Outline Complexity Theory

Transcription

Complexity Theory
Complexity Theory
Outline
Complexity Theory
VU 181.142, SS 2014
7. Logic-Based Abduction
7.1 Basic Definitions
7.2 Basic Complexity Results
7.3 Subset Minimality
7.4 Minimum Cardinality
7.5 Further Restrictions on the Solutions
Reinhard Pichler
Institut f¨
ur Informationssysteme
Arbeitsbereich DBAI
Technische Universit¨
at Wien
20 May, 2014
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7.1. Basic Definitions
Introduction
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What is abduction?
Consider the following football knowledge base.
Motivation
weak defence ∧ weak attack → match lost,
Abduction is an important method in non-monotonic reasoning.
match lost → manager sad ∧ press angry ,
It is mainly used in diagnosis, e.g., faulty systems, medicine
star injured → manager sad ∧ press sad
It aims at explaining observed symptoms, e.g., malfunctions.
Suppose that we observe that the manager is sad.
What are the possible explanations?
Reference
S1 = {star injured}
This part of the lecture is based on
S2 = {weak defence, weak attack}
S3 = {weak attack, star injured}
Thomas Eiter and Georg Gottlob.
The Complexity of Logic-Based Abduction.
Journal of the ACM, 42(1):3-42 (1995).
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S4 = {weak defence, star injured}
S5 = {weak defence, weak attack, star injured}
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Formal definition of abduction
Abduction in Diagnosis
A propositional abduction problem (PAP) is a quadruple
P = hV , H, M, T i consisting of
A diagnosis problem can be represented by a propositional abduction
problem P = hV , H, M, T i as follows:
a finite set of propositional variables V ,
a theory T , which is a consistent set of propositional logical
formulae over the variables V ,
The theory T is the system description.
The hypotheses H ⊆ V describe the possibly faulty system
components.
a set of hypotheses H ⊆ V ,
and a set of manifestations M ⊆ V .
The manifestations M ⊆ V are the observed symptoms (describing
the malfunction of the system).
A set S satisfying S ⊆ H is a solution iff
The solutions S ∈ Sol(P) are the possible explanations of the
malfunction.
T ∪ S is consistent (i.e., satisfiable)
and T ∪ S |= M holds.
We denote the set of all solutions of a problem P by Sol(P).
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Formal version of the football example
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The main decision problems
Given a PAP P we ask:
V = {weak defence, weak attack, match lost,
Solvability. Sol(P) 6= ∅, i.e., does there exist a solution for P?
manager sad, press angry , star injured, press sad}
Given a PAP P and a hypothesis h ∈ H, we ask:
T = {weak defence ∧ weak attack → match lost,
Relevance: Is h contained in at least one solution of P?
match lost → manager sad ∧ press angry ,
Necessity: Is h contained in all solutions of P?
star injured → manager sad ∧ press sad}
M = {manager sad}
Remark. If h is not necessary then we also say that h is dispensable.
H = {weak defence, weak attack, star injured}
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Not all solutions are of the same interest
Recall the solutions of our football example:
Sol(P) = {{star injured},
{weak defence, weak attack},
Given two solutions, usually the “simpler” explanation is preferable. We
define “simple” in one of the following ways:
{weak attack, star injured},
{weak defence, star injured}
subset minimality (⊆): We only accept solutions S, s.t. no proper
subset S 0 ⊂ S is a solution.
{weak defence, weak attack, star injured}}
minimum cardinality (≤): We only accept solutions S, s.t. there
does not exist a solution S 0 with fewer elements, i.e., |S 0 | < |S|.
Further preorders to eliminate undesired solutions:
priorities or minimum weight (also referred to as penalties).
Obviously, the set of solutions is nonempty.
All three hypotheses of H are relevant.
No hypothesis in H is necessary (i.e., they are all dispensable).
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7.2. Basic Complexity Results
Deciding if S ⊆ H fullfills S ∈ Sol(P) is in ∆2 P (actually, even in DP).
Sol(P) = {{star injured}, {weak defence, weak attack},
{weak attack, star injured},
Proof.
{weak defence, star injured}
By the definition, we have to check if
{weak defence, weak attack, star injured}}
T ∪ S is consistent (i.e., satisfiable) and
T ∪ S |= M holds.
We can check if T ∪ S is satisfiable with an NP oracle.
Subset minimal solutions.
Sol⊆ (P) = {{star injured}, {weak defence, weak attack}}
The second problem is clearly in co-NP, which can be checked by yet
another call to an NP oracle. In order to see the co-NP-membership, we
consider the co-problem T ∪ S 6|= M. Clearly, this problem can be
decided by the following NP-algorithm: guess a truth assignment I and
check that T ∪ S is true in I while M is false in I .
Cardinality minimal solution.
Sol≤ (P) = {{star injured}}
Hence, weak defence and weak attack are no longer relevant.
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Lemma
Recall the solutions of our football example:
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Proof (continued)
Then we define an instance of Solvability as P = hV , H, M, T i with
Theorem
The Solvability problem is Σ2 P-complete.
V = X ∪ Y ∪ X 0 ∪ R ∪ {t}
Proof.
M = R ∪ {t}
H = X ∪ X0
T = {ψ(X , Y ) → t} ∪ {¬xi ∨ ¬xi0 , xi → ri , xi0 → ri | 1 ≤ i ≤ k}
Σ2 P-Membership. Guess a set S ⊆ H and check with two calls to an NP
oracle that T ∪ S is satisfiable and T ∪ S |= M holds.
Obviously, this reduction is feasible in logarithmic space.
Σ2 P-Hardness. Proof by reduction from the QSAT2 problem. Let an
arbitrary instance of the QSAT2 problem be given by the formula
ϕ = (∃X )(∀Y )ψ(X , Y ) with X = {x1 , . . . , xk } and Y = {y1 , . . . , yl }.
Let X 0 = {x10 , . . . , xk0 }, R = {r1 , . . . , rk }, and t be fresh variables.
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The effect of the clauses {¬xi ∨ ¬xi0 , xi → ri , xi0 → ri } in T is that, for
every i ∈ {1, . . . , k}, every solution of P contains exactly one of {xi , xi0 }.
The correctness proof of the reduction is based on the following
observation: For A ⊆ X , let A0 denote the set {x 0 | x ∈ A}. Then every
subset A ⊆ X fulfills the following equivalence: For the assignment I on
X with I −1 (true) = A, every extension J of I to the variables Y satisfies
the formula ψ(X , Y ) ⇔ A ∪ (X \ A)0 is a solution of P.
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Proof of Hardness
By reduction from the Solvability problem: Let an arbitrary instance of
the Solvability problem be given by the PAP P = hV , H, M, T i. W.l.o.g.,
let T consist of a single formula ϕ and let h, h0 , m0 be fresh variables.
Then we define an instance of the Relevance (resp. the Necessity)
problem with the following PAP P 0 = hV 0 , H 0 , M 0 , T 0 i:
Theorem
The Relevance problem is Σ2 P-complete.
The Necessity problem is Π2 P-complete.
Proof of Membership.
V 0 = V ∪ {h, h0 , m0 }
H 0 = H ∪ {h, h0 }
Relevance. Guess a set S ⊆ H with h ∈ S and check with two calls to an
NP oracle that T ∪ S is satisfiable and T ∪ S |= M holds.
M 0 = M ∪ {m0 }
Necessity. We show the Σ2 P-membership of the co-problem (i.e., the
dispensability): Guess a set S ⊆ H with h 6∈ S and check with two calls
to an NP oracle that T ∪ S is satisfiable and T ∪ S |= M holds.
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T 0 = {¬h ∨ ϕ} ∪ {h0 → m | m ∈ M} ∪ {¬h ∨ ¬h0 , h → m0 , h0 → m0 }
Obviously, this reduction is feasible in logarithmic space and the new
theory T 0 is consistent. We claim that the PAP P has at least one
solution iff h is relevant in P 0 iff h0 is not necessary in P 0 .
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7.3. Subset Minimality
Subset Minimality
Proof of Hardness (continued)
Proof idea for the correctness of this reduction.
Motivation
The clauses {¬h ∨ ¬h0 , h → m0 , h0 → m0 } in T 0 have the effect that
every solution of P 0 contains exactly one of {h, h0 }.
The subset minimality criterion removes redundancy.
0
The solutions of P extended by h are clearly solutions of P .
Does there exist a subset minimal solution?
The effect of the disjunct ¬h in the clause ¬h ∨ ϕ is that any subset
A ⊆ H is consistent with T 0 (simply set h to false).
If a PAP has a solution then it also has a (subset) minimal solution.
Thus, Sol(P 0 ) = {{h} ∪ S | S ∈ Sol(P)} ∪ {{h0 } ∪ A | A ⊆ H} holds.
Sol(P) 6= ∅ ⇔ Sol⊆ (P) 6= ∅
We conclude that
So we are only interested in ⊆- Relevance (resp. ⊆-Necessity), i.e.: Is a
given hypothesis contained in some (resp. all) subset minimal solutions?
h is relevant for P 0 iff Sol(P) 6= ∅ holds and
h0 is dispensable (i.e., not necessary) for P 0 iff Sol(P) 6= ∅ holds.
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Checking subset minimality
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Theorem
Let P = hV , H, M, T i be a PAP.
For every S ⊆ H, the following equivalence holds:
S ∈ Sol⊆ (P) ⇔ S ∈ Sol(P) and for all h ∈ S, (S \ {h}) 6∈ Sol(P)
The ⊆-Relevance problem is Σ2 P-complete.
The ⊆-Necessity problem is Π2 P-complete.
Proof
⊆-Relevance. Guess a set S ⊆ H with h ∈ S and check with two calls to
an NP oracle that T ∪ S is satisfiable and T ∪ S |= M holds. Moreover,
check with another call to an NP oracle that T ∪ (S \ {h}) 6|= M.
We claim that h is in a minimal solution S 0 ⊆ S.
⊆: Sol⊆ (P) = {S ∈ Sol(P) | for all S 0 ⊂ S, S 0 6∈ Sol(P)}. Hence, any
subset minimal solution S satisfies S \ {h} 6∈ Sol(P) for all h ∈ S.
⊇: Let S ∈ Sol(P) and S 6∈ Sol⊆ (P). Hence, there exists S 0 ⊂ S with
S 0 ∈ Sol(P). But then there exists h ∈ S with S 0 ⊆ S \ {h} ⊂ S.
We claim that then also S \ {h} is a solution of P:
Clearly, if S is a minimal solution, then h ∈ S 0 with S 0 = S. Now suppose
that S is not minimal, i.e. there exists a minimal solution S 0 ⊂ S. Then
h is contained in S 0 since, otherwise, we have S 0 ⊆ S \ {h} ⊆ S and,
thus, analogously to the proof of the lemma, also S \ {h} is a solution.
T ∪ S is consistent. Hence also T ∪ (S \ {h}) is consistent.
T ∪ S 0 |= M. Hence, by the monotonicity of |=, also T ∪ (S \ {h}) |= M.
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Lemma
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Proof of Hardness
By reduction from the Solvability problem: Let an arbitrary instance of
the Solvability problem be given by the PAP P = hV , H, M, T i with
H = {h1 , . . . , h` }. W.l.o.g., let T consist of a single formula ϕ and let
{h0 , h00 , h10 , . . . , h`0 , m00 , m10 , . . . , m`0 } be fresh variables. Then we define an
instance of the ⊆-Relevance (resp. the ⊆-Necessity) problem with the
following PAP P 0 = hV 0 , H 0 , M 0 , T 0 i:
Proof of Membership (continued).
⊆-Necessity. We show the Σ2 P-membership of the co-problem: Guess a
set S ⊆ H with h 6∈ S and check with two calls to an NP oracle that
T ∪ S is satisfiable and T ∪ S |= M holds. From this we may conclude
that h is not necessary.
V 0 = V ∪ {h0 , h00 , h10 , . . . , h`0 , m00 , m10 , . . . , m`0 }
H 0 = H ∪ {h0 , h00 , h10 , . . . , h`0 }
Clearly, there exists some S 0 ⊆ S, s.t. S 0 is a minimal solution. Moreover,
since h 6∈ S, also h 6∈ S 0 holds.
M 0 = M ∪ {m00 , m10 , . . . , m`0 }
T 0 = {¬h0 ∨ ϕ} ∪ {h00 → m | m ∈ M} ∪
∪ {¬hi ∨ ¬hi0 , hi → mi0 , hi0 → mi0 | 0 ≤ i ≤ `}
Obviously, this reduction is feasible in logarithmic space and the new
theory T 0 is consistent. We claim that the PAP P has at least one
solution iff h0 is ⊆-relevant in P 0 iff h00 is not ⊆-necessary in P 0 .
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7.4. Minimum Cardinality
Minimum cardinality
Proof of Hardness (continued)
Proof idea for the correctness of this reduction.
h0 , h00
Motivation
0
The hypotheses
play exactly the role of h, h in the hardness
proof of Relevance and Necessity (without ⊆-minimality).
The subset minimal solutions may have different cardinalities. If the
failure of any component in a system is independent of the failure of the
other components, then explanations with minimum cardinality are the
ones with highest probability. Moreover, the cost of repair is normally
lower when fewer components have to be exchanged. Hence,
explanations with minimum cardinality are preferable.
For every i ∈ {0, . . . , `}, the clauses {¬hi ∨ ¬hi0 , hi → mi0 , hi0 → mi0 }
in T 0 have the effect that every solution of P 0 contains exactly one
of {hi , hi0 }. In other words, every solution of P 0 is ⊆-minimal.
For every A ⊆ H, we write A0 to denote {h0 | h ∈ A}. In summary, we
thus have: Sol⊆ (P 0 ) = Sol(P 0 ) = {{h0 } ∪ S ∪ (H \ S)0 | S ∈ Sol(P)} ∪
{{h00 } ∪ A ∪ (H \ A)0 | A ⊆ H}.
Does there exist a solution with minimum cardinality?
If a PAP has a solution then it also has a solution with minimum
cardinality, i.e.: Sol(P) 6= ∅ ⇔ Sol≤ (P) 6= ∅.
We conclude that
h0 is ⊆-relevant for P 0 iff Sol(P) 6= ∅ holds and
So we are only interested in ≤- Relevance (resp. ≤-Necessity), i.e.: Is a
given hypothesis in some (resp. all) solutions with minimum cardinality?
h00 is ⊆-necessary for P 0 iff Sol(P) = ∅ holds.
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Complexity Theory
Complexity Results
Preparation of the hardness proof
Theorem
Recall that the following problem is ∆2 P[log n]-complete.
Both, the ≤-Relevance problem and the ≤-Necessity problem are
∆3 P[log n]-complete.
CARD-MINIMAL MODEL SAT
INSTANCE: Boolean formula ϕ and an atom z.
QUESTION: Is z true in a cardinality-minimal model of ϕ?
This problem can be generalized to an arbitrary level of the polynomial
hierarchy. For every i ≥ 1, it can be shown that the following problem is
∆i+1 P[log n]-complete.
Let an instance of ≤-Relevance (resp. ≤-Necessity) be given by a PAP P
and a hypothesis h.
Determine the minimum cardinality K of the solutions of the PAP
P. This can be done by log m (with m = |H|) calls to a Σ2 P-oracle
asking questions of the sort “Does P have a solution of size ≤ k?”
CARD-MINIMAL MODEL QSATi
INSTANCE: Quantified Boolean formula
ϕ(X ) = ∀Y1 ∃Y2 · · · QYi−1 ψ(X , Y1 , . . . , Yi−1 ) and an atom z ∈ X .
QUESTION: Is z true in a cardinality-minimal model of ϕ(X )?
We need one more call to a Σ2 P-oracle asking if there exists a
solution S of size K , s.t. h ∈ S (resp. h 6∈ S).
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Proof of Hardness.
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Proof of Hardness (continued).
Let an arbitrary instance of the CARD-MINIMAL MODEL QSAT2
problem be given by the formula ϕ(X ) = (∀Y )ψ(X , Y ) with
X = {x1 , . . . , xk } and Y = {y1 , . . . , yl } and let xj ∈ X denote the
distinguished atom. Let X 0 = {x10 , . . . , xk0 }, X 00 = {x100 , . . . , xk00 },
Q = {q1 , . . . , qk }, R = {r1 , . . . , rk }, and t be fresh variables.
For every i ∈ {1, . . . , k}, the clauses ¬xi ∨ ¬xi0 , xi → qi , xi0 → qi in T
make sure that every solution S of P contains exactly one of {xi , xi0 }.
Likewise, the clauses ¬xi0 ∨ ¬xi00 , xi0 → ri , xi00 → ri in T make sure that
every solution S of P contains exactly one of {xi0 , xi00 }.
Then we define an instance of ≤-Relevance (resp. co-≤-Necessity) via
the following PAP P = hV , H, M, T i and the hypothesis xj (resp. xj0 ):
For A ⊆ X , let A0 denote the set {x 0 | x ∈ A} and let A00 denote the set
{x 00 | x ∈ X }. For every subset A ⊆ X , the following equivalences hold:
The assignment I on X with I −1 (true) = A is a model of ϕ(X ) ⇔
A ∪ (X \ A)0 ∪ A00 is a solution of P.
V = X ∪ X 0 ∪ X 00 ∪ Q ∪ R ∪ Y ∪ {t}
H = X ∪ X 0 ∪ X 00
Moreover, I with I −1 (true) = A is a cardinality-minimal model of ϕ(X )
⇔ A ∪ (X \ A)0 ∪ A00 is a cardinality-minimal solution of P.
M = Q ∪ R ∪ {t}
T = {ψ(X , Y ) → t} ∪ {¬xi ∨ ¬xi0 , xi → qi , xi0 → qi | 1 ≤ i ≤ k} ∪
Thus, xj is contained in some cardinality-minimal model of ϕ(X ) ⇔
xj is ≤-relevant for P ⇔ xj0 is not ≤-necessary for P.
{¬xi0 ∨ ¬xi00 , xi0 → ri , xi00 → ri | 1 ≤ i ≤ k}.
Obviously, this reduction is feasible in logarithmic space.
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7.5. Further Restrictions on the Solutions
Prioritization
Complexity Theory
Prioritization
Example
(the football example revisited)
Motivation
The set of hypotheses is partitioned into groups of different priorities
V = {weak defence, weak attack, match lost,
which may possibly express a kind of probability when no numerical
values are available
manager sad, press angry , star injured, press sad}
T = {...}
and which may be used to refine other minimality criteria (⊆, ≤)
M = {manager sad}
Definition
with prioritization: P = h{weak defence, weak attack}, {star injured}i
Prioritization on ⊆. Let H be the set of hypotheses and P = hP1 , . . . , Pk i
such that P1 ∪ · · · ∪ Pk = H and Pi ∩ Pj = ∅ for all i 6= j.
We define the relation ⊆P as follows:
A ⊆P B ⇔ A = B or there exists i ∈ {1, . . . , k} such that:
A ∩ Pj = B ∩ Pj for 1 ≤ j < i and A ∩ Pi ⊂ B ∩ Pi
{weak defence, star injured} ⊆P {weak defence, weak attack}
{weak defence, star injured} 6⊆P {weak attack}
{star injured} ⊆P {weak defence}
{weak defence} 6⊆P {weak attack}
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Complexity Results
Intuition of the Hardness
Theorem
“⊆” Recall that subset minimality makes things only polynomially
harder. This is due to the following equivalence: S is a ⊆-minimal
solution of the PAP P if and only if S is a solution of P and, for every
h ∈ S, S \ {h} is not a solution.
The ⊆P -Relevance problem is Σ3 P-complete.
The ⊆P -Necessity problem is Π3 P-complete.
These hardness results already hold for 2 priorities.
“⊆P ” Suppose that H is partitioned into 2 priority levels P = hP1 , P2 i.
Then the following effect may occur. Suppose that S is a solution of the
PAP and that, for every h ∈ S, S \ {h} is not a solution. Nevertheless, it
might well happen that, for some h ∈ S ∩ P1 and some X ⊆ P2 , the set
S 0 = (S \ {h}) ∪ X is a solution. In this case, S 0 ⊆P S clearly holds.
Checking if such a set S 0 (and, in particular, if such a set X ) exists
comes down to yet another non-deterministic guess.
⊆P -Relevance. Guess a set S ⊆ H with h ∈ S and check with a call to a
∆2 P oracle that S is a solution. Moreover, we check by a call to a Σ2 P
oracle that there exists no solution S 0 6= S with S 0 ⊆P S.
⊆P -Necessity. We show the Σ3 P-membership of the co-problem: Guess a
set S ⊆ H with h 6∈ S and check with a call to a ∆2 P oracle that S is a
solution. Moreover, we check by a call to a Σ2 P oracle that there exists
no solution S 0 6= S with S 0 ⊆P S.
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Further preorders on the solutions
Complexity Theory
Complexity Results
Theorem
All of the following four problems are ∆3 P-complete:
≤w -Relevance, ≤w -Necessity, ≤P -Relevance, ≤P -Necessity.
Definition
Weight-minimality (Penalization). Let a PAP P = hV , H, M, T i be given
with a weight function w on P
the hypotheses.
PFor two subsets A ⊆ H and
B ⊆ H, we write A ≤w B if h∈A w (h) ≤ h∈B w (h) holds.
Proof idea.
The hardness in case of ≤w can be proved by reduction from a quantified
version of WEIGHT-MINIMAL MODEL SAT, analogously to the
proof for ≤. Similarly, the hardness in case of ≤P is shown by reduction
from a quantified version of LEX-MINIMAL MODEL SAT. Below, we
only sketch the proof idea of the membership.
Definition
Prioritization on ≤. Let H be the set of hypothesis and P = hP1 , . . . , Pk i
such that P1 ∪ · · · ∪ Pk = H and Pi ∩ Pj = ∅, i 6= j.
We define the relation ≤P as follows:
A ≤P B ⇔ A = B or there exists i ∈ {1, . . . , k} such that:
|A ∩ Pj | = |B ∩ Pj | for 1 ≤ j < i and |A ∩ Pi | < |B ∩ Pi |
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≤w -Relevance and ≤w -Necessity. First determine the minimum weight
W of the solutions of P. This can be done by a binary search asking
questions like “Does P have a solution of weight ≤ w ?”
For this task, we need logarithmically many calls (w.r.t. the total weight)
to a Σ2 P-oracle. These are polynomially many calls w.r.t. the
representation of the weights of the elements in H.
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Proof idea (continued)
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Complexity Theory
Summary
≤P -Relevance and ≤P -Necessity. Suppose that P has ` priorities
and
P`
that each Pi contains ni hypotheses. Moreover, let |H| = i=1 ni = n.
Then we have to determine the lexicographically minimal vector
(K1 , . . . , K` ), s.t. there exists a solution S with |S ∩ Pi | = Ki for each i.
Table: Complexity results presented in the lecture
This can be done in ` stages. In the i-th stage, we determine Ki by log ni
calls to a Σ2 P-oracle, asking questions like “Does P have a solution S,
s.t. |S ∩ Pj | = Kj for all j < i and |S ∩ Pi | ≤ k?”
Solvability
Relevance
Necessity
We thus need O(` · log n) Σ2 P-oracle calls, i.e., polynomially many.
=
Σ2 P
Σ2 P
Π2 P
⊆
–
Σ2 P
Π2 P
⊆P
–
Σ3 P
Π3 P
≤
–
∆3 P[log n]
∆3 P[log n]
≤w
–
∆3 P
∆3 P
≤P
–
∆3 P
∆3 P
Remark. If the number of priorities is bounded by a constant, then
logarithmically many Σ2 P-oracle calls suffice.
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Complexity Theory
Learning Objectives
Summary
Table: Further complexity results in (Eiter/Gottlob, 1995)
Relevance
Horn
definite Horn
Necessity
Horn
definite Horn
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=
NP
P
⊆
NP
NP
⊆P
Σ2 P
NP
≤
∆2 P[log n]
∆2 P[log n]
≤w
∆2 P
∆2 P
Definition and intuition of logic-based abduction
≤P
∆2 P
∆2 P
The main decision problems of logic-based abduction:
solvability, relevance, necessity.
Restricting the set of acceptable solutions: ⊆, ≤, ⊆P , ≤w , ≤P
Intuition why non-monotonic reasoning usually has an additional
source of complexity compared with monotonic reasoning.
=
⊆
⊆P
≤
≤w
≤P
co-NP co-NP Π2 P ∆2 P[log n] ∆2 P ∆2 P
P
co-NP co-NP ∆2 P[log n] ∆2 P ∆2 P
20 May, 2014
Get practice with complexity results in the polynomical hierarchy.
Page 37
Reinhard Pichler
20 May, 2014
Page 38

Outline Complexity Theory

Transcription

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