Codings of rotations are full A. Blondin Mass e S. Brlek S. Labb - - PowerPoint PPT Presentation

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Codings of rotations are full A. Blondin Mass e S. Brlek S. Labb - - PowerPoint PPT Presentation

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Codings of rotations are full A. Blondin Mass e S. Brlek S. Labb e L. Vuillon Universit e du Qu ebec ` a Montr eal EUROCOMB 2009 September 11th, 2009 e et al. (UQ` Blondin Mass AM, U. Savoie) Codings of rotations are

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SLIDE 1

Codings of rotations are full

  • A. Blondin Mass´

e

  • S. Brlek
  • S. Labb´

e

  • L. Vuillon

Universit´ e du Qu´ ebec ` a Montr´ eal

EUROCOMB 2009 September 11th, 2009

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 1 / 15

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SLIDE 2

Palindromes

ressasser tenet reconocer kisik R R R L L L L r r r r d d

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 2 / 15

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SLIDE 3

The Fibonacci word

We define f−1 = b, f0 = a and, for n ≥ 1, fn = fn−1fn−2. Therefore, we have f0 = a f1 = ab f2 = aba f3 = abaab f4 = abaababa f5 = abaababaabaab . . . . . . The infinite word f∞ is called the Fibonacci word.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 3 / 15

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SLIDE 4

The Thue-Morse word

We define t0 = a and, for n ≥ 1, tn = tn−1tn−1. so that t0 = a t1 = ab t2 = abba t3 = abbabaab t4 = abbabaabbaababba t5 = abbabaabbaababbabaababbaabbabaab . . . . . . The infinite word t∞ is called the Thue-Morse word.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 4 / 15

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SLIDE 5

Number of distinct palindromic factors

Theorem (Droubay, Justin and Pirillo, 2001)

Let w be a finite word. Then |Pal(w)| ≤ |w| + 1. w = p q Assume that the first occurrence of some palindromes p and q ends at the same position.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 5 / 15

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SLIDE 6

Number of distinct palindromic factors

Theorem (Droubay, Justin and Pirillo, 2001)

Let w be a finite word. Then |Pal(w)| ≤ |w| + 1. w = p q q Assume that the first occurrence of some palindromes p and q ends at the same position.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 5 / 15

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SLIDE 7

Number of distinct palindromic factors

Theorem (Droubay, Justin and Pirillo, 2001)

Let w be a finite word. Then |Pal(w)| ≤ |w| + 1. w = p q q Assume that the first occurrence of some palindromes p and q ends at the same position. Then p = q.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 5 / 15

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SLIDE 8

Number of distinct palindromic factors

Theorem (Droubay, Justin and Pirillo, 2001)

Let w be a finite word. Then |Pal(w)| ≤ |w| + 1. w = p q q Assume that the first occurrence of some palindromes p and q ends at the same position. Then p = q.

Theorem (Droubay, Justin and Pirillo, 2001)

Sturmian words are full, i.e. they realize the upper bound.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 5 / 15

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SLIDE 9

Palindromic complexity

Upper bound Fibonacci word Thue-Morse word Fixed point of a abb,b ba 100 200 300 400 500 100 200 300 400 500 Length of prefix Number of palindrome factors Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 6 / 15

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SLIDE 10

The Fibonacci word is full

w = a Palindromes a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 11

The Fibonacci word is full

w = a b Palindromes a b

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 12

The Fibonacci word is full

w = a b a Palindromes a b a b a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 13

The Fibonacci word is full

w = a b a a Palindromes a b a b a a a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 14

The Fibonacci word is full

w = a b a a b Palindromes a b a b a a a b a a b

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 15

The Fibonacci word is full

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

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 16

The Fibonacci word is full

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

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 17

The Fibonacci word is full

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

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 18

The Fibonacci word is full

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 ...

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 7 / 15

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SLIDE 19

The Thue-Morse word is lacunary

w = a Palindromes a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 20

The Thue-Morse word is lacunary

w = a b Palindromes a b

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 21

The Thue-Morse word is lacunary

w = a b b Palindromes a b b b

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 22

The Thue-Morse word is lacunary

w = a b b a Palindromes a b b b a b b a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 23

The Thue-Morse word is lacunary

w = a b b a b Palindromes a b b b a b b a b a b

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 24

The Thue-Morse word is lacunary

w = a b b a b a Palindromes a b b b a b b a b a b a b a

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 25

The Thue-Morse word is lacunary

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

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 26

The Thue-Morse word is lacunary

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

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 27

The Thue-Morse word is lacunary

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!

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 8 / 15

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SLIDE 28

Complete return words

w = u u

some occurrence next occurrence

v We say that v is a complete return word of u in w, if v starts at an

  • ccurrence 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.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 9 / 15

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SLIDE 29

Codings of rotations (1/2)

The coding of rotations of parameters (x, α, β) is the word C = c0c1c2 · · · such that ci =

  • if x + iα ∈ [0, β)

1 if x + iα ∈ [β, 1) β

x

1

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 10 / 15

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SLIDE 30

Codings of rotations (1/2)

The coding of rotations of parameters (x, α, β) is the word C = c0c1c2 · · · such that ci =

  • if x + iα ∈ [0, β)

1 if x + iα ∈ [β, 1) β

x x + α

11

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 10 / 15

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SLIDE 31

Codings of rotations (1/2)

The coding of rotations of parameters (x, α, β) is the word C = c0c1c2 · · · such that ci =

  • if x + iα ∈ [0, β)

1 if x + iα ∈ [β, 1) β

x x + α x + 2α

111

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 10 / 15

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SLIDE 32

Codings of rotations (1/2)

The coding of rotations of parameters (x, α, β) is the word C = c0c1c2 · · · such that ci =

  • if x + iα ∈ [0, β)

1 if x + iα ∈ [β, 1) β

x x + α x + 2α x + 3α

1111

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 10 / 15

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SLIDE 33

Codings of rotations (1/2)

The coding of rotations of parameters (x, α, β) is the word C = c0c1c2 · · · such that ci =

  • if x + iα ∈ [0, β)

1 if x + iα ∈ [β, 1) β

x x + α x + 2α x + 3α

11111000000111 · · ·

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 10 / 15

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SLIDE 34

Codings of rotations (2/2)

Many interesting problems related to codings of rotations: Density of the letters 0 and 1, Complexity, i.e. the number of factors of length n, or palindromic and f -palindromic complexity, Applications to number theory [Adamczewski, 2002], etc. In particular, Rote (1994) expressed sequences of complexity 2n with respect to codings of rotations.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 11 / 15

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SLIDE 35

The different cases

Let C be a coding of rotations of parameters (x, α, β). If α is rational, then C is periodic. If α = β is irrational, then C is Sturmian f (n) = n + 1. If α and β are rationally dependent, then C is quasi-Sturmian. f (n) = n + k, for some constant k. Otherwise, C is a Rote sequence f (n) = 2n, for large enough n.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 12 / 15

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SLIDE 36

Main result

Theorem

Every coding of rotations is full. The proof is based on the following ideas:

1 Return words 2 Interval exchange transformations 3 Poincar´

e’s first return function

4 Many results on those dynamical systems Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 13 / 15

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SLIDE 37

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β I0 I1

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

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SLIDE 38

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α I00 I01 I11 I10

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

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SLIDE 39

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

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SLIDE 40

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-41
SLIDE 41

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-42
SLIDE 42

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-43
SLIDE 43

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-44
SLIDE 44

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-45
SLIDE 45

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-46
SLIDE 46

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-47
SLIDE 47

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-48
SLIDE 48

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

y y + α y + 2α

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-49
SLIDE 49

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-50
SLIDE 50

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-51
SLIDE 51

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-52
SLIDE 52

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-53
SLIDE 53

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-54
SLIDE 54

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-55
SLIDE 55

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-56
SLIDE 56

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-57
SLIDE 57

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-58
SLIDE 58

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-59
SLIDE 59

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-60
SLIDE 60

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-61
SLIDE 61

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-62
SLIDE 62

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-63
SLIDE 63

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-64
SLIDE 64

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-65
SLIDE 65

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-66
SLIDE 66

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-67
SLIDE 67

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

z

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-68
SLIDE 68

Idea of the proof

Let x = 0.102, α = 0.135 and β = 0.578. Then C = 0000111000011110000111000011100000111000 · · · β −α β − α −2α β − 2α I000 I001 I011 I111 I110 I100

z

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 14 / 15

slide-69
SLIDE 69

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. Labb´ e. The pictures have been produced using Sage and pgf/tikz.

  • W. A. Stein et al., Sage Mathematics Software (Version 4.1.1), The

Sage Development Team, 2009, http://www.sagemath.org. The Sage-Combinat community, Sage-Combinat: enhancing Sage as a toolbox for computer exploration in algebraic combinatorics, http://combinat.sagemath.org, 2009.

Blondin Mass´ e et al. (UQ` AM, U. Savoie) Codings of rotations are full EUROCOMB 2009 15 / 15