E NTAILMENT k

Note that Algorithm 1 still holds ifGhas an unbounded number of exchange-able pairs but only one positive (resp. negative) literal. It follows that the en-tailment problem remains NP-complete in that case. In contrast, the technique used in this algorithm does not seem to be generalizable to k ≥ 2. Take for in-stance the case where k = 2 and try to generalize Algorithm 1, replacing the literal ∼p(u) by two literals ∼p(u) and ∼q(v). Then the recursive call with in-put H +{∼p(π(u))} would be replaced by the conjunction of three recursive calls with inputsH +{∼p(π(u)),∼q(π(v))}, H +{∼p(π(u)),∼q(π(v))} and H+{∼p(π(u)),∼q(π(v))}respectively, each of these recursive calls potentially generating three new recursive calls etc, so that generalized Prop. 11 would contain an exponential number of PGsH_iand homomorphismsπ_i.

4.3 ENTAILMENT_k

We now show that, for any value of parameterk, ENTAILMENT_kfalls into the class P^{N P}, and evenP_||^{N P}, i.e., the class of decision problems solvable in polynomial time with one round of parallel queries to an NP oracle. Note that the condition on parallel queries can be relaxed by considering a constant number of rounds of parallel queries instead of a single round [BH91].

For that, we rely on Th. 2. We first deduce from this theorem a necessary and sufficient entailment condition (Prop. 12), which will be used in subsequent complexity proofs, and is also interesting for itself. Let us provide an idea of this condition on examples of Figures 2 and 5. For the graphs in Figure 2, ifp(b) is known to be true (i.e., if literal+p(b)is added toH) thenGis entailed (i.e.,Gcan be mapped toH+{+p(b)}), and if p(b) is known to be false thenGis entailed too (i.e.,Gcan also be mapped toH+{−p(b)}). Thus there are two extensible homomorphisms fromG_s toH, which can be extended to homomorphisms from GtoH+{+p(b)}andH+{−p(b)}respectively, with the formulap(b)∨ ¬p(b) being a tautology. We seep(b)∨ ¬p(b)as a propositional formula on a proposi-tional language containing the atomp(b); ifbwas a variable node associated with variablez, the propositional language would contain the atomp(z)and the propo-sitional tautology would be p(z)∨ ¬p(z). Similarly, for the graphs in Figure 5, there are three extensible homomorphismsπ₁, π₂ and π₃ from G_s toH, which mapGs to+r(a, b), +r(b, c)and +r(c, d)respectively, and can be extended to homomorphisms fromGtoH+{−p(b)},H+{+p(b),−p(c)}andH+{+p(c)}

respectively, with the proposition¬p(b)∨(p(b)∧ ¬p(c))∨p(c)being a tautology.

We will build from the set of extensible homomorphisms from any completion sub-graphG⁰ ofGcontained inG_s toHa propositional formula that is a tautology if and only ifGis entailed byH.

We define for each completion subgraphG⁰ ofGand each extensible homo-morphismπ fromG⁰ toH the setL(π)of literals that are “missing” inH for π to be extendable to a homomorphism fromGtoH. Therefore, the literals from L(π)have to be in any completionH^cofH such thatπ can be extended to a ho-momorphism fromGtoH^c. FromL(π), we define propositional formulasC(π) andD_G⁰(G, H)on a propositional language denotedP_H.

Notations 1 LetGandH be two PGs, withH being consistent, and letG⁰ be a completion subgraph ofG.

PH denotes the set of atoms occurring inΦ(H^c\H), where H^cis an arbitrary completion ofH.

For any extensible homomorphismπfromG⁰toH,L(π)denotes the set of literals lsuch thatl =∼p(π(u))for some literal∼p(u)inGandlis not inH, andC(π) denotes the conjunction of the literals inL(π)which is a proposition onPH. D_G⁰(G, H)denotes the disjunction of the propositionsC(π)for all extensible ho-momorphismsπfromG⁰toH.

Omission of subscriptG⁰ means thatG⁰is equal toGs.

For instance, in the previous example of Figure 5, with P_H = {p(b), p(c)}

andG⁰ =Gs: letπ1,π2andπ3be the extensible homomorphisms fromGstoH;

L(π₁) ={−p(b)},L(π₂) ={+p(b),−p(c)},L(π₃) ={+p(c)},C(π₁) =¬p(b), C(π₂) = p(b)∧ ¬p(c)andC(π₃) = p(c); finally, D(G, H) = ¬p(b)∨(p(b)∧

¬p(c))∨p(c).

Next Lemma 1 follows immediately from the definition ofL(π).

Lemma 1 LetG andH be two PGs, letH^c be a completion of H, let G⁰ be a completion subgraph ofG, and letπ be an extensible homomorphism fromG⁰ to H. Thenπcan be extended to a homomorphism fromGtoH^cif and only ifL(π) is a set of literals inH^c.

Lemma 2 expresses the straightforward correspondence between the comple-tions ofHand the truth assignments onPH.

Lemma 2 There is a bijectionf from the set of completions of H to the set of truth assignments onPH such that for any completion H^c ofH, any completion subgraphG⁰ ofGand any extensible homomorphism π fromG⁰ toH, L(π)is a set of literals inH^cif and only iff(H^c)satisfiesC(π).

Proof: Letf be the mapping from the set of completions ofH to the set of truth assignments onP_H defined as follows: for every completion H^c of H, f(H^c) assigns the value true to an atomp(u)inP_H if+p(u)is a literal inH^c, and false otherwise (i.e., if−p(u)is a literal inH^c).fclearly satisfies the desired conditions.

Proposition 12 LetGandH be two PGs, withHbeing consistent, and letG⁰be any completion subgraph ofGcontained in Gs. ThenGis entailed by H if and only ifD_G⁰(G, H)is a tautology.

Proof:By Th. 2 (sinceG⁰is contained inGs) and Prop. 5 (sinceG⁰is a completion subgraph ofG),Gis entailed byH iff for each completionH^cofH, there is an extensible homomorphism fromG⁰toHthat can be extended to a homomorphism from G to H^c. Let us show that the latter proposition holds iff DG⁰(G, H) is a tautology, using the bijection f of Lemma 2. ⇒: We suppose that for each completionH^c of H, there is an extensible homomorphism from G⁰ to H that can be extended to a homomorphism fromGtoH^c. Let us show thatDG⁰(G, H) is a tautology. Let v be a truth assignment on P_H, let us show that v satisfies D_G⁰(G, H). LetH^c=f⁻¹(v), and letπbe an extensible homomorphism fromG⁰ toHthat can be extended to a homomorphism fromGtoH^c. By Lemma 1,L(π)is a set of literals inH^c, so by Lemma 2,vsatisfiesC(π), and thereforeD_G⁰(G, H).

⇐: We suppose thatD_G⁰(G, H)is a tautology. LetH^cbe a completion ofH, let us show that there is an extensible homomorphism fromG⁰toHthat can be extended to a homomorphism fromGtoH^c. Letv=f(H^c). AsD_G⁰(G, H)is a tautology, there is an extensible homomorphismπ fromG⁰ toH such thatvsatisfiesC(π).

By Lemmas 1 and 2,πcan be extended to a homomorphism fromGtoH^c. In order to prove that ENTAILMENT_k is inP^{N P}, we show how to compute D(G, H) without explicitly computing all extensible homomorphisms from G_s toH, whose number may be exponential in the size of G. LetE be the set of exchangeable literals, andT_E be the set of term nodes occurring in E. The main idea is that, for any extensible homomorphism fromG_s toH, the setL(π), and therefore propositionC(π), only depend on the restriction ofπtoTE. Thus, we can defineL(ϕ)andC(ϕ)for any mappingϕfromT_Eto the setTHof term nodes inH, andD(G, H)is the disjunction of the propositionsC(ϕ)for every mappingϕfrom TEtoT_H that can be extended to an extensible homomorphism fromG_stoH. Note that a mappingϕfromT_EtoTH can be extended to an extensible homomorphism fromG_stoH iff it satisfies both following independent conditions: 1)ϕcan be extended to a homomorphismπfromG_s toHand 2)ϕsatisfies conditions 1 and 2 of extensibility, which only depend on the restriction ofπtoT_E, i.e., onϕitself.

According to Prop. 12, Algorithm 2 computesD(G, H)to determine whetherGis entailed byH.

Algorithm 2:ENTAILMENT(G, H)

Data:GandHtwo PGs, such thatHis consistent Result: true ifGis entailed byH, false otherwise begin

LetEbe the set of exchangeable literals w.r.t. (G, H) LetT_E be the set of term nodes occurring inE LetG_s=G\ E

Φ←f alse

forevery mappingϕfromT_E to the set of term nodes inHdo

ifϕcan be extended to an extensible homomorphism fromG_stoH then

Φ←Φ∨C(ϕ) returnTautology(Φ)

If the number of exchangeable pairs is bounded by a constantk, then the num-ber of mappings fromT_Eto the set of term nodes inHbecomes polynomial, which makes ENTAILMENT_kfall intoP^{N P}.

Theorem 5 For any integer k ≥ 0, the decision problem ENTAILMENT_k is in P_||^{N P}.

Proof:It is sufficient to give an algorithm that can be executed in polynomial time with a fixed number of rounds of parallel calls to an NP-oracle. We first check with a single round whether the input is a correct instance of ENTAILMENT_k, which can be done by checking each pair of literals inG. If the input is correct, we call Algorithm 2. This algorithm performs three rounds of parallel calls:

- to computeE, it is sufficient to determine for each pair ofp-opposite literals inG (whose number is polynomial) if it is exchangeable, which is in NP (first round), - |T_E| ≤ 2kw, where w is the maximal arity of a relation name, so the number of mappings fromTE to the set of term nodes inH is bounded by(n_H)^2kw, and therefore is polynomial,

- determining if such a mappingϕcan be extended to an extensible homomorphism fromG_stoHis in NP, since an extension provides a polynomial certificate (second round),

- determining if a proposition is not a tautology is in NP (third round).

It follows from Algorithm 2 that in any case where it can be decided in poly-nomial time whether the formulaΦcomputed by this algorithm is a tautology, the entailment problem is in NP. A polynomial certificate is given by a set of extensi-ble homomorphismsπfromGtoHextending the mappingsϕconsidered in this

algorithm (one for each extendable mappingϕ, so that the number of homomor-phisms π is polynomial), since computing the disjunction of the formulas C(π) and checking that it is a tautology can be done in polynomial time (we do not need to compute the setE of exchangeable literals norGs nor the mappingsϕ them-selves). In particular, it can be decided in polynomial time whether a disjunction of conjunctions of literals in which each conjunction contains at most one positive literal (or each conjunction contains at most one negative literal) is a tautology. It follows that ENTAILMENT₁is in NP, which provides a new proof of Th. 4 and of the fact that the entailment problem remains NP-complete ifGhas an unbounded number of exchangeable pairs but only one positive (resp. negative) literal.