Slopes of Tilings - LACL
Slopes of Tilings
Emmanuel Jeandel Pascal Vanier Université de Provence - LIF - Marseille
JAC 2010
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
These are Shifts of Finite Type (SFTs).
2 / 21
Periodicity Definition (Vector of periodicity) v ∈ Z2 \ {(0, 0)} such that for all x ∈ Z2 ρ(x + v ) = ρ(x).
Vectors of periodicity of the form (3k, 3l )
3 / 21
Unique directions of periodicity Definition (Unique direction of periodicity) If all vectors of periodicity of a tiling are colinear, then it has a unique direction of periodicity.
Direction of periodicity (4, 1)
4 / 21
Unique directions of periodicity Definition (Unique direction of periodicity) If all vectors of periodicity of a tiling are colinear, then it has a unique direction of periodicity.
Definition (Slope) If a tiling has a unique direction of periodicity, then if (p, q) is a vector of periodicity, we say the slope of the tiling is q/p.
θ
Slope θ = 1/4
4 / 21
The question we will (in a way) answer today. . .
Given a tiling system, what can we say about its set of slopes?
5 / 21
The question we will (in a way) answer today. . .
Given a tiling system, what can we say about its set of slopes?
Sτ will denote the set of slopes of τ . Sτ is a subset of Q ∪ {∞}.
5 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
Sτ = {1}
,
,
,
6 / 21
Results Theorem The set of all Sτ is exactly the set of recursively enumerable subsets of Q ∪ {∞}.
7 / 21
Results Theorem The set of all Sτ is exactly the set of recursively enumerable subsets of Q ∪ {∞}.
Definition Let S ⊆ Q ∪ {∞}, it is recursively enumerable if there exists a Turing machine M that halts on input (p, q) ∈ Z2 \ {(0, 0)} if and only if q/p ∈ S. • θ ∈ S ⇒ ∀(p, q), q/p = θ, M halts on input (p, q) • θ 6∈ S ⇒ ∀(p, q), q/p = θ, M does not halt on input (p, q)
7 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable.
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
It is decidable to know if there exists an aperiodic point in one dimensional shifts. 8 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
9 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ . Proof. There is a Turing machine M such that: • θ ∈ R ⇒ ∀(p, q), q/p = θ, M halts on input (p, q) • θ 6∈ R ⇒ ∀(p, q), q/p = θ, M does not halt on input (p, q)
We have to construct a tiling system τM corresponding to the Turing machine M whose slopes would exactly be the θs accepted by M.
9 / 21
Ingredients Shape of the desired tilings :
10 / 21
Ingredients Shape of the desired tilings :
This construction corresponds to the inputs θ = (p, q) such that p > q > 0: the other cases are treated in a similar way and the final tiling system is the union of all these tiling systems. 10 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
Rules on white tiles transcend black tiles. 11 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares with the same size
13 / 21
Same offset between columns
14 / 21
Shape of the periodic tilings at this point
15 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component.
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the aperiodic backgrounds
Tiles { , , , }.
17 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation :
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes This means we can make arbitrarily large squares for a same input by just multiplying the size of the squares and the offset by 2k .
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes This means we can make arbitrarily large squares for a same input by just multiplying the size of the squares and the offset by 2k . We only need to encode Turing machines inside: the only sizes/offsets possible are the inputs accepted by the TM.
18 / 21
Making sure there is a tiling with a unique direction of peridiodicity
19 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
20 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
Theorem The sets of slopes of tilings are exactly the recursively enumerable sets of Q ∪ {∞}.
20 / 21
Concluding remarks
Conjecture The sets of slopes of tilings in dimension d ≥ 3 are exactly the Σ02 subsets of (Q ∪ {∞})d−1 .
21 / 21
Emmanuel Jeandel Pascal Vanier Université de Provence - LIF - Marseille
JAC 2010
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
2 / 21
Tilings Finite set of tiles : T =
,
Finite set of forbidden patterns : F =
,
,
These are Shifts of Finite Type (SFTs).
2 / 21
Periodicity Definition (Vector of periodicity) v ∈ Z2 \ {(0, 0)} such that for all x ∈ Z2 ρ(x + v ) = ρ(x).
Vectors of periodicity of the form (3k, 3l )
3 / 21
Unique directions of periodicity Definition (Unique direction of periodicity) If all vectors of periodicity of a tiling are colinear, then it has a unique direction of periodicity.
Direction of periodicity (4, 1)
4 / 21
Unique directions of periodicity Definition (Unique direction of periodicity) If all vectors of periodicity of a tiling are colinear, then it has a unique direction of periodicity.
Definition (Slope) If a tiling has a unique direction of periodicity, then if (p, q) is a vector of periodicity, we say the slope of the tiling is q/p.
θ
Slope θ = 1/4
4 / 21
The question we will (in a way) answer today. . .
Given a tiling system, what can we say about its set of slopes?
5 / 21
The question we will (in a way) answer today. . .
Given a tiling system, what can we say about its set of slopes?
Sτ will denote the set of slopes of τ . Sτ is a subset of Q ∪ {∞}.
5 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
,
,
,
6 / 21
Example T =
,
Only allowed patterns :
Sτ = {1}
,
,
,
6 / 21
Results Theorem The set of all Sτ is exactly the set of recursively enumerable subsets of Q ∪ {∞}.
7 / 21
Results Theorem The set of all Sτ is exactly the set of recursively enumerable subsets of Q ∪ {∞}.
Definition Let S ⊆ Q ∪ {∞}, it is recursively enumerable if there exists a Turing machine M that halts on input (p, q) ∈ Z2 \ {(0, 0)} if and only if q/p ∈ S. • θ ∈ S ⇒ ∀(p, q), q/p = θ, M halts on input (p, q) • θ 6∈ S ⇒ ∀(p, q), q/p = θ, M does not halt on input (p, q)
7 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable.
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
8 / 21
Lemma For any tiling system τ , Sτ is recursively enumerable. Proof.
kp
θ kq
It is decidable to know if there exists an aperiodic point in one dimensional shifts. 8 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
9 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ . Proof. There is a Turing machine M such that: • θ ∈ R ⇒ ∀(p, q), q/p = θ, M halts on input (p, q) • θ 6∈ R ⇒ ∀(p, q), q/p = θ, M does not halt on input (p, q)
We have to construct a tiling system τM corresponding to the Turing machine M whose slopes would exactly be the θs accepted by M.
9 / 21
Ingredients Shape of the desired tilings :
10 / 21
Ingredients Shape of the desired tilings :
This construction corresponds to the inputs θ = (p, q) such that p > q > 0: the other cases are treated in a similar way and the final tiling system is the union of all these tiling systems. 10 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
11 / 21
Rows and columns • An East-deterministic aperiodic set (whites) • horizontal breaking tiles { } • vertical breaking tiles { , , , }
Rules on white tiles transcend black tiles. 11 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares only New layer : { , , , , , , }, superimposition is as follows : •
←→
•
←→ ,
•
←→
•
, , and
←→ ,
, , are superimposed to the white tiles.
12 / 21
Squares with the same size
13 / 21
Same offset between columns
14 / 21
Shape of the periodic tilings at this point
15 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component.
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the backgrounds Solution : we take A East-deterministic [Kari 92] in the first component. Transmission :
16 / 21
Synchronising the aperiodic backgrounds
Tiles { , , , }.
17 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation :
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes This means we can make arbitrarily large squares for a same input by just multiplying the size of the squares and the offset by 2k .
18 / 21
Making enough space for the computation Now that we have the shape, we need to make space for the computation : • Convert the input (p, q) in binary • strip p and q of their common last zeroes This means we can make arbitrarily large squares for a same input by just multiplying the size of the squares and the offset by 2k . We only need to encode Turing machines inside: the only sizes/offsets possible are the inputs accepted by the TM.
18 / 21
Making sure there is a tiling with a unique direction of peridiodicity
19 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
20 / 21
Lemma For any recursively enumerable set R ⊆ Q ∪ {∞}, there exists a tiling system such that R = Sτ .
Theorem The sets of slopes of tilings are exactly the recursively enumerable sets of Q ∪ {∞}.
20 / 21
Concluding remarks
Conjecture The sets of slopes of tilings in dimension d ≥ 3 are exactly the Σ02 subsets of (Q ∪ {∞})d−1 .
21 / 21
