Deciding Kleene Algebra with converse is PSpace -complete Talk at - PowerPoint PPT Presentation

Introduction Deciding Kleene Algebra with converse is PSpace -complete Talk at the PACE meeting Paul Brunet & Damien Pous ENS de Lyon February 9 th , 2014 February 9 th , 2014 Paul Brunet (ENS de Lyon) 1 / 23 KAC is PSpace

Introduction Introduction Kleene Algebra (i) : Abstraction for proving the equivalence of regular expressions. (i). Conway, J. H. (1971). Regular algebra and finite machines. Chapman and Hall Mathematics Series February 9 th , 2014 Paul Brunet (ENS de Lyon) 2 / 23 KAC is PSpace

Introduction Introduction Kleene Algebra (i) : Abstraction for proving the equivalence of regular expressions. The equivalence is PSpace -complete . (i). Conway, J. H. (1971). Regular algebra and finite machines. Chapman and Hall Mathematics Series February 9 th , 2014 Paul Brunet (ENS de Lyon) 2 / 23 KAC is PSpace

Introduction Introduction Kleene Algebra (i) : Abstraction for proving the equivalence of regular expressions. The equivalence is PSpace -complete . What if we add a converse operation to regular expressions ? (i). Conway, J. H. (1971). Regular algebra and finite machines. Chapman and Hall Mathematics Series February 9 th , 2014 Paul Brunet (ENS de Lyon) 2 / 23 KAC is PSpace

Introduction Introduction e , f ∈ R eg X KA ⊢ e = f e ≡ R el f � e � = � f � ∨ e , f ∈ R eg X KAC ⊢ e = f e ≡ R el ∨ f cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 3 / 23 KAC is PSpace

Introduction Introduction e , f ∈ R eg X KA ⊢ e = f e ≡ R el f � e � = � f � ∨ e , f ∈ R eg X KAC ⊢ e = f e ≡ R el ∨ f cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 3 / 23 KAC is PSpace

Introduction Introduction e , f ∈ R eg X KA ⊢ e = f e ≡ R el f � e � = � f � ∨ e , f ∈ R eg X KAC ⊢ e = f e ≡ R el ∨ f cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 3 / 23 KAC is PSpace

Introduction Introduction e , f ∈ R eg X KA ⊢ e = f e ≡ R el f � e � = � f � ∨ e , f ∈ R eg X KAC ⊢ e = f e ≡ R el ∨ f cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 3 / 23 KAC is PSpace

Introduction Introduction e , f ∈ R eg X KA ⊢ e = f e ≡ R el f � e � = � f � ∨ e , f ∈ R eg X KAC ⊢ e = f e ≡ R el ∨ f cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 3 / 23 KAC is PSpace

Introduction Plan Introduction 1 From Kleene Algebra with Converse to regular languages 2 Kleene Algebra wih converse Reduction to an automaton problem Closure of an automaton 3 The PSpace algorithm. 4 February 9 th , 2014 Paul Brunet (ENS de Lyon) 4 / 23 KAC is PSpace

KAC Plan Introduction 1 From Kleene Algebra with Converse to regular languages 2 Kleene Algebra wih converse Reduction to an automaton problem Closure of an automaton 3 The PSpace algorithm. 4 February 9 th , 2014 Paul Brunet (ENS de Lyon) 5 / 23 KAC is PSpace

KAC Kleene Algebra wih converse Regular expressions with converse Regular expressions with converse over X ∨ ) are Let X be a finite set, the set of regular expressions over X (written R eg X obtained with the grammar : e , f � 0 | 1 | x ∈ X | e + f | e · f | e ∗ | e ∨ February 9 th , 2014 Paul Brunet (ENS de Lyon) 6 / 23 KAC is PSpace

KAC Kleene Algebra wih converse Regular expressions with converse Regular expressions with converse over X ∨ ) are Let X be a finite set, the set of regular expressions over X (written R eg X obtained with the grammar : e , f � 0 | 1 | x ∈ X | e + f | e · f | e ∗ | e ∨ A relational interpretation of regular expressions with converse over X can be specified by a domain S and a map S 2 � � σ : X −→ P We will write ∨ −→ P S 2 � � σ : R eg X ˆ for the unique morphism equal to σ on X . February 9 th , 2014 Paul Brunet (ENS de Lyon) 6 / 23 KAC is PSpace

KAC Kleene Algebra wih converse Relational equivalence ∨ : For e , f ∈ R eg X e ≡ R el ∨ f means that S 2 � � ∀ S , ∀ σ : X → P , ˆ σ ( e ) = ˆ σ ( f ) . February 9 th , 2014 Paul Brunet (ENS de Lyon) 7 / 23 KAC is PSpace

KAC Kleene Algebra wih converse A hint from the equational theory. ( a + b ) ∨ = a ∨ + b ∨ (1) ( a · b ) ∨ = b ∨ · a ∨ (2) ( a ∗ ) ∨ = ( a ∨ ) ∗ (3) a ∨∨ = a (4) a � aa ∨ a (5) February 9 th , 2014 Paul Brunet (ENS de Lyon) 8 / 23 KAC is PSpace

KAC Kleene Algebra wih converse A hint from the equational theory. ( a + b ) ∨ = a ∨ + b ∨ (1) ( a · b ) ∨ = b ∨ · a ∨ (2) ( a ∗ ) ∨ = ( a ∨ ) ∗ (3) a ∨∨ = a (4) a � aa ∨ a (5) February 9 th , 2014 Paul Brunet (ENS de Lyon) 8 / 23 KAC is PSpace

KAC Reduction ∨ to R eg X From R eg X Let X be a finite alphabet. For e ∈ R eg X , we write � e � ⊆ X ∗ for the language denoted by e . X ′ ≔ { x ′ | x ∈ X } is a disjoint copy of X , and X ≔ X ∪ X ′ . February 9 th , 2014 Paul Brunet (ENS de Lyon) 9 / 23 KAC is PSpace

KAC Reduction ∨ to R eg X From R eg X Let X be a finite alphabet. For e ∈ R eg X , we write � e � ⊆ X ∗ for the language denoted by e . X ′ ≔ { x ′ | x ∈ X } is a disjoint copy of X , and X ≔ X ∪ X ′ . We see equations (1)-(4) as rewriting rules : 1 ( a + b ) ∨ �−→ a ∨ + b ∨ ( a · b ) ∨ �−→ b ∨ · a ∨ ( a ∗ ) ∨ �−→ ( a ∨ ) ∗ a ∨∨ �−→ a 1 ∨ �−→ 1 0 ∨ �−→ 0 February 9 th , 2014 Paul Brunet (ENS de Lyon) 9 / 23 KAC is PSpace

KAC Reduction ∨ to R eg X From R eg X Let X be a finite alphabet. For e ∈ R eg X , we write � e � ⊆ X ∗ for the language denoted by e . X ′ ≔ { x ′ | x ∈ X } is a disjoint copy of X , and X ≔ X ∪ X ′ . We see equations (1)-(4) as rewriting rules : 1 ( a + b ) ∨ �−→ a ∨ + b ∨ ( a · b ) ∨ �−→ b ∨ · a ∨ ( a ∗ ) ∨ �−→ ( a ∨ ) ∗ a ∨∨ �−→ a 1 ∨ �−→ 1 0 ∨ �−→ 0 We substitute x ∨ with x ′ in the result. We get e ∈ R eg X . 2 February 9 th , 2014 Paul Brunet (ENS de Lyon) 9 / 23 KAC is PSpace

KAC Reduction Reduction relation a � aa ∨ a w For a word w ∈ X ∗ , we define inductively w : x ≔ x ′ ∀ x ∈ X , ǫ ≔ ǫ ∀ x ′ ∈ X ′ , x ′ ≔ x wx ≔ x w February 9 th , 2014 Paul Brunet (ENS de Lyon) 10 / 23 KAC is PSpace

KAC Reduction Reduction relation a � aa ∨ a w For a word w ∈ X ∗ , we define inductively w : x ≔ x ′ ∀ x ∈ X , ǫ ≔ ǫ ∀ x ′ ∈ X ′ , x ′ ≔ x wx ≔ x w u � v u 1 · www · u 2 u 1 · w · u 2 � Example : abbabb ′ a ′ abbaa ′ February 9 th , 2014 Paul Brunet (ENS de Lyon) 10 / 23 KAC is PSpace

KAC Reduction Reduction relation a � aa ∨ a w For a word w ∈ X ∗ , we define inductively w : x ≔ x ′ ∀ x ∈ X , ǫ ≔ ǫ ∀ x ′ ∈ X ′ , x ′ ≔ x wx ≔ x w u � v u 1 · www · u 2 u 1 · w · u 2 � Example : abbabb ′ a ′ abbaa ′ = abb · ab · b ′ a ′ · ab · baa ′ February 9 th , 2014 Paul Brunet (ENS de Lyon) 10 / 23 KAC is PSpace

KAC Reduction Reduction relation a � aa ∨ a w For a word w ∈ X ∗ , we define inductively w : x ≔ x ′ ∀ x ∈ X , ǫ ≔ ǫ ∀ x ′ ∈ X ′ , x ′ ≔ x wx ≔ x w u � v u 1 · www · u 2 u 1 · w · u 2 � Example : abbabb ′ a ′ abbaa ′ = abb · ab · b ′ a ′ · ab · baa ′ = abb · ab · ab · ab · baa ′ February 9 th , 2014 Paul Brunet (ENS de Lyon) 10 / 23 KAC is PSpace

KAC Reduction Reduction relation a � aa ∨ a w For a word w ∈ X ∗ , we define inductively w : x ≔ x ′ ∀ x ∈ X , ǫ ≔ ǫ ∀ x ′ ∈ X ′ , x ′ ≔ x wx ≔ x w u � v u 1 · www · u 2 u 1 · w · u 2 � Example : abbabb ′ a ′ abbaa ′ = abb · ab · b ′ a ′ · ab · baa ′ = abb · ab · ab · ab · baa ′ � abb · ab · baa ′ February 9 th , 2014 Paul Brunet (ENS de Lyon) 10 / 23 KAC is PSpace

KAC Reduction Closure cl ( L ) cl ( L ) ≔ { v | ∃ u ∈ L : u � ∗ v } February 9 th , 2014 Paul Brunet (ENS de Lyon) 11 / 23 KAC is PSpace

KAC Reduction Closure cl ( L ) cl ( L ) ≔ { v | ∃ u ∈ L : u � ∗ v } Theorem a a . Bloom, S. L., Ésik, Z., and Stefanescu, G. (1995). Notes on equational theories of relations. Algebra Universalis , 33(1) :98–126 e ≡ R el ∨ f ⇔ cl ( � e � ) = cl ( � f � ) February 9 th , 2014 Paul Brunet (ENS de Lyon) 11 / 23 KAC is PSpace

Construction Plan Introduction 1 From Kleene Algebra with Converse to regular languages 2 Kleene Algebra wih converse Reduction to an automaton problem Closure of an automaton 3 The PSpace algorithm. 4 February 9 th , 2014 Paul Brunet (ENS de Lyon) 12 / 23 KAC is PSpace

Construction Problem Input : an automaton A an automaton A ′ such that L ( A ′ ) = cl ( L ( A )) . Output : February 9 th , 2014 Paul Brunet (ENS de Lyon) 13 / 23 KAC is PSpace

Construction Intuition a ′ a a 0 1 2 3 February 9 th , 2014 Paul Brunet (ENS de Lyon) 14 / 23 KAC is PSpace

Construction Intuition a a ′ a a 0 1 2 3 February 9 th , 2014 Paul Brunet (ENS de Lyon) 14 / 23 KAC is PSpace

Construction Intuition a a ′ a a 0 1 2 3 b ′ a ′ a b a b 0 1 2 3 4 5 6 February 9 th , 2014 Paul Brunet (ENS de Lyon) 14 / 23 KAC is PSpace