[PPT] - Synchronizing Finite Automata Lecture I: Cern y conjecture, PowerPoint Presentation

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Synchronizing Finite Automata

Lecture I: ˇ Cern´ y conjecture, Pin-Frankl’s bound and recent advances Mikhail Volkov

Ural Federal University

Mikhail Volkov Synchronizing Finite Automata

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1. Definitions

Deterministic finite automata (DFA): A = Q, Σ, δ.

Q the state set
Σ the input alphabet
δ : Q × Σ → Q the transition function

A is called synchronizing if there exists a word w ∈ Σ∗ whose action resets A , that is, leaves the automaton in one particular state no matter which state in Q it started at: δ(q, w) = δ(q′, w) for all q, q′ ∈ Q. |Q . w| = 1. Here Q . v = {δ(q, v) | q ∈ Q}. Any w with this property is a reset word for A .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 3

1. Definitions

Deterministic finite automata (DFA): A = Q, Σ, δ.

Q the state set
Σ the input alphabet
δ : Q × Σ → Q the transition function

A is called synchronizing if there exists a word w ∈ Σ∗ whose action resets A , that is, leaves the automaton in one particular state no matter which state in Q it started at: δ(q, w) = δ(q′, w) for all q, q′ ∈ Q. |Q . w| = 1. Here Q . v = {δ(q, v) | q ∈ Q}. Any w with this property is a reset word for A .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 4

1. Definitions

Deterministic finite automata (DFA): A = Q, Σ, δ.

Q the state set
Σ the input alphabet
δ : Q × Σ → Q the transition function

A is called synchronizing if there exists a word w ∈ Σ∗ whose action resets A , that is, leaves the automaton in one particular state no matter which state in Q it started at: δ(q, w) = δ(q′, w) for all q, q′ ∈ Q. |Q . w| = 1. Here Q . v = {δ(q, v) | q ∈ Q}. Any w with this property is a reset word for A .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 5

1. Definitions

Deterministic finite automata (DFA): A = Q, Σ, δ.

Q the state set
Σ the input alphabet
δ : Q × Σ → Q the transition function

A is called synchronizing if there exists a word w ∈ Σ∗ whose action resets A , that is, leaves the automaton in one particular state no matter which state in Q it started at: δ(q, w) = δ(q′, w) for all q, q′ ∈ Q. |Q . w| = 1. Here Q . v = {δ(q, v) | q ∈ Q}. Any w with this property is a reset word for A .

Mikhail Volkov Synchronizing Finite Automata

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2. Example

1 2 3 a b b b b a a a A reset word is abbbabbba. In fact, we will see that this is the shortest reset word for this automaton.

Mikhail Volkov Synchronizing Finite Automata

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2. Example

1 2 3 a b b b b a a a A reset word is abbbabbba. In fact, we will see that this is the shortest reset word for this automaton.

Mikhail Volkov Synchronizing Finite Automata

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3. Power Automaton

Not every DFA is synchronizing. Therefore, the very first question is the following one: given an automaton, how to determine whether or not it is synchronizing? This question is easy, and a straightforward solution comes from the classic powerset construction by Rabin and Scott. The power automaton P(A ) of a given DFA A = Q, Σ, δ:

states are the non-empty subsets of Q,
δ(P, a) = P . a = {δ(p, a) | p ∈ P}

A w ∈ Σ∗ is a reset word for the DFA A iff w labels a path in P(A ) starting at Q and ending at a singleton.

Mikhail Volkov Synchronizing Finite Automata

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3. Power Automaton

Not every DFA is synchronizing. Therefore, the very first question is the following one: given an automaton, how to determine whether or not it is synchronizing? This question is easy, and a straightforward solution comes from the classic powerset construction by Rabin and Scott. The power automaton P(A ) of a given DFA A = Q, Σ, δ:

states are the non-empty subsets of Q,
δ(P, a) = P . a = {δ(p, a) | p ∈ P}

A w ∈ Σ∗ is a reset word for the DFA A iff w labels a path in P(A ) starting at Q and ending at a singleton.

Mikhail Volkov Synchronizing Finite Automata

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3. Power Automaton

Not every DFA is synchronizing. Therefore, the very first question is the following one: given an automaton, how to determine whether or not it is synchronizing? This question is easy, and a straightforward solution comes from the classic powerset construction by Rabin and Scott. The power automaton P(A ) of a given DFA A = Q, Σ, δ:

states are the non-empty subsets of Q,
δ(P, a) = P . a = {δ(p, a) | p ∈ P}

A w ∈ Σ∗ is a reset word for the DFA A iff w labels a path in P(A ) starting at Q and ending at a singleton.

Mikhail Volkov Synchronizing Finite Automata

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3. Power Automaton

Not every DFA is synchronizing. Therefore, the very first question is the following one: given an automaton, how to determine whether or not it is synchronizing? This question is easy, and a straightforward solution comes from the classic powerset construction by Rabin and Scott. The power automaton P(A ) of a given DFA A = Q, Σ, δ:

states are the non-empty subsets of Q,
δ(P, a) = P . a = {δ(p, a) | p ∈ P}

A w ∈ Σ∗ is a reset word for the DFA A iff w labels a path in P(A ) starting at Q and ending at a singleton.

Mikhail Volkov Synchronizing Finite Automata

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4. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a 012 013 123 023 0123 b a a b a b b b a a a b b a

Mikhail Volkov Synchronizing Finite Automata

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4. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a 012 013 123 023 0123 b a a b a b b b a a a b b a b a b a b b a b b

Mikhail Volkov Synchronizing Finite Automata

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5. Polynomial Algorithm

Thus, the question of whether or not a given DFA A is synchronizing reduces to the following reachability question in the underlying digraph of the power automaton P(A ): is there a path from Q to a singleton? The latter question can be easily answered by BFS. This algorithm is however exponential w.r.t. the size of A . The following result (independently obtained by Jan ˇ Cern´ y and Chung Laung Liu) gives a polynomial algorithm:

Proposition. A DFA A = Q, Σ, δ is synchronizing iff for every

q, q′ ∈ Q there exists a word w ∈ Σ∗ such that δ(q, w) = δ(q′, w).

Mikhail Volkov Synchronizing Finite Automata

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5. Polynomial Algorithm

Thus, the question of whether or not a given DFA A is synchronizing reduces to the following reachability question in the underlying digraph of the power automaton P(A ): is there a path from Q to a singleton? The latter question can be easily answered by BFS. This algorithm is however exponential w.r.t. the size of A . The following result (independently obtained by Jan ˇ Cern´ y and Chung Laung Liu) gives a polynomial algorithm:

Proposition. A DFA A = Q, Σ, δ is synchronizing iff for every

q, q′ ∈ Q there exists a word w ∈ Σ∗ such that δ(q, w) = δ(q′, w).

Mikhail Volkov Synchronizing Finite Automata

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5. Polynomial Algorithm

Thus, the question of whether or not a given DFA A is synchronizing reduces to the following reachability question in the underlying digraph of the power automaton P(A ): is there a path from Q to a singleton? The latter question can be easily answered by BFS. This algorithm is however exponential w.r.t. the size of A . The following result (independently obtained by Jan ˇ Cern´ y and Chung Laung Liu) gives a polynomial algorithm:

Proposition. A DFA A = Q, Σ, δ is synchronizing iff for every

q, q′ ∈ Q there exists a word w ∈ Σ∗ such that δ(q, w) = δ(q′, w).

Mikhail Volkov Synchronizing Finite Automata

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5. Polynomial Algorithm

Thus, the question of whether or not a given DFA A is synchronizing reduces to the following reachability question in the underlying digraph of the power automaton P(A ): is there a path from Q to a singleton? The latter question can be easily answered by BFS. This algorithm is however exponential w.r.t. the size of A . The following result (independently obtained by Jan ˇ Cern´ y and Chung Laung Liu) gives a polynomial algorithm:

Proposition. A DFA A = Q, Σ, δ is synchronizing iff for every

q, q′ ∈ Q there exists a word w ∈ Σ∗ such that δ(q, w) = δ(q′, w).

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 12 23 02 13 a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 12 23 02 13 a a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 02 12 23 13 a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 02 12 23 13 a b b a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 12 23 02 13 a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a 03 01 12 23 02 13 a b b a b b a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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6. Example

1 2 3 a, b b b b a a a 03 01 12 23 02 13 a a a b b b b a a b b a a, Q . a = {1, 2, 3}; a · bba, Q . abba = {1, 3} abba · babbba, Q . abbababbba = {1} Observe that the reset word constructed this way is of length 10 while we know a reset word of length 9.

Mikhail Volkov Synchronizing Finite Automata

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7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

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7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

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7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

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7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 30

7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 31

7. Results-I

Thus, recognizing synchronizability reduces to a reachability problem in the automaton whose states are the 2-subsets and the 1-subsets of Q. The latter can be solved by BFS in O(n2 · |Σ|) time where n = |Q|. Can one do better? It is an open problem. Recently, Mikhail Berlinkov has developed a (very non-trivial) algorithm that checks whether or not an automaton with n states is synchronizing and spends time O(n) on average. The worst case complexity of Berlinkov’s algorithm is still quadratic. Berlinkov’s algorithm has been implemented by Pavel Ageev.

Mikhail Volkov Synchronizing Finite Automata

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8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 33

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 34

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 35

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 36

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 37

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 38

8. Results-II

In fact, the basic algorithm not only recognizes synchronizability but also returns a reset word provided that such exists. If one also wants to produce a reset word, one need O(n3 + n2 · |Σ|)

time. Why? One needs time to write down the word!

Clearly, the resulting reset word has length O(n3): the algorithm makes at most n − 1 steps and the length of the segment added in the step when k states are still to be compressed (n ≥ k ≥ 2) is at most 1 + # of blank 2-subsets, i.e., 1 + n

2

−

k

2

. This gives the

upper bound close to n3−n

3

. Can we do better? What is the exact bound?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 39

9. A Resource for Improvement

1 2 3 a, b b b b a a a a a a b b a b b a 03 01 02 12 23 13 a b b We see that the shortest path from a light-grey 2-subset to a singleton does not necessarily pass through all blank 2-subsets. Consider a generic step of the algorithm at which states to be compressed form a set P with |P| = k > 1 and let v = a1 · · · aℓ with ai ∈ Σ, i = 1, . . . , ℓ, be a word of minimum length such that |P . v| < k.

Mikhail Volkov Synchronizing Finite Automata

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9. A Resource for Improvement

1 2 3 a, b b b b a a a a a a b b a b b a 03 01 02 12 23 13 a b b We see that the shortest path from a light-grey 2-subset to a singleton does not necessarily pass through all blank 2-subsets. Consider a generic step of the algorithm at which states to be compressed form a set P with |P| = k > 1 and let v = a1 · · · aℓ with ai ∈ Σ, i = 1, . . . , ℓ, be a word of minimum length such that |P . v| < k.

Mikhail Volkov Synchronizing Finite Automata

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9. A Resource for Improvement

1 2 3 a, b b b b a a a a a a b b a b b a 03 01 02 12 23 13 a b b We see that the shortest path from a light-grey 2-subset to a singleton does not necessarily pass through all blank 2-subsets. Consider a generic step of the algorithm at which states to be compressed form a set P with |P| = k > 1 and let v = a1 · · · aℓ with ai ∈ Σ, i = 1, . . . , ℓ, be a word of minimum length such that |P . v| < k.

Mikhail Volkov Synchronizing Finite Automata

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10. Studying Generic Step

The sets P1 = P, P2 = P1 . a1, . . . , Pℓ = Pℓ−1 . aℓ−1 are k-subsets of Q. Since |Pℓ . aℓ| < |Pℓ|, there exist two states qℓ, q′

ℓ ∈ Pℓ such that δ(qℓ, aℓ) = δ(q′ ℓ, aℓ). Now define 2-subsets

Ri = {qi, q′

i} ⊆ Pi, i = 1, . . . , ℓ, such that δ(qi, ai) = qi+1,

δ(q′

i, ai) = q′ i+1 for i = 1, . . . , ℓ − 1.

q1 q′

1

q2 q′

2

qℓ q′

ℓ

a1 a1 aℓ aℓ P1 P2 Pℓ a2 a2 aℓ−1 aℓ−1 . . . . . . The condition that v is a word of minimum length with |P . v| < |P| implies Ri Pj for 1 ≤ j < i ≤ ℓ.

Mikhail Volkov Synchronizing Finite Automata

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10. Studying Generic Step

The sets P1 = P, P2 = P1 . a1, . . . , Pℓ = Pℓ−1 . aℓ−1 are k-subsets of Q. Since |Pℓ . aℓ| < |Pℓ|, there exist two states qℓ, q′

ℓ ∈ Pℓ such that δ(qℓ, aℓ) = δ(q′ ℓ, aℓ). Now define 2-subsets

Ri = {qi, q′

i} ⊆ Pi, i = 1, . . . , ℓ, such that δ(qi, ai) = qi+1,

δ(q′

i, ai) = q′ i+1 for i = 1, . . . , ℓ − 1.

q1 q′

1

q2 q′

2

qℓ q′

ℓ

a1 a1 aℓ aℓ P1 P2 Pℓ a2 a2 aℓ−1 aℓ−1 . . . . . . The condition that v is a word of minimum length with |P . v| < |P| implies Ri Pj for 1 ≤ j < i ≤ ℓ.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 44

10. Studying Generic Step

The sets P1 = P, P2 = P1 . a1, . . . , Pℓ = Pℓ−1 . aℓ−1 are k-subsets of Q. Since |Pℓ . aℓ| < |Pℓ|, there exist two states qℓ, q′

ℓ ∈ Pℓ such that δ(qℓ, aℓ) = δ(q′ ℓ, aℓ). Now define 2-subsets

Ri = {qi, q′

i} ⊆ Pi, i = 1, . . . , ℓ, such that δ(qi, ai) = qi+1,

δ(q′

i, ai) = q′ i+1 for i = 1, . . . , ℓ − 1.

q1 q′

1

q2 q′

2

qℓ q′

ℓ

a1 a1 aℓ aℓ P1 P2 Pℓ a2 a2 aℓ−1 aℓ−1 . . . . . . The condition that v is a word of minimum length with |P . v| < |P| implies Ri Pj for 1 ≤ j < i ≤ ℓ.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 45

10. Studying Generic Step

The sets P1 = P, P2 = P1 . a1, . . . , Pℓ = Pℓ−1 . aℓ−1 are k-subsets of Q. Since |Pℓ . aℓ| < |Pℓ|, there exist two states qℓ, q′

ℓ ∈ Pℓ such that δ(qℓ, aℓ) = δ(q′ ℓ, aℓ). Now define 2-subsets

Ri = {qi, q′

i} ⊆ Pi, i = 1, . . . , ℓ, such that δ(qi, ai) = qi+1,

δ(q′

i, ai) = q′ i+1 for i = 1, . . . , ℓ − 1.

q1 q′

1

q2 q′

2

qℓ q′

ℓ

a1 a1 aℓ aℓ P1 P2 Pℓ a2 a2 aℓ−1 aℓ−1 . . . . . . The condition that v is a word of minimum length with |P . v| < |P| implies Ri Pj for 1 ≤ j < i ≤ ℓ.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 46

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 47

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 48

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 49

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 50

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 51

11. Combinatorial Configuration

Our question reduces to the following problem in combinatorics of finite sets: Let Q be an n-set, P1, . . . , Pℓ a sequence of its k-subsets (k > 1) such that each Pi, 1 < i ≤ ℓ, includes a “fresh” 2-subset that does not occur in any previous Pj (1 ≤ j < i). How long can such renewing sequences be? A construction: fix a (k − 2)-subset W of Q, list all n−k+2

2

2-subsets of Q \ W and let Ti be the union of W with the ith

2-subset in the list. This gives the renewing sequence T1, . . . , Ts

f length s =

n−k+2

2

. Is this the maximum?

Mikhail Volkov Synchronizing Finite Automata

SLIDE 52

12. Combinatorial Configuration

The question turned out to be very difficult and was solved (in the affirmative) by Peter Frankl (An extremal problem for two families

f sets, Eur. J. Comb., 3 (1982) 125–127).

The proof uses linearization techniques which are quite common in combinatorics of finite sets. One reformulates the problem in linear algebra terms and then uses the corresponding machinery. We identify Q with {1, 2, . . . , n} and assign to each k-subset I = {i1, . . . , ik} the following polynomial D(I) in variables xi1, . . . , xik over the field of reals.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 53

12. Combinatorial Configuration

The question turned out to be very difficult and was solved (in the affirmative) by Peter Frankl (An extremal problem for two families

f sets, Eur. J. Comb., 3 (1982) 125–127).

The proof uses linearization techniques which are quite common in combinatorics of finite sets. One reformulates the problem in linear algebra terms and then uses the corresponding machinery. We identify Q with {1, 2, . . . , n} and assign to each k-subset I = {i1, . . . , ik} the following polynomial D(I) in variables xi1, . . . , xik over the field of reals.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 54

12. Combinatorial Configuration

The question turned out to be very difficult and was solved (in the affirmative) by Peter Frankl (An extremal problem for two families

f sets, Eur. J. Comb., 3 (1982) 125–127).

The proof uses linearization techniques which are quite common in combinatorics of finite sets. One reformulates the problem in linear algebra terms and then uses the corresponding machinery. We identify Q with {1, 2, . . . , n} and assign to each k-subset I = {i1, . . . , ik} the following polynomial D(I) in variables xi1, . . . , xik over the field of reals.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 55

12. Combinatorial Configuration

The question turned out to be very difficult and was solved (in the affirmative) by Peter Frankl (An extremal problem for two families

f sets, Eur. J. Comb., 3 (1982) 125–127).

The proof uses linearization techniques which are quite common in combinatorics of finite sets. One reformulates the problem in linear algebra terms and then uses the corresponding machinery. We identify Q with {1, 2, . . . , n} and assign to each k-subset I = {i1, . . . , ik} the following polynomial D(I) in variables xi1, . . . , xik over the field of reals.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 56

13. Linearization

I = {i1, . . . , ik} → D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Then one proves that:

the polynomials D(P1), . . . , D(Pℓ) are linearly independent

whenever the k-subsets P1, . . . , Pℓ form a renewing sequence;

the polynomials D(T1), . . . , D(Ts) (derived from the “standard”

sequence) generate the linear space spanned by all polynomials of the form D(I).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 57

13. Linearization

I = {i1, . . . , ik} → D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Then one proves that:

the polynomials D(P1), . . . , D(Pℓ) are linearly independent

whenever the k-subsets P1, . . . , Pℓ form a renewing sequence;

the polynomials D(T1), . . . , D(Ts) (derived from the “standard”

sequence) generate the linear space spanned by all polynomials of the form D(I).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 58

13. Linearization

I = {i1, . . . , ik} → D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Then one proves that:

the polynomials D(P1), . . . , D(Pℓ) are linearly independent

whenever the k-subsets P1, . . . , Pℓ form a renewing sequence;

the polynomials D(T1), . . . , D(Ts) (derived from the “standard”

sequence) generate the linear space spanned by all polynomials of the form D(I).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 59

14. Linearization, Step 1

D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Suppose that k-subsets P1, . . . , Pℓ form a renewing sequence but D(P1), . . . , D(Pℓ) are linearly dependent. Then some polynomial D(Pj) should be expressible as a linear combination of the preceding polynomials D(P1), . . . , D(Pj−1). By the definition of a renewing sequence, Pj contains a couple {p, p′} such that {p, p′} Pi for all i < j. If we substitute xp = p, xp′ = p′ and xt = 0 for t = p, p′ in each polynomial D(P1), . . . , D(Pj), then the polynomials D(P1), . . . , D(Pj−1) vanish (since the two last columns in each of the resulting determinants become proportional) and so does any linear combination of the polynomials.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 60

14. Linearization, Step 1

D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Suppose that k-subsets P1, . . . , Pℓ form a renewing sequence but D(P1), . . . , D(Pℓ) are linearly dependent. Then some polynomial D(Pj) should be expressible as a linear combination of the preceding polynomials D(P1), . . . , D(Pj−1). By the definition of a renewing sequence, Pj contains a couple {p, p′} such that {p, p′} Pi for all i < j. If we substitute xp = p, xp′ = p′ and xt = 0 for t = p, p′ in each polynomial D(P1), . . . , D(Pj), then the polynomials D(P1), . . . , D(Pj−1) vanish (since the two last columns in each of the resulting determinants become proportional) and so does any linear combination of the polynomials.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 61

14. Linearization, Step 1

D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Suppose that k-subsets P1, . . . , Pℓ form a renewing sequence but D(P1), . . . , D(Pℓ) are linearly dependent. Then some polynomial D(Pj) should be expressible as a linear combination of the preceding polynomials D(P1), . . . , D(Pj−1). By the definition of a renewing sequence, Pj contains a couple {p, p′} such that {p, p′} Pi for all i < j. If we substitute xp = p, xp′ = p′ and xt = 0 for t = p, p′ in each polynomial D(P1), . . . , D(Pj), then the polynomials D(P1), . . . , D(Pj−1) vanish (since the two last columns in each of the resulting determinants become proportional) and so does any linear combination of the polynomials.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 62

14. Linearization, Step 1

D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Suppose that k-subsets P1, . . . , Pℓ form a renewing sequence but D(P1), . . . , D(Pℓ) are linearly dependent. Then some polynomial D(Pj) should be expressible as a linear combination of the preceding polynomials D(P1), . . . , D(Pj−1). By the definition of a renewing sequence, Pj contains a couple {p, p′} such that {p, p′} Pi for all i < j. If we substitute xp = p, xp′ = p′ and xt = 0 for t = p, p′ in each polynomial D(P1), . . . , D(Pj), then the polynomials D(P1), . . . , D(Pj−1) vanish (since the two last columns in each of the resulting determinants become proportional) and so does any linear combination of the polynomials.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 63

14. Linearization, Step 1

D(I) =

1

i1 i2

1

· · · ik−3

1

xi1 x2

i1

1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

xik x2

ik

k×k

Suppose that k-subsets P1, . . . , Pℓ form a renewing sequence but D(P1), . . . , D(Pℓ) are linearly dependent. Then some polynomial D(Pj) should be expressible as a linear combination of the preceding polynomials D(P1), . . . , D(Pj−1). By the definition of a renewing sequence, Pj contains a couple {p, p′} such that {p, p′} Pi for all i < j. If we substitute xp = p, xp′ = p′ and xt = 0 for t = p, p′ in each polynomial D(P1), . . . , D(Pj), then the polynomials D(P1), . . . , D(Pj−1) vanish (since the two last columns in each of the resulting determinants become proportional) and so does any linear combination of the polynomials.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 64

15. Linearization, Step 1, completed

D(Pj) =

1

i1 i2

1

· · · ik−3

1

p p2 1 i2 i2

2

· · · ik−3

2

p′ (p′)2 1 i3 i2

3

· · · ik−3

3

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

k×k

(For simplicity, here we assume that i1 = p, i2 = p′.) The value of D(Pj) is the determinant being the product of a Vandermonde (k − 2) × (k − 2)-determinant with the 2 × 2-determinant

p

p2 p′ (p′)2

, whence this value is not 0.

Hence D(Pj) cannot be equal to a linear combination of D(P1), . . . , D(Pj−1).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 65

15. Linearization, Step 1, completed

D(Pj) =

1

i1 i2

1

· · · ik−3

1

p p2 1 i2 i2

2

· · · ik−3

2

p′ (p′)2 1 i3 i2

3

· · · ik−3

3

. . . . . . . . . ... . . . . . . . . . 1 ik i2

k

· · · ik−3

k

k×k

(For simplicity, here we assume that i1 = p, i2 = p′.) The value of D(Pj) is the determinant being the product of a Vandermonde (k − 2) × (k − 2)-determinant with the 2 × 2-determinant

p

p2 p′ (p′)2

, whence this value is not 0.

Hence D(Pj) cannot be equal to a linear combination of D(P1), . . . , D(Pj−1).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 66

16. Linearization, Step 2

Now we aim to prove that the polynomials D(T1), . . . , D(Ts) (derived from the “standard” sequence) generate the linear space spanned by all polynomials of the form D(I). Take an arbitrary k-element subset I = {i1, . . . , ik} of Q. We claim that the polynomial D(I) is a linear combination of D(T1), . . . , D(Ts). We induct on the cardinality of the set I \ W . If |I \ W | = 2, then I is the union of W with some couple from Q \ W , whence I = Ti for some i = 1, . . . , s. Thus, D(I) = D(Ti) and our claim holds true.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 67

16. Linearization, Step 2

Now we aim to prove that the polynomials D(T1), . . . , D(Ts) (derived from the “standard” sequence) generate the linear space spanned by all polynomials of the form D(I). Take an arbitrary k-element subset I = {i1, . . . , ik} of Q. We claim that the polynomial D(I) is a linear combination of D(T1), . . . , D(Ts). We induct on the cardinality of the set I \ W . If |I \ W | = 2, then I is the union of W with some couple from Q \ W , whence I = Ti for some i = 1, . . . , s. Thus, D(I) = D(Ti) and our claim holds true.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 68

16. Linearization, Step 2

Now we aim to prove that the polynomials D(T1), . . . , D(Ts) (derived from the “standard” sequence) generate the linear space spanned by all polynomials of the form D(I). Take an arbitrary k-element subset I = {i1, . . . , ik} of Q. We claim that the polynomial D(I) is a linear combination of D(T1), . . . , D(Ts). We induct on the cardinality of the set I \ W . If |I \ W | = 2, then I is the union of W with some couple from Q \ W , whence I = Ti for some i = 1, . . . , s. Thus, D(I) = D(Ti) and our claim holds true.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 69

16. Linearization, Step 2

Now we aim to prove that the polynomials D(T1), . . . , D(Ts) (derived from the “standard” sequence) generate the linear space spanned by all polynomials of the form D(I). Take an arbitrary k-element subset I = {i1, . . . , ik} of Q. We claim that the polynomial D(I) is a linear combination of D(T1), . . . , D(Ts). We induct on the cardinality of the set I \ W . If |I \ W | = 2, then I is the union of W with some couple from Q \ W , whence I = Ti for some i = 1, . . . , s. Thus, D(I) = D(Ti) and our claim holds true.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 70

17. Linearization, Step 2, continued

If |I \ W | > 2, there is i0 ∈ W \ I. Let I ′ = I ∪ {i0}. There exists a polynomial p(x) = α0 + α1x + α2x2 · · · + αk−3xk−3 over R such that p(i0) = 1 and p(i) = 0 for all i ∈ W \ {i0}. Consider the determinant ∆ =

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

. Clearly, ∆ = 0 as the first column is the sum of the next k − 2 columns with the coefficients α0, α1, α2, . . . , αk−3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 71

17. Linearization, Step 2, continued

If |I \ W | > 2, there is i0 ∈ W \ I. Let I ′ = I ∪ {i0}. There exists a polynomial p(x) = α0 + α1x + α2x2 · · · + αk−3xk−3 over R such that p(i0) = 1 and p(i) = 0 for all i ∈ W \ {i0}. Consider the determinant ∆ =

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

. Clearly, ∆ = 0 as the first column is the sum of the next k − 2 columns with the coefficients α0, α1, α2, . . . , αk−3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 72

17. Linearization, Step 2, continued

If |I \ W | > 2, there is i0 ∈ W \ I. Let I ′ = I ∪ {i0}. There exists a polynomial p(x) = α0 + α1x + α2x2 · · · + αk−3xk−3 over R such that p(i0) = 1 and p(i) = 0 for all i ∈ W \ {i0}. Consider the determinant ∆ =

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

. Clearly, ∆ = 0 as the first column is the sum of the next k − 2 columns with the coefficients α0, α1, α2, . . . , αk−3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 73

17. Linearization, Step 2, continued

If |I \ W | > 2, there is i0 ∈ W \ I. Let I ′ = I ∪ {i0}. There exists a polynomial p(x) = α0 + α1x + α2x2 · · · + αk−3xk−3 over R such that p(i0) = 1 and p(i) = 0 for all i ∈ W \ {i0}. Consider the determinant ∆ =

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

. Clearly, ∆ = 0 as the first column is the sum of the next k − 2 columns with the coefficients α0, α1, α2, . . . , αk−3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 74

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0. Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 75

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0. ∆ =

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 76

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0.

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 77

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0.

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 78

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0.

p(i0)

1 i0 i2 · · · ik−3 xi0 x2

i0

p(i1) 1 i1 i2

1

· · · ik−3

1

xi1 x2

i1

p(i2) 1 i2 i2

2

· · · ik−3

2

xi2 x2

i2

. . . . . . . . . . . . ... . . . . . . . . . p(ik) 1 ik i2

k

· · · ik−3

k

xik x2

ik

(k+1)×(k+1)

Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 79

18. Linearization, Step 2, completed

Expanding ∆ by the first column gives the identity

k

j=0

(−1)jp(ij)D(I ′ \ {ij}) = 0. Since p(i0) = 1 and I ′ \ {i0} = I, the identity rewrites as D(I) =

k

j=1

(−1)j+1p(ij)D(I ′ \ {ij}), and since p(i) = 0 for all i ∈ W \ {i0}, all the non-zero summands in the right-hand side are such that ij / ∈ W . For each such ij, we have (I ′\{ij})\W = I ′\(W ∪{ij}) = (I ∪{i0})\(W ∪{ij}) = (I \W )\{ij}, whence |(I ′ \ {ij}) \ W | = |I \ W | − 1 and by the inductive assumption, the polynomials D(I ′ \ {ij}) are linear combinations of D(T1), . . . , D(Ts).

Mikhail Volkov Synchronizing Finite Automata

SLIDE 80

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 81

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 82

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 83

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 84

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 85

19. Results

Thus, in the step when k states are still to be compressed, the compression can always be achieved by applying a suitable word of length ≤ n−k+2

2

.

Summing up over k = n, . . . , 2, we see that the greedy algorithm always returns a reset word of length ≤ n3−n

6

: 2 2

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

3 3

+

3 2

+

4 2

+ · · · +

n − 1 2

+

n 2

=

4 3

+

4 2

+ · · · +

n − 1 2

+

n 2

= · · · =

n + 1 3

= n3 − n

6 .

Mikhail Volkov Synchronizing Finite Automata

SLIDE 86

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 87

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 88

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 89

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 90

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 91

20. The ˇ

Cern´ y Series

Suppose a synchronizing automaton has n states. What is its reset threshold, i.e., the minimum length of its reset words? We have just presented an upper bound: there always exists a reset word of length n3−n

6

. This bound was obtained by Jean-´ Eric Pin in 1983 on the basis of Peter Frankl’s result. What about a lower bound? In 1964 Jan ˇ Cern´ y constructed a series Cn, n = 2, 3, . . . , of synchronizing automata over 2 letters. The states of Cn are the residues modulo n, and the input letters a and b act as follows: δ(0, a) = 1, δ(m, a) = m for 0 < m < n, δ(m, b) = m+1 (mod n). The automaton used as the leading example is C4.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 92

21. The ˇ

Cern´ y Series

Here is a generic automaton from the ˇ Cern´ y series:

n−2 n−1 1 2

a a a a b b a b b . . . . . . ˇ Cern´ y has proved that the shortest reset word for Cn is (abn−1)n−2a of length (n − 1)2.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 93

21. The ˇ

Cern´ y Series

Here is a generic automaton from the ˇ Cern´ y series:

n−2 n−1 1 2

a a a a b b a b b . . . . . . ˇ Cern´ y has proved that the shortest reset word for Cn is (abn−1)n−2a of length (n − 1)2.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 94

22. The ˇ

Cern´ y Automaton

Let w be a shortest reset word for Cn. It must end with a and every other occurrence of a in w is followed by an occurrence of b. Thus, w = w′a where w′ can be rewritten into a word v over the alphabet {b, c} where c := ab. Since w′ and v act in the same way, the word vc is a reset word for the automaton induced by the actions of b and c. We denote this automaton by Wn and refer to it as to the Wielandt automaton.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 95

22. The ˇ

Cern´ y Automaton

Let w be a shortest reset word for Cn. It must end with a and every other occurrence of a in w is followed by an occurrence of b. 1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Thus, w = w′a where w′ can be rewritten into a word v over the alphabet {b, c} where c := ab. Since w′ and v act in the same way, the word vc is a reset word for the automaton induced by the actions of b and c. We denote this automaton by Wn and refer to it as to the Wielandt automaton.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 96

22. The ˇ

Cern´ y Automaton

Let w be a shortest reset word for Cn. It must end with a and every other occurrence of a in w is followed by an occurrence of b. 1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Thus, w = w′a where w′ can be rewritten into a word v over the alphabet {b, c} where c := ab. Since w′ and v act in the same way, the word vc is a reset word for the automaton induced by the actions of b and c. We denote this automaton by Wn and refer to it as to the Wielandt automaton.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 97

22. The ˇ

Cern´ y Automaton

Let w be a shortest reset word for Cn. It must end with a and every other occurrence of a in w is followed by an occurrence of b. 1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Thus, w = w′a where w′ can be rewritten into a word v over the alphabet {b, c} where c := ab. Since w′ and v act in the same way, the word vc is a reset word for the automaton induced by the actions of b and c. We denote this automaton by Wn and refer to it as to the Wielandt automaton.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 98

22. The ˇ

Cern´ y Automaton

Let w be a shortest reset word for Cn. It must end with a and every other occurrence of a in w is followed by an occurrence of b. 1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Thus, w = w′a where w′ can be rewritten into a word v over the alphabet {b, c} where c := ab. Since w′ and v act in the same way, the word vc is a reset word for the automaton induced by the actions of b and c. We denote this automaton by Wn and refer to it as to the Wielandt automaton.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 99

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 100

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 101

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 102

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 103

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 104

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 105

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 106

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 107

22. The Wielandt Automaton

If u ∈ {b, c}∗, the word uvc also is a reset word for Wn and it also brings the automaton to 2. Hence, for every ℓ ≥ |vc|, there is a path of length ℓ in Wn from any given state i to 2. In particular, for every ℓ ≥ |vc| there is a cycle of length ℓ in Wn. The graph of Wn has simple cycles only of two lengths: n and n − 1. Each cycle

f Wn must consist of simple cycles of these two lengths whence

each number ℓ ≥ |vc| must be expressible as a non-negative integer combination of n and n − 1.

Lemma. If k1, k2 ∈ N are relatively prime, then k1k2 − k1 − k2 is

the largest integer that is not expressible as a non-negative integer combination of k1 and k2. Lemma implies that |vc| > n(n − 1) − n − (n − 1) = n2 − 3n + 1. Suppose that |vc| = n2 − 3n + 2. Then there is a path of this length from 1 to 2. Every outgoing edge of 1 leads to 2, and thus, in the path it must be followed by a cycle of length n2 − 3n + 1, but no cycle of such length may exist by Lemma. Hence |vc| ≥ n2 − 3n + 3.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 108

23. Back to the ˇ

Cern´ y Automaton

1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Further, v contains at least n − 2 occurrences of c. Since each

ccurrence of c in v corresponds to an occurrence of ab in w′, we

conclude that |w′| ≥ n2 − 3n + 2 + n − 2 = n2 − 2n. Thus, |w| = |w′a| ≥ n2 − 2n + 1 = (n − 1)2. The above proof is transparent idea and reveals an interesting connection between ˇ Cern´ y’s automata and Wielandt’s graphs that form a well know extremal series in the theory non–negative matrices.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 109

23. Back to the ˇ

Cern´ y Automaton

1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Further, v contains at least n − 2 occurrences of c. Since each

ccurrence of c in v corresponds to an occurrence of ab in w′, we

conclude that |w′| ≥ n2 − 3n + 2 + n − 2 = n2 − 2n. Thus, |w| = |w′a| ≥ n2 − 2n + 1 = (n − 1)2. The above proof is transparent idea and reveals an interesting connection between ˇ Cern´ y’s automata and Wielandt’s graphs that form a well know extremal series in the theory non–negative matrices.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 110

23. Back to the ˇ

Cern´ y Automaton

1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Further, v contains at least n − 2 occurrences of c. Since each

ccurrence of c in v corresponds to an occurrence of ab in w′, we

conclude that |w′| ≥ n2 − 3n + 2 + n − 2 = n2 − 2n. Thus, |w| = |w′a| ≥ n2 − 2n + 1 = (n − 1)2. The above proof is transparent idea and reveals an interesting connection between ˇ Cern´ y’s automata and Wielandt’s graphs that form a well know extremal series in the theory non–negative matrices.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 111

23. Back to the ˇ

Cern´ y Automaton

1 n 2 n−1 3 a, b b b b a a a a . . . 1 n 2 n−1 3 b b, c b, c b, c c . . . Further, v contains at least n − 2 occurrences of c. Since each

ccurrence of c in v corresponds to an occurrence of ab in w′, we

conclude that |w′| ≥ n2 − 3n + 2 + n − 2 = n2 − 2n. Thus, |w| = |w′a| ≥ n2 − 2n + 1 = (n − 1)2. The above proof is transparent idea and reveals an interesting connection between ˇ Cern´ y’s automata and Wielandt’s graphs that form a well know extremal series in the theory non–negative matrices.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 112

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 113

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 114

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 115

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 116

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 117

24. Exponents

A (directed) graph D is primitive if D is strongly connected and the greatest common divisor of the lengths of all cycles in D is equal to 1. A graph is primitive iff there exists t such that for each pair

f vertices there is a path between them of length exactly t (proved

by Frobenius in the language of matrices). (This is equivalent to saying that the t-th power of the matrix of D is positive.) The least t with this property is called the exponent of the digraph D. 1950, Wielandt: The exponent of every primitive graph on n vertices is not greater than (n − 1)2 + 1 and this bound is tight since the graph of Wn has exponent (n − 1)2 + 1. 1964, Dulmage–Mendelsohn: There is exactly one primitive graph

n n vertices with exponent (n − 1)2 + 1.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 118

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 119

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 120

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 121

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 122

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata

SLIDE 123

25. The ˇ

Cern´ y Conjecture

Define the ˇ Cern´ y function C(n) as the maximum reset threshold

f all synchronizing automata with n states. The above property
f the series {Cn}, yields the inequality C(n) ≥ (n − 1)2.

The ˇ Cern´ y conjecture is the claim that in fact the equality C(n) = (n − 1)2 holds true. This simply looking conjecture is arguably the most longstanding open problem in the combinatorial theory of finite automata. Up to recently, everything we knew about the conjecture in general could be summarized in one line: (n − 1)2 ≤ C(n) ≤ n3 − n 6 . A small improvement on this bound has been found by Marek Szyku la in 2017: the new bound is still cubic in n but improves the coefficient 1

6 at n3 by 4 46875.

The new bound is (15617n3 + 7500n2 + 9375n − 31250)/93750.

Mikhail Volkov Synchronizing Finite Automata