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State complexity of the multiples of the

Thue-Morse set

Adeline Massuir

Joint work with Émilie Charlier and Célia Cisternino

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Automata

Deterministic nite automaton (DFA) : A = (Q, q0, F , Σ, δ)

1 2 3

a b a, b

b a

Language  Regular language

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Automata

Deterministic nite automaton (DFA) : A = (Q, q0, F , Σ, δ)

1 2 3

a b a, b

b a

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Minimal automaton

Theorem

For any regular language L, there exists a unique (up to isomorphism) minimal automaton accepting L.

Theorem

An automaton is minimal if and only if it is accessible and reduced. One algorithm :

1 Eject non accessible states 2 Look for undistinguished states

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Minimal automaton

Theorem

For any regular language L, there exists a unique (up to isomorphism) minimal automaton accepting L.

Theorem

An automaton is minimal if and only if it is accessible and reduced.

One algorithm :

1 Eject non accessible states 2 Look for undistinguished states

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Minimal automaton

Theorem

For any regular language L, there exists a unique (up to isomorphism) minimal automaton accepting L.

Theorem

An automaton is minimal if and only if it is accessible and reduced. One algorithm :

1 Eject non accessible states 2 Look for undistinguished states

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Minimal automaton

Theorem

For any regular language L, there exists a unique (up to isomorphism) minimal automaton accepting L.

Theorem

An automaton is minimal if and only if it is accessible and reduced. One algorithm :

1 Eject non accessible states

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Minimal automaton

Theorem

For any regular language L, there exists a unique (up to isomorphism) minimal automaton accepting L.

Theorem

An automaton is minimal if and only if it is accessible and reduced. One algorithm :

1 Eject non accessible states 2 Look for undistinguished states

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State complexity

Denition

The state complexity of a regular language is the number of states of its minimal automaton.

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What do we want to do ?

Denition

Let b ∈ N≥2. A subset X of N is b-recognizable if repb(X )is regular.

It is equivalent to work with 0∗rep b(X ). Theorem

Let b ∈ N≥2 and m ∈ N. If X ⊆ N is b-recognizable, so is mX .

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What do we want to do ?

Denition

Let b ∈ N≥2. A subset X of N is b-recognizable if repb(X )is regular.

It is equivalent to work with 0∗rep b(X ).

Theorem

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What do we want to do ?

Denition

Let b ∈ N≥2. A subset X of N is b-recognizable if repb(X )is regular.

It is equivalent to work with 0∗rep b(X ). Theorem

Let b ∈ N≥2 and m ∈ N. If X ⊆ N is b-recognizable, so is mX .

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Theorem [Alexeev, 2004]

The state complexity of the language 0∗rep

b(m N) is min N≥0 ( m gcd (m, bN)+ N−1 X n=0 bn gcd (bn, m) )

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Thue-Morse

0110100110010110 . . . 1001011001101001 . . .

Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1 ∈2 N} .

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Thue-Morse

0110100110010110 . . . 1001011001101001 . . .

Denition

The Thue-Morse set is the set

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . .

Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1 ∈2 N} .

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Thue-Morse

0110100110010110 . . . 1001011001101001 . . .

Denition

The Thue-Morse set is the set

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . .

Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1 ∈2 N} .

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Thue-Morse

0110100110010110 . . . 1001011001101001 . . .

Denition

The Thue-Morse set is the set

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . .

Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1 ∈2 N} .

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Thue-Morse

0110100110010110 . . . 1001011001101001 . . .

Denition

The Thue-Morse set is the set

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Thue-Morse

0110100110010110. . .

1001011001101001 . . .

Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1 ∈2 N} .

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . .

Denition

The Thue-Morse set is the set

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . . Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1∈2 N} .

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . . Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1∈2 N} .

T

B 0

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Thue-Morse

0110100110010110 . . .

1001011001101001 . . . Denition

The Thue-Morse set is the set

T = {n ∈ N : | rep2(n)|1∈2 N} . T B 0,3 0,3 1,2 1,2

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Denition

Let p, q ∈ N≥2. We say that p and q are multiplicatively independent

if

pa= qb ⇒ a = b =0. They are said multiplicatively dependent otherwise.

Theorem [Cobham, 1969]

Let b, b0 two multiplicatively independent bases. A subset of N

is both b-recognizable and b0-recognizable i it is a nite union

of arithmetic progressions.

Let b, b0 two multiplicatively dependent bases. A subset of N is

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Denition

Let p, q ∈ N≥2. We say that p and q are multiplicatively independent

if

pa= qb ⇒ a = b =0. They are said multiplicatively dependent otherwise.

Theorem [Cobham, 1969]

Let b, b0 two multiplicatively independent bases. A subset of N

is both b-recognizable and b0-recognizable i it is a nite union

of arithmetic progressions.

Let b, b0 two multiplicatively dependent bases. A subset of N is

b-recognizable i it is b0-recognizable.

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The result

T = {n ∈ N : | rep2(n)|1 ∈2 N}

Theorem

Let m ∈ N and p ∈ N≥1.

Then the state complexity of the language 0∗rep

2p(mT ) is

2k + z p



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The method

Automaton Language accepted

a a

AT ,2p (0, 0)∗rep2p(T × N)

Am,2p (0, 0)∗rep2p({(n, mn) : n ∈ N})

AT ,2p ×Am,2p (0, 0)∗rep2p({(t, mt) : t ∈T })

π AT ,2p×Am,2p 0∗rep2p(mT )

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The method

Automaton Language accepted

a a

AT ,2p (0, 0)∗rep2p(T × N)

Am,2p (0, 0)∗rep2p({(n, mn) : n ∈ N})

AT ,2p ×Am,2p (0, 0)∗rep2p({(t, mt) : t ∈T })

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The method

Automaton Language accepted

a a

AT ,2p (0, 0)∗rep2p(T × N)

Am,2p (0, 0)∗rep2p({(n, mn) : n ∈ N})

AT ,2p ×Am,2p (0, 0)∗rep2p({(t, mt) : t ∈T })

π AT ,2p×Am,2p 0∗rep2p(mT )

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The method

Automaton Language accepted

a a

AT ,2p (0, 0)∗rep2p(T × N)

Am,2p (0, 0)∗rep2p({(n, mn) : n ∈ N})

AT ,2p ×Am,2p (0, 0)∗rep2p({(t, mt) : t ∈T })

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The automaton A

T ,2p (0, 0)∗{rep2p(t, n) : t ∈T , n ∈ N} States T , B Initial state T Final states T Alphabet {0, . . . , 2p1}2 Transitions δT ,2p(X , (a, b)) =  X if a ∈ T X else. For all u, v ∈ {0, . . . , 2p1}, δT ,2p(X , (u, v )) =  X if val2p(u) ∈T X else.

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The automaton A

T ,2p (0, 0)∗{rep2p(t, n) : t ∈T , n ∈ N} States T , B Initial state T Final states T Alphabet {0, . . . , 2p1}2 Transitions δT ,2p(X , (a, b)) =  X if a ∈ T X else. For all u, v ∈ {0, . . . , 2p1}, δT ,2p(X , (u, v )) =  X if val2p(u) ∈T X else.

(36)

The automaton A

T ,2p (0, 0)∗{rep2p(t, n) : t ∈T , n ∈ N} States T , B Initial state T Final states T Alphabet {0, . . . , 2p1}2 Transitions δT ,2p(X , (a, b)) =  X if a ∈ T X else. For all u, v ∈ {0, . . . , 2p1}, δT ,2p(X , (u, v )) =  X if val2p(u) ∈T X else.

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The automaton A

T ,4 T B (0, 0), (0, 1), (0, 2), (0, 3) (3, 0), (3, 1), (3, 2), (3, 3) (0, 0), (0, 1), (0, 2), (0, 3) (3, 0), (3, 1), (3, 2), (3, 3) (1, 0), (1, 1), (1, 2), (1, 3) (2, 0), (2, 1), (2, 2), (2, 3) (1, 0), (1, 1), (1, 2), (1, 3) (2, 0), (2, 1), (2, 2), (2, 3)

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The automaton A

m,b (0, 0)∗{repb(n, mn) : n ∈ N} States 0, . . . , m − 1 Initial state 0 Final states 0 Alphabet {0, . . . , b − 1}2 Transitions δm,b(i , (d , e)) = j ⇔ bi + e = md + j

For all i, j ∈ {0, . . . , m − 1}, for all u, v ∈ {0, . . . , b − 1}∗,

δm,b(i , (u, v )) = j ⇔ b|(u,v )|i +valb(v ) = mvalb(u) + j .

(39)

The automaton A

m,b (0, 0)∗{repb(n, mn) : n ∈ N} States 0, . . . , m − 1 Initial state 0 Final states 0 Alphabet {0, . . . , b − 1}2 Transitions δm,b(i , (d , e)) = j ⇔ bi + e = md + j

For all i, j ∈ {0, . . . , m − 1}, for all u, v ∈ {0, . . . , b − 1}∗,

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The automaton A

m,b (0, 0)∗{repb(n, mn) : n ∈ N} States 0, . . . , m − 1 Initial state 0 Final states 0 Alphabet {0, . . . , b − 1}2 Transitions δm,b(i , (d , e)) = j ⇔ bi + e = md + j

For all i, j ∈ {0, . . . , m − 1}, for all u, v ∈ {0, . . . , b − 1}∗,

δm,b(i , (u, v )) = j ⇔ b|(u,v )|i +valb(v ) = mvalb(u) + j .

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The automaton A

6,4 0 1 2 3 4 5 (0, 0) (1, 3) (1, 0) (2, 3) (2, 0) (3, 3) (0, 1) (0,2) (0, 3) (0,0 ) (0,1 ) (1, 2) (1,1 ) (1, 2) (1, 3) (2,0) (2, 1) (2,2 ) (2, 1 ) (3 ,2 ) (3,3 ) (3, 0) (3 ,1 ) (3, 2 )

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The product automaton A

m,2p

×

A

T ,2p (0, 0)∗{rep2p(t, mt) : t ∈T } States (0, T ), . . . , (m − 1, T ), (0, B), . . . , (m − 1, B) Initial state (0, T ) Final states (0, T ) Alphabet {0, . . . , 2p1}2

Transitions δT ,2p((i , X ), (u, v )) = (j , Y )

⇔2p|(u,v )|i +val 2p(v ) = mval2p(u) + j and Y =  X if val2p(u) ∈T X else. Remark

If i, X , v are xed, there exist unique j, Y , u such that we have a transition labeled by (u, v) from (i, X ) to (j, Y ).

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The product automaton A

m,2p

×

A

T ,2p (0, 0)∗{rep2p(t, mt) : t ∈T } States (0, T ), . . . , (m − 1, T ), (0, B), . . . , (m − 1, B) Initial state (0, T ) Final states (0, T ) Alphabet {0, . . . , 2p1}2

Transitions δT ,2p((i , X ), (u, v )) = (j , Y )

⇔2p|(u,v )|i +val 2p(v ) = mval2p(u) + j and Y =  X if val2p(u) ∈T X else. Remark

If i, X , v are xed, there exist unique j, Y , u such that we have a transition labeled by (u, v) from (i, X ) to (j, Y ).

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The product automaton A

m,2p

×

A

T ,2p (0, 0)∗{rep2p(t, mt) : t ∈T } States (0, T ), . . . , (m − 1, T ), (0, B), . . . , (m − 1, B) Initial state (0, T ) Final states (0, T ) Alphabet {0, . . . , 2p1}2

Transitions δT ,2p((i , X ), (u, v )) = (j , Y )

⇔2p|(u,v )|i +val 2p(v ) = mval2p(u) + j and Y =  X if val2p(u) ∈T X else. Remark

If i, X , v are xed, there exist unique j, Y , u such that we have a transition labeled by (u, v) from (i, X ) to (j, Y ).

(45)

The automaton A

6,4

×

A

T ,4

0T 1T 2T 3T 4T 5T

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The projected automaton π (A

m,2p

×

A

T ,2p

)

0∗rep 2p(mT ) = 0∗{rep2p(mt) : t ∈T } 0T 1T 2T 3T 4T 5T 0B 1B 2B 3B 4B 5B 0 1 2 3

(47)

The projected automaton π (A

m,2p

×

A

T ,2p

)

0∗rep 2p(mT ) = 0∗{rep2p(mt) : t ∈T } 0T 1T 2T 3T 4T 5T 0B 1B 2B 3B 4B 5B 0 1 2 3

(48)

Proposition The automaton π Am,2p×AT ,2p is deterministic, accessible, coaccessible. Proposition

In the automaton π Am,2p×AT ,2p, the states (i, T ) and (i, B)

are disjoined for all i ∈ {0, . . . , m − 1}.

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Proposition The automaton π Am,2p×AT ,2p is deterministic, accessible, coaccessible. Proposition

In the automaton π Am,2p×AT ,2p, the states (i, T ) and (i, B)

(50)

The automaton π (A

6,4

×

A

T ,4

)

0T 1T 2T 3T 4T 5T 0B 1B 2B 3B 4B 5B 0 1 2 3

(51)

The automaton π (A

6,4

×

A

T ,4

)

0T 1T 2T 3T 4T 5T

0B 1B 2B 3B 4B 5B

(52)

The automaton π (A

24,4

×

A

T ,4

)

(53)
(54)

The automaton π (A

24,4

×

A

T ,4

)

(55)
(56)

The automaton π (A

24,4

×

A

T ,4

)

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Recall

For m ∈ N, p ∈ N≥1, we write m = k2z with z odd.

For all n ∈ N, we set

Tn:=



T if n ∈ T B else.

Denition

For all j ∈ {1, . . . , k − 1}, we set

[(j , T )] := {(j + k`, T`) :0 ≤ ` ≤ 2z−1}

[(j , B)] :=

j + k`, T` :0 ≤ ` ≤ 2z−1 .

We also set

(58)

Recall

For m ∈ N, p ∈ N≥1, we write m = k2z with z odd.

For all n ∈ N, we set

Tn:=



T if n ∈ T B else.

Denition

For all j ∈ {1, . . . , k − 1}, we set

[(j , T )] := {(j + k`, T`) :0 ≤ ` ≤ 2z−1}

[(j , B)] :=

j + k`, T` :0 ≤ ` ≤ 2z−1 .

We also set

[(0, T )] := {(0, T )} and [(0, B)] :=  k`, T`:0 ≤ ` ≤ 2z−1 .

(59)

Recall

For m ∈ N, p ∈ N≥1, we write m = k2z with z odd.

For all n ∈ N, we set

Tn:=



T if n ∈ T B else.

Denition

For all j ∈ {1, . . . , k − 1}, we set

[(j , T )] := {(j + k`, T`) :0 ≤ ` ≤ 2z−1}

[(j , B)] :=

j + k`, T` :0 ≤ ` ≤ 2z−1 .

We also set

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The automaton π (A

24,4

×

A

T ,4

)

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(62)

The automaton π (A

24,4

×

A

T ,4

)

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Denition

For all α ∈ {0, . . . , z − 1}, we set Cα:=  k2z−α−1+ k2z−α`, T` :0 ≤ ` ≤ 2α−1 . For all β ∈n0, . . . , lz p m −2o, we set Γβ := [ α∈{βp,...,(β+1)p−1} Cα. We also set Γl z p m −1 := [ α∈ nl z p m −1p,...,z−1 o Cα.

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We can build a new automaton

which is accessible

reduced.

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We can build a new automaton which is accessible

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We can build a new automaton which is accessible

reduced.

(67)

0H 1H 2H 3H 4H 5H 0L 1L 2L 3L 4L 5L 0 1 3 2

(68)

0

1 2

3

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Theorem

Let m ∈ N and p ∈ N≥1. Then the state complexity of the language

0∗rep 2p(mT ) is equal to 2k + z p  if m = k2z with k odd. 2 × 3 + 1 2 =7

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Theorem

Let m ∈ N and p ∈ N≥1. Then the state complexity of the language

0∗rep 2p(mT ) is equal to 2k + z p  if m = k2z with k odd. 2 × 3 + 1 2 =7

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To go further

What about the language 0∗rep

2p(mT + r)

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The automaton π A

20

24,4

×

A

T ,4



(73)

The automaton π A

20

24,4

×

A

T ,4

(74)

The automaton π A

20

24,4

×

A

T ,4



(75)

The automaton π A

20

24,4

×

A

T ,4

(76)

The automaton π A

20

24,4

×

A

T ,4



(77)

The automaton π A

20

24,4

×

A

T ,4

(78)

The automaton π A

20

24,4

×

A

T ,4



(79)

The automaton π A

20

24,4

×

A

T ,4

(80)

The automaton π A

20

24,4

×

A

T ,4



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Denition

For all 0 ≤ α ≤ maxnlz p m , |rep2p(r )| o =: L, Rα0 = (   r 2αp + `k2z−αp, X` :0 ≤ ` ≤ 2αp−1 if α ≤ j z p k   r 2αp + `k, X` :0 ≤ ` ≤ 2z−1 else. and Rα= Rα0 \ α−1 S i =0 Ri0.

For all 1 ≤ j ≤ k − 1 and Y ∈ {T , B} and for j = 0 and Y = X , SjY = [(j , Y )] \

L

[

α=0

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Denition

For all 0 ≤ α ≤ maxnlz p m , |rep2p(r )| o =: L, Rα0 = (   r 2αp + `k2z−αp, X` :0 ≤ ` ≤ 2αp−1 if α ≤ j z p k   r 2αp + `k, X` :0 ≤ ` ≤ 2z−1 else. and Rα= Rα0 \ α−1 S i =0 Ri0.

For all 1 ≤ j ≤ k − 1 and Y ∈ {T , B} and for j = 0 and Y = X , SjY = [(j , Y )] \

L

[

α=0

Rα.

(83)

Theorem

Let m ∈ N, r ∈ {0, . . . , m − 1} and p ∈ N≥1. Let X = T or

N \T . Then, the state complexity of the language 0∗rep2p(mX + r )

is equal to

2k + z p



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