Institute for Theoretical Computer Science Chair for Automata Theory
UNIFICATION IN EL WITHOUT TOP CONSTRUCTOR
Franz Baader, Nguyen T. Binh, Stefan Borgwardt, Barbara Morawska
UNIFICATION IN EL WITHOUT TOP CONSTRUCTOR Franz Baader, Nguyen T. - - PowerPoint PPT Presentation
Institute for Theoretical Computer Science Chair for Automata Theory UNIFICATION IN EL WITHOUT TOP CONSTRUCTOR Franz Baader, Nguyen T. Binh, Stefan Borgwardt, Barbara Morawska WARU Utrecht, 26.05.2011 Outline Why unification in EL is
Institute for Theoretical Computer Science Chair for Automata Theory
Franz Baader, Nguyen T. Binh, Stefan Borgwardt, Barbara Morawska
Why unification in EL−⊤ is interesting Description Logics EL and EL−⊤ Unification in EL and EL−⊤ Unification in EL−⊤ is in PSPACE Unification in EL−⊤ is PSPACE-hard Conclusion
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– unification in semigroups with monotone operators – solving linear language inclusions – finite satisfiability of anti-Horn clauses with monotone functions
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Description logics: EL EL−⊤ concept names Nc
Great-grandfather Old ⊓ Happy ⊓ Man ⊓ ∃has child.(∃has child.(∃has child.⊤)) Old ⊓ Happy ⊓ Man ⊓ ∃has child.(Clever ⊓ ∃has child.(Honest ⊓ ∃has child.Handsome))
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Syntax Interpretation I concept name A ∈ Nc AI ⊆ ∆I role name r ∈ Nr rI ⊆ ∆I × ∆I top concept ⊤ whole domain ∆I conjunction C ⊓ D CI ∩ DI existential restriction ∃r.C {a ∈ ∆I | there is b ∈ ∆I, (a, b) ∈ rI and b ∈ CI} subsumption C ⊑ D CI ⊆ DI equivalence C ≡ D CI = DI
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concept names, existential restrictions Every concept C is a conjunction of atoms.
C1 ⊓ · · · ⊓ Cn ⊑ D and D is an atom, iff there is Ci, such that Ci ⊑ D.
Let C = A ⊓ ∃r.(A ⊓ B) ⊓ ∃r.B, At(C) = {A, B, ∃r.(A ⊓ B), ∃r.B}
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A concept term is ground if it has no variables.
C1 ≡? D1, . . . , Cn ≡? Dn, where Ci, Di are concept terms in EL−⊤ (or in EL). (Decision problem) Does a substitution σ, s.t. σ(C1) ≡ σ(D1), . . . , σ(Cn) ≡ σ(Dn), exist?
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A concept term is ground if it has no variables.
C1⊑?D1, . . . , Cn⊑?Dn, where Ci, Di are concept terms in EL−⊤ ( or in EL). (Decision problem) Does a substitution σ, s.t. σ(C1)⊑σ(D1), . . . , σ(Cn)⊑σ(Dn), exist?
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Let C = ∃r.(A ⊓ ∃s.(A ⊓ B) ⊓ B), Part(C) = {∃r.A, ∃r.B, ∃r.∃s.A, ∃r.∃s.B}
A ǫ ⊓ A ǫ B ǫ s B ǫ r
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Let C = ∃r.(A ⊓ ∃s.(A ⊓ B) ⊓ B), Part(C) = {∃r.A, ∃r.B, ∃r.∃s.A, ∃r.∃s.B}
Let C be an EL−⊤-concept term and B a particle, then
Let A, B be constants, then the unification problem {A ⊑? X, B ⊑? X} has a solution in EL, not in EL−⊤.
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The following problem {∃r.(A ⊓ B) ⊑? X, ∃r.(B ⊓ C) ⊑? X} has a unifier in EL−⊤: X → ∃r.B A unifier in EL: X → ⊤
Part(σ(X)) = ∅ .
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If γ is a unifier in EL−⊤:
γ1(X) =
SX
γ1(D) ⊓ A1 ⊓ · · · ⊓ Am where γ(X) ⊑ γ(D)
γ2(X) =
SX
γ2(D)
γ3(X) =
SX
γ3(D) ⊓P1 ⊓ · · · ⊓ Pk Pi is a particle and there is Aj, such that Aj ⊑ Pi.
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such that a unifier in EL can be defined. Step 1 Guess additional subsumptions. Step 2 Construct linear language inclusions.
allow for any particles for each variable X. Step 3 Construct an alternating automaton. Step 4 Test the emptiness of this automaton.
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, V ⊑? ∃s.Y, (guess) 7 . ∃s.Z ⊑? ∃s.Y, (from 1 and 6)
SX = {∃s.Y}, SV = {∃s.Y}, SZ = {C}, SY = ∅
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– τ(D, D) = 1 – τ(A, B) = 0, for A, B different constants – τ(∃r.C1, ∃s.C2) = 0, for r, s different role names – τ(A, ∃r.C) = τ(∃r.C, A), for a constant A – τ(∃r.C1, ∃r.C2) = τ(C1, C2) – if τ(C1, C2) = τ(C2, C3) = 1, then τ(C1, C3) = 1
– if D is a non-variable atom, then there is Ci with τ(Ci, D) = 1, – if D is a variable, then τ(D, C) = 1, where C is a non-variable atom, implies τ(Ci, C) = 1 for at least one Ci.
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Check if there is a unifier in EL−⊤ where all subsumptions guessed in τ hold, i.e. if the subsumptions allow for any particles for all variables in the goal.
∆Γ := {C1 ⊓ · · · ⊓ Cn ⊑? X ∈ Γ | X is a variable in Γ} ∆τ := {C ⊑? X | X is a variable and τ(C, X) = 1} ∆Γ,τ := ∆Γ ∪ ∆τ
C1 ⊓ · · · ⊓ Cn ⊑? X ⊑? P for each C1 ⊓ · · · ⊓ Cn ⊑? X in ∆Γ,τ
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C1 ⊓ · · · ⊓ Cn ⊑? X ⊑? P for each C1 ⊓ · · · ⊓ Cn ⊑? X in ∆Γ,τ
A particle P is of the form ∃r1∃r2 . . . ∃rn.A∃r1r2 . . . rn.A∃ω.A, for a constant A.
constant A constant B Y ⊓ ∃r.A ⊑ X XA ⊆ {ǫ}YA ∪ {r} XB ⊆ {ǫ}YB ∪ {r}∅ ∃s.B ⊓ X ⊑ Y YA ⊆ {s}∅ ∪ {ǫ}XA YB ⊆ {s} ∪ {ǫ}XB ∃s.A ⊓ B ⊑ Y YA ⊆ {s} ∪ {ǫ}∅ YB ⊆ {s}∅ ∪ {ǫ}
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For each C1 ⊓ · · · ⊓ Cn ⊑ X in ∆Γ,τ XA ⊆ fA(C1) ∪ · · · ∪ fA(Cn) , – fA(∃r.C) = {r}fA(C), – fA(Y) = YA, – fA(A) = {ǫ}, – fA(B) = ∅
A∈Γ IA
θ(XA) = ∅
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AX = (Q∀, Q∃, Σ, q0, δ, F)
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AX = (Q∀, Q∃, Σ, q0, δ, F)
i1 XA ⊆ {ǫ}YA i2 XA ⊆ {s} i3 YA ⊆ {r} ∪ {ǫ}XA (XA, 0), s (i2, 0), s f∃, ǫ s ǫ (i1, 0), s (YA, 1), s (i3, 1), s (XA, 2), s (i2, 2), s f∃, ǫ s ǫ (i1, 2), s f∀, s ǫ ǫ ǫ ǫ ǫ ǫ
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The problem of deciding unifiability in EL−⊤ is in PSPACE. Step 1 Guessing τ is in NP Step 2 Constructing ∆Γ,τ and IΓ,τ is in P Step 3 Constructing polynomially many polynomial size ǫ−AFA’s is in P Step 4 Checking two-way alternating finite automaton for emptiness is in PSPACE.
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A = (Q, Σ, q0, δ, F)
Signature:
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DFA A Unification problem ΓA q0 start q1 r1 r2 ∃r1.Xq0 ⊓ ∃r2.Xq1 ⊑? Xq0 A ⊑? Xq1
ω ∈ L(A) iff there is a unifier γ of ΓA, such that ∃ω.A ∈ Part(γ(Xq0))
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We define a flat EL−⊤ unification problem Γ: Γ :=
(ΓAi ∪ {Xq0,i ⊑? Y}) where Y is a new variable, not contained in any ΓAi .
Γ is unifiable in EL−⊤ iff L(A1) ∩ · · · ∩ L(An) = ∅
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Decision problem for unification in EL⊤ is PSPACE- complete.
Not presented here:
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