Equations over sets of natural numbers. Artur Je z University of - PowerPoint PPT Presentation

Questions Problem � Expressive power? � Complexity of the membership problem? Remark Operations {∪ , ⊞ } : Context-free grammars over an alphabet { a } . Least solutions are ultimately periodic. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 8 / 27

Questions Problem � Expressive power? � Complexity of the membership problem? Remark Operations {∪ , ⊞ } : Context-free grammars over an alphabet { a } . Least solutions are ultimately periodic. General membership problem: NP-complete Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 8 / 27

Tool: positional notation Using base- k notation. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 9 / 27

Tool: positional notation Using base- k notation. Σ k = { 0 , 1 , . . . , k − 1 } . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 9 / 27

Tool: positional notation Using base- k notation. Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0Σ ∗ Numbers ← k . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 9 / 27

Tool: positional notation Using base- k notation. Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0Σ ∗ Numbers ← k . Sets of numbers ← → formal languages over Σ k . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 9 / 27

Tool: positional notation Using base- k notation. Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0Σ ∗ Numbers ← k . Sets of numbers ← → formal languages over Σ k . Example (10 ∗ ) 4 = { 4 n | n � 0 } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 9 / 27

Example of non-periodic solution ( k = 4) Solution 10 ∗ , L 1 = 20 ∗ , = L 2 30 ∗ , L 3 = 120 ∗ . = L 12 Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 10 / 27

Example of non-periodic solution ( k = 4) Equations Solution 10 ∗ , L 1 = B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } , 20 ∗ , = = ( B 12 ⊞ B 2 ∩ B 1 ⊞ B 1 ) ∪ { 2 } , L 2 B 2 30 ∗ , L 3 = B 3 = ( B 12 ⊞ B 12 ∩ B 1 ⊞ B 2 ) ∪ { 3 } , 120 ∗ . = = ( B 3 ⊞ B 3 ∩ B 1 ⊞ B 2 ) . L 12 B 12 Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 10 / 27

What needs to be proved By general knowledge there is a unique ε -free solution. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 11 / 27

What needs to be proved By general knowledge there is a unique ε -free solution. Vector of sets ( . . . , i0 ∗ , . . . ) is ε -free. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 11 / 27

What needs to be proved By general knowledge there is a unique ε -free solution. Vector of sets ( . . . , i0 ∗ , . . . ) is ε -free. We need to show that it is a solution. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 11 / 27

What needs to be proved By general knowledge there is a unique ε -free solution. Vector of sets ( . . . , i0 ∗ , . . . ) is ε -free. We need to show that it is a solution. Example For example 10 ∗ , the rule is B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } So we want to prove that 10 ∗ = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ ∪ { 1 } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 11 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Proof. 10 + ∪ 20 ∗ 20 ∗ 20 ∗ ⊞ 20 ∗ = Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Proof. 10 + ∪ 20 ∗ 20 ∗ 20 ∗ ⊞ 20 ∗ = 10 ∗ ⊞ 30 ∗ 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ = Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Proof. 10 + ∪ 20 ∗ 20 ∗ 20 ∗ ⊞ 20 ∗ = 10 ∗ ⊞ 30 ∗ 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ 10 + = Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Proof. 10 + ∪ 20 ∗ 20 ∗ 20 ∗ ⊞ 20 ∗ = 10 ∗ ⊞ 30 ∗ 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ 10 + = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ ∪ { 1 } 10 ∗ = Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Calculations Rule: B 1 = ( B 2 ⊞ B 2 ∩ B 1 ⊞ B 3 ) ∪ { 1 } Proof. 10 + ∪ 20 ∗ 20 ∗ 20 ∗ ⊞ 20 ∗ = 10 ∗ ⊞ 30 ∗ 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ 10 + = 20 ∗ ⊞ 20 ∗ ∩ 10 ∗ ⊞ 30 ∗ ∪ { 1 } 10 ∗ = Remark Similar proof for ij 0 ∗ in base- k notation. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 12 / 27

Any regular language Theorem (Je˙ z, DLT 2007) For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then R ∈ EQ ( ∩ , ∪ , ⊞ ) . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 13 / 27

Any regular language Theorem (Je˙ z, DLT 2007) For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then R ∈ EQ ( ∩ , ∪ , ⊞ ) . Idea Let �{ 0 , . . . , k − 1 } , Q , q 0 , F , δ � recognize R. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 13 / 27

Any regular language Theorem (Je˙ z, DLT 2007) For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then R ∈ EQ ( ∩ , ∪ , ⊞ ) . Idea Let �{ 0 , . . . , k − 1 } , Q , q 0 , F , δ � recognize R. We introduce variable B i , j , q for set { ijw : δ ( q 0 , w ) = q } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 13 / 27

Any regular language Theorem (Je˙ z, DLT 2007) For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then R ∈ EQ ( ∩ , ∪ , ⊞ ) . Idea Let �{ 0 , . . . , k − 1 } , Q , q 0 , F , δ � recognize R. We introduce variable B i , j , q for set { ijw : δ ( q 0 , w ) = q } Information the indices carry: leading symbol i second leading symbol j q—the computation of M on the rest of the word Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 13 / 27

Equations for B i , j , q Example 4 � � B i , j , q = B i − 1 , j + n ⊞ B k − n , x , q ′ ∪ . . . ( x , q ′ ): q ∈ δ ( q ′ , x ) n =1 Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 14 / 27

Equations for B i , j , q Example 4 � � B i , j , q = B i − 1 , j + n ⊞ B k − n , x , q ′ ∪ . . . ( x , q ′ ): q ∈ δ ( q ′ , x ) n =1 state q ′ k − n x ���� . . . + i − 1 j + n 00 . . . 0 i j x . . . . . . � �� � state q Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 14 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; δ : Q × Q → Q , transition function; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; δ : Q × Q → Q , transition function; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; δ : Q × Q → Q , transition function; Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; δ : Q × Q → Q , transition function; F ⊆ Q : accepting states. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Trellis automata (one-way real-time cellular automata) Theorem (Je˙ z, Okhotin, CSR 2007) ∀ trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , set L ( M ) is in EQ ( ∩ , ∪ , ⊞ ) . Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Q : finite set of states; I : Σ → Q sets initial states; δ : Q × Q → Q , transition function; F ⊆ Q : accepting states. Closed under ∪ , ∩ , ∼ , not closed under concatenation. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 15 / 27

Main lemma Lemma For every trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , there exists a system of equations in EQ = ( ∪ , ∩ , ⊞ ) with least solution { 1 w 10 ∗ | w + 1 ∈ L ( M ) } , . . . , Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 16 / 27

Main lemma Lemma For every trellis automaton M over Σ k with L ( M ) ⊆ Σ ∗ k \ 0Σ ∗ k , there exists a system of equations in EQ = ( ∪ , ∩ , ⊞ ) with least solution { 1 w 10 ∗ | w + 1 ∈ L ( M ) } , . . . , 1 w 10 ∗ represents w . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 16 / 27

The construction Set of variables { X q | q ∈ Q } . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ ∃ q ′ , q ′′ : δ ( q ′ , q ′′ ) = q , Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ ∃ q ′ , q ′′ : δ ( q ′ , q ′′ ) = q , au ∈ L M ( q ′ ), Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ ∃ q ′ , q ′′ : δ ( q ′ , q ′′ ) = q , au ∈ L M ( q ′ ), ub ∈ L M ( q ′′ ). Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ ∃ q ′ , q ′′ : δ ( q ′ , q ′′ ) = q , au ∈ L M ( q ′ ), ub ∈ L M ( q ′′ ). Let 1 au 10 ∗ ⊆ X q ′ , 1 ub 10 ∗ ⊆ X q ′′ . � X q = ρ b ( X q ′ ) ∩ λ a ( X q ′′ ) q ′ , q ′′ : δ ( q ′ , q ′′ )= q a , b ∈ Σ k Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

The construction Set of variables { X q | q ∈ Q } . Actually, X q = { 1 w 10 ∗ | w + 1 ∈ L M ( q ) } aub ∈ L M ( q ) ⇔ ∃ q ′ , q ′′ : δ ( q ′ , q ′′ ) = q , au ∈ L M ( q ′ ), ub ∈ L M ( q ′′ ). Let 1 au 10 ∗ ⊆ X q ′ , 1 ub 10 ∗ ⊆ X q ′′ . � X q = ρ b ( X q ′ ) ∩ λ a ( X q ′′ ) q ′ , q ′′ : δ ( q ′ , q ′′ )= q a , b ∈ Σ k λ a (1 w 10 k ) = 1 aw 10 k ρ b (1 w 10 k ) = 1 wb 10 k − 1 Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 17 / 27

Part III Complexity of equations with {∪ , ∩ , ⊞ } Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 18 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. P polynomial time. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. P polynomial time. NP nondeterministic polynomial time (may guess). Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. P polynomial time. NP nondeterministic polynomial time (may guess). PSPACE polynomial space. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. P polynomial time. NP nondeterministic polynomial time (may guess). PSPACE polynomial space. EXPTIME exponential time. P ⊆ NP ⊆ PSPACE ⊆ EXPTIME Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Computational complexity: basic notions Fix X ⊆ N 0 . Determine algorithmically whether x ∈ X . n = log x : length of notation of x Time complexity: in t ( n ) elementary steps. Space complexity: using s ( n ) elementary memory cells. P polynomial time. NP nondeterministic polynomial time (may guess). PSPACE polynomial space. EXPTIME exponential time. P ⊆ NP ⊆ PSPACE ⊆ EXPTIME C -complete set X : every problem in C can be reduced to X . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 19 / 27

Complexity of solutions Trellis automata recognize P-complete languages. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. NP-complete sets: relatively easy. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. NP-complete sets: relatively easy. PSPACE-complete sets: requires some efforts. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. NP-complete sets: relatively easy. PSPACE-complete sets: requires some efforts. Upper bound: Theorem (Okhotin, 2001) Every conjunctive language can be recognized in time O ( n 3 ) . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. NP-complete sets: relatively easy. PSPACE-complete sets: requires some efforts. Upper bound: Theorem (Okhotin, 2001) Every conjunctive language can be recognized in time O ( n 3 ) . Corollary Every set of numbers in EQ ( ∪ , ∩ , ⊞ ) is in EXPTIME. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Complexity of solutions Trellis automata recognize P-complete languages. P-complete sets of numbers. NP-complete sets: relatively easy. PSPACE-complete sets: requires some efforts. Upper bound: Theorem (Okhotin, 2001) Every conjunctive language can be recognized in time O ( n 3 ) . Corollary Every set of numbers in EQ ( ∪ , ∩ , ⊞ ) is in EXPTIME. Theorem (Je˙ z, Okhotin, STACS 2008) EQ ( ∪ , ∩ , ⊞ ) contains an EXPTIME-complete set. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 20 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Transition function δ : Q × Γ → 2 Q × Γ ×{← , ↓ , →} . Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Transition function δ : Q × Γ → 2 Q × Γ ×{← , ↓ , →} . If q = q acc , accepts from here. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Transition function δ : Q × Γ → 2 Q × Γ ×{← , ↓ , →} . If q = q acc , accepts from here. If q ∈ Q E , accepts from here if accepts from some next conf. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Transition function δ : Q × Γ → 2 Q × Γ ×{← , ↓ , →} . If q = q acc , accepts from here. If q ∈ Q E , accepts from here if accepts from some next conf. If q ∈ Q A , accepts from here if accepts from every next conf. Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Alternating Turing machines Tape alphabet Γ, set of states Q = Q E ∪ Q A ∪ { q acc } . Transition function δ : Q × Γ → 2 Q × Γ ×{← , ↓ , →} . If q = q acc , accepts from here. If q ∈ Q E , accepts from here if accepts from some next conf. If q ∈ Q A , accepts from here if accepts from every next conf. Theorem (A. Chandra, D. Kozen, L. Stockmeyer 1981) APSPACE = EXPTIME APTIME = PSPACE Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 21 / 27

Idea of encoding Problem How to encode a configuration? Artur Je˙ z ( University of Wroclaw ) Equations over sets of natural numbers. December 13, 2007 22 / 27