes in the Thue-Morse w = a b b a b a Palindromes a b b b a b b a b a b a b a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

26 Palindromes in the Thue-Morse w = a b b a b a a Palindromes a b b b a b b a b a b a b a a a Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

27 Palindromes in the Thue-Morse w = a b b a b a a b Palindromes a b b b a b b a b a b a b a a a b a a b Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

28 Palindromes in the Thue-Morse w = a b b a b a a b b Palindromes a b b b a b b a b a b a b a a a b a a b... There is no new palindrome at this position : a lacuna! Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

29 Palindromes in the Thue-Morse w = a b b a b a a b b Palindromes a b b b a b b a b a b a b a a a b a a b... There is no new palindrome at this position : a lacuna! Hence, the Thue-Morse word is rich in palindromes but is not full. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 10 / 25

30 Complete return words some occurrence next occurrence w = u v u We say that v is a complete return word of u in w, if v starts at an occurrence of u and ends at the end of the next occurrence of u. Fact A word w is full if and only if every complete return word of a palindrome factor of w is a palindrome. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 11 / 25

31 Theorem (Blondin Massé, Brlek, Garon, L., 2008) For n 1, let L(n) be the index where the n-th interval of lacunas start in the Thue-Morse word and l(n) be its length. Then { 2 L(n) = n+2, if n is odd, 2 n n+1, if n is even. and l(n) = { 2 n, if n is odd, 2 n 1, if n is even. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 12 / 25

32 Palindrome complexity Number of palindrome factors Upper bound Fibonacci word Some periodic word Thue-Morse word Fixed point of a abb,b ba Length of prefix Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 13 / 25

33 Partie 2 Les codages de rotation sur deux intervalles sont pleins travail commun avec Blondin Massé, Brlek et Vuillon Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 14 / 25

34 Codings of rotations The coding of rotations of parameters (x, α, β) is the word C = c 0 c 1 c 2 such that { 0 if x + iα [0, β) c i = 1 if x + iα [β, 1) β x 0 1 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 15 / 25

35 Codings of rotations The coding of rotations of parameters (x, α, β) is the word C = c 0 c 1 c 2 such that { 0 if x + iα [0, β) c i = 1 if x + iα [β, 1) β x x + α 0 11 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 15 / 25

36 Codings of rotations The coding of rotations of parameters (x, α, β) is the word C = c 0 c 1 c 2 such that { 0 if x + iα [0, β) c i = 1 if x + iα [β, 1) β x x + α x + 2α Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 15 / 25

37 Codings of rotations The coding of rotations of parameters (x, α, β) is the word C = c 0 c 1 c 2 such that { 0 if x + iα [0, β) c i = 1 if x + iα [β, 1) β x x + α x + 2α 0 x + 3α 1111 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 15 / 25

38 Codings of rotations The coding of rotations of parameters (x, α, β) is the word C = c 0 c 1 c 2 such that { 0 if x + iα [0, β) c i = 1 if x + iα [β, 1) β x x + α x + 2α 0 x + 3α Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 15 / 25

39 The different cases Let C be a coding of rotations of parameters (x, α, β). If α is rational, then C is periodic. If β = 1 α is irrational, then C is Sturmian Fact n (C) = n + 1. If α and β are rationally dependent, then C is quasi-sturmian. Fact n (C) = n + k, for some constant k. Otherwise, C is a Rote sequence Fact n (C) = 2n, for large enough n. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 16 / 25

40 Main result Theorem (Blondin Massé, Brlek, L., Vuillon, 2009) Every coding of rotations on two intervals is full. The proof is based on the following ideas : 1 Return words 2 Interval exchange transformations 3 First return function 4 Many combinatorial results on those dynamical systems Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 17 / 25

41 Idea of the proof Let x = 0.102, α = and β = Then C = I 0 0 β I 1 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

42 Idea of the proof Let x = 0.102, α = and β = Then C = I 00 β α I 01 0 β α I 10 I 11 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

43 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

44 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

45 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

46 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

47 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

48 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

49 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

50 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

51 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

52 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I 011 y 0 β y + α y + 2α I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

53 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

54 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

55 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

56 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

57 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

58 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

59 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

60 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

61 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

62 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

63 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

64 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

65 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

66 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

67 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

68 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

69 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

70 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

71 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 z 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

72 Idea of the proof Let x = 0.102, α = and β = Then C = I 001 β 2α I 000 β α I β I 111 z 2α I 110 α I 100 Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 18 / 25

73 Partie 3 Conclusion Défaut palindromique Les 4 classes de complexité palindromique Conjecture de Hof, Knill et Simon Conjectures Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 19 / 25

74 Palindrome defect Brlek, Hamel, Nivat, Reutenauer (2004) also defined the defect of a finite word w as D(w) = w + 1 Pal(w). This definition is extended to infinite words w by setting D(w) to be the maximum of the defect of its factors : it may be finite or infinite. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 20 / 25

75 Palindrome complexity classes The palindrome complexity Pal(w) and defect D(w) divides the set of infinite words w into four classes : Pal(w) D(w) Examples 0 Sturmian words, Coding of rotation 0 < D(w) < (aababbaabbabaa) ω Thue-Morse word finite Fixed point of a abb, b ba Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 21 / 25

76 Palindrome complexity classes The palindrome complexity Pal(w) and defect D(w) divides the set of infinite words w into four classes : Pal(w) D(w) Examples 0 Sturmian words, Coding of rotation 0 < D(w) < (aababbaabbabaa) ω Thue-Morse word finite Fixed point of a abb, b ba Conjecture (Blondin-Massé, Brlek, L., 2008) Let w be the fixed point of a primitive morphism ϕ. If the defect of w is such that 0 < D(w) <, then w is periodic. The conjecture can also be stated for (uniformly) recurrent words in general. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 21 / 25

77 Palindrome complexity classes : Pal(w) = A substitution is in class P if there is a palindrome p and for each α Σ there is a palindrome q α such that ϕ(α) = pq α (Hof, Knill and Simon, 1995). Conjecture (adaptated from Hof, Knill and Simon, 1995) Let w be the fixed point of a primitive substitution. Then Pal(w) = if and only if there is a substitution ϕ such that ϕ(w) = w and ϕ is conjugate to a class P substitution. This conjecture was proved by B. Tan (2007) for the binary case and was generalized to f -pseudo-palindromes in the binary case for uniform morphisms (L., 2008). Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 22 / 25

78 The Thue-Morse word palindrome complexity class The primitive morphisms ϕ on {a, b} prolongable on a such that ϕ(ab) 11 group as follows : Pal(w) D(w) First primitive morphisms (21%) 0 < D(w) < 0 58 (0.5%) finite 9674 (78%) Can we characterize the Thue-Morse word palindrome complexity class? Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 23 / 25

79 The Thue-Morse word palindrome complexity class The primitive morphisms ϕ on {a, b} prolongable on a such that ϕ(ab) 11 group as follows : Pal(w) D(w) First primitive morphisms (21%) 0 < D(w) < 0 58 (0.5%) finite 9674 (78%) Can we characterize the Thue-Morse word palindrome complexity class? Conjecture Let w be the fixed point of a primitive substitution. Then w is such that Pal(w) = D(w) = if and only if there exists a morphism ϕ in class P such that ϕ(w) = w and looks like the Thue-Morse morphism. Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 23 / 25

80 The Thue-Morse word palindrome complexity class Below are the primitive morphisms ϕ on {a, b} prolongable on a such that ϕ(ab) 10 that generates a fixed point in the Thue-Morse palindrome complexity class : a ab, b ba a aababa, b abb a aabbabba, b ab a aabbaa, b bab a abbabbaa, b ba a ababaa, b bba a aabbbaa, b bab a abbaab, b ba a abbabb, b bba a aababa, b abbb a aabb, b bbaa a abbba, b baab a ababaa, b bbba a abab, b baba a abba, b baaab a abbbba, b baab a abba, b baab a aab, b baabaa a abbba, b baaab a ab, b baabba a aab, b bababb a aaab, b bababb a aba, b bbaabb a abba, b baaaab a abb, b aababa a abbb, b aababa a aba, b bbaaabb a ab, b aabbabba a ab, b baabaabb Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 24 / 25

81 Useful software This research was driven by computer exploration using the open-source mathematical software Sage [1] and its algebraic combinatorics features developed by the Sage-Combinat community [2], and in particular, F. Saliola, A. Bergeron and S. L. The pictures have been produced using Sage and pgf/tikz. W. A. Stein et al., Sage Mathematics Software (Version 4.3.4), The Sage Development Team, 2010, http :// The Sage-Combinat community, Sage-Combinat : enhancing Sage as a toolbox for computer exploration in algebraic combinatorics, http ://combinat.sagemath.org, Sébastien Labbé (LIRMM et LaCIM) Complexité palindromique Séminaire DAC, IML 25 / 25