Evaluating Datalog via Tree Automata and Cycluits

Abstract

We investigate parameterizations of both database instances and queries that make query evaluation fixed-parameter tractable in combined complexity. We show that clique-frontier-guarded Datalog with stratified negation (CFG-Datalog) enjoys bilinear-time evaluation on structures of bounded treewidth for programs of bounded rule size. Such programs capture in particular conjunctive queries with simplicial decompositions of bounded width, guarded negation fragment queries of bounded CQ-rank, or two-way regular path queries. Our result is shown by translating to alternating two-way automata, whose semantics is defined via cyclic provenance circuits (cycluits) that can be tractably evaluated.

Appendix: A Proof of Theorem 9

Theorem 9 1

There is an arity-two signatureσfor which there is no algorithm$\mathscr{A}$withexponential running time and polynomial output size for thefollowing task: given a conjunctive query Q of treewidth≤ 2,produce an alternating two-way tree automatonA_Qon${\Gamma }^{5}_{\sigma }$-treesthat tests Q onσ-instancesof treewidth ≤ 5.

To prove this theorem, we need some notions and lemmas from [12], an extended version of [13]. Since [12] is currently unpublished, relevant results are reproduced as Appendix F of [5], in particular Lemma 68, Theorem 69, and their proofs.

Proof of Theorem 9

Let σ be $\mathscr{S}^{\text {Bin}}{~}_{\mathsf {Ch1},\mathsf {Ch2},\mathsf {Child},\mathsf {Child^?}}$ as in Theorem 69 of [5]. We pose c = 3, k_I = 2 × 3 − 1 = 5. Assume by way of contradiction that there exists an algorithm $\mathscr{A}$ satisfying the prescribed properties. We will describe an algorithm to solve any instance of the containment problem of Theorem 69 of [5] in singly exponential time. As Theorem 69 of [5] states that it is 2EXPTIME-hard, this yields a contradiction by the time hierarchy theorem.

Let P and Q be an instance of the containment problem of Theorem 69 of [5], where P is a monadic Datalog program of var-size ≤ 3, and Q is a CQ of treewidth ≤ 2. We will show how to solve the containment problem, that is, decide whether there exists some instance I satisfying P ∧¬Q.

Using Lemma 68 of [5], compute in singly exponential time the ${\Gamma }_{\sigma }^{k_{\text {I}}}$-bNTA A_P. Using the putative algorithm $\mathscr{A}$ on Q, compute in singly exponential time an alternating two-way automaton A_Q of polynomial size. As A_P describes a family $\mathscr{I}$ of canonical instances for P, there is an instance satisfying P ∧¬Q iff there is an instance in $\mathscr{I}$ satisfying P ∧¬Q. Now, as $\mathscr{I}$ is described as the decodings of the language of A_P, all instances in $\mathscr{I}$ have treewidth ≤ k_I. Furthermore, the instances in $\mathscr{I}$ satisfy P by definition of $\mathscr{I}$. Hence, there is an instance satisfying P ∧¬Q iff there is an encoding E in the language of A_P whose decoding satisfies ¬Q. Now, as A_Q tests Q on instances of treewidth k_I, this is the case iff there is an encoding E in the language of A_P which is not accepted by A_Q. Hence, our problem is equivalent to the problem of deciding whether there is a tree accepted by A_P but not by A_Q.

We now use Theorem A.1 of [26] to compute in EXPTIME in A_Q a bNTA $A^{\prime }_{Q}$ recognizing the complement of the language of A_Q. Remember that A_Q was computed in EXPTIME and is of polynomial size, so the entire process so far is EXPTIME. Now we know that we can solve the containment problem by testing whether A_P and $A^{\prime }_{Q}$ have non-trivial intersection, which can be done in PTIME by computing the product automaton and testing emptiness [25]. This solves the containment problem in EXPTIME. As we explained initially, we have reached a contradiction, because it is 2EXPTIME-hard. □