Outline. Complexity Theory. Introduction. What is abduction? Motivation. Reference VU , SS Logic-Based Abduction

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1 Complexity Theory Complexity Theory Outline Complexity Theory VU , SS Logic-Based Abduction Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität Wien 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 5 June, 2018 Reinhard Pichler 5 June, 2018 Page 1 Reinhard Pichler 5 June, 2018 Page 2 Introduction What is abduction? Motivation Abduction is an important method in non-monotonic reasoning. It is mainly used in diagnosis, e.g., faulty systems, medicine It aims at explaining observed symptoms, e.g., malfunctions. Reference This part of the lecture is based on Thomas Eiter and Georg Gottlob. The Complexity of Logic-Based Abduction. Journal of the ACM, 42(1):3-42 (1995). Consider the following football knowledge base. weak defence weak attack match lost, match lost manager sad press angry, star injured manager sad press sad Suppose that we observe that the manager is sad. What are the possible explanations in terms of weak defence, weak attack, and star injured? S 1 = {star injured} S 2 = {weak defence, weak attack} S 3 = {weak attack, star injured} S 4 = {weak defence, star injured} S 5 = {weak defence, weak attack, star injured} Reinhard Pichler 5 June, 2018 Page 3 Reinhard Pichler 5 June, 2018 Page 4

2 Formal definition of abduction Abduction in Diagnosis A propositional abduction problem (PAP) is a quadruple P = V, H, M, T consisting of a finite set of propositional variables V, a theory T, which is a consistent finite set of propositional logical formulae over the variables V, a set of hypotheses H V, and a set of manifestations M V. A set S satisfying S H is a solution iff 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). A diagnosis problem can be represented by a propositional abduction problem P = V, H, M, T as follows: The theory T is the system description. The hypotheses H V describe the possibly faulty system components. The manifestations M V are the observed symptoms (describing the malfunction of the system). The solutions S Sol(P) are the possible explanations of the malfunction. Reinhard Pichler 5 June, 2018 Page 5 Reinhard Pichler 5 June, 2018 Page 6 Formal version of the football example The main decision problems V = {weak defence, weak attack, match lost, manager sad, press angry, star injured, press sad} T = {weak defence weak attack match lost, match lost manager sad press angry, star injured manager sad press sad} M = {manager sad} H = {weak defence, weak attack, star injured} Given a PAP P we ask: Solvability. Sol(P), i.e., does there exist a solution for P? Given a PAP P and a hypothesis h H, we ask: Relevance: Is h contained in at least one solution of P? Necessity: Is h contained in all solutions of P? Remark. If h is not necessary then we also say that h is dispensable. Reinhard Pichler 5 June, 2018 Page 7 Reinhard Pichler 5 June, 2018 Page 8

3 Not all solutions are of the same interest Recall the solutions of our football example: Sol(P) = {{star injured}, {weak defence, weak attack}, {weak attack, star injured}, {weak defence, star injured} {weak defence, weak attack, star injured}} H = {weak defence, weak attack, star injured} Obviously, the set of solutions is non-empty. All three hypotheses of H are relevant. No hypothesis in H is necessary (i.e., they are all dispensable). Given two solutions, usually the simpler explanation is preferable. We define simple in one of the following ways: subset minimality ( ): We only accept solutions S, s.t. no proper subset S S is a solution. minimum cardinality ( ): We only accept solutions S, s.t. there does not exist a solution S with fewer elements, i.e., S < S. Further preorders to eliminate undesired solutions: priorities or minimum weight (also referred to as penalties). Reinhard Pichler 5 June, 2018 Page 9 Reinhard Pichler 5 June, 2018 Page 10 Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results Recall the solutions of our football example: Sol(P) = {{star injured}, {weak defence, weak attack}, {weak attack, star injured}, {weak defence, star injured} {weak defence, weak attack, star injured}} Subset minimal solutions. Sol (P) = {{star injured}, {weak defence, weak attack}} Cardinality minimal solution. Sol (P) = {{star injured}} Hence, weak defence and weak attack are no longer relevant. Lemma Deciding if S H fullfills S Sol(P) is in 2 P (actually, even in DP). Proof. By the definition, we have to check if T S is consistent (i.e., satisfiable) and T S = M holds. We can check if T S is satisfiable with an NP oracle. 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 = 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. Reinhard Pichler 5 June, 2018 Page 11 Reinhard Pichler 5 June, 2018 Page 12

4 Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results Proof (continued) The Solvability problem is Σ 2 P-complete. Proof. Σ 2 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. Σ 2 P-Hardness. Proof by reduction from the QSAT 2 problem. Let an arbitrary instance of the QSAT 2 problem be given by the formula ϕ = ( X )( Y )ψ(x, Y ) with X = {x 1,..., x k } and Y = {y 1,..., y l }. Let X = {x 1,..., x k }, R = {r 1,..., r k }, and t be fresh variables. Then we define an instance of Solvability as P = V, H, M, T with V = X Y X R {t} H = X X M = R {t} T = {ψ(x, Y ) t} { x i x i, x i r i, x i r i 1 i k} Obviously, this reduction is feasible in logarithmic space. The effect of the clauses { x i x i, x i r i, x i r i } in T is that, for every i {1,..., k}, every solution of P contains exactly one of {x i, x i }. The correctness proof of the reduction is based on the following observation: For A X, let A denote the set {x 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) is a solution of P. Exercise. Give a full proof of the correctness of the reduction. Reinhard Pichler 5 June, 2018 Page 13 Reinhard Pichler 5 June, 2018 Page 14 Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results The Relevance problem is Σ 2 P-complete. The Necessity problem is Π 2 P-complete. Proof of Membership. 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. Necessity. We show the Σ 2 P-membership of the co-problem (i.e., the dispensability): 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. Proof of Hardness By reduction from the Solvability problem: Let an arbitrary instance of the Solvability problem be given by the PAP P = V, H, M, T. W.l.o.g., let T consist of a single formula ϕ and let h, h, m be fresh variables. Then we define an instance of the Relevance (resp. the Necessity) problem with the following PAP P = V, H, M, T : V = V {h, h, m } H = H {h, h } M = M {m } T = { h ϕ} {h m m M} { h h, h m, h m } Obviously, this reduction is feasible in logarithmic space and the new theory T is consistent. We claim that the PAP P has at least one solution iff h is relevant in P iff h is not necessary in P. Reinhard Pichler 5 June, 2018 Page 15 Reinhard Pichler 5 June, 2018 Page 16

5 Complexity Theory 7. Logic-Based Abduction 7.2. Basic Complexity Results Proof of Hardness (continued) Proof idea for the correctness of this reduction. The clauses { h h, h m, h m } in T have the effect that every solution of P contains exactly one of {h, h }. The solutions of P extended by h are clearly solutions of P. The effect of the disjunct h in the clause h ϕ is that any subset A H is consistent with T (simply set h to false). Thus, Sol(P ) = {{h} S S Sol(P)} {{h } A A H} holds. We conclude that h is relevant for P iff Sol(P) holds and h is dispensable (i.e., not necessary) for P iff Sol(P) holds. Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Subset Minimality Motivation The subset minimality criterion removes redundancy. Does there exist a subset minimal solution? If a PAP has a solution then it also has a (subset) minimal solution. Sol(P) Sol (P) So we are only interested in - Relevance (resp. -Necessity), i.e.: Is a given hypothesis contained in some (resp. all) subset minimal solutions? Exercise. Give a full proof of the correctness of the reduction. Reinhard Pichler 5 June, 2018 Page 17 Reinhard Pichler 5 June, 2018 Page 18 Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Checking subset minimality Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Complexity Results Lemma Let P = V, H, M, T be a PAP. For every S H, the following equivalence holds: S Sol (P) S Sol(P) and for all h S, (S \ {h}) Sol(P) Proof : Sol (P) = {S Sol(P) for all S S, S Sol(P)}. Hence, any subset minimal solution S satisfies S \ {h} Sol(P) for all h S. : Let S Sol(P) and S Sol (P). Hence, there exists S S with S Sol(P). But then there exists h S with S S \ {h} S. We claim that then also S \ {h} is a solution of P: T S is consistent. Hence also T (S \ {h}) is consistent. T S = M. Hence, by the monotonicity of =, also T (S \ {h}) = M. The -Relevance problem is Σ 2 P-complete. The -Necessity problem is Π 2 P-complete. Proof of Membership. -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. Finally, check that S is a minimal solution. By the previous lemma, we thus need polynomially many further calls to an NP oracle to check that T (S \ {h }) = M holds for every h S. Reinhard Pichler 5 June, 2018 Page 19 Reinhard Pichler 5 June, 2018 Page 20

6 Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Proof of Membership (continued). -Necessity. We show the Σ 2 P-membership of the co-problem: 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. From this we may conclude that h is not necessary. Clearly, there exists some S S, s.t. S is a minimal solution. Moreover, since h S, also h S holds. Proof of Hardness By reduction from the Solvability problem: Let an arbitrary instance of the Solvability problem be given by the PAP P = V, H, M, T with H = {h 1,..., h l }. W.l.o.g., let T consist of a single formula ϕ and let {h 0, h 0, h 1,..., h l, m 0, m 1,..., m l } be fresh variables. Then we define an instance of the -Relevance (resp. the -Necessity) problem with the following PAP P = V, H, M, T : V = V {h 0, h 0, h 1,..., h l, m 0, m 1,..., m l} H = H {h 0, h 0, h 1,..., h l} M = M {m 0, m 1,..., m l} T = { h 0 ϕ} {h 0 m m M} { h i h i, h i m i, h i m i 0 i l} Obviously, this reduction is feasible in logarithmic space and the new theory T is consistent. We claim that the PAP P has at least one solution iff h 0 is -relevant in P iff h 0 is not -necessary in P. Reinhard Pichler 5 June, 2018 Page 21 Reinhard Pichler 5 June, 2018 Page 22 Complexity Theory 7. Logic-Based Abduction 7.3. Subset Minimality Proof of Hardness (continued) Proof idea for the correctness of this reduction. The hypotheses h 0, h 0 play exactly the role of h, h in the hardness proof of Relevance and Necessity (without -minimality). For every i {0,..., l}, the clauses { h i h i, h i m i, h i m i } in T have the effect that every solution of P contains exactly one of {h i, h i }. In other words, every solution of P is -minimal. For every A H, we write A to denote {h h A}. In summary, we thus have: Sol (P ) = Sol(P ) = {{h 0 } S (H \ S) S Sol(P)} {{h 0 } A (H \ A) A H}. We conclude that h 0 is -relevant for P iff Sol(P) holds and h 0 is -necessary for P iff Sol(P) = holds. Complexity Theory 7. Logic-Based Abduction 7.4. Minimum Cardinality Minimum cardinality Motivation 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 if fewer components have to be exchanged. Hence, explanations with minimum cardinality are preferable. 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) Sol (P). So we are only interested in - Relevance (resp. -Necessity), i.e.: Is a given hypothesis in some (resp. all) solutions with minimum cardinality? Reinhard Pichler 5 June, 2018 Page 23 Reinhard Pichler 5 June, 2018 Page 24

7 Complexity Theory 7. Logic-Based Abduction 7.4. Minimum Cardinality Complexity Results Complexity Theory 7. Logic-Based Abduction 7.4. Minimum Cardinality Preparation of the hardness proof Both, the -Relevance problem and the -Necessity problem are 3 P[log n]-complete. Proof of Membership. 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 Σ 2 P-oracle asking questions of the sort Does P have a solution of size k? We need one more call to a Σ 2 P-oracle asking if there exists a solution S of size K, s.t. h S (resp. h S). Recall that the following problem is 2 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. CARD-MINIMAL MODEL QSAT i INSTANCE: Quantified Boolean formula ϕ(x ) = Y 1 Y 2 QY i 1 ψ(x, Y 1,..., Y i 1 ) and an atom z X. QUESTION: Is z true in a cardinality-minimal model of ϕ(x )? Reinhard Pichler 5 June, 2018 Page 25 Reinhard Pichler 5 June, 2018 Page 26 Complexity Theory 7. Logic-Based Abduction 7.4. Minimum Cardinality Complexity Theory 7. Logic-Based Abduction 7.4. Minimum Cardinality Proof of Hardness. Let an arbitrary instance of the CARD-MINIMAL MODEL QSAT 2 problem be given by the formula ϕ(x ) = ( Y )ψ(x, Y ) with X = {x 1,..., x k } and Y = {y 1,..., y l } and let x j X denote the distinguished atom. Let X = {x 1,..., x k }, X = {x 1,..., x k }, Q = {q 1,..., q k }, R = {r 1,..., r k }, and t be fresh variables. Then we define an instance of -Relevance (resp. co- -Necessity) via the following PAP P = V, H, M, T and the hypothesis x j (resp. x j ): V = X X X Q R Y {t} H = X X X M = Q R {t} T = {ψ(x, Y ) t} { x i x i, x i q i, x i q i 1 i k} { x i x i, x i r i, x i r i 1 i k}. Obviously, this reduction is feasible in logarithmic space. Proof of Hardness (continued). For every i {1,..., k}, the clauses x i x i, x i q i, x i q i in T make sure that every solution S of P contains exactly one of {x i, x i }. Likewise, the clauses x i x i, x i r i, x i r i in T make sure that every solution S of P contains exactly one of {x i, x i }. For A X, let A denote the set {x x A} and let A denote the set {x 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) A is a solution of P. Moreover, I with I 1 (true) = A is a cardinality-minimal model of ϕ(x ) A (X \ A) A is a cardinality-minimal solution of P. Thus, x j is contained in some cardinality-minimal model of ϕ(x ) x j is -relevant for P x j is not -necessary for P. Reinhard Pichler 5 June, 2018 Page 27 Reinhard Pichler 5 June, 2018 Page 28

8 Prioritization Motivation The set of hypotheses is partitioned into groups of different priorities which may possibly express a kind of probability when no numerical values are available and which may be used to refine other minimality criteria (, ) Definition Prioritization on. Let H be the set of hypotheses and P = P 1,..., P k such that P 1 P k = H and P i P j = for all i j. We define the relation P as follows: A P B A = B or there exists i {1,..., k} such that: A P j = B P j for 1 j < i and A P i B P i Prioritization Example (the football example revisited) V = {weak defence, weak attack, match lost, manager sad, press angry, star injured, press sad} T = {... } M = {manager sad} H = {weak defence, weak attack, star injured} with prioritization: P = {weak defence, weak attack}, {star injured} {weak defence, star injured} P {weak defence, weak attack} {weak defence, star injured} P {weak attack} {star injured} P {weak defence} {weak defence} P {weak attack} Reinhard Pichler 5 June, 2018 Page 29 Reinhard Pichler 5 June, 2018 Page 30 Complexity Results The P -Relevance problem is Σ 3 P-complete. The P -Necessity problem is Π 3 P-complete. These hardness results already hold for 2 priorities. Proof of Membership. 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 Σ 2 P oracle that there exists no solution S S with S P S. P -Necessity. We show the Σ 3 P-membership of the co-problem: 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 Σ 2 P oracle that there exists no solution S S with S P S. Intuition of the Hardness 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. P Suppose that H is partitioned into 2 priority levels P = P 1, P 2. 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 P 1 and some X P 2, the set S = (S \ {h}) X is a solution. In this case, S P S clearly holds. Checking if such a set S (and, in particular, if such a set X ) exists comes down to yet another non-deterministic guess. Reinhard Pichler 5 June, 2018 Page 31 Reinhard Pichler 5 June, 2018 Page 32

9 Further preorders on the solutions Definition Weight-minimality (Penalization). Let a PAP P = V, H, M, T be given with a weight function w on the hypotheses. For two subsets A H and B H, we write A w B if h A w(h) h B w(h) holds. Definition Prioritization on. Let H be the set of hypotheses and P = P 1,..., P k such that P 1 P k = H and P i P j =, i j. We define the relation P as follows: A P B A = B or there exists i {1,..., k} such that: A P j = B P j for 1 j < i and A P i < B P i Complexity Results All of the following four problems are 3 P-complete: w -Relevance, w -Necessity, P -Relevance, P -Necessity. 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. 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 Σ 2 P-oracle. These are polynomially many calls w.r.t. the representation of the weights of the elements in H. Reinhard Pichler 5 June, 2018 Page 33 Reinhard Pichler 5 June, 2018 Page 34 Proof idea (continued) P -Relevance and P -Necessity. Suppose that P has l priorities and that each P i contains n i hypotheses. Moreover, let H = l i=1 n i = n. Then we have to determine the lexicographically minimal vector (K 1,..., K l ), s.t. there exists a solution S with S P i = K i for each i. This can be done in l stages. In the i-th stage, we determine K i by log n i calls to a Σ 2 P-oracle, asking questions like Does P have a solution S, s.t. S P j = K j for all j < i and S P i k? We thus need O(l log n) Σ 2 P-oracle calls, i.e., polynomially many. Remark. If the number of priorities is bounded by a constant, then logarithmically many Σ 2 P-oracle calls suffice. Summary Table: Complexity results presented in the lecture = P w P Solvability Σ 2 P Relevance Σ 2 P Σ 2 P Σ 3 P 3 P[log n] 3 P 3 P Necessity Π 2 P Π 2 P Π 3 P 3 P[log n] 3 P 3 P Reinhard Pichler 5 June, 2018 Page 35 Reinhard Pichler 5 June, 2018 Page 36

10 Summary Learning Objectives Table: Further complexity results in (Eiter/Gottlob, 1995) Relevance = P w P Horn NP NP Σ 2 P 2 P[log n] 2 P 2 P definite Horn P NP NP 2 P[log n] 2 P 2 P Necessity = P w P Horn co-np co-np Π 2 P 2 P[log n] 2 P 2 P definite Horn P co-np co-np 2 P[log n] 2 P 2 P Definition and intuition of logic-based abduction 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. Get practice with complexity results in the polynomical hierarchy. Reinhard Pichler 5 June, 2018 Page 37 Reinhard Pichler 5 June, 2018 Page 38