Converse elimination in the algebra of binary relations Dimitri - PowerPoint PPT Presentation

Converse elimination in the algebra of binary relations Dimitri Surinx Hasselt University Joint work with Jan Van den Bussche Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Calculus of relations Natural set of (primitive) operations on binary relations (graphs) [Peirce, Schr¨ oder, Tarski] over some domain V id = { ( x , x ) | x ∈ V } r ∪ s = { ( x , y ) | ( x , y ) ∈ r ∨ ( x , y ) ∈ s } r ◦ s = { ( x , y ) | ∃ z : ( x , z ) ∈ r ∧ ( z , y ) ∈ s } r c = { ( x , y ) ∈ V 2 | ( x , y ) �∈ r } r − 1 = { ( x , y ) | ( y , x ) ∈ r } r + = the transitive closure of r Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Calculus of relations Derived operators di = { ( x , y ) ∈ V 2 | x � = y } = id c all = V 2 = id ∪ di r ∩ s = { ( x , y ) | ( x , y ) ∈ r ∧ ( x , y ) ∈ s } = ( r c ∪ s c ) c r − s = { ( x , y ) | ( x , y ) ∈ r ∧ ( x , y ) �∈ s } = r ∩ s c π 1 ( r ) = { ( x , x ) | ∃ y : ( x , y ) ∈ r } = ( r ◦ all ) ∩ id π 2 ( r ) = { ( x , x ) | ∃ y : ( y , x ) ∈ r } = ( all ◦ r ) ∩ id Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Fragments A fragment is a set of primitive or derived operations The most basic fragment N is { id , ∪ , ◦} N ( F ) denotes N extended with the operations in F We say that F has an operator if the operator is in F or if it can be constructed from other operators in F E.g. r ∩ s = r − ( r − s ) E.g. π 1 ( r ) = ( r ◦ r − 1 ) ∩ id Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Queries Fix a binary relational vocabulary Γ (Label set) Expressions over a fragment F are built from relation names in Γ using the operations in F ⇒ Map instances of Γ (graphs) to binary relations Well established logic: N ( − 1 , c ) ≡ FO 3 (Tarski & Givant) Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Boolean queries Boolean queries (graph properties): test nonemptiness of expression results Examples: Transitivity: all − ( all ◦ (( R ◦ R ) − R ) ◦ all ) � = ∅ S - T connectivity: S ◦ R + ◦ T � = ∅ R 2 ◦ R − 1 ◦ R 2 � = ∅ matches the pattern Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Converse elimination Which boolean queries expressed using − 1 can be equivalently expressed without using − 1 ? E.g. R 2 ◦ R − 1 ◦ R 2 � = ∅ is equivalent to π 1 ( R 2 ◦ π 2 ( π 1 ( R 2 ) ◦ R )) � = ∅ Formally: a fragment F admits converse elimination if every boolean query in N ( F ) can be expressed in N ( F − { − 1 } ) Theorem ([Fletcher et al.]) If F has projection, but neither intersection nor transitive closure, then F admits converse elimination Note: the theorem clearly applies to the example above Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Motivation Comparing different computational models/logics as is done in complexity theory and finite model theory E.g. π allows looking back ( π 2 ) but not navigation back. Is looking back always sufficient to eliminate converse? Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Remaining questions Theorem ([Fletcher et al.]) If F has projection, but neither intersection nor transitive closure, then F admits converse elimination 1 What if intersection is present? 2 What if transitive closure is present? 3 Expression complexity of converse elimination? Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Remaining questions Theorem ([Fletcher et al.]) If F has projection, but neither intersection nor transitive closure, then F admits converse elimination 1 What if intersection is present? ⇒ Converse elimination fails: ( R 2 ◦ R − 1 ◦ R 2 ) ∩ R � = ∅ is not expressible in the most powerful fragment without converse ( N ( c , + )) 2 What if transitive closure is present? ⇒ We will prove that it also fails 3 Expression complexity of converse elimination? ⇒ We will prove an exponential blowup in degree Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Converse elimination fails in the presence of transitive closure Theorem The query that checks for the existence of the graph pattern is not expressible in the largest language without converse. Formally, R 2 ◦ ( R ◦ R − 1 ) + ◦ R 2 � = ∅ is not expressible in N ( c , + ) Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Bisimulation game for N ( c ) Restricted version of the 3-pebble game for FO 3 Intuition: relax invariant condition of partial isomorphism so that converse relations need not be preserved by Duplicator Let G 1 = ( G 1 , a 1 , b 1 ) and G 2 = ( G 2 , a 2 , b 2 ) be pointed graphs. A k -round bisimulation game on G 1 and G 2 proceeds as follows: The spoiler picks an i ∈ { 1 , 2 } and a node x i in G i The duplicator responds with a node x j ( j � = i ) in G j Continue two k − 1-round subgames on: ( G 1 , a 1 , x 1 ) and ( G 2 , a 2 , x 2 ) ( G 1 , x 1 , b 1 ) and ( G 2 , x 2 , b 2 ) Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

k -bisimilarity and indistinguishability of pointed graphs The duplicator wins the game if for any newly started subgame on ( G 1 , y 1 , z 1 ) and ( G 2 , y 2 , z 2 ) we have ( y 1 , z 1 ) ∈ R ( G 1 ) iff ( y 2 , z 2 ) ∈ R ( G 2 ) for every R ∈ Γ y 1 = z 1 iff y 2 = z 2 If the duplicator has a winning strategy for the k -round bisimulation game on G 1 and G 2 , we say that G 1 and G 2 are k -bisimilar Theorem ([Fletcher et al.]) ( G 1 , a 1 , b 1 ) and ( G 2 , a 2 , b 2 ) are k-bisimilar iff for any expression e ∈ N ( c ) with degree at most k: ( a 1 , b 1 ) ∈ e ( G 1 ) ⇔ ( a 2 , b 2 ) ∈ e ( G 2 ) * degree( e ) is the maximum number of nested compositions Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Family of graphs to prove inexpressibility Let Q : R 2 ◦ ( R ◦ R − 1 ) + ◦ R 2 and define the family G m 1 and G m 2 : y 1 y m +1 G m 1 w 1 w m +1 t 1 t m +1 u 1 u m +1 G m 2 v 1 v m +1 w ′ w ′ 1 m +1 ⇒ Q ( G m 1 ) � = ∅ and Q ( G m 2 ) = ∅ Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Bisimilarity result on G m 1 and G m 2 Theorem For any a 1 , b 1 ∈ G m 1 there exists a 2 , b 2 ∈ G m 2 such that ( G 1 , a 1 , b 1 ) and ( G m 2 , a 2 , b 2 ) are m / 2 − 1 -bisimilar Main technical contribution. Proof is long and technical Corollary A boolean query in N ( c ) with degree at most m / 2 − 1 cannot be true on G m 1 and false on G m 2 simultaneously Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Proof that converse elimination fails in presence of TC Corollary A boolean query in N ( c ) with degree at most m / 2 − 1 cannot be true on G m 1 and false on G m 2 simultaneously Recall: Q = R 2 ◦ ( R ◦ R − 1 ) + ◦ R 2 Suppose e � = ∅ with e ∈ N ( c , + ) expresses Q � = ∅ . For any m define k m = |G m 1 | = |G m 2 | . For any n there exists e n without + equivalent to e on graphs G with domain size at most n . ( e k m � = ∅ is equivalent to Q � = ∅ on G m 1 and G m 2 ) ⇒ Can ensure that degree( e n ) is logarithmic in n ⇒ exists l such that degree( e k l ) ≤ l / 2 − 1 ⇒ By the Corollary e k l is nonempty on both G l 1 and G l 2 ⇒ Contradicts that e k l � = ∅ is equivalent to Q � = ∅ on G l 1 and G l 2 Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Exponential blowup in degree for converse elimination Consider the family Q n : R 2 ◦ ( R ◦ R − 1 ) n ◦ R 2 � = ∅ Degree of Q n is O (log( n )) Converse elimination for this family cannot be done in degree o ( n ) Corollary Converse elimination in the language with projection, but with neither intersection nor transitive closure is exponential in the degree Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)

Directions for further research 1 We know that converse elimination is exponential in the degree. Can we establish a similar result in the length? 2 Another interesting derived operator: Residuations [Pratt. Origins of calculus of binary relations] C / B is the maximal X such that X ◦ B ⊆ C B \ C is the maximal X such that B ◦ X ⊆ C How does N ( /, \ ) compare to N ( c )? Satisfiability and equivalence problem in N ( /, \ )? Converse elimination in the algebra of binary relations Dimitri Surinx (Hasselt University)