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Equations over sets of natural numbers Artur Je z Institute of Computer Science University of Wroc law May 22, 2008 Artur Je z (Wroc law) Equations over sets of natural numbers May 22, 2008 1 / 1 Joint work Joint work with

  1. Outline of the results 1 Resolved—expressive power How complicated the sets can be? ◮ with regular notation ◮ much more 2 Resolved: complexity How many resources are needed to recognise? EXPTIME 3 General: universality ∩ , · and ∪ , · are computationally universal 4 One variable How to encode results in one variable? Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 7 / 1

  2. Outline of the results 1 Resolved—expressive power How complicated the sets can be? ◮ with regular notation ◮ much more 2 Resolved: complexity How many resources are needed to recognise? EXPTIME 3 General: universality ∩ , · and ∪ , · are computationally universal 4 One variable How to encode results in one variable? 5 General: addition only Can we use only addition? Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 7 / 1

  3. Positional notation Base k . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  4. Positional notation Base k . Σ k = { 0 , 1 , . . . , k − 1 } . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  5. Positional notation Base k . Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0 Σ ∗ Numbers ← k . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  6. Positional notation Base k . Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0 Σ ∗ Numbers ← k . Sets of numbers ← → languages over Σ k . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  7. Positional notation Base k . Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0 Σ ∗ Numbers ← k . Sets of numbers ← → languages over Σ k . Example ( 10 ∗ ) 4 = { 4 n | n � 0 } Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  8. Positional notation Base k . Σ k = { 0 , 1 , . . . , k − 1 } . → strings in Σ ∗ k \ 0 Σ ∗ Numbers ← k . Sets of numbers ← → languages over Σ k . Example ( 10 ∗ ) 4 = { 4 n | n � 0 } We focus on properties in base- k notation Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 8 / 1

  9. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  10. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  11. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Least solution: ( ( 10 ∗ ) 4 , ( 20 ∗ ) 4 , ( 30 ∗ ) 4 , ( 120 ∗ ) 4 ) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  12. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Least solution: ( ( 10 ∗ ) 4 , ( 20 ∗ ) 4 , ( 30 ∗ ) 4 , ( 120 ∗ ) 4 ) Checking: X 2 + X 2 = 20 ∗ + 20 ∗ = 10 + ∪ 20 ∗ 20 ∗ Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  13. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Least solution: ( ( 10 ∗ ) 4 , ( 20 ∗ ) 4 , ( 30 ∗ ) 4 , ( 120 ∗ ) 4 ) Checking: X 2 + X 2 = 20 ∗ + 20 ∗ = 10 + ∪ 20 ∗ 20 ∗ X 1 + X 3 = 10 ∗ + 30 ∗ = 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ , Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  14. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Least solution: ( ( 10 ∗ ) 4 , ( 20 ∗ ) 4 , ( 30 ∗ ) 4 , ( 120 ∗ ) 4 ) Checking: X 2 + X 2 = 20 ∗ + 20 ∗ = 10 + ∪ 20 ∗ 20 ∗ X 1 + X 3 = 10 ∗ + 30 ∗ = 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ , ( X 2 + X 2 ) ∩ ( X 1 + X 3 ) = 10 + . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  15. Important example— ( 10 ∗ ) 4 Example X 1 = ( X 2 + X 2 ∩ X 1 + X 3 ) ∪ { 1 } X 2 = ( X 12 + X 2 ∩ X 1 + X 1 ) ∪ { 2 } X 3 = ( X 12 + X 12 ∩ X 1 + X 2 ) ∪ { 3 } X 12 = X 3 + X 3 ∩ X 1 + X 2 Least solution: ( ( 10 ∗ ) 4 , ( 20 ∗ ) 4 , ( 30 ∗ ) 4 , ( 120 ∗ ) 4 ) Checking: X 2 + X 2 = 20 ∗ + 20 ∗ = 10 + ∪ 20 ∗ 20 ∗ X 1 + X 3 = 10 ∗ + 30 ∗ = 10 + ∪ 10 ∗ 30 ∗ ∪ 30 ∗ 10 ∗ , ( X 2 + X 2 ) ∩ ( X 1 + X 3 ) = 10 + . Remark Resolved equations with ∩ , + or ∪ , + specify only ultimately periodic sets. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 9 / 1

  16. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  17. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Using the idea Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  18. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Using the idea ( ij 0 ∗ ) k for every i , j , k Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  19. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Using the idea ( ij 0 ∗ ) k for every i , j , k Theorem For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then ( R ) k ∈ EQ. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  20. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Using the idea ( ij 0 ∗ ) k for every i , j , k Theorem For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then ( R ) k ∈ EQ. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  21. Generalisation and how to apply it Idea We append digits from the left, controlling the sets of digits. Using the idea ( ij 0 ∗ ) k for every i , j , k Theorem For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is regular then ( R ) k ∈ EQ. Example (Application) Let S ⊆ ( 10 ∗ Σ k 0 ∗ ) k . How to obtain S ′ = { ( 10 n ( d + 1)0 m ) k : ( 10 n d 0 m ) k ∈ S } ? �� � S ′ = � � S ∩ ( 10 ∗ d 0 ∗ ) k + ( 10 ∗ ) k ∩ ( 10 ∗ ( d + 1)0 ∗ ) k d ∈ Σ k Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 10 / 1

  22. Application: complexity Definition Complexity theory (of a set S)—how many resources are needed to answer a question? ”Given n , does n ∈ S ” Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 11 / 1

  23. Application: complexity Definition Complexity theory (of a set S)—how many resources are needed to answer a question? ”Given n , does n ∈ S ” Resources: space time non-determinism Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 11 / 1

  24. Application: complexity Definition Complexity theory (of a set S)—how many resources are needed to answer a question? ”Given n , does n ∈ S ” Resources: space time non-determinism For example EXPTIME. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 11 / 1

  25. Application: complexity Definition Complexity theory (of a set S)—how many resources are needed to answer a question? ”Given n , does n ∈ S ” Resources: space time non-determinism For example EXPTIME. Definition Reduction: Problem P ≥ P ′ if we can answer P (fast) then we can answer P ′ (fast). Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 11 / 1

  26. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  27. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  28. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  29. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  30. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  31. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  32. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  33. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  34. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Example Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  35. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Example Machine abcd q e → abcd ′ e q ′ Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  36. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Example Machine abcd q e → abcd ′ e q ′ String ( 0 a 0 b 0 cqd 0 e ) k → ( 0 a 0 b 0 c 0 d ′ q ′ e ) k Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  37. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Example Machine abcd q e → abcd ′ e q ′ String ( 0 a 0 b 0 cqd 0 e ) k → ( 0 a 0 b 0 c 0 d ′ q ′ e ) k If ( 0 a 0 b 0 c 0 d ′ q ′ e ) k is accepting we want to add ( ( qd 0 ) k − ( 0 d ′ q ′ ) k ) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  38. Problem Given a resolved system with ∩ , ∪ , + and a number n, does n ∈ S 1 . EXPTIME-complete Idea state of the machine is a string—encode as a number easy to define final accepting computation recurse back transition is a local change easily encoded using regular notation Example Machine abcd q e → abcd ′ e q ′ String ( 0 a 0 b 0 cqd 0 e ) k → ( 0 a 0 b 0 c 0 d ′ q ′ e ) k If ( 0 a 0 b 0 c 0 d ′ q ′ e ) k is accepting we want to add ( ( qd 0 ) k − ( 0 d ′ q ′ ) k ) Using the trick with intersection with regular sets. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 12 / 1

  39. More results—greater expressive power Problem Regular sets are very easy. Slow growth, decidable properties etc. Can we do better? Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 13 / 1

  40. More results—greater expressive power Problem Regular sets are very easy. Slow growth, decidable properties etc. Can we do better? Idea For regular languages we expanded numbers to the left. Maybe we can expand in both directions? Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 13 / 1

  41. More results—greater expressive power Problem Regular sets are very easy. Slow growth, decidable properties etc. Can we do better? Idea For regular languages we expanded numbers to the left. Maybe we can expand in both directions? We can. But this is not easy. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 13 / 1

  42. More results—greater expressive power Problem Regular sets are very easy. Slow growth, decidable properties etc. Can we do better? Idea For regular languages we expanded numbers to the left. Maybe we can expand in both directions? We can. But this is not easy. Theorem For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is recognised by a trellis automaton M then ( R ) k ∈ EQ. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 13 / 1

  43. Trellis automata Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  44. Trellis automata Definition (Culik, Gruska, Salomaa, 1981) A trellis automaton is a M = (Σ , Q , I , δ, F ) where: Σ: input alphabet; Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  45. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  46. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  47. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  48. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  49. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  50. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  51. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  52. Trellis automata 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 (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  53. Trellis automata 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. Can recognize { wcw } , { a n b n c n } , { a n b 2 n } , VALC. Theorem For every k and R ⊂ { 0 , . . . , k − 1 } ∗ if R is recognised by a trellis automaton M then ( R ) k ∈ EQ. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 14 / 1

  54. Computational completeness of language equations Model of computation: Turing Machine Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  55. Computational completeness of language equations Model of computation: Turing Machine Recursive sets: Definition S is recursive if there exists M, such that M [ w ] = 1 for w ∈ S and M [ w ] = 0 for w / ∈ S Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  56. Computational completeness of language equations Model of computation: Turing Machine Recursive sets: Definition S is recursive if there exists M, such that M [ w ] = 1 for w ∈ S and M [ w ] = 0 for w / ∈ S Language equations over Σ, with | Σ | � 2. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  57. Computational completeness of language equations Model of computation: Turing Machine Recursive sets: Definition S is recursive if there exists M, such that M [ w ] = 1 for w ∈ S and M [ w ] = 0 for w / ∈ S Language equations over Σ, with | Σ | � 2. Theorem L ⊆ Σ ∗ is given by unique solution of a system with {∪ , ∩ , ∼ , ·} if and only if L is recursive. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  58. Computational completeness of language equations Model of computation: Turing Machine Recursive sets: Definition S is recursive if there exists M, such that M [ w ] = 1 for w ∈ S and M [ w ] = 0 for w / ∈ S Language equations over Σ, with | Σ | � 2. Theorem L ⊆ Σ ∗ is given by unique solution of a system with {∪ , ∩ , ∼ , ·} if and only if L is recursive. Multiple-letter alphabet essentially used. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  59. Computational completeness of language equations Model of computation: Turing Machine Recursive sets: Definition S is recursive if there exists M, such that M [ w ] = 1 for w ∈ S and M [ w ] = 0 for w / ∈ S Language equations over Σ, with | Σ | � 2. Theorem L ⊆ Σ ∗ is given by unique solution of a system with {∪ , ∩ , ∼ , ·} if and only if L is recursive. Multiple-letter alphabet essentially used. � Remaking the argument for sets of numbers! Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 15 / 1

  60. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  61. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  62. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S VALC( T ) (transcription of computation): recognised by a trellis automaton Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  63. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S VALC( T ) (transcription of computation): recognised by a trellis automaton Trellis automata → resolved equations over sets of numbers Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  64. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S VALC( T ) (transcription of computation): recognised by a trellis automaton Trellis automata → resolved equations over sets of numbers Technical trick: resolved equations with ∩ , ∪ , + → unresolved with ∪ , + (or ∩ , +) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  65. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S VALC( T ) (transcription of computation): recognised by a trellis automaton Trellis automata → resolved equations over sets of numbers Technical trick: resolved equations with ∩ , ∪ , + → unresolved with ∪ , + (or ∩ , +) Extracting numbers with notation L ( T ) from numbers with notation VALC( T ) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  66. Outline of the construction Theorem S ⊆ N 0 is given by unique solution of a system with {∪ , + } ( {∩ , + } ) if and only if S is recursive. Turing Machine T for S VALC( T ) (transcription of computation): recognised by a trellis automaton Trellis automata → resolved equations over sets of numbers Technical trick: resolved equations with ∩ , ∪ , + → unresolved with ∪ , + (or ∩ , +) Extracting numbers with notation L ( T ) from numbers with notation VALC( T ) Remark Least (greatest) solution—RE-sets (co-RE-sets). Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 16 / 1

  67. One variable Problem How many variables are needed to define something interesting? Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 17 / 1

  68. One variable Problem How many variables are needed to define something interesting? Idea Encoding k � ( S 1 , . . . , S k ) → p · S i − d i . i =1 Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 17 / 1

  69. One variable Problem How many variables are needed to define something interesting? Idea Encoding k � ( S 1 , . . . , S k ) → p · S i − d i . i =1 EXPTIME holds for X = ϕ ( X ) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 17 / 1

  70. One variable Problem How many variables are needed to define something interesting? Idea Encoding k � ( S 1 , . . . , S k ) → p · S i − d i . i =1 EXPTIME holds for X = ϕ ( X ) unique solution ϕ ( X ) = ψ ( X )—recursively-hard ( ∩ , ∪ , +) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 17 / 1

  71. Addition only Problem Is addition enough to define something interesting? (general case) Artur Je˙ z (Wroc� law) Equations over sets of natural numbers May 22, 2008 18 / 1

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